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2 carefully shown. What are the nonoverlap periods for d>(, 02? Sketch the effect of eliminating, successively, the inverters feeding Oj, d>2, both 4>i and 4>2. What do the nonoverlap intervals become in each case? To increase the gap between
 118 
PROBLEMS: Chapter #139
13.57 Sketch a CMOS implementation of the circuit using PI3.56 above, using minimumsize unmatched complementary CMOS everywhere except for the outputs which should provide 10 times the normal matched inverter output. What total device width is needed?
SECTION 13.8: MULTIVIBRATOR CIRCUITS 13.58 Consider the monostable circuit of Fig. 13.47 of the Text, implemented with devices from an SSI CMOS package operating at VDD = 5 V with devices for which V,h = VDD/Z, tP = 15 ns, Ron = 200 £2 and the protection diodes begin to conduct at 0.5 V and are fully conducting at 0.7 V. Using a 20 pF capacitor, find R required for a positive 200 ns pulse at o02. What is the minimum length of the posi¬ tive input triggering pulse? What happens if the input pulse is positive for lpsec? If the largesignal voltage gain of G2 is about 20 V/V, what is the fall time of 1)02 for a long input pulse? What is the maximum length of input pulse for optimal behaviour? What would you estimate the output pulse tran¬ sition times to be [Hint: tT is in the range of tP/l to 3 tP .] 13.59 For the monostable circuit of Fig. 13.47 of the Text, whose waveforms arc given in Fig. 13.50 and an expression for T given in Exercise 13.15 (on page 1110), let VDD = 5 V, V„, = 0.6 Vdd> R = 22k£2, C = 1500 pF, and R„n = 180£2. Find the values of T, AV, and AV2. By how much does Oqi change dur¬ ing the interval 7? What are the peak sink and source currents of G\! Use x>o = 0.7 V.
D 13.60 In a particular CMOS implementation of Fig. 13.47 of the Text, G2 is a simple inverter and G\ a NOR, both of which use all minimumsized devices for which (W/L ) = 2. For this process, IV, i =1 V, (i„ Cox = 2jlp C„x = 20pA/V2, and VDD = 5 V. The function of R is implemented using a simple current mirror employing two minimumsized pchannel devices and a groundedsource diodeconnected minimumwidth nchannel device of 10 times the minimum length. Find the value of C for a 10 ps output pulse, accounting for the nonzero value of Vol °f G and the actual value of Vlh of G2.
CL 13.61 The circuit shown is a OneShot intended for operation using the standard technology introduced in the introductory NOTE of this Chapter, with VDD = 3.3V. All NMOS have minimum L and lOx minimum width. Q6 is minimum size. Q4 is matched in current to Q3. <2 5 has minimum width but 10X minimum length.
+ VDD
+ VDD
+ VDD
For C = 10 pF, estimate the output pulse length. Estimate the parasitic capacitances at the output and internal node (for which C is a shortcircuit), and then the propagation and transition times for each of the gates. What is the minimumlength trigger pulse? What is the maximum trigger pulse which still allows regnerative turnoff. Between pulse inputs, what is the required powersupply current? What is the supply current immediately after triggering? What is the voltage across C during the period

119 
PROBLEMS: Chapter #1310
between inputs? How long does it take C to recover to zero volts following the end of a pulse at the output (or after a long input)? [Hint: Q$ operates for a time with drain and source functions inter¬
changed.] 13.62 For the astable multivibrator modified as suggested in P13.77 on page 1154 of the Text, provide a design for operation at 1MHz, using a 100 pF capacitor, VpD = 5 V, and V,i, = 0.44 VDD What values of resistors would you use?
.
D
13.63 Describe the operation of the circuit shown. It is implemented with a 3.3 V supply using the standard¬ ized technology introduced earlier in the introducotry NOTE. Q\, Qi is a regular minimumsize matched CMOS inverter. Qj, 04 is matched but 10 times wider than minimum. Q$ has minimum width and a length which is lOOx the minimum. Capacitor C = 10 pF. For inverter thresholds at VDD/1, how long does the output stay positive? How long negative? Suggest a way to arrange for a 50% duty cycle. + VDD
+ VDD
13.64 (For other ringoscillator problems, see also P13.2 and P13.3 above.) A ring of 5 inverters is constructed with the basic matched CMOS inverter described in the NOTE at the beginning of this Chapter. Operation in this technology is at 3.3 V. Estimate the equivalent load capacitance, the propagation delay and oscillation frequency. What would you expect the frequency to become if: a) an additional inverter loads each stage, b) the supply is reduced to 2.0 V?
 120
PROBLEMS: Chapter #1311
SECTION 13.9: SEMICONDUCTOR MEMORIES: TYPES AND ARCHITECTURES 13.65 For various RandomAccess Memories listed, complete the missing Table entries. Address Bits #
a b
Block
Col
Total
Blocks
Rows
Columns
Words
Bits/Word
10
7
22
32
1024
128
4M
1
0
8
11
c
d e f
4 3
Structures
Row
Total Bits
4M
16M
25 24
10
12
256
Bits/Block
16 16M
11
1024
256M
16M
1 4
8M
64M
13.66 A 1 Mbit memory chip is organized as 4 square blocks, each of which uses simple NOR decoders for row and column selection. How many inputs would each decoder need? For a bit address consisting (from its left) of blocknumber bits, row bits, and column bits, what are the columnaddress digits of bit 102,476 on the chip?
SECTION 13.10: RANDOMACCESS MEMORY (RAM) CELLS 13.67 Consider an SRAM cell of the type shown in Fig. 13.55 of the Text using the standard 3.3 V, 0.8 pm technology described in the NOTE at the beginng of this Chapter's problems. For each of the inverters, use a minimumsized matched design. For the access transistors, use NMOS of 3x the minimum width. What is the total area of the gates of all the devices in the cell? Assuming that the connection overhead in the cell causes the cell area to be twice as large as the gate area, what would the dimensions of a square cell be, approximately? For the Read operation, let us examine the situation in which the word line is activated to select a cell for which vB = vB = VD[/1 initially. For this situation, evaluate the available current to charge/discharge the bit line, assuming that vQ and Vq do not change, but including the body effect for one of the gating devices. Now consider the bitline capacitance, assuming 128 cells on the line and capacitance about the same as a 1 pm gate stripe 128 cells long. Using this value and the current data, how long will it take to establish a differential voltage of about 0.2 V between the bit lines? 13.68 For the situation described in PI3.67 above, consider the Write operation in which the bit lines have complementary values, either 0 or Vdd as the word line is raised from 0 to Vdd For this analysis, con¬ sider the cell to consist of 2 separated inverters with fixed inputs, either 0 or VDD , and to be concrete, with Q high and Q low. The write operation will be successful if either input is moved beyond the regeneration point at V,h = VDD/2. The excess driving current at x>q = VDD/1 or x>g = VDD/1 will be a measure of operating speed. For Vg = Vg = VD[/Z, evaluate the direction and magnitude of the net current flow in the access transistors with the bit line, B , low and bit line, B , high. Is switching possi¬ ble? Now estimate the total capacitance at Q (or Q). For the successful case (or cases), evaluate the excess drive current initially (at VDD/1), and on average. Use the average current(s) to estimate the time for regeneration to begin.

121 
PROBLEMS: Chapter #1312
D*L
13.69 A proposed CMOS static RAM uses cells such as that shown in Fig. 13.55 of the Text, having a cell supply voltage of 5V and wordline selection voltages of 0 at rest, 5 V for reading, and 5 V for writing. Digitline voltages are precharged to 2.5 V for reading, and 0 and 5 V for writing. The cell itself uses minimumsize devices for which (W/L) = 2pm/3(i;» and IV,I = IV with 25lA/V2. What is the threshold voltage at the drain of Q  or Q2 at which the cell i.„ Cox = 2.5(1,, Cox will change state? What currents supplied to or from the cell, move the cell output voltage half way from its stable state to the threshold? What must the width of (2 s ( and Q6 ) be to ensure that readout is nondestructive? What currents can be supplied during the writing process by such a device? Is the design viable? Why or why not?

13.70 Consider the onetransistor dynamic RAM cell in Fig. 13.58 of the Text, using a minimumsize NMOS transistor in the standard 0.8 im technology with a storage capacitor of 40 fF. The cell pitch (bittobit spacing) in the bitline direction is 2.5 im. A l(imwide bit line is used to couple the 256 cells in a column. Bitline capacitance per unit area is about the same as gate capacitance. Sense amplifiers and drivers add another 70 fF to the bitline capacitance. What is the total bitline capacitance? For max¬ imum cell signals t)cs of (Vpp V, ) and 0 V, subject to a possible charge deterioration due to leakage in one direction or the other of 20% of full signal, and bitline precharge to (VDp/l V,), what bitline signals will result? [Hint: Recall that V, is subject to body effect.]
—

13.71 For the situation described in PI3.70 above, where the voltage (Vpp  V,) on each 40 fF cell capacitor can deteriorate by as much as 20% in the 10ms interval between guaranteed refresh cycles, estimate the corresponding leakage resistance. 13.72 In a particular dynamicRAM technology, cell leakage currents can be reduced to 10 fA (1 femptoampere = 10~l5/l ). What is the minimum allowable capacitor for a 4 ms refresh interval with a recover¬ able cellvoltage loss of 1.5 V? 13.73 For a particular DRAM having 1024 rows and 1024 columns in each of 16 blocks, with a readwrite cycle time of 30 ns and a refresh cycle of 10 ms, what fraction of the available cycles is spent on refresh if one word in each block is refreshed in parallel in one cycle. How many sense amplifiers does such a design require? If all blocks share a single set of 1024 sense amplifiers, what does the refresh overhead become?
SECTION 13.11: SENSE AMPLIFIERS AND ADDRESS DECODERS D 13.74 Consider the bitlinevoltage equalization process involving transistor Qÿ in Fig. 13.60 of the Text. Assume bitline capacitances of 1 pF. Using a minimumsize NMOS in the standard technology intro¬ duced in the introductory NOTE of this Chapter, how long must
13.75 Consider the regenerative process occuring in the sense amplifier in Fig. 13.60 of the Text, immediately following the rise of the
 122 
PROBLEMS: Chapter #1313
13.76 A particular 1 Mb DRAM uses a square cell array with 4bit readout. How large a wordline (row) decoder is needed? How large a bitline (column) decoder is needed? For a 1024row decoder, how many address bits are used? Using the design in Fig. 13.63 of the Text, how many decoderarray NMOS are needed? How many dynamicload PMOS? How many input addressbit inverters are needed? D 13.77 For a tree column decoder such as that shown in Fig. 13.65 of the Text, how many input layers are needed for 256 lines? 1024 lines? How many transistors are used in each case? If each transistor is the minimumsize standard NMOS (as specified in the introductory NOTE) , what series resistance is acquired with each switch? For a 1pF bit line directly connected, what is the greatest number of layers that can be used while ensuring that a logic zero settles to Voo/10 = 0.33 V within 7 ns. What can you do to improve this situation? 13.78 For the NOR address decoder, part of which is shown in Fig. 13.63 on page 1132 of the Text, draw row 13, indicating the connection of its transistors to the first 4 address lines. How many transistors, includ¬ ing the load, are connected to each row line of a 256 Kbit square array?
SECTION 13.12: READONLY MEMORY (ROM) 13.79 A CMOS ROM of the general type shown in Fig. 13.66 of the Text uses a gated load structure in which the PMOS loads are turned on only at evaluation time. Minimumsize standard 0.8 im NMOS (see the introductory NOTE) are used in the array. Connections to each cell require 30% overhead in each dev¬ ice dimension. Approximately, what would the typical cell dimensions be? If the array has a 15% overall overhead for decode, sensing buffering and connection, how many bits of ROM can be installed in a chip 1mm2 in area. If the ROM is configured for a 32bitword output, how many words (expressed as a power of 2) can be accommodated in a chip of about this size? D 13.80 Design a bit pattern to be stored in a (14 x 5) ROM which provides the results of division of one twobit number, X, by another, Y. The 4bit word address is to be (jc !, x0, y i, yo). The output is to be (f <7i. <7o> n> ro) where F is the 1bit overflow (divideby0) flag. Q is the 2bit quotient, and R is the 2bit remainder. Give a circuit implementation resembling that in Fig. 13.66, but in which an installed transistor represents a logic 1 internally. (Hint: While this saves ROM transistors, it requires additional inverters). Excluding the input decoder, how many transistors do you need in the heart of the ROM? How many are used in inverters? How many transistors would be required in total, without using the extra inverters? How many transistors would be required for the 4bit input decoder using the circuit of Fig. 13.63?
.
13.81 Sketch the decoder and array parts of a 16word 8bit MOS ROM combining Fig. 13.63 with Fig. 13.66 of the Text, as modified in P13.79 above. For simplicity, represent the NMOS devices by circles at the intersection of access and output lines, and the required inverters by triangle symbols; but show the PMOS explicitly. Provide a sketch of the relative timing of the precharge and access signals. D
13.82 A ROM used to record the presence or absence of unusual properties of diverse materials in an inventorycontrol system. A designer has a choice of representing this data as either high or low signals at the output of a transistor array such as that shown in Fig. 13.66 of the Text. What choice of data representation is best if a) Transistors exist at array nodes and must be selectively "removed" to allow the digit line to go high.

123 
PROBLEMS: Chapter #1314
b)
Transistors must be created and connected at array nodes as needed to lower the digit line.
c)
Fuses must be blown to disconnect an npn emitter from a digit line to allow it to remain low.
d)
Highvoltage programming pulses must be applied to raise the threshold of a floatinggate transistor allowing the bit line to be always high.
[Hint: In each case, identify the relative cost of the production or programming technique for representing a logic zero or logic one. Obviously, it will be best ot chose a logic representation which is cheapest for the statistics of the data being represented, which in this case has mostly zeros with only a few ones.]

124 
Chapter 14
BIPOLAR AND ADVANCED  TECHNOLOGY DIGITAL CIRCUITS SECTION 14.1: DYNAMIC OPERATION OF THE BJT SWITCH L
14.1
A particular BJT inverter of the type shown in Fig. 14.1a) of the Text, uses Vcc = 5 V, Rc = 2 k£2, and Rb = 10 k£2, with a 0 to 4.3 V input signal. For the transistor, pp = 100, pR = 0.25, VBE = 700 mV at Ic = 1 mA, rx = 50 Q and n = 1. Find aF, aR, ISE, he, h, VBEsal, VCEsat, Icsai, hsm, pforced [Hint: Use Equations 4.100, 4.102, 4.109, 4.110, 4.113, 4.114.]
14.2
Consider the dynamic operation of the circuit described in P14.1 above in response to a 0 to 4.3 V input pulse. For the BJT, fy = 1 GHz, Cje = Cjc = ts, = 0.5 pF and T, = 1.5 ns. Estimate tj, tr, tf , t0ff and the time the base voltage remains at essentially 0.7 V following the fall of the input. [Hint: Think a lot about the details of Fig. 14.1.]
14.3
A BJT for which the storage time constant is 20 ns, and P = 200, is operated in a circuit for which lc Sai = 10 mA and the base turnon current, IB2, is 1 mA. Calculate the storage delay under the condi¬ tions that the base turnoff current, IBU is a) 0 mA, b) 1 mA, c) 10 mA.
D
14.4
In the following circuit, ts is measured for various values of C : For C = 0, ts  80 ns; for C = 8 pF, ts = 30 ns. Estimate values of t, and P if VBE = 0.7 V and Vce sat = 0.2V. What value of C would you chose to reduce ts to zero? For p twice the present value, what would ts be with the capacitor you have chosen? What would it have been with C = 0?
5V 500Q
vo
0.2V
SECTION 14.2: EARLY FORMS OF BJT DIGITAL CIRCUITS 14.5
With reference to Fig. 14.4 on page 1164 and P14.5 on page 1232 of the Text, provide a better estimate for Vil, as that voltage V/ for which the gain is 1 V/V, for a gate with a single fanout, assumed to consist of 450£2 connected to the lowimpedance base of a saturated transistor. [Note that a fanout of zero is not normal in logic applications, and 2 or more lowers the gain, implying a need for higher VIL]. Assume that X>BE = 0.700 V at iE = 1 mA and that P = 50. [Hint: Proceed by finding the required rn and the corresponding currents and voltages.] For two inputs operating simultaneously (for example joined), what does VIL become?
14.6
For the RTL NAND logic gate shown in Fig. P14.7 on page 1232 of the Text, evaluate VIH for each of and Pf = 50, P/? =0.1 with VBE = 0.700 V at the inputs under the condition that for V/h, Pfarced  P can be calculated approximately using the in small, saturation is quite Since [Hint: 1 mA. VBE a« = iE external emitter current.]

125 
PROBLEMS: Chapter #142
DL 14.7
Your boss is considering the possibility of raising the input threshold of the DTL circuit of Fig. 14.6. She asks you to help, by calculating, some important parameters, for operation with input B high and input A controlling: a) b) c)
d)
The base current in Q required to lower the output to VDdI2. Use p = 30. The input threshold voltage Vlh (at A ) corresponding. All junctions have a 0.7 V drop at 1 mA. The maximum base current available to Q when A is high.
The input current flowing from D ! when A is at 0 V. Now, she suggests that you add a diode and change a resistor so that V,h is raised by about 0.7 V, while the maximum base drive remains the same. What change do you make? What is the input current needed now when = 0? What is the fanout available with this redesign, for which $f„rccd  P/2 is maintained?
D
14.8
Following the general direction suggested by the twoinput NAND DTL gate in Fig. 14.6 of the Text, sketch a circuit using only a single transistor, but many diodes, to provide the function Y = ABC + D + EF using voltage and resistor values the same as those in Fig. 14.6. What is the base current which results when D alone is high? When all inputs are high?
SECTION 14.3: TRANSISTORTRANSISTOR LOGIC (TTL OR T2L) D
14.9
Modify the design of the /Cform DTL gate in Fig. 14.7 to double the turnoff current of Q 3. Using the results of Ex. 14.4 on page 1168 of the Text, what does the turnon base current of (23 become? What is the absolutely greatest fanout N that ensures that Q3 is barely saturated? What does Vol become for % that value of fanout? Use pfi = P/r/100, P/? = 50.
14.10 For the modification and current levels suggested in P14.9 above, and is = 10 ns, what does the storage delay become for N = 0? for N equal to /* of the maximum fanout? Recalculate these values in the event that p is doubled (from 50 to 100), but with the N values kept the same. 14.11 Consider the output stage of a DTL gate, consisting of a transistor with p/r = 50, operating at forced P of 10, with a load resistor of 2 kQ connected to a +5 V supply. Assume Vcesm = 0.2 V, what is the minimum value of external load resistance connected to +5 V, that still ensures saturation? For a resis¬ tor of twice that value, calculate the 10% to 90% rise and fall times of the output with a 10 pF load capacitor. D*L
14.12 Consider the circuit of Fig. 14.9 of the Text as the basis of a verylow voltage BJT logic structure. For Vbe = 0.7 V nominally, but 0.6 V at turnon, and VCe = 0.2 V in saturation, with P/r = 40 and p« = 0.1, prepare a design meeting the following specifications: a) the total collector current that can be sus¬ tained by Q3 with fiforced = Pf/2 is > 20 mA, b) resistors of a single value, R, are used, c) VCc = V is chosen as small as possible, d) NMH > 1.5 NML with fanout N = 10 for which Pforced ÿ 20. For what value of N does £>3 reach the edge of saturation with U/ high? D
14.13 Extend the structure of the logic gate of Fig. 14.9 of the Text to provide a logic gate to perform the fol¬ lowing function: Y = A B + C D while maintaining the same input thresholds. Use 6 transistors and 3 resistors in total.
 126 
PROBLEMS: Chapter #14—3
14.14 A manufacturingprocess deviation in the production of T2L gates using the circuit of Fig. 14.19 of the Text, reduces current gain such that Pf = 9 and (3« = 0.05. For input high, estimate all node voltages and branch currents, for VBE = 0.7 V, and a load of lkft connected to the 5 V supply. What is the larg¬ est possible fanout (excluding the lkft load), for which saturation of (2 3 is still possible? 14.15 Repeat the analysis suggested in P14.14 above, with input low (at 0.3 V) and a resistor of 1 kft con¬ nected from the output to ground.
L* 14.16 For the situation described in P14.14 above, and with input and output joined by a 200ft resistor, esti¬ mate all the node voltages and branch currents. Calculate VCE sat for Q l relatively precisely using a negative value for P/orcej. L*
14.17 Modify your response to P14.16 above for the situations in which a) a load resistor of 200ft is con¬ nected from the output to: a) ground, b) +5 V.
SECTION 14.4: CHARACTERISTICS OF STANDARD TTL 14.18 Using a similar analysis style to that found following Eq. 14.6 on page 1180 of the Text, find the value of R2 which raises point C of Fig. 14.23b) to 3.0 V. What does the slope of the BC segment become? If R 2 is to be kept at lkft by a desire to maintain the turnoff current level that R2 provides, what change in R\ would be needed to raise C to 3.0 V? What change in turnon current to (2 3 does this pro¬ duce (in absolute value and as a percentage)? What is the effect on gate storage delay? And so you see, once again, the nature of compromise in real design! 14.19 Using the data provided in the answers to Exercise 14.11 on page 1182 of the Text, find the noise mar¬ gins that apply at the interface between two sets of T2L logic gates, one operating at 55°C and the other at 125°C. Note that there are two pairs of margins depending on the relative temperatures of the driving and driven gates. 14.20 For the circuit of Fig. 14.26 of the Text extended to as have 4 OR inputs, what is the maximum base current provided to Q3? In a particular circuit with values shown, and no load, with 3 of the 4 OR inputs already low, the storage delay is 10 ns. What delay would you expect when all 4 inputs are brought low simultaneously?
14.21 For the tristate gate shown in Fig. E14.16 on page 1187 of the Text, find the voltage at the tristate input at which the collector current in Q6 just reaches 1 mA. For all junctions, the voltage drop is 0.700 V at 1 mA, Pf = 50 and Pÿ = 0. 1
.
14.22 For the tristate gate shown in Fig. E14.16 of the Text, estimate the possible current flow in the base of Q3 if the connection from the tristate input is low, but the corresponding link to Q 1 is broken. Use Pf = 50, VEE = 0.7 and VEE ,wt = 0.2 V.
SECTION 14.5: TTL FAMILIES WITH IMPROVED PERFORMANCE 14.23 A Schottky npn transistor consists of 2 elements: a BJT for which lE = 1 mA at VBE = 0.75 V with n = 1 and P = 50, and a Schottky diode for which 7 = 1 mA at 0.5 V with n = 1. For the emitter grounded, find the base and collector voltages and transistor base and diode currents, for input and load currents, respectively, of a)
1 mA, 0 mA,

127

PROBLEMS: Chapter #144
b)
1 mA, 1 niA,
c)
1 mA, 10 mA,
d)
10 mA, 10 mA,
e)
10 mA, 1 mA.
L
14.24 Consider the Schottky TIL circuit of Fig. 14.28 in the Text, with both inputs A and B high. For dev¬ ices as specified in P14.23 above, find the voltages at the base and collector of Q3, Q\ and Q6. Assume that the circuit has a fanout of two similar gates.
D 14.25 Consider the activepulldown circuit shown in Fig. 14.28 of the Text. Find the value of R 5 for which half the emitter current from Qi flows in the base input of Q 3, but only 1 mA is required to cause the SBD in J23 to conduct. C
14.26 Consider the output stage of the Schottky 1'IL gate shown in Fig. 14.28 of the Text. For both inputs high and a current Iflowing into the output, find the incremental output resistance for a) I= 1 mA, b) / = 10 mA. Use the device data provided in P14.23 above. L
14.27 For the lowpower Schottky TTL gate of Fig. 14.31, find the current that flows in the power supply with a) inputs both high, or b) inputs both low, both for the output i) opencircuited or ii) shortcircuited to ground. What is the power dissipated in the gate under all 4 conditions? For a propagation delay of 10 ns average, what is the delaypower product for operation with a 10 pF load at 30 MHz?
SECTION 14.6: EMITTERCOUPLED LOGIC (ECL) L 14.28 Reconsider P14.38 on page 1235 of the Text, for the situation in which currents ranging from 0.951 to 0.051.
Vn and V/H
are based on
14.29 Consider Fig. P14.38 on page 1235 of the Text, for the situation in which VBE = 0.75 at current 1=4 mA. Find R so that V,h = 1.32 V. What are the values of V0n and V0l that result? Find V/// and V11 for a current split in QR and the input transistor in the ratio of 1000 to 1. What are the correspond¬ ing noise margins?
D
14.30 Modify the circuit of Fig. P14.38 on page 1235 of the Text to create an ECLtoT2L converter by con¬ necting the upper supply connections (now grounded) to +5 V, and adding a number of 0.75 V diodes in series with the emitters of Q% and Q 3 (to lower the output voltage). Maintain the input threshold at 1.32 V by a suitable choice of R/l (still connected to ground) with 7=4 mA, and corresponding VBE = 0.75V. Select the resistors (called R 1) connected to the bases of Q\ and Q2, and the number of diodes, N, to meet T2L worstcase output specifications, namely V0L = 0.5V and V0n = 2.7 V, while keeping R 1 as small as possible.

128 
PROBLEMS: Chapter #145
14.31
Consider the circuit shown as a T2LtoECL con¬ verter: Use values of Iand R as specified and cal¬ culated in PI4.29 above. Arrange that the current from the T2L gate is 8 mA when its output is at the minimum specified high value of output, ie V0n = 2.7 V. What is the T2L output current at Vol  05 V? What are V0L and VOH of the converter cir¬ cuit? Use 0.75 V for all junctions, when operated at 4 mA.
VI
R2 D3
D2
D1
'® 14.32 For the circuit of Fig. 14.37 of the Text, a manufacturing error reduces the junction size of Q2 by a fac¬ tor of 2 and its p to 30. What is the corresponding effect on NMH ?
14.33 For the circuit of Fig. 14.35 of the Text, calculate the smallsignal voltage gain from input to OR output for x>i biased at VR . 14.34 Estimate the propagation delays expected to the OR output of the ECL gate of Fig. 14.35 of the Text, loaded with a single fanout for which CEQ  3 pF. Assume that the capacitance at the output of an unloaded gate is 2 pF, and that for the transistors, fT = 5 GHz and Cp = 0. 1 pF. What do the delays become for a fanout of 10? 14.35 For signals whose rise and fall times are 1.2 ns, what length of unterminated gatetogate interconnect can be used if a ratio of rise time to return time of 6tol is required. Assume that the signal pro¬ pagates at 2/3 the speed of light (which is 0.3 mm/ps or 300 im/ps).
D 14.36 Consider a version of the ECL gate in Fig. 14.33 of the Text, in which resistors RE, R 2. R 3 are replaced by current sources. What values would you use in each case? One other resistor must be changed. Which one? To what value? Using the techniques in Example 14.2 on page 1208 of the Text, evaluate the temperaturerelated changes associated with V0h< Vol and VR for this new circuit. Are there any other changes you would suggest?
SECTION 14.7: BICMOS DIGITAL CIRCUITS 14.37 For the circuit of Fig. 14.44e) of the Text using VUD = 5V, with kp = Zikn = k, IF, I = IV, and with Ri = R2 = rSDP at small IFiD/, I and IFJC/)I = VDD, find Vth, VOII and Vol with a) no load, b) a 5 kO load to 2.5 V. Assume P = 100, VBE = 0.7 V, and k = 400pA/V2. D 14.38 The circuit of Fig. 14.44d) of the Text has the apparent advantage over the circuits of Figs. 14.44c) and d) of using no resistors, which generally occupy far greater area than a MOS device. However the R

129 
»1 J
t
I
1
1 » ÿ
ÿi'To
r PROBLEMS: Chapter #14—6
ÿiual fullswing outputs. Show how the circuit nal CMOS inverter connected between input cilied in PI4.37 above, what are the average V of output swing? : specifications are given in the introductory re to be augmented by BJT output devices.
'. 14.44d) of the Text, using predominantly 5n transistors for which VBE = 0.7 V at full p = 50. For the resulting circuit, estimate V/i and V/// . For a load capacitor of 10 pF, at arc the available currents at u0 = Vdd/11 id by the output circuit. If the circuit is augi, what do Voh and Vol become on a short>r a fullswing signal to be established [to,
L3 V circuit shown, MOS devices are the in P14.39 above, with IV, I = 0.6 V, 2.5xpCox = 100pA/V2, with (W/L), = = 2.5(W/L)i, and f0.8 pm, (W/L)2 = (W/LhaI1(l Find Vol , ÿo//> #/.//> 1piil for a 10 What supply current flows if D0 = VDD/11
provide a circuit sketch of a 2input r having the same W/L ratio, what arc the r, what is the threshold voltage for the other s operate in saturation? Assume \ln/mup = e Text,
IRCUITS n Example 14.3 of the Text, for the condi
5FET width to 5im. Find
Voh< Vol, Vil,
current, the average static power load, the dynamic power loss at 2GHz,
ge supply
lilent
ransfer characteristic is shown in Fig. 14.52 to some of the vari20pm, pm, Ws = WL = WPD = 10pm, P 3 V, Vjs = 2 V. In particular, conisder
I. V,H, NHl , and NMt/
PROBLEMS: Chapter #147
(separate) changes, as follows: a) all VlD to 0.8 V, b) all VtD to  1.0 V, C) Wi = 2WPD to 5 Note that the latter design choice will also make propagation delays quite asymmetric.
xm.
D
14.45 Using the FETsizing ideas associated with the FL design of Fig. 14.51 of the Text, provide a directlycorresponding design for the SDFL NOR in Fig. 14.53b) of the Text. Using the results for the FL design, find Vol, Voh, Vil< Vih and noise margins for your SDFL gate. Note that the design requested, for which the devicesize ratios correspond directly to those used for Fig. 14.51, will have a problem with fanout. What can be done to increase the fanout from 1 to 4?

131

PART II
SOLUTIONS pages 133 to 425
Chapter 1
INTRODUCTION TO ELECTRONICS SECTION 1.1: SIGNALS 1.1
Results: (See "Rough Work/Notes" at the end).
THEVENIN R
NORTON
1
•
VA
''V)vs=isR S11a
(T)is = C(dvs/dt)
JK)vs(t)
t )vs Cs
S11b
R1

1
R2
 VA
WV
)vs(t)
rWls=vs/(R1+R2)
< R1+R2
S11c R 'wv
C
'\Jvs=lsR/(1+RCs)
;=isR
T" C 2
S11d
—•

133 
SOLUTIONS: Chapter #12
1.1
(continued)
THEVENIN C1+C2
NORTON

,
HI(Jyvs C1/(C1+C2)
'>)(C1+C2)dvs/dt
C1+C2
S11e
©is=vs(1+RC2s)/R >R 4=C2 vw#
Jvs/(1+RC2s)
4
S11f R2 VW
©is(t)
KJisR2
'V//VS
ÿ
©vs/R (T)
R2
>R
fo)vs/(1+RCs)
S11h
Rough Work/Notes: a)
Simply recall that the Thevenin generator is the opencircuit voltage, while the Norton generator is the
shortcircuit current: b) d)
The Norton equivalent is presented in the timedomain by the derivative, and in the complex frequency domain using Laplacetransform notation. Converting only the resistive part to the Thevenin form or Norton form avoids the use of complex fre¬ quency notation.
 134 
SOLUTIONS: Chapter #13
SECTION 1.2: Frequency Spectrum of Signals 1.2
Results: In general, co = 2nf, 2n rad/s = 1 Hz = 6.283 rad/s, and 1 rad/s = 0.159 Hz. Note that the rightmost columns arc the results from PI.3.
Line Label
Frequency (Hz)
Frequency (rad/s)
a
60
377
b
120.0
c
400.0
Period
1.67 x
10~2
16.7 ms
754
8.33 x
10~3
8.33 ms
2513.3
2.50 x
10~3
2.50 ms
d
1.01 x
106
6.346 x
106
9.90 x
10~7
990 ns, 0.99 is
e
97.3 x
106
611.4 x
106
1.03 x
10"8
10.3 ns
f
1
6.28
g
60.0
377
h
0.159
1
i
109
j
1.3
Period (s)
4x
10"
1.00
1.67 x
10~2
6.29
1.00 s
16.7 ms
6.29 s
6.28 x
109
1.00 x
10~9
1.00 ns
25.1 x
10"
2.50 x
10~12
2.50 ps
Results: These are tabulated in the two rightmost columns of the table above.
Examples:
—1 = 0.0166 seconds = 1.66 x 10
a)
For 60 Hz, period =
g)
For 377 rad/s, corresponding ,to
i
s = 16.6 or 16.7 ms.
= .00265 sec/rad, period = 2n rad = 2tc X .00265 = 0.01665 s
= 16.7 ms. j)
For 400 GHz, period = 1/(400 x 109) = .0025 x
10"9 = 2.50 x 10"12 = 2.50 ps.
Conclusion: Clearly, dealing with frequency is easier, either directly from the specification in Hz, or from the tabulated calculation derived from rad/s.

135

SOLUTIONS: Chapter #14
1.4
1.5
AT = 50  25 = 25C". Period of a 10.7 MHz wave = 1/(10.7 x 106) = 93.46 ns. Total variation in the period would be 93.46 x 10~9 x 3 x
10~6 x 25 = 7.009 x 10"12 = 7.01 ps.
0.2 V peaktopeak = 0.1 V peak a 0. \Al = 0.0707 Vrms.
a)
1000 Hz s 2tt x 103 rad/s = 6.28 x 103 rad/s, with a period of 1/1000 = 1 ms. Since this is the reference: Amplitude Ratio is 1 times; Frequency Ratio is 1 times.
2.12 Vrms, 20jxsec period. Vh 2.12 Amplitude ratio, = 29.98, or about 30 times. .0707 V„
b)
Frequency ratio,
1x10"
1
Period ratio
50 times.
20xl0~6
Th
1.0 V peak amplitude, 12.57 rad/s frequency.
c)
Amplitude ratio, Frequency ratio,
1.6
fh fa Vr
1.00 0.1
10 times.
12.57 fc_ = _
fa
2n x
103
= 2xl0"3 =
The Fourier series for a square wave of frequency
1000
/ and
= 1/500 times. 10 V peak amplitude is (from Eq. 1.2 of the
Text):
v(t) = 4(10)4t (sin 2kft + 1/3 sin 3(2nft) + l/5sin 5(2nft) + I/7sin 7(2nft) + L41sin 9(2nf t ) + 1/11 sin 11 (2nft) + ...). x
If i Energy per unit time in a voltage wave u(f) of duration X, across a unit load = — J u (t)dt. In one cycle of the square wave (period T = Vf), associated energies are proportional to: a) For the square wave: (102) T = 1007" Ws. b)
third harmonic:
1 3
fifth harmonic:
1 5
seventh harmonic:
ninth harmonic:
"12
1
40 n
For the fundamental (first harmonic):
2
40
40 n
\_
40
7
K
1 9
For the first 9 harmonics:
= 1.184 X
2
2
T/2,
T/2,
'
40
T/2.
7C
40 n
40 71
 136 
T/2,
T/2,
K 2
40
T=
T 2
1+
1111 + 25 + 49 + 8T 9
= 95.95 T Ws.
SOLUTIONS: Chapter #15
c)
Above the ninth harmonic, total energy is proportional to 100 7  95.95 7 = 4.05 7, correspond4.057 X 100 = 4.05% of the total. (See page 200 following Chapter 3, Solutions, for a ing to 1007 graphic view of this)
d)
At and above the 3rd harmonic, the total energy is proportional to 1007 1007
Of the total, this is:
1.7
40 7 .40,2  (—) Jt
x 100 = 100
100 7
For a square wave of amplitude V, o =
4V (sin —
40
_1_
K
2
40
= 18.9%.
+ 1/3 sin 3(0/ + 1/5 sin 5(0/ +
(0/
7/2.
7t
• • •
).
Assume the pass band includes both fundamentals, i.e., / > 2 kHz, and totally excludes energy outside the band. For cutoff at / = 4 kHz, power levels for unit loads are:
Pi =
P2 =
4x1.1 7t
4x1.2 n
(l2 + ()2] = 1.96 x 1.111= 2.18 W, and
H
2.33 W, where
P2  Pi = 0.15 W.
For cutoff at / = 5+, between 5 and 6 kHz,
(l2
+ l/32+ 1/52 = 2.26 W, Pi = 1.96 P2 = 2.33 W, where P2  P, = 0.07 W. For cutoff at / = 8 kHz, »i = 1.96 + (1/3)2 + (1/5) + (1/7)
(l = 1.96 (1.172) = 2.29 W, P2 = 2.33 (l + (1/3)2 j = 2.33 (1.11) = 2.59 W, where P2Px= 0.29 W.
See that the closest one can get to equal power is for filtering at a frequency between 5 kHz and 6 kHz, where the difference is 0.07 W, or
(2.26 + 2.33)2
x 100, or 3% of the average energy level.
SECTION 1.3: ANALOG AND DIGITAL SIGNALS 1.8
1 Vrms corresponds to "ÿ2 V or 1.414 V peak. a)
For sampling at the peaks, the square wave ampli¬ tude would be 1.414 V peak or 2.818 Vpp.
b)
90* from a negativegoing zero crossing is at a nega¬ tive peak. The next sample is at the positive peak. The result is the same as a).
 137 
SOLUTIONS: Chapter #16
c)
45" from a positivegoing zero crossing, the amplitude is square wave of amplitude IV and frequency / . For case a) and sampling at other frequencies:
sin 45° = 1.00 V. The result would be a
For sampling at 2(2/) = 4 / Hz, the result is a sequence of positive and negative pulses of ampli¬ tude ÿ2 V, width l/4f, spaced l/4f apart. I
For sampling at 14(2/) =
level of V~2 V = 1.414 V.
1.9
/ Hz, the result is
a dc
Five bits correspond to 25 = 32 values. Six bits are needed with values ranging from 000000 = 00 to 111111 = 63o where the subscript 10 indicats a radix 10 number, and the leastsignilicant digit is at the right. Thus 63 is the largest value. Generally, for particular cases: b5, b4, b3, b2, b\, b0 represent b02° + b]22r + + b525. Thus, 0,0 = 0000002; 7,0 s 0001112, 1510 s 0011112; 31,0 = 0111112; and 33,o = 100001.
1.10 The even numbers from 0 to 30 are: 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, a total of 16 values in all. Thus, one could use only 4 bits (for 16 values) with a rccoding. For particular values: 0 ÿ 0000; 8 ÿ (4) ÿ 0100; 14 s (7) = 0111; 28 = (14) = 1110. The biggest value is 1111s (15) ÿ 30. 1.11
<— Most to Least Significant Time Bit# Value
0 7 128
1 6 64
2 5 32
3 4 16
4 3 8
—> 5 2 4
6 1 2
7 0 1
In order of presentation (MSB first, at time 0, at the left), the digits have weight magnitude 128, 64, 32, 16, 8, 4, 2, 1. The 8digit binary number in Fig. 1.8 is 101110100: a) For all bits positive, its value is 128 + 32 + 16 + 4 = 180. b) For MSB negative, its value is  128 + 32 + 16 + 4 = 76.
For the MSB digit reversed (becoming 0, at the left), the value is a) 52, b) 52. 1.12 In order of presentation (MSB last, at time 7, at the right), the digits have weights 1, 2, 4, 8, 16, 32, 64, 128. The number in Fig. 1.8 in time order is 10110100 with value, a) All digits positive: 1 + 4 + 8 + 32 = 45. b) MSB negative: same, 45. c) MSB as a sign (1 negative): same, 45. For the MSB digit reversed (becoming 1, at the right), the digits are 10110101 with values as follows: a) All digits positive: 45 + 128 = 173. b) MSB negative: 45 = 128 =  83. c) MSB a sign: 45 x (  1) =  45.
1.13 For a 5 bit representation, the largest number is 11111 of value 25 = 1 = 32  1 = 31, and the smallest is 00000 = 0. In conventional form (MSB, at the left),
 138 
SOLUTIONS: Chapter #17
01101 = 0(16) + 1(8) + 1(4) + 0(2) + 1(1) = 13. For a system where 10000 = 3/2 V, 00001 = (3/2)/16, and 01101 = 13 X (3/2)/16 = 1.219 V. The highest available output is 11111 = 31/(3/32) Check: 31(0.09375) = = 2.906 V. The smallest (nonzero) output is 1(3/32) = 0.09375 V. 2.906 V, OK. For closest to 1.000 V: See 1.000/0.9875 = 10.67, for which the nearest integer is 11 and the nearest representation is 11(3/32) = 1.03125. This corresponds to 1110 = 0101 12.
SECTION 1.4: AMPLIFIERS 1.14 For each amplifier in turn, beginning with b): b) No dc connection to ground implies that /+ = /— = 1mA: P = 2 (1 mA x 10 V) = 20 mW.
"i
=
/;
=
Rin
„
20
Pin =~ÿX
20x10~3
uA = 2 mA. = 2000uA = 2 x 10 r ÿ
.OlxlO3 2xl03 nW
= 2°FW.
1/1 = 1mA. i0=ÿ= Rl Pout
1 1 ÿ2 x ÿ2 x
ÿ
10
~
_1—r = 50 V/V = .05 V/mV A„ = nj on. u
,i
20x10
3
A, = Ap =
1XI° — =— 2x10 xlO i;
'««<_ _ I Pin
Eff =
— P
,
= 34 dB.
= 0.5 A/A = 0.5x10"3 mA/pA = 6 dB.
_
_
_ 10 0.5 x 1025 w/w Q25 mW/„w ÿ 20 x 10~6 0,5 X 10 = = 0.025 = 2.5%. 20 x 10"3
I
 139 
,
6
25
_ M dB
j
346 2
_
dfi
SOLUTIONS: Chapter #18
Input
Supply
l+
/_
P
h
Rin Pin
mA
mA
mW
mV
HA
kiQ
lw
V
mA
a
3
3
60
1
1
1
0.0005
2
20
b
1
1
20
20
2xl03
0.01
20
ÿ
c
0.05
0.05
I
100
103
0.1
50
d
10
10
200
14.1
1.41
0.01
10
n
Pload
Ap
4,
Output
Pout
ratio
mW
V/mV
0.1
20
2
66
20
86
1
1
0.5
0.05
34
0.5x103
2
10
0.2
10
0.02
26
10
2.R2
28.2
0.1
40
0.2
0.125
0.5
0.05
10
0.013
0.01
l0
dB
ratio
dB
dB
%
40x103
76
33
6
0.025
14
2.5
0.01
20
0.2
23
10
46
0.02
26
4
36
20
20
0.01
20
0.1
20
20
mA/(lA
ratio
Eff.
mW/iW
xlO3
e
3.1
3.1
0.063
5
SO
xlO"3 xlO3
c)
Pou, = 0.1 mW, Eff = 10% = j X 100. P = 10 Pout = 10 (0.1) = 1mW. 1 x 10~3 P / = = 0.05 mA. 2(10)
2(10) i), = j, /?,„ =
p.n = p\)„ F
°
103 X 10"6 x 0.1 x 103 = 0.1 V = 100 mV.
ijE x 103 x 10ÿ210 jj)x 10r3 —
=— 1(W2
Rfoad —
X
10"3
6
=5 x
10~5 W = 50 XW.
_ ÿ yrms = 2 V peak.
2
= 20012 = 0.2 k!2. 10 x 10~3 2000 2 = 20V/V = .02 V/mV = 26 dB. /t» = 100 100 X 10'3 10 X 10 A, = = = 10 A/A = .01 lnA/pA = 20 dB. x 106 103 ii X 10 ÿ 10 log 200 = 23 dB. 7 = 200 WAV = 0.2 mW/)tW p = 50 x 10"6 200 P / = = 10 mA.
—
*
'°
d)
V+ + V_ 2(10) X>i = (Pin Rin)v' = (10 x 10~6 x .01 x 103)'/' = 10"2 V = 10 mV rms = 14.1 mV peak.

140 
SOLUTIONS: Chapter #19
ii
'"

14.1 x.... 10~3 , >11 Vx 1fWl . A 1 ill mA mA = 1.41 r  1 1.41 103uA. r V X .01 103 Rin = Av Vi = 0.2 x 103 x 14.1 x 10"3 = 2.82 V ÿu2
W =
—
= ÿT =
_

ÿ
ÿ
A1
ÿ
,
2.82
"or = 282mA
A„ = 0.2 x 103 ÿ 201og0(200) = 46 dB. 90 9 y 1A3 I , r = 20 A/A = 0.02 mAAtA At = — = P = 26 dB. 1.41 x 103 x 10~6 h
ap = jr = '
in
10_ÿ = 4
40 X 10x10
X
103 WAV = 4 inW/nW s 36 dB.
= ÿ x 100 = 20%. £//=%P 200 JJ
e)
i„ = f =
Rl = Pout Voio
— —— 10 x 103
= 0.05 mA Vpeak.
= (O.VT;2) x (.05 x
10"3/T2) = 0.0125 raW.
3
0125 X = 0.125)0.W. 0.1 x 103 Ap Ap = 0.1 x 103 = lOlogio(lOO) = 20 dB. •u< = (Rin Pin?' = (10 x 103 x .125 X 10"6)/' = (1.25 x 50 mV peak.
Pin
•u. ~
io ii Po
50 X 10"3 .05 x 10~3
Eff
P+
 
5x
1.414 =
"
10"6
.0125 x 10 20*100
=
10Y" = .0354 V = 35.4 mV rms = 35.4 x
„ ,
50mV 10 x 103 0.5
Rin
/ =
1(j
=
=
= 0.0625 mW.
°°ÿ2(10)'Q
3
= 3125ÿA = 0003125 mA = 3.1 x
10'3 mA.
1.15 Largest undistorted positive output signal is 7V peak. Largest undistorted output sine wave can be 7 volt peak. 70 Corresponding input = = 140 mV peak. Largest sine wave input (having no dc component) is 140 mV peak.

1.16 For the largest possible undipped output, center the output between +7 V and 9 V. The corresponding 9)/2 = 8 V and an rms of sine wave has a peak voltage of (7 2 = 5.66 V, with an offset of +7 8 = 1 V.
Required dc input offset = —1/50 = —20 mV. Required ac input signal = 850 = +160 mV peak, or 113 mV rms.
 141 
SOLUTIONS: Chapter #110
,  l)2 = 4, 4 = 4 (V,  l)2, (V/  l)2 = 1,
1.17 For Va =4 V, V0 = 8  4 (V or 2 V, with 0 V forbidden.
V,
 1 = ±1,
and V, = 0
Now, for a sine wave input \>, = V; cosoo/ correctly biassed, x>, = 2 + V, cos cot, where, v0 = 8  4 (2 + V, cos cot l)2 = 84(1 + V/Cosco/)2 = 84(1 + 2V,cos(»/+V,2cos2(0/)

= 8  4 (1 + 2Vj cos cor + V2 (1 + cos 2(0/ >2) = 4  8Vf cos co/  2V,2  2V,2 cos 2co/ = 4  2V(2  8 V, cos (0/  2 V,2 cos 2co/. Now, for an output signal (at the input frequency co) < 1 V peak, 8 V, £ 1, V, < 0.125 V. Now, % 2nd harmonic distortion = 1.18 X)0 = 5
8V,
x 100 =
0(
8(.125)
X
195
100 = Jx 100 = 3.125%.
 10"10 e40"' for v, > OV, dÿd,.
x>o is largest when U/ = 0, at which point \>0 = L+ = 5  10"10 c40(0>= 5 V. For bias at Vp = 52, 52 = 5 — 10~'° e40 V', for which eWV' = 2.5 x 1010, and V, = — 2'5,* 10'° = 0.598 V. 40
4(h),
Now for L: v0 = V/ = 5  10 e V0 In (5 — D/)10'° in f5 0 6M010 Solve Iteratively: \), = , with \>,0 = 0.6V. Thus U/ = —  ÿ= 0.612V, and — 40 40 In (5  0.612) 10'° e r 0.613 V. 0.613 See convergence: L = V. = \>i 2 = 10
_
—
__
Peak sinewave allowed is limited by L to 2.5  0.613 = 1.89 V (peak). d v0 40u, 40 u, Gain, 5  10"'° c = V, = 0.598. Vo = 2.5 = d v, = 0.598 = 10"'° (40) e d "0/ U/ i.e.,
= 10~'°
(40 e40(a598)] = 97.8 V/V.
SECTION 1.5: CIRCUIT MODELS FOR AMPLIFIERS R[ R(f A, R„ Rl + Rn Rl 1.19 /l\)t — >4\)o ÿ Thus = 1 + 7T„ n Rl + R, Rl lvL Rl and R„ = RL
A\>l
1
100 1 70
= Ik
For a 50012 load, gain = 100 x
= 0.429 k!2. Use R„ ~ 430 ohms.
500 = 53.8 V/V. 500 + 429
1.20
Originally, x>„ = 1667 \)s = vs
10 kQ
vl
Rl, u
vol
—•
•——AAAr—
Rl, u +
whence l
R,

Ri + m
Ri = 1667 V/V Ri + 10
MVr
Now, with a second amplifier connected, Rj/2    (2). = 909V/V Rj/2 + 10
vo2
Ri Ri + 10 (D/(2)> Rj/2
Rj/2 + 10

142 
1667 909
Ri + 20 Ri + 10 '
(1).
SOLUTIONS: Chapter #111
Thus, 1667 /?, + 16670 = 909/?,
1818° 1667
+ 18180, and /?, =
'ÿfj0 kO  909 = 1.99 = 2.0 kO.
1.21 Ro
vs
•—
Avo, Ro
v
vo
Ri
(open)
JO
Ri
Gain of an internal clement in the cascade = xAvo. For the condition stated, this must be 1. Rj + R„ R R Thus, p ' X Ay,„ = 1, or Auo = 1 + ~ R, + Ro Ki Particular Cases: See, for R0 = 0, Av„ = 1; •
for R(j
/?, , A
= 2,
for/?, =oo, Avn = 1; forA„0 = 11, R„ = 10/?,. 1.22 vs
1 MQ
1M, A1, 10k
"£a2=10
fTA1=10
vs
1 MQ
10k, A2, 0.1k
10k'fvA*. °1k
Jzl
1M, A1, 10k
jd "ÿFA1=10
ÿTA2=10
100 Q
vo
—• ? 100 Q (b)
Gain (A i A j) =
10* 100 in x 10 x lOifc + 10* 100 + 100
X
1A/ 1A2 10 X 777777 = 10 1M + 1M
,A
23
= 12.5 V/V, the best,
where:
104 106 x 10 x 100 = 102 x 10 x 101 104 + 106 100 + 106 100 + 104 106 „. „ 10 x 100r X X . = 10 JGain {A ) = = 0.05V/V. r 2 101 100 + 104 ~ 106 + 106 1 1 10' 100 10 x = 0.5V/V. Gain (A 2) = 10 x 2 101 100 + 100 104 + 106 Gain (A2 A 1) =
ÿ

w

143
x(l)
=
0.01 V/V.
SOLUTIONS: Chapter #112
100 I = 0.0001 V/V. 104 100 + 106 Note how important sourcetoload matching can be: Only one of 4 possible oneortwoamplifier combi¬ nations does any good; and one twoamplifier arrangement is worse than cither oneamplifier arrange¬ ment!
Gain (wire) = 1 x
Choices: double gain, double input resistance, halve output resistance. In general, a change in an input or output resistance by a particular factor provides less than that factor of improvement, due to the comparison process inherent in a voltage divider. Thus change of gain has the greatest affect for a given factor. Next choice would be output resistance (reduction by 2) since this implies reduced power loss in many applications. 1.23 80 dB m
108MO = 104 V/V:
104— 106 = 0.99 x 104 V/V. 104 + 106 104 For 10 kft load, A„ = 104 0.5 x 104 V/V. 104 + 104 10 1 104 = 9.99 V/V. For 10 ft load, A„ = 104 104 + 10 1001 x>s x 104 For a 0 ft load, i„ = = u,, for which gm = — = 1A/V. 4 For lMft load, A„ =
10'
1.24 Individual Amplifiers: Amp 1
Ri 106
ÿ vo
R„
10
104
2
104
100
103
3
104
1
20
With a 0.5 x 106ft source and a 100 ft load: a)
102 106 x 10 x 102+ 104 0.5 x 106 + 106
Vv
'
u„
z
~
3
 0.5 X
0.5 X
Arr
1 X 10 X = 0.066 V/V. 101 1.5
104 ,1Q2 , = x 100 x 11 = °178 v/vx 100 x 102+103 51 106 + 104 1 102 , 10 104 x 1x T = — x 1 x — = 0.016 V/V. 12 51 20 + 102 106 + 104
Ar
 144
Ar
SOLUTIONS: Chapter #113
b)
OneStage designs: Loss
Amp
c)
Least Loss
1
Input low
Output high
source
2
med
med
load
3
med
low
load
Ranking (best first): As Input Coupler: 1, 2/3, 3/2; As Output Coupler: 3, 2, 1; As Provider of Gain: 2, 1, 3. TwoStage Designs:
Input: clearly A) is #1 on list 1, and reasonable on list 3; Output: clearly A3 is #1 on list 2, but worst on list 3; But, A 2 is #2 on list 2, but #1 on list 3. Conclude: Try (Aj A2) and (A A3). d)
Highest Gain for a TwoStage Design: For (A 1 A 2):
106
Gain =
106 + 0.5 x 106
104
x 10 x
104 + 104
x 100 x
100 X 10 x i x 100 x J= — 1.5 2 11 102+103r "TT 9
= 30.3 V/V. For (A, A 3):
Gain =
106
106 + 0.5 x 106
104 x 1 x 1 —100 = —— x 10 x —100 —— 104 + 104 + 20 1.5
x 10 x
10 —21 x 1, x — 12
= 2.78 V/V Certainly (Aj A2) seems best with an overall gain of 30.3 V/V e)
Reconsidering:
Certainly, maximizing the gain is a good idea, since coupling is never perfect, (i.e. there is always a loss). Of the highest gain choices, pick the highest input resistance for the input stage and the lowest output resistance for the output stage, i.e., A 1 and A 2 respectively. chose (A) A 2). Try also:
For (A 2 A3):
104
Gain
= 1.49 V/V.
104 + 0.5 x 106
.... x 100 x
104
104 + 103
100
x 1x
1
10
,
10
x 100
*
tt
,00* xlx
in_
Now if two amps of the same kind can be used:
For (A 2 A 2):
Gain =
—
0.5 x
= 16.2 V/V.
104, .x 106 + 104
100 x
104 x 100 x —103ÿ —7 + 104

145 
100
.=
102+ 103
x 100 x 451
11
11
SOLUTIONS: Chapter #114
Note in retrospect, that the only way to possibly better the value of 30.3 V/V is to use (/I i A i), (A 2 A 2), (A 2 A 1). See that the loss at the output is too great in the (A ] A ) and (A2A 1) cases. 1.25
Amp
Ri
Ro
#
n
1
10
100
104
2
104
1000
103
3
104
100
105
A/A
With 10 k£2 source, and 10 k£2 load: There are 9 possible amplifier pairs:
a)
Gain 1, 1 =
104 1000 1001
Gain 1,2 =
104
™X r X 100 100 X 10 + 104
X
10+ 104
X
100
x 100 x4Sr 1001
2
104 r 
104 + 104
= 4901 A/A
104 x 104 103 . x 100 x . 100 x . 103 + 104 104 + 104 10 + 104 7
j
1000 x 1000 x „ x 100 x = 4541 A/A. 1001 11 2
Gain 1,3 =
—
105 104 104 x x 100 x 100 x 104 + 104 105 + 104 10 + 104
1000 x 100 x — x 100 x = 4541 A/A. 1001 2 11 104 103 104 x 103 x x 100 x Gain 2, 1 = 104 + 104 104 + 104 10 + 103
= Gain 2, 2 =
42 x 103 x
—
104 T
104 + 104
x
= i x 103 x 2
Gain 2, 3 =
11
2
x 103 x
X
100 x = 24752 A/A 2
103 x 103 103 in1 x 103 + 104 103 + 104
x 103 x
= 4132 A/A. 105 103 x 10 x s 103 + 104 10 + 104
x 102 x — 11
ÿ—
11
11
= 4132. A/A.
105 104 104 x x 100 x 100 x 104 + 104 10 + 10s 104 + 104 =
42 x 100 x 1 x 100 x 2 = 2500 A/A. 1r\4
Gain 3, 2 =
103 x
104 x 103 X 104 + 104 =
Gain 3, 1 =
101
ia5
ÿ— x 103 x — x 100 x— 104 + 104r 105 + 104r
 146
1
n3
103 +
j
104
SOLUTIONS: Chapter #115
=
x 103 x J = 4132 A/A. 42 x 100 x 4?11 11
104
Gain 3, 3 =
x 100 x
104 + 104
= ± x 100 x 2
11
105 105 x 102 x 104 + 105 104 + 105
x 102 x
11
= 4132 A/A.
Summary:
Gain
Combination
24752 4901 4541 4132 2500
(2,1) 0.1) (1.2). (1.3) (2,2), (2,3), (3,2), (3,3) (3,1)
Note the relative superiority of (1,1) is due essentially to the /?,• of amplilier 1 being 10£2.
1.26 Figure of merit: AisRÿRi Amp
Ri
A is
1
10
102
Ro 104
2
104
103
3
104
102
Ais
R„
X
Ri
Rank
103
1
103
102
3
10s
103
2
Lowest ranked are A 2, Ay. Consider: Gi,2< ÿ2,3< G12, G 33 with values (from P1.25 above) of 4132, 4132, 4132, 4132, respectively. Thus the highest available gain with the lowestranked amplifiers is 4132 A/A.
1.27
Amp
Ri
Sm
Ro
Fig. of Merit (1)
Rank
Fig. of Merit (2)
Rank
SmRoRi =
(1)
gmR„Rj2=
(2)
A iLs R(i
AiSR„Ri
1
10
102
10
104
10°
2
10'
3
2
104
103
10"'
103
106
2
1010
2
3
104
102
10"2
10s
107
1
10"
1
Now, gn =
Vi
A /v
i(>
h)
~
Ri ii
~
Ri
 147
SOLUTIONS: Chapter #116
For A , gm =
= 10 A/V. ia3
For A 2,
Sm=~ÿ = 10' A/V. = 10~2 A/V .
For A 2, gm =
Figure of merit (FM) for a transconductancc amplifier is gmR0Ri
ÿ
But gmR„Rj =
A
Ri
x R„ x Rj
= ASR0. Use as FM1. However, high Rt is obviously very important for the gm generator. Consider AisR„Ri = gmR„R2, as FM2. 1.28
Aix = f'h
«(=o
R>=ÿ
v =0
ib
Rn =
Gm = \Jm
Mr i_o
—
Ohe ,,
/
vb/rK
'"*•
= oo , as described.
but ic = p ib, and ih =
obe
—
— P
— r B
P ib "Ohe
—
P obe/rn ————— "Obe
p.
Numerically: A is = P = 200 mA/mA, and G„, =
= ——ÿy = 40 mA/V. — r 5 X 10 ji
1.29 For the gain v,A>b :
El.14, see ib = ix , ob = d,, and ie = ib + ic = ib(l + P) = (P + 1)ix , Thus Correspondingly, and (p oe = itRt = + l)Reix, ox = ix(rK) + oe = ix(rK + (P + l)/?e). l)RA(rK (P (p l)ReixY[(rn )Re)ix], or or f(P (P 1 = + + + + + 1)R.], oh = + oAh = 1)R,]] R/(re r/Re) rÿ(P = = + RA + HI + v/ob = HI + For the resistance Rj„ seen at the emitter: apply a test voltage vx at E, with the base input grounded, and Re removed. See ib =  Oj/rn: ic = pib = Thus total current from \)x is 1/[(P "OjcfP Thus = = + 1Yrpi] = /"HP + 1) = rt. Thus the + lYrn. ix "oAk PvAn Rin  v/ix resistance seen by Rt is Ri„ = r/(p + 1) = re From
Fig.
—

1.30 For the gain oA>b' See (from P1.29 above) that ib = (upsilonb oeyrK, that t)e = (P + 1)ibRe, and that upilonc = rKib + (P + 1)ibRe = ib(rn + (P + 1)/?,). Thus Vc/ob  PRiAr* + (P + 1)/?«], or dividing by (P + 1), \>A>b ~ [P/f/XP + OW/HP + 1) + Re] =  aR/(rc + Re). For the resistance seen by Rb, note that the output at C is a current source p,y, whose current is independent of \>c. Thus the output resistance seen by RL is infinite.

1.31 From Exercise 1.14, Rin = rK + (P + l)Re at the base. Now with a resistor Rs connected from a source os to the base, oh = [RbARs + Rin)]ox, or oh = u,[r„ + (P + l)/?eHÿs + rK + (P + 1)RJ. Now Thus OeA), = fivA)y,]/ivA)v], where from P1.29 above, o/Oh = (P + l)RAfit + (P + = [(r„ + (P + l)R«HRs + r„ + (p + 1)R,)] x [(p + l)RHr„ + (P + l)Re]].
This reduces to half the value without Rs when the leftmost factor becomes Vi, that is when Rs = [r* + (P + 1)RJ.

148 
SOLUTIONS: Chapter #117
This is a very logical result which can be seen directly, since when Rs = Rin , half the input signal is lost in the resulting voltage divider.
SECTION 1.6: FREQUENCY RESPONSE OF AMPLIFIERS 1.32
=
t)„
= — \>i
7(2V )
2mV
1
is fr = 1 ms. = 2000 V/V. At 1 kHz, the period v
Delay of 0.2 ms corresponds to
—0.2j— x 360 = 72°.
103
Thus, the corresponding phase shift is 72°, lagging.
1.33 The 3 dB bandwidth = 100 kHz  0 kHz = 100 kHz. For capacitor coupling, the bandwidth is 100 kHz  20 kHz = 80 kHz.
—
1.34 See for circuit a) that \>„ul = u, \)„ is the voltage across R , fed by C, i.e. a highpass output. Correspondingly, for circuit b), xtoul = v>, o„ is the voltage across C, a lowpass output. In fact, the circuits are really the same, with both output types available: highpass across R, and lowpass across C.
—
1.35 See immediately, that as frequency goes from 10 kHz to 20 kHz, gain drops by a factor of 2 from 8 X 103 to 4 x 103 V/V. Conclude that 10 kHz and 20 kHz are on the 20 dB/decade rolloff, such that

11x10
Af =/, = 20 x 103 x 4 x 103 = 80 x 106Hz. Thus /,, the
8x1C
unitygain frequency, is 80 MHz. Now, since at 1 kHz, the gain is only 11/8 larger than at 10 x the frequency, one can conclude that the midband gain is likely to be 11 x 103 V/V.
Nÿ4x10°
or>
kHz
10 20
x if)6
Thus the 3 dB frequency is fH = r = 7.27 kHz. At 11 x 103 f/r.. por tan i1 Mh = frequency /, the phase shift is tan i1 ///«• 60°, f/fH = tan 60° = 1.73. Thus, / = 1.73 (7.27 kHz) = 12.6 kHz is the frequency where the phase lag is 60°.
1.36 For each stage of the amplifier, the 3 dB frequency is at tow = 1/RC rad/s. For 2 stages, the output reaches lW~2 of midband amplitude at the frequency where each stage contributes (l/2)'/\ that is when • 1
[l2 + (oyto//)2 f
2
—
Zi
i 2
, or 1 +
2
to
= 2Vl =
toH
= (Oh (1.4142  if = 0.644 a>w. Thus, for 2 stages, the 3 dB frequency becomes 0.644/(RC) rad/s. (o
Now, for a modified cascade, where one stage has (0W = 1/RC and the other has (%i = 1AJcRC) = (%/fc, the response will be: T(s )
— (1 + j a>toH) K(1 +j ku/(>)H)
The response is 3 dB down when
or, 1
 2k(Ohto2
+ k2
to' 4
(Oh
1
+(*+1>'
k(Ql
K 1 +j (4
(4+1)
(Oh
to" 2 (Oh
+ 1)
= 2, or
 149
to
k to2
(0H
(Oh
to toH
= 2,
k2ÿr+(k2+ 1) (Oh
W
(Oh
1=0,
SOLUTIONS: Chapter #118
Thus
a)
Check:
(k2 + 1) ± ( (ft2 + l)2 + 4k2f
0)
whence
Ik2
for  1  1 ± (1 + 6 + 1)*
See (0 = cow
(1).
k  2± V~8
=
2 = (0//(0.414)y' = 0.644(0// : OK.
{k2 + 1) ± ( k4 + 6k2 + 1)'/' 2k 2

k2  1 ± (k4 + 6k2 + if' 2k2
= (On
_
_
= (0„
(  1 ± ÿ2
To find AT for which W/W/, = 0.95 from (1), solve by Trial and Success: Now, for the convenience
1
of smaller k, from (J):
co = CO//
2
jI_
1 1 1.5 — + + 4 k2 4k4
~~
2/fc2
2
~T
in general. For k = 0.1:
a)
= (0W
(  0.5 ± 50 ± (.25 + 150 + 2500)'/' J
For k = 0.2:
(0
= (ow
(  0.5  12.5 ± (.25 + 37.5 + 156.25)* )
For k = 0.25:
a)
= (0„
= (0„
[  50.5 ± 51.48 54
= 0.99(0,/.
= 0.964cow.
(  0.5  8 ± (.25 + 24 + 64)'/l ) * = 0.946(0,
/.
Thus, the required value of k is about 0.25. 1.37
/ Av
1
— 105
1+/
= j a
®
. = 1: = — = 20 dB, totalling 60 + 0  20 = 40 dB.
0 dB, and
i
/©
1000
©

I 1—I 10 102 1o3 10
Overall, 1000 ÿ 201og 10J = 60 dB, At

©
1+
ji
At / = 10: 1
. 1+j •»
10 »
10s
,nj
= 1=0 dB, and 1+

10 j io 150
s 3
dB, totalling 60 + 0  3 = 57 dB.
SOLUTIONS: Chapter #119
At
/ = 100: 1000 x 1+j
100
1+
105
= 1000 x 1 x
1 s 60 + 0 + 0 = 60 dB.
= 1000 x 1 x
1= 60 + 0 + 0 = 60 dB.
10 yioo
At
/ = 104:
At
/ = 10s: 1000 x
= 1000 X
At
/ = 106: 1000 X
= 1000 x
1000 x
1/2 x 1 s 60  3 + 0 = 57 dB.
x 1 s 60  20 + 0 = 40 dB.

See, 3 dB bandwidth = 105 Hz 10 Hz = 10s Hz. Now, phase is 6* at a frequency which is a factor of 10 on the midband side of fL and fH.
Thus, theregion for which phase extends from +6° through 0 to 6° is from 10 (10 Hz) to
from 100 Hz to
104 Hz.
—1Q510— , or 77;
1.38 a)
10"if
We see in general that: Aÿ = (if
+ 105) (ÿ#• + 1) 1
See at very high frequencies, that
See at very low frequencies,
f —> / 10 x//10
A(f)
= 1, or/ = 10 Hz, 4r: 10
/
107/ =I°  inÿ /'
A (f )
See at midband frequencies,
_ _10ÿ
(105) (1)
*
A (J)
=
107/ 105 (fAO)
= 103, where the midband extends from
/ = 10s Hz.
to >J
Check: 101j 10
See, at / = 10 Hz, A(f) =
(j 10+
See, at / = 105Hz,A(/)
, and
105)(ÿ + 1) 107 j
v
10s 105 105 + 105) 0' 75 + D
Thus the midband gain is verified to be
103V/V.

151

A(f)
108
103
WT5 = Tf
1012
io5o+i) (io4)
, and A(f)
103
Ti
SOLUTIONS: Chapter #120
b)
Now, Au =
(if
'°7 if + io5) (jf/io + i)
107/105
4+i 10s
103
1 10
1
+ jf
1+
if 105
1+
"T Jf
From this form, we see that the critical frequencies occur when:
103
~ = 1 4 / = 10 Hz, = 1 + / = 105 Hz, and _10
/
,3
such that between these frequencies, A =
10. — (1 +ye,)(l ye2) N
ÿ
= 103 V/V.
and
Note that the latter approach is more straightforward. 1.39 vo
10 KQ
—• ÿGmvl
Ri
„
At low frequencies, A =
Rs + Ri d.v = 2.54, R„ = 2007, Gm = 40 /.
C 10 pF
< Ro
x Gm R0 = Am. The upper 3 dB frequency, 1
For „ „ „ > 106, Ri 2k Ra C I= 106 X 2 7C x 200 X 10"" = 0.0126 A = 12.6 mA. Now
i.e.
(0H =
1 2 n 2007 x 10 x
KRo C).
10~12
2: 106,
_ R' _ = 40 x 12.6 X 12.6 X 25/126 2.5/l 2.6+10 Rs + Ri = 40 x 200 x 0.198ÿ0.198 + 10) = 155.3 V/V.
Now, Am = Gm R„
GainBandwidth (iGB ) =
Ri R,+Ri
X
„
1
D
"X R„ C
See that for large /, GB =
Now, Gain x 2n x
107 =
1 00
Rs C
Gm C
X
Ri Rs + Rj
40/ C
(2.57)
R.i "f
100C
2.5
, independent of 1\
_
100 100 = C Rs 10 x 10"12 x
104
=
109, whence Gain =
109 2ji
x
107
= 15.9 V/V.
See directly (but approximately), that the previous design has a gain of 155 and a bandwidth of 1 MHz. Thus this design, with a bandwidth 10 X greater, should have a gain 10 X less. Required current / = 107 x 2n x 200 x 10_" = 126 mA.
 152
SOLUTIONS: Chapter #121
1.40 The circuit described is as shown,
Rs 20 KQ
vs
•vw
In this circuit, C, is small and included within C\ RC 1 = R.\ 'I Ri = 20k £1 II 1001: £2 = 1001:£2 = 1001:£2ÿ5 +
(or ignored). The resistance seen by C j is 1) = 16.671:£2. The resistance seen by C2 is Rci (200£2 II 11:£2 + lk£2) = (ll:£2/(5 + 1) + 11:£2) = 1167£2. For a low cutoff frequency at 20 kHz, 2n(20)103 = l/(/?c2ÿ2)» where C2 = l/[2rc(20)103(l 167)] = 6.8 nF. For a high cutoff frequency at 80
—
kHz, 2tc(80)103 = V[Rc\C\), whence C, = 1/[2tc(80)103(16.67 x 103) = 119 pF. The midband gain {for
whose calcultion
Ct = 0 and C2 = °°} is v„A)s
=
jqqÿ20
X ÿX
lllV+ÿ0 2 =
x
ÿOOÿOO
=
59.5 v/v. Now for the transfer function (in terms of p = s/2n, a complex Hertz value, and (using Table 1.2 on page
32 of the Text): = 0; For p
VW =°°,
 fl+pÿ+20)
"
(, t
ÿ „ 2
at
ÿ
(1 +0.52)(1 +0.5yi bandwidth where: 1 dB —>20log \0K = dB For the 1
Now
.
1
1+///80
For
P
N°W
 jf
= j 40,
3t
and 40
'
kHZ
 tan"'0.5  tan~'(  0.50) = 0°.
— 1; K = logfo (1/20) = 0.89125
= 0.891.
= 0.891, when 1 + /2/802 = (1/0.891)2 = 1.259, and / = (0.259)ÿ80 = 0.509(80) =
40.708 kHz.
Likewise
0,
d„A>,(p) = 0; OK.}
(40 kHz), midband 59 5 59 5 UfAxO/) = = (1+ (1 + ;4(V80)(1 + 20/ 40) y0.5)(l 7 0.5) ' li>„A)vl =  = 47.6 V/V and/ (1)ÿ(40))=
Now
ICHect
1
1720/
= 0.891, when 1 + 202//2 = 1.259 and / = 200.259* = 39.304 kHz.
Thus the 1 dB bandwidth is 40.708
— 39.304
= 1.404 kHz.
 153 
SOLUTIONS: Chapter #122
1.41 R
u„ =
R +
+vl
•
' 11 0
va
ÿR ÿ
>
>0 «
4
=
R

Vn =D„ Vh.

»
L
R__
Cs Us
At low frequencies ,

At high frequencies :
At
CD
=
u„
Cs
Cs vh = 7 \), . 1 R + Cs
vb
*
x X>i.
R
J_
+ Cs
RCs  1 RCs + 1
» 1; Magnitude =1,0 = ±180°.
"0.v
y>0
—Dv
>
= 1; —— RC
Magnitude = 1,
u„ 1 aT2 /I : = 4r. with magnitude —7 = 1, and
r—
—
—
N4 OdB CD
0°. 12°.
90°. 100*
1/10RC1/RC
10/RC
 154
(0
SOLUTIONS: Chapter #123
SECTION 1.7: THE DIGITAL LOGIC INVERTER 1.42
C.O
5.5V
3
0.5Y235V 2.65V 2..5oV

From the description (and diagram), Vot = 0 + 0.5 = 0.5 V; V0n = 5 1.5 = 3.5 V. The transitionregion gain is essentially constant at [ 10 V/V]. The transitionregion width = (Yoh ~ VolY\ gain I =(3.50.5/10= 0.3 V. The transition region is centred at VM = 2.5 V. 2.35 V. Tlius 2.65 V, and VtL = 2.5 0.3/2 = Correspondingly, V/W = 2.5 + 0.32 = 2.35 1.85 and 0.85 0.50 2.65 V V. If the transi¬ 3.5 V,L = = = = = = VOL Vm NML NMl V0H tion region doubles to 0.6 V, the margins change by 0.62 = 0.30V to NMy = 0.85 0.30 = 0.55 V, and NMi = 1.85 0.30 = 1.55 V. To equalize the noise margins for lower gains, the transition centre must be moved by (1.55 0.55)2 = 0.50 V (lower) to VM = 2.5  0.5 = 2.0 V.




1.43 See with the switch open (as in Fig. 1.31(b), Von = VDD = 5 V. With switch closed (as in Fig. 1.31(c), VOL = 0.050 + [50450 + 1000)](5.000  0.50) = 0.050 + 0.0476(4.95) = 0.286 V. For input high, Ps> = 5[4.952050] = 23.6 mW. For input low, PSj = 5(0) = 0 mW. For a 5 k£2 "leak" and assuming that the lower end of the switch is essentially grounded, Von = 5(5(l + 5)) = 4.17 V, and the static power is essentially Pss = V2/R = 52/[ 1 + 5) = 4.17 mW. Now for 50% duty cycle, average dissipation is P5ÿ = 0.5(23.6) + 0.5(4.17) = 13.9 mW.
1.44 For ideal switches, Vott = VDD = 5.0 V, and V0i = 0 V. Since there are no static current paths, static power is P$ = 0 mW. For each switch with a leakage resistance of 5 kI2, VOL = 5 V (505050) = 49.5 mV and Vqh = 5  0.50 = 4.95 V. In each state, the static power loss is 52/5050 = 4.95 mW. Thus, the average power loss is 4.95 mW. 1.45 For a 1 V logic swing, RC\ = Rc\ = R = 1VAmA = 250 £2. For a 0 V upper supply, V0h = 0*0 V and Vol = 0 1V =  1.0 V. For equal noise margins, the switch should operate at (0 + 1)2 = 0.5V.. For Vie 5 V, and Iee = 4 mA, the power loss is constant at P$ = 5V(4inA) = 20 mW. Since the circuit switches currents, gate operation is unaffected by reasonable switch resistances (
—


155 
SOLUTIONS: Chapter #124
1.46 For Vdd = 5 V, R„„ = 50 12 (and ignoring leakage) with a 10 pF capacitor and operation at 100 MHz, dynamic power is PD =fCV$D = 100 x 106 x 10 x 10"12 x 52 = 25 mW. Transition times are governed by X = R„nC = 50 x 10 Xl0~12 = 0.5 ns. For the transition times e'"05 = 0.1 for t 0.51n0.1 = 1.15 ns. Thus the 10% to 90% rise and fall times are essentially 1.15 ns (ignoring the very small time for the initial 10% swing). The output reaches VDD/2 when t  0.51n0.5 = 0.35 ns. Thus the propagation delay is 0.35 ns.
——

1.47 Here, the supply is 3V, switches operate at 1.5 V, Rnn = 50 ft, and Rieuk = 5 kft. For no leakage, Vol = 0 V and V0„ = 3.0 V. For leakage, V0L = 3(505050) = 29.7 mV = 0.03 V and Von = 3F 30mV = 2.97 V. For no leakage, the static power is 0 mW. For leakage, the static power is 3ÿ5050 = 1.78 mW.
—
1.48
Left Switch
RiqWt owitch
Vot V0*
OKI OFF
•
OM OFF OV IV
OV
IV
r r n cv_ r QC O
T/2
T
3T/Z IT 5V* 3T
The time constant for each transition is x = RcC = 250 x 3 x 1012 = 0.75 ns. See V0n = 0 V and xlnO.l = 2.3x = 2.3(0.75) = 1.73 ns, and Vol = 0  4mA (250ft) =  1 V. Now tTLH hnL xln0.5 = 0.69x = 0.69(0.75) = 0.52 ns. For 200 MHz operation (independent of duty tpLH tpHL cycle) the dynamic power at each output is Ppi fC(AV)2 = 200 x 106 x 3 x 10"12 x 12= 0.6 mW. For 2 outputs, the dynamic dissipation at 200 MHz (independent of duty cycle) 2(0.6) = 1.2 mW. The static dissipation is 4mA (5V) = 20 mW. The total power dissipation is 21.2 mW. Note that if the powersupply voltage is reduced, the dominance of static power also reduces, to give 13.2 mW total for a 3V supply.
—
—
 
 156
Chapter 2 OPERATIONAL AMPLIFIERS SECTION 2.1: OP AMP TERMINALS 2.1
Each op amp has pins for
input :
2 unique
1 unique 2 sharable Thus an 8pin package can accommodate: 2 op amps, using 2(3) + 2 = 8 pins, with none unused Thus a 14pin package can accommodate: 4 op amps, using 4(3) + 2 = 14 pins, with none unused. output:
power :
SECTION 2.2: THE IDEAL OP AMP 2.2
Voltage between input pins, v = \)+  u_ = 3V7104 or 300 TV. In particular, from the negative to the positive input, one would expect 300xV or 0.3 mV. 0.3 = 100.3 mV. If the positive pin is at +100 mV, the negative would be at 100
—
2.3
o+ = 3.5 V/104 = 0.35 mV, across 1 k£2. 0.35 x IP"3 106 = 0.35035 V = 0.35 V. = o+ + \>im = 0.35 X 10 3 + 1 X 103
For \>0 = 3.5 V:
required \)/
gain is Check: Overall 6
10*
 x 104 — 103 + 106 r
t
1o?
r X 104 = 10 V/V. = — 106
0.35 V at the input produces 3.5 V at the output.
SECTION 2.3: ANALYSIS OF CIRCUITS CONTAINING IDEAL OP AMPS  THE INVERTING CONFIGURATION 2.4 For the desired connection: vl
R1 .7 kQ 4.7
•vw
R2 47 kQ
rÿ
X1
_ R2 vo
—•

ÿ
47 _ 47
_1Q y/y
For R and R% exchanged: \>o = — Vi

157
Ri
47
= 0.1 V/V. i=Zif47
Ri
SOLUTIONS: Chapter #22
2.5 Vl
10 KQ
For an ideal op amp: to obtain u0 = +10 V, D_ = 0 V, and the current in the grounded lOkQ resistor is
V 100 kQ
10 kQ
—
— wv
zero. Thus, t»/ = 
vo
,lOOkQ lOkQ = 1 V.
r1
104:

For u0 = +10 V, t)_ =  lOÿO4 = 10"3 V. 10~3yi05 = 10"4 A, to the D_ node; Whence current in 100 kQ is (10 kQ current in grounded 10 is 10~3/l 04 = 10"7 A, to the t)_ node; For gain =
(10~4 + 10"7) A = 104 A, to the input. v, =  10~3 104 (lO4 + 10"7) = 10"3  1  10"3 =  (1 + 2 x 103) = 1.002 V.
current in input 10 kQ is
2.6
Want Gain of 2 V/V with three 100 kQ resistors: There are 2 solutions:
— 
vl RI > 100 kQ m
•
100 kQ 100 kQ

aaa.
<
r. vl
Riÿ
100 kQ
17
100 kQ
•WV
100 kQ
For (a),
For (b), 2.7
200kQ
W
lOOkQ
= 2 V/V, and Rin = 100 kQ.
Vo_
lOOkQ
W
100km
=  2 V/V, and Rin =
lOOkQ
= SO k£2.
There are 2 potential solutions: a)
R i = 220 kQ, and R2 = 10(220kQ) = 2.2 MQ > 1 MQ; no good.
b)
R2 = 220 kQ, and R, =
220IcO.
ÿ
vl
= 22 kQ « 1 MQ; OK.
R1 22 kQ
•Wv<
R2 220 kQ
Wv
N +
JT1 
158 
/
vo ~
SOLUTIONS: Chapter #23
2.8
For an inverting op amp, with Rin = 100 k£2, use Ri = 100 k£2. (a) For a direct design of gain = 20 V/V, R2 = 20(1?,) = 2 MO; not allowed directly, but R 2= 1MO + 1MO in series is OK. If very large resistors are to be avoided completely, and even 1 MO is too large, consider a network for R2: There are many possible designs, but only three which use 3 1000 resistors to meet the specifications, all having R1 R  = 100 kO. vo 100 kQ vl
wv
Va
R2
R4
Vb
—•
Wr
ÿ—
(b)
,
Make R = R2 = R3 = 100 kO: Vb = 
r 1+ (c)
R4
and v„ = t)/,
v„ = u,
1+ 1+
R* R 2 II R 3
R4 50k
= 20o,
.
—
= 20 > /?4 = (201) 50 = 950 kO, but this may be too large for some applications.
Make R i = R2 = R4 = 100 k£2: See Vb
Thus
D„  Vh Vb Vh Va = —— + —, where va  Vi and, summing currents at %, —R4 R3 «2
= Vb
K4
_L
_L
r3
r2
JL r4
, and tlius
V/V.
100 . . + 1 + 1 = 20, or —— i
(d)
*2 — Vi =  Vj Ri
i
on
a3
,
d jff 3
=
100 20—2
=
= — R4 — v,
1
1
1
/?3
R2
R4
0.
= 20
l°° , . = 5.55 k£2, a very good solution! —— 1o tKln
i
Make R = R3 = R4 = 100k£2: See Thevenin equivalent at Vb (to the right) is a source vyi with source resistance R 3 II R4 = 50 k£2. Now, gain to equivalent source Cu«/2) must be 10 V/V:
R2= 10 R 1  50 = 10(100)  50 = 950 k£2, which, again may be too large. 2.9
Two possible gains: a)
With an ideal op amp: Gains are
b)
= 10 V/V, and
= 0.1 V/V.
With an amplifier with gain of A =100 V/V, and G = Gains are G = 
100ÿ10
1 + (1 + 100W100
R7/R1 l+d+RÿO/A'
10 = 9.009 V/V, 1 + 11/100
0.1 1M00 = 0.0989 V/V. 1 + 1.1/100 1 + (i + lMOoyioo See that the error in the highgain case = 10%, where G/A ~ 1/10, and = 1% in the lowgain case, where G/A = 1/1000 (We will see why in Chapter 8).
and G =
—
 159
SOLUTIONS: Chapter #24
—R '/R  2.10 From Equation 2.1: G =  ,, n l+ (l+/?*R,>a
—
'
Now, IGI "2: 0.9 (100) when
Whence A >
R2 . For f= 100 > /?
IGI =
on
1
1 + 101/A'
2: 0.9 (100), or 1 + 101/A < ~ = 1.11 or 101/A £ 0.11.
t
= 909 V/V.
Now, IGI "2> 0.99 (100) when 101/4 < v
Whence A 2>
—0.99
 1 = 0.01010.
yjjyy = 104 V/V.
100 Check: G =  j == 99.0001. 1 + 101404 1 + (1 + 100)404
2.11 For a small test voltage x> applied at the negative opamp input terminal (where /?2 returns), the op amp Av The current flowing into R2 is i = (o output voltage \)o = AvYR 2. The input resistance + A ). Now, for an input resistor R j, and input voltage v>/, the vol¬ Ri = x/i = \y[D(l + A )4?2] Ri tage at the op amp negative input is x> = V/ RjA.Ri + R i) and Do =  Av. Thus, combining, ARj AR 2 ARAl+A) RzK, (1 Rj + R  Ry(l + A) + R i 7?A 1 + + + + R2/Ri)4k /?2 /?i expected.
— .
——
—
_
2.12
vo=978mV
Since the op amp is conventionally ideal with zero input current, 10 ftA flows in Rf. Thus Rf = (10.1  978)10~Xl0 x 106) = 98.8 k£2. Rf is likely to be a nominal 100 k£2 resistor, with a tolerance of 98.8 100) 100  98.8/100 x 100 = 1.2%. [Even more likely, this is a ±5% resistor near 96.8 V/V. The input the middle of its value distribution.] The openloop gain A = —97840.1 = resistance at the negative input node is /?, = lO.ltnVAOpA = 1010 £2. Check: Using the result in P2.ll above R, = RfA 1 A ) = 98.84 1   96.8) = 1.01 k£2, as above. OK. For input resistor R, Vq/os = + /?,)] x 96.8. Now, x>qA3S =  10.00 for Rs + Ri = 96.8/?,/10.00 = 9.68A,, 8.68(1.01) 8.68 = 8.77 kQ. Ri = Rs =



—

Check: From Eq. 2.1, C =
,+(1~**'i)at
=
i+
 160
{l'*9iMJ7y96.S " ~ 10'00n'' " rCql,i"id
SOLUTIONS: Chapter #25
Note how much smaller Rs is than the nominal value of lOOÿQÿOV/V = lOjfcQ! 2.13
Vl
R1 100 kQ
Since Rin = lOOkQ, R, = lOOkQ. For equal gain/stage,
—
_•
WV— —WH•
R2
R4
R3
VO
G, G2 = ± ÿ200 = 14.14V/V,
vw—t
and R2 = 14.14 (lOOkQ) = 1.414MQ, which is too large.
_d
Use R2 = 1MQ, for which G\ = 106/105 = 10 V/V. 200 G2 = 10 = 20V/V. Use R4 = 1 MQ, Rj = R/G2 = 106/20 = 50 kQ.
Thus, use R = 100 kQ, R2 = 1 MQ, R3 = 50 kQ, R4 = 1MQ.
2.14 Now for the circuit above, with Rin = 2MQ, make R \ = Ria + Rih, each 1MQ, since larger resistors are not available, and R2 = 1 MQ. Correspondingly, G\ = 1/2 V/V, and G2 must be +200ÿ1/2) = 400 V/V. Use R4 = 1 MQ, and R3 = 106/400 = 2.5 kQ. In summary, use R
2.15
= 1MQ in series with 1MQ, R2 = 1MQ, R3 = 2.5 kQ, R4 = 1MQ.
R1
vl
1 MQ
Va
R2
Vb
•
R4
— •
vo
AAA.
AAA
Since Rin = 1 MQ, use From page 70 Df the Text,
_
V,
~
R2 R\
Thus 22 =
X1
R4
R\

R4
1+«7+«7
,+ T + K
I MQ = R2 = R4.
 (i) and
R 3 = W0 MQ = 50 kQ. If resistors £ 100 kQ only are available, one could make R3 = R}a II R3h, each of 100 kQ. Alternatively, chose R3 = 100 kQ and select a suitable value for R4: ÿ4 R4 1 (1 10) , 21kQ. 22 = + T~ oT and R4 + = 22 1 = 1 Unfortunately, R4 = 21/1 1 = 1.909 MQ is too large! One can see that there are in fact no other choices than using two resistors in parallel for R3, or in series for R2 or R4.
Rewriting (1) above:
Vo V/
«2 R1
Ra Ri
R2R4 R 1 R3
which, for R ( = 1 MQ, becomes
R 2 R4 , and for R3 = 0.1 MQ, then 22 = R2 + R4 + 10 R2 R4. + + R2 R4 u, R% Now, if cither R2 or R4 are 1MQ, say R2= 1 MQ, then 22 = 1 + R4 + 10R4, and R4 = 21/1 1 > 1 MQ. Obviously, using two resistors in series for R4 is possible also, but not as nice from a practical point of
view, since a new circuit node (for connection of the two resistors) must be found. Thus, the first solu¬ tion (where R3 = R3a II R3b) is the preferred one:
 161 
SOLUTIONS: Chapter #26
In summary, use R = R2 = R4 = 1 MO, and R3 = 100 kO II 100 kO.
2.16 For
v2
tj
1
= 0: For A = 00, and the feedback working,
x>a = 0. Since it = i2, then x>h = 0. Now, for W/, = For w2 = 0: 0. h  '4> and =  (R44I3)
>R3
Wo = W vi
a
R1
R4
b
R2 AAAr
V0
—•
Wv
1+
/?
4 #4 —— + —— Ri R3
, as derived in
Example 2.2 on page 69 of the Text. Using superpo¬ sition:
>
JT
*2 Ri
Wo = 
R2 R.
1+
R4
R4
ÿ+r7
W,
 r4 W,
SECTION 2.4: OTHER APPLICATIONS OF THE INVERTING CONFIGURATION 2.17
w „{s)
Z2(s)
w,(s)
Z,(s)
R 2 II
1
R ÿ/C 2s
R2
C2s
R 2+ VC 2s
1 + R2 C2s
1
R /C t /? i+l/C 5
Ri II Cj
/?
1 + /? 1C* jr
4B
Thus
Ri R.
w0(s) Wj(s)
1+RCS
40
1+R2C2s
20
which is independent of frequency if RC = R2C2. (a)
For C2 = 0.1 Ci = O.lftF, and R2 = 10 Ri
105Q:
104 x 10~6 .v 1 + 105 x 10"7 s
= 10
See
v„ =  10 — W/
1+
©
I*)
rlTt

vo
26
\
1 + 10"2j 1 + 10"2j
r\t\ \
20
For R2 raised to 1M£2: w„ 106 1 + 10~2s See w(104 1 + io'j
A —io\y«>* \ —
O
= 100
[b)
40
 10V/V, independent of frequency. (b)
10* »o*\ p
1 +sÿ00 1 +s40
r r
ÿvi
te
20
clE
(c)
See
w» w,
= 1
(O
40
For R2 lowered to 10 kQ:
20
1 +s400 1 +S4000
IA1 \
2.18 Using w0 = Vc
1
—— La
20
J ' wi(t)dt, see: f
W
For W/ = +1 V, the output Vq = Vc
•vi
0
— Jof ' ldt = Vc 1 x 10"3 r

162
 103 1.
10
10'

SOLUTIONS: Chapter #27
That is, the output is a negativegoing ramp with slope of 1 V/ms or 1000 V/s, proceeding from Vc = 10 V downward, reaching zero in 10 ms.
Directly: Following the 1 V step, the input current is VI
1 RC
2.19 Assuming
charging C, causing o0 to fall at a rate
_inJ,  10 V in  —r = 10 ms 10 V/s, moving =
10
— —101
103
— J
1 1 X cos 0.1 Vc = 0 V, o0 = CR j V/ dt = CR J 0.1 sin 2k 60 t = 10 ,n t 3 2rc 60 x 10= 26.5 X 10~3 cos 2n 60 t = 26.5 x 10~3sin(2rc60/ + 90°) =  26.5 X 10~3sin(27t60f  90°) J
2n 60 t indicating that the output is a sine wave of 26.5 mV peak, leading the input by 90°, or, alternatively, is an inverted sine wave, lagging by 90°. Note that the latter idea, that of a lagging inverted output is the most consistent with the STC lowpass view of the Miller integrator. 2.20
, dvi
dvi
x>o=CR —dt7— =  5 x 10"3 dt = 1 V. dVi =  200 V/s. That is, for an output of +1V, dt
20V t2
VI
—
0V
the input must fall at a rate of 200 V/s.
ti
ta
See / 1 = 0, 200 /2 = r = 20 ms, tr3 1/1 1/10" 20 —r t3 = 20x10"3
+5V__ VO
0V
t2 ta
ti
ÿ5V
I O
 VI03
I
I
For the rise,
For the fall, t)0
=  5 ms X 1 V/ms = 5 V. =  5 ms x 1 V/ms = +5 V.
20ms 40ms
2.21 (a)
o
(«)
Vl'
10 kQ
100 Q 1
•— ([ÿ
vl
— 40 ms.
®
0.1V
VO
• 10V
Immediately, upon the input rise, v0 =
10
s

lOifeft
= —10 V. Time constant: x = 10012 X 1 X 10~6 =
1000
= 100 is. In 10 is: the output rises (almost linearly) to 10 +
as the input falls, the output rises by 10 V to +1 V.

163
T 100ns

10 x
10"6
1UU X lu
x 10 = 9 V. Then,
SOLUTIONS: Chapter #28
(b)
5V
© ®
vl
_kÿ2L
vo
50mv
T0.1ms
of
100mS
At the falling edge of the input (time /j ) the output rises by 50 mV x 10kQ/100Q = 5 V, then begins to fall with a time constant, t = RC = 100 x 10~6 = lOÿs = 0.1 ms. By the rising edge of the input (100/0.1 = 1000 time constants later), the output has reached 0 V.
2.22 v1
v2
v3
Want,
R1 30 kQ
=  (Dj + 2d2 + 3d3)
= 3
Rf 30 kQ
15 kQ
i)0
But, DO
d2
3
+— + yi D _
  Rf
J.
Rl
r2 R) Thus, make R3 = 10 kQ, Rf = 30 kQ,
R3 10 kQ
R, = 30 kQ, and R2 =
= D] + 2d2  3d3. See that there are several decompositions:
2.23 Want
\>q
a)
DO
b)
Do
c)
Do
j
2d2 + 3u3 ,
b
(Dj
1>3 1
+ 2\)2) + 31)3
(o, + 2d2  3d3) , with corresponding circuits:
R1 v1 10 kQ
•Wv—
R2 10 kQ
R3 30 kQ
—Wv
R4 30 kQ VO
R5 v2 10 kQ
•WV—
R6 20 kQ
R7 30 kQ
_D
R8 v3 10 kQ
(a)

164
= 15 kQ.
SOLUTIONS: Chapter #29
2.23 (continued) R1 v1 20 kQ
•
ÿ
R2 20 kQ
AAAr
R5 v2 10 kQ #
R4
R3 30 kQ
—
Wv
30 kQ
VvV
J?
R8 v3 10 kQ
vo
—•
JT1
(b)
R1 v1 20 kQ
R7 15 kQ
R6 v3 10 kQ
v2 10 kQ
R3 10 kQ
R2 20 kQ
R4 10 kQ
—
AMrfVlAi

Wv
VO
10 kQ
w»
JJ
JJ
3(c)
For each circuit, there are many variants of which some are: Version
r2
Rs
«4
Rs
Re
«7
*8
ZR
# Amp.
a\
10
10
30
30
10
20
30
10
150
3
a2
10
10
30
30
10
10
15
10
125
3
as
10
10
10
10
10
20
10
10 1 10 1 10
110
3
b\
20
20
30
30
10
10
120
2
b2
20
10
15
30
10
10
95
2
b3
10
10
15
30
10 1 10
10
95
2
c1
20
20
10
10
10
10
105
3
10
15
Conclusion: Note that b) is obviously the simplest, using the fewest op amps, derivable directly from c) which is the bruteforce approach. Note that a) is clearly not a good choice, using an extra op amp to separate (unnecessarily) the D( and u2 inversions. Conclude b2 is best, with the fewest op amps, the fewest resis¬ tors, the lowest total resistance, and on input resistance ÿ10 kfl.
 165 
SOLUTIONS: Chapter #210
2.24 For an ideal op amp and virtual ground at v>_, 'c 1 f =  —— hti + 2t)2 =0 — Cs 10/fc
JQ' (o,(r) +
 1000
t>0 =
2v2(t)
ic
D
=
1 10_7jt
U2
, and
= o0
) dt.
SECTION 2.5: THE NONINVERTING CONFIGURATIO Vo R2 47 2.25 Normally: =1 + — = 1 + — = 11 V/V. — J 4.7 v,
/?,
With resistor exchange:
2.26 Want
vo =1.5 V/V, •0/
— = 1 + Ri
= 1+
47
= 1.10 V/V.
with three 1 k£l resistors:
Solutions:
R3 1 kQ R2 1 kQ
1 kQ
R1 1 kQ
1 kQ
vo
—• ="( a )
1 kQ
VO
="( b )
vl
2.27 Want 00=1)!+ 2\)2  3d3:
Simple approach:
Make R2 »ÿ
— R /2 —> R 1 = 1R2.

R2
+
R1
Ri
+
Now if Rj = 2/? 3, and o2 = 0:
 166 
2R2
1
+
SOLUTIONS: Chapter #211
Vo = 1 +
2R,
(Ui
Ri
In general, u0 = D + 2v>2 for more gain for u3.
 2u3.
+ 2i>2) = (\) + 2u2). Unfortunately, neat and simple, but not quite right! We see the need
Second approach: (The idea is to reduce the gain given to (v>i + 2t)2) ). Let Rf = 3R2, such that for t>3 alone, v0 = 3l)3. But, now, x>(/oa = 4. Introduce an additional input u4 = 0V connected to v>u by R4. Now by superposition,
R2R 4 1>a =
R2\\R4 R  + R2 II
D,
+
R \ II I?4 X>2 = R2 + R  II R4
ÿ
/?,+
4
2" 4
A4 RtR
ÿ 1+ÿ? 4
•u, + —\)2, R,R
r2+ R
R2+R 4
I'M
i+#4
or
=
R 2 "4
R i R2 + R  R4 + R2 R4
d,
+
/?4 1
/?2 + /?2/?4 + f?i I?4
Now, if it is required (for \)3 = 0), that =
l~E
R\ 7ÿ + ÿ + 1 Rl
i)i
ÿ2
+ 2o2 = t>„, need R\=2R2, for which
+ 2x)2>i i = — , for —R4
Now for the desired output, v0 = 4o(, , and
which
i
IT + 2+ 1
R4 — R j =2R2. Thus
kQ.
use
R2= 10 kQ —>
r4
R i = R4 = 20 k£2, and use
—  = 4 21 = 1, Ra
or
R2 = 10 kQ —» Rf = 3R3 = 30
2.28 v2
vH1
Rf
R2
RN1
Rf
vo
Va
R3
vo vp1
RP1
vp2
RP2

Try the circuit in 1) Want u0 = 10 (l) t)2). Do = —10122 = Rf/R2 t)2. Rf = 1OR 2 For R2 = 10 kQ > Rf = 100 kQ, for which u0 =  10 v2
—
Now for Ui alone, (R2 = oo), d0 = 11 + Rs/R2
Va
using
R j,
I Di = (1 + 10) \>j = llDj
 167 
R2,
R/ :
For
i)2
alone,
SOLUTIONS: Chapter #212
—
and, together, D0 = 1lx>, 10u2 We see that there is too much of Uj: It needs to be attenuated:
Introduce an additional resistor /?3 from uu to ground: 10 R*
7ÿT/?7 U, = TTU
=
Now for /?, = 10kf2 , /?/(10 + /?3) = 1011, and 11 R3 = 100 + 10/?3, whence /?3 = lOOkO, with the result that x>0 = 11 *>i X lOll  10 \)2 = 10(\)  d2), as required. In summary. /?, — R2 = 10 kO, Rf = R3 = 100 k£2.
This configuration, with Rj/R2 = RyR , is referred to as a difference amplifier. Alternatively: From the equations supplied in P2.44 on page 1116 of the Text: u0 =
Rf R.AM
We want v0 = 10 (x>i
Now for
\)/>i
Rf + 1+ Rni
Wm
— v2).
= \), see 1 +
Rpi Vp I
Now, for vN\ = \)2, see
rf Rni
Rp  II Rp 2
RP .2
RPi
Rf
Rni
+ Vp2
Rp i II Rpi
RP 2
= 10, and for RNi = 10 kQ, RF = 100 kf2.
= 10= 11
Rp  = 100 ki2. Clearly, if \>P2 = 0, then u0 = 10(\Ji  1)2).
Rp2
Rp l + Rp 2
, and for /?/.,= 10 k£2, RP2
2.29 (a) For the circuit of Fig. 2.19, using the result in Equation 2.11 on page 83 of the Text, 1+1?'/R  1 G= Now with /? 2 = 0, /?  = oo, /l =10 V/V, sec G = = 0.909 1 + (1 + R/RiYA 1 + (1/10)
.
ÿ
V/V. (b) Now for the circuit of Fig. 2.16, with /?,, R2 finite, G =
1 + R/R i
Now for G = 1.00, 1 +(1 +/?2«,/10 1 + RÿRi = 1 +(1 +/?2ÿ/10= 1 +0.1 +0.1 Rÿu and 0.9/?ÿ?, = 0.1, /?/??,= 0.1/0.9, and R\=9R2. Thus, /?2 = 10 kf2 , 1? = 90 kii. Thus
Gain (nominal) = 1 + 10/90 =1.11 V/V. 1.111 1.111 Gain (actual) = = 1.00 V/V, as required. 1 + 1.11/10 1.111
SECTION 2.6: EXAMPLES OF OPAMP CIRCUITS 2.30
Im
R
Ri
soa
> t Cxt
mA
10V
__
For U/ =0,X)K = 0, and the meter is required to read midscale. That is, IM = 0.5mA. Thus R i = (0 10)/0.5mA = 20kQ with the 10V supply. Now, for fullscale reading with V/ = +IV , IM must be 1mA. 10/7?  = 1mA, and 1//? + (1 + U//7? + (U/ 10)//?, = 1, whence VR = 1  11/20 = 9/10 = 0.45, and R = 1/0.45 = 2.22kQ. Check: for v, = 1 V: 1 9 I 10 1 _.c _.c = 0.45  0.45 = rr lM = 20 2.22 R R\ = 0 V: OK.
—
 168 
—
SOLUTIONS: Chapter #213
2.31

/?, = +5V/0.5mA = 10 kO; R2 = (15 5)/0.5mA = 20 kQ. Op amp standby dissipation is 2(15V) x 2mA = 60 mW. Op amp dissipation when loaded is 20 mA (15 5) = 200 mW. Total op amp dissipa¬ tion is 200 + 60 = 260 mW. Note also: That the bias network dissipation is 15 x 0.5 mA = 7.5 mW, and the load power consumption is 5 x 20 mA = 100 mW.
+ 16V

D R1 1kQ
i>
ÿ
6V
Thus the total supply power is 60 + 200 + 7.5 + 100 = 367.5 mW.
2.32 From Equation 2.13 (or directly by superposition): R2 R2 *4 = = IT V + U2 R 3 + R4 «i

Hence "On 0 =
gain A =
1
«7U'

ÿ2ÿ 1
+ TTÿ7ÿ
100 + 1000 100 1 + tocio 100 1), + \)2 = 10 (1)2  Ui). v. + v? 100+10 10 10 ' 1 + KK100 D0 = 10 V/V: (Note the sign!)
——
1), 1)2
2.33 For this situation, gain
Do Dj 
100 = 5 V/V. 10+10
 D.,2
To recover the gain, two possibilities exist:
Remove R{ and R2 and connect sources directly. b) Make R2 = R4 200kQ. The latter is probably best for secondary reasons. For example, it reduces (by a fac¬ tor of 2), the effect of minor variations in Rs 1 and Rs2. For the case in which RS2 is 8 k£2, there are two basic approaches: a) "Pad out" /?3, i.e., add an additional 2kQ resistor in series with RS2 and R 3, or, b) change R 4 to 180kQ to compensate. (Maka)
—
R2 100 kQ
vs1
Rs1 R1 10 kQ 10 kQ
vs2
Rs2 10 kQ
•— JWV—vw
—
R4 100 kQ
R3 10 kQ
—
w\
vo
—•
—
ww

wW*
tng
2.34 Using superposition,
R,
d3
R j + R4
i)0
1+*7
consists of 3 parts.
Thus
R! . 1)2 — + «i
l+
, or D0 =  th

169
R2
R 1eq
i)0
R4 ) R3eq
=  1)1
R2
R2 R.
Rl 1 + R4
/?4 —R2 — + i)2 ———— K3 + K4 K\
+ i)3
1+r7 1+
R4 R3
1+
*2 *1
SOLUTIONS: Chapter #214
Now,
R
R
i 4 = — , and — Ki K3
V0 = \),
R2 1 + rr
r2
— + \)2 1+
1+
+ V3
Ri
1+
R7
Ri
R2

Ri
ÿ2 (ÿ1+ÿ2) +\)2—
(/?i+/?2)+ÿ3.
/?.
Ri — Co2  Ui) + \)3. Note that the operation is the same, but with the output established Ki by U3 for (t)2  Ui) = 0. This is an interesting and important result! whence Do =
2.35 Check first that it is a balanced difference amplifier for II 1501:£2 = 300/3 = 100 k£2 = R2, (OK). (a)
= 5 (1/2) + 15 (60360) = 2.5 + = 5V.
current in R 1 = 0, and \>0
For v>! = \)2 = 0V, x>A = 15
(c)
For "Uj = 3V, x>2 = 2V, see x>A = 2
—%—
t>i
t)B
/?!
t)ÿ
=5
100 II 150 = 15 300 + 100 II 150
(b)
Vo
x R2 = vB
and V>2: see that R4i, II R4a = 300k£2
100 100 11150 15 + + 0, or 300 + 100 II 150 100 + 100 154S = 2.5 + 2.5 = 5V. Now, since vA = 5V, and u, = 5V,
For Ui = v2 = 5 V: See (by superposition), 1)4
\>
60 360
+ 15
 D + x>B
"
15
= 2.5V, and \)0 = 2.5 (2) = 5V.
= 1 + 2.5 = 1.5V. Thus \)B = 1.5V, and
=2 (1.5) 3 = 0V.
Alternatively, to use the extended result from P2.34 above, realize that R4B with R4a and the con¬ x 15 = 5V, and 150k II 300k = nected supplies have a Thevenin equivalent of U3 = ÿ
100k£2. Thus, since R\ = R2 = R3 = R4, \>0 = u2  t>i + v3. Therefore the output is the conven¬ tional difference plus the extra (reference) input. Thus (a) o0 = 5V, (b) v0 = 5V, (c) Vq = 2 3 +5 = 0V.
2.36 For an overall gain of 100 V/V, with 10 V/V from the input stage, and 10 V/V from the output stage, see that 1 + 2RÿR\ 10, and R/R3 10. For R\ = 10k£2, R2 = (10 1) R\l2 = 45k£2, and for R3 10k£2, 5 (45k£2) = 5.45 V, and R4 = 100k£2. Now, for \), = 5.0V, m2 = 4.9V, \)01 = 5.0 +

x>02  4.9
 5f~ÿ'9 10k £2



ÿQ~ÿ9
(45k £2) = 4.45 V. Check: \)0l  x>02 = 5.45
 4.45
= 1.00V = 10 (0.1), as
expected! 2.37 See that Vy = (1 + RÿR\)x>x and that the current into X is ix = (i>x
vx
, or Thus Rin = \>XAX = [X)X ~ (I + RJROVxYR 3
 VyYR 3.
Rin = /?y(l  1  R/R 1) =
 R[RyR2»
as required.
See, from the diagram that /?, = R4 + Rin = R4  RRyR2. R \RyR2 R>, 1 , or vxA)w = Also, using a voltagedivider ratio, = " ~A' ' {RJR\){RJRti ' "" 1 R 4 + /<„ Dy R4 RxRJR2 1 + Rj/R 1 . For R\=R2, these become: /?,• = R4  R3, and Djv4)h? = (1 + Rj/R i)(\>xAi\y) 1 (/?yR,)(Rÿ3)
—


170

SOLUTIONS: Chapter #215
a)
b) c)
— 1/(1 — R4/R3) and Dj'Ajiv — 2ÿ1 — Rj/Rÿ) For R4 — 2/? /?,• = 1R 3 — /? 3 = R3, = 1/fl — 2) = — 1V/Y and ÿOjvÿiy = — 2.W/V. 1/(1 For /?4 = /?3, /fj = /?4 — /? 3 = 012 v>x/biv — — R4R3) — and \)yZ\iy/ — also. 3,
<x>,
00,
For Ra = R/2, Ri= V/V.
 R4 
=  R;A x>ÿw = V(l
For \)rA)w = 10 V/V, 10 = 2ÿ(1  Rÿ). Thus (1 R/R4 = 0.8, whence RA = O.8R3.
 RJRJ = 1/fl 
 RJR3) = 210 =
1/2) = 2 V/V, and oK\)w = 4
0.2, and
2.38 Use the results from P2.37 above: Input resistance at X to the right is (/?), and to the left is R
—— R
Overall, at X, the resistance is R II (/?) = ~ R
= 00! Now the short circuit
.
current at X to
.
ground, with V/ applied is V/ZR (from W) and 0 from Y for a total of iyR Thus the Norton equivalent at X is a current source I n = t)/R with a shunt resistance Rn  <». For an impedance Z connected at X, the current flowing will be Ojdl and the voltage will be V/ZZR . In general, the transmission x>xZ\)w = 2Z/R. For a capacitor for which Z = V(Cs ), %A>w = 1ZsCR . For this noninverting Z/R and vY/ow integrator, the time constant is RC. The unitygain frequency is co„ = 1/(RC)
—
2.39 The commonmode input (at 60Hz) is 8 V peak, and the output is 0.6 V peak. The differencemode input (at 1 kHz) is 1+1 =2 mV peak, and the output is 60 rnV peak. Thus, the commonmode gain = 0.6V/8V = 0.075 V/V, and differencemode gain = 60mV/2mV = 30 V/V. Thus the CMRR = 30/.075 = 400 s 201ogio 400 = 52 dB.
lO100"20 = 10s. For a differencemode output of 2 Vpp, the required commonmode signal out¬ put = 1/100 x 2 = 20 mVpp. Thus = 10s, and vicm = — x 20 x 10"3 = 10 Vpp.
2.40 CMRR =
ZA)tli V /
Ulcj)i
ZUU
SECTION 2.7: EFFECT OF FINITE OPENLOOPS GAIN AND BANDWIDTH ON CIRCUIT PERFORMANCE , A x / = /, = 107. Thus / 3ÿ = f/A„ = 107/106 = 10Hz. At / 2.41 At and above the cutoff frequency = 100 kHz, the available gain, A = f/f = 107/105 = 100V/V. 2.42 Closedloop gain of 20 dB corresponds to a gain ratio of 102tK2° = 10 V/V. In general, f,m fA 1 + R/R\Y For a gain of 10 V/V, R/Rt = 10, and f3dB = 107/(1 + 10) = 0.909 MHz. For a gain of +10 V/V, 1 + R/R, = 10, R/R{ = 9, and f3dB = 107/(1 + 9) = 1.00 MHz. The phase shift at the 3 dB frequency is 45°, and 6° at 1/10 the 3 dB frequency. Thus 6° shift is reached at 90.9 kHz for the inverting amplifier, and at 100 kHz for the noninverting, and is less than 6° for all lower frequencies. 2.43 Amplifiers have A0 = 106 V/V and /, = 107 Hz: For maximum bandwidth, use the noninvcrting form with gain (1 + R/R\). For a single amplifier, with / 3,//) = 10 kHz: 104 = 107/(1 + R/R ()• Thus 1 + R/R\ = 107/I04 = 103, and the available gain is 103 V/V. For 2 amplifiers in cascade, each with a 3dB frequency at / j, f 3dB = (f2 l)*4 f \ = 0.64f4/ ] . Noyv, for / 3,tB = 10kHz, f\ = 10kHz/.644 =

15.54 kHz, with corresponding gain per stage of stages in cascade, a gain of
R7
I+K,
107
15.54 x
103
= 643.6 V/V, and, for two
(643.6)2 = 4.14 x 10s V/V.
2.44 Gain with feedback at low frequencies is likely to be:  R/R 1 = 100 V/V. Assume A 0 1 » 100: Now, 120 (25) = 30 kHz. For the closedfor a 3dB frequency fB < 120 kHz, 100 = 120(25), and fL =
loop amplifier, the unitygain frequency (where Rÿi = 1) is 120(25) kHz = 3 MHz. But the unitygain

171

SOLUTIONS: Chapter #216
frequency of the ncgalivcgain amplifier is
(1
+fRjRxy which
and for the op amp itself, /, = 6 MHz. 2.45 /, = 100 MHz, /? i
*>/
A„ = 20 V/V:
Now, for
is
4" ÿ7 Ior
=
R/R i D,
l+ (l+/?ÿ,/A„
'•
Thus fA = 3 MHz
= 10 V/V,
1 R2 1 Ri Ri R2 1+ = = 10.5, and = 21. Check: + 2 /?,' 2 20 2 /?! Ri f, 21 100 x 106 10 V/V, OK. Correspondingly, / vfl = = = 4.55 MHz. 1 + 21 1 + (2220) l+RfRt
10+4
SECTION 2.8: LARGESIGNAL OPERATION OF OP AMPS 2.46 The largest possible peak output with zero average is 8 V. The corresponding input has a pcaktopcak value of 2(8)/ 10 = 1.6 V.
2.47 A 6V pp triangle wave at /Hz moves 6 V in 1/2 x 1// seconds with a slope of 6/(1/2/) = 12/V/s. Now, if this just matches the slew rate: 12/V/s = 10V/p.s, and / = 10/12 X 106 Hz = 0.833 MHz. SR 2.48 The slewratelimited bandwidth of an amplifier with a sinewave peak output V„ is fR = „ ,, . Now 2kV„ fR = fb, when V„ = SR42 n fb). Now, for SR = 2 V/fts, and fb = 0.5 x 106, V„ = 2 x 106A2 n X 0.5 x 106) = 0.64 V peak.
SECTION 2.9: DC IMPERFECTIONS 2.49 Nominal 4V peak swing is reduced by less than 4V/100 = 40mV. Since gain = 100 V/V, RfR\ = 100 40 and the gain for the offset voltage is 1 + R/R\ = 101. Thus, the required offset < = 0.40 mV. 2.50 a) b)
= 100; Rin = 100 kfl = /?,+ = 10 Mil For no compensation: x>0 = (1 + 100) (lmV + 30 X 109 X 100k II 10M) = 101 (1 mV + 3 mV) = 404
Af
mV = 0.40 V. For compensation, using R} = Rt II R2 ~ 100 kfl, only the offset current and offset voltage apply. Thus u0 = (101) (lmV + 3 X 10~9 x 100 k) = 131.3 mV = 0.13 V. For case (a), bias current dominates; For case (b), offset voltage dominates. For each dominant effect halved, the output offset becomes: (a) 101 (1 + 3/2) = 0.25 V; (b) 101 (1/2 + 0.3) = 0.08 V.
2.51 vl

100 kQ
•1
Note that the offset gain = 1V/V :
= 1 X Iffi3 30 x 10"9 x 10 X 106 = 301 mV = 0.3 V. (b) For compensation with /? 3 = 10 M£2 from the negative input to ground, Do = 1 mV + 3 X 10~9 x 10 x 106 = 31 mV = 0.03 V. Thus, use 10 Mil to compensate. (a) For no compensation, X)q
10 MQ
vo
J71

172
SOLUTIONS: Chapter #217
2.52
For no offset compensation, vi
R
= 101 [2xl0"3+(R II (100 R)) 1.1 x 10"6] < 0.5V. Thus 2 x 10~3 + 1.1 x 106/? < 4.95 x 103, „ 2.95 x 10"3 2.68 ki2. and R = ———= 1.1 X lO""6 For compensation, with R?, = R, 1 x 10"6)] < 0.5V. \)0 = 101 [2x10~3 + R(2
100R
j
RJt 100R = R
2 95 x
in3
Thus R = r = 14.75 kO, (use 15 kf2). 7 0.2 x 10~6 For this design, Rin ~ 14.75kO = 15 k£2. 2.53 a)
Positive limiting is caused by current flowing out of the capacitor at the op amp negativeinput node. Contributions include: bias current of 100 nA, offset current (worst case polarity) of 10 nA, current from the input resistor when the offset voltage is + 2mV is hnVAQkkl = 200nA Total current is 100 + 10 + 200 = 310/iA . The voltage reach ouput will positive saturation in T = CV/l = 0.1 x 10~6( 10/(3 10 x 109) = 3.22s.
.
b)
For negative limiting, current must flow into the input end of the capacitor. Maximum current from input resistor (with offset  2mV) is 200 nA. Maximum offset current is 10 nA. Minimum bias current is 100 nA (flowing into the top amp). Thus the total current is 200 + 10  100 = llOnA for which nega¬ tive saturation occurs in T = 0.1 x lOÿ 10/110 x 10~9 = 9.09s. For Compensation: Make Ra  R\ = 10kf2, in which case the bias current flows through R i (and Ra) to produce voltages at each input which are equal and cancel. Adjusting Rc with Rh provides an additional ± 2mV on the positive opamp input to cancel the voltage offset, which though unknown may be rela¬ tively stable (certainly if the temperature is fixed). Even the offset current can be compensated on the short term. Typically a factor of lOx improvement is possible (at least) with saturation time increased to perhaps one minute. Because Ra ~ R j, doubling the bias current does not matter much. (Again cancella¬ tion is usually good to within a 10% difference error.) For compensated positive saturation, the choice of Ra = 101:£2 compensates bias current, while the x Vc = 2mV or current from Rb compensates for the offset voltage. Here, vc = 10* 12+10m L2
1/1001Vc = 2/1000 and Vc ~ 2V. Note that the offset current flowing in 10 k£2 produces a voltage of 10 x 10"9 X 10 X 103 = 0.1 mV. For this included, Vc = (2.0 + 0.1)1001/1000 = 2.1V 2.54 The nominal closedloop gain is G = 1 + R/Ri = 1 + 101:12(11 10Mil) is responsible for an offset of
lO/lO4 = 101 V/V. The bias current flowing in = H88 mV. Thus the bias current is
icr3 r— = 1.19liA. The input offset voltage is 0.6/101 = 5.94 mV. Note that a fraction of the 5.94 —lOxlO3 1 1 :88 x
mV voltage offset is due to the offset current flow in 10 kfl. Now, with all resistors reduced by a factor
of 10, the change in input offset voltage is
= 198 mV, due to the difference between I„ffse, 1.98*9kCl = 0.22 [lA . Now the flowing in 10 kf2 and in 1 k£2. That is, I„ff(l0k  Ik) = 1.98. I offset original offset voltage 5.94 mV at the input includes the Vos + I„ffse, X 101:12, or Vos = 5.94mV  220 x 10~9 x 104 = 3.74 mV. To summarize, for the basic amplifier, Vos = 3.74 mV, = 1.19 pA, = 0.22 pA. Now if the 10 kt2 resistor at the negative input is capacitorcoupled, the offset voltage is multiplied only by 1, but the bias and offset currents flow in 1 MI2. If they add, the output offset becomes 1M£2(1.19 + 0.22) X 10"6 = 1.41V!. Two possible compensations arc possible:
—

173 
SOLUTIONS: Chapter #218
a)
b)
c)
If the resistances R \ and R2 are reduced by a factor of 10, the output offset reduces to 0.14V. If a resistor Ri = 1MCl is connected from the positive opamp input terminal to the actual input, which is grounded, bias currents are compensated, and the output offset becomes 1M£2(0.22 X 10"6) = 0.22V. If both techniques are used (ie using a 1 k£2 resistor and two 100 k£2 resistors), the offset due to offset current reduces to about 22 mV. With the voltage offset remaining, about 22 + 3.74 = 25.7 mV would be found at the output.

174
Chapter 3 DIODES SECTION 3.1: THE IDEAL DIODE 3.1
Diodes are ideal: Thus the forward voltage drop is OV and reverse current is 0 mA: (a) (b)
Diode is polarized to conduct by the +5V and OV connections: Thus Va = OV, and la = (50)/ IkO = 5mA. As in (a), but the most negative supply is 5V. Thus Vb = 5+0 = 5V, and Ib = (+5  0 
(5))/lkfl = 10mA. (c) (d)
3.2
Diodes both conduct: Thus Vc = 10 + 0 + 0 = 10V, and Ic = (0 0 0 (10))/lk£2 = 10mA. Both diodes are polarized to conduct. Thus Vd = +5 0 = 5V, and Id = (+5 000)/ lkil = 5mA.
Ve = 5V —
(e)
Upper diode is polarized to conduct, but lower to cut off. Thus current le = 0mA, and 0V = 5V (with the upper diode conducting the meter current).
(f)
Upper diode polarized to cut off, and lower to conduct. Thus current, If = 0mA, and Vf = —5V 0  0 = 5V (with the lower diode conducting the meter current (assumed very small) through lkfl).
(a)
i)
V& = +5V; upper diode conducts; Vy = +5V.
ii)
The output is high if either input is high. For high (+5V) defined as logic '1', the output is 1, if A = 1 or B = 1; ie, Y = A + B. Thus, the function is logic OR (in positive logic).
iii)
For high (+5V) defined as logic '0', the output is high (logic '0') if either A or B is high (logic '0'). That is Y = A + B , or Y=Y=A+B=A • B = AB (for simplicity). Thus the function is a logic AND in negative logic. The AND idea can be verified by noting that the output is low (logic '1') only if A and B are both low.
i)
, Vc and VD are all 0V. Vy follows the highest input. Thus Vy = 0V. Y = A + B + C, an OR (in positive logic). Y = A • B' C = ABC , an AND (in negative logic).
(b)
ii) iii) (c)
i)
See that the output follows the lower of A or £ (just as the output followed the upper of A or B in (a)). That is, = Ve = 5V > Vy = 5V.
ii)
From above, Y = A + B, ie Y = Y= A+ B= A • B = AB That is, the function is AND or, directly, Y = A • B = AB (since both inputs must be high for the output to be high). Directly, from logic first principles, or in analogy to the circuit in (a), Y = A + B (in nega¬ tive logic). That is, the function is an OR.
iii)
(d)
i) ii)
= 5V, VB = 0V —> VY = 0V. AND (positive logic).

175 
SOLUTIONS: Chapter #32
iii)
OR (negative logic).
(e)
i)
VA = VE = 5V, Vc = OV > Vy = OV.
ii)
AND; Y = A •£ • C = A£C. OR ; K = A +E + C.
iii)
3.3
Use positive logic, that is, '1' = 5V, '0' = OV. For D5 open, P  A • E = AE. For Z>6 open, Q = B • C = BC . Now, note that the current available from node P or Q ( 100(1A) exceeds that drawn from node Y: Thus if P or Q goes high, Y is also pulled high. Thus Y = P + Q + D and, Y = AE + BC + D (in positive logic). Now, for VA = VE = 5V; VB = Vc = VD = OV; that is A = E = T and B = C = D = *0', for which Y = l*l+0*0 + 0 =1. Thus the output is logic '1', or 5V (for ideal diodes)
3.4
12V rms = 12(1.414) = 16.97V peak For a 12V battery and an ideal diode, the peak diode current is (16.97 12)/(10 + 50) = 82.8mA The diode begins to conduct (and ceases to conduct) when 16.97 sincot = 12V .or sinco/ = 12/16.97 = 0.707 or cot = tc/4 s 45". Average value of
(16.97 sin 0
1 r3w* iD = —2k V4
—
 12) dQ)/60 =
VS
.
3W4
16.97V
3tt/4  it/4 = (.0635  .0500) 2n> A = 13.5mA. Now for VB = 14V: 14)/60 = Peak current = (16.97 0.0495A or 49.5mA. Conduction begins when 16.97 sincof = 14, or sincol = 14/16.97 = 0.825, or tot = 55.6°.
0.2
io

82.8mA 0mA
\
— 0
>> Olt 45°
135° 180*—55.6*
=P2n
Average current
2n
= 0.0507  0.0445 = 0.0062A or 6.2mA.
 176
(.565 .565)

0.233
SOLUTIONS: Chapter #33
3.5
Thus, overall:
foi = Iik
ID3 = /4/f  /D4 = 2  1 =
1mA;
ID2 = I2K  ID3 = 4 
 3 = 5mA; Thus diodes are all conducting, and V„  OV.
1 = 3mA;
SECTION 3.2: TERMINAL CHARACTERISTICS OF JUNCTION DIODES 3.6
Given: V, = 0.700V Now
@
/, = 100ÿA, and
iD = Is eÿ7 —» \)p
V2 = 0.815V @ I2= 1mA. = nVT In i[/Is = nVT In iD  nVT In Is.
Vd l— t)D2 = "Vy (111 io/lDl)Here, 0.700  0.815 = n VT In 0.1/1 = 0.025n ( 2.3026). n = 0.1 15/(.025(2.303)) = 1.997 or 2.00. Now Is = iD e'V,/"Vr = 100 X 10"6 e"70(K2(25,) = 10"4 e"14 = 8.32 x
3.7
i = 1$ e Thus,
\)t

10"17A.
(1). r, and  v2 = nVT In i/i2 = u2 + nVT In i/i2 = 700 + 1(25) In 1/0.1 = 700 + 25 In 10 = 757.6mV, or
Now from (1), i\ = i2 e
"V
7
—
— 8)/4k = 2mA, and IiK = (8  0)/8k = 1mA.
1mA;
ID4 = ISK =
~ Idi = 8

+ 8V
+ 8V
are all zero volts. Thus, /4ÿ = (0
Consider the conducting state of each of the diodes: Assume D\ conducts; thus its cathode is at OV. Correspondingly, the current in the IkO resistor, IlK = (0 8)/lk = 8mA. Assume D2 conducts; thus its cathode is at OV, and I2K = (8 0)/2k = 4mA. Now, since IlK > I2K, then D\ and D2 both conduct as assumed. Now, noting that succeeding resis¬ are tors progressively larger, assume that both D3 and D4 also conduct, and that their anode and cathode voltages
= 0.1 e<815  7°°yiC25)
0.758V.
_ 9.95mA.
WnV
3.8
= Is e 7 —> v = nVT In iAs, or x> = v„ + nVr In iA„. At 10mA, u = 700 + 2(25) In (10 x 103/10) = 355mV, or 0.355V. At lOpA, u = 700 + 2(25) In (10 x lOÿ/lO) = 9.2mV.
3.9
+ nVT In iA„ = 700 + 2(25) In 0.1/1 = 585mV. At 100}iA and 95"C, o = 585 + 2.0 (25  95) = 445mV, or 0.445V.
(
At lOOjiA and 25"C, v = t)„
3.10 The leakage doubles for each 10"C rise in temperature. Now at 95°C, it is 2(95 ~ 25)40 = 27 = 128 times its value at 25°C, or 128nA, or 0.128(xA. Now at 100°C, it is 2(l00~ 25)40 = 275 = 181 times its 25°C value, or 181nA.
 177
SOLUTIONS: Chapter #34
Note that it is 2(10° noted.
95)10
= 2a5, or 1.41 times as large as at 95°C, ie 1.41 x 128 = 181nA, as already
SECTION 3.3: PHYSICAL OPERATION OF DIODES 3.11 Acceptor concentration is NA = 10"™ . Thus the hole concentration is pp0 ~ 10~m, while the electron con¬ centration is npQ = 10~", n » m. More precisely, in 10" atoms, one is ionized at a particular tempera¬ ture to produce one hole and one electron. As well, the number of acceptor  produced holes is 10" X 10"™. Thus the total number of holes is 1 + 10" x 10"™ in 10" atoms, where 1 + 10" x 10~™ =10", while np0 = 10"" directly. Ppo = 3.12 From Eq. 3.6, n2 = BTÿe E"*r t where for silicon, B = 5.4 x At 200 K, ni2 = 5.4
1031 x 200V112*862 * and n, = 1.62 x 105 carriers/ cm3. X
At 300 K. n? = 5.4 x
,
*
200)
1031, Eq = 1.12eV, k  8.62 X 10"5.
= 2.639 x
1031 x 3003el. 12/(8.62 x 10"5 x
1010,
300) = 2.2616 x
1020,
and rij = 1.50xbold {10 sup 10}carriers/bold {cm sup 3}..IPAtA00K ,n sub i sup 2 "=" 5.4 "times" 10 sup 31 "times" 400 sup 3 e {1.12/(8.62 "times" 10 sup 5 "times" 400)} "=" 2.706 "times" 10 sup 25,.//,andn sub i "=" 5.20 X 1012 carriers/ cm3. 5 2 x 1012 1 5 x 1010 'n;x = 3.46 x 104%! the increase is —For the 100 °C rise above room temperature, F 1.5 x 10'° 5 2 x 10'2 At this temperature (127 °C), the fraction of ionized atoms is — ~ '0~10, or one in every 10
—
(US) billion! Note that the reference on page 139 of the Text is to British billion.
3.13 From Eq. 3. 10C, resistivity is p = \Z[q(p\ip + np.„)]12cm. (a) For intrinsic silicon at room temperature: n = p = «, = 1.5 x 10l0/cm3. Thus p = H1.6 x 10",9(1.5 x 10'°)(1350 + 480)] or p = 2.28 x 105I2cm (b)
For ntype silicon with pn0 = 2.25 x 104.
ND
=
10l6/cm3: nno = 1016,
and pn„ = n2/ND = (1.5 x 1010)ÿ1016, or
Thus (including mobility reduction): p = 141.6 x 10"'9(2.25 x 104 x 0.8 x 480 + 1016 x 0.8 x 1350)] or p = 1.73 12cm. As noted in the solution to P3.12 above, for a 100 °C rise n, will increase by a factor of 3.46 X 104. Thus for the intrinsic material p will decrease to 2.28 x 105/3.46 x 104 = 6.59£2cm, while for the doped material, it will stay the same, at 1.73f2cm 3.14 The depletion region will be larger in the lighterdoped p region. In fact, it will be 10 x larger there than in the n region.
 
3.15 Eq. 3.18 in the Text states that the builtin voltage is V0 = VTln[NAN[/iij2], where for intrinsic silicon at room temperature. Here 700 = 251n[/V)iA/041.5 x 1010)2]. (1) Thus NaNd = (1.5 x 10lo)2e7(Xm = 3.254 x 1032/cm6 (a)
For Equal Doping:
NaNd =N2 = 3.254 x 1032, whence N = 1.804 x 10l6/cm3 = 1.80 x 1022/m3. From Eq. 3.20 in the Text, Wdep = xn + xp = 2x = [(2c/qHVND + VNA)V 0]'A
tij
= 1.5 X
10l0/cm3
(2)
where (from Table 3.1 on page 157 of the Text), q = 1.60 x 10~I9C, e, = 1le„ = 11(8.85 x 10~")F/cm = 9.74 x 10~nF/cm = 9.74 x 10~uF/cm.

178 
and
SOLUTIONS: Chapter #35
Thus Wdep = [2(9.74 x 1CTI3H1.60 X 1(TI9)(2/1.804 x 10,6)0.7],/4 = 3.07 x 10~6cm = 0.307pm. The distance depletion extends into each region is W/2 = 154 nm. From the preamble to Eq. 3.21, the uncovered charge on each side is qj = qNDxn = 1.6 x 10~19 x 1.80 x 1032 x 154 x 10~9 x 30 x 10"6 x 50 x 106 = 0.665 pC. (b)
For 10tol doping:
From (1), NaNd = 3.254 x 1032/cm6, and for ND = 10Na, say, see 10N% = 3.254 x ND = 5.70 X 1016/cm3 = 5.70 x 1022/m3. From (2), 0.405pm.
Wdep = [2(9.74 x
1032, where NA = 5.70 x 10'Vcm3 = 5.70 x 1021/m3, and
10~I3X1.60 X
Now Wdep = xn +xp =xp{Na/Nd Thus xp = 0.405/l.l 0.368pm
10'9)(M;5.70 x 10,6)(1 + 10)0.7f'4.046 x 10"5c/n =
+ 1) =*,,(0.1 + 1).
—
—
xn XpAQ = 0.037pm. The charges uncovered on each side arc equal, and of value qj = qNDx„A = 1.6 x 10~19 x 5.70 x 1022 x 0.368 x 10"7 x 30 x 106 x 50 x 10"6 = 0.504 pC 3.16
—
Is Id = /
—
15  Id = 10 and lo
ÿ>
 15  10  5.
Thus the diffusion current is Io = 5nA.
3.17 Combining Eq. 3.21 and Eq. 3.22, see
qNAND nand ÿAWdep = na +nd Ha + Hp nand (V0 Vr) qj = a 2esq na + nd + qj
=q
2e,y
l
q
na
+
l
nd
.
+ Vr ) f whence
1
For the (30 X 50)pm 2 junctions in P3.15 above, for which b)
(a)
V0 = 0.7 F:
Na =Nd = 1.80 x 1022/m3. Na = 5.70 X 102i/w3, ND = 5.70 x 1022//n3.
a)
For Na = Nd:
= 30 X 50 x 10i2[2 x 9.74 X 10" x 1.6 x ~'9 x (1.80 x 1022/2)(0.7 + VR)t = 7.94 x 10'3[0.7 + VR]'A For VR = 0.0F, qj = 7.94 x 10,3(0.7),/i = 6.65 x 10'3C or 0.665 pC, as in P3.15. For VR = 10V, qj = 7.94 x 10"13(0.7 + IO)'"4 = 2.597 pC. For VR = 11V, qj = 7.94 X 10~l3(0.7 + 11)'/' = 2.716 pC. From AQ = CAV, the junction capacitance is C = AQ/AV = (2.716  2.597X11  10) = 119 fF. qj
Now, from Eq. 3.26, •
'
Cjo  A
e.t<7 2
ÿ
nand [Na +Nd
1
Vo

(1)
Thus, Cjo = 30 x 50 x 10_12[9.74 x.10"11 x (1.6 x 10~19/2)(1.6 x 10_l9/2)(1.80 x lOÿ/iyOJf' = 0.475/F. At 10.5 V, Cj = Cjd[1 + VrW0Y' = 0.47541 + 10.50.7]'/l = 0.475ÿ4 = 0.119pF = 119 fF same as found above.
Now for a graded junction with m = 1/3:
At
VR = 10.5V, Cj = 0.47M1 + lO.SOJ)173 =
189 fF.
 179 
SOLUTIONS: Chapter #36
VR = 100V, Cj = 0.47541 + 10OO.7)w = 90.6 fF For Na = No/10 = 5.70 x 102l//n3 qj = A [2zsqNAN[ANA + ND)(V0 + VR)]'A = 30 X 50 X 10~12[2(9.74 X ir")(1.6 X 10~19) X 5.70 x 1021 x 10/11Jÿ(0.7 + VRf = 6.028 x 103(0.7 + VRf. For VR = 0,qj= 6.028 x 10'3(0.7)'/' = 0.5043 pC. For VR = 10F: qj = 6.028 x 10~13(0.7 + 10)'/" = 1.9718 pC. At
(b)
= 11V, qj = 6.028 x 10"13(0.7 = ll)'7' = 2.0619 pC. 2.0619  1.9718 xr •. Now, the junction capacitance is C = AQ/AV = j= 90.1 fF.
For
VR
aÿ./a„
Now, from Eq. 3.26 ((1) above),
Cjo = 30 x 50 x 10"l2[9.74 x 10"" x (1.6 x 10"l9/2)(5.7 x 102I)(1M1)(M).7)]71 = 0.360/jF. and Cj = CJrAl + Vr/V0]v'. At VR = 10.5F, Cj = 0.360/(l + 10.5D.7)'7' = 0.3604 = 0.090 = 90 fF, very much like the earlier esti¬ mate. Now for a graded junction with m
 1/3.
VR = 10.5F, Cj = 0.36041 + 10.541.7) = 0.360416) = 143 fF. At VR = 100F, Cj = 0.36041 + 1000.7) = 68.7 fF At
3.18 Q = CV
—> C = ~Ty— = O.lpF.
That is, the depletion capacitance is O.lpF.
3.19 From Eq. 3.27, Cj = Cy0( 1 + FR/F0rm. For VR = 2F, 1.8 = C;o(l + Wo)"16.
VR = 10F, 0.2 = C,0( 1 + lOVo)"1'6Divide these to get 9 = [(1 + 2Y0K1 + lOYo)]"1'6 Thus 1 + 2Y0)41 = lOV0) = 9171'6 = 0.253, and 1 + 2/F0 = 0.253 = 2.5W0, and 0.5W0 = 1  0.253 = 0.747, whence F0 = 0.530.747 = 0.709V. Now, 0.2 = Cj0( 1 + 100.706)16 = 0.2(15.1)' 6 = Thus Cj0 = 15.4 pF. For VR = OF, Cj = 15.4(1 + 00.71)'6 = 15.4 pF, as expected, the same as Cyo, by definition. For
3.20
PD = VB IR.
Tims, IR =
=
50
*
= 0.42mA.
For breakdown only 10% of the time, a peak current of 0.42/0.1 = 4.2mA can be tolerated. 3.21 From Exercise 3.12, in doped silicon, pp = 400cmWs.
DP = Fÿp,, = 25 x 103 x 400 = I0cm2/s . From Eq. 3.30, the diffusion length, Lp = V DPiP where %P = 1/ii. Thus LP = 10(1) 109 = 10~4cm = 10~6m = 1pm. For the diffusion profile, the excess hole tion is pn(x)= p„{0)e~x/L where a 10% level is reached when e~x/L =0.1 or x/L = 2.3. From Eq. 3.12,
Thus the conentration reaches 10% at 2.3L = 2.3pm from the depletionregion edge. 3.22 From Eq. 3.34, ls = Aqiij
Dp Lp N/j
Dn Ln Na
where from Table 3.1 on page 156 q = 1.6 x 10_19C, «, = 1.5 x 10lo/cm3.

180
concentra¬
SOLUTIONS: Chapter #37
Now, from Eq. 3.12, Dn/\ln = r/; = VT and from Eq. 3.30, L = whence D/L = jiVr/(p.VrT)'/' = (iVrA)/'. Thus Dp/Lp = (400(25 x lOÿ/lO"9)7' = 105cm/Vj = 103m/Vs
Dn/L„ = (1100(25 x Thus Is = 3 x 10"6 x 5 x
10X2 x
XT
10"9))'/'1.17 x 10scniAIs = 1.17 x 10ht/Vs.
10"6 x 1.6 x 10"19 x (1.5 x 10'° x 106)2[ lOÿlO16 x 106) + 1.17 x 103/(1017 x 106)] 102[1019 x 5.4 = + 1.17 x 10~20] = 6.03 x 10"nA
3.23 From Eq. 3.34, D„
Dn EÿkT where n2 = BT}e and LpND L„NA 5=5.4x 1031, Eg = 1.12eF, k = 8.62 x lO~5eV/K.
Is = Aqtij
t
We see that the temperature dependence originates in the find the coefficient:
At 300 K, n? = 5.4 x 1031(300)3e " l l2<8'62 x
l0"5(300))
n2 term.
Evaluate this at two temperatures and
= 1,458( 1039)ÿ433
= 1.458(10)39 1.55 1 x 10"'9 = 2.2616 x 102°. At 303K, n,2 = 5.4 x 1031(303)3e l l24862x l0 ÿ303» = 1.502(10)392.382 x 10"19 = 3.5782 x 1020.
_
coefficient is Thus the temperature v
(2.2616)(303  300)
x 100 = 19.4 >°C
3.24 From the solution to P3.22 above, in which (from Eq. 3.33),
I= Aqn2
Dn
LPND
Dn LnNA
e
V/V
1
the ratio of the hole to electron components lp/In = 10 Of the total, I= lp + /„ ,
Ip/I
In Ip+In
1
l+/„4
19/1.17 X
= 89.5% (holes), and In/I =
""
10
20
= 8.55
1
IpI„
1 =I/In
= 10.5% (electrons).
For a total current of 1 mA flowing in the junction, Ip = 895jA flows as hole current and /„ = 105pA , as electron current. Now, in general, the stored charge is Q = T/ for each current component. For holes, Qp = ipIp = 1 x 10~9 x 895 x 10*6 = 0.895 pC.
For electrons, Qn = T„/„ = 2x 10"9 x 105 x 106 = 0.210 pC. Total stored charge is Q = 0.895 + 0.210 = 1.105pC, tT = Q/I = 1.105 x 10~12/1 x 10"3 = 1.1 ns. Here, from Eq. 3.39, Q = (tTA/T)l = T Tl/VT = 1.1 x
and
the
mean
10"9 x 1 x 10"3(25 x 10"3), or Q =
transit
time
4 pF
Thus xr = nVpCj/I. For this junction, 3.25 From Eq. 3.39, generalized, Q = (%T/nVT)I. xT = 2(25 x 103)1 x 10,2/l x 10"3 = 50 ps. For a junction lOx larger, XT is the same, namely 50 ps, simply because it relates stored charge to current level at an essentiallyconstant junctionvoltage. For a larger junction, the current density changes, but the total charge does not. Total stored charge at 1 mA is Q = xTI = 50 x For 10 mA, (2io = 10(50) = 500/C = 0.5 pC.
10~12 x 1 x 10~3 = 50 fC.
 181 
SOLUTIONS: Chapter #38
3.26 From Eq. 3.38, q = Xji and i = l$e ÿ"Vr) Thus q Now, Q = dq/dv> = xTIsA,nVT)eÿ"Vr .
— xxIseÿ"Vr.
Thus Cd = xtKiiVt), as Eq. 3.39 indicates (/ being substituted for i and n added for generality).
SECTION 3.4: ANALYSIS OF DIODE CIRCUITS 3.27
The exponential v characteristic
Straight line D Slope
RT Wv
VT
 —
ID
1
1
Straight line A
t'/,(V
For the Load Lines: (b)
Vy = IV, Rt = 10012, 1T = 1/100 = 10mA. Vr = 0.9V, Rt = 10012, lT = 0.9/100 = 9mA.
(c)
Vy
(a)
= 0.9V, Rt = 900, lT = 0.9/90 = 10mA.
From the Graph: (a)
VD = 0.75V, ID = 2.5mA. ID = 10  Vo/0.1 = 10 10Vo = 10  10 (.75) = 2.5 mA. VD = 0.73V, lD = 1.7mA. Check: lD = 9  Vy/0.1 = 9  lOVy, or /e = 9  10 (.73) = 1.7 mA. VD = 0.74V, ID = 1.8mA Check: = 10  Vo/.09 = 10  11.11 VD = 10  11.11 (.74) = 1.77 mA. Check:
(b)
"
(c)
3.28 Though the diode is "like the one sketched in Fig. 3.20", use its analytical description to provide a com¬ parison with the results in P3.27a) above.
.
For the diode: x> = 0.7 + 0. 1log

Solve iteratively: Initially with 1 = 1:
— log 1=3 — 0 = 3mA, i =3 —» / = 3 — log 3 = 3  .48 = 2.52mA, 1=
1—ÿ1=3

182
SOLUTIONS: Chapter #39
i = 2.52 —» i = 3  log 2.52 = 2.60mA, whence i = 2.60 > / X 30 log2.60 = 2.59mA , whence
—
—
ID = 2.59mA, and V0 = 0.7 + 0.1 log 2.59 = 0.741V. 3.29 For the diode, 1)
— Vp„ + rDi Here, VDD = 1.00 = (1000) i + 0.65V + (200) i,
or for i in mA, ID = i = 3.30 For
VDO = 0.70V, and rD = 100, 1.00 = 0.1/ + 0.70 + 0.01/, and
Id = =
'
For
= 2.92mA, and VD = 0.65 + 2.92 (.02) = 0.708V.
oT:o0oT = "577" = 2'73mA'
VD0 = 0.75V and
_
with
V° = 0J0 + 273 (a01) = °727V
rD = 00, ID = i = q ÿ
The exponential characteristic
—
= 0.750V.
(a)
For Vr = 0.8V and RT = 1000, /r = 0.8/100 = 8mA. From the Figure, see VD = 0.7V, ID = 1mA. Check: ID = 8  Vo/0.1 = 8  10VD =810 (0.7) = 1 mA.
(b)
h=
Straight line B
Slope
= 2.50mA, with
'/»
= Q + Q Q2
125mA' and vd = 0.65 + 0.22
(1.25) = 0.675V. ÿ
(c)
(d)
Straight line A
6ÿ0.8
1.0
b=
OX
— O
70
Q1 + 0 01
(0.91) = 0.701V. 08 ~
Id =
= 0.91mA, and VD = 0.70 + 0.01
1°'75 =
0.50mA, and
VD = 0.750V.
,V
1 /XI
Check: Overall, using the logarithmic model, the supply voltage is V = 0.1(/o) + 0.7 which for ID = 1mA , d = 0.1(1) + 0.7 + 0.11og(l) = 0.1 + 0.7 + 0 = 0.8, as proviced, OK.
+ 0.11og(/o/l)
3.32 Drop across each diode = 4.0/5 = 0.8V. Thus 0.8V = 0.7V + 0.1 log (IpAinA ), or log ID = (0.8 0.7)/0.1 = 1, whence Ip 1o'°8 = 101 = 10, or ID = 10mA (as can be seen directly). Now R = (10 4.0)/10mA = 0.6kO, or 6000.
—
3.33 For R = 5000 used in P3.32 above, with each diode having a drop of D volts: See, i =
i0~55X>
= 20  10u, also u = 0.7 + 0.1 log ill.
Solve iteratively: u = 0.8V > / = 20  8 = 12, u = 0.7 + 0.1 log 12 = 0.7 + 0.1079 = 0.8079V; \) = 0.8079V > / = 20  8.079 = 11.92, t> = 0.7 + 0.1 log 11.92 = 0.7 + 0.1076 = 0.8076V; •u = 0.8076V
» / = 20  8.076 = 11.92 mA, x> = 0.7 + 0.1 log 11.92 = 0.7 + 0.1076 = 0.8076V.
Thus the diodestring voltage becomes 5(0.8076) = 4.038V, or 4.04V.

183 
—
SOLUTIONS: Chapter #310
3.34 After shunting, three identical diodes in parallel share the current equally. Original current is i = (1) io'0807ÿ' = 10mA (as can be seen directly).
Final current in each diode = 10/3 = 3.33mA. Final voltage drop = 0.7 + 0.1 log 3.33/1 = 0.752V.
SECTION 3.5: THE SMALLSIGNAL MODEL AND ITS APPLICATION 3.35 In general, r = nVT/lD. At 0.1mA, r =
= 50012. At 10mA, r =
0.1mA
= 512. 4!r~r 10mA
Now the "average" resistance would more likely be considered at (0.1 x 10)'7' = 1mA , than at (0.1 + 10>2 = 5.05mA. Thus the geometric mean is likely to be most relevant, and therefore "best". But, let us check:
See (r\
+ r2)/l =
2(25)
r =—
——
= 5012, and at 5.05mA, r =
(r i r2)v' = ÿ500
x 5 = 5012. Now, at 1mA, Note the correspondence 9.912. of the value at 1mA = ———
+ 5)/2 = 252.512, and
(500
2(25)
1mA 5.05mA with the geometric mean, as expected. Note that the arithmeticmean results appear to be less easily interpreted.
3.36 At 2mA: r
nVT 2(25mV) = ——— = 25 12. With a second identical diode shunting the first, the current is — —— 2mA ID 6
shared equally: 1mA in each. At 1mA: r =
2(25)
= 50 12, and for two in parallel, req = 501150 = 25 £2. This demonstrates that diode incremental resistance is independent of diode junction size.
__
3.37 For small 1)j, use the smallsignal slope resistance, r. In general, r =
For / = 10mA: r =
= 5£2; V(/os = = 50£2; 5Q
For I= 1mA: r = For / = 0.1mA : r =
For I= 0.01mA : r =
0.1
= 500£2;
.01
5+5100Q
nV
7* —— I
and
I)# 1)5
f
r
+ R$
= 0.00498V/V = 0.005V/V.
ÿ000 = 00476V/V = 0.05V/V. 500+ 1000
= 5000Q;
= 0.333V/V = 0.33V/V.
ÿ5°°°
5000 + 1000
= 0.833V/V = 0.83V/V.
3.38 For d0 = % = 0, see that I\ and / 2 both split equally between D\, D2 and Z23, D4 respectively. Thus all diode currents are equal. In general, for 1)5 0 (say +8), the currents in £>1, D2 redistribute with their sum = I\. Extra current in £>2 needed to drive Ri high implies that there is less current in D4 than in the balanced case, the extra 0 is provided half by that in D2 part of / 2 flowing in £>3. In general, current needed by RL for 1)0 increasing and half by that in D4 decreasing, with corresponding changes in D\ and £)3, such that load current originates ultimately in vs, as Kirchoff's current law would require.
*
*
For small 1)5 (around zero volts), with all diode currents equal to ill, •
Rt 
jrf[r + r2\1 (f3 + r4] = i
For I= 10mA, RT =
For / = liA, Rt =
10mA
50 50 + 1/1 m
II
50 50 + 1/1 l/l
200 I
*°*
200 _K)0 = I I
, 10a: = 0.999V/V. = 100, and with RL = 10k£2, 1j(/o5 = 10 +
ÿ7ÿ = 100k£2, and VoA>s = , '°f 10k = 0.0909V/V. 100k + lpA in,
 184
SOLUTIONS: Chapter #311
Concerning Linearity: Signal size (for RL = lOkO) is of little concern at 1mA, since the load current is likely to be a small part of /. But, at low /, the load current may cause the diode current to vary over a wide range. Operation is linear for diodevoltage variation of ±10mV or so: Note that the corresponding variation is Ai, where 10 = 2(25) In
or In

= .20, or AiA 1.2, corresponding to 50 a variation of about 20%. Now for a positive output signal, iD2 can increase by 20% of its normal current III, while iD4 decreases by the same amount, the output current being the sum of the two lpA changes. Thus for RL = lOkO and I= lpA, X>0 is limited to 2 x 0.20 X 10k£2 = 2mV peak, in current
which case % is restricted to
i
i
2mV = 22mV peak. .0909
An Alternative View: The largest (positive) output occurs when the drop in D2 increases by lOmV, and that in D\ decreases by lOmV. Thus, at the limit, —v0 = 2(10) = 20mV. But for lpA and a 10k£2 load, x>qA)s = 0.09, or u0 = 0.09 x>s Thus, the limit is at 1)5  .09 % = 20mV, or % = 22mV.
.
3.39 For the regulator, ID =
104
= 10mA. For each of 5 diodes, r =
nVr
—— =
2(251
ÿ
= 512; Resistance of
the total diode string = 5(5) = 2512. For ±10% supply variation, expect an output variation of x 100 = ±1%. For a 2mA load increase, x (0.1 x 10) = ± 40mV, equivalent to ± ± 4V 25 + 600 the diode current reduces from 10mA to 8mA (by a small amount) and the output drops by 5(5£2) (2mA)
= 50mV, or
X
100 = —1.25%. For both effects, assuming approximately linear operation, the com¬
 
bined drop would be [40mV 50mV] = 90mV or 2.25%. The lowest output voltage would be 4.00
 .09 = 3.91V.
From First Principles: (for a O.lV/decade current change (which is not n
10 = 9V. For diode current i and voltage o, see
ÿ
 2 precisely!)), Vs
= 90% of
 8.33 \>
(1),
.5° = + 2, whence
' 0.1 log (13
0.6
i = 13
 8.33 \>). + 0.1 log (13  8.33 (.782)) = 0.7812V, and
(2). Thus D = 0.7 + and x> = 0.7 + 0.1 log i/l X) Iterate: with = 0.8  .09/5 = .782 initially. Thus u = 0.7 v = 0.7 + 0.1 log (13  8.33 (.7812)) = 0.78123V. Thus the output drop is 5 (0.8000 .78123) = 93.9mV.
—
3.40
rj
= nVT/Iÿ
—» n = (50£2 x XmAyiSmV
jo
0
(1). Thus, at 5V, 0.2 =
(1 + W0T
(3). Now (l)/(2)
0.2/0.1 =
= 2. Also Cj =
V„ + 10 ID v0+ Vo + 5
> 0.75/0.2 =
m
= 2.0
1 + W0 1 + W0
V0 = 0 V:
and
0+5 0+1 0+10 0+5
1+ 5 V0 = 1.0: 1 + 1 1 + 10 and
For
1+5
Vp + 5
Vo + 1
(5).
= 3.75 + m = 0.82,
= 2.0 > m = 1 , A = 0.18.
= 3.75 + m = 1.21,
= 2.0 + to = 1.15 , A = 0.06.
 185
. Thus,
at
VR  , 0.75 =
(2). Thus, at 10V, 0.1 =
Explore trial solutions of (4), (5). For
70
(1 + VVVo)"
= 3.75
jo
(1 + lOVo)"
(4), and (2)/(3)
SOLUTIONS: Chapter #312
For
2 +5 2+1
V0 = 2.0:
2+10 2+5
and
See result is nearer
and
 1.56,
= 2.0 » m = 1.27 , A = 0.29.
—
F0 = 1 » Try:
0.5 + 5 0.5+1 0.5 + 10 0.5 + 5
F„ = 0.5:
—
= 3.75 > m
= 3.75 = 3 .666™ + m = 1.02, t
= 2.0 = 1.909m > m = 1.07 , A = 0.05.
See between 0.5 and 1.0 + Try:
0.7 + 5 0.7+ 1
F0 = 0.7:
= 3.75 = 3.353m + m = 1.095, I
0.7 + 10 0.7 + 5
and
= 2.00= 1.877m > m = 1.10 , A = 0.005 = 0.
Overall, conclude that n = 2, m = 1.10, Now, at
F0 = 0.7V, and Cj0 = 0.75 (1 + 1/0.7)'' = 1.99 pF.
,, = 1.99 pF. F0 = OV, C.1(0) = (1 00.7) 
For forward conduction: As noted on page 155 of the Text, Eq. 3.27 does not properly represent Cj for forward conduction. [For example, check here, where if the diode is forwarded biased at VD = F0, and that Vr =  Fo and Cj becomes infinite!] Rather, for forward conduction, one uses from experience Cj = 2Cyo = 2(2) = 4pF, here.
CT = Cj + Q = lOpF, then Q = 10  4 = 6pF. Generally, Q = (xT/nVT)ID Thus XT = nVTCdAD = 2(25 X 10"3) x 6 x lO~i2A
Now, at 1 mA, as
X
10"3 = 300 X 10"12 =
300 ps.
Correspondingly, at 10mA, CT ~ 10(6) + 4 = 64pF. 3.41 For a diode lOx the area of that in P3.40, but otherwise using the same technology, both n and XT will be the same, 2 and 300 ps, respectively. Thus, at Ip = 5mA , r = nVjAp = 2(25)5 = 10 £2, and Cj  (1t/hVt)Id = [300 x l0~n/(2 X 25 x 103)] x 5 x 10"3 = 30pF [Aside: This could be seen more directly, since Cd must be 5x larger at 5 mA then at 1 mA. Correspondingly, Cd = 5 X 6 = 30pF Since the junction is lOx larger, so must be Cj. Thus, Cj = 10 X 4pF = 40 pF.
.
.
Thus at 5 mA, CT = Cj + Q = 40 + 30 = 70 pF. For 10V reverse bias, Cj will be lOx that value given in P3.40 (which was 0.1 pF). Thus, here, C;(10)= 10(0.1)= lpF
____
SECTION 3.6: OPERATION IN THE REVERSE BREAKDOWN REGION ZENER DIODES
20 (5 0.2) = 6.70V. For no load, breakdown is sustained for supply vol1000 967 0.2 tages down to about 6.7V. For a 9V supply, and bare breakdown, the load can increase to 200£2 11.3 = 5.65mA load, the lowest supply voltage for regulation is V, where = 11.5 0.2 = 11.3mA. For
3.42 Knee voltage
= 6.8V 

ÿ
V 6.1
0.2k
 5.65 = 0.2, from which V = 6.7 + 0.2 (5.85) = 7.87V. 
186
_
Thus the lowest supply for regulation
SOLUTIONS: Chapter #313
with halfmaximum load = 7.9V. 3.43 Line regulation (Eq. 3.60) is
R
+ r,
20 200 + 20
= 0.0909V/V = 90.9mV/V.
Load regulation (Eq. 3.61) is  (rz II R) =  (20 II 200) = 18.212 or, 18.2mV/mA. 3.44 Worst case: Vs low, IL high,
*<
9 (*S) 68 (1.03) 10 + 0.4
m
8.55 7.004 =0 10.4
The lowest output occurs for 10mA load,

8.55 6.596 10.0 = 3.20mA, for which 0.148 More precisely:
.031 = 6.565V
lz =
9(.95)  6.8 (1.03) = 10mA + 2 (0.2)mA, whence R
Vz high, where
148kÿ
Vz
Vz
low,
Usc 150£1
_
9 (.95)  6.81 (.97) 10.0 = Vs low, where Iz ~ 0.148
~ 6.8 (.97)  (5
 3.2) 20£2 = 6.596  .036 = 6.56V.
5ÿ 14j'56  10 = 3.45mA, for which Vz ~ 6.8 (.97)  (5  3.45) 20 = 6.596 
8
= 6.57V.
ÿ (103) _
The highest output occurs for 2mA load, Vz high, Vs high, where Iz ~ 0» x 4o 9 45  7 004 (L03) (14'52 ~ 2 = 1452mA' for which Vz = 68  5> 20 = 7.004 + .1904 = 2= + 0 148 7.194V = 7.19V. More precisely,
lz =
94ÿ0194 0.148
= 15.24mA, for which Vz = 6.8 (1.03) + (15.24  5.0) 20 = 7.004 +
0.205 = 7.209, or 7.21V. 3.45 R
vo
—•
VwV
VDD — 24 V
j
1 6.8V Zener *
at
£V = 2 (6.8) + 2 (0.7) = 15V, but since Vz is specified at 20mA and VD at 10mA, use an lz in between, where 2 (20  lz) (5£2) = 2 (/z 10, 40  2lz = lz  10, 3/z = 50, 10) (2.512), or 2 (20  Iz) = lz and lz = 16.7mA. For a nominal load of 15mA, and nominal 24V 24  15 supply, R = — — = 0.284kf2. Use a 27012 resistor, as a stan¬


— —
dard value. For supply 10% high, resistor 5% low, and no load,
= 102'9
/z *
"
3'899,,z AISO
Vz = 15V +
"
+ 2 (2.5)) x 10~3 = 15 + .015 7Z  0.250 = 14.75 + .015 whence /z, /z = 102.9  3.899 (14.75 + 0.015 lz) = 102.9 57.51 .0585 /z, and I = 42.88mA, for which Vz = 14.75 + .015 z = 16.67) (2 (5)
(42.88) = I5.39V. For each zener at 42.88mA, Vz = 6.8V 20) (.005) = 6.911V, and PD = 6.911 x 42.88 = 296mW.

+ (42.88 
SECTION 3.7: RECTIFIER CIRCUITS 3.46 The peak output is 8ÿ~2  0.7V = 11.31  0.70 = 10.6V: The diode conducts for about Vi cycle, but more precisely, between the pqints where 11.31 sin 0 = 0.7, or sin 0 = 0.0619, for which 0 = 55} 180 2 —7ÿ— = 0.48 or, 48% of a cycle. Ignoring the diode 3.55° and 180 3.55°, that is, for 

—
360 drop, the average output = 8ÿ 24c = 3.60V. With a constant 0.7V drop for 0.48 of a cycle, the average output is 3.6  0.48 (0.7) = 3.26V. Alternatively, and perhaps better, including the diode, the peak vol¬ tage is 10.6 V, and assuming the waveform to be approximately a halfsine, the average output is 10.64t= 3.37 V The peak inverse voltage across the diode is approximately the peak input = 8ÿ~2 =
 187 
SOLUTIONS: Chapter #314
\k 11.3V. For Rs = SOD.. rD = 100 and a load of lk0, D0 for a half cycle, s  0.7) — 0 = y(Dc (1 + .01 + .05)k or u0 ~ 0.94 \)$  0.66. Correspondingly, at the peak, the peak output is 0.94 (11.3)  0.66 = 9.97V. Overall, the average output is 0.94 (3.26) = 3.06V.
'
3.47 Peak inputs are ± 8ÿ2 = ± 11.31V. Peak outputs are 11.31 0.70 = 10.61 V, and 11.31 + 6.80 = 4.51V. For conventional forward conduction: (As noted in P3.46), conduction is from 3.6° to 176.4°, or about 48% of the cycle, For reverse conduction: current flows for the parts of the cycle between which 11.31

sin 0 = 6.8V, or sin 0 =
ft 8 ]j
= 0.601, for which
0 = 36.96° or 37°, and 180°  37° = 143°, that is 37)/360 = 29.4% of a cycle. Assuming for (143 that the rectified halfwaves are sinusoidallike, for the conduction period, the forward current average is about (10.6Mt)(48ÿ0) = 3.24V, and the reverse is about current average (4.5 lA)(29.4/50) = 0.844V. Overall, the average value of the output is 3.24  0.84 = 2.40V.
—
3.48 Fullwinding peak transformer output voltage = 16ÿ2 = 22.62V. V1
(opencircuited)
3.750.7

11.31V vo with Di only
3.05V
0.7V
3.75V 1kfl
6.8+0.7

7.5V
vo with D1, D2 as zeners
6.8V
For fullwinding voltages > (6.8 + 0.7) = 7.5V, D\ conducts in the forward direction (0.7V), while D2 breaks down (6.8V). At the peak, the shorteddiode current flow is (22.62  7.5>1OO0 = 151.2mA with D0 = 22.62<2  0.7  (10002) x 151.2mA = (11.31 0.7  7.56V) = 3.05V, (or, see this as 7.5/2 0.7

= 3.05V}. But current also
(lows
= 3mA from an equivalent source of
in Rlÿrl ~
II
pp
=
250, to produce an additional drop of 3mA x 250 = 0.075V. Correspondingly, v0 ~ 3.05 .075 = 2.975V. Thus, the peak value of the output is 2.975V, relatively constant while the sum of the opencircuit wind¬ ing voltages exceeds 7.5V. The peak diode current flow is 151.2mA + 3mA/2 = 152.7mA.

188

SOLUTIONS: Chapter #3—15
3.49 Transformer Peak Voltage = 12ÿ2 = 16.97V. Load Peak Voltage = 16.97  2 (0.7) = 15.57V. Now, 16.97 sin 0 = 0.7 + 0.7 = 1.4, for which sin 0 =
16.971.4 = 15.57V
16.97V
jg'gy
= 0.0825, and 0 = 4.73°. That is the output
is zero for 4(4.73°) = 18.92° per cycle or 18 92 1 x „ = 0.876msec. Thus, the average out360 60 16.97 1.4 = 10.80  1.4 = 9.40V. Peak put <=
—
— jt
Vo

Inverse Voltage for each diode is 16.97
16.3V.
—
0.7 =
3.50 For a 12V sinusoid, the peak is 12ÿ~2 = 16.97V. Capacitor charges to the peak voltage less one diode drop (due to the flow of small capacitor leakage currents), that is, to 16.97  0.7 = 16.27V. Thus the output is a dc voltage of 16.27V. The PIV required of the diode is 16.27 + 16.97 = 33.2V. 3.51 For a constant current of 1mA flowing for one cycle, the voltage drop is V =
whence C =
x
i
—1 — —10— = 41.7(iF.
ÿ = 0.4V = c
*
mÿ
c
,
For Zi the ripple and 2x the load, need 4x the capacitance or 4(41.7) =
For 0.4V ripple with 16.97V peak input, diode conduction occurs for inputs from 167xF. 16.97V, or for an angle from sin1 (16.57/16.97) to 90°, that is, from 77.54° to 0.4 to 16.97 = 16.57V 90°, or 12.5°. Thus for 12.5/360 = 0.0347 of a cycle, diode current flows to replace charge lost through the 1mA load. The average diode current during conduction is 1/0.0347 = 28.8mA. For 0.2V ripple, the interval is from sin1 16.77/16.97 = 81.2° to 90°, or 8.8°, corresponding to 8.8/360 = 0.0244 of a cycle. Thus, the average diode current for a 2mA load is 1/.0244 x 2 = 81.8mA!
3.52 For fullwave rectification, the discharge interval is essentially halved. Using the results of P3.51 above, for the same ripple, half the capacitance is needed, namely 41.7/2 = 2O.9(0,F for 1mA and 0.4V ripple, and 167/2 = 83.4(iF, for 2mA and 0.2V ripple. Also diodes conduct twice per cycle. Thus the diode average conduction currents are Zi of the originals, ie 28.8/2 = 14.4mA, and 81.8/2 = 40.9mA, respec¬ tively. In each case, diode PIV = 16.27 + 16.97 = 33.24V. 3.53 12V applied to a 10012 load implies a 120mA load current. Now CV = IT. Thus C = (120 x 10~3 x Zi x 1/60/0.4 = 2500jF. Assume negligible transformer resistance: Thus the peak sine wave required is 12 + 0.4/2 + 0.7 = 12.9V and the transformers RMS voltage = 12.SW2 = 9.12V per side. Thus the transformer should have an 18.24V rms centretapped secondary. For diodes: PIV = 12 + 0.2 + 9.12 ÿ2. = 25.1V. Diode current flows from sin1 (12.9  0.4)/12.9 = 75.69)/360 = 0.0397 of a cycle. Average diode current for each diode = 75.69° to 90°, or (90 120/0.0397 x Zi = 1.509A, with the peak current being about twice as high, ie (1.509  0.12) 2 = 2.78A!!
—

189 
SOLUTIONS: Chapter #316
3.54
Ideal Case: Peak output = 20
— —
0.7 0.7 = 18.6V, for which the peak output current = 18.6/200 = 93mA. Ripple voltage is v ir/r  93 x IP"3 x 1/60 x 1/2 1000 x 10"6
VOC, the transformer opencircuit voltage

0.77SV.

Diode conduction is from sin1 ((20 .775)/20) = 74.0* to 90', ie 16° or 16/360 = 0.0444 of a cycle, with an average diode current = 93/0.0444 x 14 =. 1.047A, and an 0.775/2 = average DC output = 18.6

VRpeaK VOC 2VD RS IL
—
18,2V.
With Source Resistance, Rs, the peak diode current is limited, and the output voltage will drop. As the output voltage drops, the recharging interval will increase, and the ripple will decrease, both because at lower voltages the load current is smaller, and also because during the charging intervals), the load is
supported directly through the diodes. Now, the average output decreases to 18.2x where x <* 1.0, and the ripple to 0.78*. Thus the peak diode current (near the sinewave peak and where (say) the ripple is halfway) is + 0.78(jc/2)/Rs = (18.6 ll.&x)/Rs. Now, assume the current to be triangular in form, flowing for a fraCtiony of a cycle, twice per cycle. Thus average charge delivered from the supply (through two diodes) in one cycle must be that
required by the load in the whole cycle, that is, 2(14) (y) Now for y: Diodes begin to conduct at
... .
(following the input peak) at 90°+ 90 s,° where y =
180  sin"' 0.958y  sin"1 0.999y —
= 0.95 initially: * (2) 4 y = (180  sin"' .958(.95)
(3) 4
Try
Try
200
(1)
(2)
and
* = 1l.Ov (18.6  17.8*)
* = 11.0 (.119) (18.6  17.8 (.95)) = 2.21.
 sin"1 .958  sin"' .999)4360 = (180  73.3  87.4)360 = 0.0536, (3) 4 = 11.0 (.0536) (18.6  17.8) = 0.472. *
* y= 0.97: (180
=  sin"1 (.958 X 0.97) sin"1 (.999) (.97))/360 = (180  68.3  75.7)/360 = 0.100, = 11 (.100) (18.6  17.8 (.97) = 1.47.
* = 0.99: (180 y =
(3)
 sin"' .999 (.95))360 = (180  65.5  71.6>360 = 0.119, and
= 1.0: * (2) 4 y = (180
* Try
Rs 78jc/2
6~ 17 ta> = Now. for Rs = 1.012, (1) » 2('/i)y
1 8 1x
2x — 0 r= sin"1 (0.958*) and cease conduction —18 :—— lo.o 18.2* + 0.78*/2 — m— 180  sin"1 (0.999jc),
sin1
r
1 8 6 — 17 8r
—
 sin"1 (.958) (.99) sin"' (.999 X .99))/360 = (180  71.5  81.5)/360 = 0.075,
* = 11 (.075) (18.6  17.8 (.99)) = 0.807.
 190 
SOLUTIONS: Chapter #318
3.56 + 2.3V
(a)
For
* vl
Thus
• R1
10 kQ
vo
—Wv— •
R2
~=~
vl
V*
=
= 10 +
10 II /?,
=
and
10 II Ri + 10'
\0RX , 40 10 + R 
R{ = 100 + 10 R{ 10 + R i + 10 R\, 20 R\ = 100, whence R i = 5k£2. For symmetrical operation, R2 = 5kO also, note, further that for symmetrical operatiop, R ) and R2 can be combined, as shown.
(a)
vo
10 kQ
•VW+ 2.3V
K
10 Rx
• 10 kQ
 2.3V (b)
x>o > 3.0 V, current begins to flow in /? j.
—• R
>
• 5 kQ
Hard limiting at +5V can be provided using two additional diodes: D2, from u0 to +4.3V with anode at v0, and D4, from D0 to 4.3V with cathode at Do
• 10 kQ
(b)
2.3V
(c) Here, + 4.3V
R iii x 4.3 = 2.3 V and R la Ria + R lb
5kf2. ÿ
Thus
vo
10 kQ
•wvD4
R
_
ÿ3) 2.3
D1
02
ÿ
lb la In — = 5 , or —— —R\a + R\b R\a + R\b
R lb 5 (4.3) Substitute the last into the first to get:. R lb 5 2.3, or R u, = = 9.35ki2,
D3
vl
R lb =
"X
> R2b B •
R2a
'
• 10 kQ
(R ia + 9.35) = 0.535 Rla + 5 and R\„ 9.35 5 ÿgain, a resistor can be = 1 535 saved by replacing Rlh and R2/, by a single resistor equal to their sum (18.7k£2) between nodes A and B with no ground connection.
=" (c)
4.3V
 192 
SOLUTIONS: Chapter #317
Try x = 0.98: y = (180  sin"1 (.958) (.98) sin1 (.999 X .98))/360 = (180  69.86  78.24)/360 = 0.0886, jc = 11 (.0886) (18.6  17.8 (.98) = 1.13. Try x = 0.983: y = (180
 sin"1 (.958 x .983) sin1 (.999) (.983))/360 = (180  70.34  79.12)/360 = 0.0848,
x = U (.0848) (18.6  17.8 (.983)) = 1.028.
Try jc = 0.984: y = jc
[l80  sin1 (.958 x .984)  sin"' (.999 x .984) ]/360 = (180  70.50  79.43)/360 = 0.0835,
= 11 (.0835) (18.6  17.8 (.984)) = 0.996.
Try jc = 0.986: y = jc
[l80  sin"1 (.958
X .986) sin"1
(.999 x .984)
]/360 = (180  70.84  79.43)/360 = 0.0826,
= 11 (.0826) (18.6  17.8 (.986)) = 0.953.
Try jc = 0.985: y = jc
[l80  sin"1 (.958 x .985) sin"1 (.999 x .985) ]/360 = (180  70.07  79.74)/360 = 0.0821,
= 11 (.0821) (18.6  17.8 (.985)) = 0.964.
Use jc = 0.985. One can conclude that: a) The iterative process is not a very good one, but the result is probably OK. b) The output voltage decreases to 18.2jc = 18.2 (.985) = 17.9V, a drop of about 18.2  17.9 = 0.3V, or 0.3/18.2 X 100 = 1.6%, in the transformer resistance of 1 ohm, implying an "equivalent" current of 300 mA.
SECTION 3.8: LIMITING AND CLAMPING CIRCUITS 3.55 The upper limiting level is 2.3 + 0.7 = 3.0V. The corresponding input threshold level is 3.0 + (3.0/10k) X 10k = 6.0V. The corresponding lower values are 3.0V and 6.0V respectively. The gain K (for linear operation) is 10k/(10k + 10k) = 0.5V/V. At twice the upper threshold, Vin = 2 (6.0) = 12V, and the current is (12  3)/10k = 0.9mA.
 191 
SOLUTIONS: Chapter #319
3.57 vl
vo
R
M—M
—•
Consider an input \)f = A sin tor . Now, A sin 0 = 1.4V, where ©<360 = 5/100. See that 0 = 18° and A = = 4.53V peak, or 9.1Vpp.

4 53  1 4 The peak diode current = = 10mA. Use . n 4.531.4 a resistor, R —= 31312.
M—K
.

3.58 For light load, the output is a square wave of period T going from +0.5V to +0.5 (100  10) = 89.5V. As the load resistance reduces, the negative side of the waveform is no longer flat at 89.5V, but rather rises toward ground. As well, upon the positive transition, the diode current increases initially, with 0.7V or so positive output at first. For RC = 27, in one half cycle, where t = 7/2, the output falls to 7V2
e 2T =
posaibly < 0.5V
0.7V
e~m
= 0.779 of its original value. Thus, assuming the diode to have a constant 0.7V drop when conducting, the waveform initially rises to 0.7V, then drops to 0.7 (.779) = 0.55V, then falls to 90 + .55 = 89.5V, then droops (up) to 89.5 (0.779) = 69.7V, then rises to 0.7 V, and so on. In practice, the diode conducts at voltages lower than 0.7 V. Thus the upper level is not simply an exponential, but will fall more rapidly, to slightly less than 0.55V.
T.C.RC
(100100.5) 89.5V ÿ

3.59
, 0.1pF
VI
it
02
N—r—
vo
: 0.1pF
For a lOOVpeak sine wave and no load, the output would be 2(100) 0.7  0.7 = 198.6V for a 0.7V diode drop. For a pp ripple of 5% of peak, ripple voltage is (5/100) (198.6) = 9.93V, and the average output voltage = 198.6  9.93/2 = 193.6V. The . , 0.1 x KT®x 9.93 corresponding load current is / =
....
1/(20 x
= 19.86mA.
 193 
103)
SOLUTIONS: Chapter #320
3.60 Assume ideal diodes having a drop of 0V. Initially, the input capacitor C, is charged, with 0V on its internal end and input low, while the output capacitor C2 is discharged. As the input rises by 100V, charge is dumped through the connecting diode £>2 from Ct into C2. Since the capacitors are equal and the input voltage change is 100V, the output rises to 50V. When the input falls, C, is recharged through the grounded diode D\, while the connecting diode D2 opens, leaving C2 charged at 50V. Now, when the vo input rises again, diode D2 does not conduct until 100Vthe input rises by 50V. For the remaining 50V change of input, charge is shared equally by C\ and 75V C2, while the voltage on C2 rises J?y 25V to 75V. When the input falls, C\ is recharged. Correspond¬ 50V ingly, after 2 cycles, the output becomes 50 + 25 + 25 —
= 87.5V. After four cycles, the output is 50 + = 96.9V. After eight cycles
25 +
the output is 50 + 25
12 3 4
# of Cycles
25
1+
1 2
25
25
+ + —2 + —— 4 1
+4 +
_1_ = 99.8V. 16
3.61 Ultimately, an equilibrium is established with the output reaching V, at the beginning of a cycle,
V2 half
) = 0.975V, since two capacitors C supply the load. V3 at the end. Here, V2 = V, (1 Also V3 = V2 (1 5400) = 0.95V2 = 0.95 (.975) V\ = 0.92625 V. But at the start of the next cycle, with V, = (V0 + V3yZ, hence V, = (V0 + .92625 V,y2 = Vo/2 + .4631V,. way, and
—
V =
2(104631) (V, + V3y2 ='(0.931
=
2(5369)
= a931U°' ÿ = °9263V\ = 0.863V0, and the average output =
V0 + 0.863 V0)/2 = 0.897V0.
Here, including an approximation of the effect of diode drops, the average output would be about 0.897 (100  1.4) = 88.4V, with ripple = V,  V3 = (.931  .863) 100 = 6.70Vpp, the output ranging from 88.4 + 6.70/2 = 91.8V, to 88.4  6.70/2 = 85.0V.
SECTION 3.9: SPECIAL DIODE TYPES \VnV
3.62 For a junction diode, i = I$e
Thus, 300 = n (25)lnl00ÿy, or 12 = nlnlOO  nln/s Also, 370 = n(25)ln100075, or 14.8 = /ilnlOOO (2)  (1) = 2.8 = n(ln 1000  InlOO),
whence, n = 2.8ÿ6.908

T, and \) = nVf\n{i/Is)
 4.605), or n =
1.216
 /iln/s
(1)
(2)
NOW, ls = ie'ÿr = le 370(..216 x 25) = g.jg x 1Q6A Check: i = 5.18 x i0VOoa2i6x25) 0.1 A, OK.
_
At 20 A, \) = nVM'/Is) = 1.216(25)ln[20/(5.18 x 10"6)] = 461 mV. Thus the ohmic drop is R(20A ) = (800 461)103, whence R = 339 x
—
10"3/20 =
16.95 mil.
Check: 20 A and 17 mO > GA r>R = 20 x 17/1000 = 0.34 V. Thus the series resistance of this power diode is 17 x 10_312 or 17 mii.

194 
SOLUTIONS: Chapter #321
xrti V
3.63 For a junction diode, i = Ise
r, and v = nVT\n(i/Is).
For the specified diode having a 10Q scries resistance,
= lOi + 1.5(25)ln(//7s ) At 10 mA, 420 = 10(10) + 37.51n(10tfs), ln(107s) = (420  100)67.5 = 8.533, whence Is = 106081 = 1.97 X 10"3mA.
107s = e8533 = 5081,
10"3)) = 334 mV. At 50 mA, vD = 10(50) + 37.51n(5M.97 x 10"3) = 880 mV. At 1 mA, vD = 10(1) + 37.51n(l/(1.97 x
3.64
x/ny
i=IRe
': Taking base e
logarithms, u = nVT\n(i/Is)
At 0.1 mA, D = l.l(25)ln[0.1 x 10Vl0~15] = 696 mV. At 10 mA, V = l.l(25)ln[10 X lO'VlO"15] = 823 mV. 3.65 For a varactor (or variablecapacitance diode) (from Eq. 3.27), Cj = Cy0( 1 + Vr/VqY'" . Here, 33 = Cy0(l + 26.2)09, Cj0 = 33(1 + 22.2)09 = 59.1 pF. Thus, at 0 V, C = 59.1 pF
At 1 V, C = 59.1(1 + 1/2.2)"09 = 23.6 pF At 10 V, C = 59.1(1 + 102.2)0,9 = 7.1 pF
3.66 In sunlight,
lOOOW/m2 = 1000 x
lOÿW/cm2 = 100mW/cm2
At 25°C, the diode photocurrent is IR = 100 X 0.7 = 70 (lA. At 125 °C, IR = 70 x 106(1.035)(1252WO = 70 x 10"6(l.O35)10 = 98.7 pA At 25 °C, the dark current is 1.5 nA
_
At 125 *C, the dark current is 1.5 x 2(125 25>1° = 1536 nA = 1.54 pA. 3.67 In direct light, ID = 20 x 0.7 X 106 + 1.5 X 10~9 = 14.0pA In reflected light, IR = 0.5 x 0.7 x 106 + 1.5 x 10'9 = 0.35 pA
+ 10V
+ 5V (
For direct light: "Oq = 14 X 106 X 10s = 1.40 V. For reflected light: u0 =0.35 x 106 x 105= 0.35 V In the modified circuit shown, D2 is biased by light passing through a filter to adjust its intensity to about half the direct light beam applied to D .
In operation
I\ is either 14 pA or
0.35 pA, and l2 is adjusted (by varying the light bias) to balance the
two output current magnitudes.

195 
SOLUTIONS: Chapter #322
For direct light: I= 1\d ~ hFor reflected light: —I = 12 I \r
—
ÿ
Now, the magnitudes should be equal, that is 1 I\d Thus
I\d — I2
—
ÿ
 h I = I h  I\r I.
Iir)
Take the positive sign:
2/2 = Iid + hr = 14 + 0.35, and I2 = 7.17 pA. Check: For /, = U\iA , 1= 14  7.17 = 6.83pA . For /, = 0.35pA ,/ = 7.17  .35 = 6.82\iA . OK For a ± 05 V signal, R = 1/6.83 X 10'6 = 146 k£L Note that in this balanced circuit, if the diodes are matched, dark currents cancel and operation is independent of the operating temperature of the diodes (as long as they are at the same temperature).
3.68 See that for a usual op amp, the diode is connected across virtual ground and = 0 V. Thus, if the diode is not exposed to light, the current through it is zero. Note that this is the case with an ordinary diode in a sealed package, where U0 = 0 V. When light falls on the junction of D , photocarriers are produced with the photocurrent polarity (which allows photocurrent to flow from cathode to anode). When an illuminated diode is shortcircuited (as is D in Fig. Q3.68); current is extracted from the exter¬ nal circuit. For the connection shown, v»0 rises. For light at 20 mW/cm2, and the diode rated at 0.7 pA/inW/c/n2, the photo current is 0.7 x 10~6 X 20 = 14 p.A. For light applied to the circuit, = 5 k£2 X 14 pA = 5 x 103 X 14 X 10"6 = 70 mV. For no light applied, u0 = 0 V (if usual small offsets are ignored). 3.69

A
•
B (a)
A
•
14
V B 14
(b)
r*
j
As noted in P3.68 above, an illuminated photodiode attempts to conduct photocurrent (as seen exter¬ nally) flowing from cathode to anode. Such a current is shown in Fig. 3.69a) as i. In Fig S3.69b), the situation is redrawn with the external circuit indicated by the dashed lines, where current labels in b) fol¬ low from a): Here i'i is /, i2 is i, 13 is i2 is i\.
We conclude that a photodiode can generate current, which flows out of its anode, just like a photooperated battery with the anode positive. Certainly, this happens if the external circuit is a shortcircuit (This was illustrated in P3.68 above.) But as the resistance in the external circuit rises, the voltage across the diode increases until the internal junction begins to conduct increasingly. Thus an opencircuited sili¬ con photodiode when illuminated has an opencircuit voltage equal to a diode drop (of 0.7 V or so for silicon). Thus, tt behaves like a 0.7 V battery. When an external load is connected, the terminal voltage drops until at (say) 0.5 V, 99% or more of the photo current flows in the external load, and only 1% is internally shortcircuited by the (slightly) forwardbiased junction. For die specific solar panel: vo Output power level is * p I = IV = 100 x 10"3 x 14.5 = 1.45 W.
ÿ D1
Solar Panel (D1..Dn)
t
Dn
~ 12 V
TBattery
Panel power level is P = /V = 110 x 14.5 = 1.59 W.
_
Opencircuit power level is p =iv = 110 x 10"3x24= 2.64 W
0
—
(c)

196
SOLUTIONS: Chapter #323
For an open circuit, the photocurrent of 110 mA or so flows in the seriesconnected forwardbiased junc¬ tion to create the opencircuit voltage of 24 V. Since each cell has a forward voltage of 0.67 at 110 mA or so, the number of series cells is approximately 24/0.67 = 36 cells. Maximum power calculations: Maximize P = ILV = (/  i)v (1) v where i = Ise wvrT
(t>
©'

Maximize P = (/  Is e
(2)
7b.
=" (d)
Now
   
= /  Iseÿ'  x>ls/VTeÿT = I Is(1 + (\A/T)eÿ
Maximum power occurs when dPAv = 0, or / = (1 + \A?T)Ise (3), or v = (IA  l)Vr b = (/ iyiVT (4). (5) Now, Pmax = (/  i)ViVT Combining (2) and (4), PmM occurs at current i = /?e(M ~
T
= t'(l + \yVT), or / = i + ix/VT, or
= 0.367/.veM
(6)
For the specific case, x> = 670 mV, and i = 110 mA, with the initial assumption of n =1:
10"3 = /se6701(25), and Is = 110 x 10"3e "67(K!5 = 2.525 x 10~I3A = 2.525 x 10~wmA Generally, i = /se(M l), and taking In, lni/75 = IA  1, or IA = lni + 1 « ln/,s    (7), or
From Eq.(2), 110 x
"
ÿ
lnt + 1  ln/j
, or lni + 1  ln2.525 x 10~10 ' Solve Eq.(9) interatively for / = 110 mA: Try i = 10 mA: / = 110t2.30 + 23.1) = 4.33 mA Try i = 4.5 mA: i = 11011.504 + 23.1) = 4.47 mA Try i = 4.47 mA: i = 11011.497 + 23.1) = 4.472 mA For the specific case, i =
/
=
lni +23.1
(9)
Thus, maximum power occurs for an internal diode current i = 4.47 mA or 4.47/1 10 = 4% of full shortcircuit current.
At maximum load power, each diode voltage is (from Eq.(4)):
v = (IA
 )VT = (1104.47  1)25 x 10~3 =
590 mV.
For a stack of such diodes (with n = 1) with opencircuit voltage of 24 V, the maximumpower voltage would be 590/670(24) = 21.1 V. This is clearly much higher than the 17.5 volts specified. Probably n > 1, although the detail of the specifications, whether nominal or best/worst case, may also be suspect.

if n = 2 is assumed, 67(350 from Eq.(2), 110 X 10"3 = /se6702(25), and Is = 110 X 10"3e = 1.67 X 10~7A = 1.67 X
Now
I I From (8), / = r, or i = lni + 9.70 lni + 1  ln(1.67 x 10"4) Solve Eq.(10) iteratively for / =110 mA: 110 Try i = 10 mA: i = = 9.16 mA In10 + 9.70 110 Try i = 9 mA: i = = 9.24 mA
ln9.2 + 9.70
 197 
(10).
10ÿmA .
SOLUTIONS: Chapter #324
Try i = 9.2 mA: i =
Thus, mnv Eq.(4) c> For the !
Clearly,
110
= 9.23 mA ln9 + 9.70 im power occurs for i = 9.23 mA, or 9.23/110 = 8.4% of full current, for which (from xl) the diode voltage is (IA = 1)nVT = (1109.23  1)(2 x 25 x 10~3) = 546 mV tack of diodes, this maps to 546670 x 24 = 19.5 V more likely to be 2 than 1!
For the operation at 14.5 V, each diode drop is 14.524 x 670 = i = Iseÿ(Vl) = .67 x lOV0*50 = 0.55 mA. Thus the current loss at 14.5 is about 0.6 mA, which is quite small!
3.70 Input power is P = ft) = 10 X Now, i =
IseWnV\
405
mV,
and
103 X 1.9 =
19 mW and 10 = [seÿ2ÿ\ or Is = 10e,9(xy,2(23) = 3.12 x 10~21mA .
At a power level of 60/2 = 30 mW, assume the current increases by 50% to 15 mA. Thus for i = 15mA, u = nV,ln(i/7s) = 1.2(25)ln[15/3.12 x 10"27] = 1.2(25)(63.74) = 1.912 V,
for which
it) =
15(1.912) = 28.7 mW, a bit small.
Now, try / = 301.912 = 15.69 mA, for which 15.69 x 1.914 = 30.03 mW.
t)
= 1.2(25)ln[15.693.12 x 10~27] = 1.914 V, and
it) =
Thus a current of 15.7 mA produces an output of 1.91 V for a dissipation of half rated power, 30 mW. For 10 mA operation: R = (5
— 0.5)( 10 x 10~3) = 45012
For 15.7 mA operation: R = (5  0.5)(15.7 x 10~3) = 28712 From Appendix H, standard 1% resistor values to be used would be 453 12 and 287 12.
3.7 1 The detector is specified to have an opencircuit voltage of 500 mV and a shortcirucit current of 70nA. From the discussion in P3.68 above, this specification relates to operating in the lowvoltage or solarcell mode. For an emitter current of 5 mA, the output shortcircuit currents can range from 5 x 10~3 X G.6ÿ100 = 30flA to 5 X 10"3 X 1.6T00 = 80pA
In this circuit, the isolated ± 5 V supplies are labelled (1) and (2) (for primary and secondary). Op amps A  and A2 operate in the inverting mode. Resistor R ( = 100/:12) where R i establishes the operating current for Ddi (and DD2) at 50 (i.A. Initially it causes the output of A ! to rise driving DE through R 5, limits the maximum current in DE to 30 or 40 mA. As the current rises in DE, so does the current in Dd (and Dp). The current in DD\ rises until it equals that in R\ (50 (J.A). R2~ R 1 ( = 1001:12) ensures

198 
SOLUTIONS: Chapter #325
that Dp2 operates at the same current. Diode mismatch can be compensated by minor control of R2. Resistor R3 (with R4) establishes the overall voltage gain to a value Rt/R 4 (here of lOOk/lOk = 10 V/V). Resistor R 4 isolates the input and controls the range of current variation in Dp , Dp2 to ± 0.1/VIOk = ± 10 pA for ± 0.1 V signals. Notice that the current transfer gain is completely compen¬ sated by establishing the current in Dp 1 (via R\) and allowing the current in DE to seek an appropriate value (limited only by R$ on those rare occasions on turnon, or when the loop is broken accidentally.

199
SOLUTIONS: Chapter #326
200
Chapter 4 Bipolar Junction Transistors (BJTs) SECTION 4.1: PHYSICAL STRUCTURE AND MODES OF OPERATION
#
Type
Mode
1 2 3 4 5 6
npn npn PnP pnp pnp npn
active cutoff cutoff saturated cutoff* saturated
ÿCutoff, but nearly active.
4.2
Two junctions, each in two states of conduction, imply 2x2=4 modes of operation. Table 4.1 lacks the case: EBJ Reverse Biased, CBJ Forward biased.
SECTION 4.2: OPERATION OF THE NPN TRANSISTOR IN THE ACTIVE MODE 4.3
Generally, ic  he0"ÿ'. Here, 2.0 x 10~3 = Ise1(m5 and Is = 2 x 10"3e~70025 = 1.38 x 10~,SA. Now, from Eq.4.4 Is  AEqDnnÿA,NA W) and from Eq. 3.12, Dn/\in  D,/\xp = VT. Thus, D„ = Vri„ = 25 x 10"3 x 1100 = 27.5 cm2/s.
Now (from Eq.4.4) Ag = NA WI,/(qDn nf2), and calculation in cm ,
Ae = 1018 x W x 1.38 X 10_l5/(1.6 X 10"19 x 27.5 x (1.5 X 1010)2) = 1.39Wc/n2 (with IV in cm) = 1.39 x 104Wi.m (with W in \xm). For W = 2\im = 2 X 10_4cm , Ae = 1.39(2 x 10"4) = 2.78 x 10"4cm2 = 2.78 x 10"4 x 108 = 2.78 x lOfyn2, being, 167 X 167pm2. 4.4
for
example,
For a transistor whose EBJ is 100X larger, Is is lOOx larger, namely ls = 100 x 1.38 x 10~'5A or Is = 1.38 x 10"13A. For vBE = 0.70V, ic = 100(2) = 200 mA. ÿ and VBE = 25mVln(l x 10"V(1.38 x 1013) = 568 mV. For ic = 1mA, 1.0 x 10"3 = 1.38 x l03e BT3e~E,/KT , with B = 5.4 x 1031 and EG = l.UeV, For Is = AEqDnn?ANAW) and n? = k = 8.62 X l0~seV/K.
Thus Is = AEqDnBT3e~E,/kT/NA W = constant X T3e~E,/kT = K* At 300K, Is = /sf*3003e~130*l0'/300 = K* 4.09 X 10"12.
 201 
X
7ÿ1.12ÿ.62x10 ÿ
_ ÿÿiÿoxiovr
SOLUTIONS: Chapter #42
Is = K* 4oo3cI",0xlC),/'400 = K*4.9l xl0~7. Thus at 400K, Is increases by 1.20 x 105 times! to Is = 1.2 x 105 x 1.38 x 10~I3A = 1.656 x 10~8A, for
At 400K,
which vBE = 25 x 10"3ln[l x 10_V( 1.656 x 10"8)] = 275 mV
4.5
Now, in the base, npQ = n2/NA , and at 25°C, np0 = (1.5x 10lo)2/l()'7 = 225(Kcm3, and (from Eq.4.1) n,,(0) = np0e°"ÿT = 2250e7(XV25 = 3.25 x 10l5/cm3. Now (from Eq.4.2), /„ = AEqDn(  np(0y\V), and
using cm ,
/„ =
 20 x 10~4 x 20 x 10"4 x
For W = lpm = 1 x
1.6 x
10~4cm , /„
10~19 x
=  4.43 X
21.3 x 3.25 x 10I5/1V =  4.43 x 10r8/lV. lOÿ/lO'4 = 4.43 x 10"4A, or  0.443 mA.
For W = 0.1pm, /„ =  4.43mA. (from
Now
Eq.4.4),
Is = 20 x 10*4 x 20 x 10"4 x
Is = AEqD„n2/{NAW), and with all dimensions 1.6 x 10"19 x 21.3 x (1.5 x 10lo)Vl017VK = 3.067 x 10~20AV.
in
cm,
For W = 1pm = 1 x 10~4cm, Is = 3.07 x 10"16A.
10"15A. Check: Now, for vBE = 700 mV, ic = Isev,'/v'r. For W = lpm, t'c = 3.07 x 10,6e70025 = 0.444 mA.
For W = 0.1pm, Is = 3.07 x
For W = 0.1pm,
ic = 4.44mA, both
(as
expected), the same as /„ (within a factor a). Now, for the base current (from Eq.4.12),
p = 1/ Dn Na
tation of Eq.3.30. Here, the base minoritycarrier Xb = L2/Dn = (19 x 10_4)2/21.3 = 169.5 ns. Thus, f 1017 IV W2 1.7 —— r + P = 1/ 21.3 X 1019 0.6 2(21.3 x 108 x 169.5 x 10"9)
W
+
]/2_
W2
lifetime
, where, t/, = L2/Dn , by adap¬
(using calculations
in cm)
is
141.33 x 10~31V + W2(1.385 x 10"3)].
P 0.997, P = 141.33(1) + 12(1.385)]10~3 = 368, for which a P+l and for W = 0.1pm: p = 141.33(0.1) 4 0. 12(1.39)] lO3 = 6807, for which a = 0.9999. Note that for IV = 0.1pm, p is very very high. Such a transistor would be very difficult to make in prac¬ Now, for W = lpm:
tice, and would have low breakdown voltages. (See P4.6, next).
4.6
Generally, from Eq. 4.12,
P = 1/
/>
'A W 'Na x — x —
Text, Nd = 1019/cm3, NA = 10'Vc/n3, Ln
1 IV2 , where %b Ln 2/Dn . From P4.2 in the 2 D„ X/, Nd Lp = 19 pm, D„ = 21.3cm2/s , Lp = 0.6 pm, Dp \.lcm2/s .
—
Thus Tb = (19 x 10~6)2/21.3 x (102)2 = 16.95 x _.
Thus
„
P= V
—
+
10"8 = 169.5nj. IV2 ,
1017 x IV 1.7 —— +  x —— —21.3 5 5X 10"9) . 1019 0.6 2(21.3 x 108)(169.5
or
P = 141.33 x 103W + 1.385 x 10"3W2] = 141.33W + 1.391V2]103 Now, for P = 1000: 1.39W2 + 1.33W = 1 = 0 or 1.33 ± V1.332 4(1.39)( IV =
_
2(1.39)
For
P=
ÿ
2000:
1.39W2 +
— 1)
 1.33 ± 2.71
2(1.39) 1.33V  0.5 = 0 or
4(1.39)(  0.5) _ — 1.33 ± ÿ1.33ÿ — 2.(1.39)
1.33 ±2.13 2(1.39)
 202 
0.469 pm
_
ÿ
SOLUTIONS: Chapter #43
4.7
i'c = Ise*"ÿ', whence Is = ice
~ v"ÿr
= 10 X 10"6e " 65025 = 5.11 x 10~17A For a 0.500 V drop at 10 pA, Is = 10 x IQiÿe ~5CXy25 = 2.06 X 10~14 Junction size increase is 2.06 x 10I
,
Check: e
(500  650)  — =
ÿTJ.
403, OK.
At 0.65 V, this large junction has a current of 2.06 X IQ~14e 65025 = 4.03 mA. At 0.70 V, the current is 2.06 X 10~14e70025 = 29.8 mA. Check: 4.03e(700  65W25 = 29.8, OK.
4.8
Ise",/Vr,
"
and Is = ice %c = Here, Is = 10 x 10"3e ~ 69025 = 1.03 x
p = icAB = 10 x 10"V75 x 10"6 =
Also
"iii= a=TrfT Is/a =
10"14 A. 133.3 and
°"3' whence
= 1.039 x 10",4A and Is/f> = 7.73 x 10"17A.
4.9
Here, ig = 0.753 ± 0.001 mA and i'c = 0.749 ± 0.001. Thus iB iE i'c = 0.004 ± 0.002, a current ranging from 0.002 to 0.006. " 0.749 + 0.001 . 0.7490.001 3 Thus D = 124.7, for which a = .K , = 0.9973 = 375 to P varies rfrom 0.002 0.006 p+ 1
— —
ÿ
——
.
——
to 0.9920
Directly: a ranges from 0.750/0.752 = 0.9973 to 0.748/0.754 = 0.9920. Clearly, measurement of the currents which are nearly equal leads to a lot of error in P, although a is relatively insensitive (except that casual measurement of ig and i'c can easily suggest that a is negative! 4.10 From P4.3, ic = 2.0 mA, VBE = 0.7 V, n = 1, = 10l8/cm3, p„ = 1100cm2/Vs, Is = 1.38 X 10~I5A, Dn 27.5 cm/s. AE = 1.39 W cm (for W in cm) W = 2pm = 2 x 10_4cm The Minority stored charge in the base (with calculations in cm ) is (from Eq.4.9)

Un
AgWqni2 2Na
yt/Vr
c v in'0\2 n4\2/ 1 £ v I0~iy)(1.5 x 10'Y 1.39 x (2 x HTT(1.6 or Q„ = —e  x 
' 2(1018)
l.446 X
For equation 4.12, assuming recombination to be dominant,
3=1/
_
_ ,nl2 « 10IZ pC = 1.46
W2 2Dn%b
Pÿ2 120 x (2 x 10"4)2 u u .u ,r .• the lifetime xB = 1.ÿ = for which,  = 87.3 ns 2(27.5) 2Dn
"
4.1 1 For (ensured) activemode operation of an npn transistor v>cu ÿ 0V and x>Cg = x>CB + X>BE > 700 mV. In (for n = 1); 10 X 10"3 = Ise10025, or 1$  10 X 103e28 = 6.91 X 10""I5A. the active mode, ic  heÿ Also, in the active mode,
p = ~r~ = ifl
lOOpA
= 77 = 100, and ig  ic + 'a = 10mA + 0.1mA = 10.1 0.1
mA.
 203 
SOLUTIONS: Chapter #44
4.12 For the tabulated devices,
= ~§ = P
(b)
ÿT = 20ÿ /£ = /c + /c = 1 + 02 = OU
1.02 mA; a = ~ = 117
 Ic = 2  1.96 = .04wiA
1.02
p=
= 0.980.
(c)
Ic = aIE = 0.98(2) =
(d)
7C = plB = —IB =  _ÿy995 x 0.01 = 1.99 mA; IE = lc + IB = 1.99 + .01 = 2.00 mA;
P
(d)
Iq
0.04
= 49.
~°"5
ÿ= = 199. 1 a 1  .995
= 1ÿTxllO= Jr/£ 122
/c = a/£ = a=
= 40nA;
0 995
ot
H=
(c)
1.96mA; IB = lE
—ÿ
IE
=
HO
100
mA;
/B =
ÿ
=
i_
10
mA;
= 0.909.
Ic = p IB = 1000 x 0.001 =
1mA;
l° IE = Ic + IB = 1 + 0.001 = 1.001mA; a = lE
1 1.001
0.999.
Device
Ic
Ib
h
#
mA
mA 10.1
e
mA 10 1 1.96 1.99 100
f
1
a
b c
d
0.1 0.02 0.04 0.01 10 0.001
1.02 2 2.00 110
1.001
a
P
0.99 0.98 0.98 0.995 0.909 0.999
100 50 49 199 10 1000
4.13
nlE= piB
ÿ> aiE= piE
The controlled sources can be labelled a iE or
P iB where a i£ = P iB


4.14 Generally, ic = Is (1), and at O.lptA, 10"7 = Is (2). Now, For n = 1, at 1mA, 10~3 = Is e7cm5 dividing, 10~Vl0~7 = e(700u>ÿ5 = 104. Taking logarithms, 700 o = 25 In 104 = 230. Thus, \) = 700 230 = 470mV, or 0.47V. For n = 2, see v = 700  2(25) In 104 = 700  2(230) = 240mV, or 0.24V.
204 
SOLUTIONS: Chapter #45
4.15 Generally, for the base open, ic = P is = 100 JCuo At 25*C, ICBO = O.lnA, and ic = 100 (0.1 x 10"9) A = lOnA. 95  25 At 95°C, Icbo = 0.1 X 109 x 2 10 = 10"10 x 2yA , whence ic = 100 x 10"'° x 27 = 1.28xA. ÿ
4.16
aF/aK =
100, the relative junction
0 00993
P* = ——— = 1  0.00993 laR
WR
aF Thus aR = yy =
area.
150 ÿ
ÿ
ÿ
/100 = 0.00993, and
= 0.0092 s 0.01.
SECTION 4.3: THE PNP TRANSISTOR a
0 975
4.17 For a = 0.975,
p=
4.18 For Db,Is/P=
10~I3A, and for DE,Is/a= 101IA. Thus, lÿ/l/oc = 102, = 102, or p + 1 = 100, and p = 99.
= 39. For I„ = 10pA, Ic = 39 (10)pA = 3.90(iA = 0.390mA, 975 and IE = 390 + 39 = 429J.A = 0.429mA. Generally, x>bei = ÿbe\ + nVT In ictfcv Thus at ic = 0.390mA, vbe = 700 + 25 In = 676mV. 1.00
+
ingly,
=
{_
t'c = Is e64V25(l) = 99 x 10~13 c25'7 = 1.46A! device (where 10"" e1(m5 = 14.5A).
or
a/fi = 10"2.
Correspond¬
Note that this is a large transistor, a 14.5Ampere
Now,
SECTION 4.4: CIRCUIT SYMBOLS AND CONVENTIONS 4.19 (a)
Here,
iE = I= 1mA, independent of vBe
fr°m
~y(l) = 0.909mA to
or
P> V?
ranges from 0.6V to 0.8V;
ÿ  (1) = 0.997mA.
.
(b)
Vc = Vcc ~ Kc Ic ranges from 105 (.909) = 5.45V No, \E variation has no effect on Ic or Vc
(c)
For Vcb
0, at the largest value of ic, Rc
ÿ
to 10  5 (.997)
ic  Vcc  0 =
ic = a iE
ranges
= 5.02V.
10V, whence
Rc
ÿ
= 10.03kG
(use 10ki2).
4.20 Here,
ic = a (I + ic)
Now, for high
p, a =
1 and ic = 1 + 0.1 = 1.1mA. For
p = 10, a •
p, the largest allowed Rc is limited to be Rc = l(O.lmA) (9.09kfl) = 0.909V peak or 1.82 Vpp. For p = 10, vc = 0.909
10
0.909, and ic = (1.1) (.909) = 1mA. For high 9.90kQ, and oc = (9.09kf2) = 0.826V peak, or 1.65Vpp.
4.21 For
VE = 0.7V, use RE = (0 0.7  710/1 = 9.3kfl. ÿNow
and large
Rc <
p.
That is,
ic ,
0.6 10 9.3 (1 .01)

'°Q2 °
°
=
(0.1mA)
the largest ic occurs for small RE, small
x 1 = 1.021mA. For activemode operation,
I. Now for Rc R, varying by 1%, the lowest possible value of V>c = = 9.79kO. Use 9.7k£2. 10  (9.7) (1.01) (1.021) = 2.8mV. Note that operation is still in the active mode, since VE = 0.70V.
 205 
SOLUTIONS: Chapter #46
SECTION 4.5: GRAPHICAL REPRESENTATION OF TRANSISTOR CHARACTERISTICS 4.22 Generally, TC ~ 2.0mV/°C. Now, for 10mA, at 25"C, VBE = 700mV. Thus, for 10mA, at 0°C, VBE = 700 + (0  25) (2.0) = 750mV. For 10mA, at 50°C, VBE = 700 + (50  25) (2.0) = 650mV. For VBE constant at 620mV, at 25°C, i = 10e(62(K700)<25 = 0.407mA; at 0°C, i = 10 eÿ20"750*25 = 0.055mA; at 50'C, i = 10 e (62°650>ÿ5 = 3.012mA, a nearly 55 to 1 range!!
4.23 Here, A
ic = 2.19  2.10 = 0.09mA, for A VCE = 9  2 = 7V. Thus
average current of ( 2.19 + 2.10/2 ) = 2.145mA. Thus
0.1mA, r„ =
j~y = 1.67M12
4.24 At IOOliA, r„ =
**ÿ.4 2M
0.1m/l
ÿ
r„ =
4ÿ = A .09 i
= 77.8k£2, at
an
VA = 77.8 x 2.145 = 167V. Correspondingly, at
1.7MO, and at 10mA, r„ = ~ = 16.7kO
ÿ
17 kO.
= 2MO. For an increase in VrE from 5 to 50V, the increase in current is
= 22.5tlA. Thus at 50V, the current becomes 100 + 22.5 = 122.5uA.
SECTION 4.6: ANALYSIS OF TRANSISTOR CIRCUITS AT DC 4.25 In general, assume the active mode initially, with a conducting emitterbase junction, then verify that the collectorbase junction is not conducting. Since (a)
VB =
lB

4V;
VE = 4
 19.6pA; Vc
p = 50, a = 3
°
lE =
+0.7 = 3.3V;
= 0.98. = 1mA; lc = 0.98(1) = 0.98mA;
= 10 + 4.7 (0.98) = 5.39V. See this is OK; operation is in the active
mode. (b)
° ~3~5
3
= 1.606mA; Ic = 1.606 (.98) = 1.574mA; 1 57 IB = ~jq = 31.4pA; VE = 10 + 4.7 (1.574) = —2.60V. Since VE is well above VB , the transistor
VB = 6V; VE = 6 + 0.7 = 5.3V; 1E =
is saturated.
__
In practice, in saturation x>Ec ~ 0.2V. Thus, Vc = VE 0.2 = 5.3 0.2 = 5.5V, and ~5'5 = 0.957mA, with IB=IEIC = 1.606 .957 = 0.648mA. Ic = *T. /
(c) (d)
(e)
9
t
n qq
a
= 19.6iA; VB = 2V; VE= 2 + 0.7 = 1.3V; lE = g3 = 1mA; Ic = 0.98mA; IB = Vc = 8 + 4.7 (0.98) = 3.39V. See OK, active! VB = 0V; VE = 0V. Thus the transistor is cutoff. Vc = 10V, and IB = Ic = IE = 0mA. ~10 = 1.128mA; /c = 1.128 (0.98) = 1.105mA; VB = —4V; VE = 4  0.7 = 4.7V; IE =
lB =s
ÿ
~ 22.1pA;
Vc = 0  1.105 (3.3) = 3.67V.
Since
Vc > VB, operation is in the active
mode, as assumed. (f)
VB = 6V; VE = 6 0.7 = 6.7V; IE = h=
= 13.8jxA;
= 0.702mA; lc = 0.702 (0.98) = 0.688mA;
Vc = 0  0.688 (3.3) = 2.27V.
tion, as assumed.
206
Since VC>VB, this is activemode opera¬
SOLUTIONS: Chapter #47
4.26 (a)
VB = 4V; VE = 4 + 0.7 = 3.3V; lE = Ic = 0.5 X
°~~3'3V
Re 1 = 0.5mA; Vc = 10 + Rc (0.5) = VB
= 0.5mA; Thus, RE =
= 66kn Now,
0.5
 VBC = 4 0 = 4V; Thus,
(4 +
10) = 12kO.
(b)
VB = 6V; VE = 6 + 0.7 = 5.3V; /£ =
Vc = 10 + Rc
(0.5) =
VB
~ ~5'3
° /<£
 VBC = 6  0 = 6V.
= 0.5mA; Thus, RE = Thus
0.5
= 10.6kQ. Now,
= yj (6 + 10) = 8k£2.
4.27 Assume active mode, with forward conduction of the baseemitter junction. (a)
= 93(iA; Ic = 93iA x VE = 0V; VB = 0 + 0.7 = 0.7V; lB = 93 = 1023(1A = 1.023mA; Vc = 10  2k (0.93) = 8.14V.
(b)
VE = +10V; VB = +10  0.7 = 9.3V; lB = = 1.023mA; Vc = 0.93 (2) = 1.86V.
(c)
~ = 93)lA  70(iA = 23(iA; /c = (23(iA) 10 = VE = 0V; VB = 0.7V; IB = 0.230mA; IE = 11 (23) = 0.253mA; Vc = 10  2 (0.23) = 9.54V. See that VE = 0V, and that if the transistor conducts with VB = 0.7V, that the current in the upper
(d)
10 = 0.930mA; lE = 930 +
= 93(iA; Ic = 93 (10) = 0.930mA; lE = 11 (93)

100k£2 will exceed that in the lower, providing no net base current. Thus the transistor is cut off, with v, = +io (10 10) = = iov, „,d /. ic = OmA.
 
«v. rc
4.28 (a)

10k£2 (IB) 0.7 Assume active mode, and consider the basetoemitter circuit: Thus 0 (21 ) 10 \)1B 10k£2 (IE) = 10V. But IE =(3+ IB = 9.3V, whence =2Ub. Thus 10 IB +
VB = 0  42.27 (10k) = 0.423V; VE = 0.423  0.7 = 1.123V; IE = 21 (42.27) = 0.888mA; Ic = 20 (42.27) = 0.845mA; and Vc = +10  10k (0.845) = 1.55V. IB = ~ = 42.27(xA.
(b)
1B = 100k 21(10k) = 32.26)lA; VB = 0  100k (32.26) = 3.23V; VE = 3.23 0.7 + = 3.93V; IE = 21 (32.26) = 0.677mA; Ic = 20 (32.26) = 0.645mA; Vc = +10  10k (0.645) =
As before: 3.55V.
<*>
Thus,
''
"
1«+7mL 
= "4mA;
°
110 + 47.6X 90 8.26  0.70 = 7.56V; /c = =pr (0.174) = 0.166mA; lB = (10) = 8.34V.
"
0 MAmA
10k£i (°174mA) = a26V;
V' =
= 8.29 (lA; Vc = 10 + 0.166
4.29 For VBE = 0.7V, the current in the baseemitter shunting resistor is 0.7V/10kI2 = 70XA. This flows in the resistor to 10V, creating an equivalent base source of VBB = 10  lOOkfl (70(lA) = 3.0V, with RBB = lOOkO. Now, for the baseemitter loop and base current IB, 3V  (lOOkil) lB  0.7V  (P + 1) (3.3kH)
iB = 0.
Thus, IB =
For p = oo:
3 3 (p + l) + 100'
IB = 0iA; VB = 3V; VE = 3  0.7 = 2.3V; lE =
= 0.697mA;
Vc = 10  3.3 (0.697)
= 10  2.3 = 7.7V. For
p = 100: lB =
33(1qÿ+
1qo
=
vb = 30  5.35 x 10"6 (10s) = 2.465V; VE = 2.465
207

SOLUTIONS: Chapter #48
0.7 = 1.765V; lc = 100 (5.35) = 0.535mA;
Vc = 10  3.3 (.535) = 8.23V.
(ÿ}3+
= 16.87ÿA; VB = 3.0  (16.87) (0.1) = 1.313V; VE P = 10; lB = 3 3 1QQ 0.613V; Vc = 10  3.3 x 103 (16.87 X 10~6) (10) = 9.44V. For
= 1.313
 0.7 =
"
+ '> =
2
(2.3) = 4.4V.
4.30 From the solution of P4.29, VBB = 3V, RBB = lOOkO. For p = oo;
VE = 3  0.7 = 2.3V, with IE =
For f generally, /, >
2 3V
"7
J • j/C a u
»+ i)
= 0.697mA.
=
23
3.3 + 10(yP + 1) ' 100
33 + °'8 <0'697) = '5576' = T"U8' 3.3 3J+T .imP +1) 0.8248, P + 1 = 100/.8248 = 120, whence P = 120. 4.31 Here, (a)
10A
VBB = For
"
p = oo; VB
23
OJSK " 4125' <*
(9V) = 3V; RBB = 10k II 20k = 6.66K2.
= 3V;
VE = 3  0.7 = 2.3V; IE =
2 3V
= 2.3mA; Vc = 9
Vex = VCVE = 4.4 2.3 = 2.1V. (b)
(c)
p = 100: IE = 3 ~ °;L, = 2.156mA; VE = 2.156 (lk) = 2.156V; Ik + 6.6301 Vc = 9  2k (2.156) x 100T01 = 4.731V; VCE = 4.731  2.156 = +2.575V. , 23 For p = 10: IE = , , = 1.432mA; VE = 1.432V; Vc = 9  2 (1.432) 10/11 = 6.396V; 6.(yl
For
1
Ik 4
VCE = 6.396  1.432 = 4.964V.
4.32 (a)
Let 10 /
lc  i
(b)
For
Vfl =
.= (c)
93
~
0.7V,
"2 2'°191) =
For V„ = 0.7

l5m 
ÿ
ÿ
= 0191mA.
From the previous result (1) in (a):
(.590)) = 3.79V.
°590mA = 7c. and VCE = 1010 (.0191 +
1334 =
VCE = 1010 (.0212 +
(d)
 
From the supply to ground: 10 = 10k(/ + //p + /) + 100k(/ + i>P) + 0.7, or (1), or + 10 i/50 + 100 *750 = 9.3  10/  100/. Thus 12.2/ = 9.3  110/ (°2) 93 ~ , or / = / = = 0.582mA = /c, for which VCE = 1010 (.02 + J" (0.582)) IZ.Z 1Z.Z = 3.86V.
jjK
Ji
= .0212mA. Thus, Ic =
(0.571)) = 3.96V.
93~ 1
/°
( °212)
lZ.Z
= 0.571mA, and
—
Let the base voltage be x> and base current be /. Thus, \> = 0.7 + (P + 1) / (lk) = 0.7 + 51 / (2); Also (1); = o + 100k (o/68k + /) = V + 1.47 o + 100 / = 2.47 v + 100 / (2) (3). (3) Now o/68k) with 1) D 10 510 0.147 / » 2.47 ((p / 10 10 it = = + + oc (4). Then (4) with (1) —» u + 100 / = 10  510 /  0.147 t), or 2.617u + 610 / = 10 2.617 (0.7 + 51 /) + 610 / = 10, or 1.832 + 133.5 i + 610 / = 10. Thus, . 10 1832 JU8_ 0UmA) whence u = 0.7 + 51 (.011) = 1.261V, and Ic = p i = 50 743.5 133.5 + 610 (.011) = 0.55mA, VE = lk (p + 1) i = 1 (51) (.011) = 0.561V, Vc = 10  10k (51(.011) +
_
—

_
_

= 10  5.61  .185 = 4.205V. Thus, VCE = 4.205  .561 = 3.69V
 208 
ÿ) 68k
SOLUTIONS: Chapter #49
4.33 +15V 150.7V
+15V +14.3V
1 10 kQ< 1k{101J =
For 3 = oo:
VA =
100'°0200
10.7 + 0.7 = 11.4V; VE = (
P
For
= 100:
(15)
15
<10kQ?i01k
= 5V; VB = 5
= >R
 0.7 = 4.3V; Vc
= 15 
(4.3)
= 10.7V; VD =
1M) lk = 3.6V.
Ik At node A :
VM = 5V, and Rÿ = 100kl2 II 200kl2 = 66.7kl2; = 4034mA; Vb = lk (4034) = 403V; and = 403 + °7 = 473V
ioaffÿiol At node C: V = 15 — +ÿioi (®7) = 14.94; R = 10k II 101k = 9. 10kl2. Now, from the collector node C, / = 0.4034 x jjjy = 0.399mA. Thus, Vc = 14.94 0.399 (9.10) = 11.31V, and VD = 11.37 + h=
at
jq
0.7 = 12.07V.
Thus,
_ ,, . 15 — 12.07 x 100,) —101 — = 2.90V. VE = 0 + lk (Ik
SECTION 4.7; THE TRANSISTOR AS AN AMPLIFIER 4.34 Generally, gm =
Ic 10~® : For lflA, gm = —— X 10 25 Vt
= 40 x
r
106 = 40pA/V; and for
lOOpA, gm = 4mA/V;
for 1mA, gm = 40mA/V ; for 100mA, gm = 4A/V. Vp 0C Vp Vt Vp (x R 1 4.35 At the emitter, rt = — = — =  = — . At the base, rK = —— = = t— = (P + 1)re For gm gm 'c h 'B 'c4> gm pH = 100, a = 0.99. Now, for lc = liA, . = 24.75kl2 ÿ 25kl2, rK = r = 2.5M12; ÿ re = — 40 x 10"6 40 x 10"6 for lOOpA, re ~ 25012, rn = 25kl2; for 1mA, re = 2512, rK  2.5kl2; for 100mA, re = 0.2512, rn = 25012.
— ——
.
—
—
IE Vf
ot
4.36 Gain = gm RL, where gm = —— ~ Thus
lE . —— Vf
The voltage across
a K , for which the  gain  = —h —x Re = —— ex 7/r Vf
RL = Ic RL = a IE RL
K K = —— , a constant! —— cc i£ Vt
There is no gain variation possible. The bias current does not matter, way! 4.37 The input resistance at the emitter, = (p t l)re
ait re
=
151
>50xi0m 250
151
(250)
_
Thus the gain is constant. load resistor is varied this
25 x IP'3 qr = 25012. The input resistance at the base, 100 x 10" The voltage gain, basetocollector, is 37.75kl2.
Vp_ re = —— = IE =
if the
is a constant, K.
39.73V/v_
 209 
——
SOLUTIONS: Chapter #410
4.38 The collector load is the lOOkQ resistor from collector to (grounded) base. Now, IE = 1mA, a Rr 25mV 100k£l _4QQQy/y phe resistance "seen" by and re = 4ÿr = 2512. Thus the gain 2512 1/nA v>„ Mi l>0 25 Mi x the source % is re = 2512. Now, for Rs = 7512, = = 14, and I)., = = 4000 x 75 25 o,U, + Ms 14 = 1000V/V.
_
 
— —
— — —
SECTION 4.8: SMALLSIGNAL EQUIVALENTCIRCUIT MODELS 4.39
4>
gmV
+
*' ÿ
See t) = r„ =
Mh, ÿ
ib
re +rE 1  8m rc
re + i'e
re + rE 1 a
25 + 75
— 8m
8m Mbe
17
Vta'
E
.i;
Mhe
= Mb//
r/ = (P + 1) (re + rE) 10(H01
re
Mbe > 2nd gm Mbe
re + rE
ÿ gm' Vbe'
r*'
+
v
Vbe'
= gm' Vbe'
M = 8m
=V
Mhe'
r, + rE 8m
re + rE
= (P + 1) (re + rE). Now, for Ic

Thus, gm' = gm 1
= V
1mA,
re + rE
I'e
+ I'E
4.40 (a)
4> + V#
V
ÿ
Te » 25
al
See directly (with either Tgm or Ta) that Ik12 X v., , whence m„ = 25 + lJk!2
— = 0.976V/V.
h
' Vo
r 210
re + i'e
2512, and rE = 3re = 7512, (o/rcye grn i'e a
 9.9 x 10_3A/A = 9.9mA/V.
ÿ
8m
Now,
re =
= 101 (25 + 75) = 10.1k 12, and gm' =
gmV
'
. re + rE
re + rE
re + rE
SOLUTIONS: Chapter #411
(b)
lOkn
rn
= ~gm
Vo
+
Vs
ngm '
For
RS
4>
2.5kO
gmV
a
As
in (b), with Rs = 0,
P
100(1)
Rs + rn
0 + 2.5
tyi
V*
_
8m '*71 RL
~
_ 40V/V.
(Vv)
i'm Rl,
X
—Vs— = —2.5_2.5+ 10 x 40 x 1 %
= "P
(c)
'"n + Rs
=
8V/V. For hn, see U„ = ~(i(i ) RL
1kO
p)
=
V
whence
RL
see
ra + RS
Ri, whence
Rs + rn
V,
VRl + rn
= i'„.
= 40
100(1) = 8V/V. (10 + 2.5)
1 = 40V/V, or
X
(d)
Vj_
a Rl
Vv
re + rE
0.99(1*) = 7.92V/V, 25 + 100
—
40 x 25 = 8V/V, 25 + 100
= ~8m Rl x
or
re + rE with the former (using Ta) being more direct
al m gmv
(e)
'
iokn
See 2.5kn
V),
x 1x
/
=
X
Rb
II rK
Rb 11 rn + Rs
=
 40
2.5*2 = 4.44V/V. 2.52 + 10
2.5kn
4.41 Since
IE =
1mA,
25m V re  1mA = 250, and — = vs
re
= 0.99
25
= 297V/V. For
a signal
10  7.5 (.99) (1) = 2.575V. For guaranteed active operation, \>CE ÿ 0V. Thus the voltage of 0V, VQ largest allowed sinusoid has a peak value of 2.575 V 0V = 2.575V at the output, and 2.575/297 = 8.67mV peak at the input.

4.42 Now, i = I s eÿ7 in general. Thus ~r~ = e(1>l
i\ = le~l(y15 = 0.670/ mA).
x>ÿ7, and
'2
for
,
i2 = / mA initially, i = Iel
That is, the current increases by 49%, or reduces by 33%, for ±10mV variation around the operating point. For linear operation over a ±10mV input range, a current of 1.49/
(or
 211 
SOLUTIONS: Chapter #412
——
must be tolerated. For X)Co ÿ 0, with / = 1mA,
= 1.49 / = 1.49mA, and
Rc
Rc
ÿ
— —— < 1.49mA
6.7kf2. 4.43 For
— V)f
lOOpA
_
4.44 For u across the twoterminal device, the R  II r„ — x v. v* = ——
*
2001/
= 25012, and r„ = yic = 100pA, re = 2 x 106 » 2 x 106 _ ÿooov/y  _ÿ£L 250 re voltage
across
= 2 x 106£2.
the
Thus, the gain
baseemitter junction
is
RiWrK + R2
For this situation, the total current, i = 8m
1+
=o R2 +
+ 8m
+ R2
fli II
=
X>
1 _
8m (R i H rK)
Ri II rK + R2
A, II rK + R2
rnR\ = \)
rnR\ R\ + rK
_
Thus, resistance r =
R 1+/*it+Pÿ i
R \Ri+r kR i+r rR i
+ R\ R2 + rn —rz(3+1) ——x> = Ri{ rK +
Ri
R2)
(Ri
P+l
i
(a)
For R2 = 0, R\ = °o, r =
(b)
For R \ = °o, R2 = rK, r =
(c)
For /?) = 7?2 = r«, r
ÿ
ÿ2 + rn (1 + Rj/Rj)
iyR, + (p+l) rK + fit (1 + 0)
+ re
(R i + ÿ2)
0 + r„ _
f __
0 + (P + l)
p+l
it
"
+/?!)•
= r..
.
= 2re.
0+p+ 1 At '*+ 0 + 1)
— + P+
_ "
Ri
3r,
P+2
1
= 3 re.
4.45 For V/, = dv , small compared to x>„ , the gain is
„ — = gm (rQ II Rf),
A =
l>/ + u.v
h=—
whence
am Vn
= r,
Now for
gm
Rf
Rin
=
(1  ~8m (r„ H R/))
—
•0, f,
Rf
r„
1 + 8m >,A
r ~
rn II
2 8m
ro
~ 2re, that is, very small.
 212 
Rf 1 1 + — (r„ II Rf)) gm + I'll rK RfA\
1 + gm (ro I' Rf)
— , = r„ , the gain is Av = gm (r„ II r„) = 
and Rin = rK
and the input current
SOLUTIONS: Chapter #413
4.46
VJi2
Y
gm2 Vji2
gml Vnl
Y„'v
Note that the emitter current of Q2 is the base current of Q\, and therefore re 2 = rnl and and gm2 = gmrtPi+1) By considering the input rK2 = (Pi + 1) r,t j. Correspondingly, Mn j = u* 2 = base current, see: rK' = rn 2 + (p, + 1) ] = (Pi + 1) r„ i + (P, + 1) rK j = 2 (pi + 1) r„ By considering the output collector current, see: gm' D = gml Dre i + gm2 u„ 2 = gm l t/2 + (gml tÿHP. + 1) = (gml vV) [1 +
KPi + 1) ] = gml v2.
Thus g„,' = gmÿ2.
SECTION 4.9: GRAPfflCAL ANALYSIS 4.47 With p = 200 and VA = 100V, for iB = 1, 2, 5, 8 and lOpA, ic = p iB = 200, 400, 1000, 1600 and 2000pA, and r„ = VaAc = 500k£2, 250kf2, lOOkO, 62.5kf2 and 50kf2, with the current at 10V greater than that at 0V, by 10/r„ = 20, 40, 100, 160 and 200pA (ie by 10% (= (10V/100V) X 100)). For VCC = 10V and RE = 5k12, the intercept is ic = My5kCl = 2mA. 5kQ load Una
VEC
For the operating point (Q), iB = 5p.A, Vec ~ 5V from the graph, and ic ~ (5 (200)) (1 + 5/100) s 1050pA, with vEC = 10  5kO (1.05mA) = 4.75V. For a ±3pA peak input wave, operation varies from Q to A to B to Q above. At A: iB = 2pA, vEC = 8V from the graph, with ic ~ 2 (200) (1 + 8/100)iA = 432iA = 0.432mA, for which uec = 10  5kf2 (.432) = 7.84V, and ic = 2 (200) (1 + 7.84/100) = 0.431mA. At B: iB = 8pA, X)Ec = 2V from the graph, with ic ~ 8 (200) (1 + 2/100)pA = 1.632mA, and vEC = 10 5 (1.632) = 1.84V.
Thus the output wave has a positive peak of 4.75  1.84 = 2.91V, at an output current of 1.632  1.050 = 0.582mA, {Check: RL' = 2.91/0.582 = 5k£2}, and a negative peak of 7.84  4.75 = 3.09V, with a current of 1.050  0.432 = 0.618mA, {Check: RL' = 3.09/.618 = 5k«}.
 213 
SOLUTIONS: Chapter #414
Note that the positive and negative peaks are different, indicating that a (small) signal distortion results. 10 + 5
Basa Currant
 10pA SpA /
\
>7
—
For a ±10p.A peak input wave: See from the graph that for base signals of more than +5iA, iB > lOpA and the transistor saturates, and that for signals less than 5pA, iB < OpA, and the transistor cuts off. For this situation, the output is clipped for 50% of the cycle.
« +15pA
10 + 5 ÿ
 5pA
OpA
10V
VEC
SECTION 4.10: BIASING THE BJT FOR DISCRETECIRCUIT DESIGN 4.48 From Fig. 4.39b), in general, IE = (VBB
Rb
]
 VBEy Re+
Re, or P + 1 = 100 Ie = 99 /, P+l 100 P > 99, IE is within 1% of its maximum value.
Rb
Rb
.
P+l
„
„
,
For p = oo, IE
VBB VBE Re
_
.
 —= /.
. For RE = RB, P + 1 = 100 or — re
p = 99.
For
That is, for
Alternatively, one could interpret the situation to mean ±1% of a nominal value, where the largest occurs 11 12 for P = oo, and the nominal for p = 99, where —— = 777, and the minimum where P+ 1 100 P+l 100' for which P = —ÿ— 1 = 49. 10
4.49 See that
= VBB = — 3
33
ÿToo7
,
4V. Generally, 1
50
R2
Here, 100mA =
Re
Thus
R" + jr*io
Vjh 10
, whence 3R2 = R\ + R 2. or R\ = 2R2. Now, since — x 12 = VBB = — 3 2R2{Ri) 3(150) ' = 150, or 432 /?2 = 150, or R2 = ÿ4ÿ = 225£2, and Rt = 2R2 = 3 2
A 
R\ R 2 1+A2
40.7 50
nc
12
50(30)/ 10 = 150£2. Now 
A
—
= 0.03005k£2. Practically speaking, use RE = 30 12, with RB =
1+5TX1o
Ri II R2 = 150 =
Vnn
—. IE = Kb Re + p+l
+A2
L
A
45012. For a conservative design, use smaller values, such as
R2 = 98mA, and RE Ic 
= 20012 and R\ = 2(200) = 400£2.
ÿ (100) = 4V. Thus Rc = \ = lE = 100mA, Ic = 773 98mA 51 0.0408k£2, for which use Rc = 40 12. Now, for the design overall: RE = 30 £2, Rc = 40 £2, R2 = 200 200 x 12 = 4V. Q, Ri = 400 £2, where RB = 200 II 400 = 2°"ÿ°0) = 133.3£2, and ' BB 400 + 200 600 50 40.7 Thus, IE — 400.1012 (30) = 4.995V, and VCB = 101.2mA, VCE = 12 0.1012 51 30 + 133.331 Now, for
= 4.995  0.7 = 4.30V.

4.50 Assume 1E varies 5% over the entire range of p, from 20 to 00. Assume that for ±1V output, the base signal is very small. Further, assume that VB~ 0V, and that operation is for Dcb  0 Now, for a IV output signal and
5V  1V — = /c = IE = 4mA. P = 00, —lit 7777 £2
In practice one would use
RE =
1.00k£2, in which case
collector goes to 0.3V with a 0  5  0.7 , whence RB  21 1.075 +
Rb
Since
VE
VE S5
= 0.7V,
— 1A: (1)
0 7
5
RE = — = 1.075k£2. 50.7
1
= 0.7V, and the
V signal peajc. For RE = 1.075k£2, 1E = 0.95 (1mA) = 4.3  1.075 = 72.5k£2. In practice use a smaller (standard) .95
20+ 1
214
SOLUTIONS: Chapter #415
value, say 68k£2.
4.51 For
5~ VCB = 0.5V, and p = 200: lE =
Rb = 95.2
0 5V
ÿ
~
5
°'
1 056mA = 1.056mA, and lB = = 5.25{tA. Thus,
= 95.2kO. {In practice one would use a lOOkft resistor.} Now, for
—0.7 = 0,
ÿ
or
1.056mA to 0.787mA and,
IE = ——
P = 50, 5  3.6 IE 
= 0.787mA. That is, with RB = 95.2k£2, IE varies from
36+ÿr VCB from 0.5V to 95.2
Alternatively, with Rg = 100k£l: For
(0ÿ87) = 1.47V.
P = 200: IE = ——ÿ
= 1.049mA, and VCB = 100
3.6 +
0.522V. For p = 50: IE =
5
"
o s~
°'7
' "sT \J\J
201 = 0.773mA, and VCB = 100
4.52 For this situation, the "base" current is IB =
=
07 (201)
Ie
1.056mA. Thus,
Now> for y
= Q 5V>
p
Ji 1
V
Vcc = 5V
2(K))
=
(°_ÿ3) = 1.516V.
bE lE IE —— + —— = —1. 1 «p pe? + P+
=
ÿ 201
ÿ
Now for
p=
00
= 3 6ka ; =
,and pM = 200, 5 ~ °'5 " 0/7
Ra~rÿ7ÿ" ~ 1.056
3..6kLl
=
133.2kfl. In practice, use a (smaller) standard value, 130k£2, or 07 anÿ f°r P = 00 a"d 120k& as it is more commonly available. With R p = 120kI2, IB = = \20k = 85.8k£2. In practice, use a larger standard value, say RB = 91k£2, for VCB = 0.5V, Rb = g
'**
5~ 070 which VCB = 91k (5.83pA) = 0.530V, and IE = = 1.047mA. Now for p = 50, base current flows in RB to produce a voltage drop which combines with a constant voltage drop of Vp = 0.53V i„ due to Rt. Thus. 0 ™mA' = 5 "X6 R. ,E =
°'336~
'
(0.700) 0.7 = 1.78V. That is, for /?p = 120k£2, RB = 91kd, and 1.047mA, while VBc varies from 1.78V to 0.530V.
4.53 For Pÿ = 100, and using the solution for P4.52, /?p = (though larger). Now IB =
= 10.3xA, whence RB =
00A:
—7ÿ
~
10.3
P
"""
> 50, IE varies from 0.700mA to
66.95kQ. Use 68kf2 as very close
= 48.5kf2. Use 47k£2 as close (though
smaller). For these choices and for P = 00, VCB = X 47 = 0.484V, and IE = —0.700 68 3.6 5 ~ 0 484 "" 0,7 1.06mA. Now for p = 50, lE = = 0.844mA, and VCB = 5  3.6 (.844) 0.7 = 1.26V. 3.6 +
——
_
4.54 In Fig. 4.42a) of the Text, / = 1 mA. For P in the range 40 to 200, base current ranges from 1/41 = 24.4pA to 1/201 = 4.98xA The lowest the emitter can operate is at 5 V. The lowest the base is allowed to go is  5 + 0.7 =  4.3 V. Thus the largest acceptable RB = 4.324.4 X 10"6 = 176 k£2. For Rb = 176 kfi, the base will range from  176 x 103 x 24.4 x 10"6 =  4.29 V to  176 X 103 x 4.98 x 10"6 =  0.876 V. Now, at low p, r„ = (P + 1)re = (p + l)Vr/7 = 41(25 x 10'M x 10"3 = 102512.
.
Thus consider Rb = 100(1025) ~ 100 kfl, in practice.
215
SOLUTIONS: Chapter #416
The base voltage now ranges  100 X 103 X 4.98 X 10~6 =  0.50 V.
 100 X 103 X 24.4 x 10 6 =
from

2.44
V
to
As you will see in subsequent Sections, the 100r„ design allows 99% or so of the signal currents applied to the base lead to enter the transistor. From the signal point of view, this is a very efficient design.
4.55 For VBE = 07 V, the drop across R is (5   5  0.7) = 9.3 V. For / = 1 mA, use R = 9.34 mA = 9.3 k£2. From a practical point of view (See Appendix H of the Text) a resistor of 9.31 k£2 is available on the 1% scale. If a lower cost resistor were needed, 9.1 k£2 ± 5% unit would be acceptable. The out¬ put current Iremains essentially constant provided Q2 does not break down or saturate, from + 5 V or more to  4.8 V or so (if VCEm = 0.7  0.5 = 0.2 V)
SECTION 4.11: BASIC SINGLESTORE BJT AMPLIFIER CONFIGURATIONS
.
4.56 As is customary, ignore VA in the bias calculation: Directly, IE = 10.0mA Thus, lE = (100401)10 = 0.049 V, and 9.90 mA and Ic = 10.0{101) = 0.099mA. Thus VB = 0 0.5(0.099) = 0.5(9.90) 5.05 10.0 V. 0.749 Now 0.700 V. 0.049 = = = = = Vcc Rck Vc VE For this bias situation: gm = /(VVy = 9.9025 = 0.396A/V = 396 mA/V, re = VT/IE = 2540 = 2.50 £2, r„(p + 1)re = 101(2.50) = 252 £2, r„ = VA/IC = 100(9.90 x 10"3) = 10.1 k£2.






4.57 Using the result of P4.56 above: /?, =rK= 252 £2, R„ = Rc II r„ =0,5k II 10.1ÿ = 476£2, At) = DoA)5 =  p(flc II r„}{Rs + r„)=  fiRAR* + Ri) =  100(476X500 + 252) =  6.3. V/V, A, = iAb =  PV(rc +RC) =  100(10.UX10.U + 0.5k) =  95.3 A/A. For A \) with a 500 £2 load there are two approaches: a) the direct, and b) the Thevenin: (a) A\> =  P(/?c Hr„ II RlWs +rn)=  100(0.5 II 10.1 II 0.5X0.5 + 0.252) =  100(244452) = 32.4 V/V. (b) The amplifier as a Thevenin equivalent voltage gain, At = A v0 =  63.3 V7V with a Thevenin load, Thus with resistance source R„ = 476£2. equivalent Am = A moRARl +Ro) =  63.3(500X500 + 476) =  32.4 V/v. When comparing with the results of Exercise 4.31 in the Text, we see that with resistor/current scaling that the voltage and current gains are essentially constant, this is reasonable, since we are dealing with a linearized circuit model. Here, even the nonlinearity associated with bias design is eliminated by the use of the constantemittcrcurrent bias design. In general, for such designs, parameters scale by the same factor, and gains are constant.

4.58 The need for highestpossible gain for a fixed load implies a large bias current: Thus, for P = °°, VB = 0 9 1 0.8mA, whence and for ±1V swing, and Uc S 0, VE = 0 + 1 = IV and IE = IE =
Re =
0_07
 —
9 = 10.38k£2. U.o
nQ
Now, if we use Re = 10k£2 (as a standard value), we see that me falls to 0.7V (for p = «.) with vCB = 0.3V, which is often acceptable for linear operation. Otherwise, use RE = llk£2. ,7 9 For Re = 10k£2: With p = «: VE = 0.7V, Vc = +0.7V, Ic = = 30.1£2; = 0.83mA; re = rK = <*>, r„ =
100V
= 120.5k£2, —• =
.
1 lOJk il 120Jfe „ =— — = 307V/V, and for ±1V output,, „Mh = t)., At
xiv
= 3.26mV. wuk
t'90v°
icoÿuiot) x
10k (.740) = 1.60V,
re =
25mV
,00k
 'L52V' ,c =
= 33.8mA,
lr
= 91 (33.8) = 3.074k£2, r„ =
216
01 =9 100V
= 135mA,
=
SOLUTIONS: Chapter #417
90 (10& II 135&) rrrÿ 3.074/t + lOifc
,
—=
fAI, t. {Alternatively, 3
64.1V/V.
=
64.0V/V}. Now for ±1V output, o, =
J
VI*
1
= 15.6mV, and

3.07 90 (10A: il 135ife) x  = 10 + 3.07 91 33.8
—
1U I
3*(i I
(15.6) = 3.66mV.
4.59 Approximately, since the resistor through which base current flows is R$ ~ Rq, I e 's essentially fixed at = 26.912, and r„ = = 0.930mA for reasonable p. Thus, re = = 215k£2. Now, 10A 0.93/n/t 0.93 „ 1.37A II 10A 50 10A II 215A c, ,nr n\ e en i nun andi  for x x = 51 (26.9) = 1.372kl2, = 0.177 x PR = 50,
x
0.98
ÿ=
x>s
= 37.5V/V.
355
_ 1ÿ1110
150 x x 151 4.06 1110+ 10
— o.v
ÿ
p =
ÿ
+ÿ
150, rK
10AH15A = 26.9
=
ÿ

151 (26.9)
= 4.062ki2,
355 =
and
0.93mA, re = 26.9, and r„ = 215k£2. For p = 50, rih = 51 (26.9 + 100) = 10 11 2l5k x x = 0.282 x 0.98 x 75.3 = 20.8V/V. For p =K 100 + 26.9 6.47 1110+ 10 51
4.60 From P4.59 above, 6.47k£2, and

for,
Now
IE ~
ÿj7 "
\°k
ÿiflO
=
= 150; rih = 151 (26.9 + 100) = 19.2kO, and
X '98 X 153 = ~293V/V We 19 + 10 see that the design using unbypassed resistor in the emitter is relatively insensitive to p variation.
4.61 For each transistor,
IE = 1mA, re = 2512, and rK = 151 (25) = 3.78k£2. Thus
7ÿ7—\>hi— = —151
J*25
ÿ
"
10A: 3,78t =  jrr" = 109V/V. Now Rhi = 3.78kI2, and — 151 25 Oft 1 3J*k , = 0.274V/V. Thus — = .274 (109) (199) = 5934V/V. =— 10k + 3.78/fe o, vs
= 199V/V. Now, Rh2 = 3.78kO, and
4.62 Rj = (P + 1) (re
P
= 50. Thus 10kf2 = 51 (125 + RE), whence 11 50 (10k II life) 104 125 P • ,n ,oc .u n voltage gam == = 710. Thus the o>.5 =  — RE =
+ Re) =
10kl2,
= 12512,
re =
——
—
2.27Vrv. 4.63
Rb
2k = 3k "I r 151 P+l 21.58 = 21.412 for p = 150, and 8.3£2 for P = oo. Now, the gain from a 10012 source is: a (Rl II Rc) . 150 (1*113*) . „ c n p = co, "i general. For P = 150, vf/os = — = 6.14V/V. For vA>s = R +R 1QQ + 214 Ik II 3k o,A»ts =  = 6.92VIV. For RB = 0, the results for P = 00 apply, that is r, = 8.312 and 100 + 8.3 \),/os = 6.92 V/V.
IE =
3mA,
re = 3mA = 8.3312. Tlius /?, = RE
rr +
= 3k II 8.33+
„
—
*
4.64 For a base current i and 10*12 (i +
10A:12
VEB =
) 07, whence — 10££2 (101 i + 9  1010 i  0.7  0.7 10 i  0.7  1010 i 
0.7V, using KVL: 9
_ io*12 (101 i+ ÿr) = 0, or 10A: 12
 4 (0.7) = 2030 i, and / = 3.054fiA. Thus Ic = 50 (3.054) = 0.153mA, VE = 9  10kQ. (101 (3.054 X 10"3) + ÿ) = 9  3.084  0.7 = 5.216V, VB = 5.216 0.7 = 4.516V, Vc = 10kf2 (101 (3.054
0.7 = 0, 9
1yjfC
x 10"3) +
lyjK a/i
= 3.784V. Check: VB
 Vc
= 4.516
 217 
 3.784 = 0.732 = 0.7 +
10kf2 (3.054pA)m as
SOLUTIONS: Chapter #418
required. For all designs, all couplings arc via capacitors: (a)
Source coupled to B ; Load to E \ (Ground to C)
(b)
Source to B ; 10k£2 coupled from E to ground (or (better) 10/3kQ from E); Load to C .
(c)
Source to B ; Ground to E, Load to C.
(d)
Source to E; Ground to B , Load to C .
4.65 For all designs, Iq = 0.153mA, re = (a)
Since u€
= U/,, the shunt
25 = 163.412; .153
10k£2 can be ignored, and A„ =
10k II 10k = 0.968V/V. 0.163 + 10k II 10k
(b)
For  = 1V/V, and o, = u, the voltage Vb
+ Vcc
REeq 2.5
Vc
10k II
ib
L—AAAr—
RLeq
0.5V/V. For
X
Rf 10 kQ
rvJVs
a
Vo
wv
Vb
and the gain 2.5k 100 x 101 10k II 10k
2.5ki2,
Re 10 kQ
—
Thus
2u.
Ru„ = Rl II Rc II Rf/2 = 10k II 10k II 5k =
[10 kQ
ÿ
is
Rf
across
10k
=
100
Re = lOGkG, Gain =  101 s
1V/V. *
i'
10 kQ <10 kQ
(c)
For K large, the signal across Rj is essentially only due to the output voltage. Thus Ruq = = 20.4V/V. = 3.33kQ, and the gain = RL\\Rc\\Rf =
(d)
Base is grounded, and the gain = +20.4V/V.
4.66
P = 50; VA + Vcc
= 100V. Now for 100l/ 1MO, r„ = .. . = ujo 0.1
+ Vcc
25mV 0.1mA = 1.919kk2,
re 
0.1mA 20 kQ
ÿ
Vb
Gain XJVs
1 100 kQ
>
Vn
vh
Ic = 0.1mA,
250Q, RUq = 2k II 50k II 1M
1.919k = 0.885V/V, 250 + 1.919
= 101 (.25 + 1.919) = 219ki2, Gain
<60 kQ
Rinb
Vb
D.v
•»«
100 II 219 = 0.775. Thus gain 100 II 219 + 20 Uv
0.775 x 0.885 = 0.685V/V.
 218 
SOLUTIONS: Chapter #419
4.67 Iterate:
VB2 =
0.7V,
VBl =
1.4V,
IBi >
68k
X
±~ = 0.102pA. 101
Thus,
VC2~ 1.4 +
0.1 = 1.5V,
= 2.33mA, IB2 = = 23.1nA, /fll = /c2 = —rrÿ 1.5 101 + 0.7 + 0.33 x 10"6 x 1 x 10w = 1.73V. Thus, IC2 = 5
15
2 33
0.7 + 23.1 /101 = 0.33xA. Thus, VC2 = 0.7 68/k 1.73 = 2.18mA 1.5 See with R = 68kf2 included, IE2 ~ 2.20mA, IEl = 32iA. Now, re2 = 25/1.20 = 11.4£2, rKl = 101
(a)
(11.4) = 114812, rei = 25/032 = 78112, rni = 781 (101) = 78.9k£2. Thus,
= 52.1V/V,
= nno]A4S, 608„ — " II 68 0.781 + 1.148 t).5
= 0.59 1V/V, and
— = 52.1 X
= — — ub 101
L5k
"
ik
11.4
0.591 = 30.8V/V, with
68k + 0.78k) 101 = 193kf2. Now with R = 68kI2 removed, the base current in Q\ reduces slightly, and the collector of Q2 lowers by 0.1V or so, with IEi increasing by 0.07mA. Thus IE2 = 2.3mA, IEl~ 23(iA, with
Rin = (1.148k II (b)
10.912,
1.098kf2,
Now, reX~ 25/023 = 108712, and rnl ~ 110kf2. V),. vh i 098 , = 54.5V/V, — = , = 54.5 x 0.503 = = 0.503, and 10.9 1.098 + 1.087 uv uft 27.4V/V, with Rin = (1.098 + 1.087) 101 = 221kQ.
re2 ~
rn2~
="521x4h4
'
1MQ. 1  30.8
Thus resistance seen by Ms for (a) is
193k£2 = 31.4 II 193 = 27kf2.
SECTION 4.12: THE TRANSISTOR AS A SWITCH CUTOFF AND SATURATION 4.68
Ic =
ÿ
1x12
Rb =
= 4.8mA, and IB = =
2.69kI2=
:r— =  Rb
2.7k£2.
ÿ
Rb
Now,
But
Ib
= 3. Thus 4.8 = 3 (4.3)/RB, whence 50.2
$forced = p =
= Rb
Rb
rb
<ÿ
X
= 0.448 pkf2.
4.69 (a)
U/ = 0V
(b)
x>i
=
$forced ~
—> Qi cutoff, and Q2 saturated; 5V
lg
j> ~
saturated, at d 50.2 50.7 / l/tf2 l/tS2
Qi
 219 
Q2 4.8
4.3
cutoff.
= 1.12.
4.8 4.3
<, (V2,
and
SOLUTIONS: Chapter #420
4.70 (a) Assume the transistor is saturated. Working on the diagram: + 6V
+ 10V
+ 4.3V
+ 9.8V (4.3+9.8)/2 = 7.0SV
10 kQ
> 10 kQ
•
10k//10k 6 kQ
> 10 kQ
10 kQ
0>.
I CD, mA mA
VE = 7.05  5kQ(lmA) = 2.05V, VB = 2.05 + 0.7 = 2.75V, Vc = 2.05 + 0.2 = 2.25V, 5275 10225 0 775 Ic = lOJfc — = .775mA, IB = lOJfc— = .225mA, and (if0rcej .225 = 3.44. For the edge
—
t)
10 ~
°:2
oog
~V

, for which 430  100 u = 9.8 D, 10k 101 (9.8  4.24) 99\) = 420.2, and o = 4.244V, with I= = 0.561mA (at the edge of saturation). 100 10ifc For saturation: Vc = 5V, VE = 5 0.2 = 4.8V, VB = 4.8 + 0.7 = 5.5V, Ic = 1  0.1 = 0.9mA, and of saturation at
(b)
— 5 — 0.7 —
t)
= VE,
100 =
10A:

pf
=
jjy = 9. For barely linear operation, / = IE = (P + 1) IB = 101 (0.1mA) = 10.1mA.
SECTION 4.13: A GENERAL LARGESIGNAL MODEL FOR THE BJT: THE EBERSMOLL (EM) MODEL 4.71 For p/r = 150, ttE = 150C151 = 0.9934. Now, we are given Ise = 2 x 10"13A Since die BCJ is 40 times larger than the EBJ, Isc  40I$E = 40(2 x 10~'3) = 8 x 10~I2A, and aR = (l/40)oc/r =
0.9934/40
=
0.0248.
Finally,
p* =
1  aR
1
00248 = 0.0248

0.0254.
4.72 For the forward active mode, iE = 100 mA. From Eq. 4.10, and for diode connection, iE = IsEie3"ÿ* ~ 1), since vBC = 0. Thus, VBE = Vyin[iE/ISE] = 25 x 10"3ln[100 x 10ÿ/(2 x 10"13)] = 673 mV. 4.73 With this connection, operation is in the reverse active mode, with collector current flow in the forward 100 mA, BCJ direction, making iE in Fig. 4.55 of the Text negative. Now, in Eq. 4.107, IE = negligible large is and is due the zero, first of the the value term to 1 0, = vBC: Thus \>BE 10~13A). 40(2 x Thus where 100m4 = ISc = 40/SE = t)£C = W»c = 25 x 103ln( 100 x 10"3K40 x 2 x 10"13) = 581 mV.


4 74
a'pr7T=lr=
For normal saturated operation, with lB
°om
°0199' and
°"5'

1 mA, lE = 0, $forced =
220
.
SOLUTIONS: Chapter #421
VCE_ = vyin[(l + (Pforced + 1>P*H1  P/»rc«fpF)] = 25 x 103ln[(l + (0 + 1)0.0203/(1  0)] = 25 x 103ln50.25 = 97.9 mV. 4.75 For the collector open, and IB finite, (3forced = 0. From Eq. 4.114,
Pforced —
For
aF =
P/? 1+
0,
Vce sal
Pf + 1
70 = — 71
to
P/r
1+
~
280 281
•—
1
to 0.9964, with aR =
'°986 , or 0.109 to 0.111, Thus, to . 1.0986 1.0996
la«
• + P
(Pforced + iyP/{ 1 forcedÿF
Now, for p/r = 70 to 280, and, correspondingly,
P/?
= 0.9859
VCe sat = Vy In
— = 10
0.0986 to 0.0996, and
VCE sat = 251n 1 +
1
0.109
to 25 In
1 , or 58.0mV to 57.7mV. 0.111
4.76 For opencollector operation, $j„rCed = 0, and 1
= VT In
VCE sat
1+
P/t
.
For
VCE sat =
100 mV,
ps = 0.0187. Now, for the circuit shown, assume Vce p« _ _. 100 mV 50 x 10 6 „ . • ™  = 50p.A. Thus PJforceti = P1 0.05. Now, to be 100 mV. Thus —Ik t  = tc = ~r~ 1 x 103 " + Ik 51.8 1 +(1 .05/0187 , 1 + (P/+Iyp* w 98,67 mV Thus 251n = 251n = Vrl" 1 (,05/100 To555 = lMfr 1+
In
P*
= 4, 1 + ~— = 54.6, and
25
—
ÿ
t)0
=
..
„
ÿ
= 49.3mV.
2
Now for

„
—
R\  R2 = 500L1, ic = 97.33 mV, with
251n c a. sa, See Rce

„
~  lOOpA, and
P/
= 0.1, and Vce sat = 251n —j
=
q jyjoQ
= ~~ = 48.7mV.
(49.3  48.7)JmV) (]00 50) ÿ )
_
AK ÿ
4.77 For Table 4.4, p£ = 50, and pfi = 0.1. For the required table, p£ = 0.1, p£ = 50, _ qj The voltage from emitter lead to collector lead em'tter current 3
_
—

J— _ j
j
25 ln
L°2_+10°p/P/
= 58.1 mV. For
•
P/
iÿ± 25 in
Jf®.
0.001.
= 25 ln
Now, for
P/
= 0.1, VC£ sal =
= 0.05, VCE sa, = 25 ln
= 6.9 mV. For
For
p, = 0.09, VCE
102+ÿ(05) =
p, = 0.01, FC£"j(
= 0.75 mV. For
In summary:
221 
p,
,7 8 mV
M,
= 25 ln
_
For ÿ = 0 02 rel="nofollow"> yC£ Mf =
25 In
= 0, VC£
—
= 25 ln
3.4 mV. For
P/
=
ÿ~T = °5° mV'
SOLUTIONS: Chapter #422
Pforced
0.1
0.09
0.05
0.02
0.01
0.001
0.000
VEC sat (mV)
oo
58.1
17.8
6.1
3.1
0.75
0.50
4.78 Here, aF = 0.995 implies pE = ÿ 0.25. The limiting value
ÿgg
= 199> and aR ~ ~jj~ ~ °199 ~ °*2 ~ P« = 1 of forced P is, 199. For $forced *= Py < 199, 1 + (Py + 1)0.25 5 + 4py = 25 In Vce sai = 25 In 1  .005py 1  Py/199 Now for Py = 199, 180, 100, 40, 20, 2, 0: For Py = 199: VCE sat = °° For Py = 180: ÿ
= 223 mV. For py = 100: VCE M = 25 In = 168 mV. For Py 5 + 4(40) = 133 mV. For Py = 20: FCE M, = 25 In = 40: VCE sal = 25 In = 114 mV. 1  40199 For py = 2: VCE sat = 25 In = 64 mV. For py = 0: VCE sat = 25 In = 40 mV.
VCE sa, = 25 In
These results are summarized in the table:
Pforced
199
VscsatimV)
180
100
40
20
2
0
223
168
133
114
64
40
Now, for Ig = 10 mA, Ic = 1 :nA, Py = yyy = 0.1. In normal mode:
VEE
= 25 In
5 + 4(0.1)
i_AL 199
42.2mV. In inverted mode:
1 + (0.1+iyi99 1 0.1/0.25
VEC sat = 25 In

4.79
1.0055 .. = 25 In — = 12.9mV. 0.6
— —
200 Pf = 0.995, and aF = pE + 1 201 ar = = 0.667. 2+1 (1), and See iDE  0.667 iDC = 10 +1 = 11 (2). From (1), 0.995 iDE 0.995 ('d£ iDc = 10.0 (3). (2) (3) > 0.337 0.663 iDC = 10.945 iDC = 0.945, and iDc = 2.804mA. >From (1), iDE = 0.667 (2.804) + 11 = 12.87mA. Check in (2): 0.995 (12.87) 2.804 = 10.00, with aR iDC = 1.87 mA, and (XF iDE = 12.8 mA.
Here,
12.9mA on ioc
a)

b)
Now, i = Is e
x/v,T
+ \) = VF In i/Is Thus x>EB = 25 In
222

—

12.87 X 10"3 = 697mV. ÿ14 10"


SOLUTIONS: Chapter #423
Now for the collector:
Isc = 10
649mV and VEC ml = 697 «)
From
14
14
= 1.49 x 10
x
 649 = 48mV.

4' 14:
nearly the same as in b).
25
"ÿ
A. Thus, x>CB = 25 In
S
25
2 804 x
10ÿ _
iir  456mV'
4.80 For vBE = 700 mA at ic = 10 mA, with n = 1 'ÿ 1(m5 = 10 x 10_3e ~ = 6.91 x 10"lsA . Is = ice From Eq. 4.1 16: iB = From Eq. 4.117: Here, \>BC = \)BE
jre*ÿ + PR
PR
 he*""'.  0.10 = o  100, in mV, and iB =
iT =
100 pA.
From Eq. 4.116: 100 x
10'6 = 6.91 x 10_l5[(l/50)eÿ5 + (l/0.1)e(u" looy25 = 6.91 x 10,5(eÿ5)(lx50 + 10e"4)
= 6.91 X 10l5(0.185)eu'25. Thus v = 251n[(100 x 10"V(6.91 x 10"15 x 0.185)], or o = 627.1 mV. Check: 691 x 10~15[l/
SECTION 4.14: THE BASIC LOGIC INVERTER 4.81 Model the fanout as a single 0.7 V diode in scries with R = RB/n where n is the fanout. (a) For a fanout of 10, R = 45040 = 45£2:
Von = Vcc ~ (RcARc + R)(Vcc ~ VBE) = 3.0  (6404640 + 45))(3.0  0.7) = 0.85V, Vol = VCEm = 0.3 V; V,L = VBE = 0.5 V, roughly. V,„ = VBE + Rb(Vdd  VceJM) = 0.70 + 450(3  0.3/(640 x 30) = 0.70 + 0.0632 = 0.763 V NMn = VOH  V,H = 0.851  0.763 = 0.088 V, and NML = V,L  V0L = 0.5  0.3 = 0.2 V. For t>0 = 1.0 V, ic = (3.0  1.0)454  (0.7  0.7/45 = 3.125 mA, for which re ~ 253.125 = 8 £2, (b)
the gain is G =  [640 II (45040)](3031)(1/8) = For a fanout of 1, /? = 450 Q.:
 5.09 V/V
Voh = 3.0  (6404640 + 450/3.0  0.7) = 1.65 V, VOL = 0.3 V, V,„ = 0.763 V, V,L = 0.5 V, NM„ = Voh ~ V,H = 1.65  0.76 = 0.89 V, and NML Gain G ~
= VIL
 VOL = 0.5  0.3 =
0.2 V
 [640 II 450]303 1(1/8) =  32 V/V.
0.3/540 = 4.22 mA. Now, ic = IseV"/nVr, and VBE = vBEO + Vrln(»A„, = 0.70 + 251n(4.224) = 736 mV. Thus Vqh = Vcc ~ (VDD  VDE)RC/(RC + RB ) = 3.0  (3.0  .736/6404640 + 450) = 1.67 V.
4.82 For saturation, ic = (VCc
~
223 
and
SOLUTIONS: Chapter #424
Now, for Vol,
= (3031)5 = 0.1935, and p* = aR/(l  a*) = 0.19353  0.1935 = 0.24. Collector current with output low (say 0.3 V) is iE (VDD  VceÿYRc (30  0.3)540 = 4.22 mA. Now, the base drive iB depends on the fanout of the previous gate being largest for a fanout of 1: iB = (Vcc ~ VbeKRc + Rb), or iB = (3  0.7X640 + 450) = 2.11 mA. For these conditions, using Eq. 4.114, where (5
—
—
= yrln[(l + (p/orced + l)PfiXl  h«rceA>F)] = 25In[(l + (2.0 + 1)0.24X1  250)] = 66.8 mV Now, for Vil, operation is where the transistor barely turns on, and the gain is  1 V/V. For such an arrangement, G ~
 $RC/(RB + rn).


1 or RB + rK = pKc or rK = p/?c RB = 30(640) 450 = 18750. Now, r„ = (p + l)re = (p + l)VTAE, or iE = (p + l)VT/rK = 31(25 x 10"3)18750 = 41pA, for which iB 1.4 iA, and x>/ VBE at 41 (lA. Now, iE = Isc and \)BEl vBEl + VVlnigÿEi
Thus
PRcARb + rn) =


—
Thus at 41 pA, x>BE = 700 + 251n[41/1000] = 620 mV. Thus VIL = \>BE + iBRB = 0.620 + 41 x 10"6 X 450 = 0.638 V.
Now, for Vih , operation is where the transistor is turned on, but past the edge of saturation and the gain is 1 VfV. This gain results as the base collector junction conducts, shunting RE • Though it is quite possible to calculate the detail using the EbersMoll model, the process is quite tedious. A good worstcase approximation occurs when the incremented resistances of baseemitter and basecollector diodes are about equal, when the base current splits equally between them, while that in the baseemitter is enough to sustain i'c in saturation. For this situation, iE ~ (3.0  0.3)540 = 4.2 mA and for saturation iB = 4.250 = 0.141 mA.

Now at ic = 4.2 mA, x>BE = 700 + 25/«4.2/1 = 736 mV and V,H = 0.736 + 450(2)(0.41 X 10"3) = 1.105 V. A more usual estimate might ignore the basecollector current split (and a factor of 2) and sim¬ ply use V,H ~ 0.736 + 450(0.41 x 10"3) = 0.921 V. The larger value gives a more conservative view of noise margins: Now NMh = V0H  V,H = 1.67  1.105 = 0.57 V, and NML = V,L  V0L = 0.638  0.067 = 0.57 V. Now, the largesignal voltage gain is


G = (Vow " VolWil ~ Vw ) =  (167 0.067X1.105  0.638) or G = 3.43 V/V. Note that this is low because of the choice of a conservative value for V/W. For the other value G = (1.67 0.067)0.921 0.638) =  5.63 V/V. Now at v»o = 0.7 V, and assuming no effect of fanout load, iE{3.0  0.7)540 = 3.59 mA for which re = 255.59 = 6.96 £2 and rK = 31(6.96) = 215.9. Thus, the smallsignal gain = PRcARb +rn)= 30(640X450 + 216) =  28.8 V/V.




SECTION 4.15: COMPLETE STATIC CHARACTERISTICS, INTERNAL CAPACITANCES, AND SECONDORDER EFFECTS 4.83 For a groundedbase amplifier, the output resistance r„B is approximately
ÿ5 = 200 x 106£2, with r„=ÿ = x 10'3 0.1 50 x 109 Ic 1 1 1 r That is, for ru in r=— +120 x 2 x 106 200 x 106 •
r
/>
rM =
(240) = 1200MQ = 1.2GQ
 224 
•
4\)c : = A ic
—
=— + — p r„ or r„ r„i, 240 200 1 , or = . —1 = ——1   ——
= 2M£2. Mil,
.
II pr0 Here, r„b = Thus 200
240
ÿ
200 (240)
SOLUTIONS: Chapter #425
4.84 VQa = BVcbs ~ BVcbo = 50V; 4.85
Rce sal 
A V,CE A Ic
Voh = BVCE0 = 30V; V0c = BVebo = 7V. 0.1V
0.20.1 31
Gcncrally' ÿCE
~
2mA
= 2mA, VCE sal = M±H= Vc£ off + Q 05(2)) and
off
"
ÿCE "ff
~
=
+ ÿCE 5"'
"
Thus at
± .05(2) = .15  .10 = 0.05V.
Otherwise, we could use values at one of the 1mA or 3mA points, to obtain the same result since the same line is involved in all 3 cases.
= 109,' and hfe = 90. If one fe = (1211)mA .001 Mb IB 11mA assigns the increase in dc P to the effect of collector voltage on the base width, then at the particular
4.86 At 1.20mA,
value of vCE = 10V, 10
V/i =
1.20
  
(L20).
(1.20
— .99)
A/c
1,20mA
hFE —
12. , r„
lc = hfe lB +
(1.29
1.20)mA
.09

with r„ = —— . Thus 1.20 = 90 (11 x 10"3) +
lc
(1), and also 1.29 = 90 (12 X
1.29
10"3) + 10
1£(L29)
= 57.1V, and from (2), VA = no _ on cni v m3
_
1.29  90 (12) X 10"
(2). From (1),
Thys
_
n
4.87 Assuming saturation, with lB = / = 1mA, and VCE sat = 0.05 + /c(0.05), we see v„ = 5  VCE mt, and 5  (0.05 + Ic (0.05))   1.0. That is, 0.82 lc  5 .82  0.05 .05 /c, Ic = 4.75mA, and VCE sa, Ic = nori 0.82 = 5  .287V = 4.71V. = .05 + 4.75 (0.05) = 0.287V, with 5  (.05 + Ic (05)) 1.67 4mA, 0.82 Ic = 5 .05 .05 Ic 3.28, Ic = Now for I= 4(1) = 4mA, Ic 0.82 .87 = 1.92mA, and VCE sa, = .05 + 1.92(.05) = 0.146V, with v0 = 5 .146 = 4.85V.

—
p is 200 at
4.88 From the graph in Fig. 4.68: At IpA:
Thus an average TC is (200  70K 125 or 1.38/120 x 100 or 1.15 %/°C.

125 °C, 120 at 25 °C, and 70 at
 55 °C.
55)= 1.38/'C,
At I mA: p is 105 at 125 °C, 190 at 25 °C, 320 at  55 °C. (320  105H180) = 1.19/°C or 1.1SF190 x 100 = 0.63 %/°C.
Thus the average TC is
4.89 From Eq. 4.120, the stored base charge is Qn = Wlic/(2Dn ) = xFic For W = 1pm, = (10"6)2 X 1 x 10_3/(21.3 x (10~2)2) = 4.69 X 10"13C. Tlie forward base transit time XF = W2/(2Dn) = Q„/ic = 4.69 x 10_l3/(l x 10"3) = 4.69 x 1010j = 0.47 ns.
The
smallsignal
capacitance
diffusion
emitter
Cde = xFgm = tFIC/VT = 0.47 x lO"9 x
1x
103
is
(from
Eq.
= 1.88 pF.
x io_3 For IV = 5pm: Qn = 25(4.69 x 10,3C) = 1.17 x 10nC, xF = 25(0.47) = Cde =25(1.88pF)= 47 pF. It is apparent that thick (wide) base regions can lead to slower operation! 4.90 From Eq. 3.26, Cj„ = A ( where (from Eq. 3.18,
es9 2
ÿ
NaNd
(i
)X(Na +Nd )X V0)
V0 = Vrln
4.121)
NaNd ".2
 225 
11.8
ns,
SOLUTIONS: Chapter #426
Here tij = 1.5 x 10l0/cmiNA = 10l7/cm3, ND = 10l9/cm3, q = 1.60 X 1019C, ts A = lOjxm x 10pm = 100 x 10~8cm2. Thus V0 = 25 X 10_3ln[(1017 x 10,9K1.5 x 1010)2] = 0.909 V.
— 104
X
10~12F/cm, and
Thus Cje0 = Cj0 = 100 x 10_8t(l.04 x 10"12 X 1.6 x 10"l9/2) X 1017 x 10l9/(1017 + lO'ÿ.OOf1 = 10~6(1.093 x lO14)" = 0.104 pF, and Cje = 2CJe0 = 0.208 pF. Now for
( = Cjc ), use Eq. 3.26 with ND =
10ls/cm3 in the collector.
V0 = Vrln(10ls x 10l7/C1.5 X 1010)2) = 0.67 V, and Cjo = 10 x 100 X 10"8[1.04 x 10"12 X (1.6 X 10",9/2) x 10'7 x 10l5/(1017 + 10l5>0.67]/' = 0.157 pF and Cj = CjA 1 + VR/V0)'" = 0.157/(1 + 2.00.67)04 = 0.090 pF. Thus = 0.090 pF =ÿ0 fF.
Thus
Now, from P4.89 above for W = li»i, = 1.88 pF, and from directly above, CK = 1.88 + 0.208 = 2.09 pF. For operation at 1 mA, gm = lcWT = 1 x 10"3/25 x 10°3 = 40 mfiJV.
CJe =
0.208 pF. Thus
Thus, fT = (]/2n)g„/(CK + CV) = 40 X 10"X(2.09 + 0.090) X 10~l2K27i) = 2.92 GHz.
4.91 Here, fT is 10 GHz at 10 mA and 7 GHz at 1 mA, where gm is 400 mA/V, and 40 mA/V respectively. For each case, Cre + = gmA2itfx)
Thus CKio + Cn = 400 x lO'ÿn x 10 x 109) = 6.37 pF and CKl + 0.909 pF.

+ Cje> and Cje °
Now, Cn = Cje
Subtract (1), (2), 9C = 6.37 X 0.91, and C = 0.607 pF(/mA). + CM = 0.909  0.607 = 0.302 pF. Thus, pF at 10 mA, and 6.1 fF at 10 pA. 6.07 = From (2), Cje
At 10 pA, fx =
= 40 x

\Qr\2n x 7 x 109) =
(2)
40CKI000 _2 = 206 MHz 2n(0.302 + 0.0061) x 10"
—
4.92 We note that the area of all junctions is increased by 4, and that the current density is unchanged. Now, using the results of Ex. 4.440 on page 315 of the Text: At 4 mA:
From Eq. 4.120, xF is unchanged, at 20 ps. From Eq. 4.121, Cje increases by 4, to 4(0.8) = 3.2 pF From Eq. 3.26, Cjeo increases by 4, to 4(20) = 80 fF From Eq. 3.26, Cÿo increases by 4, to 4(20) = 80 fF. From Eq. 3.18, V0c is unchanged at 0.9 V.
See Eq. 4.124, mcaj is unchanged at 0.33
Cje increases by 4 to 4(40) = 160 fF C,iincreases by 4 (since both area and charge are 4 x larger) to 4(0.84) = 3.36 pF increases by 4, to 4(12) = 48 fF From Eq. 4.130, fx at 4 mA is unchanged, at 7.5 GHz. For operation at 1 mA, gm = 40 mA/V, Cje = 160 fF, 40 X 10"3 ,„ YY TU Thus, rr = 6.3 GHz. fx 2ji(800 + 160 + 48) x 10~15
—
226
= 48 fF, Qt = 3.2/4 = 0.80 pF.
SOLUTIONS: Chapter #427
4.93 For this transistor, = 2 pF and CK = 10.7 pF at 1 mA. If we assume that Cje 0ÿ=2 pF, Then CJe = 10.7  2 = 8.7 pF at 1 mA, and 8.7/100 = 87 fF at 1 mA/100 = 10 pA, and 8.7 pF at 1 pA.
.
.n
.
,
At 10 pA, fx ~
40 x 10~Vl00 pp = 15.6 MHz.
271(2.0 + 2.0 + .087) x 10~12 40 ™ x lO'VlOOO At 1 pA, fT = = 159 MHz 2tc(2.0 + 2.0 + 0.009) x. 77112 10" {Note that if Cje is assumed to be 0, rather than equal to Cÿ, fT values are 30.2 MHz and 3.17 MHz}. n_mn
ÿ
The proposed change would reduce the capacitance Cÿ (and Cje) by a factor of 100 since the junction areas are reduced by that factor. Continue to use Cjc = 8.7 pF at 1 mA for lack of a better choice. Now at
ic = 10 pA,
x 10~3/100 pp = 501 MHz. 8.7/100)
40 271(2100 + 2100 + x 10~12 Now at 1 ic = lpA , 40 x 10~3/l000 pp = 131 MHz Jfx = 271(2100 + 2400 + 8.7/1000) x 10"12
fT
~
 227
SOLUTIONS: Chapter #428
NOTES
 228 
Chapter 5 FIELDEFFECT TRANSISTORS (FETs) SECTION 5.1: STRUCTURE AND PHYSICAL OPERATION OF THE ENHANCEMENTTYPE MOSFET 5.1 In general, a channel is induced for x>Gs ÿ t>s + V, = 0 + 1.5V. Hence Uqs 2: 1.5V, here. In general, the drain end of the channel is pinched off for uCD £ V, = 1.5V. Now for vGS = 3.0V and Vs = 0, X)G = 3.0V, and the drain is pinched off for uD > 3.0 1.5 = 1.5V. Hence uDS £ 1.5V, here. In general, saturation occurs for a given x>Gs, when vDS £ vGS  V,, or \)DS > 3.0V 1.5V = 1.5V, for which x>D 2: 1.5V, here (which is, of course, when the drain end of the channel is pinched off). In general, triode operation occurs for uDS < x>Gs  V, , for which Up ÿ 1.5V, here.
SECTION 5.2: CURRENTVOLTAGE CHRACTERISTICS OF THE ENHANCEMENT MOSFET (a)
\)ds
= x>d
 
— \>s =2.1 — 0 = 2.1V; x>G$ — x>G — x>s — 3 — 0 — 3V; x>Gs — V, = 3 — 1 — 2.0V ÿ \>ps
» saturated mode. (b) (c)
(d) (e) (f)
(g)
5.3
—
— — —
—
—
2 2 = 4V; Dcs V, = 4 2 = 2.0 ÿ 2 = 1.9V; x>Gs = x>G x>s Vds = v»o —\)s —0.1 vDS » triode mode. 1 = IV <  V, = 2V » cutoff mode. x>sd = Vs  \>d = 0  3 = 3V; vSg = Vs ~ % = 0 1 = 3V;  i>g = 2  0 = 2V; + V, = 21 = IV < msd » \>sd = Vs  \)p = 2 saturated mode. 3 = 3V; Since saturated, X)ps ÿ Dcs V, = 3 2= IV; V, = 2V > n channel; vGs =0 vD = VpS + vs > I 3 = 2V. 1 = 4V; \)JG = Vs  x>G = 3 0 = 3V; vSG + V, = V, = 2V » p channel; vSp = t)j  vD = 3 3 2 = IV < Ojo » saturated mode. V, = 2V > p channel; t)s = 3V, t)0 = 3V;  vG < V, = + 2V for cutoff; ÿ IV. IV, and 3 2 £ 2 = uG % = t»G
—
—
—
Id (mA)
—
Vds s l*OJ ~
Triode region
—
Vi ÿ
*
Ic
/
vcs ~ Vr Saturation region
Since V, = IV, the uGs are relabelled as shown. See iDa = 2.25 mA, \)GSb = 4 V, Ddsc > 3 V, ij),) = 1 mA, iDe = 5 mA, iDf = 0.75 mA, \>QSg = 5 V, t>DSH = 2 V.
—
SOLUTIONS: Chapter #52
ijj (mA)
vDs
(0 (a) (c) I
8 2
05 H 2
i
as
—
O O o >
Mi¬
CaHW(e)
5.4 ÿ
vcs
Triode
 V,
VDS s l'CS
*~
Kr
Saturation region
>
l'cs ™ V'/
4

vos
j
+ 4 5 q.5 3 2..5
V, + 3 ,Lj
 veJ = v, + 2
i
". l'CJ
o
1
2
3.4
6
5
*
(SXt)
7
•
3
8
\9
tel
3.5 2.5 2.
3 2.5
'2 \.s
V. + 1 2. 1.5 l.s I £ I 05 I " 0S il»(V)
4 Vta*r,(cutoff)
The axes labels are indicated by (a), (b), etc, at the top right and left, and at the right bottom.
5.5
Vs = 0V, VG = 3V. Now, for saturation VDS S VGS  V,. and for V5 = 0, VD £ 3  1 = 2V. Thus the device is in triode operation for VD < 2V. 2d]). In general, in triode mode, = &(VV/L)[(uG,s  V,) vDS For vDS = 2V, iD = 20 X 1(T6 X 10 [(3  1) 2  22/l] = 200 ( (2) (2) 22/2) = 200 ( 4  2) = 400)lA. Check, in saturation, iD = 1/2(20)(10) (vGg  V,)2 = 100 (3  l)2 = 400(xA.

For vDS = IV, iD = 200 [(2) 1 \2/l] = 300xA. For x>ds = 0.5V, iD = 200 ((2) 0.5  0.52/2) = 175pA.
For x>DS very small, iD = k'(W/L) (vGS  V,) VDS, whence 1 vDs 1 2.5kQ. rDS ~ (3 k'(W/L) (vGS  V,) 200  1) = iD 2 (3  1) 0.04V. 100 1 1 or 10% when increases by Now rDS = 1.10 (Dg5 V, )x>ds ((Vas V, ) \>os t)oy2) (vGS  V.)  vDS/2 = 0.909 ((vGS ~ V)), for which, vDS =2 (1 0.909) (vGS  V,) = 2 (0.091) (3  1)
Now, rDS increases by 1% when D5y2 = ((vcs
for which vDS  V,) %) 777, 100
—

=
—
—
= 0.36V. 5.6
For triode operation at low vDs> 1d = k (W/L ) ((vGS  V,) vDS), whence 10" 1 Vds , or rDS rDS 
k'(W/L) ((vGS  V,)) iD 5k CI . whence vGs = + 1.
20(202) (t)GS
 1)
5000Q t)Gs
 1'
rDS
For rDS = lktt: vGS = y
I
+ 1 = 6V, and for rDS = 1MI2: uGS =
 230
lu /Ca6
+ 1 = 1.005V.
SOLUTIONS: Chapter #53
For x>DS near OV, and less than a 10% increase in
Uds
ÿ
2 (\)OT
rDS,
(x>cs
 1) (0.0909).
 1)  \)DS/2 > yy( vGS ~ 1).
or
For rDS = lkft , x>Gs = 6V, and \>ds ÿ 2 (6  1) (.0909) = 0.909V, at which iD = k'(W/L) [(oGJ  V,)uOJ  vÿs/1] = 200 ((6  1) 0.909 (0.0909)2/2) = 200 (4.545)  .413) = 826iA. For rDS = 1MQ, VGS = 1.005V, and \)DS < 2 (1.005  1.00) (.0909) = 0.91 raV, at which iD = 200 t( 1.005  1) (0.91 x 10~3)  (.91 x 10"3)2/2] = 200 [ (5 x 0.91 x 10"6)  0.912 x 10"6/2] = 100 (9.1 x
106  .83 x 10~6) = 827 x 10~6iA = 0.83nA!
or, easier,
iD =
~ 0.91 nA, as a quick approx¬
imation. 5.7
iDj = V2k'{W/L) (vGS  V,)2, and for triodc operation, For ipÿ = 1.00iDxs :
0,
2 (Mas  V,) ± V 22 (mGs ~ V,)2  4 (uGS V,)2 , whence \)DS = 

or Vim = (Vgs
— V,)[l ± ÿ1  1] = (uspilonGS  V,)
For iD, = 0.99 iDs : 2 (Ugs V,) Vps
—
— Vps — 0,99 (y>Gs — V,)2, or 1 ± V 1  (,99)2 J vDS = (vGS  V,)
Vps
whence applies). Thus dds = 0.859 (uGs  V,).
For iD, = 0.90 iDs : vDS = (yGS For io, = 0.50
iGs 'ÿ
Dds
— (Vm
 V, ) ~
V, )
— 2 (vDS) (vGs — V,)
(vGS
1 ± V 1  ,92 1 ± Vj
. .
 V,) (1 ±
+ 0.99 (oG.s
— V,)2
= 0,
0.141), (where the negative value
Thus uDS = 0.564 (uGS  V,). Thus t>os
— 0.134 (oGs — Vt).
Now, for oGS = 2 V, and V, = 2V, X>DS = 2V, 1.72V, 1.13V, and 0.264V, for 100%, 99%, 90%, and 50% of saturation value, respectively.
5.8
Here, r„ =
—
= 50k"; Va = r°
= 50kfl
2.1 + 2.2
= 107.5V; and X, =
)'A
0.0093V'. 5.9
Cox p Assuming p.,, = j p„, Kp = y x y x 20 x p  166.7pAIV2. = 5V, iD = K (oG5  V,)2 (l+X vDS) = 166.7 (5  2)2 (1  ,01(5)) =
Now Kp = yPp
For vGS = vDS (1.05) = 1.575mA.
166.7 (32)
5.10 For the substrate Vn of the PMOS connected to +5V. while the source voltage is varied: IV,I = IV,ol + Y 2
N



5.11 In each of the circuits, consider the saturation transconductance factor is K = V2\an Cox(W/L) = 1/2 X 20 X 10"6(202) = 100WV2 = OAmAA/2. Accordingly: (a) vGS = % = 5V; saturated operation, for which iD = K (\)GS  V,)2 = 0.1 (5  2)2 = 0.9mA. Thus Ia = 0.9mA.
 231 
SOLUTIONS: Chapter #54
(b) (c) (d)
v>,sG = x>sd ', saturated operation, for which 0.4 = 0.1 (x>SG Vsc = 0. which is not possible, or 4V. Thus Vb = +4V. V'g.s = 0; cutoff, for which Ic = OmA.

2)2, or x>SG + 2 = ± V~4 = ± 2. Thus
0G,S = \), X>DG = 3V; saturated operation, for which iD = 0.9 = 0.1 (v  2)2, or v = 5V. Thus Vj = vG vGS = 25 = 3V.
 2 = "ÿ9 = 3, or v
Section 5.3: THE DEPLETION  TYPE MOSFET all 5.12 For conditions, K = V2\inCox(W/L) = 1/2 x 20 x depletion mode). (a) (b)
(c)
(d)
(e)
the
transconductance
saturatin
106 X 2002 = 1000 x 10~6 = 1/nA/V2; V, = —4V
is
= 4 0 = 4Y. cutoff. iD = OmA, and vDS = 5  0 = 5V. "0(7 = t)C5 + o.s = 2 + 0 = 2V, and Oy = o? + % = 0 + 3 = 3V. This implies saturation for 4)2 = 4mA. which iD = (x>cs  V, )2 = 1 ( 2 x>as vGvs = 00  0V, and vDS = vD  vs = 5  0 = 5V, for which vDG = 5  0 = 5V > "1 V, I". This implies saturation operation, for which iD = 1 (0 4)2 = 16mA. t)Ciy
—
—

= vs + vDS = 0 + 2 = 2V, for which VDG = 2 vG = x>gs + x>s = 0 + Q = OV, and This implies triode operation, for which iD = K [2 (v>GS  V,) %  v%s]  1 (2 (0 = 12mA.
0< 4) 2
IV, I .
 22)

Vs = vG  vGS = 0  1 = IV, and vD = vs + vDS = IV + 5V = 4V, for which vDG = vD vG = 4V 0 = IV, I . This implies operation at the edge of saturation, for which i'd = 1 (1 4)2 =
—
—
25mA. (0
factor
(for an nchanncl
VGS =vGX)s = 2 0 = 2V, and vDS =vDvs = 5 0 = 5V, for which vDG = vD  vG = 5 2 = 3V < IV, I This implies triode operation, for which, iD = K [2 (vGS  V,) Uns =1 [2 (2  4) 5 52] = 60  25 = 35mA.

.

 vs = 0 0 = 0V.
See that operation is in the
(i)
= vgs + Vs = 2 + 0 = 2V, and vDS = vD triode mode, but with ip = OmA.
(h)
2 = 2V. See that vDG =vD v, = vSG + vG =  2 + 0 = 2V, and \)DJ = vD  vs = 0 0 = 0<V,  > triode operation, for which iD = K r2 (vGS  V,) %  x>bs = 0
—
=1
—
[2 (2   4) 2 22 ] = 24  4 = 20mA.
5.13 Depletion MOS: For simplicity, use IV,I = 2V. (a) (b) (c)
(d)
Saturation with Dcs = 0, for which la = in = K (dgÿ  V, )2 = 0. 1 (0   2)2 = 0.4mA. Saturation, for which 0.4 = 0.1 (X3GS  2)2, and vGS + 2 = ±ÿ~4 = ±2V, whence vGS = 4V or 0V. Clearly, 4V is not possible. Thus Vh = 0  0 = 0V. Saturation, for which 0.9 = 0.1 (vGS  2)2, and x>GS + 2 = ± ÿ~9 = ±3, for which x>GS = 5 or +1V > Clearly 5V is not possible. Thus \)Gs = +1V, and Vc = 0 + IV = +1V. 18 1 I= Devices are connected symmetrically: Ij = = 0.9mA. Also see saturated operation,


—
—
—2)2.
(e)
K=V2\lnCox (W/L ) = 1/2(20 X 10 6)(10) = 100 X 10 6 = 0.1/nA/V2,
for which 0.9 = 0.1 (Ucs Iÿere, Vps  VGs Thus 2) vGS  vGs 2 Cues
I
ÿ
—
—
Thus mgs = +1V (as in (c)), and Vj = +1V. = 0 < IV, I implies triode operation, for which iD = 0.4mA = 0.1 or 4 = 2 viGs + 4 Ugj vGs, vGs + 4 vGs —4 = 0, whence
—
4 + 4 V~2  4 ± V 42 _ 4f4)i  L±Z —± = 2 ± 2 Vr2. VgS ~ —L±—Z.—2LJLLL 2 2 2<~2 = 0.828V, and Ve = 5  .828 = 4.172V.
 232 
But it must be positive. Thus = 2 +
SOLUTIONS: Chapter #55
convenience, let K = 1/21'(W/L). Operation is triode mode in both cases: Thus iD = K 2 (uG5  V,) v>SD  Vsd , V, being negative for a pchannel depletion device, when X>SG is used.
5.14 For
For vD = 4.8V, v)yD = 5.0
 4.8 = +0.2V, \>G = 5.0V, vGS = 5.0  5.0 = 0.0V;
 V,) (+0.2) 0.22 ], or 0.1 = K [0.4V,  0.04 ]    (1). For vD = 4.95, vSd = 5.0  4.95 = +.05V, vG = 0V, x>SG = 5  0 = +5V; 0.1 = K
\l
(0
[2 (+5  V,) (+.05) .05 ], or 0.1 = K [+0.5 + 0.1V,  .0025 ], or 0. 1 = K [o.4975 + 0.1V, ]   (2). 2
0.1 = K
Now (l)/(2) + 1 =
0.4V,
 0.04
+0 4975 ,. Q1y
ÿ
or "°4
Vt
~
04 = + 0.4975 + 0.1V,, or 0.5 V, = +0.5375,
whence V, = 1.075V. Thus with \)Gy as defined, the depletion threshold is 1,075V. From (1). 0.1 = AT [ 0.4V, .04], K
K (Mas ~ V,)2 =
bss
(1 
0.296mA.
=
ÿÿÿ04 = ÿ4ÿ =
p)2 = "P" (Vgs  V,)2.
Thus
IDSS
=K
"ow. 4, =
V,2 =
0.256 x (1.075)2 =
SECTION 5.4: MOSFET CIRCUITS AT DC
Vd° ~ V° = = 0.4mA. In 2 7.5x£2 Rd saturation, iD = K (vGs  V,)2, or 0.4 = 0.4 (vGS  l)2. Thus vGs 1 = 1. or uGS = 2V. _2 ~ Vss 5 3V 117 andA Rs = = ——— = — = 7.5kl2. 01 2 = 2V, x>s = vG  vGS = n U.4 Ij) U.4
5.15 K = Kn = 1\in
Cox
~ L
=
x 40 = 0.4mA/V2. (20) 2 D
Now Vy
lD =
inn
V°  1z2l  o.4mA, and Vy 5 7.5(0.4) 2V. Thus = = + VGS = 2V, whence 0.4 ) 7.5 (2  l)2, and K = 0.4mA/V2.
5.16 Now, ID =
=K

Thus
for
Vpp ~
Rd
K=
p
= 0.2mA/V, in the source circuit:
ÿ — = i0 = 0.2 (Vy  l)2,
+ 5 = 1.5 (Vy + l)2 = 1.5 Vs2 + 3Vy + 1, or 1.5V/ + 2Vy 4 = 0, whence 2 ± 5.29 —2 ± V 22  4(4) (1.5)1 2 ± V 4 + 24 „ „ „ 2.43 or +1.10 (too small). Thus Vy =
 = 2 43 — 2'00 2.43V, and VD = +2.43V, with a corresponding change of — — — = 21.5%. Vy
2
5
—
=
=
11X
or
=

5.17 For Fig. 5.25, VDD = 10V, VD5 = VGy, and operation is in saturation. Thus VD = VD0 ID R, or VG = 10  OAR. Also ID =K (VGy  V,)2, where K = V2\in Cox x W/L = Zi x 20 x 40 = 0.4mA/V2, and VGy = VD. Thus 0.4 = 0.4 (VD
 l)2, or VD
1 = 1, or
VD = 2V. Thus R =
10
—9
Q4
= 20kl2.
= 20)tA. VD = 2V, ID = Also ID = V2k\W/L) (vGS  V,)2 = 0.5 x 10"3 (\)Gy  l)2 = 20 x 10~6, or vGS  1 = (40 x 10"3)'/' = 0.2, whence uGy = 1.2V. Now, since VG = 5V, Vy = 5 1.2 = 3.8V, see Rs = *?'** = 190kl2. Thus, to 20pA
5.18 For
one significant digit, /?y = 200kl2.
 233 
SOLUTIONS: Chapter #56
5
5
or  l)2 and ID = QÿC,y = 500 (uGS  l)2, or 5  vGS = 100 x>GS  200 \)GS + 100, or 100 vj;  199 ÿ + 95 = 0,' or v&s  1.99 VGS + 0.95 = 0, whence 1.99±V 1.992 1.99 ± V .1601 1.99 ±.4 ,  4(.95) , = =  = U95V' or < l Thus' Vs = 5 U95V — vGs = 2 2
Now, ID = 500 (VGS
ÿ
Q
= 3.805V, ID = 5,19
Vc = 0.5 (\)
OQC
loJTlOM l)2,

X
= 19.025pA, and VD
2 = 5  .15 x 19.025 = 2.15V.
5V = 2,5V N°W f0r Ucs =
0.5 (t)  l)2 =
(u 
90
5"
U~2'5
= '• and
= K {Vgs
~
V,) =
l)2 = 5  2 \), D2  2t) + 1 = 5  2\), t>2 = 4,n> = 2V. Thus
Vs = VG + vGS = 2.5 + 2.0 = 4.5V, and iD = We actually find
iD =
5 ~ 45 iK
(2V) = 1.8V, in which case
= 0.5mA, VD = 0 + 4k (0.5mA) = 2V.
iD is reduced to 90% or 0.45mA.
—
For K = 1/2k(W/L) = 0.5mA/V2 and V, varying, 5 lkfl (0.45mA) uG.y = 2.5V, whence vgs = 5 0.45 2.5 = 2.05V. Now, 0.45 = 0.5 (2.05 V,)2, or V,  2.05 = ± (0.90)* = +.949, whence V, = 2.05 ± .949 = 1.101V. That is, V, could have raised by 10.1%. ,, 0.45 2 For V, = IV and K varying, again vGS = 2.05V, but now, 0.45mA = AT (2.05 l)2, or K = ÿ


0.408mA/V2. Thus K could have dropped by
= 18.4%.
ÿ
Note that the effect of V, is essentially direct, a 10% change in current resulting from a 10% change in V,. However, the change in current is only about 10/18.4 or about 54% of that in K, due to negative feedback included in the circuit. (See Chapter 8.)
5.20 For the Depletion Device, K = l/2\lpC„x(W/L) = 1/2(8 X 106)(50(V2) = 0.1m/l/V2 and x>Gs = 0. Thus iD = Ipss = K{\)GS  V,)2 = lmA/F2(0  2)2 = 4mA. Thus V5 = 15  lkfl (4mA) = 11V = VG, and VD = 0 + 2k12 (4mA) = 8V, whence x>SD = 118 = 3V>V,. Thus the device operates in saturated mode. Triode operation begins for uJD = V, = 2V, in which case VSs = 8 + 2 + 4 = 14V, with operation being saturated for Vss > 14V.
VG = jjj x 5 = 4V. Now vSG = 5  lkfl Assuming saturation, iD = K {\>GS  V,)2 = 1 (uGS  2)2,
5.21 Here K = V2\ipCnx{W/L) = l/2(8)(250) = ImA/V2. See

— —
(iD) 4 = 1 iD, or iD = 1 uSG = 1 + t)GS. or 1 + t»Gis = (vGs 2)2 = v2f 4t)<7c + 4, or vG$ 5x>Gs 4131 5 ± ÿ 52 5 ± 3 61 J , or x>GS 0, whence uGJ = =
—
—
2
 vGS = 4  0.697 = 3.30V, and VD = 0 +
+3=
= 0.697V
(or 4.305V (too large)).
Now
5~330
x 1 = 1.70V. Thus VDS = 1.70  3.30 = 1.60V, and VGD = 4 1.70 = 2.30V > V,. Thus, operation is in saturation.
Vs = VG
5.22 For operation at ip = 150xA in saturation, ip = ( 1/2)(X„ Cox (W/L)(x>Gs  V,)2 or 150 X 10"6 = 0.5(20 x 10"6)(3ÿ2)(X)GS  l)2, or t)GJ  1 = 1*, whence Now Rt =
ÿ52 ÿ = 150 x 10"
20kQ.
lfR2 = R, = 20k Q, then vD2 = + 2 V
\>GS
2
V.
(also).
If M3 is joined to M2 with corresponding elements connected, then the current in R2 will tend to double. If R2 is reduced to 10 k£l, v>DS2 = Vpsi = 2 V.
 234
SOLUTIONS: Chapter #58
VA =
90 91
— 24 04
9616_ 9 091
—
i
= 127V. From (1): 9.091 = 2
= 0.240mA/V2. Check: At 0.25mA, gain = 2 <244
10 II
127 1/4
= 2 <06 x
9.091 X
1 (10 II 127)
5®l = 4.804 V/V
10 + 508
+ OK. For an output distortion of 10%, as stated, from Eq 5.35, K (VGS V,)
10 u„
= 5o,ÿgs
VD = (< + 1)V, gm = 2K II Rg II r0
fa
(vGS
lD = ZmA, =
(50)/?,
50 + /?,
 V,)2
Rg
<* (/?c II /?c II r„) = 2 X (104 II 104 1 50) = 100V/V, /?, ~
Rl = r„ : \)>, = 2 (104 II 50 1 50) = 50V/V, /?, = 104 , or 2 (104 II 50 II Ri) ~ 2 (50 II /?,), /?, = 1 + 2 (50 II Ri) 1+ 2
I= K (x>GS
104 = , r kO. 1  gam 1 + 2RV I
For
/?,
2K (Vas
 V,) x>„,
or
1 (VD
=
 l)2.
 Vt) = 2 (V0  1) = 2 <1 mA/V, r„ = Vÿ/7 = 504 kO, whence — =
= gm Rl = 2R<~1 V/V, and /?,
Rl=Rg: d>( = 2
10
 5 (0.5) = 2.5V.
K = V2k'{W/L) = 2/2= 1niA/V1.
5.28 Generally.
u2t
= 104, Rj +
100 Rj
Ri +50
104
For I= 1mA and
,
104
"
1 + 100
= 99kO.
= 196kO. For RL = /?,: Vb, =
104, Rÿ + 50 Rt + 100 Ri2 = 104 Ri + 50 x 104,
101/?,2  9950 Ri  50 x 104 = 0, /?,2  98.5 /?,  4950 = 0, whence /?, = 98.5 ± 171.8 = 135kO, and o/o, = 2 (50 II 135) = 73.0V/V.
985 ± V
9852
4 ( 495°)
_
5.29 Note that while the lower end of r„ is not actually grounded, the signal there is small. Assume it to be Rc H Rl II r„ 10 II 10 II 100 . „ „ „ lk£2, gain  = zero. For Rs = 11rt  = 2.38V/V. For Rs = = 4.76 2 1/gm+Rs 1A + I Vi v„ 4 76 4 76 Oft, gain y = = 4.76V/V. For Rs = 3.76kG, gain y = = 1V/V.

—— —
—
——
5.30 For the Tmodel, the equivalent resistor in the source is r, = Vgm = \A).725mA/V = 1.38 kfli. The out¬ put resistance of the follower (with body effect ignored) is Rou, = r„ II rs = 47 II 1.38 = 1.34kf2. 47 Thus, the noload gain is Go = V/V. = j 33 + 47 For load RL, the gain is G = (Rl/[Rl + Roul)) x G0 for which GRL + GRout = G0RL, and Rl = GR„„AG0 G). For G = 0.95 V/V, RL = 0.99(1.34X0.971 = 0.95) = 63kG.

For G = 0.90 V/V, RL = 0.90(1.34X0.971  0.90) = 17.0kO.
5.31 For the situation in which Is is fixed, the gm of the transistor is independent of V, : This follows from the fact that: iD = ]/2k'(W/L)(vcs  V,)2, gm = Oi/yOucs, or gm = V2(2)k'(W/L)(vGS = V,) = k\W/L){2iDA\W/L)f. Thus gm = (2k'(W/L)iD)y\ depending only on 'o
 is
Now, for
Vgmb =
x=
0.2, gni/, = 0.2 (0.725) = 0.145 mA/V. Thus the additional load on the source is 6.90 kQ.
IX). 145 =
 235 
SOLUTIONS: Chapter #57
SECTION 5.5: THE MOSFET AS AN AMPLIFIER 5.23 Here iD = V2k'(W/L)(vGS  V,)2 = 0.5(2 x 10"3)(5  I)2 = 9 mA, and md = VDD  iDRD = 12  0.5 x 103(9 x 10~3) = 7.5 V. For \)GS = 5 + 0.5 = 5.5 V, iD = 1(5.5  2)2 = 12.25 mA, and iD = 1(4.5  )2 = 6.25 mA. That is, current reduces by 9
— 6.25
= 2.75 mA, or increases by 12.25
Total variation in drain current is 2.75
+
for vGS = 5  0.5 =
—9
4.5 V,
= 3.25 mA.
3.25 = 6.0 mA.
Thus Vj = idRp = 6.0mA x 0.5kCl = 3.0 V. [Note that the gain is
 2/(2 x 0.5) = 
3.0 V/V.]
5.24 Here, from P5.23 above, K = V2k'(W/L) = 1/2(2) = 1niA/V2. For v>sv = ± 0.5V, and VGS = 5 V, the largest value of x>GS is 5 + 0.5 = 5.5 V for which iD = 12.25 mA. For saturation, the smallest value of vDS is x>GS  V, = 5.5  2 = 3.5 V. Thus the largest value of Rp that can be used is Rp = (12 35/12.25 = 694£2. Now, for a 1 kfl load resistor, and dgs = 5.5 V, operation is in the triode mode where iD = K(2(vgs V,)\)DS vis), and iD = (VDD DDSyRL. Now with vDS =v, for simplicity, (12  uyi = 1(2(5.5  2)u  u2), or 12 U = 7u u2, or u2 8\> + 12 = 0. Thus v = (  8 ± ÿ82  4(12)>2 = (8 ± ÿ64 482 = (8 ± 4>2 = 2 V. Thus = 2 V. Now for zero signal, X)Gs = 5 V, ip = 9 mA, and x>DS — 12  S/l = 3 V, with operation just at the edge of saturation. Correspondingly, the negative output swing for a + 0.5 V input is 3— 2" = IV. For a 0.5 V signal, •oGs = 4.5 V, ip = 6.25 mA and uÿy = 12  6.25(1) = 5.75 V, with output swing for 0.5 V input being 5.75 3 = 2.75 V. Thus the ratio of peak voltage outputs is 2.75 to 1 or 2.75 V/V.
—





—
—
—
5.25 K = V2k'(W/L) = ImA/V2, V, = 2V, VDD = 12V, RL = 0.5kO, x>Gs = 5V ± 0.5V. Thus iD = K (\)GS V,)2 = 1 (5 2)2 = 9mA, and x>D = VDD  RD iD  12  0.5(9) = 7.5V. See, from Eq. =2<~lf9 = 6mA/V. From Eq 5.43, g,„ = 2K (Vcs  V,) = 2(1)(5  2) = 5.44, that gm =2ÿ~K 6mA/V, in correspondence. From Eq 5.40, = g„, Rp = 6mA/V (0.5k£2) = 3.0V/V. For a From ±0.5(3.0) signal. 5.35, a Eq output input, expect ±0.5V = +1.5V ip = /if [(Vcs V,)2 + 2K (Vas  V,) + /ft)2]. Thus for vgs = ±0.5V, iu = 1 (5  2)2 + 2 (1) (5 2) 0.5 + (1) 0.52 = 9 + 3 + 0.25 = 12.25mA, for which VD = 12  0.5 (12.25) = +5.875V, and iD = 1 (5  2)2 + 2 (1) (5  2) (0.5) + (1) (015)2 = 9  3 + 0.25 = 6.25mA, for which Dp = 12 0.5 (6.25) = 1.5 = 6.0V, and 7.5 + 1.5 = 9.0V, as calculated from a 8.875V. This is to be contrasted with 7.5 model. linearized

<~ID


—
j
<~K<~ID
from Eq 5.4 or (10) (300/3) = 500pA/V2. g„, =2 5.26 Kp = jHp Cox jr = gm = 2 "ÿ0.500 M = 2 ÿ2 = 2.83mA/V. Generally, gain = gm RL = 10V/V implies that = 3.53kf2. Operation is reasonably linear for x>gs « 2 (VGS  V,), but Ip = Rl = 2.83 gm K (Vr* V,)2 > (VGS  V,) = ilp/K Thus, linear for x>gs <<2 (VGS V,) = 2 Hlp/K , that is Dgs < ~ 0.6V < 2 wOJ = 2 ÿ8 = 5.66V. For 1% nonlinearity, vgs= 0.06V peak. For 10% nonlincarity peak.
.


us.v
5.27 Gain = gm (RL II r„), where gm = 2 V/f Ip , and
10k II
9.091 4.808
Va_ 1
/„
(1), and at 0.25mA, 4.808 = 2
1ovA 10 + VA 10 VA (4)
10 + 4
14 1/4+10 . whence 2 VA + 10
VA
J
&
=
ÿK/A
'i?
.

Now, at 1mA, 9.091 = 2
10/: 
1/4
ÿK(l)
(2). Thus (l)/(2)
9.091VA + 90.91 = 9.616V), + 24.040, and
SOLUTIONS: Chapter #59
Now the follower output resistance is 1.34/:£2 II 6.90k £2 = 1.12 k£2 amd the noload gain is 6.90 II 47 0 814V/V " 1.38 + 6.901147 For a gain of 50%, 0.5 = 0.814 x RlARl + 112) or 0.50/?,. + 0.56 = 0.814RL, or 0.314/?,, = 0.56 or Rl = 1.78 k£2
SECTION 5.6: BIASING IN MOS AMPLIFIER CIRCUITS

5.32
+9V

10kQ
10MQ
10MQ
10k£2

For this device, V, = 2V, K = V2k (W/L) = 0.5mA/V2. Assume saturation: For V, = 2V, VG = 1/2 (9) = 4.5V, 7 = K(x>GS  V,f = (1), and x>GS = 4.5  10 7 05 (t)Gs 2)2 (2). Sub¬ stitute (2) in (1) > 2 / = (4.5 10 /  2)2 = (2.5  10 7)2 = 6.25 50 7 + 100 I2, or 100 I2  52 / + 6.25 = 0,
—

.
, 52 ± V 522  4 (100) (6.25) , whence I= = 0.189mA, (or too large a ÿ (100) value). % = 10k (0.189) = 1.89V, \)GS = 4.5 1.89 = 2.61V, vD = 9  (10K£2) (0.189mA) = 9 1.89 = 7.11V, and vDS = 7.11 1.89 = 5.22V » OK, saturation. Operation remains in saturation until vGD k V, = 2V, ie, for vGS = 2.61V, and vDS k 2.61  2 = 0.61V. Thus the peak negativegoing output signal allowed is 7.11 (1.89 + But note that the largest positivegoing output sig¬ 0.61) = 4.61V. nal (for cutoff) is 1.89V. Now, For V, = IV, / = 0.5 (vGS  if, and x>GS = 4.5  107. 27 = (4.5 10 7 l)2 = (3.5  107)2 = 12.25  707 + 10072, and 10072 727 + 12.25 = 0.
— —
722
72 ±
JL(1°°) (1225) = 0.276mA. \>s = 10 (.276) = 2.76V, vGS = 4.5  2.76 = 1.74V, Vp = 9  2.76 = 6.24V, Vps  6.24  2.76 = 3.48V. Now, saturation prevails while vGs £ 1.0V. Thus the maximum negative swing is 6.24 4.5 1 2.74V. The largest positivegoing output signal Thus 7 =
200
(for cutoff) is 9
 6.24 = 2.76V.

+
=
5.33 j.O\/

The design is required to endure the following variations: V, from 1 to 2V, K from 0.3 to 0.5mA/V2, and ID from 0.5 to 1mA: Largest current occurs when V, smallest (IV) and K largest (0.5mA/V2). (1). Thus 1 = 0.5 (\>gs " l)2. or vGS  ± ÿ2 + 1 = 2.414V Smallest current when V, largest (2V) and K smallest (0.3mA/V2). Thus 0.5 = 0.3 (Ogs  2)2, and vGS = €67 + 2 = 1.29 + 2 = VGG  2.414 1mA, (2). From (1), 3.29V where 7? =RS. „= R Vco ~ 3.29 0.5mA. See From (2), = VGG  2.414 = 7?, and 2 VGG 
R
6.58 = 7?. Subtracting, VGG + 4.166 = 0 —» VGG = 4.166V. For 9 ) = 8.6M£2 VGG = 4.166V, 7?gi = 10M £2, 7?G2 = 4.166ÿ (Use 8.2 M£2), and Rs = R = 4.166  2.414 = 1.75k£2 (Use 1.8 k£2).
"jÿ166
4.166  3.29 = 0.507mA. Now for /„ = 1mA, V, = IV, and a 0.5V signal, x>p > VGG  V,. l./*> 1.75 9 ~ (3 + °5)  5.33k£2. (Use 5.1k£2). That is, i)D > 4.166  1.0 = 3.166, and 7?„ = Check:
j66
1mA
236
SOLUTIONS: Chapter #510
5.34
k'(W/L) = 1.0mA/V2, iD = Vlk{W/L) (vGS  V,)2, 9~ ÿ . Thus = l/2(l)(t>GS = 2)2 = 0.5(t)GS 2)2, and iD =
+9V
For V, = 2V, and
—
.
10MQ
9  DCS = 20 (0.5) (uGS
20kn
20«i2
,
 2)2, 9  Ugs = 10 vG.s  40 VGS + 40, V
 39 oGS + 31 = 0, whence ÿ = 39 ± 392 4 (10) (31) 9 2 79 = 2.79V. vDS = 2.79V, and ID = —ÿ— = 0.311mA. For nega¬ 10
tive peak outputs of up to 2Vp, operation remains in saturation. For

• 2 and iD iD = 0.5 (t)G5  l)2,
V, = IV,
ÿ
9~"GS 20 + 1 = 0, whence mGs =
— vGs = 10 (uG5 — l)2 = 10 "0Gs — 20 \>Gs + 10, and 10 vGs — 19 192 — 4 (10) (1) 19 ±17.9 1>845V, 91.85 , „ Thus VDS = 1.85V, ID = — = 0.358mA, with a 27Io)= 20 = 9
dGj
19 ±V
A
IV peak signal allowed. 5.35 For
 
RG2 = 10MI2 from
•
'0 = 2q
„ . = "to™
Thus 9
gate to source, Vps
— 2 vGs = 10 (y>Gs —
38 ± ÿ
Thus x>DS = 5.22V, and
—
3S2 4 (10) <31)
= 2 x>GS . For V, = 2V, iD = l/2( 1.0)(\)GS  2)2 and 2)2 = 10 uG5 40 x>Gs + 40, and 10 dGj 38 vGs + 31
—
20— = 2'61V'
2(10) 9  5 22
1D = —
—
38 ±14.28
— = 0.189mA.
For negative peak outputs, 5.22 2.61 + 2 = 4.61V
—ÿ
is allowed for operation in saturation. For V, = IV,
iD = 0.5 (vGS  l)2 and iD =
•
Thus 9  2 t)GS = 10 (vGS

 l)2 = 18 18 ±16.85  40 = —182—=
— 20 uGs + 10, and 10 x>Gs2 — 18 x>Gs + 1=0, whence \>G$ 9  3 48 1.74V. Thus "Ops = 3.48V and ID = ——— = 0.276mA, with 3.48 1.74 + 2 = 3.74V negative output 2 in 10 dG5
on
m
Jin
10
n
1
u
peaks allowed, while saturated operation prevails. 5.36
+5V
Here, V, varies from 1 to 2V, K = 1/2k(W/L) varies from 0.3 to to 1mA. The largest current (1mA) occurs for the smallest V, (IV), and largest K (0.5mA/V2). Thus
0.5mA/V2, ID varies from 0.5 iomh
 l)2.
5
— 2 414
, , > 2.59kfl. uGJ = 2.414V, and R > 1mA The smallest current (0.5mA) occurs for the largest V, (2V) and smal¬ lest K (0.3mA/V2). Thus 0.5 = 0.3 (v>GS  2)2. x>GS = "*L67 + 2 = 1 = 0.5 (t)GS
5
 3 29
2 59 + 3 42 < 3.42kfl. Use R = = 3.0kft. Because of feedback, the effect of variation is reduced. The circuit automatically allows a negative signal = V, > IV but the gain is smaller than in P 5.33, since R here (3.0kI2) is less than RD there (5.2kQ). Raising R to (say) 3.3kH would be allowed here, and
3.29V, and R <
ÿ
would improve the gain by 10%. 5.37 In P5.36 above, the minimum negativegoing signal is IV. Here it should be 1.5V. That is, we want x>os  vGs + 0.5V for the case in which V, = IV.
 237 
SOLUTIONS: Chapter #511
For smallest V, (and also the largest current), from the results in P5.36, \)CS = 2.414V, and vDS = 2.414 + 0.5 = 2.914, with 52 914 in R > = 2.09k£2. If we use RG2 = 10MO, RGl =
RGI
ÿ
(0.5) = 2.07MQ. Use RGi = 2.0MO. Now for V, = 2V (and also the smallest current), from the solution for P5.36 we see, vGS = 3.29V,
and vDS =
3 29
x 2.0 + 3.29 = 3.95V. Thus R <
5
—Q 3 95 < 2.1k12.
5 Notice that a solution barely exists, using R = 2.1kQ, essentially as a consequence of a demand for large signal swings with a limited sup¬ ply voltage.
5.38 Here, Vp = 4V, IDSS = 32mA. For
+9V RGI
RG2
Thus (1 
vp
)2 = 1/4, 
/c = 8mA, 8 = 32 (1 
"p
)2.
± 1/2 1 = 1.5 or 0.5, and VGS vp =
= 0.5 Vp = 2V. Now for a negative swing of 2V to the edge of saturation, x>D > vG +  V,  Now for the largest possible value of RG, \)g will be lowest and \>G lowest. The lowest possible x>G is 0V, with Rg i = oo and RG2 = 10M£2, in which case Rs = 2V/8mA = 0.25kf2, and VG ÿ 0 + 4 = 4V for lowest swing or4 + 2 = 6V for no signal, with RD £ 96/8 = 0.375kf2. Note that for a 2V positive out¬ put swing, VG rises to 6 + 2V = 8V, and the transistor is not yet cut off. OK. Note that this design with no biassing supply is relatively sensitive to device variability, all as a result of wanting a large signal swing with a small supply.
.
.
5.39 Here, iD = Vl\LnCnx(W/L){x>GS  V,)2 = 1/2 X 20 x 10"6(42)(2.5  l)2 = 45pA Thus, assuming the Early effect to be negligible, Q2 operates at 45 pA. For a 5 V supply R = (5 2.5)445 X 10"6) = 55.6 k£2. > 1.5V. Transistor Q2 remains in saturation for i)0 > (vGS V, ) = 2.5 1 = 1.5 V or



.

In triode mode, «0 = \i„Cox(W/L)[(\)Gs V, )x>Ds  1)0/2] Now for iD reduced to half, for vDS =x>, 452 = 20(42)[(2.5  1)\)  d2/2], or 0.5625 = 1.5o  \>2/l, or o2 3u + 1.125 = 0, whence

u = (   3 ± ÿ32 4(1. 125)YZ = (3 ± 2.121)2 = 0.439 V. Thus the current reduces to 1/2 normal for Vps = 0.439 V. Check: iD = 20 x 2[(2.5  1)(.439)  .4392/2] = 22.5 pA.
—
5.40 For a 1 Mil output resistance at 100 pA output, VA = 106 x 100 x 10"6 = 100 V. This requires that the output transistor have a channel length L = lOOV/lOV/pm = 10pm. Use this for both transistors. For the edge of saturation at x>0 = 0.5 V, uGs = 05 + V, = 1.5 V. Now for Qt, 25 = 20(W/10)(1.5 l)2, or IV, = 25 X 10410(0.5)2) = 100pm. Correspondingly, IV2 = (10025)W, = 4(100) = 400pm. The output current will be exactly 100 pA when Q, and Q2 operate identically, with x>DS = vGS = 1.5 V and v0 is 1.5 V above the negative supply. For operation at 0)0 = 5 V above the negative supply, i0 = 100 x 10"6 + (5  1.5yi06 = 103.5pA.

 238 
SOLUTIONS: Chapter #512
S541(a)
Topology A: This design is relatively straightforward, except that it pchannel devices operating at twice the current density of the nchannei. Widths are as indicated: 8 transistors arc needed. The total width of the NMOS is XW„ = 4 + 2+ 16 + 40 + 4= 66pm. The PMOS width is TWp + 2 + 4 + 8 = 14p.m. If the same current density is used in the PMOS, the total PMOS width would be double, namely 2(14) = 28pm. Notice, incidentally, that neither of these designs, nor the others to follow, compensate for the Pn/py, ratio r. For such compnesation, multiply all the PMOS widths by r. Topology B: If both ends of Iref(  10pA) are available: + VDD
+ VDD + VDD
x16
VSS
VSS
ix16
x40
 VSS
VSS
This is a design for equal current densities in the PMOS and NMOS transistors: #T = 7, ZW„ = 62 pm, XWp = 28 pm.
239
SOLUTIONS: Chapter #513
Topology C: The attempt here is to create the 5 (lA output separately without making all the other transistors twice as large. +VDD +VDD
+ VDD
+ VDD
S541(c)
This uses 9 transistors but it is smaller:
= 36
(J.ni
and
ZW,,
= 16 (Im. "
SECTION 5.7: BASIC CONFIGURATIONS OF SINGLESTAGE IC MOS AMPLIFIERS 5.42 For Fig. 5.44a): Input: Operates at vGs = 2 V. Output: Source saturates at 5  0.5 = 4.5 V; amplifier saturates at
t)GS
— V, = 2  1 =
IV.
Overall: Input: 2 V. Output: 1V to 4.5 V.
For Fig. 5.44b):
Input: Operates at  2 V. Output: Operates from + 4.5 V to  1V where the drain falls V, below the gate. Overall: Input  2 V . Output: 1V to + 4.5 V.
—
For Fig. 5.44c):
Input: Saturation at O/ = 5 + 1 = 6 V. Output: Sink saturates at  5 + 0.5 =  4.5 V. Driver saturates at 5 + V,  VGS = 4 V. Input: For output at  4.5 V, input is limited to  4.5 + 2.0 =  2.5 V. Overall: Input:
— 2.5 V
5.43 Eq 5.101 indicates
For I= 2.5(iA:

Av =
Au =
to
+ 6 V; Output:  4.5 V
to 4 V.
\VA I Let I . = IREF. . Iref 1
»
1)° (1°Q)
ÿ2.5
For / = 25(iA:

Av =
=  200V/V. For I= 0.25p.A: A„ =
V
_
,= 63.2V/V. 25
jn 1)001
ÿ
*,° (100)
Vo.25
= 632VA'.
5.44 For the diodeconnected NMOS (call it Q4) of half the width of Q\, its width is also half that of Q$. Thus, Q4, Qi have the same k , and the same V,. Thus x>sg 3 = ÿ>sc2 = 5/2 = 2.5V.
Iref = 1  (1/2) x 20 X For fit. i = 112.5 =
(2.5
 l)2 =
112.5xA, with a total supply current = 2 (112.5) = 225iA.
(1/2) X 40 x 100/10 (l)GS
 240
 l)2 = 200(bGJ  l)2.
Thus x>0 = vGS = 1 +
SOLUTIONS: Chapter #516
5.48 (continued)
rl'
S
'ik loss
J
o >  V, saturated o <  V, j triode t) < 0 triode
>  V, u < j V, ÿo < 0 t>
IVtl
saturated
j triode triode
5.49 Assuming no backbias effect: (a) Devices identical, V, = 5/2 = 2.5V, i = K (\>cs  V,)2 = 1 (2.5  2)2 = 0.25mA. Thus /, = 0.25mA. Here for (b) Lower transistors operate as one with K = 2mA/V2. V2 = t), i = 1 (5  \) 2)2 = 2 (o 2)2, (3  o)2 = 2 (\>  2)2, 3  x> = ± <2 (v 2) = ± 1.414 o ± 2.828. Thus, 3  t) = 1.414 o 2.828, or 3  D = 1.414 v + 2.828. Correspondingly 2.414 x> = 5.828, o = 2.414V, for which i = 2 (2.4142)2 = 0.343mA. Thus V2'= 2.414V, I2 = 0.343mA. (c) / 3 operates at Ipss if the upper transistor is in saturation, in which case, i = K (VGS  V,)2 = 1 (  2)2 = 4mA = /3. Also i = K (uGS V,)2, or 4 = 1 (uos  2)2, vGs  2 = ± 2, whence \)GS = 4V (which is too high to allow the upper transistor to saturate). For o = V3, Thus, the upper is in triode mode, the lower in saturation. / = K (2 (0  2) (5  \>)  (5 = K (m 212. 20  4 \)  25 + 10 \)  o2 = \)2  Av + 4,
—

2o2  lOo +
9 = 0, o =
+10 ± y
102  4(2) (9)
2(2)
10 ± 5.29 = 3.82V (or an impossibly low 4
value), for which / = 1 (3.82  2) = 3.31mA. Thus /3 = 3.31mA and From Symmetry, V4 = 2.5V, and 14 1 (2.5  2)2 = 0.25mA.

(d)
From Symmetry, 16.25mA.
(e)
V5 = 5/2 = 2.5V, and /5 = 1 (2 (2.5
 2)
(2.5)
V3 = 3.82V. 2.52) = 5 (4.5)
 2.52 =
5.50
+5V
—
•
1
d
Hÿi
Here V, = IV, KD = 90pA/V2, and KL = 10iA/V2. Ignoring the body effect, and assuming both in saturation,:' =K (Dr.? V, )2. For the driver, gmD = 2Kn (t)ÿ V,) = 2KD V i/KD 2 V KD i. For the load, g„,L = 2 y KL i. Ignoring r„ , (and backbias effects), gain 1 2 V K.j) i 1  ÿ0 = 3V/V. =Vo/ = gmD J gml 2 iKl I For u0 = V00/2, / = /ft (2.5  l)2 = 10 (1.52) = 22.5pA. Now, l)2 = 90 (v, 4)2, or v,) for Qd, 22ÿ5_jf_ 90 ((5 V/ 4 = ± 22.S90 = ± 0.5, O/ = 3.5 or 4.5 (not possible). Thus V/ = 3.5V for v0 = VddI2. Equation 5.78 applies while devices are both in saturation: ie for v0 down to IV) I , or vG = IV, where V/ = +4V (at cutoff), and v0 up to Do + 1 = V/.
.


 243 

—
i

SOLUTIONS: Chapter #517




For Qd i = 90 (5 u,  l)2. For QL: i = 10 (uc  l)2. Thus 90 (4 v,)2 = 10 (o0 l)2, or 3 (4 D/) = DG 1, or oG = 13 3\>/. Check: For D/ = +4, oG = 13 3(4) = IV as found before. Now for D, = u0+l, Do = 13 3(d0 + 1) = 10 3d0. 4d0 = 10V, d0 = 2.5V, for which d; = 2.5 + 1 = 3.5V. Check: For D/ varying from 4 V to 3.5 V, d0, varies from 1 V to 2.5 V, respectively. That is a 1.5 V change results from a  0.5 V change, consistent with a gain of 3 V/V.
—

—
—



5.51
Eq +5V
V, = V,„ + y [ V 2
5.25:
+
V
X = .J _ .
2 v 2
Vja
at which = 5  V,. Thus V, = 0.9 + 0.5  ÿ06), or V, = 0.513 + 0.5 V 5.6 V,. Iterate: Try V_= 2V, V, = 0.513 + 0.5 ÿ6  2 = 0.513 + .949 = 1.46, V, = 0.513 + 0.5  1.46 = 1.53V, V, = 0.513 + 0.5 ÿ5ÿ6  1.53 = 1.52V. Thus the upper output voltage is 5  1.52 = 3.48V, at which V, = 1.52V, and % = yA, 2 V 24>y + VjB) = 0.5/(2 0.6 + 1.52) = 0.172, and g„ = 0, since iD is zero at the upper limit. At oG = 0V, VJfi = 0 and V, = 0.9V. Thus i = K (dgs  V,)2 = 10 (5  0  0.9)2 = 6.168mA, for which gm = 2K (oGS  V,) = 2 (10) (5  0  0.9) = 82pA/V, and % = yA. 2 V 2 Oy + Vfo) = 0.5/(2 V 0.6 + 0) = 0.323.
Vnn  V, = 5  V,
(V 0.6 + 5  V,
5.52 From P5.51: At V0 = 2.5V, VSB = 2.5V, V, = V,„ + y (V 20r + VSB V2d>/) = 0.9 + 0.5 (ÿ 0.6 + 2.5  "H).6) = 1.39V. Now, gmL = 2K (dgs  V,) = 2 (10) (5  2.5  1.39) = 22.2iA/Vt i = K (dgs  V,)2 1 = 10 (5.0  2.5  1.39)2 = 12.3pA, % = = ÿ = 0.142. Thus, gmh =vgmL = 2 0.6 + 2.5 2 V 24>y + V™ .142 X 22.2pA/V = 3.15pA/V. Also = 2 AT
—,  —


*

5.53 Assuming V, includes the backbias effect, iL K (dC5  V,)2 = 22.5 (0 2.0)2 = 90iA, gmL = 2K (dGj " V.) = 2 (22.5X2) = 90pA/V, gmB = % gmL = 0.2 (90) = 18pA/V, r„ = VaAl = 50/90pA = 556k£2, gmD = 2 VAT iL = 2 ÿ90x90 = 180iA/V. Thus, gain (around Vo = 2.5V), is g„, (r„ II r„ II l/gmB) =  180iA/V (556k/2 II l/18pA/V) = 0.180 (278k II 55.6k) = 8.34V/V. This again applies reasonably well until the load enters triode operation at DG = Vdd 1 V, I = 5 2 = 3.0V, or until the driver enters the triode region at oG = D. Now for the lower level, it = 90(iA, that is 90 = 90 (dGj l)2 > dGs = 2V. Now for dGs = 2V, triode operation begins at vDS = uGS  V, = 2  1 = IV. In actual fact, as dg falls from the middle, that is from 2.5V to IV, input must rise by (2.5 1) /8.34 = 0.2V. Thus, the output range is from 3.0V to about 1.2V.



—
SECTION 5.8: THE CMOS DIGITAL INVERTER 5.54 For the NMOS, Kn = l/2\i.„Cax(W/L) = 1/2(100)(24) = 100pA/V2. Since the inverter is matched, Kp = 100pA/V2 also. With a 3.3 V supply, VOH = 3.3 V and VOL = 0 V. Now V,H = (l/8)(5VDd  2V,) = [5(3.3)  2(0.8)}ÿ = 1.86 V, and V,L = 3.3  1.86 = 1.44 V. Thus NMh = V0H  Vm = 3.3  1.86 = 1.44 V, and NML = V,L  V0L = 1.44  = 1.44 V. Now, by symmetry, Vlh = Vw = VDD/2 = 3.2/2 = 1.65 V. Peak current from the supply is iD = 100 x 10"6(3.2k2  0.8)2 = 72.25 pA. Current is half the peak at dgj = D when one of the transistors is in triode operation, and one is in saturation, where 72.254 = 100(d 0.8)2 or u = 0.8 + 0.362* = 1.40 V.

Thus by symmetry, halfcurrent operation occurs at D/ = 1.40 V and 3.3  1.4 = 1.90 V.
 244
SOLUTIONS: Chapter #518
Current is onetenth the peak value where v = 0.8
+ ((72.2540)400)'/' = 1.07 V.
Thus 1/10 peak current occurs at U/ = 1.07 V and 3.3  1.07 = 2.23 V. For output resistances at the output limits: For triode mode operation, ip = k'(W/L)[(x>GS  V, )\)DS  x>p/Z\.
iD = k'{W/L)(x>GS  V,)\)pS, and = W(W/L)(x>gs V,)= 1/(100(2)(3.3  0.8)) = 2 X 103O.
For small x>DS, rDs
Thus the output resistances in each extreme state are 2 kil. For output currents of peak value = 72.25pA and triode operation with full input signals: 72.25 = 200((3.3  0.8)x>DS  \>Ay4], or u£y4  2.5x)DS + 0.36125 = 0, or x>t,s  5mds + £.7225 = 0 and VDS=(  5 ± ÿ52  4(0.7225))4 = (5 ± 4.70)4 = 0.15 V. Thus, the peak current of 72.25 pA flows at v0 = 0.15 V and 3.15 V 5.55 From Eq. 5.102, tPHL = tPW =
„=
L6C 50x lO"15 , = 16 x 10"6(24)3.3 = kn(W/L)n VDD 100 X
and the average propagation delay is tP = (tPLp + tPHL yi = 0.121 ns. Alternatively: For saturation mode operation initially, iDi = (1/2)100(24)(3.3  0.8)2 = 0.6125 mA, and at the mid point: iD2 = 100(24)[(3.3 0.8)(3.34) (3.34)2/2] = 200(4.125 1.36) = 553 pA.



Thus the average current is (553 + 612.5)4 = 583 pA „ , 50 x 10~15 x 3.34 . ,__9 and t„p ~ 7= 0.142 x 10 'j = 0.142 ns. 583 x 10"6 For operation a) with an ideal (0ns) input, dissipation is entirely dynamic, Pj = fCVpD, or 50 X 10 = = 1.125 mW at a frequency of K4(121xl0"12)) = 2.07 GHz. Pd = 4 4(0.121 X 10~9) tp b) with transition times of 2tp, an additional power is lost due to dcvicetodevicc current whose peak value is 72.25 pA (from P5.54 above). Average power loss per transition is the product of the half the peak current and the supply voltage for the duration of the active part of two transitions, while both dev¬ ices conduct as the input goes from 0.8 V to 3.3  0.8 = 2.5 V. Thus the average power loss in two transitions/cycle is (72.254) x 3.3 X 2((2.5  0.8)2 X tp/33y4t/> = 61.4 pW. Thus the total power loss is 1.125 + 0.061 = 1.186= 1.19 mW. See that for this logic, at this supply voltage, at this frequency, that the dominant loss is due to load capacitance charging, (ie fCVpp). The delaypower product is, DP = 0.121 x 10~9 x 1.19 x 10"3 = 0.14 pj. For the approximation on page 435 of the Text, DP = CVpD = 50 x 10l5(3.3)2 = 0.54 pj. Note that this is essentially 4x the earlier value, simply because in the Text, / is assumed to be VtP. f = (2t„)~x would have been a better choice.
—
—
SECTION 5.9: THE MOSFET AS AN ANALOG SWITCH 5.56 Now, for lOmV to ground with a 3.3V, 2. lkll source, iD = (3.30  0.01)/ 2.1k& = 3.29V/21kO = 0.157mA. Now in the triode region, iD = AT 12 (x>Gs  V,) Vps  >ds ~ AT (2 (x>Gs  V,) \>ps)< or * 0.157 x 10"3 = 1/2 x 20 x 10"6 x W/10 x 2 (5  1) (10 x 10"3), or 0.157 = 10 x 10~6 x 2 (4)W. Thus W = 0.157/80 x 10"6 = 1963pm, which is quite a large device! Now, if 0.10V were acceptable: (.30 .10X21 ,n6 ,nn ~—Lx 10 W = = 190pm.
J
800
ÿ
5.57 Af„ = 1/2 (20 x 10"6) (50L/L) = 500pA/V2. Assuming pp = l/2p„ with Wp = 2Wn, then Kp = Kn = 500pA/V. Now, for operation in the triode mode,
245
SOLUTIONS: Chapter #519
i'd  /i { 2(t)Cj  V,) x>DS  v£s ) ~2K (\)qs  V,) vDs, and ros  "Odsÿd = M2K (\>GS  V,)). Now, for V/ = 5V with VGn = +5V and VGp = 5V, only the n channel device conducts with 5  2)) = 1250. Now, with 5kI2 load, ac loss in the switch is 125/(125 + res l/(2(500 X 10~6) (5 5000) = .0244 or 2.4%. Now, for V) = +5V, with Kp = Kn , the result is the same and the loss is 2.4% in the switch. Now, for V/ = 0V, with VGn = +5V, VGp = 5V, both switches conduct (equally), with
—
—
—
—
rDS = 1/(2 (500 X lO"6) (5  0  2)) = 333.312 each. Thus the total switch resistance is 333.3/2 = 167£2, and the ac switch loss is 167/(167 + 5000) = 0.0323, or 3.2%.
SECTION 5.10: THE MOSFET INTERNAL CAPACITANCES AND HIGHFREQUENCY MODEL 5.58 Gate oxide capacitance ranges from 1.75 fF/pm2 for 20 nm oxide to 0.35 fF/pm2 for 100 nm oxide. For a 1 pF capacitance of area W2 in the thinoxide technology, 1 x 10~12 = W2 x 1.75 x 10~15 and IV = 23.9pm.
For the thick oxide, W = [(1 x 10"12K0.35 x lO"15)]* = 53.5pm. For the minimumsize MOS, the area is IV x L = 2.4 x 1.2 = 2.88 pm2. Over the range of oxide thicknesses, from 23.922.88 = 198 transistors, to 53.522.88 = 994 transis¬ tors would be required.
5.59 Here, Lov = 0.15 pm, Csb0 = Cjbo = 40 fF for a 10 pm wide device, V0 = 08 V. From Table 5.1 on page 364 of the Text, for tox = 20 nm, Cox = 1.75 fF/pm2. a) For (W/L) =  (100 pm/2.4 pm), and Lov = 0.15 pm
Cov = WLov Cox = 100 X 10"6 X 0.15 X 10"6 X 1.75 x 10"15/1012 = 26.25 fF Basic CgS = 2/i(WL)Cox = 22 X 100 X 10"6 x 2.4 X 10"6 X 1.75 x 1015/1012 = 280 fF, in saturation. Including overalap, Qgs = Cgs + Cov = 306 fF. Basic Cgj ~ O.OfF, in saturation. Including overlap, Cgd = Cgd + Cov = 26 fF. assuming the source and drain areas are proportional to the device width, then Csb0 = Cj/,0 = (4(K10)100 = 400 fF and since \VDB\ = IVÿl = 2 V, from Eq. 5.11, Csb = Cdb 40041 + 2.00.8)14 = 214 fF. {This can be seen directly from Ex. 5.41, since IVSBI = \VDB\ is also 2 V there.} Now

b)
For (W/L) = (10/24):
C„v = 10 x 10"6 x 0.15 x 10'6 x 1.75 x lO'VlO"12 = 2.62 fF. C'gs = 22 X 10 x 10"6 x 24 x 10"6 X 1.75 X 10~l5/l012 = 280 fF Cgs  C„v + Cgs = 282.6 fF, and CgJ = Cov + C'gd = 2.6 fF, and Cdb = 214 x 10200 = 21.4 fF, and Csh = 21.4 fF. 5.60 For the wide transistor at 100 pA: From Eq. 5.44, gm = (2k'n(W/L)IDf = (2 X 100 X 10~6(1002.4)100 x lO"6)'"4 = 912pA/V. and Cgs = 306 fF and Cgd = 26 fF. Thus fT = g„A2n(Cgs + Cgd)) = 912 x W~%2k x (306 + 26) x 10~15) = 437 MHz. For the longer transistor at 100 pA:
10_6(1024)100 x lO"6)ÿ = 91.3pA/V . and Cgs = 283 fF, and Cgd = 2.6 fF. Thus fT = 91.3 x 10~6/{2rc(283 + 2.6) x 10~15 = 50.9 MHz. gm = (2 x 100 x
246
SOLUTIONS: Chapter #520
Now at 10 (XA, a reduction by a factor of 10, gm will reduce by "ÿ10 = 3.16, and so will fT to 138 MHz and 29.9 MHz for the two transistors. 5.61 From Exercises 5.41, 5.42, CRI = 30.6 fF, Cgj = 2.6 fF and fj = 1.38 GHz. For drain and source grounded, the input capacitance is C  30.6 + 2.6 = 33.2 fF, whose impedance at fT is Z = lA.2nfTC) = 1/(2n x 1.38 x 109 x 33.2 x 10~15) = 3.47 k£2.
Now for a voltage gain of  2 V/V, the input capacitance becomes At /7/IO, Zin = 1/(2jc x (1.3840) x 38.4) = 30.03 kQ.
Cin = 30.6 + 2.6(1 
 2) =
38.4 pF.
SECTION 5.11: THE JUNCTION FIELDEFFECT TRANSISTOR (JFET) 5.62 For V+ = 4V, operation is in saturation, and ip = lass (1
X)qs
Q — ) = 10mA. —7}Vp— )2 = 10 (1 —A
For V+ = 2V, operation is at the edge of saturation and iD = 10mA, also.
For V+ = IV, operation is in triode mode, and
iD —
DSÿ
Vp
[1 2 (0  2) 1  l2 J = j ( 4  1) = 7.5mA. For t'o = —— = 5mA = —— [2 1. 2 = 4 Mas Vas ]. _— 4 \2 (2) Mas vas ~

ÿ
2ÿ— — V424f2t = mds = 
4 ±2ÿ2
4+
Check: iD = y
*
iD = 5mA, 5 = 4
Thus mgs =
(mgs
~ ÿds<
— Vp) x>DS — (Mas)2 j,
"°ds
~ 4 ÿ>ds
or
+2 = 0, and
r
= 2 ± ÿ2 = 3.414 or 0.586V. Clearly V+ = 0.586V.
[2 (2)(.586)  .5862 ] = 5mA.
5.63 For triodemode operation: i0 = Now, for
£2
loss r (mgs  Vp) Mas ~ Vds 1 — [2 J
[2 (mgs + 2) 1  l2 j, whence 2 = 2 (mgs + 2) 1 = 2 mgs +41.
= 0.5V.
[2 (0.5  2) 1 1 ] = 2.5 [2 (1.5) 1] = 5mA. Now, for ia = 1mA, 1 = [2 (mGs + 2) 1  l2 J, or 0.4 = 2 mgs +41, whence MGS = 0.43 Check:
2
iD = j
= 1.3V.
5.64 For triode operation: „
„
.
iD
=
2lass
For small mds, »o = —rpr
Vp
vp
[2 (oGS  Vp) mDs  i>ds ]• ~
Vp , 1 "°as ... = y7~rrrvp) vds, and rDS = t— = z:(y>GS  Vp) 2wss 2'dss 'd Vp2
Now for
IDss = 10mA, Vp = 2V: 22 200" For
= 2) — 2V, the switch is cut off!) = 2 x 10 (1
22
(VasVp)
= .n nrr = 10012. For 2 ÿ = _2V' rDS = = "• (0f COUrSe' Since at 2) 2 x 10 (2 For mgs = 0V, rDS = „
—
247
mgs
IV, rDS
=
SOLUTIONS: Chapter #521
5.65 For mds = 2V, and Vp = 2V, the JFET is just at the edge of saturation, for which
iD = K (mgs
_ (»GS Vp )2 = [Dss V,?=ÿL "p
_
j
vgs
5 = 10
or
1

and
2
= ÿ = 707 Thus Ugs = 2 (0,707 1) = "°586V
1+
1
Check: ip = 10
2
.586 2
= 5.00mA.
For mgs = 0.586V, x>DS = 7V, 5.10 = 5.00 (1 + A. (7  2)), or 0.10 = 25A., whence A. = 0.1 = 0.004V'1. 250V 1 —±— 1 Thus VA = = = 250V, and r„ (at 5mA) = = 50k£2. 0.004 X 5mA
f
1
More painstakingly: In saturation, iD = IDSS
2
ÿgs
(1 + Xmds). Thus, at mds = 2V, 5
2
1
VGS
(1 + X 2), and at x>DS = 7V, 5.1 = 10
2
Divide: Thus
= 10
2
ÿr = 5

(1 + A, 7).
? + ÿ = 1.02, or 1 + IX = 1.02 + 2.04A,, whence A, = —— — = 0.00403V '. 7 — 2.04
1 + 7.X
w
2
1
Thus 5 = 10
1
VGS 2
(1 + 2 (.00403)), or
= 2 (.2957) = 0.591V, for which V„ =
1
Vqs
2~
1 2(1.00806)
1 = 248V, and r„ = .00403
= 0.7043, whence
Idss
(1
UGS
uG,s
)2X
(10 (.7043)2 (.00403))"1 = 50.02kD.
5.66 Now, iD =
2 (\)GS
(a)
For pchannel; mgs = 0; vGD = 5V
(b)
For nchannel; mdg = 5V
1

trie mode, or iD = IDSS jJ in Viode
 V,,) (dD5)  Vds
—» saturation.
—> saturation.
VGS
y = ± Vi, whence \)GiS
Thus
iD
For nchannel; whence
(d)
Mps
=
— 4mA. 2
ÿ°GS
, or 1 = 4
= 2 (± 14 1) = 1 or 3 (cutoff). Thus Vh = \)G

4
(2 (0 iD < IDSS —> triode. Thus 1 = 4±V 42 = 4(1) 0.268V, or very large.
=
2) \)Dj
Thus
in
1
saturation.
ÿG5
2
 mgs =0
 %), or \)GS  4

, or 1=
+ 1=0,
Vc = +0.268V.
The pchannel device is operating with the gate somewhat forwardbiassed in the triode mode. Thus or
Vd
iD =
(vCs
~
Vn
Vp) mds
 vbs), or 1 = ÿ (2 (Vd  2) (Vd) {Vdf) = 2V}  4Vd  Vd\ 164<l) 4 ± 4.472 *t
 4 Vd 1=0, whence Vj =
=
5.67 For the lower device, assumed to be in saturation, iD = loss 12 ~lD Thus, ip = 4 1 ~ÿT = (2  ('d )2, 4  4 i0 +

Id
=I dss
1
= IDSS
IV. (c)
lu
Thus
1
+ 5 ± V 52  4(4) = +5 ±3 2
2
_
2 i
ÿ
, and Mas = i0 (Ik 1.1) = i0.
i'd = iD , ip  5iD + 4 = 0, whence
or 4mA (not acceptable).
 248 
ÿÿG.V
=
ID = 1mA and dGs = —IV.
Now, the
SOLUTIONS: Chapter #522
upper circuit is the same. Thus, the since the gate is at OV, source is at +1V and V0 = OV. Now, if both resistors are raised to 2kX2, ID(reduces, but it is the same in both cases, and V„ = OV is 2ip retained. Here vGS = 2iD, and ip = 4 j = (2  2iD)2 = 4  8i0 + 4t"o, for which 2 9 ± V 81 4(4)(4) 9 ±4.123 4ip 9ip +4 = 0, and ip = = 0.61mA. Now for /D = 0.61mA, vGs = 2(4) 1.22V, but V0 remains at OV.


5.68 Current in the 1M£2 network can be ignored. Thus
ID = / = 10mA.
Now, ID = IDSS
1
vGs
, or
2
1
10 = 10 «.
05
ÿ
ucs
..
= OV (as could be seen directly). Since vGS = 0, vG
Va
2 IDSS
50V
1
vGs
_ 2 (10)

0,
(10)
= lOmA/V. For /?/, = <*>, u„A), = g,„ r„ = 10 x 103 (5 x 103) = 50V/V. For Rl = r„, X)„A)j = £m (r„ II Ri) = 10 x y = 25V/V. Now /?,• = 1M12 II (1M12/(1  gain)) in general, or /?, = 1 II 1/(1 50) = 19.2kf2, or 1 or I 1/(1 25) = 37k£2, in the two cases.
—
—
SECTION 5.12: GALLIUMARSENIDE (GaAs) DEVICES  THE MESFET 5.69 Here, from Eq 5.108 and 5.109, gm = 2 P (VGS  V,) (1 + A, VDS), r„ ~ (VGS  V,)2), and the
highest available gain is p. = gm r„. For x>GS = +0.2V, gm = 2 (10"4) X 100 (0.2  1.0) (1 + 0.2(3)) = 200 x 10"4 x 1.2 (1.6) = 38.4mA/V, r0 ~ l/(.2 x 100 x 10"4 (.2   1.0)2) = 34712, and p = 38.4 x 10"3 X 347 = 13.3V/V. For X>GS = 0.2V, gm = 2 (10"4) X 100 (0.2 1.0) (1 + 0.2(3)) = 20 X 10"3 X 0.8 X 1.6 = 25.6mA/V, r„ = 1/(20 X 104 (.8)2) = 78112, and p = 25.6 x 10"3 x 781 « 20.0V/V, For \>GS = OV, gm = 2 (10"4) X 100 (0  1) (1.6) = 20 X 10~3 (1) (1.6) = 32.0mA/V, r„ = 1/(20 X 10"4 (1.0)2) = 50012, and p = 32.00 x 500 = 16.0V/V.

5.70 +3V
4 Vdd
P = 100 x 10~4A/V2 = 10mA/V2 for a 100pm device. From Eq 5.107, iD = P ('OgjV/)2 (1 + tops) assuming operation is in saturation, and % = VDD  iD RL, iD = (VDD  1)dsWlFor \)GS = +0.2V, (3  Day >0.1 = 10 (0.2  l)2

.+ VDS
VGS
(1 +0.2 Vps)> or 3 vDS = 1 (1.2)2 (l+0.2uD5) = 1.44 + 0.288 D/jj, and 1.288 \)ds = 3 —• 1.44 = 1.56, x>ds = 1.56/1.288 = 1.211V. Now this exceeds (0.2   1.0) = 1.2V, OK. For uGS = 0.2V, (3  \)05}0.1 = 10 (0.2  l)2 (1 + 0.2 tioy), or 3 0.64 + "Ops ~ 0.64 (1 + 0.2 x>ps) 0.m\)DS, 1.128t)os = 2.36, vDS = 2.36/1.128 = 2.092V.
—
—
—
For vGs = OV, 3 Dus = 1 (l)2 (1 + 0.2uoy ) = 1 + 0.2uay, 12x)ps = 2, and Vp$ = 1.67V. Voltage gains: For x>GS = 0.2V to 0.2V, "gain" = (1.211  2.092)/(0.2 0.2) = 2.2V/V. For VGS = 0.0V to 0.2V, gain = (1.67  2.092)/(0  0.2) = 2.13V/V.

5.71 Now, Pi = p2 = 10mA/V2. Assume that the dc output is stabilized at half the supply voltage. That is, Uasi = Vpsi = 5V. Now, ID2 = P2 (Vgs2 ~ V,)2 (1 + A, x>DS2) = 10 (0  l)2 (1 + 0.1 x 5) = 15mA, and 1) (1 + 0.1 X 5) = 30mA/V, r0i = Ipi = ID2 = 15mA, with vGSi = OV as well. Thus = 2 (10) (0 l)2) = lkft, A„ = 30 (lkO II IkO) = 15V/V. 1/(0.1 (10) (0   l)2) = lka r02 = 1/(0.1 (10) (0

249 
—
SOLUTIONS: Chapter #523
5.72 For V0 = +3V, vDS2 = 10  3 = 7V, ID1 = 10 (0  l)2 (1 + 0.1 (7)) = 17mA = IDi. Now for fij: 17 = 10 (t)Gsi ~  I)2 (1 + 0.1(3)), or (oC5i + l)2 = 17/(10 (1 + .3)) = 1.308, \)cs, = ± 1.144 1 = 2.144 (cutoff) or +.144V. Now gml = 2 x (10) (.144   1) (1 + 0.1(3)) = 29.7mA/V, r01 = 1/(0.1 (10) (1.144)2) = 0.764kQ, r02 = 1/(0.1 (10) (0  l)2) = lkQ. Thus, the gain: =  gm r01 II r02 = 29.7 (lkQ II 0.764kO) = 12.9V/V.


 250 
Chapter 6
DIFFERENTIAL AND MULTISTAGE AMPLIFIERS SECTION 6.1: THE BJT DIFFERENTIAL PAIR 6.1
Eq.6.7, 6.8. iE i (a)
(b)
(c)
1+
rj
_ vWr w
'£2 =
m >
l+e
*
1+ e
zjj

w
\>"ÿT
=
v7v~
1 + eÿT '
iC2 = a iE2~ iE2  0.99/, when  „ = 0.99, or ev/Vr = 1/0.99 1 = 0.0101, or vd = VT In 1+ 6 .0101 = 4.595 Vr =  115mV. That is, i)fli must be lower than X)B2 by 115mV. ic\ =ol iEi~ iE  = 0.95/, when
T
1+ e
T
= 0.95, or e
Vj/V'
= 1/0.95 1 = 0.05263, or o
2.94, or v>d = 2.94 VT = 73.6mV. That is, vBl must be higher than x>B2 by 73.6mV. For ic i = 9.0 i'c2 with iE\ + ic2 = 1,1 ic2 = 9j'c2. or ic2  //10 = 0.1/, for which
Therefore,
I—zr = 0.9/, e~v/Vr 1+e
iE\ = 0.9/.
= 1/0.9 1 = 0.1111, vd/VT = 2.197, or vd = 54.9mV.
That is, X>Bi must be higher than vB2 by 54.9 mV.
Case
*>fli V
%2 V
V
1>CI V
~°C2 V
a b
0 2
0 2
0.7 1.3
6 6
c e
2.0 2 1
1.3 0.3 2.8
f g h
4 4.0 1
1 1.0 3.5 4 0
2 10 10 4 3.5 10
6 6 10
d
(a)
(b) (c)
(d) (e)
(f)
—
4.7 +3.3 2.8
3.5
—
2 3 8 10 3
—
—
—
—
6V > "ÿrc2 ~ 10 6 = 4V, = 0V, X)E\2 = —0.7V > VB 2 = 0.7 + 0.7 0V (or lower), and i'c2 = 4V/4k£2 = 1.0mA. Therefore i'ci = 2.0  1.0 = 1.0mA, and t>ci = 10  4(1) = 6V. For equal current split, \)B\ = vB2 = 0V. 2.0 \) > X)gi2 = 2.0 0.7 1.3V. Now t)ct = 6V > id = (10—6)/4 = 1mA, and ic2 Vji = x>B2 = 2.0  1.0 = 1.0mA, and oC2 = 10  4(1) = 6V. x>EI2 = 1.3V. Thus one of Ogj, mB2 = 1.3 + 0.7 = 2.0V. Therefore, \>B\  2.0V. Now, since X>B2 = 1.0V, ici = 2.0mA, and Dei = 10  4(2) = 2V. Also i'c2 = 0mA, so x>c2  10V. t)j
—
— —
—
—
i)£i2 = 0.3V. Thus one of vBi, \>B2 = 0.3 + 0.7 = 1.0V. Therefore vB2 = 1.0V, Q2 conducts 2mA and X)C2 = 10  2(4) = 2V, with X)a = 10V. x>E\2 = 2.8V. Thus one of oB1, x>B2 must be 2.8 + 0.7 = 3.5V. Therefore vB2 = 3.5V and iE2 2mA. Thus \>C2 = 10  2(4) = 2V, possibly. Therefore Q2 is saturated with uC2 = 2.8 + 0.2 = 3.0V, with extra current flowing in the base of Q2. But Q\ is cut off and \)ci = +10V. X>B 1 vB2 —4.0 V + \>£ 2 = —4.0 0.7 —4.7V, vE2 = 8V > iE2 = (10—8)/4k£2 = 0.5mA. Thus ic 1 = 2.0  0.5 = 1.5mA, and uc, = 10  1.5(4) = 4V.
—
— —
—
—
—
 251 
—
SOLUTIONS: Chapter #62
(g)
(h)
Uei2 = 3.3V. Thus one of t)B), vB2 is at 3.3 + 0.7 = 4.0V. Thus vBi = 4.0V, with Q\ conducting, Qi cut off, 0C2 = 10V. For iC\ = 2mA, dCi = 10 2(4) = 2V. But x>Ei = 3.3V. Thus Q ( is saturated with t)Ci = 3.3 + 0.2 = 3.5V. "0Ei,2 = 3.5  0.7 = 2.8V with Q\ cut off and Q2 conducting. Thus t)Ci = +10V, and v»C2 possibly as low as 10 2(4) = 2V. But vE2 = 2.8V. Thus uC2 = 2.8 + 0.2 = 3.0V.


Case a
b
e
0.2 0.2 0.2
f g h i
2.0 2.0 2.0 2.0
c
d
—
/ mA 0.2 0.2
ÿUfll
V 0.00 0.01 0.00 0.037 1.00 1.00 0.01 0.00 1.00
Vb 2 V 0.00 0.00 0.05 0.00 1.05 1.00 0.00 0.05
0.951
% V 0.700 0.695 0.664 9.675 1.714 1.758 0.752 0.722 0.228
Vci V 9.60 9.52 9.90 9.35 9.30 6.00
5.21 9.05 3.00
VC2 V 9.60 9.68 9.30 9.85 9.90 6.00 6.79 2.95 9.00


—» Dei
(a)
x>c2 9.60V » iC2 = (10.0 9.6)/4k = 0.1mA, that is iE2 = /  0.1 = 0.2 0.1' = 0.1mA = 9.60V. Since i'ci = 'C2> vbi = u«2 = 0.00V. Note that for iE = 0.1mA, vBE = 0.700V.
(b)
= = Oÿ3"1ÿ and '«> = vd=vBl vB2 = .01  .00 = lOmV. Thus iE2 = } i + 0.200  .0803 = 0.1197mA, v>B£1 = 0.700 + 25 In (0.1197/0.100) = 0.7045V. Thus x>E = 0.010 .7045 = 0.695V, t)Ci = 10  4(.l 197) = 9.52V, and \)C2 = 10  4(.0803) = 9.68V.
(c)
= 0.176mA, iE\ = 0.200  0.176 = 0.024mA, t)Ci =  x>B2 = 0.05V, iE2 1+ 10  4(.024) = 9.90V, uC2 = 10  4(.176) = 9.30V, x>BE2 = 0.700 + 25 In (0.176/0.1) = .714V. Thus vE = vB2  vBE2 = .050  .714 = 0.664V.
(d)
Assuming x>B\ > 0 and iE2 < 0.1mA, iE2 = 0.1 e i675700)ÿ5 = 0.0368mA, iE 1 = 0.200 0.0368 = 0.1632mA, uBE, = 700 + 25 In (0.1632/0.100) = 712.2mV. Thus uB1 = 712.2  675 = + 0.037V, and t)c, = 10  4(. 1632) = 9.35V, vC2 = 10  4(.0368) = 9.85V. ic2 = (10  9.90)/4 = 0.025mA. Thus iCi = .200 .025 = 0.175mA, and uCi = 104 (.175) = 9.30V See from (c) that x>BE 2 = 0.664V, and vBE 1 = 0.664V + .050V, and since uBi = 1.00V, x>B2 = 1.00  0.050 = 1.05V, and % = 1.05 0.664 = 1.714V. Inputs equal: current splits equally and j'ci = i'c2 = 2.0/2 = 1.0mA, Dei = 10  4(1) = 6.00V, uC2 = 10  4(1) = 6.00V. Thus uBE = 700 + 25 In (1.00/0.1) = 757.6mV, and % = 1.00 .758 = 1.758V. 20 20 / = 2ÿ92 = °803mA' and 'ei = vd = dBi  db2 = lOmV. Thus iE2 = ÿ t ÿÿ1035 ÿ 2.00 .803 = 1.197mA, X)BE 2 = 700 + 25 In (0.803/0.100) = 752mV, % = 0  .752 = 0.752V, DC = 104 (1.197) = 5.21V, X>c2 = 104 (0.803) = 6.79V. 20 = 1762mA, and vBE 2 = vd = vB,  vB2 = 0  .05 = 0.05V = 50mV. Now iE2 = ÿ
(e)

+ÿ492

urf = t>B
—

(f)
(g)

(h)

+ eso25

700 + 25 In (1.762/0.1) = 772mV, that is vE = vB2 vBE 2 = 50  772 = 722mV = 0.722V. 1.762 = 0.238mA, and vct = 10  4 (.238) = 9.05V, and VC2 = 104 (1.762) = Now (El = 2.0
2.95V.

252
SOLUTIONS: Chapter #63

t>o = 3.00V + iCi = (10 3)/4 = 1.75mA. Thus iC2 = 2.00  1.75 = 0.25mA, VC2 =104 (.25) = 9.00V, vBE = 700 + 25 In (1.75/0.1) = 771.6mV, vBE 2 = 700 + 25 In (0.25/0.1) = 722.9mV. Thus x>E = 1.00  .772 = 0.228V, and x>B2 = .228 + .723 = 0.951V.
(i)
,
SECTION 6.2: SMALLSIGNAL OPERATION OF THE BJT DIFFERENTIAL AMPLIFIER , \>j/l Vr a Ie 6.4 Using e* = l+x+x2/2 in Eq. 6.11: 'ci = =a/ r +e u] qI Vd vd . vd Vd 1+ / 1+ / 1— + 8 2 VT 2 VT 2 2 VT 8 Vr 8 Vj VT 4
a/ 2 a/
1+ 1+
2
xx2 vd Vd 2 VT 8 Vr2 vj Vd + 2 VT 8 Vr2
Now for
1
Vd_

vd/2 = lOmV, i 1 
 0.08  0.016 = 0.904.
1
16 Vr3 2
vd 2 VT
1
Vd 64 Vr4
vd 2 VT
Vd 2 Vr
1
4
8
Vr2
64
Vr4
8»Vr
2
vd
1 4
2Vr
•1 J
"1—
{Rather than 1.000 for a linear system.}
Alternatively (and directly) at 2
1+
16 Vr3
*
vd/2 =
2 e 1075
a/
lOmV: j'ci =
g 1025
of which the signal part is 0.380 —r— 2
.
2
10 25
That is, we see that the higherorder approximation implies a reduction in output from that derived from the linear one.
1.380
ÿ2
Vd 2 Vr
 8VdV/
Vd 8 Vr2
1
2
1+
ÿXX2
1
_
a / «L + 2 2Vt
1
2
10 25
current by
«I
+ c1075
1
.4
about 10%
2 (1.492) 1.492 + 0.670
whereas from Equation 6.12:
'ci
qI
qI aI aI Thus the i+ = 1.400  of which the signal part is 0.400 ÿ 2 2 Vr linear approximation produces a result which is high by (0.400 0.380)/0.380 = 0.053, or about 5%. For ÿspecified errors, using the result (1) above, but only the term in \)d, we see that the error is about
vd /VT = e. Now for e = 10%, y/Vr = (2 (0.1))'/* = 0.447, and x>d!2 = 0.447 (25) = llmV. 2 For 5%, vd/2 = (2 (.05))'/l (25) = 7.9mV. For 1%, vd!2 = (2 (.01))" (25) = 3.5mV. Check with the original (Eq. 6.11): 'ci =
2e 3.575
q/
e
3.575
+ C3.S75
=
iE i =
\>./i
'
1.150 + .869
— (1 + vd/2 VT) = ~9l + 3.5/25) = 6.5
e
2(1.150)
q/
2
2e
q/
»,/2Vr
vr + e x>. a vTr ' T
_
~
q/ '
. For vd/2 = 3.5mV, , whereas, from Eq. 6.12: i'ci =
(1.140). Thus the error is (0.140
 0.139)/0.139 a 0.7%.
For a differential input of 0V, iCi = iC2 = 112 = lOOpA, and y>C) = 30.1 (10) = 2V = x>c2. For IE = a lOOpA, = 250 = 40V/V. For out= = 25012. For differential output, ÿ ' ree + 250 vd IE 0.1mA puts individually, the gain magnitude is 20V/V. For vCB = 0.4V and very small signals, v>ci = VC2 = 0.4 = 2.4V. That is, the upper limit of the input range is +2.4V. 2V, and t)fii = vB2 = 2.0
——
—
 253 
SOLUTIONS: Chapter #64
Rl re
lOOJfcQ 0.25k Q. 400V/V. Differential input resistance is (p + 1) (re + re) = 151 (0.25 + 0.25) = 75.5k£2. To double the input resistance, add 0.25kf2 resistors in series with each emitter, at which point the gain (for differential output) is (100k+100k)/(250+250+250+250) = 200V/V.
6.6
For I= 200pA,
re =
25mV/100(xA
= 25012. For differential
output, gain is a
6.7 For each transistor, IE = 200jiA and re = 25mV/0.2mA = 12512. Thus the differential input resistance is 2 (201) (0.125k£2) = 50.25k£2. 0i
iokn
Vol
iokn
Vo2
iokn
iokn VI1
For differential output: Differential gain from bases = 200/201 (10k/0.125k) = 79.6V/V. Differen¬ tial gain from input sources = 50.25/(10+10+50.25) x 79.6 = 56.94V/V. Commonmode gain = 0V/V. CMRR = 56.9/0 = <», as ratio and in dB. Commonmode input resistance = (P + 1) (R) = 201 (0.5M£2) = 100MI2. For singleended output:
...
10k£2 200 201 0.125 + 0.125 2"wv ÿ
(J r) 400pA
\A
 w
CMRR = 28.5/0.00995 = 69.1dB.

50.25 70.25
56.94 2
'OOW5V/V' 9.95 mV/V. 2864V/V
ÿ
From P6.7 above, for outputs taken differentially, Aj = 56.9V/V. For Acm : For matched loads, it is 10k£2/lM£2  10k£2/lMI2 = 0V/V. For ± 1% loads, it is 10k(1.01)/lM  10k(0.99)/lM = 0.02(10k)/lM = 2 x 10"5V/V ÿ 94dB, for which CMRR = 56.9/(2 x 10"5) = 2.85 x 106V/V ÿ 129dB. For ± 10% loads, correspondingly, Acm = 2 x 10~4V/V ÿ 74dB, and CMRR = 2.85 x 105V/V ÿ 109dB
6.8
Note that the collector resistors are ideal (that is both are exactly 10k£2). For DC Bias: Assume that junction voltages arc adequately modelled by rel, re2~ 25mV/200pA = 0.125k£2 nominally. Now for / in (2i and 0.400 /
iokn

in Q2, 10 (1.1) V2
P

0.5MQ
(200)(0.S)
P

+ 10 (9)
onl
2oo79j
ZUU (1. JJ
•
+ °'125 f = °'125 (04 " ,)
Now multiplying by 100: 6.11

/ + 12.5/ = 5  12.5/ + 1.636 4.09/, 35.2/ = 6.636, / = 0.1885. That is iE\ = 188.5xA, and iE2 = (200)(1.1)
2U.511A.
 254 
SOLUTIONS: Chapter #65
,
For signals: Now rei = 25mV/l 88.5iA = 132.60, re2 = 25/211.5 = 118.20, rK = ((200) (.9) +1) 132.6
= 24.0kO, rn 2 (200 (1.1) +1) 118.2 = 26.1k£2. Now for commonmode input, base currents split, with 26.1/(24.0 + 26.1) = 0.52 of the input change in the base of Q\ (and 0.48 in the base of Q2). Correspondingly, for a total base current i, ici = Pi 9>i = (200) (0.9) (0.52)/ = 93.6/, and ic2 = p2 ih2 = (200) (1.1) (.48)/ = 105.6/. Thus ici/ici = 105.6/93.6 = 1.128. Note that the dc current ratio is
211.5/188.5 = 1.122, essentially the same. From either point of view, there will be about a 12% total x lOÿV/V. mismatch in the two output voltages. Since = Acm ~ ÿ (SQQkQ,) X 479'7 lOfcfl 10IcCl 79>7V/V CMRR = 66.4 x 103V/V ÿ 96.4dB. .133 + .118 12 X 10
_
_
"
_—
Ibias = 400(xA, Iei = Iei = 200pA nominally, and re = 25mV/0.2mA = 125£2, Re = 9 (125) = 1.125k£2, and Re + re = 1.25k£2. Using the dc analysis from P6.9 above with iE\ =/, see that 10 (1.1/ )/(200(.9)) + 1.25/ = 1.25(0.4 /) + 10(0.9) (0.4  /)/(200(l.l)). Multiplying by 100, 6.11/ + 125/ = 50 125/ + 1.64  4.09/, or 260.2/ = 51.64, and / = 0.1985. That is, iE\ = 198.5pA and iE2 =
6.10 For

= 1015. Thus there is about a 1.5% mismatch, such that Acm ~ 1m£2 x 198.5 79 7 , = 53.3 x 104 1.5/100 = 1.5 X lOÿV/V with Ad = 79.9V/V, the same as before, and CMRR = 1.5 x 10"4 & 114.5dB. Note that there is a nearly 20dB improvement due to the balancing effect of the emitter 201.5uA, with
iE i
resistors.
SECTION 6.3: OTHER NONIDEAL CHARACTERISTICS OF THE DIFFIERENTIAL AMPLIFIER 6.11 For Ibias = 200(iA, reX = re2 = 25mV/0.1mA = 250£2. For the Basic Amplifier, differential gain = Rc "i" Rc = 4 Rc~VIW. Now ±5% variation in Rc produces an output offset of 0.1mA (1.05/?c — 0.25k £2 + 0.25k £2  0.95Rc) = 0.1•01(0.1)/?c = 0.01/?c Corresponding input offset (to Areduce output to zero) is Rc Rc 2.5mV, or from equation 6.49: IV05I = Vj= 25mV (2 (5/100)) = Vos  Volgain = —— 4 Rc Rc Rc = 2.5mV. For emitter resistors Re = 9re: Here, the differential gain = n (1+9) = 0.4RC. Now 0.25 .01Rc to compensate an output offset of 0.01/?c> we nee(l Vos = = 25mV. 0.4Rc
—
——
ÿ
<,)
6.12 From P6.ll above, to compensate for ±5% Rc variation, one needs a 2.5mV input offset with no emitter resistors. This involves an increase in one of the collector currents to 105pA and decrease in the other to 95iA. Now with Re = 9 re = 9 (250) = 2250£2 nominally, but actually ranging from 0.95(2250) = 2.1375k£2, to 1.05(2250) = 2.3625kO, equivalent offset can reach .105(2.3625) .095(2.1375) = .2481 .2031 = 45mV. Total maximum offset is approximately 2.5 + 45 = 47.5mV. For uncorrelated variation: For nominal Re (from P6.ll above), acquire 25mV due to Rc variation. For nominal Rc and varying Re, to achieve lOOfiA in each transistor, we need an offset of 0.1(2.3625) 0.1(2.1375) = 22.5mV. (See the worst case is again 22.5 + 25 = 47.5mV, as an alternative approach). For no correlation, Vos = (22.52 + 252)y' = 33.6mV. For collector resistors trimmed: Collector currents will be both 100xA and, as above, Vos = 22.5mV.

Vos = (22 + 22 + 22 + 22)'/l = (16)'7' = 4mV. Vos = (0.52 + I2 + 22 + 42)* = (.25 + 1 + 4 + 16)y' = (21.25)* = 4.61mV.
6.13 For equal 2mV offsets:
For unequal offsets:
6.14 For the offset totally compensated, the collector currents in both transistors will be 100/2 = 50pA. Assume pi is 5% high and p2 is 5% low, while Rsi is 5% low and Rg2 is 5% high. Thus the total offset
 255
SOLUTIONS: Chapter #66
is Ib\ Rs\  hi Rs2  ,50/(105 + 1) ? 100 (.95)  50 /(95 + 1) x 100 (1.05) or 50 x IP"6 x 100 x 103 .95 1.05 = 50 x 10"3 (.896  1.094) = 50 x 10~3 (.198) = 9.9mV. 100 1.06 .96 Thus the offset can be as large as 9.9mV.
6.15 For each transistor, re = 25mV/150pA = 0.1667kft. For \>Bc = lQmV, t»c = 60kO/.1667kO X lOmV = 3.6V. Thus, the lowest collector voltage is 15V  150pA x 60k£2 3.6V, or 15  9  3.6 = 2.4V. For bare saturation, the base voltage can exceed this by 0.7  0.4 = 0.3V. Thus, the highest usable commonmode input is 2.4 + 0.3 = 2.7V.
SECTION 6.4: BIASING IN BJT INTEGRATED CIRCUITS 6.16 As seen from the emitter, IE = 100iA and re = 25mV/100pA = 250 fl.
Thus the resistance between the
terminals is 25012.
For two in parallel, the current divides (say equally) with each re = 25mV/50iA = 50012. The parallel resistance is then 50012150012 = 25012, as before. One can see this directly since the junctions are bigger, but the current is the same.
For two in series, the current in each tance is 250 + 250 = 50012. 6.17 From Eq.6.63,
For 0.1% error,
Iref
is the same; the resistance of each is the same, and the total resis¬
_!_ For 1% error,
1 + 2$
= 0.999, p =
= 0.99, 1 = 0.99 + 1.98/p, p = 1.98/0.01 = 198.
1.998 = 1998. 1  0.999
6.18 At 1mA, VBE =700 + 25 In (ImA/lOmA) = 642.4mV. Required r = (642.4mV)(0.1)/lmA = 64.212. Use r = 60 12. For P = 90, IR = 1mA and I0 = 1mA, IB2 = 1/90, IEl = (1  1/90) = 0.9889, 1mA and /ci = 0.9778 mA, lE2 = 91/90(1) = 1.0111, and IC2 = 1.0000mA. VBEt = 700 + 25 In (0.9778/10) = 641.9mV, VBE2 = 700 + 25 In (1/10) = 642.4mV, VrX = 60 (.9889) = 59.3mV, Vr2 = 60 (1.0111) = 60.67mV, VBEX + VrX = 641.9 + 59.3 = 701.2mV, Vbe2 + Vf2 = 642.4 + 60.67 = 703.1 mV. Thus rx must be increased 6419 to = 61.8912. For P = 90, I= 0.5mA: lB2 ~ 0.5/90,
7°3;*~ 0.9889
and IE = 0.5  0.5/90 = 0.5 (0.9889) = 0.4944mA, and Icx = 90/91 (.494) = 0.489mA, and VBE\ + VrX = 700 + 25 In (0.489/10) + 0.4944 (61.89) = 624.55 + 60.67 = 655.1 mV.
655.1  625.1 Assume la ~ 0.5mA > VBE2 = 700 + 25 In (0.5/10) = 625. lmV. Thus IE2 ~ 60 mA, and /C2 = 90/91 (0.500) = 0.4945mA. Gain = .4945/.5000 = 0.990A/A. For P = 90, IR = 2.0mA: IB2= 2/90, IEX = 2 2/90 = 2 (0.9889) = 1.9778mA, and Icx (1.9778) = 1.9561mA, and VBEX + VrX = 700 + 25 In (1.9561/10) + 1.9778 (61.89) = 781.6 mV. 7816 659 8 = 2.030mA, IC2 IC2 ~ 2mA + VBE2 = 700 + 25 In (2/10) = 659.8mV, lE2 =
—
—
= 0.500
= 90/91 Assume
= 90/91
(2.03) = 2.008mA. Gain = 2.009/2.00 = 1.004A/A.
For p = 70, 0.5mA: lB2 = 0.5/70, IEX = 0.5  0.5/70 = 0.5 (0.9857) = 0.4929mA, Icx = 70/71 (0.4929) = 0.4859mA, VBEX + VrX = 700 + 25 In (0.4859/10) + 0.4929 (61.89) = 654.9 mV. Assume lC2 ~ 0.5mA 654.9625.1 = 0.4967 mA, IC2 = 70/71 (0.4967) = + VBE2 = 700 + 25 In (0.5/10) = 625. lmV, IE2 = 60 0.4897 mA, gain = 0.4897/0.5000 = 0.979A/A.
 256 
SOLUTIONS: Chapter #67
And at 1.0mA; IB2~ 1.0/70, IEl ~ 1.0  1.0/70 = 1.0 (0.9857) = 0.9857mA, /c, = 70/71 (0.9857) = 0.9718mA, VBEl + Vri = 700 + 25 In (0.9718/10) + 0.9857 (61.89) = 702.7 mV. Assume 702 7 642 4 = 1.005mA, /C2 = IC2 = 1.0 + VBe2 = 700 + 25 In (1/10) = 642.43mV. Thus IE2 =
'60
70/71 (1.005) = 0.991, gain = .991/1.00 = 0.991A/A.
And at 2.0mA: IB2 ~ 2.0/70, lE\ = 2.0 (0.9857) = 1.9714mA, 1C\ = 1.9714 (70/71) = 1.944, VBEX + Vrl = 700 + 25 In (1.944/10) + 1.9714 (61.89) = 781.1. Assume IC2 = 2.0mA + VBE2 = 700 + 25 In (2/10) 7811 ~ 659 8 = 2.021mA, IC2 = 70/71 (2.022) = 1.993mA, gain = 1.993/2.0 = = 659.8mV. Thus IE1= 60
0.996A/A.
619
T~ = 1 + 2f' Io = 100Mj4
1+
1/150 =
98684ÿA Thus
which rQ = 150V/100(xA = 1.5MS2. To compensate,
ÿ ÿ
x
l0
is low by 100  98.684 = 1.316)lA, for
'jO6 =
X
ÿ~6' or
=
(1ÿ1ÿ) =
1.974V. For a net error of <1%, output current must range from 99pA to 101iA. At 99i.A, r„ contributes 99 98.684 = 0.316J.A, for which Vou, = .316 x 1.5 = 0.474V. At 101(xA, r„ contributes 101  98.684 = 2.316ta, for which V,mt = 2.316 x 1.5 = 3.474V.
—

6.20 For a change from 25°C to 75°C, VBE drops by (75 25)2 = lOOmV. For lOOmV to be a 5% change, the drop in R must be 100mV/.05 = 2V. That is, Vcc = 2.7V, and R = 2/100pA = 20k£2. 6.21 (a)
1mA
1mA
0.5mA
Need 9 BJTs.
 257
2mA
SOLUTIONS: Chapter #68
(b)
0.5mA
(c)
Need 10 BJTs. For both ends of Ir available, T& is not needed. Require 9 BJTs.
6.22
Io_ Ir
PL p« +
or
2i P+l
[o_
p'+t&tI EI
P+ p+l
1
Ir
1+
before.
as
P2+P
For two outputs, I0 3i , and Ir = P' + ÿ
Io
_ =
ÿ"1+
P+l 1
P2 + P
6.23 1
Jin
J
/a
jio Iq
h
Pi+1 P2+I P3+I P2 P2+I P. 1 1 + Pi+1
P3+1
P2+1 1 + Pl(p2+D Pl+1 + P2(Pl+D P2 (P.3+1)
 258 

Kp,+l) + KPz+l)
(1)
noted
SOLUTIONS: Chapter #69
For the optimal location: See a) Q3 provides a secondorder effect, and any P would be OK. b) For Q i with lower P, icl is fixed and iEt will increase, causing VBEl to increase, and Io to increase, c) For Ql with higher P, l0 is larger for a fixed VBE. Conclude that one should make pi = (1  k) P,
P2 = (l+k)p, p3 = p.
Substitute in (1)
I_o_
___
P + kP+1
Ir (lk)p(p + kp+ I)
Pkp+1 (p + JkP)(P+l)
(l+k)p (Pkp+1)
_
p + kp + 1 +P&P+ 1
(1 k)(P + kP + 1) (1 + k) (p  kp + 1)
p (k + 1) (p + 1) (P  itp + 1)
1
(lk)(P + kP+l) (1 + k) (P  kp + 1)
2(8+1)
p (* + 1) (p + 1) (p  *p + 1)
_! (1 k)(P + kp+l) (1 + k) (pkp+ 1)
=
_ 2_
p (1 +*)(pJfcp+ 1)
P2  k p2 + p + k P2  k2 p2 + k P p2 + Jtp2 + p  itp2  k2 P2  k P + 2
(p + tp)(ptp+ 1) (P*P)(P + *P+ l) + 2
p2 + p + kp  k2p2 2r2 ' P2 + P  *P + 2  Jfczp
This becomes one, if kp = 2  kP, or kp = 1, or k = 1/pi Note that for k = 1/p, pi = (1  1/P)P = p  1, and p2 = (1 + 1/p = P + 1, with
P3 = p.
6.24 a
—P3+1 LP2+IJ L_
j
J
Pl+1
P2+2 JR
.
P3+I P2+I 1 P2 P2+2 P3+1 P2+l P2+I
ii + 1
P2 •*" 2 P2 + 1
p'
___
Pi + 1
,c
P2+1
=
P.i (P2+2) P2+2 + p2 (P3+I)
=
_ 1 P2+I
1
Pi + 1 1
P2+ 1
p2 + 1
+1
1
P2 + 2 P2 + 1
=
1
Pi+1
P2 P3 + 2p3 P2 P3 + 2p2 + 2
1 1+
2 (P2  P3 + 1)
P2 P3 + 2P3
For optimal placement of transistors: See (a) that Q\, being diodeconnected, Pi does not matter, (b) that for Q2 with low beta, iC2 being fixed, iE2 increases, VBE increases, and iE 1 increases, (c) that for p3 high,
I0 increases.
Thus,
use
pi = P,
P2 = (1  k) P, P3 = (1 + k) p.
 259 
Correspondingly, 'R
=
SOLUTIONS: Chapter #610
_
(1 + k) (1 k) p2 + 2 (1 + fe) p P2 P3 + ~ p3 2 (1 P2 + 2P2 + + lc) (1  k) p2 + 2 (1  k) P + 2 if 2&P = —2£p + 2, or &P = !4, or k = Zp
.
p2  k2P2 + 2frp + 2p p2  k2 P2  2/fcp + 2p + 2 '
'S 1S
Umty
6.25 Assume very large P: For the lOOpA reference: VBE = 700 + 25 In (0.1/1) = 642.4mV. For the lpA output: VBE = 700 + 25 In (103/l) =527.3mV. For the lOpA output: VBE = 700 + 25 In (10~2/1) = /642 4 527 31 x 10"3 584.9mV. Thus, for lpA output, RE = K = H5k£L For lOpA output, _6
—
E
(642.4  584.9)
103
10"5
SECTION 6.5: THE BJT DIFFERENTIAL AMPLIFIER WITH ACTIVE LOAD 6.26 Each transistor conducts 50pA, for which r„ = 150/50pA = 3M£2, and re  25mV/50iA = 50012, Gm = 2/(500+500) = 2mA/V, (3M12II3M12) = 2 x 10"3 x j x 106 = 3000V/V. =~ 5QQ 5QQ R„ = 3MS1W3M£2 = 1.5M12, Rin = (75 + 1) (500 + 500) = 76k£2. For a 76kl2 load, A„ = 2 x lO"3 (1.5A/12H76&12) = 144.7V/V.
ÿ
6.27 For each of the transistors, with an emitter resistor RE = re, the output resistance increases to R„ = r„ (1 + gm Re') ~ 3 X 106 (1 + 500/(500)) = 6M12. Thus, the output resistance is approximately 6M12 11 6M12 = 3M12. The overall transconductance is 2/(500+500+500+500) =, ImA/V. The opencircuit voltage gain is
 ImA/V X 3M12 = 3000V/V.
—
6.28 / = lOOpA > Collector current for all transistors is 50xA, for which r„ = 75V/50pA = 1.5M12, and re = 25mV/50pA = 50012, and r„ = (75 + 1) (0.50) = 38kl2. For the cascode transistors, Ro = r„ (1 +g„, rK) = r„ (p + 1) = 1.5 X 106 (76) = 114M12. Also r = 10 (75) (1.5 X 106) = 1125M12. Thus the output resistance (in Mil) is 114 1 1 114 1 1 1125 1 1 1125 = 114/2 1 1 1125/2 = 51.8M12. Overall, gm = Gm ~ 2/(500+500) = 2mA/V, A„ = 2mA/V x 51.8M12 = 103.5 x 103V/V.
„

SECTION 6.6: MOS DIFFERENTIAL AMPLIFIERS 6.29 Here, gm =
ÿ
—— Vcs ~ "t
ipi = iD2 = 1/1= K (VGS
 V,)2
and gm ÿ
ÿ 2IA.K), or gm max
(Kl/Z)
iD = K (Vgs  V,)2, and
K(Vgs±ÿV,) =
2
(2), where K = Vlk = V2kn(W/L), and
= 2K (VGS
gm = 2K (vGS V,) = 2K (VCs
0.9 (2K) (VGS
Vu or „  V,) = ± ÿ, 2 VGs  v, _ .."T = ± 2(.01) = ±0.02. .. VGS
0.1 (Vgs
6.30
(1), with a maximum at
for each transistor
 V,),
when
 V,), ±
VGS
 V, = EL 2
(from (1) with (2)).
J" ~ F/)
For
a
10% gm
VGS±~~Vt = 0.9 VCs
= ± 0.2. For 5%:
ÿ
„Vgs  vi l7
In general,

drop,
0.9 V,, or
= ± 2(.05) = ± 0.1. For 1%:
 V,)2 = K(x>GS  V,)2, where K = '/: \lpCox{W/L) = '/2 x 10 X 120/6 = 100pA/V2. For equal current division, 25/2 = 100 (Vgs  l)2. Vgs  1 = ± (l/8),/4, and vCs = 1.354V. Thus VGS = 1.35V. Now, gm = IK (Vgs  V,) = 2 (100) (1.354  1) = 70.8pA/V, and r„ =  = = 4M12. iD = V2kp(W/L)(vas
/£)
260
LJ/L
SOLUTIONS: Chapter #611

2 (4 x 106) Maximum gain will occur for outputs taken differentially: (a) For ideal loads: A,, =  = 2 (1/70.8 x 10"6) 6 6 2 (4 x 10 4 x 10 283.2V/V, (b) For loads with VA = 50V: AX)= = 283.2/2 = 141.6V/V. 2 (1/70.8 X 10"6)
>
II
6.31 With balanced loads, ip =
 V,)2, with i)cs = 1.354V (from P6.30 above).
= 100(dCiS
 l)2, uGs = (~f + 1 = 1.3354V. For iD = 1.1 (25/2) = 100 (vGS ÿ°cs = (~3~)/l + 1 = 1.371V. Thus the input offset = 1.371  1.335 = 36mV. = 0.9 (25/2) = 100 (X)Gs
iD  l)2,
Now for
O
6.32 From Exercise 6.15,
k'n = \inCox = 20pA/V2, (W/L ) = (12015) = 20, and V2k'n(W/L) =
1.25V, g„,
VGs
 V,
f<w = Vgs
 V,
V,
100iA/V, and A Rp 1.25  1
=
100
Rn A(W/L)
_
1.25 2
(W/L)
 1 .. _j_
100
IV.
=
For
an
RD
200pA/V2, with mismatch of ± 1%,
= ±1.25mV. For a (W/L) mismatch of ±1%, Vos =
= ±1.25mV. For V, tolerance of ±0.6mV, V0s = ±0.6mV.
Thus the worstcase offset is 1.25 + 1.25 + 0.6 = 3.1mV. The likely offset = V 1.87mV.
1.252 + 1.252 + 0.62 =
6.33 Here, I0 ~ Iref = 100pA, and K = V2k'(W/L) = 1ÿ(200) = 100\iA/V2. Roughly: iD = K (vGS V,)2, 100 = 100 (vGS  l)2 > vGS = 2V. Now, for Qu Q4> x>DS = 2V and \)ps 2 (1 + )= 1+ = 1.1. This could be ignored, but let us include it in a more basic calculation:

——
—
and 100 = 100 (u  l)2 (1 + 1/20), or 1 = More precisely: For Q\, Q4, Vp$ = vGs = .05i)3 i)2 t) i) (i)2  2 4 1) (1 + .05u). Thus  2 + 1 +  O.li)2 + .05u = 1 or .05 i)3 + 0.9 D2  1.95 1) = ÿ i) i) 0, or .05 u2 + 0.9  1.95 = 0, = (0.9 ± 0.92 + 4(1.95) (.05))/(2(.05)) = (.9 ± 1.095)/0.1 = 1.954V. Thus VGSi = 1.954V. Now for V0 = V03 = VD4, see IQ = 100iA, from symmetry arguments. Va 90 For each device, r„ ~ ~P~ = —~r = 200k£2, and gm = 2K (vGS  V,) ~ 2 (100 X 10'6) (2  1) = 100i/\ //> 200(xA/V. For the whole mirror, (from Eq. 6.116) R„ul ~ r„3 ro2 = 200 x 10"6 x 200 x 103 X 200 X 8MQ. 12V, Thus, 106£2 the standard for with X 103 = 8 output being at 2 + 2 = 4V, the extra = V0 = current = (12  4)/8M12 = 8/8 = l(iA. Thus lo  101p.A for V0 = +12V.
lo  Iref
 100pA.
VD1 = 2.0 + 2.0 = 4V. Correspondingly, for Q\, with x>Gs = X> ~ 2.0V and Vps ~ 2V, lOOpA = 100pA/V2 (u — l)2 (1 + ÿ), 1 = (d22d + 1) (1 + 0.1d) = i)2 — 2o + 1 + O.li)3 0.2d2 + O.lt). or + O.li)3 + 0.8u2  1.9i) = 0, ÿ( O.li)2 + .81)  1.9 = 0, i) = ~° 8 ± = L916V Thus> uC5i = 1916V = 1.92V. Now
6.34 Assume
From results of P6.33 above,
 l)2 (1 +
~ 2.0V. Thus
—
2()4(1
for Q2, ip = 100 (1.916
VGs
= 91.9pA. Now, for Q3 with VG ~ 2 (1.92) = 3.84V, 10 =
91.9jlA. From P6.91 of the Text, R„ ~ (gm r„)r„ V0 = 12V, l0 = 91.9 + (12  3.84)/8 = 92.9pA,
261


8M£2 using the results of P6.33 above. Thus for
SOLUTIONS: Chapter #612
6.35
10,; I
Generally speaking, see that currents in Q3 and Q5 will be the same, provided the output voltages are the same. Further, the current in each will slightly exceed 100/2 pA since the voltage at node X will rise slightly due to doubling of the equivalent K of Q3, Q5. Since the current in each of Q 3, Q5 is only slightly more than half what it was, each output resis¬ tance will be twice as large. In particular, with outputs joined, the output resistance will be only slightly smaller than before: Now to check these ideas: iD2 ~ lOOpA = (100 + 100) (\>GJ  l)2. Thus \)CS = 1 4 = 1.707V, rather than 2.0V previously. Thus node X will rise about 0.3V, and iD3 will increase by 0.3/200kf2 = 1.5)0,A. The current in Q3, Qs will be about 5O.Sp,A, for which gm = 2K (Ugs  V,) = 2 (100 x 10~6) (1.707 1) = 141)oA/V. Now for outputs joined, r035 = 20/(100 + 1.5) = 197kQ.
I
'
""
OS

Consider r02 in two parts of 400kf2 each, with R03 = R05 = 197k x 400k x 141pA/V = 11.1MH. Together, R035 = 11.1/2 = 5.6MI2. When outputs operate independently, the output resistance decreases a lot since node X is grounded via l/gm of the other output. From Eq. 6.116 (full version) /?03 = r03 + VgmS + gm3 r03 l/gm5 ~ 2 r03 = 800kQ. An improved circuit would split both Q3 and Q2 into two parts, with each pair of halfsize transistors driven from the gate of Q 4, Qt respectively. The total device width needed would be same as in the original design, that is 4W fqr each original transistor of width W. The version with only Q3 duplicated uses a total width of 5W and has poorer performance!
IB = 200p.A, ID for each transistor = lOOpA, r„ = VA/ID = 20/ lOOpA = 200kf2, iD = V2k'(W/L) (VGS  V,)2, 100 x 10"6 = 1/x 200 x 10'6 (vGS  l)2 ) VGS = 2V, 2 gm = k\W/L)(vGS  V,) = (200) (2  1) = 200pA/V . Gain A„ = = 20V/V. Gain
6.36 For
reduces by a factor of two for a load of 100kf2.
SECTION 6.7: BiCMOS AMPLIFIERS 6.37 (a)
b)
Here, ic = lOpA, g„, =
10 x
ÿ
in6
= 400pA/V, r„ = lOOV/lOpA = 10MI2, /?, = \Vgm = 100/400 x
10~6 = 250k£2, Av = 400 x 10"6 x 10 x 106 = 4000V/V. For iD = lOpA = 'A X 20 X 20/2 (vGS  V,)2 = 100 (vGS  V,)2. Thus vGS  V, = (1CH00)'/' = 0.316, gm = k\W/L) (vGS  V,) = (200 X 10"6) 0.316 = 63.2pA/V, r„ = 20/1OpA = 2MG, /?, = 00, = 63.2 X 10~6 x 2 x 106 = — 126.4V/V, that is much, much less!
P2
100
6.38 For / = lOpA, gm2 = 400pA/V, rn2 = =
2MI2. From Eq. 6.116, Rnut ~ gm2 888 X
106ft. Thus
— = 63.2 x 10~6 x ÿu,
. = 250ka, r02 = 10M£2, gml = 63.2pA/V, r0i = 10"6 x 10 x 106 x (2 x 106 II 0.25 X 106) =
gm 2 400 X 10 r02 (r01 II rn2) = 400 x
101
x .888 X 109 = 55.6 X 103V/V.
6.39 For I= lOOpA, see by scaling from P6.38 above, gm2 = 4mA/V, rn2 = 25kf2, r02 = 1MS2, r0i = 200kf2, and \)G5iV, = (100/100)* = 1. Thus gml = 2 (100 x 10~6) 1 = 200pA/V, = 200 x 10"6 x 100/101 x 88.8 x Rout ~ 4 xlO"3 x 106 x (200 X 103 II 25 x 103) = 88.8 x 106, 106 = 17.6 x 103V/V.
 262 
SOLUTIONS: Chapter #613
6.40 From P6.37 at lOpA: For the BJTs, gm = 400pA/V, rK = 250kO, r„ = 10MO, rÿ = 10 (100)107 = 10 x 109£2. For the MOS, gm = 63.2pA/V, r„ = 2MO. For the circuit as shown, R„u, = 2 X 106 x 63.2 X 10~6 X (10 X 106 X 400 X 10"6 X 250 X 103) II (10 X 109) = 126.4 x (4000 x 250 x 103) II 1010 = 126.4 x .909 x 109 = 115 x 109O. With Q3, Q6 not used, Rom ~ 10'° II (107 x 400 x 10~5 x (.25 x 106) II (10 x 106) = (10 II .976) 109 = 0.889 x 109O. With Qs, Qi eliminated, Rou, = 2 x 106 x 63.2 x 10"6 x (10 x 106 II 10 x 109) = 1.264 x 109O.
SECTION 6.8: GaAs AMPLIFIERS 6.41 For a 1pm long, 1pm wide GaAs device, V, = 1.0V, p = Is = 105A, n = 1.1. (a)
Symmetry would indicate that Va = 5/2 = 2.5V, P = Pio = 10 (100) = ImA/V2, id = P (uGj V,f (1 + A, dds) = 1 x 10"3 ( 0  l)2 (1+0.1 (2.5)). Thus /„ = 1.25mA. Assume operation is in saturation. Lower transistor (Qi) operates with Mas  0. Upper one (Q2) l)2, or x>gs2 ~ (1/1.5Y' ~ 1 = —0.18V. Thus, io\ = l)2 = (0 is 1.5 X larger: Thus 1.5 (t)cs2 20 x 100 x 10~6 (0 l)2 (1 + 0.1 (5 .18)) = 2.964mA. See that the drain of Q2 is at about 5 lkll (2.96) = 2.04V, while the gate is at OV. See that operation is indeed in saturation. Check: l)2 (1 + 0.1 (2.96   0.18)) (Dc,y + l)2 = 0.752, dcj = 0.867 2.964 = 30 x 100 x 10"6 (oOT 1 =  0.132V OK. Vh = 0.13V, and Ih ~ 2.95mA.
 

(b)
100pA/V2, X = 0.1V1, VA = 10V,
—
—

(c)
See 2, operates at vGS = 0 with Vci = OV. Icl = 20 x 100 X 10"6 (0   l)2 (1 + 0.1(5)) = 3.0mA. Then /53 = lS2 = 3.0/2 = 1.5mA = Ic2, and VC2 = 5  2(1.5) = 2V. Thus, 1.5 = 10 (100 x 106) (Oas  l)2 (1 + 0.1(2)), or (\)CJ + l)2 = 1.5/(1(1.2)) = 1.25, x>Gs = 1.12 1 = 0.12V. Thus Vcl ~ 0.12V. Check: lc2 = 1 (0.12 + l)2 (1 + 0.1(2)) = 1.505mA. See that Q 1 is near cutoff, though it is larger. Assume Vd is near +5V, say at 5V, in which case l)2) (1 + .1 (5  0)) = 0.12mA. Now iD\ = 2 x 100 x 10~6 ( 0.8 (1 vDS). Let vDS = t), which is small, such that the X V,) (t X + iD2 = P 2 )Gs ~ x>DS  \)o5

(d)

I
x
_
and Id = 0.12mA. (e)

1) V = 10 x 100 x 10~6 [2 (0 ~~2± 4 4 (.12) 0062V Thus + 0.12 = 0, and D =
term can be ignored. Thus 0.12
o2), D2  2 D
J103
 


ÿ
1.5mA. 6.42 Here, Pi = 100 x
062 = 4.94V
iDl = 0.5
10"6 x 5 = 0.5mA/V2:
(0
  l)2 (1 + 0.1 (1)) = 0.55mA, for vDSi = V, .
For
V0 = 3V, VS5 = 5V, Vs2 = Thus (dot + U2 =
—4V, iD2 = 0.55 x 10"3 = 20 (100) x 10"6 (dgs   l)2 (1  0.1 ( 3   4)). 0.25, dgj = ± .5  1 = 0.5V. Vbias =  4  0.5 = 4.5V.
(c)
Lowest VG = 4V + j V, \ = 3V. For V0 = 3V, r„ of Q2 does not matter, the current being established by Q 1 0.55mA.
(d)
F,»m
(b)
_
 =5
l)2 (1 +0.1 (5)) = See that Q\ is turned on with Ve near OV. For Q2, iD = 10 x 100 X 10"6 (0 l)2 Thus, (0)) (0.2 (1+0.1 x 10"6 2.88mA. x 1.5mA. For Q2, iD = 20 100 = Q2 is in triode 1) [2 (0.2 2.0mA Correspondingly. ignored. 1.5mA small assumed and mode with x>DS = D, = 2,4 2 ~ 4('?5) D D2], and D2 2.4d + .75 = 0, D = = 0.37V. Thus Ve = 0.37V, and le =

(a)
— u2], or 0.12 = (2d 
ÿ
Eq. 5.122:
at a
value
I0 =
' 0.1 X 0.5(0 1)' * 20kft
lr = 20kQ. From Eq 5.121: gm = 2p (vGS  V,) (\ + X x>DS). Thus r =1) 0.1 X 2.0 (—0.5 gm2 = 2 (2) (0.5  1) (1 + 0.1 (1) ) = 2.2mA/V . Thus R„ut = gm2 ro2 roi = 2.2 x 10"3 X 20 X
—
 263 
SOLUTIONS: Chapter #614
103 x 20 x 103 = 880kO. (e)
Now for the output raised from 3V to +1V, ie by 4V, AI=
4V j = 4.5jA. 880 x 10
6.43 10
—
VBIm2
J
Add Q3 with IV3 = 20pm, P3 = 2mA/V2, p2 = 2mA/V2, P, = ImA/V2, V, = IV, A. = 0.1V',
.VO
(a)
»{q3 02
5.3V
(b)
n
(c)
T €V T
(d)

Use the results of Ex. 6.24. Since Q2, Q3 are the same size, V/«v,.v i = 5.3V permits vDSI = IV. Thus Vbias 2 = V/)(us + 1.0 = l)2 4.3V. Check: Now, for uDJi = IV, i = ImA/V2 (0 (1 + 0.1(1)) = 1.1mA. Now for Q2, 1.1mA = l)2 (1 + 0.1(1)), or 1 = 2 (t)cs + l)2, and 2mA/V 2(x>gs  1 = 0.3V. OK. 1 VvT .707 = = Vcs Now, Q3 remains in saturation for vDS S vGS  V, or vDS > 1 = 0.7V. Thus, the output can be as low as 4.3 + 1.0 0.3 = 3.3V.
,

Now as Q3 is operating just as Q2, VQ can go as low as 4.3 0.3 + 1 = 3.0V, at which point l0 = 1.1mA.
+
.
1 1 lOkfi, r02 , r = 10.2kf2 = r03, gm 2 = 2 (2) = = uz l)2 x2x(0.3l)2 0.1 0.1 x 1 x (0 (0.3 + 1) = 2.8mA/V = gm3. Thus, R02 = gm2 r02 r0i = 2.8 x 10.2 x 10.0 = 285.6k£2, and ÿ03 Sm3 ÿ03 ÿ02 = 2.8 x 10.2 X 283.6 = 8.16MO 4 For a 4V change in output voltage, AI= = 0.49pA. r01 =
—
—
(e)
—
6.44 See Qi and Q3 have the same width. Thus for VDD = 5V and VA = 2V, uDj3 uosi = {5—2)12 = 1.5V, and VE = VB = 5  1.5 = 3.5V. See 5 X 10~3 = IV x 100 X 10~6 (0  l)2 (1 + 0.1 (1.5)). Thus IV, = iv3 = 43.5p.rn, and IV2 = 43.5/2 = 21.7pm. Note that the diodes are sized to give 0.75V drop each 1 r„ = Now, for //2  V») (1 + A, vDS). Thus = 2.5mA. < gm 2P (vas " V,)2 P 1 1 roi = = 2.3k£2 = 'r03, = 4.61kQ. Also gmi = 2 (4.35) UJ' and r02 = 0.1 (2.17) (0 + l)2 0.1 (4.35) (0 + l)2 (0 + 1) (1 + 0.1 (1.5)) = 10mA/V = g„(3, and gm2 = 2 (2.17) (0 + 1) (1 + 0.1 (5  2)) = 5.64mA/V. 2.30 5.64 x 4.61 10 x 2.30 x + 2.30 t)/, 23.15 5.64 x 4.61 + 1 Now, from Eq. 6.132: a = — = = 0.926V/V. From 2 30 25 1 10 x 2.30 + + 2.30 2 Eq. 6.133: R„ = r—— = = 31.1kQ. From Eq. 6.134: R„ = r0i {gm3) ~ = 2.3 X 10 X M 2 2 1 .926 1a = 26.5kfl.
,  3OOA
 264 
SOLUTIONS: Chapter #615
04J 'DSSeq
Thus t)C52 = 0.7V, with X)DS2 = 3 l)2 x (1 + 0.1 (0.7)), or W{ =
ÿ ÿ

ÿ
jL = 4.67p.m.
Also 0.5 =
x W2 (0.1) (0.7  l)2 (1 + 0.1 (2.3)), or W2 = M X = 45.2pm. For x>DS = 6V, the current loss will increase, with 1)2 (1 + 0.1 *0), i>dsi = 3> increasing as well. Thus i = 0.467 (0 and i = 4.52 (v + l)2 (1 + 0.1 (6  v)), (1 + O.lv) = 9.68 (V2  2v + 1) (1.6  O.lv), 1 + O.lv = 15.5 v2  31v + 15.5  .968 V3 + 1.94 V2  .968 v, 0.968v3  17.34v2 + 32.07v  14.5 = 0, v» 18v2 + 33.13

j"j~Cg
>
 0.7 = 2.3V, 0.5 = W, (0.1) (0 
Ql
v  14.98 = 0.
18\>2 + 14.98  v3

= 0.543v2 + 0.452 0.030V3. Now, for v = 0.7: v 33.13 = .543 (.7)2 + .452  .03 (.7)3 = 0.266 + 0.452 .01 = 0.708; for v = 0.71: v = 0.543 (0.71)2 + 0.452 0.030 (0.7 1)3 = 0.274 + 0.452 0.011 = 0.715. See that the process does not converge: Thus reformu¬ v3 + 33.13v 14.98 , o (0.056\)3 1.84v 0.832)1/1 Now, for v 0.7: v (0.056 (.73) late it: v2 = = = = + 18 0.832)" 0.832)" (0.4752)" = = 0.689. See that now it will diverge for = (0.0192 + 1.288 + 1.84 (.7) lower values: Also for v = 0.8: v = (0.056 (.83) + 1.84 (.8) 0.832)" = (0.027 + 1.472 0.832)" = 0.818. Again we see that it also diverges, but that a solution lies between 0.7 and 0.8, but nearer to 0.7. Now, for V = 0.75: v = (0.056 (0.75)3 + 1.84 (0.75) 0.832)" = (0.024 + 1.38 0.832)" = 0.756; for V = 0.74: V = (0.056 (0.74)3 + 1.84 (0.74) 0.832)" = 0.743; and for v = 0.73: v = (0.022 + 1.343 0.832)" / = 0.7302. Solve this cubic iteratively:

D
=


.




Now conclude v = x>ds\ = 0.73V, for which Q\, iD = 0.467 (l2) (1 + .1 (0.73)) = 0.501mA, and for Q2, iD = 4.52 (.73  l)2 (1 + 0.1 (6 0.73)) = 4.52 (0.0729) (1.527) = 0.503mA. This calculation is very sensitive due to squared term. Use iD = 0.502mA.


analysis:
SmallSignal
for
Kut = rQ2 gm2 r0h
r„ =
which
ÿ
P (uG5  V,)1 ' r02 =
8m = 2
p (Vcs  V,) (1 + X X>DS )•
Here
1

0.1 (45.2 x 0.1 x 10"3) (0.7 1) 1) (1 21.4kft, gm2 = 2 (4.52 x 10"3) (0.7 =
24.6kn,
1 + 0.1 (3 0.7)) = 0.1 (.467 X 103) (0 + l)2 3.34mA/V. Thus R„ul = 24.6 X 103 x 21.4 x 103 x 3.34 x 103 = 1.76MQ. Now, the expected current increase as vDS rises from 3V to 6V is (63)/(1.76 x 106) = 1.7pA. The earlier estimate of change from 0.500mA to 0.502mA by 2pA is quite consistent. The smallsignal scheme is certainly more straightfor¬ ward!

r0i =

6.46 From the results of P6.45 above, for VDD = 6V and V0 ~ 6/2 = 3V, the upper composite would conduct 0.5mA, while the lower one would be biased at V/ = 0V to correspond. For each composite, the output resistance is 1.76M£2. For the amplifier, the output resistance is 1.76/2 = .88MQ, and the gain is 1) = 0.934mA/V, and the gain is gm Rout Here, gm = gm i = 20, (vGS, V,) = 2(0.467 X 10"3)(0 x 106 822V/V. 0.88 X x 10'3 0.934 =


6.47 Though it is not required explicitly by the specification, make all devices the same size. For Vcm ~ 0V, VDD = 5V, and Qi and Qj matched, the voltages across each will be the same, and 5 ~ ~05 V/M, = X)DS3 = = 2.25V. Thus 0.5 x 10"3 = W, (0.1 x 10"3) (0.5   l)2 (1 + 0.1 (2.25)), or W  =
0.5 0.1
——
°
X
11 r X ——— = 16.3pm = W2 — 1.225 (0.5)
265

Wj. Use / = 1.0mA: Now for 0.5mA in Q\, the
SOLUTIONS: Chapter #616
voltage at the sources of Q, and Q2 is 0.5V, and that at the source of Q3 is 0.5 + 2.25 = 2.75V, with 2.75 0.5 = 2.25V at its gate. Thus vDs2 ~ 2.25  0.5 = 1.75V. .Now, 0.5 x 10~3 = 16.3 (0.1 x Vi 1 10"3) («cs  l)2 (1 + 0.1 (1.75)). Thus Man = ± 0.5 1 = 0.5109  1 = 0.489V { 175 l63 (rather than 0.500V). Thus, there would be an llmV offset to maintain VQ at 0.5 + 1.75 = 2.25V. Note that since the drain voltages of Q\ and Q2 are different, as required by g.3 operating at vGSi & 0, while Q  and Q2 are matched and share a source connection, there must be an input offset voltage. Otherwise, for gate inputs connected, and the drain supply exactly half that of the source bias supply, the output will rise until Q 3 enters the triode mode and vGS2 = 0. For the offset acceptable, and nominal operation with iD ~ 0.5mA and vDS = 2.25V, and 3)G5 = 0.5V, gm = 2p (vGS  V,) (1 + A, x>DS) = 2 (0.1 x 16.3 x


ÿ
r> r>3\ (0.5 lO3) 0.1 (2.25)) = 2.00mA/V. r„ " (1 +.mm  1) i
/
1
=
0.1 (1.63 x 10"3) (0.5 + l)2
24.5ka
gm2
r02 + 1
gm 3 ÿ03 gm3
r03 + 1
(Ugs
Now
from
Eq.
6.139
 2.00 x 24.5 2 x 24.5 + 1  2 x 24.5 2 x 24.5 + 1 2 x 24.5 + 1
~gm I r01 gm 1 '*01 + 1
* P '  V,)2
.
of
the
Text:
49 50 50
3)„ 3),
= 49 (50) =
49 50
2450V/V. Alternatively, for W\ = W2, but W2 smaller, so vDS\ = vDS2 = 2.0V, and VGS3 = 0V with 33GJi = vGS2 = 0.5V, all operating at 0.5mA nominally. Find: 0.5 x 103 = Wi (0.1 x 103) (0.5 + l)2 (1 + 0.1 (2.0)), 1 1 0.5 X ; (1 = 16.7p.m. Now for Wy 0.5 x 103 = W33 (0.1 x 10~3 v(0 + l)2 1 v + 1.2 (0.5) x 0.1 (5  2)), or W3 = = 3'85ÿm' Now gm3 = 2 (0,1) (3,85) (0 ~ 1) (1 + 0,1 (3)) =
2=oTx
ImA/V, r03 =
—
of "k * TT l
—
0.1 (0.1) (3.85) (l2)
with
2.004mA/V,
,
r0i = '02 =
2.00 (24.0) 2.00 (24.0) + 1 1 (26) 1 (26 + 1) 2.00 (24.0) + 1
= 25.97k£2, and gm, = gm2 = 2 (1.67) (0.5
r= 23.95kf2. 0.1 (1.67) (0.5+ l)2 48 = 1296V/V. 49 49
Thus
  1) (1 + 0.1) (2)) = the
gain 5
is
—— 3),
26 27
SECTION 6.10: MULTISTAGE AMPLIFIERS 7
1E9 = 28.6ÿ = 0.5mA. Thus lc3 = 0.5mA, Ke = 100£1, Iqg = 2.0mA, Iqa ~ Ics = 1.0mA —> re = 25£2. Vbi ~ 15 — 3(1) = IGi lc2 0.25mA 15 ~ 12 7 12V, V£7 = 12 + 0.7 0. = 12.7V, lE1 = = 1mA + = 25a VBi = 15 + 1 (15.7) = 0.7V, V0 2.3
6.48 DC bias for
VBE = 0.7V, P = °°,
and ±15V supplies,
— —
= ov, IES =
015 = 5mA > 3k
re = 5Q.
_
3k 20* + 20k . 15.7* X x AC for p = oo: Rin = <*>, Roul 3kQ. II 5Q  5a Gam = rrr100 + 100 25 + 25 2.3* + .025* 3k = 200 x 60 x 6.75 x .998 = 80.9 x 103V/V. 3* + .005* 15.7* AC for p = 50: Rin = 51 (100 + 100) = 10.2kQ, Rout = 3kQ. II 5 + = 3kQ II ,313kQ = 28312 51 For gain: Secondstage Rj2 = 51 (25 + 25) = 2.55kf2, thirdstage Ri2 = 51 (2.3 + .015) = 117.6kQ, _ 40*112.55* 50 , , 3*11117.6* ,. 50 „ ..... x x x fourthstage Ri4 = 3k (51) = 153kQ. Thus the gam = — X } ÿ 2 50 2397 3* Z925 50 J434 50 50 157 II 153* ~ 2.325 3.005 3.005* 51 51 0.2 51 .05 2.325 51
*
—
_
*
 266
_S_
—
SOLUTIONS: Chapter #617
6.49 To raise icy from 1mA to 2mA, and ia from 5mA to 20mA, reduce R4 to 2.3kQ/2 = 1.15k£2, R5 to 15.7/2 = 7.85k£2, R6 to 3/4 = 0.75k£2. Now Ri4 = 101 (25/20 + 750) = 75.85kQ, /?5H/?,4 , 7.85* II 75.85* *6 „ 750 xr == 612™' A< ÿ7T«7 = '0Msy/y Now'
"'
_
_ 
12.5 + 1150
25—
50 (59.2) X (6.12) x 0.998 = 8099V/V, and R„ =
R6 II (reS + Rrfp + 1)) = Rl 750 II (1.25 + 7850/101) = 750 II 78.97 = 71.411 For load RL, loss is = 0.8. Thus RL = R + ?14 0.8Rl + 57.12, 0.2RL = 57.12, Rb = 28612. For a 28612 load, the upper swing is limited by Qy saturat¬
A2 A3 A4 = 22.4 x
A = A,
ing. Assume 0V between the emitter and collector of Qy, and look down from the collector of Qy. See an equivalent load resistance at the emitter of Qg of 28612 11 75012 = 20712 connected to a supply of (.286/(.286+.75)) (15) = 4.14V. At its base, see a resistor of 101 (207) = 20.9kl2 to a supply of 4.14 + 0.7 = 3.44V. The equivalent circuit is as shown: Looking down from the collector of Qy, see 20.9 II 7.85 = 5.71kl2 to 3.44 + (20.9/(20.9+7.85)) (15+3.44) = 11.84V.
I
Thus,
i
Q$ does not saturate). The maximum positive output is 10.47  0.7 = +9.77V. For negative output, with g8 cutoff, the output is (286/(286+750)) (15) = 4.14V. Thus with a 28612 load, the output can swing from 9.8V to 4.1V.
+15
7.85kQ
.,s;
«j"ÿ9kQ 3.44 • 3/
>20.S

VA =+15 t
5ÿ71 (15 ~
~
1184) = 1047V (for which
6.50 For the resistor values: 10)5 = 2 k!2; 10  0.7)0.5 = 18.6 k!2; R6 = (0 The reference resistor, Rq = (0 10)1= 12 k!2; R} = 7A2/2.) = 2 k!2; 10.7 k!2; = R2 = 3<0.S2) = R$ ( + 0.7 R4 = (2 0.7yi = 1.3 k!2. For the emitter resistances: For Qi, Q\, re \ = re2 = 25 mV/0.25 mA = 100 12. For Q4, Qs, re4 = re5 = 25 mV/1 mA = 25 12. For Qy, rel = 25 mV/1 mA = 25 12. For Qs, rei = 25 mV/5 mA = 5 12.
—
—


For the input resistances: r„ = (P+ l)r, =51re, in general; rK\ = r„2 = 5.1*12; rn4 = rre5 = r„7 = 51(25) = 1.275 k!2; rn8 = 51(5) = 0.255 k!2. Now Rn = Rid = rK1 + r„2 = 2(5.1) 10.2 k!2; Ri2 = rn4 + rn5 = 2(1.275) = 2.55 k!2; Rn = rnl + (P + l)(R4) = 1.275 + 51(1.3) = 67.575 k!2; Ri4 = rnS + (P + l)/?6 = 0.255 + 51(2) = 102.26 k!2. As noted on page 557 of the Text; d„A>«/ = (RÿRiiXhAi) = (240.2)(teÿt,) = 0.196/eg4( , where ie84f = ie/ib% x ib/icl x ic/ibl X lbyic5 X ic/ih5 x ib/ic2 x jc24(.
___
Here, ie/ib% = p + 1 = 51; ih/icl = R&RS + Ri4) = 10.7/1:10.7 + 102.26) = 0.0947; ic/ibl = p = 50; k/ics = RARy + Rid = 2/(2 + 67.6) = 0.0287; ic/ib5 = p = 50;
. . lbAc2
Rj+R2
(Rl+R2) + Rn
12+ 12 12+12 + 2.55
_
non. '
. . _ Rÿ _
„
Thus overall, \y„A>id = 0.196 x 51 x 0.0947 x 50 X 0.0287 x 50 x 0.904 x 50 = 3070 V/V. Clearly this method fails if p = 00, although it is generally OK for large (but finite) p, in which case the currentdivider factors become smaller and smaller as P rises.
267
SOLUTIONS: Chapter #618
NOTES
268

Chapter 7
FREQUENCY RESPONSE SECTION 7.1: SDOMAIN ANALYSIS: POLES, ZEROS, AND BODE PLOTS V(> (.S ) Zshunt Here 7.1 Using the voltagedivider rule: T(s) = V:(s) •shunt + V„(s) VC2s R/C\s R\ 1 ,, „ T(s) = 11*.= Cj 1/C2ÿ + /? /(I + V i(j) 1 + R\C\S R + 1/C 1 + R) C s 1 or T(s)  —— . That is, we see a singletime„ „ w =6 1+R(C, + C2)s 1 +Ri C2s/{\ +Ri C, s)
—
——
constant response with a time constant x = Ri (C + C2), where the resistor R i "sees" C j + C2, the parallel capacitance of Cj and C2 when the source is shorted.
Type of STC: See that the circuit passes dc directly via R i, with the signal reduced at high frequencies. Thus it is a lowpass (LP) STC circuit. See for s = 0, V0(syVi(s) = 1, and for j = <*., V„(sYVi(s) = C/(C, + C2) < 1. Now, for R, = 104n, C, = (0.5/10)iF, and C2 = 0.5xF. There is a pole at ox, = 1/(7?  (C, + C2)) = M;104(0.05+0.5) 10 ) = 182 rad/s, and a zero at 0)z = K/?iCi) = K104(0.05)10"6) = 2000 rad/s. From the Figure, or directly, see T (co) IT(
a) b)
+
10 100 ik iok
+
Ri
1 + (co R\ C i
Thus,
1 + (co /?i (Ci + C2))2
IT"(0)1 4>(co)
= 1 =0dB,\T(oo)\ 1 =
C, C, + C2
1 ÿ 0.1 20.8dB. 0.1 + 1 11 0.540 + 0.5 d>(co) = tan1 co R, Ctan"' co R\ (Ci + C2).
1/10 1/10+1

= tan
0.540 0.540 + 0.5
4> (coz) = tan1 1  tan"' For
 tan"1 1
= tan1 1/11  tan"1 1 = 5.19*  45* = 39.8*.
*i (C, + C2)
,
R C,
= tan"1 1  tan"1 11 = 45*
coz)'/' = (182 x 2000)* = 603 rad/s, com = (co„ "p wz
73.2* = 56.4*.
269
(com) = tan1
 84.8° = 39.8*.  tan"1
= 16.8*

SOLUTIONS: Chapter #72
7.2 (a)
10'4 ÿ + 10> ro (s + 1) (s + 100) (s + 105) (s + 106)
_ 10'4 (10) (j) (1 + j/tO)
(100) (105) (106) (1 + s) (1 + s/100) (1 + s/105) (1 + sAO6) 102 s (1 + j/IO) or OT T(s) S = (1 + j ) (1 + j/100) (1 + s/105) (1 + j/106) '
_
_
(b)
(11)°(1)°()1)1()1) = ov/v, and
As s + 0, T(0) =
90'. As s (c)
(d)
—>
/
oo, T (oo) =
\2
i
 = —J = OV/V,
= 180°. Poles at s = 1, 100, 10s, 106 rad/s. Zeros at s = 0, 10, oo, oo rad/s. As the frequency rises, the zeros increase the gain while the poles reduce it. We see that the gain increases from 0 to the first pole at 1 rad/s, then begins to rise again at 10 rad/s until 100 rad/s where it flattens, beginning to fall again at 105 rad/s, and more at 106 rad/s. Thus, we see that the gain is greatest from 100 rad/s to 105 rad/s. Thus at (0 = 103 rad/s,
rao3)3 _ 10 (10 ) (102)
103 (10) (1) (1)
= 1030 (1q3)
_ 90
= 90 + 90  90  84  0  0 = 6°.
_
_
tanl102 tanl103 tanl Also at
(0
10
_ tanl 102_
= 104 rad/s, I T(104) I =
10
t
1
103
00 ) (10 )
104 (102) (1.1) (1)
= 0.9x10 , and
(e)
phase is 0', occuring from about Gain at 103 rad/s, T (co = 103) =
_
At
_ (102)(103)(1 + (lOVlO)2 )'/l_ a + (loM)2)14 (l + (10V100)2)14 (i + (io3/io5)2)'/i (i + (ioMo6)2)14
102 (103)) (lO"*)) 103 (10) (1) (1)
.
10,5
103 to 104 rad/s.
ÿ
rad/s,
_ 102 2 31 _ l03V/Vs60dB. n2v/v = in2 + 3 + 231 _ 10'(1q5> (lp4> T (to 10) i
=
10* (lQ5) (104) 10s (103) (1.414) (1)
£n
1Q5 (1Q3) (1 + (ioMoW (1 + (lOVlO6)2)*
,0,707x0,.57dB,
It certainly would have been better to prepare the Bode plots earlier, certainly by the end of part (a). It is actually possible to sketch the pole and zero locations immediately, and to sketch the shape of the magnitude plot. However the absolute magnitude requires a calculation like the conversion done in part (a) (see P7.4 following). Probably easiest after (c), most useful before (d), and usefully possible before (b).
(See the Bode Plot on the following page.)
 270
SOLUTIONS: Chapter #73
7.2
(Continued)
,0
10V>
3
0)
rad/s
0)
rad/s
,<4
6 7.3
102 J
(1+5/10) (1 + 5) (1 + j/100) (1 + 5/105) (1 + 5/106)
From P7.2 above, T(s ) =
'
102 102 (1 (10M0)Y
+ For co = 100 rad/s,  T (to) ,5\2y/. 1<>"\2\/i (KXKIOO)2/' 1002)" (1 + (lOOÿOY) (1 + (lOOtOY)" (1 + (1 + 102 x 102 x 101 0.707 x 102 + 2 + 1 ~ 2 0.707 x 103 a 57dB. (to) 90 tan1 10 tan" = + = = 102 (2'a ) (1) (1)
100  tan"1 1 tan"'
For to = 2 x
10"3  tan1 10"4 = 90 + 84.3  89.4  45  0.1  0 = 39.8*.
105 rad/s, 102 x 2 x 105 (1 + ( 2 x
I
T(co)  =
10
•
(1 + (2 x
105)Y (1 + ( ———
100
,*2)!/i (i +
102 x 2 x 10s (2 x 104) 2x
10s.2.y,
105 (2 x 103) (1 + 4)'/i
(1 + 0.22)*
''
(ÿr)2)'7' (i + (ÿrÿ)2)'7' 106 4x10"
4 ( V 5) ( V 1.04) X
108
= 0.4385 X 103 a 52.8dB.

O (to) = 90 + tan"1 (2 X 104) tan"1 (2 x 90  90  90 63.4  11.3 = 74.7*.

10s)  tan1 (2 x 103)  tan1 (2)  tan"1 (0.2) = 90 +
 271 
SOLUTIONS: Chapter #74
SECTION 7.2: THE AMPLIFIER TRANSFER FUNCTION 108 J 10) 7.4 From P7.2, T(s ) =  0 + (s + 1) (s + 100) (s + 10s) (s + 106) 108 s (i + 10) .„3 1 x hr~,f—— x 10 x — S (s + 1) (j + 100) , & lo5 x 106 (I w,
\
Thus, A„ = 10', Fl(s) =
103 s (s + 10) AL(s) = , , , ' (s + 1) '(s + 100) A
7.5
,
N
(' + , F„(s) = 1) (s + 100)
'
(s +
.
f
\
Ah(s) =
From P7.4, FL(s) = (. + 1)C»
(s
1 J
+ 1F)<1 +
5
W
„+»)(,+ •) 106
_s_
10s M
5_, =s+100
10' 1
FH(s) =
with the dominantpole responses indicated by =. Thus A(s)
=
1
105 „ 103 s (s + 100) (1 +
7.6
+
+ 10)+ 100)
1) (s
103
„
+ I00)
s (s

110s
A
_
1+
10®
ÿ
105
ÿr) 10s
For the lowfrequency response of the transfer function given in P7.2, there are 2 lowfrequency poles, at 1 rad/s and 100 rad/s, and zeros at 0 and 10 rad/s. For the 3dB frequency: (a)
(b)
.,
Dominant Pole: CO/, ~ 100 rad/s. Root Squares: CO/. = (1002 + l2  2 (102)
f = 102 (1 + 10"4  2 x
2)'/j = 99 rad/s. _1 (co2 + 102) (co2) (co2 + l2) (co2 + 1002) 2
„
X3C y*
10
2co4 + 2(10)2 co2 = co4 + 1002 co2 + co2 + 1002, co4  CO2 (9801)  10000 = 0, co2 = [  9801 ± V 98012  4(10000))>2 = [9801 ± 9803]/2, that is co2 = 9802, or CO/, = 99.00 rad/s. Note above in (1) that the zero at zero frequency must be
7.7
included in the calculation. Why?
For the upper 3dB frequency of the transfer function in P7.2, two poles, at
105 and 106 rad/s.
For the 3dB cutoff: (a)
Dominant pole:
cow = 105 rad/s.
(b)
Root squares: H
ÿ7 WH = l/V v — (10s)2
CD// = 0.995 X (c)
'
= V —U (1 + —~r)y' +— (106)2 10s 102
10s rad/s.
'
10s (1 + 1/100)" '
ÿ(1 + a/10 ) (1 + J/106)
ÿT2 (co) = 77 = 1/2, (1 + (ci/10 ) ) (1 + (0/106)2) 1+ + 1/1012) + cd4/1022 = 2, CD4 + CO2 (1012 + 10'°)  1022 = 0, cd4 + 101 x 10'° co2 1022 = 0,_ , 101 x 1010 + ÿ1012 x 1020 + 4 x 1022 , Exactly: F,,(s) =
' CD2 (1/1010
7,
CO2 = 
, 101 x 1010 ± 102.96 x 1010 ., ,nl0 ,„5 „nnn x 10 cd22 = rad/s. = 0.9805 X 10 and (oH = 0.990
 272 
or
7.8
Old T (s) =
Modified
Note
____
(j + 10) 1 x 103 x " " ' (s + 1) (s + 100) (1 + j/105) (1 + s/106) ,y (s + 10) 1 T(s) = x 103 .x. (s + 1) (s + 100) (1 + s/105) (1 + sAO6) j
the
polezero
103 s (s + 10)
cancellation
106 (1 + 1/4)* = 0.894 x 106 rad/s. 1 (1 + (oylO6)2) (1 + (o>{2 x
with
106))
(s + 1) (s + 100) (1 + sÿO6) (1 + si2 x
Exactly,
SOLUTIONS: Chapter #75
106))2
to4 + 2 x 1012 in2  2 (1024) = 0, 10'2( 2 + 3.46) _ Q ?3 x 10,2 Thus 2
the
, and (oH =
= 1/2, 1 +
result
lW (1/(1 x
(1 + s/105) ' (1 + sA2 x 106))
that
T(s)
105))2 + (1/(2 x 106))2 =
or = 2, 2 (1024) 2 x 1012 ± ÿ 4 x 1024  4 (2 x 1024) or
1012
+
J_. 4
= Q g56 x 1q6 rad/s
7.9 (1)
Using
C2, Ret/ (a)
(b)
For opencircuit time constants: = R\ + ÿ2i ancl = C2 (ÿ1 + ÿ2)
C\,Req=R\, and
Ti
For
Rl = R2=\0kCl, Ci=C2 = lOOpF, t, = 104 x 100 X 10"12 = 106s, 2 — =  ~rx22 = 100 x 10"' (104+ 104) = 2 x I0~6s, coH = — X, + x2 1 x 10"6 + 2 x 10~6 = 1/3 x 106 = 333 x 103 rad/s, or = 0.333 x 106rad/s. Alternatively, the sum of squares approach yields (% = 1/(12 + 22)* x 106 = 0.447 xl06rad/s. For R] = lOkO, R2 = lOOkO, C, = lOOpF, C2 = 10pF, X, = 104 x 100 x 10'2 = 10_6s, x2 = 10 x 1012 (10 x 104 + 1 x 104) = 10 x 10~8 (11) = 1.1 x 106s, For

J = 0.476 X 106 rad/s. Alternatively, the sum of squares = (1 + 1.1) 10~6 approach yields c% = 1/(12 + l.l2)* x 106 = 0.673 x 106 rad/s. R, = lOkQ, R2 = 100k£2, C, = C2 = lOpF, x, = 104 x 10 x 10"12 = 0.1 X 10~6s, x2 = 10 x 10~12 x (10s + 104) = 10 x 10"12 x 104 (11) = 1.1 x lOÿs, c% = 1/((0.1 + 1.1) 10~6) = 0.833 x 106 rad/s. Alternatively, using the sum of squares, (% = 1/fO.l2 + l.l2)* x 106 = 0.905 x 106 rad/s. CO//
(c)
 R\ C\.
Vo
+ vo C2 S R2
Jc AA •—
Rz
Ri
vt
vi =
AA
« vo
jc = vo (R2 C2 S + 1) C1 S iRl = VO KR2 C2 S + 1) C1 S + C2 S] S 1] vo [R1 R2 C1 C2 S R1 C1 S + R1 C2 S + R2 C2 +
2+
 273 
SOLUTIONS: Chapter #76
T(s)=V"
Exactly:
(2)
Vj
—
1
s2 R R2 C\
——
C2 + s (R 1 C + R 1 C2 + R2 ÿ*2) "t* 1
—J——
. Now, response is 3dB   „ (Rx C + R\ C2 + Ri C2) + (1 to R\ R2C\ C2) down when (O2 (Rx C\ + R\ C2 + /? 2 C2)2 + (1 co2 R\ R2 C\ C2)2 = 2. Now, for particu¬ lar cases: T(j(o) =
(a)
j co
—

ÿ
—

/?, = R2 = 10k£2, C, = C2 = lOOpF. Now, co2 (3 x 104 x 10"10)2 + (1 co2 x 104 x 104 x 10"'° x lO"10)2 2 = 0, 9 x 10"12 co2 + (1  co2 10"12)2 2 = 0, 9 co2 10"12 + 12 co2 10"12 + co4 1Q24 2 = 0, co4 + co2 (7 x 1012) 1024 = 0, co2 = _ . inS , .n,2 (~7 ± 7.28) 7 x 1012 ± 10l2ÿ72 4 10 , and cow = 0.3J4 x lO6 = iol2i= 0.14 x 1a12




A1,
rad/s. (b)
(c)
Rx = lOkO, R2 = lOOkQ, C, = lOOpF, C2 = lOpF. Now, co2 (lO4 x lO"10 + 104 x 10"" + 10s x lO"11)2 + (1  co2 x 104 x 10s x lO"10 x 10"11)2  2 = 0, or co2 (10"6 + 10"7 + 10"6)2 + (1  co2 10"12)2  2 = 0, or co2 (2.1 x 10"6)2 + (1  10~12 co2)2 2 = 0, 4.41 x 10"12 co2 + 1  2 x 10"'2 co2 + co4 10"24 2 = 0,
CO4 10"24 + 2.41 x 1Q— 12 co2 — 1 = 0, co4 + 2.41 x 1012 co2  1024 = 0, CO2 = 1012 (2.41 ± V 2.412  4yi = 1012 (2.41 ± 3.132>2 = 0.309 X 1012. Thus (On = 0.601 x 106 rad/s. Rx = lOkfl, R2 = lOOkO, C, = lOpF, C2 = lOpF. Now, co2 (104 x 10"" + 104 x 10"" + 10s x 10")2 + (1  co2 x 104 x 105 x 10"" x 10"")2 2 = 0. Thus co2 (1.2 x 10"6)2 + (1  co2 (0.1) 10"12)2  2 = 0, 1.44 x 10"12 co2 + 1  2 co2 (0.1) 10"12 + co4 (0.01) IP"24  2 = 0, co4 + 1.24 x 10+14  10+26 = 0, CO2 = (1.24 X 1014 ± V 1.242 X 1028  4 ( 1026)>2 = 1014 (1.24 ± ÿ 1.242 + ,04>2 = 0.00801 x 1014 = 0.801 x 1012. Thus c% = 0.895 x 106 rad/s.
_
7.10 Using shortcircuit time constants:
For C (, Req = R\ II R2, and x x = Cx
Rx R2/( Rx + R2).
For C2, Req = R2, and x2 = C2 R2. (a) For Rx=R2 = 10kf2, C, = C2 = lpF, x, = 1 X 10~6 X 104/2 = 0.5 x 10"2s, and tOi = 1/0.5 X 102 = 200 rad/s; x2 = 10"6 x 104 = 10"2s, and co2 = 100 rad/s. Thus CO/, = 100 + 200 = 300 rad/s. Using the sum of squares idea, CO/, = (1002 + 2002)'/l = 224 rad/s. (b)
For R 1 = lOkft, R2 = lOOkft, C, = lpF, C2 = 0.1pF, x, = 10"6 X 101:12 II 1001:12 = 0.909 X 10_2s, x2 = 10s x 10"7 = 10"2s. Thus CO/, = (1/0.909 + 1/1) x 1/10"2 = 210 rad/s. Alterna¬ tively, CO/, = ((1/909)2 + (1A)2/1 x 100 = 147 rad/s.
(c)
For R, = 10k£2, R2 = 100k£2, C, = C2 = 0.1pF, x, = 10~7 x .909 x 10+4 = 0.909 x 10"3s, x2 = 10"7 X 10s = 10"2s. Thus CO/, = K0.909 x 10~3) + 1/10"2 = 1100 + 100 = 1200 rad/s. Alternatively, CO/, = (1/9092 + 1/102)'7' x 1000 = 1054 rad/s.
SECTION 7.3: LOWFREQUENCY RESPONSE OF THE COMMONSOURCE AND COMMONEMITTER AMPLIFIERS 7.11 For Midband Gain:
«
 *
all capacitors are ac short circuits.
 ÿ0mT10«T22M * * '
""ÿ>
2
10"3 (2« »
13.14 V/V.
Thus,
1="
—/?( —II — j
R(j2
x Au = —R 'I— + "CI "02 10"S "67 103 *
HI*
*
*
106fl, and X 6.975 lOOkO = + 6.875M£2 Rc 1 = CC\: 10~6) x 10kl2 For 20ki2 = 2.28Hz. + = 30kf2, and fpx = K2rc x 6.975 X 106 x 0.01 Cc2: RC2 = 10"6) = 53.1Hz. fp2 = K2tt x 30 x 103 x 0.1 x For
274
SOLUTIONS: Chapter #77
For
fPs
Cs: R's = Rs II Vgm = lOJfcQ II (l/(2 x 10~3)) = 0.5/: 111OA: = 4760, and = H2K x .476 x 103 x 1 x 10"6) = 334.3Hz(with fa = V[2n x 10 x 103 x 1 x 10"6) =
15.9Hz.
7.12 Use the largest capacitor at the source, where the resistance level is least. Now, Cs = lA2n x .476 x 103 xlO) = 33.4iF. Use Cs = 30pF. Now, Cc, = V{2k x 6.975 x 106 x 1) = 0.0236pF. Use CC\ = 0.02jxF. Now, CC2 = V(2n x 30 x 103 x 1) = 5.3(xF. Use Cc2 = 5pF. Actual critical frequencies are: fpS = J/(2jt X .476 x 103 x 30 X 10~6) = 11.1Hz, fpl = 1/(271 x 6.975 x 106 x 0.02 x 10"6) = 1.14Hz, fp2 = V(2n x 30 x 103 x 5 x 106) = 1.06Hz, /rf = 1/ (2n x 10 x 103 x 30 x 10"6) = 0.53Hz. 7.13 Modify
Equations 7.35 through 7.36 for the addition of
 
MO MO nrT5T Vgm +ZS

g <«) tTVT». om V. gm + Ys
Here,
r$
in series
with C$ : 1
"
Rs
Zs
V(s
Cs + rs)
_ s (Rs + + 1 Cs ... "S 'd (s) 1 + srsCs Rs Rs (s Cs rs + 1) " s Cs (Rs + rs) + 1 s Cs (Rs + rs) + 1 Rs (s Cs rs + 1) vs ( . g 8m s Cs (Rs + rs) + 1 s Cs (g„, Rs rs + Rs + rs) + Rs gm + 1 8m + 1) (s Rs Cs rs + (Rs + r?) (*) = V, n g r R r gm + rs rs + Rs Ks 1
s
J
_
ÿ
x
(Rs
_
ÿ
)
Cs .v . See that there is a zero at o)2 = V(CS —' gm 1
_
+ Rs _ (gm Cs Rs rs + Rs + rs)
at to„ "p
+ rs)), and a pole
[—
1 + gm Rs Rs/gm rs (1 + gm Rs) + Rs , Rs .. .. „ == rs +  as rs + Vgm II Rs = rs + Vgm I+ gm Rs I+ gm Rs + Rs
quency
is
noted.
Now the equivalent transconductance is
gm (Rs + rs) gm
(Rs
, where the resistor associated with Cs at the pole freRs + rs (1 + gm Rs)
=
Cs
_
,
1 — —II— Vgm
+ Rs rs
=
8m  = gm 1+
Rs rs Rs + rs
, as noted. t
Rs rs + Rs + rs
Now for the situation in P7.ll, with gm = 2mA/V,
Rs = 10k£2, the gain is reduced by a factor of gm Rs rs Rs + rs , „ two, when = 2, or gm Rs rs = Rs + rs, or ——  = 1/2, or 1 + —— gm Rs rs +RS + rs + Rs rs Rs 10 x 103 10 x 103 „ pole: For the new y — =  = 52612. — rs = 55 = gm Rs ~ 1 19 x x 10 1 103 2 10~3 X RP=rs + (Vgm) II Rs = 526 + (1/(2 x 10"3)) II (10 x 103) = 526 + 476 = lOOOfl This
should have been obvious since the gain was to have been reduced by a factor of 2, and 1lgm = 50012. Thus fpS = V(2n x 1000 x 1 x 10"6) = 159Hz, and fjS = 142.11 x 1 x 10"6 (10/:£2 + 0.526/fc£2)) = 15.1Hz.
 275 
SOLUTIONS: Chapter #78
SECTION 7.4: HIGHFREQUENCY RESPONSE OF THE COMMONSOURCE AND COMMONEMITTER AMPLIFERS 7,14
r< =
'ÿ3 a2,5/ 0.15 xlO"1
9.1 II 10 II 500 15.45 166?7— =  25ÿ5 For Ccl:
Rct
,nne £ 25 6 = = ~17"95 = "18V/V' 166/7 = 10**2 + 40*0 II 25.2**2 = 10*0 + 15.5*0 = 25.5*0,
fp j = 271 x 25.5 x 103 x For
CC2:
25171140 = rn = 166.7 x 151 = 25.17k*2, AM "  25.17 1140+ 10 , 4.93 xlO3 "a6°7 X
= 166.7a X
ÿ
= 6.24Hz.
1 x 10'6 RC2 = (9.111500*0 + 10*0) = 18.94kO,
/ 2 = zr = 8.4Hz. 27t x 18.94 x 103 x 1 x 10"6
'
For CE:
Rce
50 + 40*1110* 167 +  II 8.2* = 220.3 II 8.2* = 2140, 151
—
.,n
r = 74.4Hz, p = fPE 27t x 214 x 10 X HT6 Overall:
r = 1.94Hz. fzE = 271 x 8.2 x 103 x 10 x 10~6
(
fL = 6.242 + 8.42 + 74.42  2 (1.94)2
= (39+71+55357.5)'/' = 75.1Hz.
7.15 Use data from P7.14 above, where for
0.214kO.
Cc i, Rci = 25.5kO; for Cc2, Rci = 18.9kO; for CE, RCE = Now, for fpE = 20Hz, CE = —j = 37.2(xF, for which (.214 x )x
=rr = J fzE 2ti x 8.2 x 103 x 37.2 x 10"6
0.521Hz.
Now,
10 20 for /_ Jpli
= 0.521Hz,
5 = Cc, =  = 12.0JJ.F. Now, for fp2 Jp = 2Hz, Cc2 = 271 x 25.5 X 103 X 0.521 In x 18.9 x 103 x 2
4.2(iF.
Alternatively, for
—r= p = 0.521Hz, Cc2 = fn2 2tc x 18.9 x 103 x .521
16.2uF, and for /„ Jpi1 = 2Hz,
Cc 1 = ÿ= 3.12pF. This arrangement is better, since it makes the pole  zero x x x
27t 25.5 10" 2 cancellation independent of (3, although it takes a larger total capacitance.
7.16 Here, as in P7.14 above, rc = (25 x 103X0.15 x 10~3) = 166.7 12. Now, rE = 350 £2, and the resistance looking into the base, (call it r'n) is rK = 151(166.7 + 350) = 78.0**2. 150 (9.1 II 10 II 500) x 103 78.0 1140 26.4 150 4.72 x 103 NOW x A" ° ° 78.0 II 40 + 10 166.7 + 350 "xT liT 516.7 6.58 V/V.


"W 

Rci = 10**2 + 40**2 II 78**2 = 10**2 + 26.4**2 = 36.4**2, and f,,i = 1/(2tc x 36.4 x 103 x 1 x 10~6) = 4.37 Hz For Cc2: RC2 = 9.1 II 500**2 + 10**2 = 8.94**2 + 10**2 = 18.9**2 and fp2 = K27t x 18.9 x 103 x 1 x 10~6) = 8.42 Hz For CE: RCE = 350 + [167 + (40**2 1110**2X51] II 8.2**2 = 350 + [167 + 53] II 8.2**2 = 350 + 214 = 564*2 and fpE = K2tc x 564 x 10 x 10~6) = 28.2 Hz with fx = 1X271(8.2 X 103 + 350)10 x 10"6) = 1.86 Hz For Cc i:
 276
"
"
SOLUTIONS: Chapter #79
For fL :
fL = [4.372 + 8.422 + 28.22  2(1.86)2]14 = [19.1 + 70.9 + 795.2  6.9f = 29.6 Hz
7.17 From Eq. 5.115, fT=gm/[2n (Cgx + Cs
id
=/dss(1~)2 (1). yp
1 = 4 (1  ~)2 > 1 + ~ = (i)"
=
1/2. Thus
ÿ(1/2) 2
=
2n,A/V.
Sm =
~rr (1  ~TT~) = Vp Vp
144.7MHz. (b)
/t (c)
—
y =
1/2, V,JS = 1. Now 2* 10"' = 2it(2 + 0.2)x 10~12
 
Thus frT
___
(vGS  1) = ÿ2,  v,)2, 200 = 100 (uG5  l)2 k\W/L){MGS 10"6) <~2 = 283J1.A/V, Cs.s = 0.15 x 10"12 + 20 X 10"15 + =  V,) = (200 X 0.1 x 10"12 = 0.27pF, and Cgd = 20 x 10"I3F = 0.02pF. Thus 283 x 10~6 —rr = 155.3MHz. fr = 2k (0.27 + 0.02) x 10"12 10xl0~3 fr  rr = 9.65GHz. 2tc (0.15 + 0.015) x 10"12
iD = V2k\w/L)(»GS
2
7.18 For convenience, K = V2\i Cox (W/L ) = VI (.05 x 1012) x 1 x 10"'5 x 27/3 = .225 x 10~3A/V2 = 225[iA/V2; gm = 2K (vGS  V,) 2 (225 x KT6) (2.5  0.5) = 900pA/V; Cgd = Ld W Cox = .3 x 27 x 1 x 10"'5 = 8.1fF. Cgs = 23 WL Cox + Ld W Cox = 2/3 x 27 X 3 x 1 x 10"15 + 8.1 x grn 900 x 1Q6 pr 10"15 = 54 + 8.1 = 62.1fF. Thus fT = „ 2.04GHz. „ , ~ \ = o .. 10"

.
—
7.19




1) = 240fF, for gain of 1V/V, or = 200 + 20 (1 (1 gain) = 200 + 20 (1 1/1) = 100) = 2220fF, for gain of 100V/V. Cd = Cdh + Cgd (1 1/gain) = 100 + 20 (1 1401F, for gain of 1V/V, or = 100 + 20 (1  1/100) = 120.2fF, for gain of 100V/V.
C, = Cgs + Cgd



7.20 Gain, gate to source, is —g,„ (r„ II RD II /?/,) = 1 x 10"3 (50 II 10 II 30) x lpF + 0.5pF (1 6.52) = 4.76pF, Cd = 0.5 (1  1/6.52) = 0.577pF. Input pole: fpg = l/(2tt (4.76 X 10"12) x (1001:12 II 1A/12)) = 0.368MHz.


103 = 6.52V/V. Cg =
Output pole: fpd = V2.K x 0.577 X 1012 (50 II 10 II 30) X 103) = 42.3MHz. Upper 3dB frequency fH\ = fpg = 0.37MHz. For Rs reduced to zero, the output pole dominates, and fn2 = fpd = 42.3MHz. Now for /W3 = 0.9 (42.3) = 38MHz, with Rx nonzero and for
fPg'=fg Now
J_
//
+ JL
11
0.000559)'/". Thus fg = 86.4MHz. Now fg = \A2k (4.76 X lQ~i2)Rs), whence
1
1
38.0' fg2
fp
1
= (0.000693
38.02 ' fg
42.32
Rs = lA2n (4.76 x 10"12) (86.35 x 106)) =
38712
7.21 Using the results of P7.20 and Equations 7.63 and 7.64, tO/»i = 1/
£
J
Cgs + Cgd (1 + g,„ Rt) + Cgd (Rl'/Rs) Rs
1 [ 1.0 + 0.5 (1 + 6.52) + 0.5 (6.52/90.9)] 90.9 x = 2M rad/s, or
103 x
1 12
10"
[4.76 + .036] X 90.9 x
10~9
ÿg" ÿgd + Sin + Cgd — £7 rr7: Lgd Rl cs* . 4.796 x 10 2 , p:=r = 1.47Grad/s, or /„2 = 234MHz. f,  —— np
x fpX
=
n 1/fmn
0.365MHz.
(0;)2
)
=
.„n
1x

10"12 x 0.5 x 10"'2 x 6.52 x 103
(FG/R )
ÿyiAA/iu
t
2k Cgd
277

SOLUTIONS: Chapter #710
1 x 10~3 (2k (0 5 x 10~12))
= 318MHz' for which f»~fp I = 0.365MHz. Now, for Rs = lk!2, fz = 318MHz, /„  = K2w (476 + 0.5 X (6.52/1)) 10"12 X 103) 19.8MHz, f„2 = K(A 76 ~ + 3 261 X 10"'* r = 392MHz, for which J  ' fH ~ 19.8MHz. Z x 6.52 x 103) 2k (1 x 0.5 x 10"24
iD =K (vGS  V,)2 I= K (2  l)2 = K ; gm = 2K (oGS  V,) = 2 (1) (2  1) = 8m 2 X 10"3 „ 2 X 10~3 o A A/ TU r Thus fT 2mA/V. 10* = Thus C„ . = ————— > in9 —— — gs + C„rf 5 = gd = 2jc 2k (Cgs + Cg(/) (Cg!l + Cgd) 2% (109) 0.318pF. Now, if Cgd = 0.2 Cgs = Cdh, then 1.2 Cgs = 0.318pF, Cg, = 0.265pF, Cgd = Cdh =
7.22 Generally,
—

Rr
Ri„
v
—
0.053pF. Thus C,„ = 0.265 + 0.053 (1 10"3) = 1.5kl2, and 3/(2 X
3) = 0.477pF. For input source, resistance, 3/gm = capacitance = 4 (0.053pF) = 0.212pF, , rrr" = 500MHz. Problem P7.50 of the Text provides the topol. fn ~ 2n (0.212 x 1012) K (1.5 x 10"3) ogy for which this high performance is possible.
————— Rr
7'23
Rr
=
KTmS
*
Rf/A = 0.95 Rf/A + 9.5, (R/A)(l  0.95) = 9.5, whence Rf
a'5
=
Rr/A
ÿTHo'
= 4(9.5)/0.05 = 760kl2! See that
must be very high, even in such a lowimpedance circuit!
Rf
7.24 /p = K2K (C„ + Cÿ) rj. For lc = 2mA, gm = 2mA/25mV = 80mA/V, rK = Vgm = 200/80 = 2.5kl2. Now, 12.7 x 106 = V{2n (C„ + 0.5) x 10"'2 x 2.5 x 103). C« = 5.01 0.5 = 4.51pF, and fT = P„ cop = 200 (12.7 X 106) = 2.54GHz. For Ic = 10mA, gm = 10/2 x 80 = 400mA/V, rK = 2/10 x 2.5 = 0.5kl2, CK = 10/2 x 4.51 = 22.55pF, /p = 1/(2k (22.55 + 0.5) X 10"12 x 0.5 X 103) = 13.8MHz. For C„ = C„ = 0.5pF, Thus =
4"51ÿF 2mA
Ic c
0 5x2 — = 0.222mA. That is, /r is maintained at 2.5GHz for currents > 0.22mA or so. — —rr. 4.51 J
24 x
7"25 91
103
" 166™'
= 1667
25 17 II 40
2517k"' ÿ = 25.17 .»40+ 10 * = 0.607 X 25.6 = 17.95 = 18V/V. Now, C„ + CM =
500
X
151 =
x =" 1' 150 x gm 166.7 151 0 948pF Cjt o 94g _q 30 o 648pF 2k fr 2k x 109 Input Pole: CT = 0.648 + 0.3 (1  25.6) = 8.63pF, Rr = 25.17*12 II 40*12 1110*12 '
_
_
_
+ 50 =
6.12kl2,
 Xrz
r = 3.00MHz. 2k x 8.63 x 10_12x 6.12 x 103 Output Pole: CT = 0.3pF, RT = 9.1*12 II 10*12 II 500*12 = 4.93kl2,
/„, p
ÿf„2 r = 108MHz. Jp = 2k x 0.3 x 10"12 x 4.93 x 103 The upper 3dB frequency is fu = 3.0MHz. 7.26 For R
 350 in series with Q, using data from P7.25 above, the total equivalent emitter resis¬
166.7 + 350 II 8.2* = (0.167+0.336) k!2 = 0.503kl2, and at the base, x Rin = 40*12 II (151 (0.503kl2)) = 40 II 75.95 = 26.2kl2. Thus Au = lQ 4.72 (9.1 II 10 II 500)*12 _q ?24 x 9 38 ÿ.79V/\. = 26,2 x 32.4 0.503 0.503*12 tance
becomes
_
_
_
 278 
SOLUTIONS: Chapter #711
CK  0.65pF, Cp = 0.3pF. Input Pole: Cr = 0.65 (1  (0.336/0.503)) + 0.3 (1  9.38) = 0.216 + 3.114 = 3.33pF, ÿfpl = Rt = 26.2k £2 II 10k£2 = 7.24k£2, Jp r = 6.60MHz. 2rt x 3.33 X 10"12 x 7.24 x 103 Output Pole: As before, fpi = 108MHz. Thus fpi dominates, and fn~ 6.6MHz. Now, from P7.25:
7.27 The output pulse is positive with amplitude = 50 (50 x 10~3) = 2.5V and duration of 50is. Its 2.2 2.2 A V = 7ns. Its sag (or droop) = 2jc x 50 x 106 2n fH = 2it X 50 X 50 X 10"6 = 0.0157, or 1.6%. transition times are
= 271 A tp
SECTION 7.5: THE COMMONBASE, COMMONGATE AND CASCODE CONFIGURATIONS 150
1
a/r, aZr. 151 125 0.3 = 1.26 .3 7.28 re = 25/0.2 = 12512, fT = Cu = ~*C*~ 2ti fT 2k (Ck + Cp) 271 x 1 x 109 9.1 II 1011 400 .125 II 8.2 < re II Re ÿ Rc II Rl II r„ = ÿ = 0.96pF. Am = ax .125 II 8.2 + .1 .125 re re II Re + Rs x TTF = 20.65V/V = 20.7V/V. fpl = = C"3) 2rt Cn (r, II Re H Rs) .2231 .125 1 1 = 3GHz. fpl = 27t Cp (Rc II Rl II r„) 271 (.96 X 10"'2) (.125 II 8.2 II .100) x 103 r = 112.6MHz. Thus fH ~ 113MHz, with AM = 20.7V/V. r,2„ (0.3 x 10"'2) (4.71 x 103)

"
4ÿ7
_
150
7.29

a/re
IE = 0.15mA, re = 25/0.15 = 167£2. Thus C„ = 2k fT
1
167 151 ~Cp = 271 X 109
 0.3 = 0.946 
0.3 = 0.65pF. 1) = 0.65 + 0.3 (2) = 1.25pF, At the input: CT = C« + CM (1 10 (151(167)) 15 = II II 25.2 = 4.85k£2, Rt = 15m II 10m II
J/„, = r, = 26.3MHz. Jp
271 (4.85 X 103) (1.25 x 10"12) At the emitter of Q2: CT C„ + CM (1   1/1) = 0.63 + 2 (0.3) = 1.25pF, Rr = 16712,

= 953MHz. 2k (167) (1.25 x 10~12) At the collector: Cp ~ Cp = 0.3pF, Rp = 9.1A: £2 II 10k£2 = 4.76k£2,
fpl ~
1
fp3
2n x 4.76 x 25.6MHz.
Midband
103 x 0.3 x 10~12
Am = 
gain
15 II 25.2 15 II 25.2 + 10
= 111MHz. Thus fH ~
150 151
15k II (151 x 167) 15k II (151 x 167) + 10k
9.1 II 10 0.167
279
9.40 19.40
150 151
1 + 26.3 150
151
1 1 + —1112 9532 2
9.1kQ II 10k£2 0.167k£2
4.76 = 13.6V/V. 0.167
SOLUTIONS: Chapter #712
SECTION 7.6: FREQUENCY RESPONSE OF THE EMITTER AND SOURCE FOLLOWERS 25mV 150 Vt 1 730 re = = 595mA/v= 151 (l67> = 252k"> c* = ir = o~i5ÿ4 = 167" 8m = x
ifr ik
i5ox£_L'
—2nx 109
 0.3 = 0.947 .3 = 0.65pF. v 8.2 1110
eC
M
0.963V/V.
151 (0.167 + 8.2 1110) 151 (0.167 + 8.2 11 10)+ 10
w
8.2 11 10 + 0.167
X
4.5 4.667
"
.
X
151 (4.667) 151 (4.667) + 10
~
CT = Cp + Cn (1  (4.5/4.667)) = 0.3 + 0.65 (1  0.964) = 0.323pF, Rt = 10*0 II (151 (0.167 + 8.2 II 10)*12) = 10 *12 II 705*0 = 9.86kO, For fH:
rp = fH~fP\ r> = 50.0MHz. 2n x 9.86 x 103 x 0.323 x 10~12 For
10.23kO, fL =
7.3'
r. = Vgm For
llF,
fL: CCi =
RCi = 10*0 + 8.2*0 II (.167 +
) = 10*0 + 8.2*0 II .233*0 =
ÿ= 15.6Hz. 2n x 1 x
 lkn, Am
r
10~6 x 10.23 x 10
.(10'fi"
a
p—— = fH: CT = 1 + 1 + 1x10
 »*»v/v. 1.09pF, F
RT =
lOOkO,
1.46MHz.
 = fH = 2n x 105 x 1.09 x 10"'
SECTION 7.7: THE COMMONCOLLECTORCOMMONEMITTER CASCADE VBE ~
0.70V,
1670, gm2 =
X
7.32 Since
IR
Cn =
= 2.27kO,
11[In
5 95 y In3 3*D
2K X 10
.
»
=
vT
=
150JA lÿlA. Now IEl
For
AM. Rin2 =
= 151 (2.27) = 343kO. For Q,:
0.3 = 0.65pF.
18.5kfl. A„ 70*n 25.2k Cl = 100„

IE2 ~ 160  10 = 150ÿA, re2 =
= 5.95mA/V , rn2 = 151 (167) = 25.2kO, IB2 =
= 1 + 10 = 1ljxA, re i = For Q2:
~ 0.70/70k0 = 10)tA. Thus
9.1 II 10
Ziifl
X
150
ÿ
.
X
Cn = 0.3pF. 18.5
X
(18.5 + 2.27) 151
m.
.24.4V/V. x 3136+ 100 20.77 For ///: At the base of Q2: Cr = 0.65 + 0.3 (1 + 4.76 x 5.95) = 9.45pF, Rt = 70* 12 II 25.2*12 II (2.27* £2 + = 70 II 25.2 II 2.93 = 18.5 II 2.93 = 2.53kl2, fpX = 0.167
x
,993
x
ÿÿ)
1
2n x 2.53 x
= 6.66MHz. 103 x 9.45 x 10_l2
At the collector of p = J Q2. f„2
pr = 111MHz. i
27t x 4.76 x 103 x 0.3 x 10"12 0.3 Q = + 0.3 (1  (18.5/18.5 + 2.27)) = 0.333pF, RT = 100*12 II (151 CT of (2.27 + 18.5)) = KXM2 II 3.1M12 = 96.9kl2, fp3 p = w = 4.93MHz. 27t x 96.9 x 103 x .333 x 10~12
At the base
\
v,
fu
~
1
4.932
1
1
6.662
1112
+ T +
= 3.96MHz.
 280
SOLUTIONS: Chapter #713
ForfL : For CC\ Rci = 10k + 9.1k = 19.1kft, /„ = For
CE: RCE
.
7°*
167 l» 70. +
" ÿ l0°™» a
'
j = 8.33Hz.
167 II
70
ÿ
0.019 = I860,
r = 85.6Hz, and f, Jz = 0Hz. fP2= 1% x 186 x 10 x 10"6 fL  (8.332 + 85.6Y' = 86Hz.
Jp
7.33 a) For R = 14kO:
iff
x
2I7
IR =
ÿ2  50ha, IE2 = 160  50 = llOpA, re2 =
= 438mA/v> r«2 = 343k£2 /fl2 = x
= 493ft, gml =
_
„
F" 2':
„
C" "
For G,: C„=
4.38 x
10"3
2n x 10' 2 01
„
2n X109
= 227ft,
gm2
=
"131" = 0/73, /£1 = 073 + 50 = 50•7,AA, r" =
= 2.01mA/V, r* = 151 (0.493) = 74.5kft.
.
"°'3 = a40pF'
0.3 » Use 0.30pF. K
x 0.993 x AM: Rin2 = 14kft II 34.3kft = 9.94kft, AM = ' (9.94 + 0.493) 151 4.76 1570.4 ,, 9.94 ~ ' (10.4) 151 + 100 10.4 1670.4 0.227 For fH: At the base of Q2: CT = 0.40 + 0.3 (1 + 4.76 x 4.38) = 1DO 14 II 34.3 (0.493 + Jjp) = 14 II 34.3 II 1.16 = 9.94 II 1.16 = 0.963kft, For
*
.
ÿ
1=
6.95pF,
RT =
pr = 23.8MHz. Also/.,2 = 111MHz. /„, p =27t x 0.963 x 103 x 6.95 x 10"12 At the base of Q\: CT ~ 0.3 + 0.3 (1 — 9
'
9/+9q 493 )
RT = 100 11(151 (0.493 + 9.94)) ÿ lOOkft II 1.58A*ft = 94.0kft, fp3 pr = 5.39MHz. pj = Itt wO/inv inJuAH/t x 94.0 x 103 x 0.314 x in12 10'2 271 =
0.314pF,
J
Thus fu =
1 5.39
1 23.8
vy
1/.
= 5.26 x 106Hz. (.4931:
For
=
fL: fpX = 8.33Hz. Now,
For
+ jÿ) II 14k
CE : RCE = .221k II 14 +  = 0.234kft,
r = 68.0Hz. Thus fL = (682 + 8.332)14 = 68.5Hz. /„2 p =271 x 234 X 10 x 10"6
b) For R =
IR = 0, lE2 = 160pA, re2 =
rn2 = 151 (0.156) = 23.6kft, IB2 =
= 0.156kft, gm2 =
y
X
= 6.37mA/V,
Jg. = li07pA, IEl = 1.07pA, rtl =
= 23.4kft, Ifr3 r„, = 151 (23.4) = 3.5Mft. For Q2: C„2 = — ÿ  0.3X10"'2 = 0.714pF. For Qt: CKi = 2n x 10~* fi 17 y
—
0.3pF.
„
.
„

.
= 23.6kft, A„ = 23J_ 7097 = = Zili x o 993 x For AM:
0.156
Ffn2
r„2
47.0
7197
4.76 —
nnn. X 0.993 X
y/V
281

23.6
X
(23.6 + 23.4) 151
SOLUTIONS: Chapter #714
For fH: At the base of Q2: CT = 0.714 + 0.3 (1 + 6.37 x 4.76) = lO.lpF, RT = 23.6 II (23.4 + 100/151) = 11.92kI2, /pi = l/(2n x 11.92 x 103 x 10.1 X 10"12) = 1.32MHz, fp2 = 111MHz.
At the base 23.4))
=
ÿ+\3 4 ) = 0.449pF, RT = lOOkft II
of Qx: CT = 0.3 + 0.3 (1 99kI2,
fp 3
=
23 l/(27t x 99 x
103 x 0.449 x 10"'2)
(151 (23.6
3.58MHz.
=
+
Thus
(23 4 + (100/151)) For fL: /p, = 8.33Hz. Now For CE: RCE = 0.156 + = 0.315kQ, fp2 = 1/(2tc x 315 x 10 x 106) = 50.5Hz. Thus fL = (50.52 + 8.332)'71 = 51.2Hz.
Summary (also using P7.32)
K(kf2)
Am (V/V)
14 70
18.7 24.4 15.0
oo
/z(Hz) 68.5
86.0 51.2
/h(MHz) 5.26 3.96
1.24
See that R is important in improving gain and bandwidth. The worst idea is to use R = oo. Perhaps a good design would be at R = ÿ 14 x 70 = 31kI2. (You can check this for interest).
SECTION 7.8: FREQUENCY RESPONSE OF THE DIFFERENTIAL AMPLIFIER 7.34 For each, IE = 300/2 = 150pA, re = 166.712, gm = 150/151 X 1/166.7 = 5.96mA/V, rn = 166.7 x 5% 151 = 25.2k£2, CK = *  0.3 x 10"12 = 0.65pF. 271 X
For Gain:
/tf = 2tc
10Q 109
•{f ((4 + 4)110)
____
X
2 (0.1667) 1
.0 + 2(25.2)
=
F°r 1
(y II 25.2) X 103(0.65 + 0.3 (1 + 5.96 x 4 II (y) x 10"12
7.75MHz. 7.35 As in P7.34:
B>'7'96)'
2n (4172> 492 x 10 9
re = 166.712, gm = 5.96mAAÿ, r„ = 25.2kI2, CK = 0.65pF, Cp = 0.3pF.
—
150 (4 For R• connected to the collector of the input transistor: Gain = x 151 2(0.1667) (0.1667) 503 150 151 2 (2.86) x = 12y/y 60.3 151 0.333 2 (0.1667) 151 + 10 150 2.86 Gain from base to collector of the input transistor = = 8.53V/V, (from which base 151 0.333 there is a Millermultiplied Cp, and a voltage divider with two C„ in series). Thus, CT = 0.65/2 + 0.3 (1 8.53) = 3.18pF. RT = 10k 12 II (2 (0.1667) 151 = 10k 12 II 50.3k£2 = 8.34kI2, fH ~ 1/(2tc x 8.34 x 103 X 3.18 X 10"12) = 6MHz.
J*

For Rl connected to the collector of the groundedbase amplifier: Gain = +7.12V/V. Gain from base to collector of the input transistor is  (150/151) (4)/0.333 = 11.92V/V. At the input base Cr = 0.65/2 + 0.3 (1 11.92) = 4.20pF, RT = 8.34kI2, fH 1/(2tc X 8.34 xlO3 x 4.2 x 10"12) =
4.54MHz.


7.36 Parameters as in P7.35: Gain = +7.12V/V, Cr = 0.65/2 + 0.3 = 0.625pF, Rj = 8.34k£2, fp\ l/(27t x 8.34 X 103 x 0.625 X 10~12) = 30.5MHz. Now check the output pole, where CT = 0.3pF,
 282
SOLUTIONS: Chapter #715
Rt = 4k II 10k = 2.857kft, fp2 = fH ~ HVS0.5 + 1/186) = 26.2
1/(271 x 2.857 x
103 X 0.3 x 10'2) = 186MHz.
Thus,
7.37 From P7.36, rt = 166.712, gm = 5.96mA/V, rK = 25.2kft, C„ = 0.65pF, C„ = 0.3pF. With re [(4 + 4) llr 10] added in each emitter lead, and load taken differentially, AM  X 4 (0 16671 x 4 (25.2) 4.44 0.993 100.8 _ 4(0.1667) 10 + 4(25.2) 110.8 1 4 44 x 0 993 At each input, gain from base to collector of Q = x ' = 6.68V/V, that is CT 2(0.1667) 2 0.65/2 + 0.3 (1 6.68) = 2.63pF, RT = (10/2) II (2(25.2)) = 4.55kft, ÿ
—


and fH = 1/(2ji x 4.55
X
103 X 2.63 X 10"12) = 13.3MHz.
7.38 For the current source, r„ = 200V/300pA = 667kft, Cr = 0.3 + 0.5 = 0.8pF. Thus Voul = 150 ,n3 x 1ÿ1 4A x 10J ,, „ Al O x 5V = 14.9 mV. For IV peak on the load ends, x 5V = IV > Zc = 2 Zc 2 x 667 x 103 10k£2, and / = 1/(271(0.8) X 10"12 X 10 X 103) = 20MHz.
—— 
For saturation: Quiescent collector voltage = Vcc ~ //2 (4k) = 10 0.15 X 4 = 9.4V. For 5V peak on the bases, saturation begins for a collector voltage of 5.0V, for which the peak load signal is 9.4 5V = 4.4V. This occurs at a frequency of (4.4/1) X 20 = 88MHz.
—
10i4150 x 10"6) = 166.7ft, (15(yi51)x(M66.7)_03 =
7.39 Here, IE = 150pA , re = 25 X 271 x
M
109
(15(yi51)2.7 x = 2(166.7)
For fH:
103
X
2x 166.7x 151 2x 166.7 x 151 + 10 x
_ 103
2.68 x 103 333.3
503 = 60.3
CT= 0.651+ 0.3 = 0.625pF,
10)103 = 8.34Jtft. Now, fH = K2tc x 8.34 x 103 x 0.625 x 10"12) = 30.5 MHz. Note that the gainbandwidth product GB =6.71 x 30.5 x 106 = 204.6MHz
Rt
7.40
= [(2 x 166.7)151] II (10 x 103) = (50.3 II

IE = 150pA, re = (25 X 10_3y(150 X 10"6) =
166.7ft, Add rE = re = 166.7ft to double the input 2'7 x 103 x
, 151 150151 x W66.7 x Now, CK = 0.3pF = 0.65pF, AM = ÿ x 1q9 4 (166 ?) 4 x 166.7 x 151 2.68 x 103 100.7Jkft = J.ooV/V. 667 110.7kft 4 x 166.7 x 151 + 10 x 103 For fH: CT = 0.65/4 + 0.3 = 0.4625pF, RT = (4 (166.7) (151)) II 10k = 100.7k II 10k = 9.10kft, fH = 1/(271 X 9.10 x 103 x 0.4625 X 10"12) = 37.8MHz. resistance.
XT

_

Note here that GB = 3.66 X 37.8 = 138.3 MHz. Notice that this is considerably smaller than GB in P7.39 above. Why? (Hint: Although one might expect feedback through rE would allow gain and bandwidth to be exchanged (see Section 2.7 of the Text), we note here that Cp, which dom¬ inates, is outside of the feedback loop.].
7.41 From Eq. 5.1 15, fT = gmÿ27i(Cs.v + Cgd)) = 1 x 10"3/(27t(200 + 20)10~15) = 723 MHz.
 283 
SOLUTIONS: Chapter #716
The midband gain AM = + RpAYgm + Vgm) = + 5 x 103/(2(1 x 10 3) ') = 2.5 V/V. At the input: CT = Cgs/1 + Cgj = 2002 + 20 = 120/F, and RT = 10 kJTi, for which, fPin = K2ji x 104 x 120 x 10"15) = 132.6MHz At the output: Cp = Cji, + Cj„ = 100 + 20 = 120fF, and Rp = 5 kQ, for fpom = K27C x 5 X 103 x 120 x 10's) = 265 MHz. Overall, fUB = (132.6"2 + 265"V = 118.6 MHz.
which
re = VT/IE = 25 x 103/(150 x 10"6) = 166.612, or 0.16kG, and gm = P** = (PP + 1Yre = ct/re = (150251)166.7 = 5.96mA/V. Since fT = g„/(2n(CK + Cÿ)), CK + C„ = 5.96 x 10~3/(2rt x 109) = 0.949pF, and Cn = 0.949  0.3 = 0.649pF.
7.42 For all transistors,
For the gain: Now, for differential signals, consider halfcircuits: (P + 1)(re + re) = 151(2)(0.16)103 = 50.3k G.
Thus, the
ÿ
gain
Input resistance at the base is
(150251)2.7 x 103 50.3 „ = — = 0.834 x 8.05 X * == U„A)V =—— —— 2(0.166 x 103) 10 + 50.3 x>/2
ÿ
noo.
>
6.71 V/V.
For the cutofffrequency: At the input, Cp  C/2 + C,, = 0.6492 + 0.3 = 0.625pF, and Rp = 10&£2 II 50.3kG = 8.34kG. Thus, fPin = K2re x 8.34 x 103 x 0.625 x 10"12) = 30.5MHz. At the output: Cp = Cp = 0.3pF, and Rp = 2.1kG. Thus, fPou, = H2n x 2.7 x 103 x 0.3 x 10"12) = m.5MHz. Overall, f3dB  (30.5"2 + 196.5"2)" = 30.61 MHz 7.43 Comparing this with the solution of P7.42 above, the emitter resistances remain at 0.16 k£2, but gm = °ÿre reduces gmp to (5051)0.16 = 5MmA/V (while gmn stays at 5.96 niA/V. For this reason and the change in fp of the pnp, (CK + Cf)p = 5.88 x 10"X2n: x 0.3 x 109) = 3A2pF. Thus Cnp = 3.12  1 = 2.12pF. 5°'3. x (5(y5P2J x '°3 = 0.834 x 0.794 = 6.63 V/V. For the gain: = 2(0.16 X 103) 10 + 50.3 For the cutofffrequency: At the input: CT = (0.649 •• 2.12) + 0.3 = K 1/0.649 + 12.12) + 0.3 = 0.191pF {where " •• " indi¬ cates a series connection}, and Rp  10&12 II 50.3kG = H.34kG, whence fPin = 1/{2jc(3.34)103 x 0.797 x 10~12) = 23.9MHz. At the output: Cp = Cp= 1pF, and Rp = 2.1kG. Thus, fPout = \A2n x 2.7 x 103 x 1 x 10"12) = 5&.9MHz Overall, f3JB = (23.9"2 + 5ÿ.9'YA = 22.1 MHz. Notice, in comparsion with the results of P7.42, that the details of the specifications of the pnp are quite important!
284
Chapter 8
FEEDBACK SECTION 8.1: THE GENERAL FEEDBACK STRUCTURE 8.1 From Fig. 8.1, see x, = xs  xf = 1.00  0.99 = 0.01V. Thus A =x)/xi = 3.00/.01 = 300V/V. Thus P  Xf/x„  0.99/3.0 = 0.33V/V. The openloop gain is A = 300V/V. The amount of feedback is (1 + A p) = 1 + 300 (.33) = 100. The closedloop gain is Af =x„/xl! = A/(l + A p), where, directly, Af 3.0/1.00 = 3.0V/V, and, indirectly, is 300/(1 + 300 (.33)) = 300/100 = 3.0V/V. For 'the p network disconnected, 1), = vs, and t)„ tends toward A D, = 300 (1.0) = 300V. This value would not be meas¬ ured, since much before it is reached, the output would typically limit (or saturate). 8.2

See from Af = AA. 1 + A P), that 8 = 102/(1 + 102 p), or 1 + 102 P = 102/8, whence P = (102/8 1)/102 = 0.115. Now P = /?iA/?i +Rz) = 0.115, or (Rl+R2YR, = 1/0.115 = 8.696 = 1 + RJRX. Thus Rÿ/Rx = 8.696 1 = 7.696. Amount of feedback is 1 + A p = A/Af = 102/8 = 12.5V/V. a 201og]0 12.5 = 22dB. For V, = 0.125V, V„ = Af Vs = 8 (.125) = IV, Vf = p V„ = 1 x .115 = 0.115V, V, = V„/A = 1/102 = 0.01V, or V) = Vs Vf = 0.125 0.115 = 0.01V, as expected. For A increasing by 100%, A becomes 2(102) = 200, and Af = A/(l + A p) becomes 200/(1 + 200 (0.115)) = 8.33, rather than the former value of 8. That is, Af increases by (8.33  8)/8 x 100 = 4.12%.



SECTION 8.2: SOME PROPERTIES OF NEGATIVE FEEDBACK 8.3 For the original design, Af = A/( 1 + A p) = 103/(1 + 103 x 10~2) = 90.9. For the fabricated design, Af = 0.5 x 103/(1 + 0.5 X 103 X 10"2) = 83.3 results. Now, the desensitivity factor, 1 + A p, as designed, was 1 + 103 x 10"2 = 11, and, as fabricated, is 1 + 0.5 (103) (10~2) = 6. Thus for (small) changes around the original design, one would expect the original 50% reduction in A to result in a 50/11 = 4.5% reduction in Af. Using the changed desensitivity factor, a 50/6 = 8.33% change would be expected. Now, concep3 At/A, J i 2 tually (for small changes), the sensitivity, = = = .091 a 20.8dB.
—— —
d A/A 1+A p 11 A A/Af (90.9 83.3)90.9 7.690.9 zr r = Actually, = A A/A = 3 3 A/A 5001000 (103 0.5 (103)yi03 manufactured closedloop gain is (as calculated above) 83.3.
dAf/Af _
8.4

„
= 0.167
_
a 16.7% The
resulting 6
We know that A p = 89, and Af = A/(l + A p). Thus, ideally, 99 = A/(l + 89), for which A = 99 (90) = 8910, and p = 89/8910. After a time, Af = 98 = A/(l + A (89/8910)), or A = 98 + 98 (A ) (89)/8910 = 98 + .9789A. A = 98/(1 0.9789) = 4645, lower by more than a factor of 2. Check:
Af
4645 ™
—
=
1 4* 4645 (oSvoyiO)
= 98.000. Thus gain A has reduced by (8910  4645)/8910 X 100 = 47.9%.
8.5
For the closed loop, 1 + A p = A/Af = 104/102 = 100, and P = (100  1)/104 = 99 x 10"4. Thus the closedloop 3dB frequency is (104) 100 = 10 6Hz. For the basic amplifier, GB  104 X 104 = 108Hz. For the feedback arrangement, GB = 100 x 106 = 108Hz, the same! (as expected!)
8.6
For / mb of A reduced to 2 x 103Hz, the 3dB closedloop frequency reduces to 2 X 103 x 102 = 2 x 10sHz. On the surface, it appears that the desensitivity idea is not working, since the percentage change of the openloop and closedloop 3dB frequencies are the same. For the original amplifier at 104Hz, the openloop gain is down by 3dB, to 104/1.414 = 0.707 x 104, a percentage reduction of about 30%. Correspondingly, the percentage reduction in closedloop gain = 30/1 + A P = 30/100 = 0.3%, from 100 .3/100) = 99.7. For the manufactured amplifier, for which /3jB = 2 x 103Hz, the gain at to 100 (1

285
SOLUTIONS: Chapter #82
10 Hz is
104 (1 + (104/(2 x 103))2)'7'
Am = (l2 + (f/fn)2 t
104 (1 + 25)'a
= 0.196 x 104. Now
0.196 x 104 = 96.06. This corresponds to a drop of about 4% in gain 1 + A (P) 1 + .196 X 104 (99 x 10"4) for a change of / (from 104Hz to 2 X 103Hz) of 80% in frequency, an improvement of about 20 times. Note that at 104Hz, (1 + A P) is 26, correspondingly. Thus the desensitivity factor is still at work, maintaining the gain for frequencies above the cutoff.
Af1
8.7
—
With a lownoise preamplifier of gain A2, S/N = (V/V„ ) A2. For an improvement of 40dB, A2 = lO40ÿ0 = 100V/V. New S/N = 3dB + 40dB = 37dB.
(20mV, 1V)
8.8
0V/V
slope
ÿ
91V/V 50V/V
.1mV, 0.1V)
I I I II I I I I < 0.1V, and A = 103V/V, Af =
103
= 90.9V/V. Now for v„ = 0.1V, o.v = 0.1/90.9 a 1 + 103 (.01) 102 l.lmV, and u, = 0.1/103 = O.lmV. For 0.1 < t)„ < 1.0, and A = 102V/V, Af =  = 50V/V. 1 + 102 (.01) Now for u„ = 1.0V, t)5 = 1.0/50 = 20mV, and t>( = 1.0/103 = lmV. For x», > lmV, \)„ limits at IV, and A = 0, for which As = 0, as well. For
'
SECTION 8.3: THE FOUR BASIC FEEDBACK TOPOLOGIES 8.9
For the circuits shown, the feedback type is: V„/R2 i If (a) ShuntShunt: [3 =  —— . r = —— =
—— ÿ
ÿ
_
R2
If
_
(b)
ShuntSeriesK: Now, assuming R2» r, P =
(c)
SeriesShunt:
(d)
SeriesSeries: Now, assuming (Rt + R2) »r, p =
P=
Vf V„
+ R2
_
l0
/o
r
R2
Ri _
R1 R{
I„ r/R2
V„
R! + R2
vy
R
—— ~ R \ + R 2 o
r
For (b), assuming R2~ r, I j =
r + R9
 286 
hr
rR,
Rj + R2
o
assuming
(d),
For
_
Vf
_
Rj + R2
r
_
Rj
+ R2) r + R\ + R2 r(Rj
R\ + R2
I„
/0
Vr = (r  (/?, + R2)) I„
(Rl+R2)r,
Ri
ÿ
__
ÿ
A, A,
8.10 Assuming gm is very high, the basic gain A is high, and rs

SeriesSeries: V/ = r i„ , and P = x>fA„ r. ShuntShunt: if = tvft/r, and P = ifA)„ 1/Rp.
(a)
(b)
SOLUTIONS: Chapter #83
f
*
=
r
+_ÿo
+ a i+ a2
x I<"
and
R
r + Ri + R2
— Vgm ~ 0.
—
SeriesShunt: X>f =v„, and P = VfA)„ = 1. SeriesSeries: V)/ = i„ r, and P = vfA0 = r.
(c) (d)

8.11 Here, V, = V„  Vf = IV 0.1V = 0.9V. Thus A = 1,/Vj = 2A/0.9V = 2.22A/V = 2.22S, and = 0.1V/2A = .05V/A = .050 = 50ml2.
p = VfA„
SECTION 8.4: THE SERIESSHUNT FEEDBACK AMPLIFIER 8.12 For the p network, P = 2/(2 + 18) = 0.1, as suggested. Also, 7?n = 2kl2 18kCl = 1.8kl2, and R22 = 2kl2 + 18k£2 = 20kl2. For the A circuit, there are losses at the input and output. For a zero impedance source and no load: Thus A =
Rl I' Rll R22
Ria
Ria+R: + Ru*A"XRL* :Rm =
10 10 + 0 fl.8
00
II 20
* 10° * ~ II 20 + .01 = 84JV/V
Now for A = 84.7V/V, p = 0.1V/V, A P = 84.7 X 0.1 = 8.47, and 1 + A p = 9.47. Thus A} = A/(l + A P) = 100/9.47 = 10.6V/V. Now, Ri = Rs + Ria +/?,,= 0 + 10kO + l.8k!2 = 11.8kO, and R„ = RL II Roa \\R22 = 00 II 1012 II 20kl2 = 10O. Thus with feedback, Ri{ = Rj (1 +AP) = 11.8 (9.47) = 11.2kl2, and, Ro{ = RA1 +Ap) = 112 X 10.6 x 10/9.47 = 1.0612. For a 0.1V rms, lOkCl source and 10012 load, x>out =0.1 X ÿ
100/(100 + 1/.06) = 0.963V rms.
8.13 For the feedback network, 10kl2 + 190kl2 = 200kl2.
Fur fc A
lÿ R0 Roa }
=
circuit:
p = 10/(1290 + 10) = 0.05V/V, Rn = 10kl2 II 190kl2 = 9.5kl2, and Rz2 =
*=ÿ+
+ K„
»
x
= 227V/V . Ri=Ria+Rs+Ru t II Rl II R22 = 1 II 1 II 200 = 0.499kl2.
J
=
R
=
20
+
10 +10 + 9.5 10
+
9.5
" =
900
X
39.5kl2
J",7 = Af1 = 1 +A„P A, = 1— + 227 X .05
= 18.4V/V, Rtl = Rt (1 + A P) = 39.5 x 12.35 = 12.35 488kl2, Rof = R,A\ + A P) = 0.499/12.35 = 40.412. Resistance seen by the source is R,{ Rs = 488 0 0404 x 1 10 = 478kl2. Resistance seen by the load is R„f II (Ri) = 40.412 II (1A12) = = 0.0421kl2 Overall,
t
„g
ÿ
= 42.112. Overall gain, V„/Vs = Af = 18.4V/V. 8.14 For the P circuit (between nodes F and B): P = 100kl2/(lM12 + 100kl2) = 0.0909, /?u = 100kl2 II 1M12 = 90.9kl2, R 22 = 100kl2 + 1M12 = 1.1M12. For the A circuit (between nodes A, B to F): For Q\, Q2, lE = IOOjxA. Thus re = 25/0.1 = 25012, rK = 250 (120 + 1) = 30.25kl2. Thus Riti = 2 (30.25) = 60.5kl2. For Q5, assuming V„ ~ OV, lE 2mA, and re 25/2 = 12.512, with rK = 12.5 (121) = 1.51kl2. Input resistance to the right at node C is Ric (120)


 287 

SOLUTIONS: Chapter #84
+ lk£2 +
10k£2 II 1.1M£2] = 1.32M£2. Now ignoring device r„, the gain from S to F, is 1.32 X 106 60.5 10k£2 II 1.1M£2 J20 u,v/v A = °r " 1143V/V' 100 + 90.9 + 60.5 121 250 + 250 10k£2 II 1.1M£2 + lJfc£2 + 12.5£2 with Ri = (100 + 90.9 + 60.5)k£2 = 251.4k£2, and R„ = 10k£2 II 1.1M£2 II (RQ5 + lk£2). Now, since r„ of the devices has been ignored, the resistance driving node C is infinite, and thus the output resistance at the emitter of Q$ is also infinite (however strange that may be!). {Note that if VA = 200V, [12.5
_

= 2M£2, and roc = 1M£2, and R0s y ~ 106/121 = 8.3k£2, which with lk£2, will reduce Ra by about a factor of 2, not a large effect}. Ignoring the latter effect, R„ = 10k£2 1 1 1.1M£2 = 9.91k£2. Now, A = 1143V/V, p = 0.0909, Ap = 103.9 and 1 + A p = 104.9. Thus, — = , n = r02 = roi
lOOflO6)
"\)5
1143
Aa
1 4 A p
= 10.9V/V. Now, Ri( = Ri (1 + A P) = (251.4k£2) (103.9) = 26.1M£2, and Rin = Rl(  Rs = 26.1  .1= 26M£2. Now, R„f = R„/(1 + Ap) = 9.91k£2/103.9 = 95.4Q, and Rout = R„f II (~RL) = 95.4 II ( 10k£2) = 96.3£2. For source resistance and load resistance considered separately in the reduction of x>„A>s to 'A, the required Rs = 26M£2 and the required RG = 96.312. Clearly in practice, reduction of gain
through loading will predominate.
__
8.15 Assume = 0 and that current splits equally between Q\,QA. Now for Qi> 22> <24. 'o = (200pA)/2 = 100(1A = K (x>GS V,)2. Thus, \>GS 1 = (100/100)* = 1, x>GS = 2V, gmi = 2K {mgs V,) = 2 (100 X 10"6) (2 1) = 200pA/V, r0i = VA/ID = 20/100 x 10~6 = 200k£2. For Qs, iD = 200pA, (vGS 1) = (200/100)* = 1.414, t)OT = 2.414V, gm5 = 2 (100 x 10"6) (1.414) = 282iA/V, r„s = 20V/200pA = 100k£2. For the P circuit, P = 1.0, R\\ = 0£2, R22 = °°. For the A circuit,




—

— = 100k£2 II 3.55k£2 = Ri  Rs + Ria +R\\ = 1M£2 + oo + 0 = oo, R„ = r„5 II gm5 = 105 II 282 x 10* ' 11 r<"> 2 100m 2 (IQQkDAL) ÿ_ 193V/y Thus x 336kfl a = = 100k£2 10"6) ros+l/gms Vgm\ + VgmA 2 ( 1/200 X + 3.55*£2 A Ap = 19.3 X 1 = 19.3, and 1 + AP = 20.3. Thus Af = = 19.3/20.3 = 0.951V/V = D.A,. 1 *f* A p Rof = RoAl + A P) = 3.36/20.3 = 16612, that is R„u, = 16612. Now Rl( = oo (20.3) = oo, that is Rin = <*>£2.
_
_
Now for lk£2 load, the overall gain, tvb, = 1/:£2/(166 + lk£2) X .951 = 0.816V/V. OV, FG5 = 2.41V, VG2 = 5 2 = 3V, Vjj = Vs2 = 0 Concerning offset: For V„
= 25ÿ. in
oimmc in <•<>,=
2 41


=
= IOhA, in
r»j =
 2 = 2V, that is =
I2.95ÿA, in
2
= 22iA. Thus, assuming an ideal mirror, the net current offset at the gate of Q 5 is 25  10 + 12.95  22 = 5.95pA. Input offset to compensate is Vos = I0/gm . Here G„, = = —= J— =gml = 200pA/V. Thus Vos = 5.95xA/200pA/V = 29.8mV. r04 =
'onni r> 2Q0&12
rs\
+ rs2
'
—
i/gmi + Vgm2
VgmX

100 = 8.16 See that 100xA in the drain of Q\ forces ipi = lOOpA. Since / = 200(iA, thus iD2 = 200 100iA. Since iD 1 = iG2. for \)s = 0, then x>0 = 0, and iKi = 0. Thus i03 = lOOpA also. Now lOOjiA = lOOpA (y>Gs l)2, whence vGs = 2V. Also gm = 2K (t)Gj V,) = 200pA/V, rs = Vgm = 5k£2, and r„ = 20/100pA = 200k£2. For Q2 seen as part of A, the feedback is a wire for which P = 1, R\i = 0£2, and R22 = <*>£2. Now, the A circuit consists, at the input, of i)v connected via 1M£2 to the gate of Q\ and the gate of Q2 grounded Thus At the output, RL only is connected to t>„. (through /? 1 1 = 0£2). 200ÿQ 10*n 11 200kil x A = = 38.1V/V, Ri = 00, and R„ = rQ3 II RL = 200k£2 II 10k£2 = 9.52k£2. 5ÿ£2 + 5k£2 5k£2 Thus AP = 38.1 X 1 = 38.1, and 1 + Ap = 39.1, and Af = { = 0.974V/V,


+ Rif = Ri (1 + A p) = 00, Rnf = RJil + A p) = 9.52/39.1 = 243£2, that is Rout = 243 II ( 10k£2) = 249£2, and m„A)s = 0.974V/V.
 288 
SOLUTIONS: Chapter #85
Now for Qi seen as part of p, the feedback is a resistor rs2 = 5kl2, for which P = 1, and /?n = 5kl2, and R\2 = oo£2. Now the A circuit, at the input, includes R\\ = 5kl2 to ground from the source of Qt, and
otherwise is as before. Thus A = 38.1V/V, and P = 1, as before, with the same results. Now for Q2 and Q3, both 10 x wider with I= 1.1mA: iDX = 100XA as before, while iD2 = ip3 = 1mA. But K2, K3 are each ImA/V, and vGSt = vGS2 = vCs3 as before, but gm2 = gm3 = 10 (200iA/V) = 2mA/V, rs = Vgm = 0.5kl2 and r03 = 20/lmA = 20kl2. Now using the first idea (ie Q2 part of A), x A = = 484.8V/V, Ap = 484.8, Ap + 1 = 485.8, R0 = 10kl2 II 20kl2 =
ÿoTnÿto mÿ"2°kh 0.5x12 6.66kl2. Thus — = A, = , Atn = 4ÿ1 +A P 485.8 •uv 5x12 + 0.5k £2
J
= 0.998V/V,
Rof1 =
6.66kl2/484.8 = 13.7512,
Roul =
13.75 II ( 10kl2) = 13.812. See a great improvement in performance as a unitygain buffej:.
8.17 Following the idea in P8.16 above, see t)„ = 0V and iD\ = i04 = iD2 = iD3 = 100i.A, for which gm 1 = gm2 8m3 = 8m4 = 200J.A, rs = 5k£2, and r„ = 200kf2 for each. Here, consider the A circuit to consist of J2i with source grounded through /?11( and Q2 loaded by R22 and RL. Correspondingly, the P circuit consists of Q2 and Q4 with r,2 = rt4 = 5kf2, where P = 5/(5 + 5) = 'A, RiX = 5kf2 II 5kI2 = 2.5kI2, ' ÿ 200kil n„nr on C .n kT .n.n tl Thus, A = Now, R„ —— II 200A:£2 = 26.0V/V. — X 10A:i2 II 10ArC2 = lOkQ. + 5kfl R22 5k£2 5k 12 5«i2 + 2.5a: 12 = 10kl2 II 10kl2 II 200kl2 = 4.88kl2, A p = 26.0 X 'A = 13.0, A p + 1 = 14.0, Af = A/(A P + 1) = 26/14 = 1.857V/V, Rof = 4.88kl2/14 = 0.349kl2, Rom = 0.349 II ( 10kl2) = 36212.

———— —  
—
ÿ
SECTION 8.5: THE SERIESSERIES FEEDBACK AMPLIFIER 8.18 For the A circuit: At the input, R$, /?,«, and 7?u are in series such that /?,ÿ = 10kl2 + 20kl2 + 10kl2 = 40kl2. At the output, Roa, Ri, and R22 are in series such that R0 = lk!2 + lk!2 + 0.2kl2 = 2.2kl2.
= — X — = 4?"X — 40 Uj Ri R0
= 204.5mA/V. Now yp = 50V/A = 0.05V/mA, and A p P= 2.2k kl 204.5 X .05 = 10.23, and Ap + 1 = 11.2, A{ = 204.5/11.2 = 18.3mA/V. Ru = 40kl2 x 11.2 = 448kl2, Rin = R,{ Rs = 448  10 = 438kl2, R„f = 2.2kl2 X 11.2 = 24.64kl2, Rout = Rnf  RL = 24.64  1 = 23.6kl2.
Now A=
8.19 Using the results of Example 8.2 as much as possible: As before p = Vf'/I„' = 11.912 , with /„' now in the emitter of Qy A I„'/Vi' 20.51/.99 = 20.7A/V, with /„' corrected for being in the emitter of £>3. 13.65kl2, R„ = R22 + RL + R03. Here R22 = RE2 II (RF + REl) = 100 II (640 + 100) = 88. 1Q. RL Ri = and 60012, = R03 = re3 + Rc2/(P + 1) = 6.25 + 5kl2/101 = 55.812. Thus R„ = 88.1 + 600 + 55.8 = 74412. Now A = 20.7A/V, p = 11.912, A P = 20.7 X 11.9 = 246.3, A P + 1 = 247.3, Af = A/(l+A p) = 20.7/247.3 = 83.7mA/V. Thus Ri( = 13.65 x 247.5 = 3.38M12 = Rin (since Rs = 0), Rof = (744) (247.3) = 184kl2, and R„ul = 184kl2 0.6kl2 = 183kl2, as seen by RE.



8.20 For Q\, Q2, Q3, Q 4, iE ~ (200J,A)/2 = 100p.A for balanced operation. Thus 25012, rn = 101 (250) = 25.25kl2, and r„ = 200V/0.1mA = 2M12.
re =
25mV/0.1mA =
For Qs, iE  1mA. Thus re = 2512, rn = 101 (25) = 2.525kl2 and r„ = 200V/lmA = 200kl2. Now, consider the 10kl2 at the base of Q2 (included to compensate for the dc drop in Rs) to be part of the A circuit. Thus the P circuit consists only of the 1012 resistor, and P = x>fA0 = 1012, for which /?n = 1012 and R22 = 1012. For the A circuit: For the input series connection, /?, = 10kl2 + 2(25.25kl2) + 10kl2 + 1012 = 70.5kl2, and for the output series connection, R„ = /?22+ Ri + (re + (1/(P + 1)) (>*04 II J" 02)) II A" 05 = 10 + 103 + (25
.ÿ)»(200 2
x
W
X
,0» = 105ka No
(2M£i 11 2Mn 11 (101 (likn + 10Q + 25£2) ) ) X
2 (250)
 289 
1 (11:12 + 1012 + 2512)
SOLUTIONS: Chapter #86
= 0.716 X 376.2 x .966 x 103 = 0.260A/V = 260mA/V. A p = .260A/V x 10V/A = 2.60, A p + 1 = 3.60, Af =
Kj
= /?„ (1 +AP) = 10.5k£2 x 3.6 = 37.8kO, whence Ri{ = /?, (1 + A P) = 70.5 (3.6) = 254kO, whence Rin = /?if
/?o/
=

Agj =
3*60
= .0722A/V or 72.2mA/V,
= /?()/  ÿ = 37.8  1 = 36.8kO. Now = 254  10 = 244k£2.
SECTION 8.6: THE SHUNTSHUNT AND SHUNTSERIES FEEDBACK
AMPLIFIER A„t); D„ 8.21 For the basic amplifier: R,„ = r = —— = A„ Rj = 900V/V x 20k£2 = 18M£2. For the purposes of
—
D/iv
Ij
ÿ
shuntshunt analysis, convert the input into a current source Is = Vj/lOkO with a shunt Rs = lOkO. For the
p circuit: PP = 4 =  7 = 7 = 10_5A/V, Ru = 100k£2, and R2i 100k£2. 100 xlO3 V„ Rf
For the A circuit: The input network consists of lOkO II 100k£2 II 20kf2 = 6.25k£2. The gain is Rm = 18MI2 (for input current flowing into the 20kf2 input resistor). The output network consists of a scries resistor of Roa = lkf2 and a shunt load of /? 22 Rl = 100k£2 II lk£2. Now V„' .... 10# II 100# 1# II 100# 9.09 ,n6 tn6 1G 10 x 18 x10' V/A ° it + mi loot W720 x 18 xltf x 10* II loot H 20*
a°T
,1
Af
O 99
*
+ 0.99 =
— ~28 10 * 29
= 2.80 X 106V/A = 2800V/mA = 2.8V/pA, A p =  2.8
= ij
"
*
~A 1 "t u
X

106 x 105 = 28, 1 + A P = 29,
ÿ
=
= 96.6V/mA. For A , /?, = 6.25k£2, /?if = 6.25kO/29 = 216£2, Rin
= 216 II (10kQ) = 22112, R„ = lkft II 100k£2 II lkQ = 498£2, R„f = 498/29 = 17.1612, Rou, = 17.16 II (lk£2) = 17.5£2. Now, for a load of lk/2 = 50012, the gain reduces to 500/(500 + 17.5) X (96.6V/mA) = 93.3V/mA. To compensate at the input, seen as a fixed current source, we require that the source resistance increase from 10k£2 in order that more of the available input current enters the amplifier. For a source Is, Rs and
Rin = 22112, we
want the same output voltage for
the original and new loads. That is,
Rs ——221 — Is x + Rs
x %<3, or Rs = (221 + /?s) (1013). or Rs = 221/( 013) = 17k£2. Thus it is not possible for normal input circuits to compensate if Is is fixed. Alternatively, if the input voltage is fixed at Vs = Is (10k£2), then we may lower Rs (from lOkQ) so that the output is the same for the original and « • <"*> x 933 =
—221 t~
l(Ja £2
*
* If STiff *
if
"
ÿ
96 6 , 221 + 10k£2 = (221 + Rs) 1.035, whence Rs = 10.221/1.035 221 = 9.65ki2. 221 + 10#
8.22 At low frequencies, with the Ri, R2, R3 loop viewed as defining the voltage at node A, operation is as a voltage regulator, with the reference voltage being Vbe\ Here, R\ acts as a resistivewire connection to the input with a voltage comparison being made across the baseemitter of £?i Thus the feedback is of the seriesshunt variety. As a result of it, V& ~ 0.7V, lB2 = 0.7/700 = 1mA = IE2, VB = 0.7 + 1(1) = 1.7V, Vc = 1.7 + 0.7 = 2.4V, IC\ ~ (52.4)/2.7k£2 = 1mA, with rel = re2 = 25mV/lmA = 25i2, and = rn2 ~ 2.5k£2. At high frequencies, feedback is of the shuntseries variety, with I„ as output and Is = V/Rs as input. I' ~~R$ 1 For this, the P network consists of Rx and /?5 with P = =  = 0.166A/A, Rÿ + Rs /„' 1+5 Ru = R5 + R3 = 6kQ, R22 = R5 II R3 = (5 X l)/(5 + 1) = 0.833kf2. The A circuit consists at the input, of Rs II /?n II rn1 = 10kf2 1 6k£2 II 2.5ki2 at the base of Qt fed by /,and /„' emerging from the emitter of Q2 and connected to R22 to ground. The output resistance associ¬ ated with /„' is R02 = re2 + R/(P + 1) = 25 + 2.7k£2/101 = 51.712. Now, using a currentdivider
—
290
—
__
SOLUTIONS: Chapter #87
,
ÿ0
10 II 6
_— —
2.7 approach p, A = x 100 x with device H , „ — x I0l = = vv 10 l/l 6 + 2.5 2.7 + lOl (.025 + .833) /,' 3.75/6.25 x 100 X 2.7/89.4 x lOl = 183A/A. That is, A p = 183 X ( 0.166) = 30.5, l + A P = 31.5, / _ioi II = = 5.81A/A. Now A/ = = 5.81/104 = 0.581mA/V. Now A/ =~ •
>_
<
•
Da
———————
———
ÿ
—
f/j Rs V$ Is 31.5 Ri = R3 II /?,, II rKl = 10kl2 II 6kl2 II 2.5kl2 = 1.5kl2. Thus, R,{ = 1.5kl2/31.5 = 47.612, Rin = 47.6 II (10kl2) ~ 5012. Also R„ = R22 + R02 = 833.3 + 51.7 = 88512, and R„f = 885 (31.5) = 27.9kl2. This is
the resistance seen (for example) by a lowresistance load inserted between node B and the emitter of Q2. Since V& is assumed large and r02 °°. then Rou, is also infinite (independent of the feedback detail). But, even if rÿ2 were very low, Rou, would still be extremely high (see Eq. 6.78 on page 519 of the Text), since R„f is quite high. 8.23 For the dc loop, a seriesshunt configuration,
Ri = rKl = 2.5kl2, R„=R3 + re2 +
P = 1.0 (via R 1) with i?n = 10kl2, and R22 = °°12.
700 2.7*12 II (101 (2512 + 1.7*12)) —2512 x ~ij25
4 = 0.977V/V. —43 44.4
9.4612! For a
= 700 II 1051.7 =
or R„ = (70012)11(1*12 + 2512 +
42012, and the output of the A circuit consists of a 70012 load. Thus A =
.. .....
For A ,
—jjjy x
.
„.
A
A = ~434V/V TIlus' A P = 433, A,0.,, P + 1 = 444, Af = T+T3 =
420
= 9.4612. Thus the resistance seen by C3 is Rof1 = —— 44.4 IOO11F p capacitor, f n = —r = 168Hz. Note that before your exposure 2„ x 9.46 x 100 x 10"6 At node A, R„ = 42012,
to the effects of feedback, you may have considered the resistance seen by
700 II 1*12 II 10*12 = 0.4kl2, with /„ =
r = 3.98Hz! 12jc x 0.4 x l3 x 100 x 10~6
R3
to
be (roughly)
8.24 When C5 is removed, we end up with a complex feedback network, consisting of R3, R2, R\, R$ in a shuntseries feedback loop. Refer to P8.22 above for the basic calculations. For p, l„' sees two paths to I/, one through R$ ami one through R3. The resistance in the R3 path is lk!2 + 0.7kl2 II 10kl2 = 5 1.654 .. 0.7 , or If/I0' = p = (.2486 + .0492) = 1.654kl2. // = // 10 + 0.7 5+ 1.654 5 + 1.654 0.298A/A and /?,, = ((/?5 + R3) II R 1) + R2 = (5 + 1) II 10 + .7 = 4.45kl2, R22 = 1.654kl2 II 5kl2
=  ÿ.1U 4.4j Z.J = 1.24kl2. Using current ratios directly, A = pII 4" /j 2,7
2.7 + 101 (.025 + 1.24)
x 101 =
X 100 x  Irr 5.58
(.298) = 34.4, A p + 1 = 35.4, Af =
ÿ
130.4
X
100 X
x 101 = 115.4A/A. Now, AP = 115.4 x
= 3.35, A/ =
= 0.335mA/V.
8.25 Using results from P8.23: For C\, Ri„o ~ (r„i II R\) (Ap + 1) = (2.5*12 II 10*12) X 43.4 = 86.8kl2,  —7 86.8) x 103 Rsource sm ce ~ Rs 3 = 10k II 5k = 3.33kl2. Now, for 1Hz cutoff, C = 4 H R5 —7 = 2ic x 1 x (3.3 + 1.77p.F. For C2, RT~R5 + Rs H R,nD = 5k + 10k II 86.8k = 14kl2. For 10Hz cutoff,
—
!ÿ = 1.14JJ.F. C =2n xlO (14 x 103)
291 
————
SOLUTIONS: Chapter #88


8.26 For Fig. Q8.10 d) in P8.10 above, with gm 2mA/V, r„ = 10k£2, Rs = lOOkO, r lk£2, for RL = R £ 0, find Io/Vs and Roul (facing RL). This is a seriesseries feedback circuit, with the output current moni¬ tored by r and a corresponding voltage fed to the source of the transistor. Thus P = Vj 'A„ ' = (/„' ryi,,' = r = lk£2, for which R u = r = lk£2, and R22r = lk£2. The A circuit is as shown, where: r„ V gm V,' A = 2mA/V x (10/11 + Rl)
V;'
V,#
r.+fli+r
where /? = Rl in kf2 and /?, = oo.
1
RS
For /?/, = 0, A = 2
ro A = 2
A*
V
TTTTo
5. R22=r io'

i RL
= °95mA/v For
= 100kfl' A = 2
TTTToo
= 0.09mA/V. Generally, p = j— = r = lkf2, Thus AP = 1.82 x = 1.82 to 0.95 to 0.09 for Ri from 0 to 10k£2 to 100k£2 respec¬ tively, and 1 + AP = 2.82, to 1.95, to 1.09 correspondingly, with Ar =
Jo_ = Vg
_ 182 = 2.82
for Ri = lOkQ, or
See the gain is 3dB down from the 0ohm value, when
,645mA/V for
RL = 0, or 1.95 = ,487mA/V 0.09 = ,082mA/V for RL = 100k£2. 1.09
10 11+ Rl
10
= .456 + .456 (2a:). Tlius 11 +Rl
......
———— = j— (.645) = .456. 1+
2jc
= 1.82mA/V. For /?L = 10k£2,
*o
Rn=r
the form
jy
.456
11+/?/,
This is of
10 11 = = 0.419, for which /?/, =
*" ** 2(1 .456) .419 12.85k£2. It is a further 3dB down (ie 6dB altogether) at RL = 31k£2. For the output resistance of A, see that since the transistor gate and source are joined through the (floating) signal source, the MOS out20 R + 31 , )= put resistance is just r„ Thus R„ = RL + r + r„  11 + R Now 1+AP=1 + 1 (— 11+a 11 4 a Q1 D ) = (/? + 31)kO. Now the resistance seen by /?t is and R„f R„ (1+AP) = (11+/?) (
.
.
.

ÿ
—
——
ÿ
/?„„, = R„f /?=/? + 31/? = 31kft (!). The limit on the value of /?/, used depends on the degree to which transconductance constancy is required, and on the size of output voltage that can be tolerated while maintaining linear operation. Certainly for loads from 0kf2 to 12.9k£2, the transconductance varies only by 3dB, and by another 3dB for RL up to 31k£2. 8.27 Note the solution of P8.23. See that R i = lOkft = R ia + Rib with a tap at x tap, without feedback, the resistance
(/? + 2.5) II (10
is
. Let R lu = R . Now at the
— /? + 0.7 II (1 + 0.05)) = (/?
/?4
or Rox = + rK i) II (Rlh + R2 II (/? 3 + re2 + B+ 1 _ (2.5 + /?) (10.42/?) Now, + 2.5) II (10.42  /?) =
R„x ~ (Rla
12.92
9 /?«« = 0, when (2.5 + /?) (1) + (10.42 /?) != 0, 1042 + 2.5 2R = 0, or R = 12.92/2 = 6.5kO, for dR (2.5 + 6.5) (10.4  6.5) mo, r, 2.72 £1 in #u u L Rio D 0.65, Rox which =  = 2.72k£2, and,D = 61.312, with = nee R„f = ÿ
+
1 = 26Hz. 61.3 X 100 x 106 For fff = 168Hz, as in P8.23, C3 = lOOpF x 26/168 = 15.5pF, or 15.5% of the (lOOpF) capacitor needed with R 1 untapped.
the lowfrequency loop closed. For this situation, for lOOpF, fH =
292
2jix
SOLUTIONS: Chapter #89
SECTION 8.7: DETERMINING THE LOOP GAIN 8.28 At 1kHz, AB K =
127 , = 63.5V/V. — 20 x 10~3
At 10Hz, Ap =
r = 1550V/V. — 2 x 10"3
Assume the amplifier to F
be directcoupled, and that the feedback network employs a single capacitor to ground which limits the loop gain at higher frequencies by maintaining p lower there. Further, assume that the loopgain fre¬ quency response around 1kHz is relatively flat, and (separately) that at 10Hz, it falls as frequency rises. Since a singlepole response is postulated, it must fall (proportional to 1//) at 20dB/decade, from 1550V/V at 10Hz to reach 63.5V/V at 10 x 1550/63.5 = 244Hz. Thus the corner (3dB) frequency for the feedback network seems to be at 244Hz. Now, as frequency is lowered below 244Hz, P increases and A P increases, reaching 1550 at 10Hz and only 2 x 1550 = 3100V/V at 1Hz. Thus there must be an asso¬ ciated pole of P at about 10Hz/2 = 5Hz with the loop gain at dc being about 3100V/V. „Since the loop gain A P is 3100 at dc, and 63.5 at 1kHz, while P is frequencydependent, it is likely that at 1kHz, P is at most 63.5/3100 = 0.0205, and it may even be less if feedback is not unity at dc. Now assuming the
1—~rrr = , = 48V/V. For the 1 + 63.5 +A p ' closed loop, the lower 3dB frequency is a 3dB frequency of the P network, occuring in particular where P is 3dB high, at which the closedloop gain is essentially 3dB down (for A P » 1). For a closedloop pole at 244Hz with C = luF, Reaq = r = 650Q. 1 worst case, that is that A is at least 3100: Then at 1kHz, Ar =
10"6
2n x 244 x 1 x
8.29 At high frequencies,
R
p = —L— = —
9
g
= 0.0408V/V, and
AP = 0.0408 (1550) = 63.3V/V. At p has a zero when the magnitude
2 + 47 R  + R2 very low frequencies, P = 1V/V, and A P = 1 (1550) = 1550V/V. Now
z— = 32.5pF. The associ2rc x 2 x 103 x 2.45 frequencies, f„ = is at Jp ated pole = 64  0.100Hz. See at high P 2n (/?, + R2) C 2n (49k£2) x 32.5 x 10"6 that the gain will be . = 24.1V/V, and at low frequencies is essentially 1.00V/V. See 1 + 1550 (.0408)
of the reactance of C equals the resistance of Rt, for C =
7
that f lis the frequency at which P begins to rise, ie at the frequency at which P has a zero, ie at 2.45Hz, 1 A ie f 1= 2.45Hz. Generally speaking, Af = ÿ ~ p" Thus a pole of the closedloop response will + /ÿp occur at a zero of p. For the capacitor reduced from 32.5pF to lOpF, resistors must be raised by 32.5/10 = 3.25 times, to R\ = 2 (3.25) = 6.5k£2 (use 6.8k£2), and R2 = 47 (3.25) = 152.8k£2 (use 150k£2).
p
8.30 Loop gain = 1.2V/10mV = 120V/V. From the
network,
120/.0099 = 1.21 X 104V/V, is the basic opamp gain.
8.31 IAp (f)1= 10, [4f (/)(=10. Now Af =
A + ÿ
p
ÿ
p= , 100£2
ÿ
+ lOkiz
= 0.0099. Thus A =
Assume that at a particular frequency there is no phase
ÿ shift in A , but that Q = a + jb, and Af = x + jy. Thus x + j y = r = ./ a + VA + j b 1 + A (a + j b) a + ~j b a + Thus x = j y ~it Now x2 + y2 ~ lo2> and )2 + b2 (a + l/A) + b2 • 2 (a + VA )2 + b2 + VA~2 (q (a + VA) + b 1Q0 a2 (a2 + b2), whence A2 (a2 + b2) = 102. Combining: ((a + 1/4) + b ) a2 + b2 ~ z z j, (say). \ + A (a2 + b ) A A2 (a 2 + b2)2 + (a 2 + b2)  1 = 0, a2 + b2 = < or
' —
.

— —
ÿ—
_
—
.
_
——
Zn
A2 (a2 + b2) = (201)2,
4A2 =
l±ÿ2+4~40400,
A2 =
Thus
10100, A
1±
2!
+4A 2
= 100, 1 ± ÿ 1 + 4A2 = 200, 1 + 4A2 =
= 100.5. Now A2 (a2 + b2) = 102, a2 + b2 =
 293 
ÿ10
(100.5)
Thus
SOLUTIONS: Chapter #810
1(31 = (a2 + b2)y' = 10/100.5 = .0995. Correspondingly, at this frequency, A = 100.5V/V and ipi = 0.0995. An alternative approach is simply to ignore the possibility of phase shifts in either L or Af: Directly, A (3= 10 and Af = 10. Thus, Af = A/(l + A P) = A/f 1 + 10) = 10, whence A = 110, and P =AyA = 10T 10= 0.091.
_
8.32 In the circuit of Fig. Q8.15 in P8.15 above, open the loop at the gate of Q4 and inject a signal x>. Now, for Q i, Q2, 23. 24. 'd = 100(1A, that is iD = K (x>cs V,)2, or 100 = 100 (vGS l)2, whence x>GS = 2V, for which gm = 2K (vGS  V,) = 2 (100) (1) = 200(1A/V, r = Vgm = 5k£2 and r„ = 20/100(tA = 200kf2. Also for 2s. 200 = 100 (vGS  V,)2, and (x>GS V,) = <2= 1.414V, whence gm = 2 (100) 1.44 = 282(iA/V, rv = Ugm = 3.546kfi, and r„ = 20/200(iA = lOOkfl. Now, for no load, vnL.  2 (200k II 200k) 100* aQ= = 20 x 0.966 = 19.3V/V. Since p = 1, "J = H " (5k) 100k + 3.55k 2 " Af 1+Ap 200/:) = = 0.951V/V, the same as originally found. Now, for lkQ load, A p = H 1 + 19.3 2 (5k) 4.36ÿ1 1001:1111: 0.99 , „ , A .u o . x 4.36V/V, and 20 1, with 0.813V/V, where — —— = = = = = P TT7T Af1 1 + 4.36 1001: II Ik + 3.551: 0.99 + 3.55 the earlier calculation yields Af = 0.816V/V.


ÿ
.....
.—
.
— —
*
—
——
ÿ
SECTION 8.8: THE STABILITY PROBLEM 8.33 At frequency CO, = tan1 oVlO3 + 2 tan1 coTO5 = 180°.
Now at to = 105rad/s: <& = tan1 100+2 tan~'l = 89.4 + 2 (45) = 179.4°. Thus the phase shift is 180° at (slightly above) 105 rad/s.
103
A (10 ) =
ioVio3 10
l+
10J
( ioVio5 )2
i+
102 (V 2) 0? 2)
( io5/io5 )'
= 5V/V.
Now for P < 1/5, AP < 1/5 X 5 = 1, when 4> = 180 °, with no margins. For a 20dB gain margin, P = 1/10 x 1/50 = 0.02 or less. For P = 0.02, AP = 1, when A = 50. Now 20 logi0 50 = 34dB. Since the midband gain is 60dB, the gain drop is 60 34 = 26dB, implying a 3dB frequency of 26/20 or 1.3 decades above 103rad/s, that is at 103 x 101 3 = 2 x 104rad/s.

Check: At 2 x
103
103
104rad/s, 1+
104 103
2x
'2
V,
1+
104 10s
2x
2
(20.02) (1.04)
which O = tan_120 + 2 tan1 0.2 = 87.14 + 2 (11.3) = 110°, for a phase margin of 70°. 8.34
P=0.02
Not to Scale:
(10 P.0) for
C0» 105
ÿ
Re (10.0) (1.0)
(O.1.0)
A
(02x10
 294 
(00
= 48, for
SOLUTIONS: Chapter #811
From P8.33: A = 103, co;, i = 103rad/s, (Op 2 = 0)ÿ3 = 105rad/s, Pcritical  0.2. Now A P (co) ÿ ÿ r. Now from the solution of P8.33: Know at 105rad/s, .= =(1 + /aylO ) (1 + y'aylO5) (1 + ./0/tO5) /tp= 5p, (Ap) = 180°, and at 103rad/s, Ap = 0.7 X 103 p, (A p) = 45°, and at 2 X 104rad/s, A P = 48p, 4> (A P) = 110°. The plot is for P = 0.2, 2, and 0.02. Note that for 3 poles, the maximum shift is 270°. 8.35 For
p = 1.0, A p =1 at 0) = 103 x 106 = 109rad/s, at which the total phase shift is tan1 (109/103) + tan1
(109/108) = 90° + 84.3° = 174.3°. Thus, oscillation does not occur, although there is only 5.7° phase Now for P = 0.5, A = 2, when A P = 1, whence A = 1o6 = 2, for to just above 108rad/s. At 108rad/s, 1+ + a/108
margin.
A
a
( aylO3)3
(
106 103 (1 + 1y
= 7.07.
A =
10s
2 x 105 (1 + l2f 1038,rad/s, O = 90° + tan
]'
At
3 x 10srad/s, A =
105 (1 + 32f 106
~
1.
At
2 x 108rad/s,
x 2\/, = 2.05. Now at 2.1 2.1 x 103 (1 +2.V) 90° + 64.5° = 154.5°, for which the phase margin is 180°  154.5°
= 2.24. At 2.1 2.1 x
106 3x
108
108
X
10 rad/s, A =
.
= 25.5° Now withÿ an input capacitance, the p network provides a 3rd pole at
X„l°8 2\0.477
*
(0„
1
= 45412. Thus for the P network consisting of two 4.4 x 108 x 5 x 10 equal resistors R, R/2 < 454, or R < 91012. To ensure stability, even smaller resistors would be needed, say 10012.
SECTION 8.9: EFFECT OF FEEDBACK ON AMPLIFIER POLES
1 + 4 x 103 x 0.125
2ÿ X
ÿ
ÿ
= 4 X 103V/V. With feedback, Ar = 1— A = 5 x 103 + P f„ = 7.98V/V. fuB JpK + AP) v rel="nofollow"> = 5 x 103 x 501 = 2.5MHz. The unityJ 3dB =/„(l 501
A = gain, 8.36 The lowfrequency amplifier 6 v H
=
'
gain frequency = 2.5 X 7.98 = 19.95MHz, or 20MHz. Thus the pole is shifted by the amount of the feedback factor, that is by about 500 times.
8.37 From P8.36 above, for A = 4000V/V and fp = 5kHz. To achieve fpf = 10 Hz, the amount of feedback,
106
r = 200. Correspondingly, the loop gain is A P = 199, and the feedback factor, p 1+ A P = 5 x 103 4000 199/4000 = 0.04975. Use P = 0.05, for which the lowfrequency gain is Af = YÿjTÿoo x 05 19.9V/V.
=
104A"
For loop closure with a feedback factor P, the poles are: s = 1/2 (1 + s/105) (1 + s/iltf/K)) (cd,, 1 + (Dp 2) ± 1/2 [(co,,i + (Bpz)2 4 (1 + A„P) ©pi o)p 2]ÿ See that the poles are coincident at 5 x 105 when 5 x 10s = 1/2 (10s + 106/K), 10 = 1 + 10IK, or 10/AT = 9, K = 10/9 = 1.11. The poles are coin¬ cident when ((opl + (Dp2)2 4 (1 + AP) (Op, (Op2 = 0, or (10s + 106/K)2 = 4 (1 + AP) 10s x 106/X, or
8.38 A(j) =
= TÿT = 2778 Thus A P = L778' and P = 1778 x 10"4 1111115 the 4 (9) dc openloop gain of the amplifier is 10 K = 1.11 x 104V/V, with poles of 10sHz and 9 x 10sHz. The 1+A P =
4 (1(YK)
4 (9)
—\
Lowfrequency closedloop gain is Af = 1+ PA
 295 
ÿ
2.778
= 4 x 103V/V.
SOLUTIONS: Chapter #812

8.39 Note that for convenience of notation we will use co as the variable which denotes frequency in Hz! For a maximallyflat design, Q = 0.707, where from Eq. 8.38 in the Text, ((1 +Ap) (On i (On 2)ÿ
——— ((Opi + (0/;2)
Q =
(((Op! + (Op2)2

(14 Q2)f, or
(1) and (0,,, + (Op 2 = co/2, with poles at
4 (Op 1 (Op 2(1 + An$)f'. (°/''
That is, poles are at lp6 + 2P x 1q6
ÿC°/'2 (1 ± j (4 Q2  l)'/l). or
Wpl
—

1/2 (to„i + (0„2) ± 1/2
Wp2
'
'
± 1/2 (((Opi + (Op2)2
x (1 ± j (4 (.707)2 if') = 10.5 x
106 (1 ± j) Hz (as seen directly from Fig. 8.32 in the Text, and the text following), and w„ = Q (CO/,, + (Op 2) = .707 (21 X 106) = 14.85 x 106Hz = 14.85MHz (that is ÿ2 x 10.5MHz). Now from the characteristic equation (Eq. 8.37 of the Text), s2 + s ((Opi + (Op2) + (1 + A0 p) (Op,i (Op2 = 0, and (1) above, s2 + s ((Op! + (Op2) + Q2 ((0pi + (Op2)2 is the denominator of the transfer characteristic, which for s = ji0, is (  (o2 + jo) (21 x 106) + (21 x 106)2/2). Now at the 3 dB frequency (coc): [((Oc (21 x 106)) + ((21 x 106)2/2  (O2)]2 = 1/2. Normalizing to 21 x 106Hz, (oc2 + (1/2  (D2)2 = 1/2, or (O2 + 1/4 + ( (O2) + (0* = 1/2. Thus (0* = 1/4, (O2 = 1/2, (Oc = 0.707 (normalized), and, in general, (Oc = 0.707 x 21 (G («
.
S
XT
joÿ
22.7V/V. 8.40 A„ =
ÿUS
___
iq3
103, (0„i p = 105rad/s, and A (s) = 577. p = Wnj p = 103rad/s, (0„2 (1 +5/103) (1 +5/IO5)2
,f
_ 10ÿ_
10ÿ_
(1 +ÿ03)(1 +J/105)2
_ pJOÿ_ i+
' (1 + s/103) (1 + j/105)2 + p 103
(1 + j/103) (1 + sAO5)2 (1 is Denominator + .s/103) (1 + 2s/105 + s2/1010) + p 103 = 1 + 2sAO5 + s2/1010 + sAO3 + 2.s2/108 + j3/1013 + P 103 = (P 103 + 1) + j (2/105 + 1/103) + s2 (lÿO10 + 2/108) + iVlO13. Normalize to ÿ/105, that is 103 P + 1 + s (102) + s2 (201) + 100 j3. Now divide by 100 and set equal to 0: s3 + 2.01 s2 + (1), which is in the form (s + a) (s + b jc) (s + b + j c) = 0 = 1.02.S + 10 P + .01 = 0 (s + a) (s2 + sb + jsc + bs + b2 + jbc  jcs  jbc + c2) = (s + a) (s2 + 2 sb + b2 + c2) = (j3 + 2s2 b + sb 2 + sc2 + as2 + 2abs + ab2 + ac 2) = s3 + s2 (a +2b) + s (b 2 + c2 + 2ab) + a (b2 + c2) (4), (2). Now, sec a + 2b = 2.01, or a = 2.01 2b  (3), b2 + c2 + 2ab = 1.02 (5). Now, (4) + (3) 4 b2 + c2 + 2b (2.01 2b) = 1.02, b2 + c2 + ab2 + ac2 = 10P + .01 = jc (6). 4.02Z>  Ab2 = 1.02, c2 + 4.02b  3b2 = 1.02
  —
—

Special Cases: (i) Two poles are coincident when c = 0 (7), (3) » a = 2.01  2b
(8), b2 + 2ab = 1.02 (9). (5) 4 ab2 = x 2xÿ (10) (9) 4 (8) b2 + = 1.02 4 b3 + 2x = 1.02b (9) 4 (8) x/a +2ab = 1.02, x + 2a2b = 1.02a b3 = 2.01/2 b2xl2 (9) 4 (7) b2 (2.01  2b) = x, 2.01b2 2b3 = x (10) + (11) 4 2.01/2 b2  x/2 + 2x = 1.02b, 2.01b2  2.04Z> + 3jc = 0 (4) 4
b = 2.04 ±
ÿ2042 ~ 4 (+3*)2.01
2 (2.10) 1.49 (10P + 0.01), or p =
°1731~
(11)
_ Q 50? ± V .258  1.49. The two are identical when 0.258 = 1.49.x = 0 01
= 0.0163, at which b = 0.507 and a = 2.01  2 (.507) = 0.996.
296
SOLUTIONS: Chapter #813
Denormalizing, the poles are at about 1 x 105, 0.5 X 105, and 0.5 x (ii) Now the ju) axis is reached when b = 0, for which: (3)
105 rad/s.
—» a  2.01, and
(4) +
c2 = 1.02, c = 1.01,
where from (5), x = 1Op + .01 = ac 2 = 2.01 x 1.02, whence (3 =
2.05  .01 = 0.204. In this case, the 10
poles are approximately at 2 x 105 rad/s and ± j 10s rad/s. (iii) Now, Q = 0.707 for the complex pole pair implies that b = c , (4) 2ab2 = jc, (6) > b2 + 4.02b  3b2 = 1.02, 2b2  4.02b + 1.02 = 0,
4.02 ± 'V
2b 2 + 2ab = 1.02, (5)
4.022  4(2) (1.02)
4.02 ± 2.83 = 1.71 or 0.298. Now, from (3), for a = 2.01  2b 4 2 (2) positive, b = 0.298 for which a = 2.01  2 (0.298) = 1.41, and x  ab 2 + ac2 = lab2 = 2 (1.41) (0.298)2 = 0.024. Accounting for the initial normaliza¬ = 0.250. Thus x = 10p + 0.01 = 0.250, and p =
b=
—
tion, the pole locations for Q = 0.707 are at 1.41 x 10s rad/s, and at (0.298 ± j 0.298) x 10s rad/s. Now for Q = 0.707, for which p = 0.024, using the normalized frequency, w = ayl05rad/s, see T(w) =
103
103 (1 + jl00w)(l
1  w2 + 2jw Thus  T =
((25
1. Thus (squaring),
103
+ j lOOw

(1 + jlOOw) (1 w2 + 2jw)
+ jw )2 + 24
 j 100w3  200w 2 + 24
25
103
20 lw2)2 + (102w
+ 24
103
 201w2 + j
(102w
 100w3)   
(12).
Now at the closedloop unitygain frequency, T \ =
3\2x'/i  100wJ)z)
106 = (25  201w2)2 + (102w  100w3)2 = 625  10050w2 + 40401w4 + 10404w2 20400w4 + lOOOOw6, whence w6 + 2w4 + 0.035w2  100.06 = 0. Solve w6 + 2w4  100 = 0 by trial. For w = 2, 26 + 2 X 24  100 = 64 + 32  100 = 4. For w = 2.02, (2.02)6 + 2 X (2.02)4  100 = 67.9 + 33.3 100 = 1.2. Use w = 2 as an approximate solution. Now for w = 2, from (12), d>(w) = tan1 , 204  800 = 5% 102(2) 100(2') ÿ p Q2 = 779 25  804 25 — 201(22) with Q = 0.707, occurring at about 2 x 10s rad/s, is about 180  37.4 or 143°.
_
,
At(s)
.
10J
(1+S/103 K1+S/105 f +103 p
(.3+.31)10 (for p 024)
(for p0.204)
200.5x10®
(for p0.16)
2010® (for p0)
103
(for p0)
1.42x10® (tor P.024)
2x10®
(for p0.204)
(for p0.204)
297 
.„
SOLUTIONS: Chapter #814
SECTION 8.10: STABILITY USING BODE PLOTS 8.41 From Eq.8.48: At the unityloopgain frequency, Af (J(a) =
1/B e~,@ ,
ÿ
ÿ
_.Q
and \ Aj (y'ro) =
1/p
1 1 + e ye
((1 + cos 0)2 + sin2 Q)y' (1 + cos2 0 + 2cos 0 + sin2 0)14  j sin 0 1 . Now as noted on page 726 of the Text, for a margin of 45*, © = 18045 = 135*, and cos 0))
1 1 + cos 0
ÿ (2 (1 +
„
i/p
lAr I (2 (1 + cos 135*))'/"
= 1.307/p.
There is no peak when (2 (1 + cos x ))Y' = 1, or 1 + cos x = 'A, cos x =  'A, x = 120°, and phase margin = 180  120 = 60°, (and, of course, greater). For a peaking factor of 2: 2 = 1/(2 + 2 cos ©)'/l, or 2 + 2 cos 0 = 0.25, cos 0 = 1.75/2 = 0.875, or 0 = 151°, for which the margin is 29*. For a peaking factor of 10: 2 + 2 cos 0 = 1/100; cos 0 = 1.99/2 = 0.995, 0 = 174.3°, for which the margin is 5.7*.
8.42
20dB/decoda
78*
$ margin
45*
« margin
0* <J> margin
See that margins at 108Hz, are likely to be zero for 1/p s 40dB, or 1/p = lO4®20 = 100, or p = 0.01. Moreover, the phase margin is about 78* at 107Hz, where Pi contributes 90* and Pi , P3 each 6* to the total shift. The corresponding 1/p = 60dB, or P = 0.001. The phase margin is 45° at 3 x 107Hz, for
p = l/lO5020 = 0.0032. For p = .001, A,; = , At a = ÿr = 1+Ap 1 + 10 x 10~3 909V/V. For P = .0032, Af = 104/(1 + 104(.0032» = 306.5V/V. For P = 3 x 102, 1/p = 100/3 = 33.3, where 20 log 33.3 = 30.5dB = 30dB. See from the ligure that / = 108 (10 X 10/60) = 1.67 X 108Hz, and which 1/p K
2
50dB, and
K
the phase margin =  1/6 (90) = 15*.
104 , for which d> = tan"1 SAO6 2 tan" fAO6) (1 + j fAO8)2 ' (1 + / f/108. For 0 = 180*, check / = 108. See d> = tan"1 108/106  2 tan"1 108/108 = 89.43 2(45) =
—
8.43 For the situation in P8.42, A =
179.4°. For / = 1.1 X
108 O =  tan"1 110 2 tan"1 1.1 = 89.5 2 (47.7) = 185*.
298 
—
SOLUTIONS: Chapter #815
For / = 1.01 x 108, 0 = lan"1 101 2 tan"1 1.01 = 89.4 2 (45.29) = 180°. Thus margins are zero at / = 1.01 x 108Hz. ÿ3

1(]7
y
a y
n7
—
Now, at / = 3 x 107, $ = tan1 7 2 tan1 7  88.1  2 (16.7) = 121.5 for a margin of 108 10 180  121.5 = 58.5* (rather than 45°). For / = 4 X 107Hz, 0 = tan1 402 tan1 .4 = 88.6  2(21.8) = 132.2°, for a margin of 180° 132.2° = 47.8°. For / = 4.2 x 107Hz, 0 = tan1 42 2 tan1 .42 = 88.6 2(22.8) = 134.2°, for a margin of 180° 134.2° = 45.8°. Thus margins are 45° at f ~ 4.3 x 107Hz.
107
10?
107, the phase margin is 180 tan1 ÿg  2 tan1 = 180  84.3  11.4 = 84.3° (not 78° as suggested). For / = 1.2 x 107, the phase margin is 180  85.2  13.7 = 81.1°. For / = 1.4 x 107, the phase margin is 180  85.9  15.9 = 78.2°. Thus the margin is 78° at/ = 1.4 X 107 Hz. Now at / =
n4
1n4
Now, at / J = 1.4 X
1
p can be 1/698 107Hz, 11 \A I= 7 = T77T7Trxxr = 698V/V. Thus H 14.04(1.02) (1 + 142)" (1 +.142)
= 0.00143 (where the 0 margin = 78°). Now at J/ = 4.3 x 107Hz, \A 1=
_ _ 104 104= (43.01) (1.185) — 5 ~ , ,432) . ai2\ (1+ 43Y" (1 + 4i2\/j /1
/*
sai r\i\ f\ 1 qc\
= 196.2V/V. Thus rp = 1/196.2
= 0.0051 (where the phase margin is 45°).
104K with poles at 10sHz and 106/K Hz. (A) For 20dB of feedback, IO2020 = 1 + A p. Thus 1 + 104/C (P) = 10, P = 9 X 10~*/K. From the rateofclosure rule, the 1/p and A lines 1o5 should intersect at 106/K Hz, where A ~ 104AT X —g— = 103 K2, and
8.44 The available amplilier has a gain of
1°"ÿ= 9XJ°4.
? 1X 1X for Now p = K K ÿ X K = 1.11, for which the bandwidth is 106/AT = 0.9MHz, 9 x 104 A gain is As and the lowfrequency = \ aQ ~ +
where /lp = 1, or

p=
'2°
.
*+ÿ
*
'P* 1 io4a: (9 x io4/ÿ) +
=i°ÿ)=900V/v. 10
The second pole is at 106//f = 106/ 1.11 = 0.9MHz. Now from Eq. 8.36, the closedloop poles are at 1/2 (G)pI + ©p 2) ± V (CO,,, + co/)2)2  4(1 +A„ P) (Op Oip2. Here A0 = 0.9 x 104V/V, P = 1 x 103V/V, (of,i = 105Hz, tof,7 = 9 x 10sHz. Tlius the closedloop poles are at
_ (1 + 9)
X
10s
± ÿ(0
+9)
x 105)2
_ (1 + 09 x 104 x 1 x 103) (105) (9
X
105), or
5X
10s
± [(5 x 105)2  10 x 9 x 1010]'7', or  5 x 105 ± 105 (25  90)'7', or  10s ( 5 ± 8.06;).
»«* i+ pwr.p
= 0.1  Krtr. Now, using the rateofclosure rule, the intersection of the 1/p and A lines will be at IQ6/K Hz, where A = 103 K2. Now, at 106/K Hz, A P = 1. Thus 103 K2 (0.1  10~*/K) = 1, 100 K2  0.1 = 104A x (B, Now, for
i?— 10f>/K
K
 1 = 0, for
which K ~ 0.10. That is, K = 0.10, dc gain = 10V/V, p = 0.1
—
lOÿ/O.l = .099, A =
104/sT = 103, and bandwidth = 106/0.1 = 10MHz. The second pole is at 106IK = 106/0.1 = 10MHz.
299

SOLUTIONS: Chapter #816
SECTION 8.11: FREQUENCY COMPENSATION 8.45 A = 104, fp \ = 106Hz, fp2 = fpi = 108Hz. For an added dominant pole at fp0, the second pole would be at 106Hz. Now for Af = 10 (and A very high), (3 = 1/Af = 0.1. Thus 1/(3 = 20dB, while A = 80dB. Thus the dominant pole must drop the response by 80  20 = 60dB, using three decades. fpo = /,./103 = 106/103 = 103Hz. Similarly for A} = 1, fp0 = 102Hz. In both cases the closedloop 3dB
frequency would be
106 Hz.
s.p
8.46 For the existing pole lowered, from fpl = 106Hz to fp /, consider the effective second pole at / where tan1 //10s = 22.5", or / = 0.414 x 108Hz. Now for a gain of 10, must lower fpl by a factor of 1000 below 0.414 x 108Hz to fp{ = 4.14 x 104Hz. Now for a gain of one, similarly need fp { = 4.14 X 103Hz. In each case, / 2dB ~ 40MHz.
8.47 Now /„ =
=
Th" iS'
2n X C
2* Cx 10'
X
= 10S'
Now, for Miller compensation, y
l06 '
a"d
"
2ÿ7*7 = i.xCxlO1' 'frCq""Cy 10° 'imeS
" C * 2n X 10s
Jpl
X
10s
=
'•59"F
, or ———+ C2 Ri + Cf (gm R\ R2 + R\ + R2))
2k (C\ R{ 1 A
q4
10"12 x 106 + 10 x 10"12 x 104 + 6.28 Cf (100 x 106 + 106 + 104) ' 104 = <=lj, or 10"' + 10"3 + 6.87 Cf ; (1012) =1. 10"5 + 107 + 6.8 Cf (1.01 x 108) 10 x

(1

0.1
0.001), or
1 271 x °W
102 x 106 x 104
~ *p2/=
Cf
= 0.146pF.
_
= 0.159pF.
Check (From Eq.8.58):
Cf
Thus Cf1= pr 6.87 x 1012
~
ÿ = gm

R2R \ fpi
_ grn Cf_ ~ _ 1OOl04 X .146 x 1Q'2_ 2k (C,
C2 + Cf
(C, + C2))
2k ((1.59 X 10'12)2 + 0.146 X
10"12 (2) (1.59 x 10~12))
k = 77.6MHz, and /d3 p 10"9 — x 10"9
remains at 10MHz, with /„. Jp reduced to 104Hz, or 1.59 X + 0.292 10kHz. Thus the closedloop cutoff frequency raises to 10MHz. Note that the pole split lowered the dominant pole by a factor of 10 and raised the upper pole by a factor of 8 or so. Had the upper pole remained double at 10MHz, and assuming a double pole behaves as a single pole at a frequency for which each contributes 4572, that is at 107 tan"1 (22.5°) = .414 x 107Hz, the cutoff would have been at 4.1MHz. Thus pole splitting allows the same phase shift at 10MHz as formerly at 4.14MHz. Thus it seems that the dominant pole could be raised by the factor 10/4.14 = 2.4, to 2.4 X 104Hz for roughly the same margins. For this situation,
=
ÿ t = 0.066pF, for which ———— , and CfJ = /„/ = 2tc gm R2 Cf R\ x x x 10 x 10 100 2.4 2k
fP2
=
grn
looao4 x 0.066 x 10"12
Cf
C, C2 + Cf (C, + C2)
0.066 1.59 x 10"9 + 0.066 x 10"9
_
2tc (1.59 x
39 gjyjjjz
10"12) (1.59 x 10"12) + 1.59 x 2 x 10"12 (0.066 x 10"12)
Thus the poles are at 24kHz, 10MHz and 39.8MHz. Again,
the closedloop cutoff frequency will be at about 10MHz, whereas the original frequency for which the phase margin is 45° would have been about 4.1MHz.
 300
Chapter 9
OUTPUT STAGES AND POWER AMPLIFIERS SECTION 9.1: CLASSIFICATION OF OUTPUT STAGES 9.1
Peak voltages applied are: 1.414V, 14.14V, and 141.4V. Peak load currents for a lk£2 load are: 1.4mA, 14mA, 141mA. With a 50mA bias current, corresponding operating modes are A, A, AB, respectively. For a load of 0.25ki2, the peak load currents are 5.7mA, 57mA, and 566mA, with operation in modes A, AB, AB respectively. For an nearlynormal large output at zero bias current, class B operation is apparently possible.
SECTION 9.2: CLASS A OUTPUT STAGE 9.2
/ max
0 —3  0.7 i~5££2 =
F°r a lkH load, this will support a negative output peak of 1.53V,
and for 10k£2, a peak of 15.3V. In the latter case, saturation will occur earlier at 3 + 0.3 = 2.7V. For positive inputs, the positive peak is 3.0  0.3 = 2.7V, independent of load. Thus for a lk£2 load, the largest sine wave is 1.53V peak, and for a 10ki2 load, it is 2.7V peak. For a negative output at 2.7V with / = 1.53mA, R > 2.7V/1.53mA = 1.76kf2. For a second device connected in parallel with Q2, I doubles to 3.06mA, and load resistances down to 1.76/2 = 0.88ki2 can be accommodated with a 2.7V peak signal.
9.3
The output signal is voltagelimited by the saturation of Q 1 to t)„ = VGc ~ 0.2  0.7 = VGG  0.9 = 5 0.9 = 4.1V peak, or currentlimited to v0 = 10 //?£. Thus for large enough /, the largest possible zeroaverage undipped output is 4.1V peak. For /£2 £ /£, there is 101 I 91 available to the load. For a 4.1V peak, 9 / > 4.1/10012, / > 4.56mA. Thus the minimum / required is 4.56mA.
9.4
(a) The largestpossible sinewave output is 9 is 8.7V/10mA = 87011 /o
 0.3 = 8.7V peak.
The smallestpossible load resistance
7/m2
43.5mW. Supply power = 2 (9) x 10 = 180mW. Conversion efficiency = q gÿQ = 43.5/180 x 100 = 24.2%. Load power =
(b) For a signal of 8.7/2 = 4.35V across a load of 870/2 = 43512, the load power =
(A
3v/2)2
—Q435—
=
21.75mW. Supply power = 2 (9) (10) = 180mW. Conversion Efficiency = 21.75/180 X 100 = 12.1%. (c) The loss in (2 3 and R is 9 V x 10mA = 90mW, the supply power = 2 (9) 10 + 9 (10) = 270mW. For (a), efficiency = 43.5/270 x 100 = 16.1%. (d) For (b), efficiency = 21.75/270 x 100 = 8.06%. 9.5
For matched FETs, no load, and v0 = 0, I 02 = fass = 10mA = lD 1 and uGSi = 0V. Thus U/ = 0V. For operation in saturation, vDG >  = 2V. Thus the negative limit of v0 is 9 + 2 = 7V, for which IL Upc
= 7V/lkl2 = 7mA, and /OI = 10  7 = 3mA. Generally, iD = IDss (1  T7)2.
vp
3 = 10 (1

"~Z
1+
Z
=
(ÿr)ÿ = 548, x>Gs = 2 (.548  1) = 0.905V.
Thus the corresponding
IU
input is 7 .90 = 7.9V. For positive outputs, the input limit for saturation is 1)/ = 9  2 = 7V, for which Do = t). 10 + = 10 (1 + 3 5 " W2)2> or 10 + t) = 10 (4.5  t>/2)2 = 202.5  45\) + = 10 (1 ~
7HT
~ri Thus, 2.5vr 2.5'U2.
"ÿ5r">2

46 ±13.8 46 ± ÿ 462  4(2.5) (192.5)  = 6.43V = X)0 „n  =  46\) + 192.5 = «0, v> ,
2d \jL»D)
 301 
D
ÿ
SOLUTIONS: Chapter #92
Check: iDi = 10 (1 
——
)2 = 16.5mA. Compare with 6.43/lk£2 + 10 = 16.4. OK.
*"4r
43)2
C6 = 6.43V: Load power = v ' ÿ = 41.3mW. Supply power = 9 (10 + 6.43) + 9 (10) = 147.9 + 90 = 238mW. Efficiency = 41.3/238 X 100 = 17.4%. Now for a dc output of x>0 = 7.0V. Load power = 72/lkO = 49mW. Supply power = 9 (10  7) + 9 (10) = 27 + 90 = 117mW. Efficiency = 49/117 x 100 = 41.9% Largestpossible relativelyundistorted sine wave output is 6.43V peak, for which Load power = (6 42/W) = 20.7mW. Supply power = 9(10 + 6.434c + 10) = 198.5mW, where 6.43/ n is the average Now for a fixed (dc) output signal of
— —
value of the 6.43V halfsine current pulse. Efficiency = 20.7/198.5 x 100 = 10.4%.
SECTION 9.3: CLASS B OUTPUT STAGE 9.6 For Rl =oo; iD = 0 and x>GS = V, as V/ varies. For D, £
IVol= \V/\~l
IV,
va
= 0. For 1 S o,  ÿ 11V,
For v,  £ 1IV, 1 1)0 = 10V. For RL = 10k&: For Do = IV, i'o = 1/104 = lOOpA = 1 (y>o$ l)2, Das = ± (l/lO)" + 1 = 1.32V, D, = 2.32V. For u0 = 9V, iD = 9/104 = 1 x 10"3 {mgs l)2, dcs = ± (.9)* + 1 = !95V. v, = 10.95V = 11V. For v0 = 10V, iD = 10/104 = 1mA = 1 [2 (Ocs 1) \>ds~d/mL Say x>ps = 0.1. Thus 1 = 2 (mgs  1) (0.1), vGS = 1/(2(0.1)) + 1 = 6V, 9.5V, t'o = 1mA = 1mA (2 x>i = 16V. For Vo 0.52), vGs 1 = 1 + .25, vGS = (vGS  1) 0.5 2.25V, D, = 2.25 + 9.5 = 11.75. For Qh Q 2 in saturation, the largest possible sinewave output is 9V peak or 18Vpp. The corresponding input voltage is 20Vpp for no load, and 22Vpp for 10k£2 load. Equivalent gain is 18/20 = 0.9V/V, or 18/22 = 0.82V/V respectively. Supply power is OmW for no load, and 10 (9 2)/10k = 6.36mW. for lOkft load.
—

n«. ».
4*
—
in. »i

MJ. II
..
!>••>
See page 308 for expanded version. Ml.HI
9.7

Load power is OmW, or
—
(<W 2)2
respec
tively. Efficiency is «> (for no load), or 4.05/6.36 x 100 = 63.7% for a 10k£2 load.
= +10mV, Dÿ = 0.710V, iL  10 x 10~3/100 = 0.1mA, ibase = 0.1/50 = 2pA, for which the 2 X 10"6 2il4 amplifier input =m = r = 0.2mV. Thus D/ = 10 + 0.2 = +10.2mV. F v voltage is D,„ 10mA/V 10 X 103 * 20 x 10~® 103 For \)0 = +100mV, Mbase = 0.800V, iL = 100  10"3/100 = 1mA, ibaK = = 20llA, Min = ÿ jÿ_3 = 2mV. Thus D/ = +102mV. For v0
ÿ
.
For D0 = +1V, vbase = 1.7V, iL = 1/100 = 10mA, ibase = 10/50 = 200pA, Min = Thus V/ 9.8
ÿ = 4.05mV
1U/C aZ
200 x
10"6
"
= 20mV.
= +1.02V.
Assuming DGe sat = 0V, the largest possible undistorted output is 6V peak or 6ÿ2 = 4.24V rms. Corresponding output power = (6/ÿ)2/16 = 1.125W. Current from the supply is a half sinewave of 6/16 = 0.375A peak, whose average value is 0.375/rc = 0.119A. That is supply power = 12 (0.119) = 1.43W. Efficiency = 1.125/1.43 x 100 = 78.7%. Power loss in both transistors is 1.43 1.125 = 0.305W. Power
 302
SOLUTIONS: Chapter #93
loss in each transistor is the same = 0.305/2 = 0.153W. For 4V peak output: Output power = (4W2)2/16 = 0.5W. Supply power = (12 X 4/16 x 1/n) = 0.95W. Total device dissipation = 0.95 0.5 = 0.45W. Efficiency = 0.5/.95 X 100 = 52.6%. For a +14.5V Supply, and 6V peak output: Output power = 1.125W. Supply power = 14.5 (.119) = 1.73W. Device dissipation = 1.73 1.125 = 0.60W. Efficiency = 1.125/1.73 x 100 = 65%.
SECTION 9.4: CLASS AB OUTPUT STAGE 9.9 For rout < 512, reN \ \reP  512 = re/2. Thus re = 1012, and IE = 25mV/1012 = 2.5mA, the quiescent current, for which VBBI2 = 690 + 25 In (2.5/10) = 655mV, and VBB = 2 (655) = I.31V. For 5V peak output and 5012 load, lE  5/50 = 100mA, vBE = 690 + 25 In 100/10 = 748mV. Thus x>E = 5.00 + .748  .655 = 5.09V. Largesignal gain = 5.00/5.09 = 0.982V/V. For small changes around 0V and a 5012 load, gain = 50/(5 + 50) = 0.91V/V. For small changes around +5V, and a 5012 load, re = 25mV/100mA = 0.2512, gain = 50/(50 + .25) = 0.995V/V. 9.10 For each device, biased at current 1,1 = 200 (uGs  l)2. and gm = 2 (200) (\)cs  1) mA/V with rs = 100 1lgm. For a 10012 load, gain = = 0.99, 100 = 99 + .495rt, rs = 1/.49512, or l/rt = .495 A/V 00 1 1 f y/2* 1
1
= 495mA/V. Thus 2 (200) (X)GS 1) = 495, = 4.48V.
uGj
 1 = 495/400 = 1.237, \)GS = 2.237V, VBB = 2 (2.237)
SECTION 9.5: BIASING THE CLASS AB CIRCUIT 9.11 For each junction, Vj = 0.675V, to maintain an output quiescent current of 2.5mA for which 1B = 2.5/31 = 81(xA. Correspondingly, the quiescent current of the biassing junctions is 2.5/4 = 625xA. Thus I= 625 + 81 = 706pA. Now for a shortcircuit output, the maximum available current = 706pA (30 + 1) = 21.9mA. For junction bias reduced to 0.1 (625xA) , available base current = 706 0.1 (625) = 643.5J,A, for which iE = 31 (643.5) = 19.9mA, and v0 = 50 x 19.9 = 1.00V across 5012. Thus the peak output for

5012 is 1.00V. 9.12 Iq = 2.5mA for which vBE = 675mV and iB = 81pA. For a IV positive output across 5012, iE = 1V/50 = 20mA, for which iB = 20/31 = 645xA. Now for a normal bias network current /, (/  645)// x 100 = 20, / 645 = 0.2/, 0.8/ = 645, I= 806xA. Thus the resistornetwork current level = 1/2 (20%) 806 = 80.6pA. Here VBB = 2 (0.675) = 1.35V, and /?, + R2 = 1.35/80.6xA = 16.75k£2. Now the normal 80.6 81 = 644nA, for which VBE\ = 690 + 25 In (0.644/10) = current in the bias transistor is 806 621mV, and IBi = 644/30 = 21.5pA. That is /?, = 621/80.6 21.5 = 10.5kl2. Use lOkQ, for which lR1 = 621/10 + 21.5 = 83.6(J.A, R2 = 1.350 0.621/83.6 = 8.7kl2. In practice, use 8.2kl2 with a variable (or 80.6 = fixed) series resistor. Now for u0 = IV, iEn = 20mA, iBn = 645(iA, and iE\ = 806 645 80.4pA. That is, rei = 25mV/80.4(xA = 31112, rKi = 1 (311) = 9.64kl2, and for the multiplier with v (16.78) 10 tV, X 1 + 31 input u. current i is = jjj. ot r, 9.64 10 %Mk 8 + + 16.78/13.6 = 1.23kl2. Now, at the peak output of IV, reN = 25mV/20mA = 1.2512, and rKN = 31






°,
(1.250)
= 38.75£i. Gnin
at
the peak

 .. .
—
=
= 0.976
X
X
0.564 =
0.55V/V. For signals around 0V, with a 5012 load, R0 = 512 as before, but iEi rises to 644pA for which rel = 25/.644 = 38.812, and req reduces to about 38.8/311 or 0.12 of the previous, or 0.12 (1.23kl2) or about 0.91 X 0.92 0.84VAt. X 150Q. Thus the gain =
(a,(fÿ'50
.

9.13 From the sequel to Eq. 9.33 of the Text, neglecting f) for the bias situation, k = 1 + R2!R 1 = (/? + R2YR\ Now, for a junction voltage x>BE, the emitter current of the multiplying transistor = / Vt k Vrf Vf Now for a small incremental voltage , where IE = I VBE/R\. Thus re = =

R\+R2

ivbe/R\ — — Ie
 303 
ÿ
SOLUTIONS: Chapter #94
u applied across the multiplier, a current
........
+ , and ir ———Rlrn + R2(Rl+rn) (/?,
Ri
r„)t)
Rj rK rK + R2 (R i + rK)
i = iB + i'c
Thus
iB flows in the Rh R2 network, where iB =
*
rn
_~
in
„
.
the transistor collector
. .
is
Mi v R rK + R2 (/?, + r„)
— —
,
P R\ + R\ + rK
=
Ri+Ri /?2 (P+W.MP+IK = 
+ R2yR\.
. • «>"ÿ » = W  V.M and

Now, k = 1 + RyR i, or R /R\ = k 1, or R2 = (k rt (k R,) + R,2 (k  1KP + D , with re = VTAJ  VBE/R i). Thus, req = Ki + re
k =(R\
Now for k
re =
R\ rK + R2 (R\ + rn) (p+1)Ri + rn
R\ fx + "2 («i + rn) l)r, (P+ + re (R, + R2) + R, Rÿp + 1) R2
(P+l)r,
= 2, / =
1mA,
ÿ
R2 + R i II rK A) II r% p i x x«x — ic = p 1 R2 + Rl\\rn + re
P > 50, Ri = rK = (P + l)/e, re

1) R\, and $+ R2 = k R{.
ÿ — — /UIKa J .O, with Rx in ohms, req = 102 + 51 1 — f = 2.94 r,, where 52 J
=
{1
~ , Y _ 70ÿ51 y 25 = re  700/51 > re = 25 + 700/51 = 38.70,
req = 38.7 (2.94) = 11412.
SECTION 9.6: POWER BJTs 9.14 At 30°C, the junction drop at current Iis 630mV. At 10 times that current the drop would be 630 + 25 In (10///) = 687.6mV. Now at T'C, the junction drop is 500mV. The new temperature, T = 30°C + (687.6 j23.g°c. For a total dissipation of 45W, the thermal resistance, junctiontoambient
ÿ50Q)mV _
—
—
is (123.8 30)/45 = 2.08°C/W. For a junction temperature of 180C°, total dissipation could be (180 30)/2.08 = 72.1W, and the new current would be 72.1/45 X 10/ = 16/, that is 16 times the original test current. At 30°C, at this current, VBE = 630 + 25 In 16 = 699.3mV, or 700mV. At 180°C, VBE = 699.3  2 (180  30) = 399.3mV, or 400mV.
9.15 Pmax ÿ (150  55)/l.rC/W = 86.4W. For 86.4/2 = 43.2W, the junctiontocase rise is 1.1 (43.2) = 47.5*C, and for Tj = 150, Tc = 150  47.5 = 102.5°C. For TA = 30°, the thermal resistance of the heat sink required is (55 30)/86.4 = 0.289°C/W in the first case, and (102.5 30)/43.2 = 1.68°C/W in the second. For a heatsink length L , the rating is 3/L °C/W. In the first case 3/L = 0.289 and L = 3/.289 = 10.4cm. In the second case 3/L = 1.68, and L = 3/1.68 = 1.79cm. Now for a potential error of 20% in all thermal measurements, but with 86.4W applied, TE should be 150  86.4W (1.1°C/W) (1.2) = 35.95*, for which the thermal resistance of the sink must be (35.95 — 30)/86.4 = 0.0689°C/W, for which L = 3 (l+0.2)/0.0689 = 52.3cm. Note the dramatic impact of measurement error on the adequacy of a design!
—
—
9.16 For a device dissipating W watts: 40° + 1/10 x W + 0.5W + 2W = 150°, IV (0.1 + 0.5 + 2) = 110°, W = 43.1W. We now conclude = 110/2.6 = 42.3Watts. For a heat sink twice as long, W = 
* ———
ZiD "f" 1/2U
that the heat sink is already quite large, the major problem lying in the transistor itself, with its dominat¬ ing thermal resistance. For an infinite heat sink, the maximum rating would only be 44W!
IE = 5A, IB = 0.2A, P = (5 — 0.2)/0.2 = 24, and re = 25mV/5A = 5mi2. Thus rK = (24 + 1) 5 X 103 = 125 x 103 = 0.125 12. For Rih = .7212, rx = .72  .125 = 0.59512. At IE = 3A, re = 25/3 = 8.3 m!2, rK = 25 X 8.3 = 208.3 m!2 = 0.20812. Thus, Rih = .595 + .208 = 0.8012.
9.17 For
 304 
SOLUTIONS: Chapter #95

SECTION 9.7: VARIATIONS ON THE CLASS AB CONFIGURATION 9.18 For //, max = 100 mA and a standing current /, the maximum base current occurring in the pnp transistor, is (100 + i)/81. Thus IE2 > (100 + 0/81, or > 1.5mA. Thus, use / = 2 (100 + 0/81 = 2.47 + .025/  (2). Now for the quiescent state, / = lE3 = 1Ea =  (1), or / = (100 + i)/81 + 1.5 = 2.73 + .0123/ h\ = hi = /• Now, for (1) above / = 2.47 + .025/, or / = 2.47/(1  .025) = 2.53mA, for which IE2 = (100 + 2.53)/81 < 1.5, and thus use (2), for which I= 2.73 + .0123 (/), for which / = 2.73/(1  .0123) = 2.76mA. Now, since the output transistors are 5 times larger than the bias transistors, VR3 = VR4  25 In 5 = 40.2mV at a current I= 2.76mA. That is, R3 = R4 = 40.2mV/2.76mA = 14.612; Use 15£2. Now for outputs near zero volts, IE3 = IE4 = 2.8mA, and re = 25mV/2.8mA = 912. Thus, Rout = (9 + 15)/2 = 12£2. For a gain of 0.90 (dominated by the output coupling)
Rl = 0.90, where RE = 12/(1  .90) —Re12 +—
= 12012. Near ±10V, where the situations are essentially the same for a particular Re, one transistor is likely to be cut off and lE = 10/RE. Thus rc = = 25Rl Q with Rl in k£2, and Rou, = (2.5RE + 1
Rl 15) £2. Thus, r = 0.90, or RL = 0.90RL + 0.00225 RL + 0.0135. Thus RL = Rl + (2.5Rl + 15) x 10"3
0.0135/(1  0.90225) = 138£2. Thus for loads in excess of 140£2, the gain of the output stage can exceed 0.90, for outputs of ±10V.
9.19 For a standing current of 10.0mA in the output: (a) lE2 = IE4 10mA, and IE\ = 10/100 = 0.1mA, and IE3 = 10/100 = 0.1mA. That is VBB = 700 + 25 In (10/100) + 700 + 25 In (0.1/1) + 700 + 25 In (0.1/1) = 2100 57.6 57.6  57.6 = (2100 172.8)mV s 1.93V. Now, for all P increased by 10 and VBB = 1.93V, all currents will increase to maintain the voltage. The current, say 10k mA, will be such that the currents in Q2, Q\, £2 3 will change to 10k, 10k/(1000) = .01k, and 10k/(100 x 10) = 0.01k respectively. Now 172.8/25 = (In 10k/100 + In 0.01k/l + In 0.01k) = 6.91. Try k = 10: In 1 + In .1 + In .1 = 2.3  2.3 = 4.60. Try k = 3: In 0.3 + In .03 + In .03 = 1.20 3.51 3.51 = 8.22. Try k = 5: In 0.5 + In 0.05 + In 0.05 = 0.693  2.995 2.995 = 6.68. Thus, the standing current increases by more than 5 times!. (b) Iea ~ I 1 = 9mA, IE\ = /£3 = 1mA. Thus, VBB = 700 + 25 In (9/100) + 2 (700) = 2.04V. ei ~ 10 Now for all p increased by 10, the output current increases slightly, but the baseshunt current, established by resistors and a VBE which changes only slightly, stay essentially the same. Thus for a factorof10 change in IB2, from 0.1mA to .01mA, lE\ changes from 1.00 to .91 mA for the same output current. Thus it is likely that the standing current changes by a few tens of %. A great







improvement!

9.20 For I0 = 25mA = IEl, IBl = 25/100 = .25mA. Thus /C5 = 1 .25 = .75mA, vBE j = 700 + 25 In (0.75/1) = 693mV. Thus REi ~ 693mV/25mA = 27.70. Use 27Q. Without Qs, the peak load current could be 1mA x 100 = 100mA. 25
r*2 9.21 For both devices having P = 50, IE2 = 10mA, IB2 = 1E\ = 10/50 = .2mA, re2 = p5o)(10) = = 51 (2.45£2) = 125 £2, rel = 25/.2 = 125£2, r„, = 51 (125) = 6.375k£2. For \)2 at the base of Q2, ih2 = \)2/127.5 = ie 1 and ic2 = P t)/r„2 = 50 d2/127.5. Now voltage Ui at the base of Q\ = t)2+ iei re\, or Dj = D2
50 51
V2
+ D2/I25 x 125 = 2
\>2,
snd ic\ = 50/51 ie\ = 50/51 x 1)2/125. Thus, gm
50 v2
'"C  zjk) 125
("51 +
"
2039mA/v For
r' =
eq
=
ic 1~+ ic 2
iwm '102ka
—
= 10
= 250M£2. Now, for a rise in output of t), with the input (50) 10.2 = 5.1M£2, r„\ ~ 100/.2 = 500k£2, shortcircuited, the total current is approximately / = (W1M£2) + (uÿSOMfl) + (t)/(0.5M£2 I! 5.1M£2) 50
 305 
SOLUTIONS: Chapter #96
+ (v/10.2kft). Thus IIR0UI = 1 + .004 + 109.8 + 98.0 = 208.8pA/V, and Rout = 4.79km. Thus the gain = gm t„ R„u, =  202 X 10"3 X 4.79 x 103 = 977V/V, and Rin ~ (1MQ/(1 + 977)) II (51 (2 (125))) = 1.022km II 12.75km = 0.946km. For both devices having P = 150, lB2 = IE\ = 10/150 = 0.066mA. Thus rn2 = reX = 25/.066 = 375m, r„i 150 t)2 150 t)2 + 375 151 375 150 1 3 /3 (1 + ±~) = 206.6mA/V, that is, = 151 (375) = 56.6km. Thus gm eq = = 2(375) 151 U2 + D2 almost the same as with
For j22,
P = 50.
15q°°— = 10.07km,
ra2 =
Rou, = 1ML1 II co
= 10 (150) (10.07) = 15.1MO, rm = 100/.066 = 1.501Mil 

10.07km = 1MH II 10km II 100km II 10.07km = 4.75km.
103 X 4.75km = 981V/V, and Rin ~ 1M£2/[(1 + 981) II (151 (2 (375)))] = 1.018kfll II 113.3km = 1.01km. Overall, for p = 50 to 150, gm ranges from 204mA/V to 207mA/V, the gain ranges from 977V/V to 981V/V, Rin ranges from 946m to 1010m, Rou, ranges from 4.75km to 4.79km. That is, there is very little effect. Now the gain = 206.6 x
~  2 (125  25) = 500mV . Thus Rx = 1Q,ÿ °/5 = 95km, and iB = 100(lA Thus iB2 = 100  1 = 99ftA, and R2 = .500/99ttA = 5.05km. Now, at 25°C (with 700 + 504.7
9.22 At 125°C, and 100pA, VBE =700 100liA/100 = luA.
J
Ql

(100
25 X 10 = 504.7mV. Now i2 = 100 X 10"6 e = n 0.0405lA. For doubling, that is t2 = 0.0810(xA, vB = 504.7 + 25 In 2 = 522mV. Thus the supply vol50 x 106 tage = (0.522/5.05) (5.05 + 95.0) = 10.34V. At iC2 = 50pA, VBE = 700 + 25 In = 682.7mV.
iB2 ~ 0, vB =
Now at 100°,
VBE = 682.7

25)2
1QQ )< 1Q6 = 532.7mV, for which the supply voltage =
(0.533/5.05)(100.05) = 10.56V.
SECTION 9.8: IC POWER AMPLIFIERS 9.23 For the circuit shown in Fig. 9.30,
= 0.458mA. Thus Ibias =
IRX ~
~
1.15pA. To reduce this to 0.5xA, raise R\ to (1.15/0.5) X 50 = 115km with 57.5km in each half. Now for the same gain and to maintain the same assumptions for the gain calculation, raise R2 and R 3 by the same factor [(= 1.15/0.5 = 2.3)] to 2.3km and 57.5km respectively. Because of the change, the current in 2 jo, fin an(l fi 12 aH reduce by a factor of 2.3, that in Q9 reduces, but not by as large a factor due to /?6, Ri•
9.24 For the calculation of A, include QX2, fin, Qi> fis, fi9, as driven by the output resistance of fi6 and Q4. With a 27V supply, bias current = (27  3(0.7))/50 = 0.5mA. Thus Ica = lc6  /cio = lew = /ci2 = 0,5mA, Ic9 = 10 Icu  5mA. Now reX2 = 25mV/0.5mA = 50m, roX2 = 100V/0.5mA = 200km = r„xx. Now Qs operates as a follower with P = 100 X 20 = 2000, while Q2 operates as a follower with P = 100, where at the output the only load is r„7 II r„9 ~ r„7 12 where rnl = 100V/5mA = 20km. Thus the net load on the collector of £12 is roX2 II roXX II ((101) (20km/2)) = 200km II 200km II 1M£2 = 90.9km. The gain from the base of QX2 to the collector of Qx2 is about 90.9km/50m = 1818V/V. Follower gain for no load is nearly 1V/V. Thus the overall gain A is 1818V/V. Equivalent input capacitance is Cr = 10 x 10~12 (1 + 1818) = 1.82 X 10"SF. Corresponding input resistance is RT = ro6 II ro4, where at 0.5mA, r„ 87.5Hz. = 100V/0.5 = 200km. Thus, the cutoff frequency =  (200 x 10, —58) = /2) (1.82 x 10 2k

—
————
 306 
—
*
—
SOLUTIONS: Chapter #97
9.25 For equal sharing, each conducts 50/2 = 25mA. For VEB3 = 0.70V, R3 = 700mV/25mA = 2812. Note that a specification of 1.0V at 1A is given for Q 3. However a lot of this VEB is likely due to resistive effects in the base. Thus we use 0.7V as above. (Note, that we get a higher result for R3 if we use the O.lV/decade idea, in which case VEB5 = 1.000.1 log 25/1000 = 0.84V). Now at /„„, = 1A, VBE = 1.00V, and /R3 = 1.00/28 = 35.7mA. As well, IB5 = 1A/30 = 33.3mA. Thus /c3 = 35.7 + 33.3 = 69mA. For a load change from 50mA to 1A, a factor of 20, the current in Q3 varies from 25mA to 69mA, a fac¬ tor of 2.76. For Q , Q2 operating at 1mA, VBE\ = 0.700V. For Q3, Q4 operating at 2mA,  VBE\ = 700 + 0.1 log (2/10) = 0.630V. Thus RS = R6 = (0.700  0.630)/2mA = 3512. 9.26 For ±12V supplies and 2V saturation, outputs of ±10V are available. Thus a 20V peak signal is possible. Input provided is 0.1V peak. Required input resistance = 10kI2. Thus, R3 = lOkQ, and R4 = 10V/0.1V X 10k£2 = 1MI2. For the highest possible input resistance, use R4  10MI2 and R3 = 100kf2 for a 100ki2 input resistance. For the positive side, 10/0.1 = 1 + R2/R\, or R2 = 99/? j . Use R2 = R4= 1M£2 and R\ = 1MQ/99 = 101k£2, a 100k£2 and lkf2 in scries, as a quick solution. 9.27 There are several choices: (a) One is to drive A\ as shown, but with R3 connected to the output of AI( with R4 = R3 and 1 + R2/R i = 20/2 = 10, R2 = 9RU or /?, = R2/9. Use R = 10kf2, R2 = 90kI2, R4 = R3 = lOOkO. (b) Modify (a) above to merge R3 and R i into R 3 = 10ki2, with R2 = 90kI2, and R4 100k£2 using only 3 resistors in all.

SECTION 9.9: MOS POWER TRANSISTORS 9.28 K = 1/2 ip = K (vas  V,)2 = Cox W/L = 1/2 x 30 X 10"6 x 10s/5 = 0.3A/V2. At lowon\)GS, y 10 0.3 (vGS  V,)2. At high mgs, i„ = 1/2 Cox W Um (vGS  V,) = 1/2 x J. ' _2 x 10s x 10"6 x 5 x J x lu 104 (vGS  V,) = 1.5 X (X)GS  V,). These currents are equal when 0.3 (vGS  V,)2 = 1.5 X (\)C5  V,), or \)GS  V, = 5, or \)GS = 5 + 2 = 7V. For x>GS = 7V, iD = K (vGS  V,f = 0.3 (7  2)2 = 7.5A, for which gm =2 (0.3) (7  2) = 3A/V. For vGS = 3.5V, iD = 0.3 (3.5  2)2 = 675mA or .675A, for which Sm = 2 (0.3) (3.5  2) = 0.9A/V. For x>GS = 14V, iD = 1.5 (14  2) = 18A, for which gm = 1.5(Ogs ~ V,) = 1.5(14  2) = 18 A/V. Note that from the velocitysaturated relationship at 7V, the gm would be 1.5A/V (rather than 3A/V). In practice, the transition between modes begins to occur at lower values of uGs and gm .

9.29 For 5mA bias: iD = 5 = K (\)GS  V,)2 = 200 (\>GS 2)2, vGS = (5/200)'/' + 2 = 2.158V. Thus Vl4 = 4 (0.7) + 2 (2.158) = 7.12V, R = 2(2.158)/5mA = 863£2. Total FET TC = 2 (3) = 6mWC. Total BJT TC = 4 (2) = 8mV"C. Total TC = 14mV/"C. Thus 6/14 = 0.429 of Vl4 must appear across Q6. For outputs around zero: gm = 2K (x>Gs V,) = 2 (200 (2.158 2)) = 63.2mA/V. Thus, R„ul of fol¬ lower = (1/gm) II (/gm) = (l/63.2)/2 = 7.912. The gain for 10012 load = 100/(7.9 + 100) = 0.927V/V. For outputs around +20V: lL ~ 20/. 100 = 200mA, whence 200 = 200 (x>GS  2)2, uGS = (200/200)'/l + 2 = 3V, and gm = 2 (200) (3  2) = 400mA/V. Thus Rou, = (l/400)/2 = 1.25. The incremented gain = 100/(1.25 + 100) = 0.988V/V. Alternatively, note that the input must be 20 + 3 2.158 = 20.84V at the peak. The corresponding overall gain = 20/20.84 = 0.960V/V.
—
—

9.30 See that Q3, Q4 are 100 times larger than Q\, Q2. Thus for 10mA in Q3, Q4 require I= 10/100 = 0.1mA in Q\, Q2. For IQ = 5mA, 5 = 100 (vGS  l)2. Thus vGS = (5/100)'7 + 1 = 1.224V, gm = 2 For (100) (1.224 1) = 44.8mA/V, Rout = 1!gm II Vgm = = 0V, the gain with
vo

100£2 load is 100/(100 + 11.2) = 0.899V/V, with VD = 0V, VB = 1.22V, Vc = 1.22V, VA = 0V. See that DGpi = VGD2 = 0 and all transistors operate in saturation mode. For V0 = +10V, with 100Q load, iD3 = 10/0.1 = 100mA, that is 100 = 100 (\)GS  l)2, x>as = 1 + 1 = 2V. Thus VD = +10V, and VB =
 307
SOLUTIONS: Chapter #98
+12V. Now Q continues to conduct 0.1mA with VGS = 1.224V, that is = 12  1.224 = 10.78V. For Qi, Vq  10V, VG = 10.78V, and operation is in triode mode with vGD  0.78V, where, 0.1 = 1 (2 (Ugs 1) (v>ds) vis). Now, since vGG = t>G5 ÿ>ds = 0.78, and vGs = 0.78 + x>ds< 1 = 20 ((.78 + vDS 1) vDS  10 vis) =  4.4 vDS + 20 vis  10 ÿds, or 10 44 ± V4 42 4(_n (101 = °605V> and vas = .78 + .605 = 1.38V. vis ~ 4.4 vDS  1 = 0. Thus vDS = L (IU) Correspondingly, Vc = +10  0.605 = 9.40V. Thus Q\, Q3 operate in saturation, Q4 cuts off, and Q\ is in triode mode. Overall, the gain is V[/VA = 10/10.78 = 0.928V/V. Incrementally, for Q3, gm = 2 (100) (2  1) = 200mA/V, and Rou, = 1!gm = 1/200 = 512. Thus the gain vdA>a = 100/(100 + 5) = 0.952V/V.
—
—
—
_
win?
9.6
(contined)
VO
(tÿ. 10)
RL * lOkCl
(1. 0) (2.32. 1)
10
(1. 0)
(11,10)
 308 
Chapter 10
ANALOG INTEGRATED CIRCUITS

SECTION 10.1: THE 741 OPAMP CIRCUIT 10.1 For ±15V supplies, IRef  [+15 15 — 0.7 — 0.7]/39kO = 0.733mA, or 733pA. For ±5V supplies, Iref [+5 5  1.4]/39k£2 = 0.2205mA. For ±5V supplies and IRef  73mA, R$ = (10  1.4)/0.733
—
—
= 11.73kQ. One could use 12kQ as a standard value.
10.2
For ±15V supplies, IREF = 733pA. Thus \>BEU = n VT In (Ms) = 25 In (.733 x lOÿ/lO"4) = 625.5mV. Now Dbeio = "o = 625.5 + 25 In (i1.733), with i in mA, where iR4 = 25 In i1.73  5000i, or i = .005 In i/0.73. Iterate: For i = 0.1 (mA), i = .005 In 0.1/.73 = .01mA. For i = .01 (mA), i = .005 In .01/.73 = .0215mA. For i = .015 (mA), i = .005 In .015/.73 = .0195mA. For i = .018 (mA), i = .005 In .018/.73 = .0185mA. Thus i 18.3xA. For ±5V supplies, IREF = .2205mA, x>BEU = 25 In ((.2205/1014) x 10"3) = 595.4mV, and vBEio = 595.4 x 25 In i/,2205, where //?) = 25 In i1.2205 = 50001, or i = .005 In H.2205.

Iterate: For i = .01mA, i = .005 In .01/.2205 = .0155mA. For i = .015mA, i = .0133mA. For i = .014mA, i= .0138mA. Thus i ~ 13.9(iA (reduced from 18.3pA). For i = 18.3pA as before, x>BB\o = 625.5 + 25 In (.0183/.733) = 533.2mV, (or, vBeio = 595.4 + 25 In (595'4 (.0183/.2205) = 533.2mV) and, for ±5V supplies, \)BEn = 595.4mV. Thus R4 = =
"f3ÿ2)mK
3.40kfl
10.3
Replace Ra by a transistor Q25 whose collector is connected to the emitter of Qi6, emitter to a resistor R 12 connected to VEe< and base to either the base of Q n or the emitter of £>?. From page 820 of the Text, note that the voltage VB\i ~ 618mV + 550(iA (100Q) = 618 + 55 = 673mV, and IR 9 = 673mV/50k£2 = 13.5lA. Now at 13.5pA, VBE2s = 25 In (13.5pA/10~14A ) = 525.6mV. For the con(625.3  525.6)mV „ 730 xlO"6 „ . , , nection to the base of gn. Vbew = 25 In = ÿ— = 625.3mV. Thus R\2= 
...
ÿ
ÿ
of Qj, VE1 = VBE6 + R2 (IE6)
and, (from page 706), VE1 = (5256  0. Xo 517 + 9.5 x 10"6 x 1 x 103 = 517 + 9.5 = 526.5mV, for which Rl2' = 13.5pA avoid an unusual load on Qy, it would be best to include a resistor R i2 = lk£2 say (= Rj, R2), in which case the current extracted from Q 7 is about 9.5iA. The latter has the advantage of using a smaller resistor and, as well provides a (small) signal component of a reinforcing polarity from the collector of
7.4k£2. For the connection to the emitter
~52ÿ5),nV
Qs via 0710.4
For inputs limited to the supply range, the worst case is for one of In+ and /„— connected to +15V (say 15 /„+) and the other (say /„) to 15V, in which situation the collector junction of Q2 has 15 0.7 series, a combined reversed are with in and of the EBJ Q4 Q2 = 29.3V reversed across it. OK. Also

—
rating of 7 + 50 = 57 > 30. OK. Now for inputs outside the supply range: For /„+ or /„ positive, the basecollector diode of Q2 (say) conducts, reversing the EBJ of Qg, Q9. The greatest allowed voltage is 7 + 2 (0.7) = 8.4V above the positive supply. For /„+ or /„ negative, the mostnegative input stresses the CB junction of Q2 (say) to 50V when Vin = 30 0.7 50 = —20.7V below the negative supply.
—
—
 309 
SOLUTIONS: Chapter #102
SECTION 10.2: DC ANALYSIS OF THE 741 10.5

From the preamble to Eq. 10.1 on page 816 of the Text,
/c 10
VBE\\
~
Vbew = /cio ÿ4. or VT In
REF
 VT
IreF In —— = lc io °r Vx In — = Icio 84 (1). For I 730pA, R4 = 500012, and k = rep 's 10 _fc * X 10 } 0.5, 25 x IO"3 In °'5(73,0 = /cio x 5000. Now, for /C10 = i in pA, In 365// = .2/, or / = 5 In 'cio 365//. Use a process of trial and success: i = lOpA » / = 5 In 365/10 = 17.98, / = 15pA > / = 5 In 365/15 = 15.95, / = 15.5pA i = 5 In 365/15.5 = 15.8, / = 15.7pA —> / = 5 In 365/15.7 = 15.7. Thus, /cio = 15.7pA. Normally for k  1, 25 In 730// = 5/, or / = 5 In 730//, / = 19 » / = 5 In 730/19 = 18.24, / = 18.5 » / = 5 In 730/18.5 = 18.4. Thus for I rep = 730.00pA, /cio = 18.45pA (or so). Thus for k = 0.5, from „ 25 x 10~3 , 0.5(730) (1). R 4 = 4.04kl2. Use 4k£2. r In — 18.45 18.45 x 10"*
—
—
—
—
/iN
10.6
ÿ
Anttn
TT
For (2i through Q4, Is = 10"I4A, n = 1, and lc = 19/2 = 9.5pA. Thus VBE = VT In Ms = 25 In 9'5 x 10"6 = 516.8V. Accordingly, the voltage on the bases of Q3, Q4 is 2 (516.8) = 1.034V. 14
—
10.7
Use the fact that VBEB  517mV and IE6 ~ IE5 = 9.5pA. Correspondingly, VE2 = 0.517 + 9.5 x 10"6 x 1 103 = 0.5265V, IE1 = (/C6 + lcsVP + .5265/50ktt = Ic«2p + 11.3pA, and 1B1 = Ic
X
= les + hi = Ice + hi = Ice + IcA2P2) + 11.3/p, whence ICÿC6 = 1 + W2) +
P tC6
~ 1 + 142p2)
+ i (11.3/9.5) = 1 + 1/2P2 + 1.19/p. Thus ICÿC3 = 1/(1 + K2p2) + 1.19/p). For p = 200, 1.19
1 Icflci = 1 + = + 200'rl = 1/(1 + (.0025 + 1.19)/200) = 0.994. 2 (200) 10.8
For high P, ignore the base current of £ rel="nofollow">7. For Ics  /, Ice =191, for Iin pA. Now, VBE5 = 25 In Ms = 25 In 1014 /, and VBE6 = 25 In (19  1) 1014. ForRi shorted: 25 X 10"3 In (1014 /) = 25 X 10"3 In ((19  /)1014) + 1 x 103 (19  I), whence / = 19 25 In (//(19  /)) with / in pA. Try / = 10 —» / = 19  25 In (10/(19  10)) = 16.37. Try / = 12 > / = 19  25 In 12/7 = 5.52, try / = 11 W = 19  25 In 11/8 = 11.04. Thus /C5 = ll.OpA, for which /c6= 19 11.0 = 8.OpA, and /C6//cs = 8/11 = 0.727. For R2 shorted: VBES + /?,/ = VBE6, 25 In (I/Is) + /?,/ = 25 In ((19  I)/Is), R\I =25 In ((19 /)//), / = 25 In ((19  /)//), for Iin pA. Try / = 8pA > / = 25 In (19  8)/8 = 7.96. Thus /C5 = 8.0pA, for which /c6 = 19  8 = ll.OpA, and Ice/lcs = 11/8 = 1.375.
10.9
For 8pA in Qlt and llpA in Q2,
11x10ÿ
8 x 1O®
—
= 512.5mV, VBE2 = 25 In —jj = VBEi = 25 In 1 X 10 1 X iU the npn + pnp devices) = 2 (520.5  512.5) = 16mV, the negative
520.5mV. Now the offset (due to input being higher with R2 shorted. The offset is 16mV for R\ shorted. 10.10 At present,
VBEi7 = 618mV, and IEl6 ~ 16.2pA.
for /en = 550pA, IBll = 550/200 = 2.75pA, for
10.11 For the current in (2i4>
220. increased to
Thus, the revised Rg' =
618 x 103 g = 38.1kfL — 16.2 x 10
lR9 = 4 (2.75) = 1lpA, Rg" =
1.5 (154) = 231pA,
 310
VBE = 25 In
618x10""
"
11 X 10
Now,
= 56kfl.
x 10"6 = 569.1mV, and — —231 3 X 10 14
SOLUTIONS: Chapter #103
VR6 = 27 x 231 x 1CT6 = 6.23mV. For R6 = R,= 012, VBB = 2 (569.1) = 1138.2mV. For R6 = R1 = 27, VBB = 2 (6.23 + 569.1) = 1150.7mV. Now VBB = VBEls + VBEi9, and with high P, VBB = 25 In (180X 10"6 VfiElgÿlo) . „c , (VbE18ÿ10) (180~ x, • r m ..a i/ yBE 18ÿ10 or for in pA, 25 In = oc

. 
kF1*— '
ltr" , (Vbeisÿio) (1). + 25 In ——s
ÿ
,1n
Now for R6 = R1 = 012,
VBB =
\XR) x (xyR) = 5876, or (x/R f

VBEu = v,Rw = R.
Now for R6 = R2 = 2712, xyR) x (x>/R) = 9766.
VBEI9 = 25 In 42.8
VBB =
1151mV,
. , Thus (x/RY  1on. 180 (xyR) + 9766 = 0, and
X
108 =
_

ÿ
= 42.8pA. Thus /£I8 = 180  42.8 = 138.2mA, VBEls = 25 In 583.7/42.8 = 13.6k£2. Check: 1138.1mV, as required.

From (I). 10"'6 In"' 1138/25 = (180 "H80± 1802 4(5876) 180  94.3 180 (xyR) + 5876 = 0, = 2 R 2
1138mV,
1OO
9
y ino
ÿ_,4
554.4mV, and
= 583.7mV. Thus /?10 =
VBB = 583.7 + 554.4 =
VBEls = u, /? l0 = R. From (1) 10~16 In"'
1151/25 = (180

±ÿ1802
180 180±/81  4(9786) 1— .  = —a = 2 2 1)
See the result is imaginary, that is it is not possible to provide the desired operation by varying R\o. To check this fact, note that the largest real value of xyR = 180/2 = 90mA, at which VBE = 25 In (90 x 10+8) = 573mV, for which VBB max = 2 (573) = 1146mV < 1151mV required.
SECTION 10.3: SMALLSIGNAL ANALYSIS OF THE 741 INPUT STAGE 3 6 x 106 10.12 From Eq. 10.4 of the Text, Ri(i = 4 (Pw + 1) re, whence re = A ' —— = 4.97k£2, for which lE = 4 (180 + 1) oc 25 x 10 = Gm i, = —— r = 5.03mA, reduced from the present value of 9.5mA. For this change, em. 2r( 4.97 XlO3 180481 = O.lmA/V. 2(4.97)
————
 2.63kf2, and rK = (P + 1)re = 201 (2.63) = 528.6kf2, = — =   = 0.378mA/V. Thus, /?06 = 105 X 106 = 5.26 (2.63) x
10.13 Generally, R0 = ra [1 + g„, (RE II r„)]. Here, re
with r„ = 5.26MI2, where g,„ 6
t
103
re
201 x 106 [1 + .378 x 10~3 (R2 II 528.6k 12)].
 1) r = 0.99640.378 x 10"3) = R2 II 529k 12 = (4%55.26 .378 xlO"3
2.635kf2, and
R0i = RogÿRqa = 10.5/2 = 5.25MI2. For Gml = l/5.26mA/V, and for the new situation, A„ = \ X 103 X 5.25 X 106 = 1000V/V. For the old, /?06 = 5.26 x 106 5.26 [1 + .378 X 10"3 (lk£2 II 528k£2)] = 7.25MI2, and R0l = Roe II R
5— 5.26
10.14 Use Rx = R2 = 2.65k£2 = R. From Example 10.1, Fos =
a"d r =
25
xlO"3
iJxlo1'
53 x 5.26 x 103 x 9.5 x 10ÿ U5.2 (2.65 + .053 + 2.63) X 1013 1.67 of the original, or 67% larger, although the resistors are only (1 
= "3ka ™here
= 0.497mV. This is .497/.3 =
A R (J/Gm\)
—, with NR = .02 (2.65) = 5312, R + A R + re T"™
ÿ
ÿ
=
2.65/2), or 33% larger.
10.15 From Ex. 10.9 of the Text,
Gmcm =
2«o
x
—— , where P„ = 50, re5 = 25mV/9.5MA = 2.63k£2, and
r—
R
+ re5
 311 
SOLUTIONS: Chapter #104
R„ = Ro9 II RoW = 2.43M12 (From Ex. 10.10), and /?, = R = lk!2. For CMRR = 80dB, Gm/Gmcm = 104, or Gmcm = 10"4 Gml, where Gml = (l/5.26)mA/V, or Gmcm = . mA/V, 5.26 x
AR =
'
50 AR x (1 + 2.63) x 2 (2.43 x 106)
1
104
103 x
3;63 x
2
(2 43
x
'
ÿ X
103 '
1061
— 5 26 x 106 x 50=
corrcsPonÿ'ng tolerance is 67/1000 x 100 = 6.7% or
±3.4%.
10.16 Add resistors of value RE in series with the emitters of both Qs and Q9. Now, for the pchannel dev¬ ices, VA = 50V, P = 50, (rather than 125Vand 200 for 2io) Now, since /Cio = /c9 = Ars = 19fiA, re = Vt 25mV = L29k"' r" = (P + 1} = 51 (L29) = 65'8kU 8m = °/r' = (50/51)/L29 = =
~h
19M (51x50) 0.760mA/V, r„ = VA/Ic ~ (50/(19) x 106) = 2.63M12. From Equation 10.7, want
10"3 (RE II 65.8 x
gm (Re I' rK)) to be 31.1M12. Thus 31.1 = 2.63 (1 + .760 x
(1 + (65.8) (.760)
R0 = r„
, 103)), Re
T Re + 65.8 x 103
 1 = 10.83. Thus 50 RE = 10.83 RE + 712.6, and RE = 712.6/(50  10.83) = 18.2kl2. Thus use resistors RE = 18.2kl2. In this case, Rm = R0l0 = 31.1M12, and R0 = 31.1/2 = 15.6M12. Now R , pp 50, Pp —ACr\ where n —A R = .02, whence A R = .02 (lk!2) = 0.02kl2. Now /C5 = = 50' » wl,crc Gmcm =
= 31.1/2.63
—
9.5pA, and
re
— "T
2Ro
~ 25mV/9.5lA = 2.63kl2.
Tlius
Gmcm
50 — 2 (15.6 X 106)

02 x 10~3 (1 + 2.63) x 10"3

0.0088pA/V. Now Gm\ = l/5.26mA/V as noted below Eq. 10.6 in the Text. Thus CMRR =
K5.26 x 10 0.0088 x 10"
)
_ 2L6 x 103 _ 867dB_
Gm ——
I
mem
SECTION 10.4: SMALLSIGNAL ANALYSIS OF THE 741 SECOND STAGE 10.17 The situation is one in which the base of Q25 is joined to the emitter of Q7, with lk!2 connecting the emitter of Q25 to VEE, and the collector of (2 25 connected to the emitter of Ql6 joined to the base of <2 17 Here, the collector current of Q 25 is the same as that in Q6, namely 9.5(iA, for which r025 = VA/Ic = 125/(9.5 x 10"6) = 13.2M12. Now lew = 730iA, IB\\ = 730/200 = 3.65pA. Thus the emitter current of Ci6 = 9.5 + 3.65 = 13.15[iA, for which rei6 = 25mV/13.15jxA = 1,90kl2. Also rel = 25mV/730xA 34.212. From Eq. 10.12, Ri2 = (P + 1) [r<16 + ro25 II t(P + D (G 17 + ÿs)]] = 201 [(1.90 x 103 + (13.2 X 106) II (201 (34.2 + 100))], or Ri2 = 201 (1.9 x 103 + 27.0 x 103) = 5.81M12, (rather than the 4.0M12 found previously).
Rn = (P + 1) [r„6 + r„25 II ((P + 1) (rell + /?«))], or Ri2 = 4 x 106 = 201 [1.90 x 103 + 13.2 x 106 II ((201 (34.2) + /?8))], 19.9 x 103 = 1.90 x 103 + 6.87 x 103 + 201 Rit Rs =
10.18 For
6.87) x 103 „  1.9— = 55.412. 201
(19.9
4
„
Now G„2 = VI
*c 17 —— v,2
"
R9 /?9 II /?,7+ rcl6 , where —=flil7 r«l7 ± "8 a
ÿ(17
= (P17 + 1)
(re i7 + Rg). Here, rel6 = 1.90kl2 (from the solution to P.10.17 above). Thus, /?,• i7 = 201 (34.2 + 55.4) 18.01 200 x 18.01 + 1.90 jo.OmA/V. This is to be compared to QQjÿyy = 18.01kl2, and Gm2 m = 34.2 + 55.4 6.5mA/V found previously. Now Rnl = R„\w II Rn 17. where R„ns = r„ 3fl = 90.9kl2, and R„ 7 = r„7
_
(1 + gmn (Rs 'I r*n))< where r„17 =
J xTo~ÿ
=
_
suc'1 l'iat
= 227 x 105 (1 + 200/201
x (1/34.2) (55.4 II (201 (34.2))) = 00.59M12. Thus Rol = 90.9kl2 II 590kl2 = 78.8kl2, and the opencircuit voltage gain is Gm2 /?n2 = —10.0 x 103 x 78.8 x 103 = —788V/V, compared to —526.5V/V found previously). Thus the change of bias network produces a gain increase of (788/526.5  1) = 0.50,
—
312
SOLUTIONS: Chapter #105
or 50%!
SECTION 10.5: ANALYSIS OF THE 741 OUTPUT STAGE 10.19 For the basic design, Ro2 = 81kft. Now, for RL = 2kft, Ri3 = pÿ (74kft) = 4(81kft) for which, P23 = R3 (81/74) (4) = 4.38. Now the secondstage gain is A2 = Gm2 Ro2 —— , where (from the bottom Ri3 + Rol of page 716 of the Text) Gm2 = 6.5mA/V, and R„2 = 81kft. 4(81xl03) Thus A2 = 6.5 x 10"3 x 81 x 103 x = 6.5 x 81 x 4/5 = 421V/V 4 (81 x 103) + 81 x 103
—
R.o 2
(as contrasted with 515V/V available with high P23).
(using the data on page 833 of the Text), Rol 27ft.
+
P23 + 1 re23 Now R0 + re20 + R7, where P20 + 1
 81kft, p23 = 50, re23 = 139ft, p20 = 50, re20 = 512, R2 =
+ 139 Ro =
51
+ 5 + 27 = 65.9ft.

10.20 Assume the base current of Q4 to be i. Thus the collector current in gis is 180 i and the base current of <2 is is (180 <)(P Thus ÿthe load current is (P + 1) i + 180 i = pi + 180, while the 180  1 current in R6 is (P + 1) i = (P + 1) i, for which VBe 15 = /?6 (P + 1) with Ic 15 = 180




(180 i) (IP"6) 10,14 i3 (In (180 For P = 400, i in pA: 10"6 X 27 (401) i = 25 x 10"3 10"3 (In (180 /) + 18.42), i = 2.31 In (180 i) + 42.6. 1.
Thus 27 (P + 1) i x

10"6 = 25 x 10"3 In

, with i in pA.
 i) + In 10"), or 10.83 X 10"3 i = 25 x
= 100, i = 2.31 In (100) + 42.6 = 52.2pA; Try i = 52, i = 2.31 In (127.8) + 42.6 = 53.8pA;
Try
1
Try i = 53.8, i = 2.31 In (126.2) + 42.6 = 53.8pA, for which IL = 400 (53.8) + 180 = 21.7mA. Note that for low P, below about 100, the output will be Plimited: For p = 200: 10"6 X 27 (201) i = 25 x 103 (In (180  i) + In 108), i = 4.60 In (180 i) + 87.74. Try i = 50, i = 4.6 In (130) + 87.7 = 110;

Try i = 110, i = 4.6 In (70) + 87.7 = 107.2; Try i = 107.2, i = 4.6 In (72.8) + 87.7 = 107.4, for which lL = 200 (107.4) + 180 = 21.6mA. For P = 100: 106 (27) (101) / = 25 x 10"3 (In (180  i) + In 108), i = 9.17 In (180  1) + 168.9 Try i = 150, i = 9.17 In (30) + 168.9 = 200; Try i = 175, i = 9.17 In (5) + 168.9 = 183.6; Try / = 179, i = 9.17 In (1) + 168.9 = 168.9; Try 1 = 178, i = 9.17 In (2) + 168.9 = 175.2; Try 1 = 177, i = 9.17 In (3) + 168.9 = 178.9; Try 1 = 177.5, i = 9.17 In (2.5) + 168.9 = 177.3, for which lL = 100 (177.3) + 180 = 17.9mA.
 313 
SOLUTIONS: Chapter #106
SECTION 10.6: GAIN AND FREQUENCY RESPONSE OF THE 741 10.21
Using the results of the solution of P10.18 above, the overall gain is A„ Gml (Rol \\ Ri2) X
lAldB
(~Gm2 Roi) ft
120
. assuming that Ri3 Rl + R0 * p
p
is very large, where Gm\ = l/5.26mA/V, Roi = 6.7MO, Rn = 4.0MQ, Gm2 = lO.OmA/V, Ro2 = 78.8kQ, p. = 1.0, Rl = 2kO, R„ = 39 + 27 = 66£2 (from page 833 of the Text). Thus A„ = + m
10~3
(6.7 x
103) X 1 X
106 II 4 x 106) 9
y
(10.0 X
in3
=—ÿ= 362.6 X — 2 x 103 x 66
_\
5.26
x
10"3 X 78.8 x 103 111.2dB. ÿ
For the pole associated with Cc, at the base
1
1010016 104 105 10®
of Q i6> Q = Cc (1  gain) = 30 x 10"12 (1  Gm2 Ro2) = 30 x 10"12 (1 + 10 x 23 X 10"9F, Ri = Rol II R„2 = 6.7MI2 II 4.0M£2 = 2.5 X 106£2. Thus fpp =
10"3 x 78.8 x 103) =
1 23 x 10"9 x 2.5 x 106 = 2.77Hz, for which f, = 363 x 103 x 2.77 = 1.00MHz. Note that f, is the same, since the increased gain is in the Miller compensation stage! 2ji x
10.22 Using the gain result on page 832 of the Text, see that for customized compensation with capacitor Cc, 1 gain) = Cc (1 + 515) = 516Cc, where fp = C, = Cc (1 „ , or 516 Cc = 2nfp 2n Cj Ri Ri 1 , whence Cc = 1.23 x 10"'7/„. 2n fp (2.5 x 106) (a) 45° phase margin occurs when the first pole contributes 90° and the second 45°, where f  f 2 tan 45° = f 2, at the frequency of the second pole, say at f, = 1MHz. Now for At = 103, for which
J.

ir\6
lÿHz, for which Cc = 1.23 x 101°/103 = jq6ÿq3 = 0.123pF. The corresponding 3dB frequency is at 103 (1 + Ap) = 103 (1 + 243 x 103 X 10"3) =
P = 10~3, the
first pole should be at fp =
Atl = 104, for which 244kHz. Now for J
P = 10 , the
a6
first pole should be at /„ Jp =
J —106/104 — ;
j
=
104Hz, for which Cc = 0.0123pF. The corresponding 3dB frequency is at 104 (1 + 243 x 103 x
10"4) = 253kHz. (b)
60° phase margin occurs when the first pole contributes 90° and the second 30°, where f = f2 tan 30° = 0.58 f2,otf = 0.58 f, = 580kHz. Now for Af = 103, for which P = 10~3, the first pole 58 0.58kHz, for which Cc = 1.23 x 10"lo/.58 x 103 = 0.212pF, and should be at/„ Jp = 10ÿ10 = X 103 (244) = 142kHz. Now for Af = 104, for which P = 10 , the first pole should fidB = 058 co v 10® be at fp = " = 5.8kHz, for which Cc = 0.0212pF, and f3dB = 5.8kHz (25.3) = 146kHz.
*
10.23 SR = 21/Cc, where I= 9.5pA or
Cc = 0.123pF
and .0123pF, for which SR =
2
(9
51 X 10"6
~
123 x 10"12 154V/isec. and 1540V/isec. respectively. Now, from Eq. 2.33 on page 100 of the Text, the fullpower
bandwidth is fu=
2k V,O max
ÿ ÿ 2k x 10
= 2.45MHz, and 24.5MHz respectively.
314
SOLUTIONS: Chapter #107
10.24 The output stage is a Class AB type. Its standing current is defined by the current in Qs, being /C5 = 500pA, and the fact that Q6 is four times larger than Qs, being /C6 = 5 X 500pA = 2.5mA. For the gain: /, = /2 = /3 = /4 = 50pA/2 = 25pA, rel = re2 = 25mV/25pA = lkQ, r02 = r04 = 200V/25 X 106 = 8MO. I5 = 500pA, re5 = 25 x 10"3/500 x 10"6 = 5012, rn5 = 121 (50) = 6.05kO, r05 = 200V/500iA = 400kO. /6 = 1.0mA, re6 = 25mV/1.0mA = 2512, rn6 = 121 (25) = 3.025kI2, rM = 200V/1.0mA = 200kf2. For outputstage operation, assume the gain to be controlled primarily through the Qs, Qi connection. Now with /7 = 1.0mA, re7 = 2512, r<>6 = 200kI2, and with a lOkO load, RL' = lOkLl II 200k Q. II 2001:12 = 9.90k£2. At the base of Qlt Rin = 121 (9.09kO) = 1.1MO, and the load on Qs  r*05 'I Rin 400kf2 II 1.1MI2 = 283kO. Thus, the gain from the base of Qs Qb to the output 293 x 103 9 09 x 103  = 5860V/V. Now at the base of Qs, the resistance is RT = = 50 x — 10 + 9.09 xl03 r04 II r02 II rn5 II rn6 = 8 x 106 II 8 x 106 II 6.05k£2 II 3.025kO = 2.017kO. Thus the gain of stage 1 = 2C2 017 x 1031 = 2.02V/V. Correspondingly, the overall gain = 2.02 x 5860 = 11.8 x 103V/V. At the 1kQ base of Q5, RT = 2.017kO, CT = C( 1  5860) = 5861C. Since f„ = l, Cr = 5861C = 271 Rf Cj

—
—
r whence C = 13.5pF. 
2n x 2.017 x
103 x 1 x 103
SECTION 10.7: CMOS OP AMPS 10.25 For Q8, Qs, Qi, ID = 25pA, K = 1MO. 25 = 100 (x>GS lOOpA/V.
For Qh Q2, (x>gs  l)2, or
1/2 (10 X 10~6) (200/10), = 100pA/V2, r0 = VA/ID = 25V/25pA = I)2, or vGS = (25/100)" + 1 = 1.504V, gm = 2K (uG5  V,) = 2 (100) (0.50) =
lD = 25/2pA = 12.5pA. K = 100pA/V2, r0 = 25/12.5pA = 2MO. 12.5 = 100 = (12.5/100)" + 1 = 1.354V/V, gm = 2 (100) (.354) = 70.8pA/V. Id = 12.5pA, K = 1/2 (20 x 100/10) = 100pA/V2, r0 = 25/12.5pA = 2MO, 12.5 = 100
X)GS
For Qs, Q4, (Vgs ~ I)2, or vGS = 1.354V, and gm = 70.8pA/V. For Q6, ID = 25pA, K = 1/2 (20 X 10"6) X 200/10 = 200mA/V2, r0 = 25V/25pA = 1M12, 25 = 200 (Vcs ~ l)2, or vGS = (25/200)l/! + 1 = 1.354V, gm = 2 (200) (.354) = 141.6pA/V.
„ „ .
,
Qi
Qi
Qs
Qa
Qs
Q6
Qi
Qs
b (MA)
12.5
12.5
12.5
12.5
25
25
25
25
IVG5I(V)
1.35
1.35
1.35
1.35
1.50
1.35
1.50
1.50
8m (BAA/)
70.8
70.8
70.8
70.8
100
141.6
100
100
r„ (MO)
2
2
2
2
1
1
1
1
For Gains: A i =
106 (lO"6) 70.8 2 x 106/2 4 ro2 = 35.4V/V. =—Vgm\+Vgm2 —— = 2 10"6) (1/70.8 x 2 r„ II
8m6 (ro6 'I roi) = 141.6 x (70.8) = 2506V/V.
r
106 x
1x
For no load, At 2 =
106/2 = 70.8V/V. Overall, the openloop gain = A\ A2 = 35.4

For the Input CommonMode Range: Input High: VC5 = 5 1.5 + 1 = 4.5V, VGsi = 1.35. Thus, V; £  1.35 = +3.15V. Input Low: VD3 = 5 + 1.35 = 3.65V, VCD1 = IV, V, S 3.65 1 = 4.65V
4.5
 315 

SOLUTIONS: Chapter #108
For the Output ComnwnMode Range: For triode operation of the output devices, the output range is ±5V. For saturatedmode operation, F„ < 5 1.5 + 1 = 4.5V, V0 £ 5 + 1.35 1 = 4.65V. 10.26 For I kef = 12p,A: Reduce all currents in the Table on page 842 of the Text, by the factor 12/25 = 0.48.
Qi
Qi
Qi
Qa
Qs
Q6
Qi
Qs
ID (FA)
6
6
6
6
12
12
12
12
IVGSI (V)
1.28
1.28
1.35
1.35
1.4
1.35
1.4
1.4
gm (FA/V)
42.5
42.5
34.6
34.6
60
69.2
60
60
r„ (MO)
4.2
4.2
4.2
4.2
2.1
2.1
2.1
2.1
For Qs, Qs, Q7, K = 1/2 (10 x 106) (150/10) = 75iA/V2, ID = 12pA = 75pA/V2 (t)GS (12/75)" + 1 = 1.4V, gm= 2 (75) (1.4  1) = 60pA/V, r„ = 25/12 = 2.08MO.
 l)2, vGS =
For Qi, Q2, K = 1/2 (10 X 10"6) (120/8) = 75pA/V2, ID = 12/2 = 6pA = 75 (uGJ  l)2, vGS = (6/75)" + 1 = 1.283V, gm =2 (75) (.283) = 42.5iA/V, r„ = 25/6 = 4.17MQ. For Qa, K = 1/2 (20 X 106) (50/10) = 50pA/V2, ID = 6pA = 50 (t)GS  l)2, vGS = (6/50)" + 1 = 1.346V, gm = 2 (50) (VGS  1) = 2 (50) (.346) = 34.6pA/V. r„ = 25/6 = 4.17M12. For Q6, K = 1/2 (20 x 106) (100/10) = 100lA/V2, ID = 12J.A = 100 (\>GS  l)2, vGS = (12/100)" + 1 = 1.346V, gm= 2 (100) (.346) = 69.2pA/V, r0 = 25/12 = 2.08MK2. Now A = gml (r02 II rM) = 42.5 X 106 (4.2 II 4.2) X 106 = 89.25V/V, and A2 = gm6 (ÿ llr07 = 69.2 X 106 (2.1/2) X 106 = 72.7V/V. Thus A0 = A A2 = (89.25) (72.7) = 6488V/V. Also v, CM max = Vdd ~ IVGS5I + IV,I  IVGSII = 5  1.4 + 1  1.28 = 3.32V, 0/ CM min = VS5 + IVGJ3I  IV„ =  5 + 1.35  1 = 4.65V, \)0 max = VDD ~ KW + IV,I = 5  1.4 + 1 = 4.60V, o0 rain = Vss + \vGS6\  IV,I = 5 + 1.35  1 = 4.65V.
,
,
10.27 For fie, (W/L)6 = 50/10, K = 1/2 (20) 50/10 = 50pA/V2, ID = 25 = 50 (\>GS  l)2, vGS = (1/2)2 + 1 = 1.707V, gm = 2 (50) (.707) = 70.7pA/V, r„ = 25/25 = lMft. A2 = 70.7 (1/2) = 35.4V/V, A\ = 62.5V/V (from Example 10.2), A0 = A, A2 = 62.5 (35.4) = 2212V/V. Now for I6 = h = 25pA, vGs6 = 1.707V. But \)Gis4 = 1.50V. Thus, the input offset =
—
Ai
62.5
= 3.3mV.
10.28 From the solution of P10.26 above, and the development following Eq. 10.50 in the Text, R\ = ra2 II r„4 = 4.2/2 = 2.1M12, R2 = ro7 II ro6 = 2.1/2 = 1.05MI2, Gml = gml = 42.5pA/V, Gm2 = gm2 = 69.2iA/V, 1
Q
1
— ——
49
C
y
10"ÿ
, Cc = = 6.76pF. For a zero at «, R = 1/Gra2 = 1lgm2 = 1/69.2 x 10"6 = /. = 2k x •" 2n x 106 Cc Got2 14.5kQ. For C2, the lOpF output capacitance, and C2 > > Cj, the second pole is at f2 ~ = 2K C2 , fio 9 v 1rr6 :rr = 1.10MHz. Excess phase at 1MHz is tan (1/1.1) = 42.3*. For 6° excess phase at 2k x 10 x lp12 6 = = 6". or// = 1.1 X .1051 = 0.U6MH*, for which Cc =
——
«m A
ÿ ÿ
58.3pF is required. Slew rate, SR = 21/CG where 2/ = 12J.A. For the case, of 42.3* excess phase, SR 10 v irr® 17 x 106 pr = 0.206V/Xs. = 1.78V/is. For 6* excess phase, SR = x 1012 58.3 6.76 X 1012

—
316
SOLUTIONS: Chapter #109
10.29 For IB = 5 xA, and using the rearranged form of Eq. 10.58 on page 849 of the Text: Rb = [2<2 X 20 X 10"6 X 2m x 5 X lO"6)'7'] x {(2m/lf 1] = (l/m'7')(l X 105)(m71 1) = 1 x Now gm 12 = (2i„ Cox{W/L)x2IbY\ = (2 x 20 x 10"6 x 2m x 5 x 10~6)'7' = 20 x 10 and with the currents in 2 12 and Q9 equal, with (X„ = 2.5 \ip, Sm9 = SmizKMh.XW/L),# = 20 X 106m'7,(2.5m)ÿ = 12.65 X lOÿA/V.

(a)
(b)

105 X (1  Vmv').
105(1  (1/if) = 29.3 k£2, and gm2 = 20 x 10"6ÿ2 = 28.3iA/V, and gm9 = 12.7 X 10"6 = 12.7lA/V. For m = 5, RB = 1 X 105(1  (i/5)'7") = 55.3 kO, and gml2 = 20 X lO"6ÿ = 44.7iAW, and Form = 2, RB = 1 x
8m9 = 12.7p.AV. For the loop gain: (Note that the cascode transistors Qi0 and (2 11 provide unity current gain.} Assume a fixed bias current IB and inject voltage t> at the gate of Q9 and measure the return as vÿg.
Thus <9 = gm90, and ogvl3 = i9(l/g„,u) = i/8mi3 i12 = "ÿgsiAÿSmii + Rb)< and Vg.v8 = 'l/gm8  (Vgm8)Cl/(l/gml2 + Rb ))( Vgm n)(gm9v)
Thus the loop gain is L = vGSg\) = (8m
 1).
Thus gmi3RB = 2(1  Vm\ and L = V{\A,VmA + 2  2An*)  2  1/m'7'. Overall, L = 2  1/fon171. Now for m = 2, L = 1.29 V/V, and for m = 5, L 1.55 V/V.
—
Note that the loop gain is less than 1 for small in where 2  1/m1/4 = 1, 1/m'7' = 1, or m = 1. See that as m gets larger, RB gets larger, a marginal disadvantage, but that as m approaches 1, RB approaches 0, implying lack of control and sensitivity to minor variations in device parameters. Clearly m must be large enough (for example) to exceed the uncertainty in mirror gain due to the effect, for example, of
.
SECTION 10.8: ALTERNATIVE CONFIGURATIONS FOR CMOS AND BICMOS OP AMPS 10.30 For the Wilson mirror (Fig. 10.26) in the Text, VBIAS2 ~ IV,I \c + VGS 3C + Vast = Vss, or VB/As2 = VjS + V, 2 VGS. Thus the minimum voltage between VBiAS2 and Vss is 2VGs  V,. Now from Eq. 10.62, R„ = gm4C r„4C r„3 = gm r$ = k'(W/L ) (VGS  V,) r2. Now for the cascode mirror (Fig. 6.32b)), the bias situation is essentially the same, with V, + VGB 3c + Vgs 3 between VB/As 2 and Vss , ie 2VCs  V,. As well, R„ is the same: R„ = gm4 c ro4 c r„4 = gm r§, as before. Thus as measured by the output resistance and outputvoltage overhead, the cascode and (modified) Wilson arc the same. However, if Q 3G were eliminated, VB/AS2 could be reduced to 2VGS  2V, above Vss, but the input offset voltage would be effected, as well. 10.31 For 21 = 10iA, lD = 5lA, Kn = V2\inCnx(W/L)n = 1/2 x 20 x 60/8 = 75 = K, Kp = 1/2 x 10 x 120/8 = 75 = K, r„ = 25/5iA = 5.0M£2, 5 x 10"6 = 75 X 10"6 (uGS l)2, vGS = 1 + (5/75)7' = 1.258V, gm = 2K (\)GS V,) = 2 (75) (.258) = 38.7pA/V. Now R„ = 1/2 (gm r2) = 1/2 (38.7 X 10~6 x 5 x 106 x 5 x 106) = 484MH, and A = gm R„ = 38.7 x 10~6 x 484 x 106 = 18.7 x 103V/V.

,
10.32 For the double cascode, Ro4 cc ~ (gm4 cc ro4 cc) R„4C ~ gm4 cc ro4 cc gm4C r„4c ro3, Rolc ~ (Sm2C ro2c) ro2 Now R„ = R„2G II R„4 cc For conditions as in Ex. 10.28, 21 = 25pA, I= 12.5lA, Kn = K„ = 1/2 (20) (60/8) = 75pA/V2, 12.5 = 75 (oGS l)2, oGS = (12.5/75)'7' + 1 = 1.408V, gm = 2 (75 x 10"ÿ) .408 = 61.2pA/V, r„ = 25/12.5 x 10"6 = 2MQ, Ro4cc = 8m r? = (61.2 x 10"6)2 x (2 x 106)3 = 29.96GI2, Ro2C = gm r„2 = 61.2 X 106 X (2 x 106)2 = 2451VK2, R„ = (0.245 II 30)GQ. = 243MG, Cm, = gm = 61.2 X 10'6A/V. Thus A, = 61.2 x 10ÿ x 243 x 106 = 14.9 X 103V/V. Total voltage from input

 317 
SOLUTIONS: Chapter #1010
to supply is 1 + 1.41 + 1.41 + 1.41 = 3.23V.
10.33 Here, 2/ = IB = 10xA, IV, I = IV, and K = V2k\W/L) in general. For (26, Qf. h = lOjiA, K6 = 1/2 X 20 X 8/8 = 10pA/V2, 10 = 10 (t)GS
 l)2, or vGS = 2V. Thus Vbias's = 2 + 5 = 3V. For Qic, Qic • he = 10  10/2 = 5pA, Kxc = 1/2 x 20 x 60/8 = 75pA/V, 5 = 75 (uGS  l)2, vGS = (5/75)'/l + 1 = 1.258V, also g„, = 2 (75) (.258) = 38.7pA/V. Thus, VBMJ2 = 3  1 + 1.258V = 2.74V. Use 2.75V.
For Qs: h = 10pA, K5 = 1/2 x 20 x 150/10 = 150pA/V2, 10 = 150 (vGS 1.258V, Vmsx = +5  1.26 = 3.74V. Use 3.75V.
 l)2, vGS
= (1/15)* + 1 =
For Qi, Q2: h = 5pA, K{ = 1/2 x 10 x 120/8 = 75pA/V2, 5 = 75 (\>GJ  l)2, dgj = (5/75)" + 1 = 1.258V, gml = 2 (75) (uGS 1) = 2 (75) (.258) = 38.7nA/V. For Q4C: I4C = 5pA, K4C = 1/2 X 10 x 120/8 = 75pA/V2, 5 = 75 (vGS  l)2, oGS = 1.258, gm = 2 (75) .258 = 38.7pA/V.


Output resistance: rol = ro2 = 25/5 pA = 5MO, ro6 rol = 25/10pA = 2.5MO, r„lc = ro2C = ro4C = r„s = 25/5 = 5MO. Thus, Ro4C = gm4C ro4c ro3 = 38.7 x 10"6 x 5 x 106 x 5 x 106 = 967MO, and Foic = gmic r„2C rol II r„2 = 38.7 x 10~6 x 5 x 106 X (2.5 x 106 II 5 x 106) = 322MO. Correspond¬ ingly, R0 = 967 "I I" 322 = 242MO. Gain A0 = gmi R„ = 38.7 x 106 x 242 x 106 = 9.36 x 103V/V.
10.34 For all (Qi, (22. Qic. G20 63c. Q4c)> /o. K, Va> an£l
ro = 25/10 = 2.5M£2 are the same. From Eq. , or 1 = 2k CL /, = 2k x 10 x 10 "12 x 106 = 62.8pA/V. Now ID1 = 10 + 10 = 10.69, /, = * 2k CL 20pA, and rnl = 25/20 = 1.25MQ. Thus Ro4c = r„2 = 62.8 x 106 x (2.5 x 106)2 = 393MG, RoJC = 62.8 x 10"* x 2.5 x 106 x (1.25 x 106 II 2.5 x 105) = 131MQ. Correspondingly, R0 = 131MI2 II 393MI2 = 393/4 = 98.25MG. Also, A0 = gml R„ = 62.8 x 10"6 x 98.25 x 106 = 6170V/V. The dominantpole ÿfrequency, fD = —— —— = r = 162Hz Now SR = = 2n CL R„ 2k x 10 x 10~12 x 98.25 x 106 CL 2 x 10 x 10"6 Km I ÿ
J
10.35 Here, IB = 21 = 800pA, 1D\ = 1D2 = 400pA. Assume p„ C„x = 2p;) C„x = 20pA/V2, V,l = IV, VA = 25V and K = V2\imCnx(W/L). Thus K\ =K2 = 1/2 X 10 x 600/10 = 300pA/V2, 400 = 300 (x>GS  l)2, 1) = 693pA/V, and the output pole is at /, = dG5 = (4/3)ÿ + 1 = 2.155V, gmi = 2 (300) (2.155 693 x 10"6 Snt 1 = 7T = 55.15MHz. For the parasitic pole located at the folding «ic ~ e node, IF\r 2k Cl 2k x 2 x 10 400 x JO"6 ÿow at ÿc emitter of Qtc, the total capacitance is C„1G + 400pA and gm\c = —[q5 =

———
.

Cu6 = CK+ C„, with the corresponding pole at/,lc =
2k (Ck +
X, 10 , and Cn + C„ = 2k /, ic
. For
this parasitic pole to be 10 x higher than the output pole ftiC = 10 x 55.2 = 550MHz, a BJT unitygain frequency which is relatively easy to achieve.
SECTION 10.9: DATA CONVERTERS  AN INTRODUCTION 10.36 For a 100kHz sampling frequency, the highest frequency signal component that can be sampled "ade¬ quately", as noted by Shannon, is at / = 100kHz/2 = 50kHz. This means that for a square wave at 50kHz, the fundamental would be adequately represented, but the waveshapespecific harmonics would not. Note that for sampling at frequency fs = / , that for an input signal at / , output is at / , for input at 0.5/, output is at//2, at 2//2, output is dc, and at 1.1/ 12, output is at 1.1//2 for 9 or so cycles with a break and corresponding phase reversal occurring at a rate of (1.1  1.0)/2 = 0.1/ 12. The figure illus¬ trates input and output waves at various frequencies /, with sampling at /., = 100kHz. For sampling in
 318 
SOLUTIONS: Chapter #1011

e~,/RC = 0.99, or e~l/KC = 0.01, t/RC = 10 x 10~9 X 10~9 10 ,R = 4.6, RC = = 21.712. Thus the switch resistance should not exceed 4.6 4.6 x 100 x 10"12 21.712. a 10ns interval, with source resistance R and capacitor C, 1
————— „
—
Input and Output waves at various (rsquenciss fi with sampling at fs«100kHz vl at 50kHz
v«
at 25kHz
_\ I
ft
I
L
_J
vo
v. at
II I
lOOkHzJjJÿqjyÿJ f»
I
I
f*
I I I I I I I I
I
I
I
I
vo .vl at (at 100kHz)
vo (Note that the scale has . c anfl#
'

I I
_n_rn_rLn_r" I . . Note the phase shift
..
Note that the, top and bottom waves are the same (either high or low) at the time of the mark I on fs
(and ambiguity) at
every tenth cycle for fi 1.1ft/2
10.37 See that the required resolution is 0.1V in 2(5) = 10V or 1 in 100. Thus 2" > 100. Now, 27 = 128 > 100. Thus need 7 bits. For a 10bit converter, 210 = 1024, and the resolution would be 10V/1024 = 9.77mV.
SECTION 10.10: D/A CONVERTER CIRCUITS 10.38 Notice that there are two interpretations for this question depending on whether Rf = R/2 is included in the specification. If not, for R = lk!2 and n = 8, 2n~lR=2*~lR =21 R = 128kl2. But if R/2 is lk!2, the largest resistor required is 2 (128) = 256kl2. The LSB current is V/(21R) = 7.81VpA. Correspond¬ ingly, the allowed error is 1/2 (7.81V) = 3.91V iA (or 1.95V iA for the second view of R). Here, the V V MSB current is V/R nominally, or — , for a switch resistance AR. Now  V < a + An R R +A R V , 1 w 1 2~8, R=R+AR2~*R or  R„ A R„ < n = ' Rn + A Rn = R +A R R 2 (27) R ' 28 R ' + _g 28 2 28 AR AR = 2~SR + 2"8A R. Thus, = 2"8. Now, for an MSB resistor of lk£2, A R =
—
J
J—
—
—
Ra
V,
R,
— —
ÿ

[
< I8 103 = 3.9112, or for
an MSB resistor of 2kfl, A R < 7.8112. Now for both resistor error and switch resistance each contributing a half, switch resistances less than 3.91/2 = 1.95 12, (or 7.81/2 =
319
SOLUTIONS: Chapter #1012
3.91 12) areraccep able. Also, for a perfect Rf (say it is trimmed to the correct value), resistor tolerance 100 ÿ • n /2 = 0.39%. For Rj also variable, allowed resistortolerance = 0.39/2 ~ 0.2%. allowed is 18 10.39 The resistance of an R  2R ladder as seen from the supply is 2R II 2R = R. For a 10V reference, and 1mA, R = lOV/lmA = 10kl2. See that the current in the LSB switch is  , = that in the MSB 128 2"~l 1 1 1 1 1 1 1 1 12, 1 switch. Thus = 1 128 2R 2R +AR 2R 1 + AR/2R 256 ' 1 + AR/2R 256
——
.or '
ÿ
MAR=
= 1 Jj. rhat is
<}>
=
ÿ
„ 78.40. M
Now if 2R is reduced by 78.412 to compensate, doubling of the nominal switch resistance of 7812 to 15612 would again produce an 1/2 LSB error.
10.40 For device junction area 1% in error, in both Qref and Q\, the output current error may be as much as 1 >ref = 50, for Iref For n bits, the LSB current = 2""' ' Now,' 100 2"1 x j, when 2" = which n < 5 bits. If the absolute value of the output current is not critical, 6 bits is available,
SECTION 10.11: A/D CONVERTER CIRCUITS 10.41 The requirement is for ±1V signals s 2V range, with 2 bits. Now, the range is divided into 22 = 4 parts, each of 1/2 volt. Thus, use 3 comparators, with references at 1/2V, 0V, +1/2V, as shown, first in a parallel connection, or alternatively, as 2 in cascade. Output codes for the two circuits are shown in the Table.
V,
B
B0
H2
//,
Ho a)
+0.75
1
1
1
1
1
+0.25
1
0
0
1
1
0.25
0
1
0
0
1
0.75
0
0
0
0
0
+ 1/2V VI »
1/2V b)
VI » ±10V 19k£2
±0.5V
 320 
SOLUTIONS: Chapter #1013
10.42 (a) During
During 4>fl,
Vx
Vrizf , saturates at ±10V with the sign reversed from —VQ  Vref< Vy = 0 + 
that at node Y. (d)
Following d>B,
VA  VREF V* = VREF, VY = , VQ
stays saturated.
Specifically, for i) ii)
VA > Vref, (a) V0 = 0, (b) VQ = 0, (c) VG = 10V, (d) VG = 10V. VA < Vref, (a) VG = 0, (b) VG = 0, (c) VG = +10V, (d) VG = +10V.
Thus the circuit operates as a comparator of
VA against VREF.
 321 
SOLUTIONS: Chapter #1014
NOTES
 322
Chapter 11 FILTERS AND TUNED AMPLIFIERS SECTION 11.1: FILTER TRANSMISSION, TYPES, AND SPECIFICATION 1U
r(s) =
7T5T T<Ja) "
lrl = lryM)l =
*
= 90 " ,a"
log in dB, A = 20 log in dB. For co = oo, in = 1, 4> = 90  90 = 0', G = 20 log(l) = OdB, A = OdB. 2co„ 2 For co = 2co„, III = =ÿ = V = 0.894,
(5)" ((2co„ )2 + co2)" log(.894) = 0.969dB, A = G = 0.969dB.

For co = C0o , in =
= 4 = 0.707, — (co2 + co2)" V2
= 90  tan1
2co„  = 26.6*, G — co„
CO
in
$
v/v
o
G dB
A dB
l
0
0
0
2co0
0.894
26.6
0.969
+0.969
«>„
0.707
45
3
+3
a>
0.447
63.4
6.99
6.99
(oy5
0.196
78.7
14.2
14.2
coyio
0.0995
84.3
20
20
coyioo
0.010
89.4
40
40
coyiooo
0.001
89.94
60
60
11.2 Amu a ld8 OdB
For co = oo, IT! = 0 dB, co = co>p, in = 1dB,
f
TXi 3T Amin
s
co = tov , ID = 50 dB.
50dB
/ \7 V cos
ÿ
(0
(OP
 323 

20
= 20
= 45*. G =  3dB, A = 3dB. — cofl
rad/s
oo
i
0
SOLUTIONS: Chapter #112
11.3
Here, ±5% transmission variation s 0.95 ± 0.05. Thus
Amin =
Amax =
I201og0.05l = 26 dB. The selectivity factor (high pass) =
I201og(0.9)l = 0.915 dB. Now,
—— = J
s
11.4
See T(s)=
= — — 1/c s +—TIT s + 10

t,
= 4/3.2 = 1.25.
103rad/s.
for a highpass filter with co„ =

Now,
I7Ycd)I
=

'tr0'MO = a944 «ÿ 05 dB lrl F" = »944' <»! (cÿaow 0.891 (co2 + 106), co2 = 8.195 x 106, or cop = 2.86 X 103 rad/s. For = 20 dB, 171 = 0.1. Thus — = 0.1, 100 co2 = to2 + 106, 99co2 = 106, co2 = 10.1 x 103, co, = 100.5 rad/s. The selec(co2 + (103) ) 2 86 X !°3 .or , ffactor u \
.
.
tivity
— = 28.5.
,U(highpass) =
ÿÿ
___
103 Hz, wp = 2 n fp = 6.283 x 103 rad/s. Formerly, for the 6 283 x 103 same shape and 0)o = 103 rad/s, co„ r X p = 2.86 x 103 rad/s. Thus the revised co„ is co„ = —2.86 x 103 103 = 2.197 x 103 rad/s, for which X = 1/(2.197 x 103) = 0.455 x 10"3s, and f3JB = /„ = = 2n ,
Now, for a modified filter for which fp =
2 197 x
103
Jf
, ,
, . = , ,,. = .275 s 20 = = 0.35 kHz. At 100Hz, \T(f)\ v 2 it (350 + 100 ) (fo+f) 11.2 dB. Thus, A ,oo = 11.2 dB.
'
—
100
10
.275 = 
SECTION 11.2: THE FILTER TRANSFER FUNCTION _ _ s2(s +0.1)_ _ J2 (s 0.1) 115 7fs1 W= (j + 1) (s + 0.5 + j 0.8) (s + 0.5  j 0.8) (s  1) (s  (0.5  j 0.8)) (s  (0.5 + ;0.8))
_
_s (s
+
1) (j2 + 0.5s
or T(s)
11.6
(s
+ 0.1)
 ;0.8s + 0.5s + 0.25  j 0.4 + j 0.8s + 0.4; + 0.64)
s +0.1s
= —rs3 + 2s2 + 1.89s + 0.89
(jlat y
~
_ j as s

Following the preamble to Equation 11.9: a9 s (s2 + 1 x 106) (s2 + 4 x 106) (s2 + 36 x 106) (s2 + 144 x 106) ÿ s10 + b9 s9 + hg s8 + + b0
_
where the filter order is N = 10. From Fig. 11.4:
1
x103 2 x 103 . (V 4x
103
 324 
6x
103
12 x10
(s
+
_
s (s + 0.1) 1) (s2 + s + 0.89) '
_
SOLUTIONS: Chapter #113
SECTION 11.3: BUTTERWORTH AND CHEBYSHEV FILTERS 11.7
For a Butterworth filter of order N, E = (10Ama/10  if' = (10o yi° 1f' = 0.349. Now. at the edge of the stopband A (coj =  20 log (1 + e2 (to/toP)WTV' = 10 log (1 + e2 (w/(ap)2N), or 40 = 10 log (1 + .3492 (1.6)2"), 1 + .1218 (1.6)2" = 104, (1.6)2" = 8.21 x 104. Try N = 10, or 2N = 20, whence (1.6)20 = 1.209 x 104 rather than 8.21 x 104. Try N = 12, 2N = 24, (1.6)24 = 7.92 X 104. Try N = 13, 2N = 26, (1.6)26 = 20.28 X 104. That is, 13th order will clearly do the job. For N = 13, A(cqv) = 20 log (1 + £2 (cq/ooP)2N)~V' = 20 log (1 + .3492 (1.6)2b)~y' = 20 log (1 + .1218 X 20.28 X 104)/l = 43.93 dB. Now, for Amin = 40 dB exactly, with N = 13, Amin = 10 log (1 + e2 (cq/CDp)2*), or 40 = 10 log (1 + e2 (1.6)26), 1 + e2 (1.6)26 = 104, e = (104/(1.6)26)'/i = 102/(1.6)13 = 0.222. Now .222 = (10A ~/U) if, 10 = .0493 + 1 = 1.0493, A max = 10 log 1.0493 = 0.209 dB. Alternatively, if A max is raised from 0.5 dB to 0.6 dB, e = (1006™ if = .385, and Amin = 10 log (1 + e2 (G)/cOp )w) = 10 log (1 + ,3852) (1.6)25) = 44.8 dB. Now, we can check whether the filter order could be reduced for Amin = 40 dB. See that 40 = 10 log (1 + ,3852 (1.6)w), or 1.62" = 6.75 x 104. Taking logs, 2N = log (6.75 x 104)/log 1.6 = 4.829/.204 = 23.6. Thus, J 2th order would suffice !
11.8
Now, the (lowpass) selectivity ratio is f/fp = 30/20 = 1.5, A mnx = 1 dB, £ = (101710 if = 0.5088, and A (/.,) = 10 log (1 + E2 (J/f,,)2N) or 20 = 10 log (1 + (.509)2 (1.5)w), 1 + .259 (1.5)2" = 100, (1.5)2" = 991.259 = 382.3. Try N = 6, 2N = 12, (1.5)12 = 129.7. Try N = 7, 2N = 14, (1.5)14 = 292. Try N = 8, 2N = 16, (1.5)16 = 657. Thus use 8th order, N = 8. The poles all have the same frequency (D„ = 2n
ÿ—)I/8 = 136.7 krad/s. The first pole (or natural mode) p\ is fp (1/fe)1*1, or w„ = (2tc) 20 x 103 (— ,5Uoo given by p\ = (0o (cos (90 — 11.25) + ysin (90 — 11.25)) = (0„ (.1951 + j'0.9808). Combining p\ )+ with its conjugate p% yields the factor (s + 3:0.3902(0ÿ + (0„). Likewise p2 = co„ ( cos (90 2(8) j sin (90  33.75)) = (0„ (0.5556 + j 0.8315) with factor (s2 + s 1.1111C0„ + co2) and p3 = (D„ ( cos (33.75) + j sin (33.75) ) = co0 ( .8315 + j 0.5556) with factor (s2 + s 1.663 a>„ + co2), and p4 = co„ (cos (11.25) + j sin (11.25)) = co„ (0.9808 + j 0.1951) with factor (s2 + s 1.9616 co„ + co2). Thus ÿ
co08 .
—(s2 + 0.3902 co„s + to2) (s2 + 1.111 co„ s + co2) (s2 + 1.663 co„ s + co2) (s2 + 1.962 co0 s + co2) Generally, \T\ = (1 + E2 (a/U3p)2N )'A Now with fp = 20kHz, N = 8, £ = .5088. At 25kHz, T \ = (1 + (.5088)2 (25/20)l6)14 = (1 + .2589 x 35.53)"14 = 0.313 = 20 log .313 = 10.1 dB, or A = lO.ldB at 25kHz. At 40kHz, 171 = (1 + .2589 X (40/20)l6r'/' = 00768 s 20 log .00268 = 42.3 dB. Thus A =
XCs)
42.3dB at 40kHz.
11.9 Here, at cqv = 2co,, , cq/a>p = 2, N = 3. For Butterworth and Chebyshev, E = (10ÿ ""ÿ°lf = (101710 = 0.509. For Butterworth, A(cov) = 10 log [1 )w] = 10 log [1 + .5092 (2)6]  12.45 £2 (co/co+ dB. For Chebyshev, A (co,5) = 10 log [1 + £2 cosh2 (N cosh"1 cq/cOp)] = 10 log [1 + .5092 cosh2 (3 cosh"'2)] = log [1 + .259 cosh2 (3 (1.317))] = 10 log [1 + .259 cosh2 3.951] = 10 log [1 + .259 (26.0)2] = 22.5 dB.
if'
2ri/j
io
 325 
SOLUTIONS: Chapter #115
follows D, , such that
At very high frequencies, C is a short circuit, and
1 Cs
Ri From
Principles'.
First
/? 2 Cs
=
+
1
.V
Ri + R\ R  + R  R 2 Cs
J
+1
h.
1 Ri
Ri
+ R  R 2 Cs
Ri
Ri +
Vÿ/V,
= +1V/V.
/?. +
Cs R2 + 1/Cs
_1_ Ri
Ri + R2 _1_
+
R,1 «2 R
s
C
+ 1/7?2 C
.
Thus the zero is more pre
1 1 ~ = 110kHz. Note that (RX\\R2) 2 nx 159.2 x 1012 (104 II 10s) AL fP =A„ fz, ie 11 (10kHz) = 1 (110kHz). Now for AH = 1V/V, f/fp = 100, AL = AH f/fp = 100WV, for which R = lOkO, and R2 = R (1001) = 990kl2.
fz  YnC
cisely at
,
,
11.13 The Bandpass requirement suggests the combination of Fig. 11.13a and 11.13b as shown, with 1
CiRx = 2 7t(100),
—
VI
Hh
+ R 2b
11.14 T(s) = 
Rla + R 1h
1 + R\b C
I+ Rib
X
•
s
+
vo
1 ,  = 15.9kl2. For gain = 2 7t x 0.1 X 10 x 100 1V/V, R2 = 15.9kii, as well, with an input resis¬ tance at midband of about 15.9kQ.
r1 R la
s
'
R2
Ri
(Rla
C2 s C1
S
C2 s
\\Rlb)C2 1
+ (Ki.*i*)C,
5
1
C2 R2 = 2 71 (1000) The midband gain is Rffl.\ = 1V/V, with R\ ~ Rin ~ lOkO. Now, if use R\ = R2= 10k£2, C\ = 0.159iF. Chose C\ = O.lpF, 2 71 (100) (104) Now and Ri C2— O.OlfxF.
C2
J11—AA
, „
and
Rla + c s 2 Rla + C s
(Ry,
(Rib
+ /? 2ÿi Rib C2S Ria + R\b + Ria Rib C\ s
+
+ VC2 s )
Rla +
+ 1/Ct s)
ÿ
Rib C 1 Rib C2
s
Rib C,
...... s+
Rla + 1 R C s + 2/, 2
R Ih
Ry, + R2b
s
R 2b
R 2b
1 ÿ2b C2
s
ÿ
R la R ia
R 1h 1 + Rib
R2h C2 Rib C 1
+
1 (R2a II R2b)
+
1 Rib) (RIa II C,
C2
C\
or
S
s
+
S
+
ÿ
R lb
1
Rib C2
, with zeros
at
1 1 1 , and with a highfrequency and , with rpoles at Rib) (Ria II \\R2b)C2' (Ria Ci Rib C2 Rib Ci gain of Rja/Ria, a lowfrequency gain of (Ry, + Rib)!(R\a +R\b), a midband gain of (Ria + Ru)lRu or R2a/(Ria + Rib)< depending on the relative locations of the poles and zeros. For a midband gain of 10V/V, gains at low and high frequencies of 1V/V, and 3dB points at 100Hz and 1000 Hz, the corresponding Bode plot is as shown:
1
and
 327 
"
SOLUTIONS: Chapter #114
11.10 Required that A niax = 0.5dB, Amj„ > 40 dB, (p/(Dp = 1.6 for a Chebyshev filter. From Equation 11.21: e = V 1{yt™**o _j V 10OS/1° 1 = V 1.1221 = V .122 = 0.3493. From Equa¬ tion 11.22, at the stopband edge, where CD = (pt, A (co.v) = lOlog [1 + e2 cosh2 (N cosh1 cp/(Dp)] = lOlog [1 + .34932 cosh2 (N cosh1 1.6)] = lOlog [1 + cosh2 (1.04697/V). Now for N = 10, A = 10 log [1 + 0.122 cosh2 (1.047 x 10)] = 75.8dB, much greater than required. For N = 6, A =10 log [1 + .122 cosh2 (1.047 x 6)] = 39.4dB < Amin. For N = 7, A = 10 log [1 + .122 cosh 2 (1.047 x 7)] = 48.5dB > Amin. Thus, use N = 7, for which Amin = 48.5dB. Now for N = 7 and Aÿn = 40dB, 40 = 10 log [1 + £2 cosh2 (7 cosh1 1.6)]. Thus 1 + e2 cosh2 (7.329) = 104, or e2 = (104  l)/0.5802 x 106 = 0.01723, or £ = 0.1312, for which Amax = 10 log (1 + E2) = 0.074dB is possible. Check: Amj„ = 10 log [1 + E2 cosh2 (N cosh1 CD/to,,)] = 10 log [1 + .01723 cosh2 (7 cosh1 1.6)] = 40dB; OK. Now for A max raised to 0.074 + 0.1 = 0.174dB, E = (lO4™*10 if = 0.202, and A,ÿ = 10 log[l + 0.2022 cosh2 (7.329)] = 43.8dB, an increase of nearly 4dB of stopband attenuation in return for a O.ldB increase in passband ripple!!
_
oll22
11.11 Consider the question to refer to the initial specification in PI 1.7 and PI 1.10 above, for which Amax = 0.5dB and Amjn ÿ 40dB, with (o/(ap = 1.6. Here co;, = 103rad/s, for which cp, = 1.6 X 103rad/s, and, the dc gain is 1.
Now for the Butterworth filter, N = 13, with £ = 0.349, the poles are on a circle with radius (D„ = C0p (1/e)1/W = 103 (1/0.349)1713 = 1.084 x 103rad/s, at an angular separation of tt/N = 13.85*, at angles (from the negative real axis) of 0°, ±13.85°, ±27.69*, ±41.54°, ±55.38°, ±69.23°, ±83.08°. Now p j = to„ ( cos 83.08° + j sin 83.08°) = 1.084 x 103 (  0.1205 + j 0.9927) = 103 ( 0.131 + j 1.076). Correspond¬ (~ c°s 69.23° ± j sin 69.23°) = 1.084 x 103 ingly, p\, P13 = 103 ( 0.131 ± j 1.076) rad/s; p2, P\i( .3546 ± j 0.9350) = 103 ( 0.384 ± j 1.013) rad/s; p3, pu = («„ ( cos 55.38° ± j sin 55.38°) = 103 ( 0.616 ± j 0.892) rad/s; p4, p\0 = (0„ ( cos 41.54° ± j sin 41.54°) = 103 ( 0.811 ± j 0.719) rad/s; p5, p9 = (D„ ( cos 27.69° ± j sin 27.69°) = 103 ( 0.960 ± j 0.504) rad/s; p6, p8 = co„ ( cos 13.85° ± j sin 13.85°) = 103 ( 1.052 ± j 0.259) rad/s; p7 = 1.08 x 103 rad/s. Now for the Chebyshev filter, N = 7, with £ = 0.349, and C0p = 103 rad/s, the poles are (for k = 1 to 7): pk
= 103
sin
= (Op i sin 2/t— 1
= 103
(90°)
2kl N
sinh
sinh j sinh1
N
sinh
(l/0.349 )
1
+j
Ffe
+j
cos
cos
2kl N
2k— I
(90 °)
cosh cosh
_1_ sinh 1 l/t N
sinh
(1/0.349 ]
[0.2563sin ((2itl) (12.86°)) + 1.032; cos ((2Jfe—1) (12.86°)) ].
Now, pi, Pl = 103 [ 0.057 ± j 1.006] rad/s; p2, p6 = 0.231 ± j 0.448] rad/s; p4 = 103 [ 0.256] rad/s.
103 [" 0.160 ± j
0.807] rad/s; p3, ps =
103
[
SECTION 11.4: FIRSTORDER AND SECONDORDER FILTER FUNCTIONS 11.12 For infinite input resistance, the circuit must be driven as shown. Use R\ = 10kS2. The dc gain is 1+ = 11 V/V. Thus R/Ri = 10, and R2 = 10 R\ = 100kf2. For a 3dB frequency of 10kHz,
AAR2
CR22 = l/C0o, vo
or C = r = 159.2pF. 2 Ji x 10 x 10
Thus, the zero frequency is approximately at 1 _ f = 2n i Jz C R\ 2 7t x 159.2 x 1012 x 100kHz.
_
vi
 326 
104
SOLUTIONS: Chapter #116
Now, at low frequencies,
frequencies,
R la
RiId
+ÿ „2
= 1; at midband,
= 1; at high
(Rla
+ Rib)
R la 10kHz 10. Thus Ria + Rib = R ia + R\b = 10 R \a = 10 Ria, and Rib =9R\a, Ry, = 9/?ÿ, R\a = Ria and R\h  Rib • For Rin = lOkO at midband, R\„ ~ 100Hz" 1000Hz 10Hz lOkO and R\h ~ 90k£2. l 1 1 For the zero at 10Hz, = 10, C, = = 0.177iF. 2 n Rtb C i 2 it /? 5 x 10 2xcx 10x90 x 103 1 Use C = 0.2iF, for which R\b = " 2 it C( X 10
.
j2nf
o
O X X
1
2no ÿ
= 79.58kQ = /?y,, and 2 it x 0.2 x 10"6 x 10 79.58/9 = 8.84kO. Now for pole the at 1kHz:
,
= 103, C2 =
1
2 it X 103 x 79.58 x 103 check the second pole, to be at fp2 =
10.
R\a  Ry, = 1 2 it C2 R2b
= 0.002pF. Now,
1 2 it (Ria II Rib)Ci Here Ria II Rlh = (8.84 II 79.58)kf2 = 7.956k£2, and fpl = r = 100Hz, with /„> Jp at 1kHz. 2 it (7.956 x 103) x 0.2 x 10"6 Correspondingly, fzi = 10Hz and fz2 = 10kHz. The polezero plot is as shown:
100 10
ÿ5
10'
11.15 The required response is as shown in the Bode plot below:
.
IAI iL 20dB
\
OdB
10X N
I
o >Hz
1kHz
For the circuit in PI 1.14 to have a gain of 10V/V at dc,
f 1(JkHz
Ria + R2h —— = 10 —R\a +R lb
'la



(1). For a gain of
(2). Now for a lower gain at midband, C2 must provide = 10 R la a zero at 100Hz, and C\ a zero at 1kHz, with Ry, shorted there (while C i is still (relatively) open). R la (3). Thus in the midband, the gain is  =1
10V/V at high frequencies,
—R la + Rib —
The pole associated with C2 must be at 100Hz/10 = 10Hz. The pole associated with C\ must be at (1kHz) X 10 = 10kHz. Now, Rin > lOkQ, Rla > lOkCL For Ru = lOkfi, Ry, = 10 (10) = lOOkfl From (3), Ry, = R,„ + Rib, or Rib = Ry, R\a = lOOkfi  10k£2 = 90kf2. From (1), R y, + R y, = 10 Rla + 10 Rib, or Ry, = 10 (10) + 10 (90)  100 = 100 + 900  100 = 900kf2. That is (tentatively), = 1kHz, or Ria = 10kn, Rib = 90kn, Ria 100k£2, Ry, = 900k£2. For a zero at 1kHz, 2 it Rib Ct 1 C, = .00177(iF. Now, for a onesignificantdigit capacitor, use Ct = .OOlftF, '1 = 2 it x 103 x 90 x 10Y 1 or InF, to maintain /?,„ > 10kI2. Conclude: R a, = = 159kQ, Rÿ = (10/90) 2 it x 103 x .001 x 106

— ——

 328 
___
SOLUTIONS: Chapter #117
159 = 17.7k£2, Ru = 10 (17.7) = 177kQ, Rv, = 10 (159) = 1.591MO. Now for a zero at 100Hz associated with C2, use C2 = 1 1 = ~ j 1 j2nf 2 rc x 102 (R2a II Rib ) 2 n x 102 (0.177 II 1.591) 106 .999 x 10"8 = lOnF. Check: The pole from C2 is
_
104
""
v A
100 Pi
*22no
Pi v A
103 
_
—— —— = 3= 10Hz, and 2 Jt C2 Rn, 2 Jtx 10 x 109x 1.591 x 106 7
the
pole
C
from
is
l/[2 71 C t (/? „ II R tb )]
10
=
142 n x .001 x 106 x (17.7 II 159) x 103] = 0.0099 x 106 = 10kHz. Overall, there are zeroes at 100Hz„ 1kHz, and poles at 10Hz, 10kHz, for which the polezero plot is shown.
11.16 For Fig. 11.14 modified:
_ —R1 vb
V +
1/CR
4>
ÿI
R2
V90"
180°
vo 2R R vi, and T(s ) =  1= = vh + vh x>i = 2\)b v' 0 R VCs t>; R + VCs + s 1/RC RCs  1 2R  R VCs , or T(s) See a pole at 1/RC and a zero at 1/RC. Now at ~v"' = s + 1/RC RCs + 1' "" R + VCs co„ = 1IRC, phase shift is +tan 1 (1) tan~'(l) = 90°. 1 Now, for /„ = 104, C s 1.59 x 109F, and l
ÿ
,


—

.


104Hz, for 4> = 90°, R =
tan(0/2). Now, at/0 =
104£2.
2 ji X
1 1.59 x
104 x
ÿ 109
tan (
 90/2) = 104(1) =
For ld>l = 6°, R = 104 tan(6/2) = 5240; for 12°, R = 104 tan(12/2) = 105 Id; for 30°, R = 104 tan(30/2) = 2.68k£2; for 60°, R = 104 tan(60/2) = 5.77k£2; for 90°, R = 104 tan(90/2) = 10k£2; for 120°, R = 104 tan(120/2) = 17.3k£2; for 150°, R = 104 tan(150/2) = 37.3kH; for 168°, R = 104 tan(168/2) = 95.1kfl; for 174°, R = 104 tan(174/2) = 190.8k£2.
11.17 For co„ = 103rad/s, and 3dB bandwidth is of m„/Q = 200 rad/s, see Q = a\ Q
co„
103/200 = 5.
Q] S
= 1. Thus, a 1 = ox/2 = 10/5 = 200, and T(s) =
*2 +
5 CO„
Q
+ C0„
Now peak gain is 200s . At a), s2 + 200s + 106
200 co 200 j co , 20002 co2 = 200 co2 At Amin = 20dB, T(j

A

 329
SOLUTIONS: Chapter #118
+6 x 106 ± V (6 x 106)2  4 x 10'2 _ 6 x 106 ± 5.657 x 106 _ = = = .1716 x 10", and 5.829 X 106, where CO = 414 rad/s and 2.414 k rad/s. Check: 0.414 x 2.414 = 0.999 krad/s = 103 rad/s, as expected. or
11.18
(O
peak
peak
m,
r_i
OdB
T?
Amin
Amax=3dB
Amax=3dB
7
a)
£08
COP
1/2
1
COrad/s
COrad/s
The response characteristics described above are as follows: For both designs, we require A max = 3dB at (Dp = 1 rad/s, with maximum gain = 1V/V, and Aÿ,, to be measured at cqv = 0.5 rad/s. From Fig. 11.16b) for the response shown first, arrange that the peak amplitude is such that a 2 = 1, that a2 Q 1 , 4 Q4 = 4 Q2  1, 4 Q4  4 Q2 + 1 = 0, is that ~ a2' Q =1 2\'/i 4 Q2 (1  1/4 QT 1 a2 s +4 ± V lfi 4 (41 r in general, Q2 = 2 (4) = 1/2, and Q = ~iV2 = 0.707. Now, T(s) = —5 s2 + s 0vQ +(°o2
_
—
—— —
*
For a2 = 1, Q = 0.707, (0 =
1
,,rr
((to2  1)2 + 2(o2r
s2 + 1.414 s + 1
= .707, or
(0„
+ 2co2 +1—2 to2 = 2, or
'j/2)2 + 1.414 (j/2) + 1
\TT7((2.828) + 3 )
= 1, and co„ = 1. Thus
(0„
T(s)
=
= 1rad/s. Now, for A,™,, =  \T(j 'A)\ , T (j/2) =
, with Q = 0.707, and co„ =
(j/2.)2
(b)

O'lr + 7'1 (0,/707 + (oj
1 4 [ (1/4)+ 1 +.707./]
1
Amin =
2.828./ + 3

log
20
= 20 log 0.2425 = 12.3dB.
From the second response, see a2 = 0.707, and at the peak,
,
21 = 1iy, 2 (24 r, Q2 = 2 4(2 2 (2 4Q — 2 ± ÿ 22  4 (0.5) 2±ÿ2 1.707 for Q rel="nofollow"> 1.
/
*
= 4 (22
*
—— =
1, or (0.707
Q4
Q2 +
Q ~
1  K4 Q )

1,
~
2
Now, in general, T(s) =
Q)2 = 1
0.5 = 0, Q = 02 J
,
s2 + s co„/Q + (O2 ' .707 (j l)2 and for (2 = 1.707, a2 = 0.707, co = (Dp = 1 for 171 = 0.707. T(j 1) = (/I)2 + 7'1 o),/! .707 + (0o 0.707 .707 2 to2 + 1 + .343 to22 = 1 or Now, 0.707 = 2 <)] 2 v/, l)2 [((o)2 .5858 1) j (0o (co0 + + .5858 .707 s2 or T(s) = T(s) = Thus 1.657, rad/s. 1.287 to2, co2 = C0g = 1.657 co0 = s2 + s 1.287/1.707 + (1.287)j, .707 s2 with co„ = 1.287 rad/s, and (a3dB = 1 rad/s. Now T(j/2) = s2 + 0.754s + 1.657 ' 1 1 .707 .707 (1/4) r, and Amin = 20 log (7.962 + 2.1332)'/j 7.960 + 2.133/ """ 5.628 + 1.508/ 1/4 + .754 j/2 + 1.657 = 20 log (.1213) = 18.3dB, an improvement of 6dB over the arrangement in (a).

.

.
_
<
"
 330
SOLUTIONS: Chapter #119
11.19 Use the description in Fig. 11.16d for the notch. Here f„ = 60Hz, and to„ = 2 n(60) = 377.0 rad/s. s2 + 3772 Now T(s) = . Now, assuming the required 1Hz band to be centered, arrange that
s2 + s 2ZL + ill1
attenuation
is
20dB
60
_ Q'380.1)2 + 3772_ _ J_+ or 1/?
le
at
(/380.1)2 + )380.1 (377/(2) + 3772
(l + (61Wj2)2
, or
10 bandwidth is
1+
jo)
61.4
61.06 9.95

_ 6.136.
Q Now the 3cjb
6.14
Q'to)2 + 3772
;(oÿ
+ 3772
'
= 0.707, l2 +
= 0.707,
61.4©
= 0.707,
'
2
..,1V"" ,
3772  w' 3772  a)2 + j co 61.4
1 .707
61.1 ©
, ©2, = (2  If = ± 1 
...... .......... 3772©2
=
= 2,
3772  to2   (1). Now, 3772  to2 = ± 61.1co, ©2 ± 61.1co  3772 = 0, to = ± 61.1 ± 756.5 , of . which .
—'
1
= —— = 9.78Hz.
0(o)2 +
_

Q
Thus, rad/s. 1 61.06j
380.
or
2347 + 143298y/(2 ÿ
For the 3dB frequencies:
_1
60.5Hz, 2347
.61.06, 61.06  9.95, q = 102, (ÿiÿL)2  99,
60
Q
10*
=
.
,
..
the relevant solutions are
377
61.12
±61.1
„ 756.561.1 ,756.5 + 61.1 . rad/s,' and — = 347.7 2 2
2 408.8 rad/s, or 55.3Hz and 65.06Hz, for a 3dB bandwidth of 65. 1
— 55.3 = 9.8Hz.
(As noted.)
0'©)2 + /©
= 0.891,
6.14
+ 3772
'
= .891,
,
1+
1 61.4©
. — 3772  ©2rJ
_ 10wo
Qm)2 + 3772
For the ldB frequencies: 20 log x = 1 » jc = .891V/V, and
3772  ©2 3772  ©2 + 61.4 ;©
4
= .891, l2 +
61.4©
3772©2
= 1.2596, 61.4© = ± (3772  m2) (.2596)* = ± (3772  ©2) .5096, ©2 ± 120.5©  3772 = 0, ± 120.5 ± ÿ 120.52 + 4 (377)2 ± 120.5 ± 763.6 © for
2
'
1 .89!
which relevant solutions 2 2 7366 ~ 1205 120.5 + 763.6 = 442rad/s or 70.35Hz, and = 308.1 rad/s or 49.0Hz. =
For the 1% frequencies (from (1) above)
61.1©
3772  ©2
)2  1 — .99
are
= ± .142, or ©2 ± 428.8©

± 428.8 ± V 428.82 + 4 (3772) ± 428.8 ± 754.3 , . . . . 1=of which appropriate solutions are =591.6 rad/s or 94.1Hz, and 162.8 rad/s or 25.9Hz. __2
377
n = 0,
©
 331 
SOLUTIONS: Chapter #1110
SECTION 11.5: THE SECONDORDER LCR RESONATOR 11.20 Equation 11.29 indicates that:
1+
0)1,0)2 = ©„
Now BW =
1 2Q
4Q2
106
o)„
= 2 71 (20 X 10 ), —— Q
whence Q = = 50. 20 x 103 50 Q Now Q 0)„ CR, and C 796pF. 0)„ R 2 Ji x 106 X 104 1 1 Accordingly, L = 2 = 264iH. (2 7t x 106)2 x 796 x 10"12 © C For a 1mA rms input at the 1MHz resonant frequency, v„ = 104 x 1(T3 = 10V rms. 11.21 Here, O)2 = ÿ = (2 71 x 99.9 x 106)2 = (6.2769 x 108)2 = 3.940 x LC
l . Now 1017, L = C x 3.94 x 1017 99.9 X 106 f» „2 (2 .. x 106), Q = 99.9 2 = 97.9MHz. Thus = = 24.98. ,n
the response is 3dB down at Now
Q
=
CR,
o)„
C
_14:" ÿ98_
Q
=
©„ R
2 7t x 99.9 x
1 = 0.0048xH. 530.6 X 10"12 x 3.94 x 101w For offtuning by 100kHz, ©1 = 2 7t (99.9 x
Eq. 11.40 for the notch (with a2 = 1),
ÿ
Q
4 x 10'
530.6pF,
=
106 X 75
and
L
=
106 + 0.1 x 106) = 200 71 X 106 = 6.2822 x 108rad/s. From r + 0)* 1 T(s) = , and T(a>i) = , s coy«2 s2 + siWo/Q) + O)2 1+ jr + ffl„ 1 1 j 19.95
Now, 7 (00,)  = j 1.579x 10 j 6.2832 X 10B (6.2769 x 10g>24.98 { t + (6.2832 X 108)2 + (6.2769 X 108)2 .07913 X 1016 .050. Thus the attenuation expected would be 20 log (.05) = 26dB. 2\'A = (lz + 19.95"7

16
11.22 For a maximally flat response, Q = W2 = 0.707. a2 s
s2 + s
(©yg) + ©2
where a2 = 1, and T(jwi) =
Now, for the highpass filter, T(s) = 0)z
or + 1.414;© ©„ + ©„
.
For a 3dB frequency of
100kHz:
_ (10s)2 *2 x 10'° = ( (f2  1010)2 + (1.414 X 105)2/2 f, or 2 X = (105)2 + 1.414; 10s /„ +f2 1020 = / 4  2 x 1010 f2 + 1020 + 2 x 1010 f2, or / 4 = (2  1) 1020 = 1020, or f„ = 10sHz (as could be seen directly). Now, LC = V[2 n x 10s)2 = 2.533 X 10~12. For an ideal coil, Q = ©„ CR, and C = 0.707 ~ ,4 X v o X in5 104 10 71 v 2w
coil available, Q = parallel V
_
©„ L
resistance is R„ p
.. .
1
w,m L ~ ,„2 v r ~ (2 w © ;xC o 7t
, or
r =
 —CO
0
ÿ 71 x
m
2 7t x
x 10T
X ÿ
50
•
C
1
10s x
112.5 x ,nl2 10"
v 1n5\2 v X 1n < v
= 22.5mH. ow for the
— = 28312 (in series with L).
50 112.5 x
10"'2
The equivalent
= 707kf2. Since R„ » R, one can
ignore it.
SECTION 11.6: SECONDORDER ACTIVE FILTERS BASED ON INDUCTOR REPLACEMENT 11.23 For the InductanceSimulator of Fig. 11.20, L = C4 R\ Rj RÿR2. Use R\ = R$ = R2 = 10kf2, and R$ = 10k£2 to accommodate the lack of capacitor choice. Now for L = 10H and R$ = lOkfl, C4 = X x 0.1 = = .L = — —j = 10 x 10"8 = 0.1)tF, and for L = 0.1H, C4 = ÿ 10 104 x 104 x R5 104 R5 104 x 104
aL 1°
 332
SOLUTIONS: Chapter #1111
InF. Alternatively, for a fixed C, where lOnF. Now select
10H, R5 =
—
2nfC
R\ = R3 = R2 =
ÿ Use O.OluF = lOkQ, C4 = 7 = ,0159uF. ÿ ÿ =
2rcxl04xl03 C4 R\ R3 R/R2 = 104 C4 R$. For L = For L = 0.1H, R55 = —7—r = lkQ. 104 x .01 X 10"6
lOkQ in which case L =
y = —7——r = lOOkO. — 104 — C4 104 x .01 x 10"6
11.24 From the solution to PI 1.23 above, for L = 10H, Rt = R2 = R2 = 104O, R5 = 105O, C4 = 10~8F, and for L = 0.1H, Rt = R2 = R3 = 104O, R5 = 103f2, C4 = 10"8F. For /„ = 1kHz, co„ = 2 n x 103 = 6.28 x 103 rad/s. Using R$ = 20k£2, with o)„ = (1/LC)'7', or C = ÿ7, and Q = co0 C/?. For L = 10H, C = 4
— L
0)
1
= .00254pF, and Q = co„ CR = 6.28 x 103 X .00254 X 10"6 X 20 X 103 = 0.319. 10 X (6.28 x 103)2 For L = 0.1H, C = yj = .254iF, and Q = to„ CR = 6.28 X 103 X 0.254 X 10~6 x 20 0.1 x (6.28 x 10") X 103 = 31.9.
'
(a)
(b) (c)
Use Rl=R2 = R3= 104£2, Rs = 103£2, C4 = lOnF, and C = 254nF, to obtain a Q of 31.9. Use /?i = R2 = R3 = 104£2, R5 = 105f2, C4 = lOnF, and C = 2.54nF, to obtain a Q of 0.32.
For a design with equalvalued capacitors: For the simulated inductor L = C4 R\ R3 R/R2, and for /?! = R3 = R2 = lOkUt, L = 104 C4 /?5. Now with C = C4, the resonant frequency is co„ =
(—1—
L C4
f
=2 nX
103rad/s.
10"4 1 1 2 533 X 10~12 7, and ÿ5 = #5 x 104)/! = 77 x —7 = =, or C4 = ji C} 103)2 (2 x x 2 k 103 C} 2 533 _X 10'2 y/»# Now, as g = co„ CR, to raise Q, keep C4 relatively large. Now, if R5 is limited to (_: 
ÿ2 TU (C4 Thus
.0
D
1
12 103a, C4 = ( 2 533 x 10 )'/• = ,0503xF.
103
—
(Use ,05pF). Now for
C4 = .05iF, R5 =
2,533 X
(.05 x
12
ÿ,= 10"6)2
1.013k£2. In which case, Q = 0)„ C4 R = 2k x 103 x .05 x 10"6 X 20 X 103 = 6.28. It is apparent that if a smaller value of R$ were allowed, C\ = C could be higher, and Q would be raised. For example,
with C4 = C = O.lftF, /?5 =
9
v 1ry~
.
J, = 25312, and Q = 2 w x (0.1 X 10*)6
103 x .1 x 10"6 x 20 X 103 = 12.6.
11.25 For a 5thorder Butterworth with 3dB bandwidth of 104Hz, e = 1, and (Op = 2 ji x 104rad/s, with a pole K radius co„ = (np {W)w = 2itx 104rad/s. Now poles are at 90° = 82°, and 82°  = 46°, and
ÿ
®Aq
1 1 pair has a Q such that cos 82° = = — , or O = pole 0°. Thus the first complex „ F v v co„ * 2 cos 82° 2Q ÿ 3.593. For the second complex pole pair, Q = —— = 0.7198. For the 5th pole, Q = 0.5. Use a
—
ÿ
2 cos 46
cascode of two circuits of the form shown in Fig. 11.22a with one of the form shown in Fig. 11.13a on the right. For a straightforward design, seven op amps would be needed. To achieve a lowfrequency gain of 10, arrange a) a gain of 1 in one secondorder section using a wire, b) a gain of 5 in the other secondorder section using 2 series 33k£2 in the feedback path and 2 parallel 33k£2 to ground, c) a gain of 2 in the firstorder section with one 33k£2 in the feedback and 2 parallel 33k£2 at the input.
—
Now for the firstorder section (Fig. 11.13a), for which co„ =  7 and O = 0.5, R 1 = 2 n 104 1 33k£2 II 33k£2, R2 = 33k£2, C = ÿUse C = 400pF F 7 = 482.2pF. r II 80pF. v 2 7t x 104 x 33 x 103 Now for one first secondorder section (Fig. 11.22a), for which (0o =
—7 and Q = 3.593, use C4 = 2 re 104
C6 = C = 400pF II 80pF, and /?, = R2 = R3 = R5 = 33kI2, with R6 = QR = 3.593 (33) =
 333 
118.6kfl.
SOLUTIONS: Chapter #1112
Use 120k£2. Now for the other secondorder section (Fig. 11.22a), for which
0)„
=
—r and Q 2 7t 104
= .7198, use C4
= C6 = C = 400pF II 80pF, and R, = R2 = R3 = R5 = 33kf2, with R6 = QR = .7198 (33) = 23.75k£2. Use 24k£L
11.26 From Table 11.1 for the BP filter, T(s) =
sÿ(> ÿ6
__ Now, J 2
R2
.
, C6R6 , CÿCÿR\R3R3 21
ÿ
®
— , co2
C6ÿ6

C6 = C, with R6 to control Q (and co„), and Rt = R2 » R3 = R , with 1 1 r6 ... (1), and Q = (0o C R6 = , R5 to control co0. Thus ft)2 = —5, co0 = , cV rr5 c2rr5 Vrr5  (2). For capacitors using single digits 1, 2, 3, 5, the largest range of C to be accommodated by Rs in establishing co„ is 2/1 = 2. Since from (1), R5 = — — , the compensating range of R$ will be 22 = C R . 4. Now from (2), /?6 = v /? R$ Q, and for Q varying from 0.5 to 50 (that is by a factor of 100), and /?5 by a factor of 4, R$ must vary by a factor of "ÿ4 (100) = 200.
C4
Cg /?i
R3 /?5
Use C4 =
SECTION 11.7: SECONDORDER ACTIVE FILTERS BASED ON THE TWOINTEGRATORLOOP TOPOLOGY 11.27 From Fig. 11.16c) for a BP filter, see that the 3dB bandwidth is 1o4
100
500
5
®o
, Here, co„ = 2 it x 10 , and 1
ÿ
—=
= 20. Now for Fig. 11.24a, CR = — , = 1. = 2Q  1, and co„ «2 «i ~K ÿ *r. For C = InF, R =  = =— p 3 = 15.92kQ. From Eq. M 11.59, TbAs) 1 x x x 10"9 2 it 104 s2 + s ((M2) + (0o2 500 x 27t, or Q =
Thus at (0„,
Vhp = Q (2 V;
ÿ •/ (D"
w2 + j(o„
Vt
 = _KQ Now> from Eq
(coyg) + co2
f
nM k = 2 (1/(2). Thus
— 1/(2) = 2(2 + 1 = 2 (20) + 1 = 39V/V. Now using Miller's theorem at resonance (R2 + R3) R 2 4" R 3 R2 + R3 R 2 + R 3 (
•

An
00
ÿ
An
R3 = 4MQ. Also R = 2(2 X 1 = 39 as well, and R3 = 39 R2. Thus R2 + 39 R2 = 4M£2, whence R2 2 = lOOkQ and R3 = 3.9MQ, unfortunately too large. To reduce R 3, replace the original R2 R3 circuit by
a network.
R2 (V, Vb„) — —R+ R3 II1 —R2 — — (1 —39) r = Vj I1 — 4M« (40) f = 0. Thus the positive input node of the + R3 J J l
In the original situation,
= Vf
Vx = V)
2
leftmost amplifier is a virtual ground (as might have been obvious already). Thus use the network as shown:
 334 
SOLUTIONS: Chapter #1113
Note that since
100 kQ
—
WV
tage
100 kQ
1
>
VI 100 kQ
15.9 kQ
15.9 kQ
—
100 kQ
100 kQ
Va = Vh
Vx = 0, the Rx II 100
vol¬
100 + R. II 100
, 100 + Rx II 100 = 39 38 Rx (100) 100)
= 100' iotnr = 100 + Rx, 37 Rx =
" Rx
100,
Rx = 100/37 = 2.7kft. Use R2 = R 3a = R3B = R i = Rf = lOOkl). The centerfrequency gain is 39V/V.
Wv
2.7 kQ
11.28 From Ex. 11.22, for f„ = 5kHz, /„ = 8kHz, Q = 5, dc gain = 3V/V, and C = InF, one used R = 31.83kQ, R\ Rf = R 2 lOkO, R3 90kO, Rfi 25.6kO, Rf 42.7kS2, RB 00, Rÿ lOkO. Here,
—
—
—
—
—
—
f„ = 5kHz, /„ = 7.5kHz, Q = 10, and dc gain = 3V/V. From Eq. 11.67, 2.25. For RL = 10kl2, R„ = 22.5kf2, RB = «>. From Eq. 11.66, dc gain =
—
= (—)2
Rl
—
©„
K RF/R 1co
"
= = (ÿ)2 5 ~~K RF
*l
3Rl = K
3 x 10m Here K = 2 HQ, with Q = 10. Thus K = 2 1/10 = 1.9, and RF = 15.79k£2. 1.9 1 Use C = InF, R = 5 5 = 31.83kQ. Use Rt  Rr  R2 = lOkfl, and R3 = R2 (2Q 2 it X 5 X 10J X 10 1) = 10 (2 (10)  1) = 190kfl.


11.29 Required to design a bandpass filter with /„ = 10kHz, f 3BB = 0.5kHz, /?,„ = lOOkQ using C = InF, with a centerfrequency gain of 39V/V. Thus, for the TowThomas circuit of Fig. 11.26, chose the OR  . Here, negative bandpass version, for which Ci = 0, R2 = R3 = o°, and R 1 = centerfrequency gam whence Q = from Fig. 11.16c, f 3dB = = 0.5 = 20. Thus R\ = 39 but the input resisQ J 3JB 39 R 1 = 195kf2. Use R = 200kQ for a slightly higher value of Rin = tance = Ri > lOOkfl Thus R =
/"
—
rÿ= 769pF. 104 x 20 x 103x
(200) = 102.6kO. Now a)„ = and C = RC 39 2 nx = lOOkQ, and QR = 4MO. 1
As well, use r
SECTION 11.8: SINGLEAMPLIFIER BIQUADRATIC ACTIVE FILTERS 11.30 Eq. 11.73 and 11.74 indicate that co„ = (C 1 C2 R3 R4) v' and
I Q =(
(C
C2 R 3 R4)*
——
li
(1/C, + 1/C2)
f . Try R3 = R4 = 1M£2 and C{ = C2 = C.
Now for co„ =
1
. = lOpF, v Q = 10s x 106 [l/lO11 + 1/10"]' = 10" x [2 X lO11]"1 = 1/2 (as expected). Now since the required Q is 0.707, C must be raised to allow resistor values which are dif¬ ferent, yet no larger than 1M12. Thus we could try C = 20pF and proceed to find R3 R4, then each 1014 separately, as follows: See R3 R4 = (Ct C2 to2)1 = (20 x 10~12 x 105)"2 = (20 x 10~7)"2 = ——— =
105 rad/s, C =
0.25
X
1012.
105 x 106
Now for Q =
lW~2, <2 =
5
*
R3
1
—pr), or R3 = 20 x 10 u
(
 335 
ÿ
5 ÿ
2
x
10" = .707M12, for
SOLUTIONS: Chapter #1114
25 x 1012 = .354MQ. Thus, C, = C2 = 20pF, R3 = .707MK2 and R4 = R3/2 = .354MO —— .707 x 106
which R4 =
is a possible solution. Check: co„ = (20 x 10~12 x 20 x 10"'2 X .707 x But this is possibly not the solution with the largest possible resistors!
___
106 x 0.354 x 106 )_/' = 10s.
C\ = lOpF, C2 = 20pF, for which R3 R4 = (10 X 10"12 x 20 X 10"12 x 10s X 105)1 = i5\l i5 1 1 , or R3 = 10 x 10" (1.5) = 0.5 x 10"12, and for Q, <2 = 20 x 10"12 10 x 10~12 TT R3 0:5 x 1012 1.061MI2, for which R4 = — r = .471MO. But this solution is not directly acceptable since R 3 1.061 x 106
Alternatively, try
is too large (although a series combination would clearly suffice). For example, one could use C, = lOpF, C2 = 20pF, R3 = 1IVK2 + 62k£2 in series, R4 = 470kI2. Alternatively, tryrC = lOpF, C2 = 50pF, fof which R3 R4 = (10 x 50 X 10"24 x 1010) = 0.2 X 1012, and 5\—1
1rr5 1 , or R3 = 4t (1.2 10 x 10" 12
1
10"12
R3 0.2 x 1012 = 0.236M£2. .849 x 106 50 x
— IT
Could use C, = lOpF,
X
10") = 0.849MO, for which R4 =
C2 = 50pF, R3 = .849MI2, R4 = .236M£2. Of the
three solutions, the lirst, with equal capacitors, is the most straightforward. 11.31 Now, for
R3 R4y'A, R3
R4 =
\
—104 , = 20.
5 x 10z q j = 2.533 x
104 x 10"9)2 (ÿ=5) = 6.366 X 105 = 10"9 (2 re x
9
.20
104 x 2 re (C,
Q=
____
= 2 n x 104rad/s, Q =
(0„
c2 r3 R4)
Use
C\ = C2 = InF =
108.
6370, and
From Q = co0 9
y
10yF. From (0„ = (C 1
R3
[1/C, +
C2
1/C2]', R3 =
108
R4 = 6.366 x 10s. = 39812.
Check:
(1/C, + 1/C2)
R3 li (10~9 x 1Q9 x 6.37 x 10s x 398)'/' i"9\ (1/109 + 1/10") 6.36 x 10s
159.2 x 10"7 2 x = 19.97: 6.36 x 10s 10" OK. Now, for the center frequency gain: For VA (at the common node A of C,, C2, R/a and 1 Vo Rf(la)), with the voltage at the amplifier's negative input being zero volts, see VA = 0 = sC3 R3
———
——— . Writing a node equation at node A s C2 A3 Vi VA R/a
S
Cl+
1
R3
"ÿ7 + S Cl V° + V" C2R3 +
°*  a/R4 °r
+
c,
+s
(
1
Ci r3

s
a V„ s
C2 R3
C2 R3 R4
VA R/1a)
Vi R/a '
~
°r
Vj
)+
1 c2 R3
, for which the centerfrequency gain is
1
C 1 C2 R3 R4 a R 3 C2
—
r4 (C, + C2)
Here, 1 =
a R3
C2 whence a = R4 (C, + C2) '
, „ R4 398 398 (109 + 10"9) in_3 = 318.4kf2, and = i 5 = 1.25 X 10 \ Now, a 1a 1.25 x 10"3 637 x 103 (109) = 398.512. As a result, at very high frequencies (where the capacitors arc short cir
R3 C2 1
a C 1 R4 1
C2 R3
C1R3
398 1.25 x
C2 R3R4
+
1
1
a C, R4 R4 (C, + C2)
s
+
V„
R3C3+ ~s C2 r3 r4 s
5
+ s C, —— A3
(1a) V„
C,
V„
~
yields:

,
—
— 10"3
 336
SOLUTIONS: Chapter #1115
cuits), Rjn ~ 318kO, and at very low frequencies (where the capacitors are open), 398Q = 318kf2, as well.
s
= T (s) = —— 'i
11.32 From the derivation in PI 1.31 above:
1
s2 + s
C, R3
+
Rin =
318kO
+
a C  R4
1 C2 R3
C[
C2R3 R4
R4
fl4 , — (1), where the voltage at the joinpoint A of C\, C2, —a , and —(1— a) is VA .Note that the current in
R3 establishes
loss. Thus
the voltage across C2, since their joinpoint is virtualground point
VA =
 V„ —. Now the . v — s C2 R
input current
impedance is Z, (s ) =
3 '
V,
R/a
—— = !+
_ZM. s
. Thus
s2 +
1 C,
S
r3
r4 x — a
+
1
c2
2
+
Thus Zj(s)=
s
2
1 C,
R3
.
+
s
»
*3
C1
1
1a
C,C2R3R4
quencies, M
.
as s
R4 X —> 0, Z, = — a
noting that open, v
R
Thus Zj(s) = ©»
+
1
1a C ic2R 3R 4
RiCi
+
1
1 C \C2R3Ra
c2
(c,c2R3R4r
\/(CiC2R3R/!/'< Q = S
©„/?3
, and the input
c,c2r3r4
c2
Now (from Equations 11.73, 11.74), (0o =
or l/Q =
R/a
1
+ Ri C, c2 S
'
1
1
s
1 RjPi
s2 + s
c 1C2R 3ÿ4
+
C2 R3
C2 R 3
1
2 S
'
S
no current
Yi(s) = VZi{s)
c
1
 VA —— —VjR/a
is /, =
with
(la)co027
a
s
i i + C, c2
R3 ©o
+ — S + CD„ ©o
.
Now, at very low fre
+ s (—) + (la)a)„
R4
,, , as can be seen quite directly with capacitors a (1a)
R
R = —. Now, at very high frequencies, as s — rel="nofollow"> +— a (1a) 1a — a
 337 
°°,
Z, = R/a, with
SOLUTIONS: Chapter #1116
2
R4 a
capacitors shorted. Now at the center frequency, s = jco0 , Z,j wfrQ *4 .a j coyQ am
az Q2
R4 a
*4 0+7 .« , ,  g2) 2)
1 1jaQ
a (1 + a2
9
.. ,
2
•
•'"q" + co°
~G" +
,
(1_a)
.
,
« which, for a g large, &
~
is
9
*4 a
, of relatively low magnitude, and with a 90° (inductorlike) phase.
ag
11.33 For this Butterworth, N = 7, Amax = 3dB, Ajc = OdB. Here fp = 5kHz, and 0)p = 2ji(5") = 105n rad/s, and e = (lo4™ÿ0 if' = (lo™ 1 \A = 0.998. (It should ÿctuallyÿbe 1.000. Why? Why is it not?). 1/7 , or tD„ = 105ji 1 Thus for each filter stage, co,, = C0p = 3.14 X 105 rad/s. The 7 poles .998 are located at 0°, ± 180/7 = ±25.7 , ±51.4°, and ±77.14
ov2<2 . For each pole pair, cos 0 = = 1/2g,
whence g = 0.5/cos 0 = 0.5/cos 0 = 0.5, 0.5/cos 25.7 = .555, 0.5/cos 51.4 = 0.801, and 0.5/cos 77.14 = 2.25 respectively. From Ex. 11.28, see that for the Sallen and Key circuit of Fig. 11.34(c), that the dc 1 gain is 1, and from Eq.11.77 and Eq. 11.78 that (0„ = (R \R2C3C4) V2, and 1/g = (o„C4
_1_ *2
1
*1
.
Use C = 3.3nF for C\, C2 for all sections.
Now for the firstorder section, co„ = 1IRC, and R =
0.965k£2.

1 a\C
For the lowestQ Sallen and Key Section, HQ = 1/.555 =
1
3.14 x
96512 10s x 3.3 x 10"9 =
_1_
1 3.14 x 10s x 3.3 x
10~9
or
J_ , or
1 (1). Also co„ = {R\R2C2)~VI implies that R\ R2 = = .001868 HRi + NR2 = .555 X rrr tni 965 coiC 1 1 9652 Thus in (1) C2 X At Therefore R\ + 9652 R2 or Rx = + = R} = 9652, R2 9652 R2 .555 965
'
r,
965 =0 .555
or for
(2),
"
R2 in kilohm, R2 + .9312 R2 1.739 = 0, R2 =
.9312 ± ÿ .93 122 + 4(1.739)
.9312 ± 2.797
965
= 0.933kQ, for which R 1 = .933 = 0.998k£2.
_
Now for the second Sallen and Key (by analogy from (2) ), R2
_ i205
0
=
—.9312
ÿ
.93I22 + 4(1.205) ±
Now for the third Sallen and Key,
—931
ÿ
+ 4(.429)
_Q
+ .9652 R2 
.. . „.
„
.965
= 0, R2 + .9312 R2
gni
'.9652
= I.727kf2, for which R i = .727 = 1.281k£2.
Ri + .9652 R2 for which
= 0,
Ri + .9312 R2  .429 = 0, R2 =
R\ = .348 = 2.68kf2, with all capacitors of 3.3nF
value.
SECTION 11.9: SENSITIVITY 11.34 From Eq. 11.77 and 11.78, co„ = (C3 C4 R, R2)~A, and Q = ÿ
Now for co0, see
.. „ = «/, (C4 K, RJ* ,,
ÿ7
„
_VJ
 338 
and
(C3
C4 R, R2)a
C4
5C; =
d co0
i
i
l
R2'
co„ C3 x x =
—
ÿ
SOLUTIONS: Chapter #1117
Cj_
=  1/2. Now C3 = C3a + C3b and C3/, = /:3 C3a, nominally. However, note that C3a and C3(, 3 C3 3 G)„ co C3a x ; are independent from a sensitivity point of view. Thus = 1. Now Sc" = ÿr— 3 C C0„ 3C3a 3a d co„ d C3 C3a 1 C3a C3a C3rt 1 _1_ , and x r—: X X 1X OX, 2(1+ k3) 2 C3 2 (1 + k3) C3u 2 C3 C0„ 3 C3 3 C3a
—
—
_ CO
1
 ~ ~2
Sc°
k4
ÿ
C3h C3
2(1+ £4)
.
Now for Q, see
Q.
ÿ3
1
~~k3
C.3«
2 (1 + k3)
2(1+ k3)
C3a
.
to
Correspondingly, Sc" =
 2(1 1+ k4) , and Sc"
When k3 = k4 = 1, all these sensitivities become 1/4. (C3 C4 /? 1
3(2
3 C3
1/2, where Sg =
C4
—
—
R2f
1
C4
/?!
= 1/2, and S# =
Thus
*ÿ
1/2
=
c3
/?2

1/2
Q.
Thus
Sgt
*3
, and Sg = —
2(1+ k3)
+
1
1
=
3Q
, both being 1/4 for A3 = 1. Sec also 2(1+ A:3) 3 *4 1 , , both being 1/4 for k4 = 1. „ „ , Sg = 2(1+ n k4) 2 (1 + A:4)N, "
C4
11.35 From Equations 11.77 and 11.78, (On = (C3 C4 /? /?2)~'/l. a°3 li 3 to,, (C3 C4RxR2f 1 1 Now for cd„, see = 1/2 (C3 C4 R2y'/' x /?fM = + 2= 3 R, /?, C4 R2 R 3 co„ co„ CO lDn , and Sfl " = r— X x  =  1/2. Likewise sR " =  1/2 as well. 0)o 3 Ri 2 «i C0„ 2R\ 3 /? 1, /? 2 ± k. ± /? , and '* = ±1? X Now for a fixed temperature, Ru R2 = (\ ± k)R with R 3£ :1 Thus = Sr" = 1/2 (+ +( k )) = 0. + 0)
—
S*
S*' Sr™ 5*
S?"
Now for small A, /?j
= 7?2 = /?o (1 +«
(T„
(1 + a (T„
 1/2 x
 T))
3R
C4 C3 __1 2 R\ for /?„
ÿ
ax
/?2 = (1 ± A:)/? with
&
£±
2ÿrk> « I,
ÿ
=Q
Now for
x
ÿ (2/?)
RiT Ri + R2
(R\
c3
Ri (1) x (Ri+R2)2 R'{' +
R\+R2 R 1 + R2
= R"
£±
lir 2/?
ÿ 4rÿ"< w'lh 5ÿ = +/:/2. a a 1 = 2a
Now, Sg = Sgt
Sk' + Sgÿ Sk \ where here, R\, R2 = (1 ± k)R, and —
Thus
1'/.
R.
_
3Q 3 /?, (/?ÿ
R2f
R\ + R2
c3
"ÿ and
k/2. Likewise
'sg = °
*432x4
„ilhs# = •
2R
Q
Bu'
=
_
®
j
ÿ
x=iL a
* R\
= St1. Thus iSy " = Sr " x Sp' + Sr ÿ x Sp'
R\ Ri R\ + R2
C3 R 1 R2 V2
1
„co
, which around T = T„ , is a T0.
C4
Now for Q, see Q =
3R 3T
R,
l+a(T„T)
a T 1 + a (T„ ~T)
—
 T)), whence Tt1 = — a R„ and Sp ' =
a T
aR„ X
=2
*
*'
2
= ± R, with
Sk''R' = ± R
Thus Sg = k/2(k) + k/2 (k) =  k2. Also Sg = Sg Sp ' + S#2 Sr", where /?=/?„ (1 + ' dR R ÿ2= (1 ± £) /?„ a, with (ToT), and =  (1 ± k) R„ a x
(1 ±k)R0 (1 + a (T„
d 1

 T)
 339 
SOLUTIONS: Chapter #1118
a 7 l+a(T„T)
Thus
a T 1 +a(T„T)
S$ =  k/2
+ k/2
a T I+ a (T„  T)
= 0.
SECTION 11.10: SWITCHEDCAPACITOR FILTERS 11.36 For <&! high, Ct is discharged to zero (that is, virtual ground ( VG) via the second switch. Then, for <£, high, the first switch closes, charging C\ to IV, for a total charge transfer of Q = CV = 0.1 X 10/l x 1 = O.lpC. For 1MHz operation, the corresponding average current is / = 0.1 x 10~12 x 106 = O.luA. Equivalent input resistance is
—V =
produces an output change of V =
IV
'
= 10MQ. For a 2pF feedback capacitance, O.lpC charge
ÿ
2
— = —ÿ— 10"' O
tXl x
_12
= 0.05V. Thus the output change per cycle would
be 50mV. For \), = +1V, C\ charges positive when 0 occurs. When d>2 occurs, VG tends to go posi¬ tive, forcing the output negative to compensate. Thus the output changes in the negative direction. For 20 V saturation at ±10V, the maximum output change is 20 volts, requiring  = 400 cycles. The aver50mV
20
—
—
= .05V/(is, or 50V/ms, or 50,000V/s. For a 0.1V input, the output slope becomes +5V/ms or 5000V/s.
age slope is —20V in 400 (is, or
11.37 Require f3JB = /„ = 105Hz, Q = 2pF. Now co0 = X
10"12.
— A/ C* Tc
T Now for —7C 3
Thus
lAl = 0.707, and A„
= 1V/V for fc = l/Tc = 106Hz, C, = C2 =
V C3 C4 = oo0 Tc V C2 C, = 2 it x 105 x 10"6 x
T
C2 = ~~ Cu with C\  C2, then C3 = C4 = 1.257pF. C4
2x
lO"12 = 1.257 C4 Thus
Now Q =
—. *
ÿ5
1257 x 10 12 Check: C5 = to„ Tc C/Q = 2 7t X 10s x 10"6 X 2 x 10'12 x = 1.777pF. r .707 Q 1.414 = 1.7772 x 10"12, as before. Now A„ of the lowpass function is R/R6 = C6/C5. Thus C6 = 1 X C5 = 1.777pF. In summary, C, = C2 = 2.000pF, C3 = C4 = 1.257pF, C5 = C6 = 1.777pF. The output of the first integrator is the bandpass output. Its center frequency is f„ = 10sHz. Its 3dB bandwidth
c, = —i=
is
ÿ— 1.41 x 105Hz. Its maximum gain (from Eq. 11.100) is — = = —Q = —0.707 L5
1777 1.///
= 1V/V.
SECTION 11.11: TUNED AMPLIFIERS = 25£l. For RE = re = 2512: Rin = (p + 1) (re + fl£) = 201 (25 + 25) = —5 x 10 —Rl = 100V/V. Gain from base to 10.05kl2. Gain from base to collector, Abc =  = ——
11.38 For
IE =
1mA, re =
1niA
—
D
emitter, Abe =
— "i" re Re
r\
=
re + Re
c
= 0.5V/V. 25 + 25
 (100)) = 5 + 101 = 106pF.
25
+ 25
Equivalent input capacitance Ceg = 10 (1  0.5) + 1 (1
The total tuning capacitance is C = 200
1HLC = (1 x 10"6 x 306 x \0~n)Vl = 57.17 x 106rad/s. s 9.1MHz.
B =
1
1 — = —= CR 10~12 (10.05 x 103 II 10 x 103)
306 x CenterFrequency Gain A =
—
r
ÿ
)q

05
x
652krad/s. Q =
= ~50.1V/V . 5 x 103
re : Rm = 201 (10) (25) = 50.25kl2, Ahc = 
+ 106 = 306pF. Now co„ =
—B—
=
57.17 x 106 = = 652 x 103
— (2g) = 20V/V, Abe =
9 (251
87.7.
= 0.9V/V, 20) = 200 + 1 + 21 = 222pF, R = 10k£2 II 50.25kl2 = 8.34kl2, co„ C = 200 + 10 (1  0.9) + 1 (1 = (1 x 106 x 222 x lO'V = 67.1 x 106rad/s, B = 222 x ÿ.2 x g 34 x 1Q3 = 540 x K>3rad/s, Q = For RE
— 0.54
9
— x v(20) = 16.7V/V. = 124, A = — 50.25 + 10
'
 340
SOLUTIONS: Chapter #1119
11.39 Here, lE = 1mA. Thus re = 2512 and rK = 201 (25) = 5.025kl2. Gain from base to collector = 5000/25 = 200V/V, C„, = lOpF + lpF (1 200) = 21 lpF, For direct connection (as in PI 1.72 of the Text): C = 200 + 211 = 411pF, R  10kl2 II 5.025kl2 = 3.34kl2, to„ = {LCT'a = (1 x 10"6 x 411 x lO"12)"" = 49.3 x 106rad/s, B = (CRT1 = (411 x 10"12 x 49.3 5.025 3.34 x 10T' = 0.728 x 10f,rad/s, Q = x (200) = 66.9V/V. = 67.7, A = B 0.728 5.025 + 10.0

5 025 x 1O3
Rin = 5.025kl2 is transformed to —j— = 20.1kl2, C,„ = 211pF
For a tapped coil with k = 0.5:
is transformed to 21 1 x 0.52 = 52.8pF, C = 200 + 52.8 = 252.8pF, R = 10kl2 II 20.1kl2 = 6.68kl2, co„ = (1 X 106 x 252.8 x lO'V1 = 62.9 x 106rad/s, B = (252.8 x 10~12 x 6.68 x 103)"' = 0.592 x 106rad/s, Q = 62.8 = 106, A = 20.1 x 0.5 x (200) = 66.8V/V. 0.592 10 + 20.1 5 025 x 103 = —j = 502ki2,
CT = 21 1 x 10"'2 X (0.1)2 = 2.1 lpF, C = 200 + 2.1 = 202. lpF, R = 10kl2 II 502kf2 = 9.8kf2, co„ = (1 x 10"6 x 202.1 x 10~{2yv' = 70.3 x 106rad/s, B = (202.1 x 10~12 x 9.80 x 103)"' = 0.505 x 106rad/s, Q = 70.3 = 139.2, A = 502 0.505 10 + 502 0.1 x (200) = 19.6V/V. For a tapped coil with k =0.1:
Rr
Comparison:
Q
A V/V
Mrad/s.
Basic Circuit
49.3
68
66.9
Re = re
57.2
88
50.1
Re =9 re
67.1
124
16.7
k =0.5
62.9
106
66.8
k = 0.1
70.3
139
19.6
Thus the tapped coil gives the best results in general.
11.40 For the coil, Q„ = IOC
= 200, rs = or C
=
i— — (02 L
j
=
* 25.1 _
2 n x 200 x
6.281 X 25.1  6.28
rs
(00
Here, at 10MHz, (0„ L = 2 it X
L
107 X 2 X 10"6 = 125.712.
For Q„
= 0.6312, and Rp = 200 (125.7) = 25.1kl2. For resonance at 10MHz, (0„ = {LC)~V\
=4r = (2 x 107)2 x x 10"6 n
103 x 126.7 x 10'12
2
126.7pF. Bandwidth B =
 , for —— 2nCR
which R

6.28 lk!2. The resistor to be added is 6.281 II (25.1)kl2 =
8>38k!Qi
 341 
SOLUTIONS: Chapter #1120
11.41 For a single LC circuit, Q =
/„
(D„
1. Here, 50 x
11.110, fB =
0.25,
106 = —j  10. —" = —— 10 Jb
2W = 1.25, 1/N log 2 = log _
1.25, N =
ÿ
fB = IL V 2W1 = 10s V
103 =
1.2599 1 =
For synchronous tuning of N stages, from Eq.
iy V 2™ 1, or 1 = 2 V 2W 1, 2m 1 = (i)2 =
,ÿ
log 1.25
.0969
= 3.1. Thus, use 3 stages: Check:
10s ÿ .2599 = .51 x 105Hz.
OK.
For the 30dB bandwidth, (from problem 11.77 on page 972 of the Text, part b)), for synchronous tun\T(j co„)l ing: IT (jto)l = [1+4 (2ÿ1) (6 f/fh)2]N/1 IT For 1 stage : N = 1, [1 + 4 (2  1) (bf/fh)2]Yl = = 10+302° = 31.6, or 1 + 4 (8f/fh)2 = 1000,
bf/fh = (
1Q°Q ~ 1 4
)'/ÿ
_ 15.8. Thus the skirl selectivity is
f"" =
f I)
= 2 (15.8) = 31.6, and the 30dB
bandwidth is 31.6 (100kHz) = 3.16MHz. For the synchronouslytuned cascade, with Af Zi= 3, [1 + 4 (21/3 2.94. Thus  1) (Sf/fhfY1 = 31.6, 1 + 4 (.2600) (8f/fh)2 = 9.995, that is Sf/fh = 9.995  1 4 (50kHz) is (2.94) 5.88, 30dB bandwidth 5.88 and the the skirt selectivity factor is 2 = = 0.294MHz.
11.42 Using Equations 11.115 and 11.116 of the Text: /0 = f„ +
= 10.735MHz, /02
= ,0707MHz, Qi = (h =
=
= 10665MHz
f'JT2= WJ
= ÿ~2 x 10.7
= 10.7 +
1 1 „ Now /„ 151.3. = = , and Ct = — , = J (2 7t foi) L 0.1 2n
ÿ
T.
1
Now R = 2 nf„ L Q, and R\ = 2 n x 10.735 x 106 x 3 x 10~6 X 151.3 = 30.616k£2, R2 = 2 n X 10.665 x 106 x 3 X 106 x 151.3 = 30.416k£2. Since the voltage gain at resonance is proportional to R (see Ex. 11.37), the relative peak gain of each of the two is
 342
ÿ —/v 2 = JU.416
= 1.007.
Chapter 12
SIGNAL GENERATORS AND WAVEFORM  SHAPING CIRCUITS SECTION 12.1: BASIC PRINCIPLES OF SINUSOIDAL OSCILLATORSIO 12.1
Look ahead to the diagram in the solution to P12.2 following only if you need to! Convert all resistive co„ L Rip elements to equivalent values across the tank inductor L. Now, Q = Thus Rip = R Ris ,2ÿ2 w, is the equivalent resistor across L due to inductorwire resistance. Now the amplifier input
Rls
.
resistance reflected through the turns ratio is R( = n2 Rin Thus the total load on the tank is R = R„ 11 R,p 11 Rcp 11 Ri = R„ II 0)j Ll/Rk II Rcp II n2 Rin. Loop gain (from the active end of the coil and 1
R„
+
£
L'
G" "
.
1 1 Ris + — + 2 D , at n2coi2 L2 Rcn Rin
with n
oscillation occuring for n RSC
n
k + ~iT + R,
cp
1
Vz.C
r a frequency co0 = 7
Gm Rn
n
Gm = n
Gm J 1, or  = — = R n 1 LC , and,1 . Now co„2 = LC' — — l} 0)2= L2 l Ris C l or for
back) is G,„ R/n. Oscillation will occur when this is unity, that is when
R
——
R.cp
n2 Rit
1 n R:in
12.2
 343 
(a)
Showing the inductor's parasitic resistance in its parallel form, the topology shown results for a noninverting amplifier. Three classical circuit forms result depending on which of A, B, C is grounded in a practi¬ cal implementation. Grounding A pro¬ duces the design that is usually drawn from the description given.
(b)
For an inverting transconductance element, the topology shown results, again with three classical variations depending on where ground is connected. The one with C grounded, in which the device acts as a follower, is quite common in practice.
SOLUTIONS: Chapter #122
(c)
Rep
Gmv
Rip
]
RO
This may be redrawn to resemble (a) more closely to illustrate that the change of amplifier sign is compensated simply by shifting the tap from one end of the autotransformer to the other. For the negativegain version, the loopgain magni¬ tude (as measured across L) is exactly as before, with the same conditions for oscil¬ lation. Note that the amplifier inversion is accounted for by the coiltapconnection reversal.
12.3 Here, at resonance, the load on the tran
sconductor
is
R
—
1, or
Gm
R,
*4
=
R„ II
n (,Rcp II Rip llflfo ), and the voltage across the tap is Gm R, with the loop gain equal to For oscillation to occur, n Gm R.
n
Gm
R
Ro
Gm =
— Rit
1
n
1 R
1
n
, or Gmi =
n_ with Rq) RP Riin frequency co0 = (LC)7'.
n Rfl
n
oscillation at a Recall that for the connection in P.12.1
Gm \ 
above,
_n_
Ro
l n
Rin
n R,cp
n R,ip
'
Note that for a particular coil and capacitor, the relative value of the required Gm for each topology depends on the relative size of R„ and Rin of the amplifier. For Rin = R„ , the same Gm is required for 1 1 a given n. Thus, for example, for R„ = Rin = Rcp = Rip = R, Gml = Gm2 = Gm = + nR R R (3/t + 1/n ), or Gm R = 3n + 1/n
—
.
12.4


For an input resistance of lOkft, R i = 10kQ, and Rj = 5 (/? i) = 50k£2. Negative clamping occurs with at 0.6V correspondingly. D\ conducting with 0.6V drop, the negative opamp input at 0 volts, and 2.5 = 1.9V. 0.6 = 10.6V, and Thus at the edge of D\ conduction, Vr2 =10 = 0.6 I' Ri 10 6 Rf   = 0.5, or 50k \\R3 = Correspondingly, R2 = —j— R3 = 5.58 R3. For a limiting gain of 0.5, 1.9
—Ri— ÿ
ÿÿ
= 5.56kfl, and 0.5 (10k) = 5kH. Thus, R3 = 50k II R3 II (50k) = 5k II (50k) = R2 = 5.58 (5.56) = 31.0kQ. Overall, use R, = lOkO, R, = 50kO, R3 = 5.6kO, and R2 = 31.0k£2. 10 = 103Hz. For a gain of 5, P = Now, for the amplifier, A = 1000, f, = 106Hz, and thus f3dB = 1QQQ 1/5, A P = 1/5 x 1000 = 200, and f3JB ~ 200kHz. Alternatively, from page 80 of the Text, f3JB = 106/(1 + Rf/R i) = 106/(1 + 50/10) = 167kHz. For a 3dB frequency of 167kHz, 2° phase shift occurs at
tan"1
167
= 2°, or/ = 167 tan 2* = 5.83kHz.
 344 
SOLUTIONS: Chapter #123
12.5
For low input voltages, the gain is R 2 II R3
10 II 10
Rt
7.5
7.5
h
10 = 1.33V/V. For the zeners conducting, the gain is 7.5
Ri
= 0.67V/V. Assume that the zener can be characterized by a linear
Vz
— for voltages below the knee.
resistor of value R 2 = 
hK
Here,
(8.7, 7.8)
6.8 10M output 1.32V/V
(0.6,0.7)
(M. 7.8)
6.8 = 68kf2, and /?z2 = IOOjxA = 680kQ. For the poorer zener, for
=
voltages
beyond
corresponding inputs of
0.7V
(with
"0 7
= 0.53V), 1.33
Rz 1 — 68kO, and the gain is Rz 'I (R 3 + Rzi) 10 II (10 + 68)
l.l*V/V •J1WV
7.5 1.18V/V. For the better zener it is 10 II (10 + 680) = 1.32V/V. For out7.5 puts greater than 6.8 + 0.7 = 7.5V, the gain is 0.67V/V as calculated earlier. For the better zener, the corresponding input is 75 about T2r = 5.68V, and for the poorer 1.32 7.5 zener, it is about = 6.36V. 1.18


(077.8)
......  
SECTION 12.2: OPAMPRC OSCILLATORS 12.6
1 1 = 1.59kQ. For oscillation, 2 nf„C 2 it x 104 x 10 x 10"9 R2 ÿ 2 R\. For 2V peaktopeak output, peak output is IV. At the threshold of oscillation the voltage at the positive op amp input must be 2/3V. Thus the voltage across the regulating diode plus its series resistor (R2) must be 2/3V, as must be the voltage across R2, with 1/3V across Ri Thus the current in 1/3 2/3 /?,=/ = + lD. Now ID = /, eV,/1Vr where VD +ID R2 = 2/3V. Now, VD = 0.70 + 0.05 In R1 Rz 1 (1). Also ID = IDn mV, and 0.700 + .05 In ID + ID R2 = 0.667, or lD ~~~ (.05 In ID .033)
Now, (0„ =
033 A
1
1
_ 067 A
2
„ For C = lOnF, .R = ÿ
CR'
(2)> with
> 2 Ri
.05 In 0.1  .0333) = .818kf2. Try

—

Rz
(3) with (1)( try

_ Jd o linA> whence R2 =
0.1
(
(~ 05 In .01  .0333) = 19.7kf2. Try ID = ,3333 6666 .0333) = 8.11k£L Use R2 = 8.2kQ. From (2), .01 =
ID =
10pA: /?2 =
(— .05 In .02 8.2 .02 Ri .3333 (3), (3.65) 3.65kO. Use 3.6kf2. Check: from or Rt = = 7.3 OK. R2 = 8.2 > 2 /?, = 2 .010 + .0813 .3333 .6666 Check: From (2), ID = = .0926 .0813 = .0113mA. Now the drop across the diode 3.6 8.2 plus the extra R2 is 0.0113 (8.2) + .05 In .0113 + 0.70 = .0927 .2241 + 0.70 = .569V. Current in the 569 feedback R 2 = ' . = .0693mA. Voltage across R 1 = (.0693 + .0113) (3.6) = 0.200V, and the output
20xA:
R2—


o »2t
_
voltage is .290 + .569 = .859V. Too low! Now, reduce R\ slightly to R\ = 3.3k£2. From (2), Id = .3333 .6666 jqjq Q823 = .0187mA. Now the drop across the diode plus the extra R2 is .0187 8.2 3.3 .654 (8.2) + .05 In .0187 + 0.70 = .1534 .199 + 0.70 = .654V. Current in the feedback R2 = 8.2 .0800mA. Voltage across R\ = (.0800 + .0187) (3.6) = .355V. Thus the output voltage is .355 + .654
_

 345 
SOLUTIONS: Chapter #124
= 1.01V. Use C = lOnF, R = 1.59kO, R, = 3.3kO, R2 = 8.2kO. 12.7
1 + RjT?, , where oscillaFrom Eq. 12.11 of the Text, using an ideal amplifier, L(jto) = 3 +y(co CR  l/(co CR)) tion occurs when From Eq. 2.20 of the Text, for an amplifier with  0, and ILI = 1 at co„ = 1+R*fl V, (/co) = co, = 2 rerad/s and Aa »(1 + R/R i), :• For oscillation at co = 0.9 coa = vi i+
———————
——
m— — oy{l +R/R0
0.9ICR with this amplifier: 1 + RVR\
L (j 0.9 co0) = 
j 0.9 co„
1
1 + Rÿ/R\
3
2.27 /
.1899 co„
(0„
„ ;D ° + — y (.2ii) + — co, (1 + Rj/Ri) co, (1 + RÿR i)
Oscillation will occur when 0.211 = 2.27co0/(co, (1 + R/R i)), or
(0„
for a nominal frequency co0, oscillation will occur at 0.9 co„, when
= co, (1 + R2/R\) (.09295). Now
+ i = ——1899 COp 1
Rj/R
1, or 1 +
•
3+
co, (1 + RÿRy) 2.018, 3.018, R2/R 2.018 or (.09295) Thus or R,. for/, = 1MHz, the fre¬ = = R2 = R2/Ri = 3 + .1899 106Hz, (.09295) x x 106 (3.018) nominal 0.252 with frequency, f„ = be 0.9 will quency of oscillation =
2—re RC
.....
0.9
,
,
.
,,
n the closedloop opamp 3.018,
gain
—
and 1 + R2IR\ = 1 +   = Ri 3.02 3.018 value which seems a must be ——ttt, t— = 1 + 0.755 j j .25 x 106 1 x 106/3.018
= 0.280MHz. Thus, for operation
at 0.252MHz,
.
—
quite reasonable. 12.8
From Eq. 2.20 of the Text, for an op amp with A„ large and a unitygain frequency co,, the gain of the 1 + R*/R i
, and of the network is: noninverting topology is: G(co) = ; J— i+ — — co, (1 + R2/R{) 1 + RrfR i 1 R T(co) = . Overall, L(co) = ( 1+ )( 1 + ) 1+ R+ jcoRC co, (1 + R/Ri) j(oC y co RC 1 + RijdRi

—
1+
i_
is* — co,(1 + R2«i)
coRC

!+RC co, (1 + RyR,)
Oscillation will occur when the net phase shift is 0*, when co =
co, (1 + R2*Rl) RC
RyR, =
Rr, ÿ
RC
, provided the gain is at least
one, that is
(°n co, (1
+ R/R ,) 1 + R27? i
_ ÿ
ÿ
— co RC
> 1+
n+ R '/j
=
or
1, or
RC co, (1 + R'/R j)
, or (RÿR,)2 + (R/R,)  1/(RCCO,) = 0. Now for co = 4/RC =
, co, (1 + RÿR) = 16/(RC), and R2/Ri =
 346
1
1
RC (167(C)
16
SOLUTIONS: Chapter #125
R2 = Ri/16 = .0625 Ru for More generally, for
= . — RC
W/ , —1 = — a nC
to
—
®/
= v
,/
(1 + /ÿi)
a
.
with (1 + R2/R 0 = 1 + 1/(a (1 + R2/R ,)),
= 1 + 1/(ojc), ax 2  ax 1 = 0, x2  x  \/a = 0, _ , 1 ± V l 4(— l/a) 1 ± V4/a + 1 r , or \1 + R2IR\  which x  for ÿz. or x
—
1
+2
R2/Ri = 1 ± 2+ 1 Z
—5
, „ „ Now, for example: For a = 1, 1 +
I
= 1.618, R2 = .618 R\ and co = to, (1.618)* = 1.27to,. For a = 2, 1 + R2IR, =
= 1.366, /?2 = .366 /?, and to = co,
= 1.2707, R2 = .2707 R, and © = (0, (
(ÿV' = .826®,.
For a = 4, 1 + /?2?fl, =
Z
L2J°7 )'/l =
0.564®,. For a = 16, 1 + R2IR{ =
1±
1.059, /?2 = .059 Ru and ® = to,
1
JL—
2'25
=
= 0.257®,. 16 Generalizing, we see that for frequencies significantly lower than co, for an amplifier with idealized singlepole rolloff, oscillation is possible for R2 nearly zero, at a frequency which is the geometric mean of 0), and 1/RC. For excess phase shift due to additional poles near and above ®,, more compensating phase shift will be required from the RC network, implying operation at a frequency lower than in the simple case. Note that this type of oscillator, operating at co„ with a network for which 0) = 1/RC, allows an estimate of to,, and associated excess phase, if R2 is adjusted to the maximum value for which oscillation is sustained. For 1/RC « ®„, the oscillation frequency co„ can become a sensitive function of various things, such as construction, but not of excess phase. For 1/RC very near C0„ , the frequency of oscillation is quite sensitive to excess phase. By varying R, find ®„ in the range two to four times 1/RC in order to evaluate co, . ®„ < 1/RC is an indication of excess phase, where operation also requires a higher value of R2. 12.9
From the right, label the components C\, R\, C2, R2, C3, ft 3, with joining nodes Nj, N2, A/ 3 respec¬ tively. Assume a virtual ground at the opamp input into which a current flows from C\. At N\, c'ci = /C2s) = \>2 C2s = i (R\C2s + 1, D] i/C\s. At N2, t)2 = i'ci ÿ1 + = (fti + 1/C\s ), ici CyC 1). At N3, O3 = ('ci "t ici) ft2 "t ÿ2 = i (1 + R\C2s + C/C) R2 + i (R\ + l/C\s) = i (R\R2C2s + R2C2/C 1 + R2 + R 1 + 1/Ci), ic3 = "02C$s = i (R\R2C2C2s2 + R2C2C$s/C 1 + R2C$s + R\Cÿs +C/Cu At x, x>x = (j'ci + 'c2 + 'o) ft 3 + *>3 = i (1 + R,C2s + C/C1 + R\R2C2C},s 2 + R2C2C$s/C\ + R2Cjs + R\C$s + CyC1) /? 3 + i (R\R2C2s + R2C2IC\ + R2 + R \ + 1/C\s) i (R\ + R2 + R$ +
—
—
'
—
R3C/C1 + R$CyC 1 + R2C/C\ + s (R\R2C2 + R1R3C3 + R2RiC3 + R\R2C2 + R2R2C2C/C 1) + /?,/?2/?3C2C3 s2 + 1/Ci s. Now the loop gain L = i ft/A)*. Thus, for C\ = C2 = C3 = C, L(s) = ft//[ft, + 2J?j + 3 ft 3 + sC (ft,ft2 + 2 R2Rÿ + 2 ft,ft3) + ft,ft2 ftjCV + l/(Gs)]> and L(s) = RfCs/[l + (R,+ 2 R2 + 3 R3) Cs + (RjR2 + 2 R2R3 + 2 R,R3) C2s2 + R,R2 R3C3s3]. Now substitut¬ ing s = ju) and multiplying top and bottom by j, L(ja)) = +RfC(o/[j  (R 1 + 2R2 + 3ft 3) Ceo  (R\R2 + 2ft2ft3 + 2R\Rj) C2(02 j + R\R2 /?3C3co3]. For oscillation, the phase angle = zero, that is 1 = (ftift2 + 2 R2R 3 + 2 R 1R3) C2 to2, where the frequency of oscillation is co„ = ! , which for R< = R2 = with = R is co„ = —ÿj—, CR ÿ5 C V R, R2 + 2 R2 R3 + 2 R, R3 ,(0„, RfIC
_________ ___________ _________

1
—
D
RI5 = 6.2R. Now for sensitivities (with R 1 ~ R2~ R2 ~ R): d(0„ 1 1
dR
,
"
da>„
2
u;
C («, R +2R2 + 2Ri Rf2 d(0„ 1 3
.
.1.
ID
D T>
j_
_
T>
—
(3R) =
4
j
2 ÿ
/D
347

3
2 1
(o„
"
R
3«
3
Ri + 2R2
d(0«
Nowi"'» = jr; xt
2
co„0

Ri
1
3Ri + 2R' 3
w
........ .....
3 , which, x — ...... . for /?,. = /?, is 3/? , + 2/? 2(5) co„ R
SOLUTIONS: Chapter #126
d Rr = 0.3. Similarly SR ' = 0.5. Also = 1+ 3/?, ' R 2R3 (/?i/?2 + 2/? 2/? 3 + 2RiR3)~l + (1) RiR2R3 (R \R2 + 2/? 2/? 3 + 2/?,/?3)~2 (/f 2 + 2/? 3), which for R2 dR>/ = /?3 = /?, is = 1 + R2 (/?,/? + 2/? 2 + 2/?,/?)1  /?i/?2 (/?, R +2R2 + 2/?, /?)2 ( 3 /?) = 1 + R 1
dRi
™
~
9R2 + AR2 + 12/?, R + 3R /?, + 2/?2  3/?, /? (3R 1 + 2R)2 d R r. , and for /?, = R, is 27/25 = 1.08. Thus, SRr = Rf x 1 for = R' is
+ 2/?)"'  3RtR 91? 2 + 12/?, R +6R2
(3/?,
(3/?, +2/?)2 R
(3/?, +
2/?r2
=
dRi
/?/•
9 Rr 1.08 X 0.174. Also with /?, = R 3~ R, 2 Rz (/? /?2 + 2R /?2„+ 2/?2)"1 /?2 /?2 = J ' dR 2 = + 6.2R ' (/? /?2 + 2R R2 + 2R2)2 (3/?) = 2 + R (3/?2 + 2/?)"1 3/? /?2 (3/?2 + 2/?)"2, which for R2 = /?, is 2 + /? (5/?)"' 3/? 2 (5/?)"2 = 2 + 1/5  3/25 = 2.08. Thus is 2.08 x is = 0.335. Similarly, 3.08/6.2 = 0.497.
5$
12.10
,
_
Xi
Here, /? = R2 = R3 = R, and C, = C2 = C3 = C. For a current i in /?,, at node N2, \)2 = i /?, = //?, 'C2 = WCl/CV) = i/?Cs\ At node /V3, ir2 = /«, + ic2 i (1 + RCs ), 1)3 = 1)2 + /?2 (//j2) = '/? + /? (i) (1 + /?Cs) = i (2/? + /?2 Cs), i'C3 = U3C2 s = i (2/? Cs + R2C2 s2). At node x , 1ÿ3 = iR2 + iC3  i (1 +RCs) + i (2RCs + R2C2s2) i (1 + 3RCs + R2C2s2), X>x = x>3 + /?3 iR 3 = i (2/? + R2Cs ) + /?/ (1 + 3/?Cs + R2C2s2) = //? (3 + 4/?Cs + R2C2s2). Now, / flows in /?f  C,. Thus the opamp output Rf (1/Cs) voltage = = 1 Rf + 1/Cs
C1
R3
/va¬
ns
Ra
AA
=±=C3
Loop gain L(s)
N2
—
R: _A/_
V
X I
Rf

C2
Rr
1
Rf Cs + 1 " R (3 + 4RCs + R2C2s2) ' Now L(s) =  (Rf/R) [3 + 4R Cs + R2C2s2 + 3 Rf Cs + 4R RfC2s2 + R2RfC3s3]~l =  (RfR) [3 + s (4 RC + 3RfC) + s2 (R2 C2 + 4R RfC2) + R2 Rf C3 s3]1. Substituting s = jm0 and requiring the ima¬ ginary part to be zero, and L(to„) = 1, see (4RC + 3/?/ C) co„ = R2RfC3io3 —  (1), and 1 = (Rf/R) 4R + 3 Rf 1 1 ,2ÿ1 w,f)~' (3 R2C2(Q2 4R (2). From (1), From (2), R

ÿ

RfC
co0 = RC
* *1*!
Rf
~W
=
)  3/? = (4R + 3Rf) (£ + 4) + 4R RfC2) (a2  3) = R (R2C2 + AR RfC2) ( R RfC Rf J . o . AR2 AR2 3R. Thus R{ = + 16R + 11/?, = 0, AR2 + 16R Rf + 11/?/ = 0, + 3R + 16R + 12Rf  3/?, ((/? C2
_
O
Rf
whence
Rf
=
16 ± ÿ
162
—
Rf
4 (4) (11)
2(11)
R =
16 ± 8.94 /?. Note that 22
implying that oscillation is not possible. Why?
J
Rf/R
is always negative,
12.11 For each section, a maximum of 90* phaseshift is possible. Now with a positivegain amplifier, the net¬ work must shift by 360°. Thus, realistically, for operation on a phase slope of more than 360/90 = 4%, five sections are needed. Label the sections 1 to 5 from right to left with the RC join nodes called N\ through /V 5 respectively. Now for D, =\) = \)(1), ic 1 = 1x7? = x/R (1), x>2 = ic/Cs + x>i = xÿRCs +u = u (W?Cs + 1), i'c2 = x>/R + ici = (yRCs +x>yR + \/R =\yR [VRCs +2], t)3 = ic/Cs + 1>2 = t) [(\XRCs + 2YRCs + VRCs + 1] = x> (VR2C2s 2 + 3/RCs + 1),
 348 
SOLUTIONS: Chapter #127
ic3 = V/R + iC2 = v K VR2C2s 2 + 34?Cs + 1)4? + (1/RCs + 2)4?] = \>4? [L4?2C2s2 + 44?Cs + 3 ], U, = ic/Cs + \>3 = v [( ]/R2C2s 2 + 44?Cs + 3YRCs + VR2C2s2 + 34? Cs + 1] = v (VR3C3s3 + yR2C2s 2 + 64? Cs + 1), iC4 = U*4? + iC3 = x/R (1/R3C3 s3 + 54?2C2s2 + (/RCs + 1 + VR2C2 s2 + 44?Cs + 3) = x/R (VR3C3s3 + (ZR2C2s 2 + 107? Cs + 4), 1)5 = ic/Cs + u» = u (VR4C4s4 + (/R3C3s3 + 104?2C2j2 + 44?Cr + VR3C3s3 + 54?2CV + (/RCs + 1] = 1> (VR4C4s4 + 1/R3C3s3 + 154?2C2j2 + 104?Cs + 1), ics = v/R + ic4 = t>4? (VR4C4s4 + l/R3C3s3 + 15K2C2s2 + 104?Cs + 1 + VR3C3s3 + (/R2C2s2 + 104? Cs + 4) = t>4? (VR4C4s 4 + 84?3C3s3 + 1VR2C2s2 + 204?Cs + 5), i)(; = ic/Cs + \)5 = o (l/7?5C5s5 + 84?4C4j4 + 2VR3C3s3 + 204?2C2s2 + (/RCs + 1/4?4CV + l/R3C3s3 + 154?2C2j2 + 10RCs + 1) = V (VR5C5s 5 + 94?4CV + 284?3C3j3 + 35W2C2s2 + 154?Cr + 1). = K (14?5C5s5 + 94?4CV + 284?3CV + 354?2C2s2 + 154?Gr + 1)_I. Now subsli
Now L(s) =
tuting s = j co, the condition for zero phase angle is 1/7? 5 1/7?4
C4 to4  284? 2 C2 to2 +
15 = 0. Now 1/7? 2
C5 to5  284? 3 C3 co3 +
C2 002 = 
~28 ± ÿ
2jf2
~
4 (>5)
154? Coo = 0, or 28 ± 26 91
_
=
2 27.45 or .545. Now at 00 = 1/7?C, the phase contribution of each section is 45°, and the total phase is about 225*. Thus operation must be at 00 < 1/7?C. Thus choose the 27.45 solution, where 1/7?2 C2 002 =
—
.
X 1 For this condition, L I = K (94?4 C4 oo4 27.45, or co00 = = RC RC V2X45 354? 2 C2 to2+ I)"1. Now for  L  = 1, K = 9/(.191)4  35/(.191)2 + 1 = 6762.5  959.4 + 1 = 5804V/V.
12.12 The modified amplifier has 4 sections, with all 4 capacitors of value C, and all 3 resistors of value /?. The feedback resistor is /?/. Perhaps from the results of Exercise 12.5, we could reason that L = co2 C2 R Rf , (incorrectly!), but we must check: Label tlie network nodes, from the op} v 5 + j (4cd C7?  l/co C7?) v amp negative input toward the left, N0, N\, /V2, 7V3, 7V4. Note that N* = x. Now for a current i flowing into node 7V0 from C, = OV, ici = i, V 1 = ic/Cs = i/Cs, ic2 = U/R + i'ci = i(VRCs + 1) o2 = (1/7? Cs i 1)/Cs + i/Cs, ia m/R + icz = [ 14?2C2i2 + 2/RCs + 1/7?Cs + 1] = + + ic/Cs v>i / [VR2C2s2 + yRCs + 1], \)3 = ic/Cs + \)2 = i (VR2C2s2 + 2/RCs + 1)/Cs + i (VRCs + 2)1Cs = i (1/R2C2 s2 + A/RCs + 3)/Cs. Note that this corresponds to the result of Exercise 12.5. Now, ic4 = x>/R + iC3 = i [VR3C3s3 + 44?2C2s2 + 2/RCs + 14?2C2s2 + 34?Cs + 1] = i [l/7?3C3s3 + 54?2CV + 64?Cs + 1], \)4 = ic/Cs + x>3 = (i/Cs) [1ÿ?3C3j3 + 54?2C252 + 64?Cs + 1 + 14?2C2j2 + 44?Cs + 3] = (i/Cs) [1/7?3C3j3 + (ZR2C2s 2 + 107?Cs + 4]. Now, = 1)4, and D„ = Rf i. Thus L(s) = vr/ox = (/R2C2s2 + 104?Cj + 4] [1ÿ?3C3j3 + RfCs R C , ,, . Now for s = /to, the phase of L(jto) = 0 when 64?2C2co3 = r = , , ,Hl3C s4 vCR2C2«3 + lORCs2 + 4/s ~ C 1'224 3 1 where ILI must be 1 (or more). For 1 = , , r, Rr 4/co, or co = RC 2 RC 1/7? 3C co  104?C co2 1 = 1/7?3 C4 co4 + 10 / 7? C2 to2 = 10 (2/3) 7? 1 (2/3)2 7? = (6.666  .444) 7? = 6.22R.
——— ———————
'
—
—
—
___
'

.

Note that the result for u3 above allows the solution of the 3capacitor circuit in Exercise 12.5 of the Rri Rfs2RC2 , u„ and L(j(o) = , ,, Text, namely L(s) = = w = 7—7l/7?Cs + 4 + 37?Cr \)3 (i/Cs) [VR C s + 44?Cj + 3]
—
— co27?C27?/ — , as stated there. 4 + j (3co7?C  1/co RC)
'
1
R — 6
, with Q = , from Eq. 12.13 For the filter with C4 = C6 = C and 7?, = 7?2 = 7?3 = 7?s = 7?, /„ = L 7t La 11.53 and 11.54 on page 917 of the Text. Now for f„ = 10kHz with C = lOnF, 7? = j\
 349
SOLUTIONS: Chapter #128
—J5— —2n x 10 x 10~v x 104
r
= 1.59kQ. While the solution to satisfy the distortion specification is quite M
complex, we can simplify the process by assuming that the signal at \>2 is a square wave of 0.7V peak ÿ amplitude, for which the fundamental is = 0.89V and the 3rd harmonic is 1/3 (0.89) = 0.297V, (from page 4 of the Text). For this situation, since the gain from t)2 to u, is 2 times, the peak output at Ui will be 2 (0.89V) peak = 1.78V peak. Now the 3rd harmonic component at \)2 will be less than 1/3 of the fundamental there (since the wave at u2 is not very square. Thus a rejection of = 33.3 =
20 log 33.30 = 30dB will be enough. Now for the secondorder bandpass filter, T(s) = (oVQ).v w'th Sa'n of 1 at co„. Now for gain of 1/33.3 at 3 to„ T(j(a) = jj2 + (co,/(2) j + co22 (cuyQ) to (to/Q)y'co
—
_ — ___________________________
co2 + ;co (co„/Q ) + co2
co (w(/Q ) + j (co„/e) 3co„
T(3co0) =
(3cd„
9/Q2 + 64,
in4
—9
= 64> Q2 = 1 78 "
= 19.9ki2. Use R =
— 07
1niA A
10—9
64
jp
whence 100!Q = ((3/Q)2 + (9
coye)2 + (9co2  co2)2/'
ÿ2
that
(co2  co2) '
 1)2/, or
= 156. Thus, Q = 12.5, for which R6 = 12.5 R = 12.5 (1.59)
= 1.1M2.
SECTION 12.3: LC AND CRYSTAL OSCILLATORS 12.14 For the FETs, operation is at I ass = 4mA, with gm 2 Iqssÿ I I = 2 (4)/2 = 4 mA/V. For each FET, Ye 100V r„ = — = — = 25k£2. For L = 10p.H and Q = 100 at 1MHz, the equivalent parallel resistance is
— —4ftiA—
I
Rp = (2 7i x 106 x 10 X 10~6) x 100 = 6.28k£2. Now at resonance, Rp loads the tank consisting of L and C  with C2 in series. Now the voltage across C\, ie from the follower output to ground, is
C2
of that across Rp
C\ + C2
. Since the power supplied to Rp comes via the capacitors, the current must
be correspondingly increased. Thus the corresponding load on Rp is
C2 C + c2
x Rp =
= 25kI2 II 25kQ
load on Qi is 6.28
C2 C, + C 6.28
2
C2 c, + c2
, and the voltage gain of the amplifier is
x 6.28. Thus the total
c, + c2 2)
II 10k£2, or Rl = 5.56
Rl Rl + k&n
Rl Rl + 1/4
V/V. Now,
looking back from the C\C2 node to the gate of Q\, the Thevenin gain equivalent is a gain of 5.56 Rl II RL = „. = 0.957V/V, with a resistance of ,, = .239kfl. Now 0.25 + 5.56 /gm 1 gm 5.56 + .25 + Rl when loaded by the Rp L C\ C2 network, the gain to the gate end of C2 becomes
...
(0.959) x
6.28
0.957 (6.28)
,
XT let x = Now
6.01 ±5.49 2 (6.28)
2
c2 c, + c2
C2 Ci + c2
6.28
x
T"
i + C2 .
C
C2
+ 0.239
1 = 6.28 1 + C/C2
Now for oscillation, this must be unity, that is
1 i + cyc2
+ .239.
. x 2  6.01 x — . Thus 6.28 + C/C2
—1 1
_
915
(5'5ÿ
—
TU
of
TO
iCn,
oo„ 0, x = + .239 =n
Q414 (unlikely) Therefore
 350 
1
1 + C/C2
ÿ
6.01 ± 6.012  4 (6.28) (.239) . —= 2 (6.28)
= .915, 1 + C/C2 = 1.0929, C/C2 =
SOLUTIONS: Chapter #129
0.0929, or C/C, = 10.76.
Alternatively, ignoring the effect of R„ and using Eq. 12.21, C/C, = gm R = 4 x (25 II Now fy, reflected through the capacitor network becomes
22.22.
II 10) =
x Rn =
c, + c2
2
22.22
£5
6.28 = 5.75k£2, and R = 25 II 25 II 10 II 5.75 = 2.82kf2, and C/C, = gm R = 4
1 + 22.22
C/C
(2.82) = 11.28. Now R„ reflected becomes
1 + C/C which R = 5.55 II 5.30 = 2.71kO, and C/C, = 4 (2.71) = 10.84.
= jc . Thus x = 4 x 5.56
21 X
6.28
4 x 5.56 x 6.28 x (•
6.28 = 5.30k£l, for
C2
Alternatively, from Eq. 12.21, C/C, = g„, R, and here R = 5.56  6.28 '
2
11.28 1 + 11.28
c, + c2 1+x
1 +JC
. Now let C/C,
y . Thus 5.56 jc + 6.28 jc
y 5.56 + 6.28 ( 1 +JC (139.7 6.28jc) A X , r )2 = 139.7 (ÿ—)2, or jc = ( x2 ( ) As a means for solution, tryJ jc = 10: jc 1 + jc 5.56(139.7 (139.7 6.28 (10.6)) 1L34< Try = 10>6: = ÿ Try ÿ = 11 5.56 11.6 5.56 (139'7 (1° 8)) ia8: = 10.83. Thus C2 = 10.83 C,. and, from Eq. 12.20, (0„ = = (*TTT>2
JM)2 *
___ 1lit
c, c2 L (———~") C, + C2
,/
*
,
——. 106)2
.

"A?
c, c2
5.50
C, (10.83 C,) 1 , or C, = 11.83 x C + 10.83 C, 10.83 C, + C2 (2 n x 106)2 x 10 x la CO L = 2.767nF, and C2 = 10.83 (2.767nF) = 29.97nF. (2.5 x 10~9x 10 x 10V5 1 Check: for C = 2.5nF, / =  (LC) " =  = ,1.006MHz.
__ _
2
— 2 71
2— 71
For loopgain > 1: With
C2
_
set 5% lower than calculated, the loop gain will be
Ci + C2 C2
C, + 10.83 C, (.95) = 1 097 = U' °r 10% ,arger10.83C (.95) For diode limiting: There are several views of the limiting mechanism: One is to find the gate conductingdiode resistance, Millermultiplied by the follower action which reduces the existing load at 25m V Since the loaded folthe gate (Rp ) by 10%. This value is about 10 Rp. Now rg at IG is about lower gain is originally
10"3 IG
25 x
=
,+ ÿ
Ig
= (1 + 1/10.83)' = 0.915. Thus, 10 (6.28) =
25 x 10~3 1 (11.76) = 4.68pA. Now for and Iq = , .915) 10 (6.28) (J3
'
.
IG = 4.68xA, VG = 0.7 + 25
ÿ
—
= 0.57V. Now for x>GS raised by 0.57V, iD = 4 (1  2 )2 = 6.60mA, of which 4mA 1 is absorbed by the current source. In
2'
1 y = 5.5611 5.71 Now for a load resistance of 5.56 II (6.28 II 6.28) ( 1 + C/C2 c, + c2 12 1 5.56 II 5.71 = 5.56 II (5.71 (.831)) = 5.56  4.745 = 2.56kft, the peak
1 (10.84) (0.95) output swing is 2.56k£2 (6.60 4) = 6.66V. A second, simpler, view of finding the peak output is sim¬ ply to use X)GS < 0.7V, say 0.6V, and proceed, as above, to find iG = 4 (1 = 6.76mA, and a 1+

peak output of (2.56) (6.76  4) = 7.1V, or, using a much simpler view of the load as 5.56 II 6.28 = 2.95k£2, find a peak = 2.95 (6.76  4) = 8.1V.
 351 
SOLUTIONS: Chapter #1210
For trioderegion limiting: For VGG = 0, V, = +3V, and X>GS ~ 0, triodemodc operation for Q x begins for vgd = IVpl = 2V, that is for vGI = 3  2 = IV. Now for \)Gi = IV for a 2.56KO load, id = 1/2.56

= 0.39mA, iD = 4 + .39 = 4.39mA, and 4.39 = 4 (1
)2,
\)Gji
 1) = 0.2V, for
=2
—
which Vps i = 3  1 = 2V, and VGG  = 3  1 0.2 = 1.8V. Triodemode operation begins for Q2, for 3V, when O51 = vG2 = *>„ = IV. Now it is apparent that the Vdg2 ~ Vds2 = 2V, that is for V_ = worst case occurs for Q\, although that for Q 2 is easier to calculate. For simplicity, use the \)GiS = 0 characteristic. Now to reduce the loop >ain from 1.1 to Lthe trioderegion resistance must reduce the
—
___
C>
\2 (6.28)()2 = 5.56 II (6.28 (.831)) = 5.56 II 5.22 = C + C2 2.69kf2 to 2.69/1.1 = 2.45kfl to reduce the loop gain to 1. Now RT II 2.69 = 2.45, or RT~= 2.45 II —2.69 = 27.5kQ. Thus, the triode slope resistance must be about 27.5k£2. Now, iD = K (2 (vGS  VT)
equivalent load resistance from 5.56
Vds
,2 \ ~ Vps).
for vGS ~ 0,
which
iD
rv... for
_
v \/2 and Vp is iD = IDSS V/, — " IDSS = K VT   " . ÿ V "ÿDS
4 (2
1
j
Now, (ÿ)2). 2— 2 DS
1
j
Id
=r~~
(2 (1 "
= 4 (1 
33 Gs
V yp r

33 DS
V Yp r
442 Vds), and RT =
.ÿds .2. iyY). P
»
T NOW
x
o
V*DS
d iD
= (4  2
UDi)~'k£2. [Aside: Note that RT = 0 at % = 4/2 = 2 = IV), I , since VA is assumed infinite.] Now 1/(4  2 vDS) = 27.5kf2, for 1 = 110  55 vDs, or vDS = (110  l)/(55) = 1.98V. Accordingly, for x>DS
= 1.98V with 3V supplies, a peak signal output of about 3  2 = IV would result. 12.15
(os
 
= (L Cey* = 2 n (2.015) x 106 = 12.661 x 106 rad/s
(2.018) X 106 = 12.679 X 106 rad/s 106 12 66x x tu iz.00 x L = 253 2 L 50 x 103
(2). Q = co„ L/r
prom
L
 50 x 103. C,
ÿ
'
1
10~12 C, 10~3 + (ÿ)2 L, L 
4x
=0)/L. L = 1.555 X rrom (4), (*;, l, (ÿ) cjo/ Li . From i/cv = ÿp) or 1/C, 1
= 2 71
Now co„ = co5. Thus r = 1 1 (— + — ), where C„ r r
+ C„ 1 x = —j =— ~L r*L C„ C0,t Cs co» 1
1
= 4 x 10'12F. Thus L = (12.679 x 106)2
LCS C„ Cs + CP
(1). co„ =
2.018 1
(4). Now, from (1), L
1555 .00297
* '°

1 <0;
3
= 0.523H, and r
= 0.0119pF. Note, as suggested, (12.661 x 106)2 (.523) that the topology of Fig. 12.16 is the same as that in Fig. 12.13, in which oscillation is possible with the crystal acting as an inductor. In both cases the amplifier input voltage across C2 is C/C2 times that across Cj. Now, for an amplifier with input resistance Rin and input voltage V2 (across Rin and C2), the voltage across C 1 is V\ = C/C2 V2, where, for no power loss, V2/R must be the same in each case. y2 y2 rQ /q \2 y2 ~= , whence Req = (C/C2)2 Rm > 1/100 Rin, for C2 = lOpF and C, > Thus „ Ri, R„ Rr yPS lpF. Now, the equivalent load on the amplifier is RL = {R\+ Req) II R} and its gain is  (gm/j + gnm ) Rr 1000 1000m Rl = 2 Rl For this gain, Rin = 1  2Rl 1 + 2Rl , and Rl = /?,+ 1 + 2Rl x (C/C2)2 1000 >(4). Now, the loop gain, L = 2 RG C/C2, must exceed 1 for oscillation. Thus for 2 RL C/C2 = C1 , 1000 (5). Now, to solve (4) and (5), try appropri¬ X ("7T) ) I' 1000 =1 1,2 (R 1 + 1+2Rl C2 ate values of C/C 1 as a measure of the required gain. First, for C2 = lOpF )tnd C 1 = lpF, C/C \ = 1 1 , 1000 , 2Rl 10/1 = 10. From (4) and (5), or directly, 2 (*<+ , no X—2) 1000 °
= 253.2 (.523) = 132.512, with Cs =

cos2 L

ÿ
.
\+2Rl
1, or
Rl = 5k£2, whence from
1000
——
'• " ir
*
10
1000, or /? 1 + jy = 5, R\ = 5 Rx + 100 (1+2 (5)) for larger Cx, say 2pF, C/C2 = 1/5 and 2RL x 1/5 = 1, or RL = — , or R, = 2.5  40/6 = 4.17k£2. This implies that for R, =
(4), 5 =
0.9 = 4.1k£2. Use 3.9kf2. Second, 2.5kf2, whence 2.5 = Rx +
10z
 352 
SOLUTIONS: Chapter #1211
C/C2 exists: Ignoring Rf, from (4), RL = 10ÿ x (ÿ)2, and from (5) 2RL 1 + 2Ri C2 _ 1, or r> _ I ,CK, C2 c2 1000 2000 o , = = —— Substituting, —— = Rl = — x (—)2, x = 2C, 2C/C2 2C, C, 1 C2 + C/C, 1+2_£l
0, a critical value of C — 2
(
C
... ..
1
C
C
c7)2' c7 + ( c7)2 =
(C/CO2 '
0r
— — —
("§7)4 + ("§7)3 =
ÿi
200° Thus' CÿC1 < (200°)1/4 = 6.69. Try
C/C, = (2000  (C/COY4. For C/C, = 6, C/C, = (2000  6Y4 = 6.5. For C/C, = 6.5, C/C, =  6.53)w = 6.44. For C/C, = 6.3, C/C, = (2000  6.33)l/4 = 6.46. Thus C/C, = 6.4 for which
(2000
Ci =
= 1.56pF is an estimate of the critical value for which R, = OkQ. Otherwise the largest value of R occurs for the smallest value of C,, that is for C, = lpF, and is R, ~ 4.1k£l
,
SECTION 12.4: BISTABLE MULTIVIBRATORS

12.16 As specified, there is at most a 0.1V drop in the output device(s) with a voltage of 4.9  0 = 5.0 0.1 = 4.9V across the series connection of R and R2. For corresponding operation in triode mode, iD = N (2 (Vgs ~ V,) VDS nig], or = 1 [2 (5  1) 0.1 0.12] = 0.8  .01 = 0.79, R, + R2 = = K, + K2 0.79 6.20k£2 (or more). Now assuming that the threshold of the symmetric input inverter is at 2.5V, VTL and 4925 05 , for example. = VTH each lie 0.5V beyond the threshold. To establish this 0.5V offset, ' R1 R2

4;9
,

—— —
Thus,
= 4 = 4.8, or R2 = 4.8 R,. Now R, + R2 = (1 + 4.8) R, = 6.20, for which R, = — 0.5 R 1
5.8
=
1.07kQ, and R2 = 4.8 (1.07) = 5.13kf2. In practice, 5% resistors of higher values such as 2.7kQ and 13k£2 would do the job, although 1% values of 5. 11k£2 and 24.9kO would be better. In general, highervalued resistors would save power, ensure a full output swing, and improve output transition times, while lowervalued ones would tend to reduce regeneration time.

Now for device variation, the highest value of Vlh of the input device occurs for Kp 20% high, K„ 20% low, Vm 20% high, and V,p 20% low. Now at Vlh, Kn (V„,  V,nf = Kp (5  V„,  V,,)2, or 0.8 (4.2  V,h) = 9.45  2.25 V(/l. Thus, 3.25 V,„ (Vn, ~ 12)2 = 1.2 (5  V„,  0.8)2, or V„,  1.2 = V,n  3.28 = 3 28D 0 1 , or VlH = = 10.65, or V,h = 3.28V. With V„, = 3.28V and input at Vm,
(ÿ)2
R1

R2
+ 3.28 = 2.94V. Now for the reversed set of extremes, by symmetry, V,h = 2.5  (3.28  2.5) = 1.72V, for which Vm = 5  2.94 = 2.06V, and V/i = 5  3.96 = 1.04V. Notice that for the two extreme versions that the two threshold (3.27)
>
i
+ 3.28 = 3.96V. Likewise V,L =
(4.9
3.28)
J. 1J
both lie within the powersupply range (0 to 5V). Concerning the rise and fall times, a complete calculation is somewhat complex, involving estimates of the transition times of the internal node, the Vil to Vm range of the second inverter and its gain, as well as switching of the output inverter. Alternatively, a quick estimate of output rise and fall times is defined by the current available from each output transistor as K (5  l)2 = 16AT = 16mA. Thus t,, (4 5 0 5) x 1 x 1012 between 0.5 and 4.5V, for a lpF  5= 0.25ns. A closer estimate F load, exceeds 16 x 10"3 could be found by including the effect of current reduction due to triode operation over a large part of the switching range. As an approximation, the current for x>Ds = 0.5V is iD = 1 (2 (5  1) 0.5  0.52) = ÿ 3.75mA. Thus a better estimate of the average current would be = 9875mA, and of the 4.5  0.5 x Ix 10"12 „ . transition time as 5= 0.40ns. 9.9 x 10'3 The delay requested includes a lot of factors, the most important of which are first the time for the first inverter to reach its threshold, and then for the second to do so. The former is dominated by the input sets do not overlap, although
—
... ..
 353 
— —
SOLUTIONS: Chapter #1212
itself which to rise (for example) from OV to VtH = 3V, by 3V, takes (3V/lV/ps) = 3ps!! Now the out¬ put of the first inverter changes at a rate defined by the inputsignal rate and by its load capacitance. Likely the former dominates. Now, due to R\, R 2 the signal at the gate of the first inverter pair is R2 5.13 ~ = 'nPut Now the gain of the first inverter at the middle of the 1 07 + 5 13 rx+ switching region is gm r„ the middle iD = K (vGS  V,)2 = 1 (2.5  l)2 = 2.25mA, gm = 2
whereÿat
(2.5
 1) = 3mA/V, and r„ = A= ~~ = 13.3kO. l[) Z.Zj
Thus, the gain is g„, r„ = 3
X
13.3 = 40V/V.
Thus the rate of change of the output of the first inverter is limited to lV/ps x .83 x 40 = 33.2V/Xs, before regeneration. Thus the output of the first inverter will move from 0 or 5V to 2.5V in somewhat 2'5 V more than = 75.3ns. After regeneration, the remainder of the transition willbe faster, with
—
33.2K4U
the output current driving the llpF load. Shortly after the transition begins as the circuit regenerates, the available current will be at least the 2.25mA middle bias current, for which the remaining transition 1 1 x 10~12 (2.5 : = 10ns. Now the final (isolated) conclusion about : = 0.5) time will be about T = 2.25 X 10"3 regeneration that can be easily made, is to estimate the time it takes for the input of the first inverter to cross the remaining half of its active region as driven through R2 by the output changing by all (or part of) the output range. Now as implied earlier, the input active region is approximated by 5V/40 = .125V, and half is about 62mV. Now for a 5V change in output, the current change in R2 is about 5V/5.13k£2 = 0.97mA, most of which supplies the lOpF input capacitance. For the input to move 10 x 10'2 x 62 x 10 62mV, takes about = 0.64ns. Thus, regeneration, once it begins as the output
——— —
——
of the second gate begins to move, can be very fast.
 
12.17 Now for v>0 (at node B) high, 04 = 1.4V. For
L£ = 1 R1
(i). Now for D0 low,
+ ~~ = 1, or Al ÿ ÿ
ÿÿ
t<2
+ ~~ = 1

R\ 13k£2, and
IDl =
—— —
1mA, —7: K
2
"1
= 1, whence
VA = VD4 + VD3  VD2  0.7V, and for O02 = 1mA,
(2). Now, adding (2)
+ (2) + (1) +
——
—— 
A2
A2

"2
ÿ
——
R2
+ 0 = 3, R2 =
= 1, and R\ = 1 .892 = 13kO. Now, the negative ~T~ a1 3 13 3 value of/?., not available without active components, implies that the specifications are too tight. For

2. Now combining the old (1) with the new (2) »
1 1'6
R2
"
27'4 R2
= 5, R2 
39 — 5
13'7
07
+ —— = R\  7.8kf2, and R 1 =
example, allow the current in Iq2 to increase, say to 2mA, for which change (2) becomes
R2
—14 —14 := 2.87ki2. Note that while this is a solution, that the latter formulation indi:= 11.487 111.&7.8 cates the possibility of a solution for which R\oo, where the term in the denominator reaches zero. Examining the circuit with R 1 removed, one sees this possibility directly. Thus, for R \ = 00, either 13.0 1.4 j for w,jc}1 1l.6kf2, or = 1, for which R2 = 13.7kfl. Now, to ensure Ri R2 that the associated current exceeds 1mA, the smaller value (or one even smaller) should be chosen. For 14 convenience, make Rt=oo and R2 = 10k£2, for which ID\ = = 1.16mA, and lD2



_
_
— ——
*
......

13 + 07 = 1.37mA. Now for D2 conducting, the current in D3 and D must exceed 1mA. Thus R = ÿ 15  0.7 0.7 _ 741ÿ Use R3 = 5.6k12. Now, the maximum current in D4 is 5 + 1.16 = 4
3
J 5.6 1 + 1.37 Finally, note that the input thresholds are defined by the possible voltages 4mA, required). (< as 3.6mA at node A , namely Vth = +1.4V and VTL = +0.7V. For inputs lower than 0.7V, the output is high, and XtA = 1.4V. Now, as U/ rises and just exceeds 1.4V, x>o falls, and \)A falls to 0.7V. For "0/ > 1.4V, Do = 13V. Now as t)/ falls, at t)/ = 0.7V, the output reverses again, and v0 goes to +13V.
 354 
SOLUTIONS: Chapter #1213
12.18
vo
vo
2.125
5V
A 5V
4.625
3.27V
i
0.375
0V
ov
VQ
2.675 2.5
1.17V
VI
3.83V
For the Q2, Q4 inverter, since K3 = K4 and IV,I arc equal, Vlh = VDDI2 = 5V. As noted on page 934 of the Text, the transfer characteristic slope exceeds 1 in magnitude for *0/ between 5/8 VDD V,/4, and 3/8 VDD + V,/4, or VDDI2 ± (VDO/8 V,/4) = 2.5 ± (.625 .25) = 2.5 ± .375 = 2.875V and 2.125V, for which D0 is (VDD/8  V,/4), and VDD  (VDD/8  V,/4), or .375V and 4.625V respectively. While a relatively complete transfer characteristic is provided for interest, the value V,/, = 2.5V is the most essential feature. We will use this to estimate the overall characteristic Dc/Oj. Essentially, for the vol¬ tage at the internal node (/I), Dÿ < 2.5V, D0 is high, which for Dÿ > 2.5, d0 is low, in both of which cases, positive feedback via Q5, Q6 forces node A away from 2.5V in the direction to which it tends. Correspondingly, the thresholds at the input are the voltages D/ for which the voltage Dÿ is 2.5V with W = V/£, when Q6 conducts (for D0 high (5V)), and D/ = V/h when Q5 conducts (for dg low (0V)). For V//, = D, OK.



 



—
SECTION 12.5: GENERATION OF SQUARE AND TRIANGULAR WAVEFORMS USING ASTABLE MULTIVIBRATOR 12.19 Here, for a 6.8V zener, the voltage values at d02 are ± (6.8 + 2 (0.7)) = ± 8.2V. Now the D03 output amplifier is a unitygain follower. Thus the voltage at the RC common node is a triangle (approxi¬ mately) of ± IV peak amplitude. Now for the notation on pages 1003, 1004 of the Text, L+ = 8.2V, (3 1
Ri+R2
and
P (8.2V) = IV, whence p = 0.12195, and
= 0.12195 (/?, + R2) and
= .1389 R2.

rc In 2. 10"6 X 50 1 + .12195 = 0.204M£2. Use R = 200k£2, for which RC = '.245RC. Thus R = 1 .12195 .245 x 1000 x 10"12 = 200us, R 2 = 200k£2, Rt = 0.1389 (200) = 27.8kQ, or, better, R2 = 240kQ and Rt = 33.3kQ, for 13 — 82 82 ÿ ~ ÿOSmA. Thus R3 which use 33kf2. For Rj, the current exceeds (1mA + 2 ( = ÿ Now, from Eq. 12.31, with L+ = L~,
T\ =
—
2qq/ÿq
4.4kl2. Use 3.9kf2. Now, using the notation on page 1004 of the Text, D~ = L+  (L+ PL )e~'A, )e or D" = 8.2  (8.2  l)e~,/RC , in general. Now ~ =7.2 (ÿ)e~l/RC = ÿ1 e~,/RC. Thus at / = 0, dt RC RC 50 7.2V . = 0.036V/is, and at t = 50 (is, slope is .036 e 200 = .036 (.779) = 0.028V/is. slope is 9 200 X 10I "s 0 0.036 + .028 q q32v/jxs, and the slope change is .036 .028 = 0.008V/(Xs. Thus the average slope is a
_
—
For the slope change reduced by half, reduce the voltage swing across C, so that 0.036 e~5(yRC = 0.036 .008/2 = 0.032, or e~5(yRC = .8888, 50IRC =  In .889 = 0.1178. Thus RC = 50/. 1178 = 424/ps, for 424 x 10"6 which R = = 424k£l. Now for this design, at / = 0, D" = D = L+ L+  PL) = pL" = 1000 x 10112


 355 
SOLUTIONS: Chapter #1215
12.21 (continued)
VI SV
, Tt»CÿR«qC2{RSII{Rl+R2))"(100x10*)10 «10pt I+0.83V,
4.17Vf
0.83V
0.83V
1.67V
2.88+(10/110)(10+10)4.70V
1.67V 1.67V+(100/110)52.88V
?l«10ps!
+1.671.82—0.15V 10V
ClR3»10x10"x10S »1m« 9.3V
T3»R4Cl«104x10 x10*ÿ0.1m# 10 V
Since x>A is limited to 0.7V, and x>B can be made lower, the output remains at 10V in a stable In this state, for this design: Vq =
10 (10+ 10)
10/?5
R 5 + /?!
+ /? 2
10 (10) = 0.833V, 10+10+100

—
state,
~
l0(Rs + Rri
VBB = /? /?  /? + 2 5+
= 1.67V, VA = 0.7V, VD = 10V. To trigger the circuit, v>ÿ must rise to 0.7 or 120 100 by (1.67 0.7) = 0.97V, and t)c to u where t> = 0.97, or x> = +0.97 (1.1) = 1.067V, that is 10 + 100 , see from 0.833 to +0.233V. Thus the positive step at t)/ required is D/ = 1.07V. From waveform that the pulse ends when 10 (10 0.7) e~'Ams = 1.67V, or 10.7 e 1/10 3 = 8.33, e~'A0'3 = 0.778, or t = 10"3 In .778 = 0.25ms. Thus the pulse length produced is 0.25ms. For a ratelimited input: i = C ,n4 II (11 x 104)= 1.07, and' dV , and i (R5 II (/?, + R2)) = 1.07V. Thus 1000 x 109— , vx 104
—

dt
.1.07

at
.
dt
= 117.6V/s = 118V/s. For recovery: See that for a positive input pulse
0.91 x 104 x 1000 x 10"9 shorter than the output pulse, that the longest recovery time constant, controlled by RA is t3 = /?4 C = 0.1ms, where x>A = 9.3 + (9.3 + l.61)e~'A0\ and recovery is complete when = 0.7 = 9.3 + 10 t/10"4 iao, or t = 10"4 (.093) = 9.3)is. , or e (10.97)e 10.97 Thus it appears that relriggering could occur 10is or so after the output falls (assuming zero recovery time for the amplifier). In practice, recovery of the amplifier itself (from limiting) might require some¬ what more time.
_
 357 
SOLUTIONS: Chapter #1214
 8.2(1, while at 50ps, \T = +8.2p. Thus 8.2P = 8.2  (8.2 + 8.2P)e"50/424, p = 1  (1 + p) (.8888), P = 1  .8888  .888P, P = .11 12/1.8888 = 0.0589, and the peak voltages of the improved triangle wave are at
± 8.2p = 8.2 (.0589) = ± 0.483V. Now to achieve a ± IV output, two resistors must be added to the
output amplifier to produce a gain of 1 +
values, one could make R {
Rz
ÿ
ai
=
1 —.483 —
= 2.07, or R2 = 1.07 Ri . To arrange such
— 10k£2 and Rz, as lOkQ in series with 68012.
12.20 The required circuit is as shown:
VoT
R8 VoS
RA
For a 6.8V zener and 0.7V diodes, the voltage at £ is ± (6.8 + 2 (0.7)) = ± 8.2V. For 1mA drain, RA +
ÿ2x RB = 1mA = 8.2kl2. For ±1V square waves, —— R,i + Rb
8.2 = 1, that is RA = lk!2 and RB = 7.2k£2.
For ±1V triangle waves, the thresholds at node B must be ±1V. Now, for R i = 10kl2, Rz =
—8j—2 R\ =
 50ps, the current in R must be / = 82k£2. Now for C = lOOOpF, to charge to IV in „ 8.2V 0V .... Now .. the current in 1000 x 10"12 x 1 • a for d d r is i 1mA  = 410kl2. RA  = 20uA. Thus Rn = /?3
50X10
—
M
.
,,
and Rb, 1mA for the zener, 0.1mA for Rc, and 20pA for R . Thus £3 =
ÿ
1382
; —— = 1.589kl2. 1 .02 1 I 7+ + +
Use R 3 = 1.5kl2. Note how easy this circuit is to design! This is because it has a desirable direct rela¬ tionship between particular components and specific functions!
SECTION 12.6: GENERATION OF A STANDARDIZED PULSE THE MONOSTABLE MULTIVIBRATOR 12.21 See that for B more positive than A, vD = +10V and VA rises very slowly until 1)4 > V>fl, at which R 1 + £5 point oD goes to 10V, vB falls to 10 x  , and A goes slowly to 0.7V.
— —K\ + /c5 + Rz
 356 
SOLUTIONS: Chapter #1216
12.22 Vl
™j Li 4.32.51.8V 0V
X2C1(Rl+R2)0.2ma
2.5Jÿxi«C1R2«10x10 x10* 0.1ms
T2C1{R1+R2)0.2ms
t/R, C,

For node A , assuming the Dlt D2 resistances to be zero, Now, Vyi = 2.5V = 0  (0 4.3V)e 2ÿ 10" 0t/ where e" or t = 10"4 In 1.72 = 54.2is. Thus the pulse is 54.2ps long. Normally, V; would 4.3 be very much shorter than that, a few tens or hundreds of ns, at most, the delay through the two invert¬ ers. For very long pulses at the input, the loop is held open, and the output fall time is an amplified version of that at node A For gain of 40 x 40 = 1600V/V, and 5V output swing, the input active e~'A0ÿ. region is about 5/1600 = 3 x 1 x 10~3V. Now, = 4.3c >/K' C' = 4.3e~'A0ÿ, and = 1 9x)A At t = 54.2(is, e~'A0 = x = 2.5 x 104V/s. Thus the output fall time = dt 4.3 10"4 4.3 10"4 3.1 x 10"3V j = 124ns. ~ 2.5 X 104
_
.
ÿA
—

SECTION 12.7: INTEGRATEDCIRCUIT TIMERS 12.23 From Eq. 12.39, T = CR In 3 = 1.1 x 10 X 10"9 X 104 = 0.11ms. The input pulse must be shorter than 0.11ms by an amount which guarantees the relationship for component variation. For longer inputs, both comparator outputs are high, and the flipflop is set and reset at the same time. 12.24 Extending Equation 12.38 for the case in which the capacitor voltage is not quite zero, but rather x>, x)c = Vcc (Fee  v) e~,/RC = 5  (5  x>)e~t/RC Now for x>c = 2/3 Vcc = 10/3V, 10/3 = 5  (5 3.333 + 5 1.666 , 5 u 9/ u)e~—t/RC —t/RC 1 RC ÿ TT77 = RC In (3  0.6x>), and = 0.6 RC 5x)* 1.666 5  \) oXJ 1 0.6RC For w small, = 0.2 RC. Now for T = 1.1 RC, for a 0.1V change in X), (3  0.6u) 3x> 0.2RC (0.1) the change in T is 0.2 RC (0.1), or x 100 = 1.8%. 1.1JK7
_—
_
.
A
n/1
 358 
—
A
ÿ
f
SOLUTIONS: Chapter #1217
TH = 0.69 Cÿ8(RA + RB ) = 0.69 X 10 X 10"9 (2 X 104) = 1 10"9 (104) = = 69ps. Thus the period is T = 138 + 69
12.25 From Eq. 12.41 on page 1013 of the Text,
TL = 0.69 C RB = 0.69 x 10 X 138 x 100 = 66.7%. For RA = 207ps, and the frequency 1/207 x 10"6 = 4.83kHz, with duty cycle lOkfl and RB = lkfl, TH = 0.69 C (RA + RB) = 0.69 x 10~8 (10 + 1) x 103 = 75.9is, and TL = 0.69 C (Rb) = 0.69 x 10"8 (103) = 6.9(is, for which the frequency is = 12.1kHz, and duty cycle is 75.9/(75.9 + 6.9) X 100 = 91.7%. For the same frequency, T = 207 X 10~6 = 6.9 x 10"6 + .69 X 10"8 900 x irr6 (Ra + 1) 103, Ra = , 1 = 28k£2. For 10kHz, T = lOOus, 100 x 10"6 = .69 C (RA + 2RB), 0.69 x 10"5 100 x 10~6 proRa + 2 Ra = = = 1.45 X 104 = 14.5k£2. There is no combination of resistors which will F .69 x 10"8 138is, and
ÿ
duce 10% duty cycle. Use 90% duty cycle and an inverter!
SECTION 12.8: NONLINEAR WAVEFORMPING CIRCUITS 12.26 For a sinewave of peak output u, ie u sin cor, the zerocrossing slope) is uco volts per second. Thus the triangle wave reaches t)to X 774, or D (2rt/) x (1/4/) = 27tu/4 = rru/2 = 1.57 t), at the peak. Though the choice is arbitrary, let us assume a sine wave peak of 0.7V, with 1D peak = 1mA, such that the triangle input peak = 0.7 X 1.57 = 1.10V, with the drop across R being (.57) (.7) = .400V. Thus R = 0.400/lmA = 40012. Now, in general, o, = 1.10 (0/90), or 0 = 81.8 t),, over the range 0* to 90°. For v0 = 0.7V, © = 90°, u, = 1.10V, I= Also X) = 700 + 50 In i/1, and i = =
11ÿ0,7
1mA. For v„ = 0.65V, i = 1 e*650"700*50 = 0.368mA, u, = .65 + .368 (.4) = 0.79, 0 = 81.8 (.797) = 65.2°, and 0.7 sin 65.2° = 0.635V. For "o„ = 0.60V, i = eÿ60ÿ700*50 = ,135mA, v>( = .60 + .135 (.4) = Q50 mA> v. 0.654, 0 = 81.8 (.654) = 53.5°, 0.7 sin 49.5° = .563V. For \)a = 0.55V, i = eÿoTOoxso = 0.55 + .05 (.4) = 0.57V, 0 = 81.8 (.57) = 46.6°, 0.7 sin 46.6° = 0.509V. For \)0 = 0.50V, i = 018mA) e (500700)50 = 0.50 + .018 (.4) = .507, 0 = 81.8 (.507) = 41.5°, 0.7 sin 41.5° = .464V. For = 0.45V, i = c(«o700)50 0Q7 v, Q 45 + 0Q7 ( 4) = 453) © 81 8 ( 453) 37 QJ sin 41.5° = .422V. For = 0.40V, i = eÿ700*50 = .0025, u, = 0.40 + .0025 (.4) = .401, 0 = 81.8 (.401) = 32.8°, 0.7 sin 32.8° = .379V. For v>„ = 0.35V, i = e"5a7ooy5o 0009, \), = .35 + .0009 (.4) = .350, 0 = 81.8 (.350) = 28.6°, 0.7 sin 28.6° = .335V. For = 0.3V = \>, , 0 = 81.8 (.3) = 24.5°, 0.7 sin 24.5 = .291V. For !>„ = 0.2V = vh 0 = 81.8 (.2) = 16.3°, 0.7 sin 16.3° = .197V. For = 0.1V = D( , 0 = 81.8 (.1) = 8.18°, 0.7 sin 8.18° = .099V. For x>„ = 0V = v, , 0 = 0, OV. In summary:
_
_
_
_
_
_
_
0°
90
65.2
53.5
46.6
41.5
32.8
24.5
16.3
8.2
0
u„,V
0.7
0.65
0.60
0.55
0.50
0.40
0.30
0.2
0.1
0
0.7 sin0V
0.7
0.635
0.563
0.509
0.464
0.379
0.291
0.197
0.099
0
e, mV
0
15
37
41
36
21
9
3
1
0
e%
0
2.4
6.6
8.1
7.8
6.4
3.1
1.5
1
0
 359 
SOLUTIONS: Chapter #1218
Note that the output wave is generally "fatter" than the sine wave.
12.27 Here, i = O.lt)2, with a match at u = 2, 4, and 8V, using 0, 3, and 7V supplies. For u = 2V, i 0.4mA, R i = 2/0.4mA = 5k£2 as before. Now chose V2 = 3  0.6 = 2.4V, and V3 = 7  0.6 = 6.4V. 1 For \) = 4V, i = 4/5 + = 0.1(4)2 = 1.6mA. Thus, n n , = 1.6  .8 = .8mA, R2 =
4~°'6"<2,4 R + 0.1
R 2 + 0.1
j
0.1 = 1.15k£2. V 0.8
For \) = 8V, i = j +
j8" 3p t
t 4 = 0.8mA, R3 = 1 / 0.8
+
8°+6164
= 0.1(8)2 = 6.4mA. Thus 1 / (R3 + 0.1) = 6.4  1.6
 0.1 = 1.15kO. J Now for the errors: At 3V, i = — = 0.6mA, rather than 0.1 (32) = 0.9mA, with an error =  0.3mA.

At 5V, i = j +
ÿ
ÿ
ÿ
= 1 + 1.6 = 2.6mA, rather than 0.1 (5)2 = 2.5mA, for an error of +0.1mA.
Thus at 7V, the error is 0.3mA, and at 10V, it is 0mA, just as in Ex. 12.22.
Now for 1mA diodes with n  2, with V2 = 2.4V, V3 = 6.4V. Now at D = 2, / = 0.4mA and R = 5kf2 as before. Now at v = 4V, i = 1.6mA, with loi = 1.6 0.8 = 0.8mA. For 0.8mA, u02 = 700 + 50 In 4 " 24 ~ 689 = 1.139kfl. Now at u = 8V, i = 6.4mA. Here /«, = = = 689mV. Thus R2 =
—
1 1.6mA, and IR2 ~
o
_ 94 — 07
0.8
—9 4 —\ —
8
§5
77ÿ
= 430mA ÿdi = 700 + 50 In 4.3/1 = 773mV, 1D2 = im 4.24mA (OK), Thus, ID3 = 6.4 1.6 4.24 = 0.56mA, uD3 = 700 + 50 In .56 = 671mV. Thus 8 " 6 4 ~ 671 = 1.659kO.
=
R3 =
.56
Now for the errors: At 3V, i = 3/5 + j'd2, = 3  2.4  IR2 = 0.6  lR2. Try iD = .05mA, x>D = 700 + 50 In .05 = 550mV, \)R2 = .05 X 1.139 = .057V, and V© + vR2 = .607V (rather than .600V) (OK). Thus i = .6 + .05 = .65mA, rather than 0.9mA, for an error of 0.25mA. At 5V, the required i is 0.1 5  2.4  VD2 5 52407 19 (5) = 2.5mA, and the actual i = j + = 1 + = 1 + 1.67 = = 1 + yyyy
——2'4ÿÿ
2.67mA. Now for D2, and Ip2 = 1.67mA, Vp2 = 700 + 50 In 1.67 = 726mV, iD2 = = (7)2 7V, of 2.64mA / an with +0.14mA. At error required = (0.1) 1.64mA. Overall, i = 1 + 1.64 = = 7 2.4  VD2 7  6.4 VDS 7 , which for vD2 ~ .75V, t02 = 4.9mA, and the actual i = — +   + — 1.139 5 1.659 7 ~ 2,4 76 = 700 + 50 In 3.38 = 761mV, iD2 = = 338mA Now = 3.37mA.
 ——

ni"o°'75
Now ipi =
,ÿ'4 ——1.659
= 0.
Thus
i
= 1.4 + 3.37 = 4.77mA, with an error of
 360
 0.13mA.
At 10V,
_____
SOLUTIONS: Chapter #1219
 VD2 10  6.4  VD3 Now = required i = (0.1) 102 = 10mA. Actual i = iD2 1.139 1.659 10  2.4  .78 ~ 79 10 ~ 5 99mA for wh.ch ÿ = 70Q + 5Q ln 5 99 79V> fo2 = 5.99mA. 1.139 1.139 10  6.4  0.72 Now iD3 = 1.74mA, for which i)02 = 700 + 50 ln 1.74 = .73V, i/>2 = 1.659 10  6.4  .73 = 1.73mA. Thus i = 2.00 + 5.99 + 1.73 = 9.72 with an error of 1.659 0.28mA. 10  2.4
10
_
_
_
—
12.28 For each of the top two circuits to which D and v2 are applied as in Fig. 12.44, iD =  = Is e — " r R *>/ , for D/ > 0. For the lower circuit to which we apply o3: Do = where Do = Do, Do = nVT ln R Is D/
—
u /n V
Dj X \)2
d3 +nVf ln R Is Thus D0 =  nVT
nVT ln R Is D t)2 R Is d3
ÿ1 ÿ2 X
D3
d3
D
Now
(R
Oi
R Ti7lnJT~lnRT)(ÿ) = ÿln »Vr
1
R Is D, D2
(ln
iD4 = Is
e
In
nVT
p ÿ2
R
Is)2
d3 R Is
Pi P2
Is d3
, and Do = = Is R Is D3

D2 , add one unitygain inverter at the input of the d3 d3 lower logarithmic circuit (for d3), and a second unitygain inverter at the output of the antilog circuit.
.
Thus to obtain
x>o" =
D
Now, as a check, for a 1mA diode with n = 2, lk£2 inputs, and voltages of 0.5, 1, 2, 3V applied: at 1mA, d = 0.700V; at 0.5mA, D = 700 + 50 ln 0.5 = 665.3mV, at 2.0mA, D = 700 + 50 ln 2 = °'5 °'5 : See dd = 1/1 ( .6653 .6653 734.6mV; at 3.0mA, o = 700 + 50 ln 3 = 754.9mV. Test
*

+ .7549) = .5757V. Now d4 = 2 e (700+575.7V30 754,9
_
0832mA, Do = .0832 = 1/12 as required. Test
+ 665.3) = 844.5. Now i4 = e<"7<XH"8445>50 = 17.993, D0 =
of course the supply voltages are high enough! Test
1x1
—1

See dd = j (754.9 17.993V = 18 as required, provided
—1 x—1 : see directly OK, with x>0  1.
SECTION 12.9: PRECISION RECTIFIER CIRCUITS 12.29 This is called an absolutevalue circuit.
vO
top part
bottom part,
ÿ
vl
_
AA R
 361 
SOLUTIONS: Chapter #1220
12.30
R3»R
Note that while 7? 2 is a constant load on the output, since its leftmost end is always at ground, the nonoutput end of R 3 is connected to the ac input. Thus R3 » R2 for least effect, and usually R2> R for relative efficiency. For equivalent offset current effects, Rj = R\ li V?2 = R® with R »R0.
vO
R2 = R
Ri = R
12.31
+15V B 10kn
R« 10kQ
X
198k£l
Here, lOOVrms = 141.4V peak, 140Vrms = 198.0V peak. Design for +10V at node B for 140Vrms input. Now 7? 5 supplies a current which cancels the effect of U/ as t>/ goes negative, until it reaches Ri 15V 100V rms. Thus > 7?5 = 0.1061 R\. Chose R \ so the change of input from 141.4V peak
'
ÿ
—
to 198V peak produces an extra 1mA. Thus
198  141 4 = "1
1mA, or 7? 1 = 56.6k£2
series with 620Q. For \>B = 10V full scale with 1mA, R2 = lOkQ. Also 6.00kQ  use two 12k£2 resistors in paraUel.
 use 56k£2 in
R5 = 0.1061 X 56.6k£2 =
jjj,
12.32 For t>/ = +5V: = +10V, and since vB = +5V, = 0V, t>£ = (1 + 10) 5u = +10V, vA = r+1 = 0V. For V/ = 5V: x>d = 0.7V. For D/ = 0V: D0 = 0V, and since VB = 0V, Uc = 0V, 20 (5) 5.7V, +10V, 5V, +10V, and since vB = = uc = 0V, vE = \>A = vD = +10.7V. Th


resistance is (ideally) infinite. The circuit could be called a fullwave doubter rectifier.
 362

SOLUTIONS: Chapter #1221
12.33 Note that the gain is generally VqA)/ = 1 + R/R i, for /?2//e i v, < (+0.7 +0.7) = 1.4V, 1 + (/?2 II /? 3)//? 1 for ((/?2 H ÿ3)ÿ i) t)/ between 1.4V and 1.4 + 6.8 = 8.2V when the zener conducts, and 1 + {R2 II /?3 II QYRi = 1V/V beyond. In region A , Ri t>o 1 for 0 < O/ < 1.4  = 1.4 j— = 14mV, =1 100 U/
(ISOmV. 1UV)
(14mV. 1.41V)
B IDM 04 It)
—
A (Ban M 101)
+
20 40 10 10100120140100
»' ÿ»
R2
= r7 14
,
100
1
= 101V/V, and Vq reaches 101 x
= 1.414V. In region B, for 14mV < V/ < 1000 1 i>o 14mV + 6.8V x = 1+ = ISOmV, 100 11100 V/ (R 2 II 100 II 100 1 + = SIVA', and v0 R 1 reaches  51 + 1.414 = 8.35V. In region 6 1000 *>o C, for 150mV < V/ < 00,  = 1V/V. Due to the v/ bridge connection, saturation is symmetric for posi¬ tive and negative inputs.
————
12.34 The circuit is a dc restorer and rectifier, using the lower and upper amplifiers, respectively, to create ideal diodes. For a lOOmV peak sine wave at the input, the voltage at the intermediate node (B) is a lOOmV sine with lower peaks at 0V and upper peaks at 200mV. Correspondingly, the output t)„ would be a dc level of 200mV, which would remain if the input is lowered or was removed. To return the output to zero from a peaktopeak input o, one could use a resistor R 1 to ground at the output, or (better) to a negative supply connected to the output. For a ground connection, and a return to 0V in 10//, the time constant RC is such that IIf < R\ C < 10//. For 95% recovery, one must wait 3 time 3.3 10 For input average drift at a constants (since e~3 = .05). Thus, 3 R\ C = 10//, and R\ = 3/ C fC' low rate, the drift signal is coupled via the input C to node B where the resistance is infinite. Thus the average voltage at B would rise, and the output at C would rise to represent the maximum peaktopeak value of the combined signal at / and // 100. To correct this, add a resistor R2 from node B to ground (or to a negative supply). Now, from a filtering point of view, we want a highpass filter with a pole at /„, for which / is clearly in the passband and // 100 is rejected as much as possible. Now the transfer , and T(j(ti) = /ÿQ) whence 17*1 = function from U/ to \>B is T(s) = 1 +R Cs 1 + jw RC 1+ RCco 1 1 = 1.108, RCco„ = (1.108  if' = 0.33, = 0.95. Now, at 0)„, 17*1 = 1+ 1 (RCa>„y 1+
—
—_
'
(RCw)
or co0
T
3
= 2 n /, where R2 = =— 2 k fC KC 1
{?—) I= 100
1+
.48
fC'
114
Now at
1
2nf 3(2tc//100)
f
—r—,
100
14
1
for T(s) = 1+ 1 33.3
ÿ
1 RCs
1+
1 2nf
3s
20 log .03 = 30.5dB. This
implies that a "volt or so" of drift will be reduced to "1/30 volts or so", or to around 30mV, still significant, but about as good as can be done.
 363 
Chapter 13
MOS DIGITAL CIRCUITS SECTION 13.1: DIGITAL CIRCUIT DESIGN  AN OVERVIEW 13.1
MOW
o.ov
V0L = 0.0V, V0H = 3.3V, V„ = 1.2V, Vm = 1.8V, Vth = 1.5V, VM = NMl = Vn  V0L = 1.2  0 = 1.2V, NMh = V0H  V,H = 3.3  1.8 = 1.5V. Here
1.5V.
13.2
i
A
I
B C D
u
ÿ'
UPL\A
[
>mL
J
<
tpu.H
•fPUl I
"jjvuv. H
1 CYCLE
h
In one cycle, each inverter makes 2 transitions. There are 5 tPUi and 5 tPHL in one cycle. There are 10 transitions altogether in one cycle. [Count this from the diagram, or calculate as 5 x 2.]. At 100 MHz, the period is 1/ÿ100 x 106) = 10"8s = 10/ts. Thus each of 10 transitions takes 1040 = 1 ns on average. Thus tP = 1 ns. If tpm = 1.2tPm and (tPm + tPm yi 1, then //>///, (1 + 1.2) = 2, tPm = 7/1.2 =
—
 365 
SOLUTIONS: Chapter #1222
12.34 (continued)
Front end of a tin* wava
Vo
1/f 200mV
TC3.3/1
 AÿV\
nondatum
 364
Longtermdritt anvalopa of a faw tana of mtllivolta
t
SOLUTIONS: Chapter #133
Check:
Finally, 1.3875V
13.6
For
NML = V,L
 VOL = 1.3875 — 0 =
1.3875V,
and
NM„ = V0H
 V,„
= 3.3
 1.9125 =
Both transistors operate in saturation, sharing the same current i, with V/ = v0 = V,/, Thus i = V2kp(VDD  V„,  IV,pl )2, and i = V2kn(V,h  V,„)2.
Vlh:
Equating and taking square roots: Vdd  Vlh  IV,pl =
From part of the solution to P.36.6 above, for V where VM = [Vdd ~ W,p\ + ÿkn/kpVm]/[ 1 + ÿlkn/k,,l where i = Vlkp(VDD VM IV,p I )2, and i = 1/lkn(VM Generally, gm = di/dv, where for v  VM:
 V,„)2.
gmp = kP (VDd ~ VM  IV,p I) = kpÿ2,/kp = Alikp, and gmn = kn (VM Also rop = VAp/i, and r„n = VAn/i.
 (gmn + gmp)(ron II rnp) = ÿ pfk~ + Vl~][ JAp An x vAp + Ml! = <2A(
Gain =
 V,„ ) = ÿ2ikn. Vi]
Thus the transfer slope at VM is 2(Vk/kp + l)(VApVAn)t(VAp + VAn)(VDD  VM  IV,pl )] Now, for Vln = IV,p I = 0.6, IVAI = 20V, kn = kp = (1.20.8)(100) = 150 pA/V2, the gain is
 2(
=
 38.1 V/V.
 0.6)2 = 82.7(Vl ,
Check: From first principles, numerically: VM = 1.65, i = 1/2(150 x 10"6)(1.65 gm = 1/2(2)150 x 10ÿ1.65  0.6) = 1575p/W. r„ = 20452.7 x 10~6 = 0.242MQ., «ain =  gmr„ =  1575 x 10"6 x 0.242 x 106 = 38.1 V/V, OK.
and
the
13.8
See kn = (20) x 8/2 = 80pA/V2, kp = (10) x 16/2 = 80pA/V2. From Eq. 13.6 and 13.7 of the Text {or directly from the triode relation, that is, iD = k ((vGS V, )\)DS  vfcs ~ k (uGs ~ V,)x)Ds> for small For inPut hi8h' = 3,125k£1 For vDS}, see rDS = vDSAD = = v>)k 80 x lO"6 (5  1) input = 3.125k Cl, the same, since the inverter is matched. F low, rDc = 80 xlO (5  1)
13.9
Maximum currents are the same for both the p channel and n channel devices. For the output connected to an opposing supply, I max  k/2 (vGS V,)2 = 40 x 10~6 (5  l)2 = 640iA. 6 For the output at Vddÿ2, I k (Cues ~ V,)Vds  uds42) = 80 x 10"6 ((5  1) 2.5 2.5/2) = 40 x 10 (0  6.25/2) = 550pA. For the output 0.1Vdd from the limit, I= 80 x 10"6 ((5 — 1) 0.5 0.52/2) = 80 X 10~6 (2 0.25/2) = 150 pA.

—
—
—
—
—
13.10 For all the inverters, V0L = 0V, VOH = Vdd > as VDD varies from 3.75 to 6.25V, and asV,„ = V,p = V, varies from 0.75 to 1.25V, with k„ = kp. From Eq. 5.94: V/p = 1/8 (3 Vdd + 2 V,). From Eq. 5.93: VIH = 1/8 (5 Vdd  2V,), or V,„ = VDd ~ V„, generally (for symmetry). Now for V/L to be largest, VDD = 6.25 and V, = 1.25, whence V,L = 1/8 (3 (6.25) + 2 (1.25) = 2.66V, for which Vm = 6.25  2.66 = 3.59V. For V,L smallest, VDD = 3.75 and V, = 0.75, VIL = 1/8 (3 (3.75) + 2 (0.75)) = 1.59V, for
 367 
SOLUTIONS: Chapter #132
0.909 ns and tPLH = 1.2(0.909) = 1.091 ns. 13.3
See
that
the
static
dissipation
is
Thus
zero.
the
dynamic
power/inverter
PD = (300 X 106 X 3.3V5 = 990 X 10~6/5 = 198pW. Now PD  fCV2, and C = PdA/V2) = 198 x lOÿlOO x 106 x 3.32) = 0.182 pF. For this logic, tP = (K100 x 106))10 = Ins and DP = 1 x 10"9 x 198 x 10"6 = 0.198 pj 13.4
For the gates, tp = (30 + 10)/2 = 20ns. Total delay through 5 gates for 2 transitions each is T = 5 x 2 20 = 200ns. Frequency of oscillation = 1IT  l/200ns = 5MHz.
X
0110ns 0 VI V2
0
I
iEhjÿlOns 1_ , ©
_o__®
©
V3
©
.©
L
©
V4 VS

90ns
1 /
30ns
®
©
"L©
GL
1
©is
0 ÿ
© / © ©I
©a
r
©©AO
—i
i
30+10+3070ns
30+10+30«70ns
10+30+10+30+10+30+10s130ns
Note that
[3©lor matched
gates.
SECTION 13.2: DESIGN AND PERFORMANCE ANALYSIS OF THE CMOS INVERTER 13.5
is
pn/p.;, = 10040 = 2.5, then from Eq. 13.10, for {W/L)p =2.5(W/L)n = 2.5(1.2p/»/0.8pm) = (3.0 (inv0.8im).
Since
_
Since the generic process uses a supply of VDD = 3.3V, VOL = 0V and From Eq. 13.8, with V,„ = IV(/)I = 0.6V and k„ = kp for matching.
VDD\V,P\ + VM~V,„
TTluk
=
IT*A.j,/ l\.p

3.30.6 + ÿ(0.6) TTTi ' 1
3ÿ
1

matched
a
VOH = 3.3V.
1,65V
[Of course, this result could have been written down directly.]
From Eq. 5.94, V,L = (3Vco + 2V,y8 = (3(3.3) + 2(0.6)>8 = 1.3875V. From Eq. 5.93, Vm = (5VPD  2V,V8 = [5(3.3)  2(0.6)ÿ8 = 1.9125V. From Eq. 5.95, NM„ = (3VDD + 2V,y8 = (3(3.3) + 2(0.6)>S = 1.3875V. From Eq. 5.96, NML = (3V0D + 2V,y8 = 1.3875V.
366
device
SOLUTIONS: Chapter #134


which V,H 3.75 1.59 = 2.16V. Now VOL is always OV, and V0H ranges from 3.75V to 6.25V, V,L from 1.59V to 2.66V, VlH from 2.16V to 3.59V. See that the noise margins between gates (with dif¬ ferent supplies and different V,) vary widely: Consider NM„ = VOH V,H. Highest is 6.25 2.16 = 4.09V. Lowest is 3.75 3.59 = 0.16V (See that this is very bad). Consider NM, Vm = V„ Highest is 2.66 0 = 2.66V. Lowest is 1.59 0 = 1.59V.





13.11 From Eq. 13.12, for an inverter with a fanout of 1,
C = 2CgdI + 2Cgd2 + Cdbx + Cdb 2 + Cg2 + Ci4 + Cw. For the matched inverter, (W/L)n = (1.20.8), and (W/L)p = (10040X1.20.8) = (3.00.8). Thus, Cgdi = 0.5 x 1.2 = 0.6/F, Cgd2 = 0.5 x 3.0 = 1.5/F, Cdbx = 2.5 X 1.2 = 3.0/F, x 3.0 7.5 2.5 = x 1.8 fF, 0.8 X 1.2 = 17.3/F, = Cg3 = Cdb2 C„4 = 1.8 x 0.8 X 3.0 = 4.32/F Cw = Cg3 = 1.73fF. Thus C 2(0.6 + 1.5) + 3.0 + 7.5 + 1.73 + 4.32 + 1.73 = 22.5 fF. Since the inverter is matched

™
"
13.12 For the assumption iDN(0) = V2K{W/L)n(VDD
of
 V,)\
constant
Thus tpm =
ÿ
Alternatively,
CVDD
From
current.
From Eq. 13.17, tpm = CVpc/lAosiO). Now for V, = 0.2VDD, iDS(0) = Vlk'n(W/L)n(\
Eq.
13.14,
in
saturation,
 0.2)2V%D = kÿW/L)nV$D{0.32).
1.6C or fpHL = , ,,2 kn(W/L)nVDD 2(kn)(W/L)„ V)jd(0.32) 4
n/i
from
For this case, in which IpHL

—
______ 5.101,
Eq.
Vdd

tpm
=
2C kn(W/L)n(VDD
Wdd~4V, V' F21n+ VDD  V,) Vdd ~ V,
3.3V and V, = 0.6V,
06 2C inn v Irv6, X HTX 1.20.8)(3.3 0.6) 3.3 0.6 100 2C X [0.22 + 0.41] = 3111 Cs.
3(3.3) 4(0.6)


3.3
100 X 10ÿ(1.5)2.7 Now, ignoring the fact that here V, = 0.60.3VOD = 0.182VDD, (rather than 0.2 Vdd)1.7C From Eq. 13.18, tPHL = = 3434 C s. 10ÿ(1ÿ0.8X3.3) 100 X J I
.
1.6C
= 3232 C s. ow„ 100 X 10ÿ1.20.8X3.2) Assuming the estimte from Eq. 5.101 to be the most accurate, it is interesting to see that of the simple approximations, the one found here, is best. It is certainly, the easiest to obtain from first principles.
From the constantcurrent calculation above, tpm =
_6

13.13 For simplicity, use k = k'n{W/L)„, and substitute V, = 0. 1VDd (1) (0.405) From Eq. 13.14, iD(N)iD(0) = V2k{VDD Vcf = From Eq. 1315, iD(AV)iD(M) = k[(VDD  V,)VD[A ~ (VW2)2/2] = kV$D(0.SV2 ÿ

Thus, iD(av) = kV$D (0.405 + 0.325X2 = *Vj&>(0.365) 1 17C
From Eq. 13.18, tPHL = CVDD/240.3651Vo2o) = tzt. kvdd Now for the current in (1) sustained for the half transition,
 368
 1/8) = £FD20 (0.325)
SOLUTIONS: Chapter #135
tpHL
= CVDD/2/(0A05kVoD) =
h'HL =
,
V
2C
Hvdd  V.)
2C kVDD(0.9)
kVDD
From Eq. 5.101 on page 434 of the Text,
+ l/21n Vdd  v,
3VDD  4V,
0.1 34(0.1) l/21n + 1 0.1 1
2C
.....
(0.111 + 0.478) = v~
kVDD (0.9)
1.31C
' kVDD
Obviously, the sustained saturation result is reasonably good and certainly simple to obtain from first principles. Notice, as well, the impact of change of V, from 0.2 Vqj to 0.1 Vpp, being a change in the coefficient from 1.6 to 1.2 in the simplest constantcurrent formula. 13.14 From P13.8 above, kn = k,, = 80pA/V, IF, I = IV, VDD = 5V. For Vin = VDD/2 = 2.5V, Ipeuk = iD = 80/2 (2.5  l)2 = 90 it A. Assume (for the present purposes) that rise and fall times arc measured from 0% to 100%. As the input goes from IV to 4V, the current flow is triangular, with a peak value T/4 of 90pA. Over an interval of 3/5 x 774, the aver¬ age current is 90/2 = 45pA. This happens twice T/4 per cycle. Thus the average current per cycle from 2 (35) x 774x45 self conduction is = 13ÿA< T/4 T/4 The average current due to capacitance load is CVDD f = 0.5 x 10'2 x 5 x 20 x 106 = 50 pA. Total average current = 50 + 13.5 = 63.5 iA. With load, PD 63.5 x 10"6 x 5 = 317.5 pW. Without load, PD = 13.5 x 5 = 67.5 pW.
V
/

13.15 From Eqs.13.18 and 13.19, tPHL Here,
k'n
1.7C , and tpui kn(W/L)n VDD

1.7C
kp(W/L)pVDD
= 2k'p = 20pA/V, and (W/L)n = V2(W/L)p = 8pm/2p/n. 1.7 x 0.5 x 10~12 2.12 ns. 20 X 10* (8/2)(5) = 47.2 MHz. Ignoring the transitiontime peakcurrent
For V/jo = 5V, IV, I = 0.2VDD = IV, and tPHL = tPLH = tP. 
For a 5stage ring oscillator, / = [10(2.12)]"' flow per gate, PD = / CV$D = 47.2 x 106 x 0.5 x 10"12 (52) = 590pW. The DelayPower product DP = 2.12 x 10"9 x 590 X 10"6 = 1.25 X 10"12/ = 1.25pJ.
1.3V, IV, I = 0.8V, (a) Qn conducts for x>, from 0.8 to 1.3V, and Qp conducts for v, 0.8), or 0 to 0.5V. For x>, = 1.3/2 = 0.65V, neither transistor conducts and iD = 0, (1.3 from 0 to (b) Output voltages range from 0 to 1.3V. Von = 1*3V; V0l  0.0V, (c) V/L = 0.8V, and V,// = 1.3 0.8 = 0.5V. [Note that V,L > V,H !] Between V,{, and V///, no current flows.
13.16 Here,
VDD =

—
I.JV
vl
0V 1.3V
vO
0.8 1.3
0V
0.5V
[r 0.39ps
 369 
CO
<
SOLUTIONS: Chapter #136
(d) (Note in considering the transfer characteristic, that the small capacitor at the output holds the out¬ put while neither transistor conducts, (e) iD = 1/Ik (1.3  0.8)2 = 1/2 x 20 x 10~6 (0.5)2 = 2.5pA peak. For \)q at 0.8V, iD = k [(uG5  V,) vDS  dAv2] = 20 x 10"6 [(1.3  0.8) 0.8  0.82/2] = 1.6pA. Aver¬
age current is
— —ÿ——
——
1
= 2.05pA, and the time for a 0.8V change (from 0 V to 0.8 V) is = x 1 10~'2 X 0 8 t" = 0.39 p s. Conclude that propagation delay is more than 0.39ps. Now, for the out2.05 x 10"6 put moving from 0.8V to 1.3V, the average current available is (1.6+0)/2 = 0.8pA. 13 08 ÿ = 0.625ps, and the total transiThus, the time to reach 1.3V from 0.8V is about 1 x 10"'2 x —:0.8 x 10"6 tion time is about (0.39 + 0.625) = 1.02py . Considering the driven stage with input at 0.8V, the avail¬ able output current is Op.A. At an input of 1.3V, the available output current is 2.5pA. Thus, the aver,. 0 + 2.5 1.0 x 10~12 x 0.8 , „c . and, the propagation delay age current =  = 1,25pA, — = 0.64 ps. =
—
—
.
(f)
.
Frequency of oscillation of 5 gates is (at most) =
t = 156kHz. 10 (0.64 x
10"6)
13.17 Here, kn = (20) x 18/2 = 180pA/V2, kp = (20/2) x 4/2 = 20pA/V2: IV,I = IV, VDD = 5V. See V0H = 5V Vol = 0V. For V,h = d, D/ = Do = D, both devices are in saturation, and 90 (D l)2 = 10 (5 D l)2. Taking the square root, 3(d 1) = 4 D, 3d 3 = 4 D, 4d = 7. Thus, V,h = D = 1.75V. For t) v0)2), 1) (5 (5 uG) Vil = U, Qp in triode, Qn in saturation, 90 (o  l)2 = 10 (2 (5 (1). Now, taking derivatives, 18 (x>— 1) = 2(4 v) 9 (\> l)2 = 2 (4 v) (5 v0)  (5  v0)2 9 Do —9 Do d Dq (  Tÿ) + 2 (5 Do) (1) 2 (5 Do) —r—— . Now, with = 1>18d18=182d d D dD dD (2). Substitute (2) in (l) >  10 + 2 Do  10 + 2 D0, 20 D = 4 D0 + 6, d = .2 d0 +0.3 9 (.2d0 + 0.3 l)2 = 2 (4  ,2d0  0.3) (5  D0)  (5  D0)2, 9 (,2d0  0.7)2 = 2 (3.7  .2d0  0.3) (5  Do) (5 D0)2, 0.36 Do + 4.41 2.52 D0 = 37  9.4 d0 + .4 d3 312 ÿ V 3122 (.96) (7.59ÿ 25 + 10 Do  D3, 0.96 DI 3.12 D0  7.59 = 0. Thus D0 4.87V, and VtL = D = .2 (4.87 + 0.3) = 1.275V. For V,H = D, Qn in triode, Qp in saturation,
—
—
—
—





 

 —


— —
—
+ÿ4
=ÿ
 
— —
—
_
(1). 2 (4  l)2 = 90 (2 (D  1) (Do) —Do), (4  D)2 = 18 (D  1) (Do) — 9 Do jp9 Do 9 Do 9 Do . For —— = 1 —> 2 D — 8 = 18 — 18 D + 18  D) (1) = 18 (D  1) 9d + 18 Do  18 Do r9d 9d '
10 (5  D
(2). Now, substituting (2) in (1), (4  1.8 D0 Do + 18 Do, 20 D = 36 Do + 26, D = 1.8 D„ + 1.3 D0)2 (2.7 1.8 9 d£, 1) 1.3)2 1.3 (1.8 = 18 (1.8 D0 + .3) D0  9 Do, 7.29  4.86 18 = D0 D0 + (10.26)  7.29 = 0, Do + .509 D0  .362 (20.16) 9 5.4 32.4 Do, 3.24 + d0 Do d0 Do + Do = d0 + ~'509 ± 5092 + 4 ( 362) = 0.400V, and D = 1.8 (.4) + 1.3 = 2.02V. Thus VlH = 2.02V. = 0, Do =


For tp : For D0 = 5V, /„ = 90 (5  l)2 = 1.44mA. For D0 = 0V, 1p = 10 (5  l)2 = .160mA. For D0 = 1.75V, /„ = 90 (2 (5 1) (1.75) 1.752) = 0.984mA, and Ip = 10 (2 (5  1) (5  1.75)  (5  1.75)2) 0 5 x 1012 x f5 CV 1 751 = 2,89ns' For = lp3 = 10 (26  10.6) = 0.154mA. For discharging, tPHL = 934ÿ2 ÿ 44 + 0.5 X 10'12 (1.75) . charging, r = 5.57»s. .rLH =


—

.
SECTION 13.3: CMOS LOGICGATE CIRCUITS 13.18 Y = A(B + C). The corresponding pullDown network (PDN) is shown asD. Y = A(B + C) = A + B + C = A + B C . The corresponding pullUp network Usbu 1.
 370 
(PUN) is shown as
SOLUTIONS: Chapter #137
Now, the PUN dual to D i is
U2+ VDD
J *_h
*~~l'j
.jp
+ VDD
ÿ
=" ( D1 )
(U1)
See that U2 is similar to t/, but not identical, D and C being interchanged with respect to the connec¬ tion nearest to the power rail.
£>2 is a PDN obtained from
U
+ VDD
Y
J i
t!
yj
(U2)
=" (D2)
See that D2 and D\ are not identical, the transistor A (the one with input A) being near ground in D2, but near the output in D\. For this logic function, there arc 2 PUN and 2 PDN which can form 2x2=4 different gate topologies in all.
13.19 In Fig. 13.15, the PDN shown is to be called Di and the PUN is to be called Uj. Here, PDN D2 is dual to U1.
j
r1 r •£btfc
ÿti
J
iJ5 £JS ÿ=ÿ
(D2)
 371 
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£iÿ
=" (D3)
SOLUTIONS: Chapter #138
Z)3 is D2 redrawn more symmetrically. + VDD
U2 is the PUN dual to Dh drawn directly. U3 is U2 redrawn.
+ VDD
+ VDD
+ VDD
•~J_*~~fJ pI p1 1 1 Hid1 nd *ndÿ ~1
A
I

B
A
1
B
_
•A
ÿ
B
Y
(U2)
Using U\, D\ and U3, D3, there are 4 possible XOR circuits that can be constructed. When the relative placement of the inputs with respect to the supplies is considered, there are two versions of each of the perfectly square networks (like U3, £>3) depending on proximity to the supply of each of the two series layers (of paralleled transistors). Thus for each of these U and D networks there are 2 variations. However for networks U\ and D2, in which the two series nodes are not joined (but, correspondingly, not for D3 or U3), there are 4 variations: AB CD
AD CB
CB AD
CD AB
with respect to proximity to a supply. Now, for all networks like Uh D\, there are 42 = 16 possible arrangements, but for half of these (that is, 8), there are twice as many combinations each (the diagonal exchanges in each group of 4 above). Thus the number of combinations is 8
13.20 ÿ
Refer to the transistors by their variable names PA , Na, etc. Now Na,Nb,Nc, Nd are all of unit size, the same as N in the unit inverter, where (W/L)n = (1.20.8). Now for matching, PA, PB, Pc< Pd are all 4x larger than P in the inverter, which N, size of the 2.5x is turn in 4(W/L)p = 4(3.00.8) = (120.8). Total area of the NOR is 4[(1.2)0.8 + (12)0.8]
VDD
n.J re J
TjJ
= 4(11)(0.8)(1.2) = 44(0.8 X 1.2) = 42.24pm2. This is 44 X the area of a single NMOS. The area of an inverter is (1 + 2.5)(0.8 x 1.2) = 3.36pm2. Thus the NOR is 42.240.36 = 440.5 = 12.6 x larger than a single inverter.
_
aJ1bJ1cJ1dJ! _
13.21 For a 4input NAND gate, the N devices are in series and the P devices are in parallel. For mobility matching, each P device is the same size as P in the inverter, namely (W/L)p = 2.5(1.20.8) = (3.00.8). For current