CHAPTER 15
15.1
The bypass capacitors do not affect R
IN :
R IN  R G  1 M |

A V1  
g m1 R L1
1  g m1 R 5

R L1  620  78 k 22 k  r 2   o 2 1 1. 6k  598  2. 39 k  151 1. 6 k  597 
A V1  
0. 01 597 
o 2 R L2
150 3. 54 k 
 1. 99 | A V 2  
=
= 2.18
1  0. 01 200
r  2   o2  11. 6 k
2. 39 k   151 1. 6k 
A V 3  0. 950 | A v  1. 99 2.18 0. 950   4 .12
R


 r 3 
 o 2 R E2
R OUT  3300   th 3
 | R th 3  R I3 R o2 R I3 r o 2 1 
  R I3  4 . 31 k


1
R

r

R
 o 3


th 2
2
E2 
4. 31 1. 00

R OUT  3. 30 k  
k  = 64 . 3 

81

15.2
Ac equivalent circuit
The Q-points and small-signal parameter values have already been found in the text.
v o  v o  v 3  v 2  v 1 v th 
    

 |
v s  v 3  v 2  v 1  v th  v s 

v o 
 o3  1 R L3  81  232
 0 . 926
  
v 3  r  3   o3  1 R L3 1500  81 232

 o2 4700 R IN 3
 v 3 
| R IN 3  r  3   o3  1 R L3  1500  81 232  20 . 3k
   
r 2   o 2 1 R E 2
 v 2 
 v 3 
150 4 . 70 k  20 . 3 k
 2. 35 |
   
2. 50 k  151 1. 60 k
 v 2 
R IN 2  2. 50 k   151 1. 60 k = 244 k |
 v 1 

  1 |
 v th 
v th 

  0. 990 |
 v s 

g m1 R I1 R IN 2
v 2 
   
1  g m 1R S 1
 v 1 

|
R IN 2  r 2   o 2 1 R E 2
0 . 01 620  244 k 
 v 2 
 2. 06
   
1  0. 01 200  
 v 1 
vo
 0. 926
vs
2. 35 2. 06 10. 990   4. 44
1
15.3
0. 01
2
V GS  2 | V GS  9000IDS  V GS  1.80V
2
0.01
2
I DS 
1.8  2  200 A | V DS  15  15k  9kI DS  10.2 V
2
50 10.2V
2
g m  2 10mA / V 0. 2mA   2mS | r o 
 301k
0. 2mA
43k
Q2 : V EQ 2  15
 3.18V | R EQ 2  160k 43k  33.9k
43k  160k
3.18 - 0.7
151


I C 2  150
=1. 35 mA | V CE 2  15  4.7k 
1.6kI C  6. 49 V
33.9k +1511.6k

150

M1 : IDS 


g m2  401.35mA   54.0mS | r  2 
150
80  6.49 V
 2.78k | r o 2 
 64.1k
54.0mS
1.35mA
120k
 8.53 V | R EQ 3  120k 91k  51.8k
120k  91k
8.53 - 0.7
 81

 80
= 2.72 mA | V CE 3  15   2.2kI C  8.93 V
51.8k + 812. 2k
80

Q 3 : V EQ 3  15
IC 3
A V1
60  8.93 V
80
 25. 3k | r  3 
 734
2.72mA
109mS
 0. 9902mS  301k 15k 33.9k 2.78k  4. 312
g m3  402.72mA   109mS | r o3 






A V 2  54. 0mS 64.1k 4.7k 51.8k 734  812.2k 250   180


AV3 
812. 2k 250
734  812.2k 250
 0.961 | A V  4. 312180 0. 961  748
v be3  v b 3 1  A v3   A V1 A V 2 v s 1  A v3   5mV | v s 
15.4
2
*Problem 15.4/15.5 - Multistage Amplifier
VCC 12 0 DC 15
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 11 0 AC 1
RS 1 2 10K
C1 2 3 22U
RG 3 0 1MEG
M1 5 3 4 4 NMOSFET
RS1 4 0 9K
C2 4 0 22U
RD 12 5 15K
C3 5 6 22U
R1 12 6 160K
R2 6 0 43K
Q2 8 6 7 NBJT1
RC 12 8 4.7K
RE2 7 0 1.6K
C4 7 0 22U
C5 8 9 22U
R3 12 9 91K
0. 005
 165V
4. 312180 1  0.961
R4 9 0 120K
Q3 12 9 10 NBJT2
RE3 10 0 2.2K
C6 10 11 22U
RL 11 0 250
.MODEL NMOSFET NMOS VTO=-2 KP=.01 LAMBDA=0.02
.MODEL NBJT1 NPN IS=1E-16 BF=150 VA=80
.MODEL NBJT2 NPN IS=1E-16 BF=80 VA=60
.OPTIONS TNOM=17.2
.OP
.AC LIN 1 2KHZ 2KHZ
.PRINT AC VM(3) VP(3) IM(VS) IP(VS) VM(11) VP(11) IM(C6) IP(C6)
.END
VM(3)
1
Results: A V  VM(11)  879 | R IN 
 1.00 M | R OUT 
 51.8 
IM(VS )
IM(C6)
15.5 *Problem 15.5 - Use the listing from Problem 15.4, but remove C2 and C4.
Result: A V  VM(11)  2.20
15.6
100k
 1. 63V | R EQ1  100k 820k  89.1k
100k  820k
1. 63 - 0.7
101


 100
= 319 A | V CE1  15  18k 
2kIC  8.61 V
89.1k +1012k

100

Q1: V EQ 1  15
I C1
g m1  40319 A   12.8mS | r 1 
100
70  8.61V
 7. 81k | r o1 
 246k
12.8mS
319 A
43k
 3.18V | R EQ 2  160k 43k  33.9k
43k  160k
3.18 - 0.7
101


 100
=1. 27 mA | V CE 2  15  4.7k 
1.6kI C  6. 98 V
33.9k +1011.6k

100

Q2 : V EQ 2  15
IC 2
g m2  401.27mA   50.8mS | r  2 
M 3 : V EQ 3  15
100
70  6. 98V
 1.97k | r o2 
 60.6k
50. 8mS
1.27mA
1.2M
 8.53 V | R EQ 3  1.2M 910k  518k
1.2M  910k
8.53 = V GS3 + 3000I DS3 =1+
2I DS3
+ 3000I DS3  I DS3  1.87mA | V GS3  V TN3  1.93V
0.001
V DS3  15  3000I DS3  9.39V | g m3  20. 0010.00187   1.93mS
89.1k
vth  v s
 0.899v s | R th  89.1k 10k  8.99k
89.1k  10k
 o1 R C1 R B 2 r  2 
100 18k 33.9k 1.97k
A V1  0.899
 0.899
 9.03
R th  r 1
8.99k  7.81k
A V 2  g m2 R C 2 R G3   50. 8mS 4.7k 518k  237
AV3 
g m3 R E3 R L 

1.93mS 3. 0k 250 
1  g m3 R E 3 R L  1  1.93mS 3.0k 250
 0. 308 | A V  9. 032370. 308   659
vgs 3  v g 3 1  A v 3   A V1 A V 2 v s 1  A v3   0. 2V GS3  V TN3 | v s 
0.21.93 
9.03 237 1  0.308 
3
 261V
15.7
*Problem 15.7/15.8 - Multistage Amplifier
VCC 12 0 DC 15
VS 1 0 AC 1
RS 1 2 10K
*For output resistance
*VS 1 0 AC 0
*VO 11 0 AC 1
C1 2 3 22U
R1 12 3 820K
R2 3 0 100K
Q1 5 3 4 NBJT
RE1 4 0 2K
C2 4 0 22U
RC1 12 5 18K
C3 5 6 22U
R3 12 6 160K
R4 6 0 43K
Q2 8 6 7 NBJT
RC 12 8 4.7K
RE2 7 0 1.6K
C4 7 0 22U
C5 8 9 22U
R5 12 9 910K
R6 9 0 1.2MEG
M3 12 9 10 10 NMOSFET
RE3 10 0 3K
C6 10 11 22U
RL 11 0 250
.OP
.AC LIN 1 3KHZ 3KHZ
.MODEL NMOSFET NMOS VTO=1 KP=.001 LAMBDA=0.02
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PRINT AC VM(3) VP(3) IM(VS) IP(VS) VM(11) VP(11) IM(C6) IP(C6)
.END
VM(3)
1
Results: A V  VM(11)  711 | R IN 
 8.29 k | R OUT 
 401 
IM(VS)
IM(C6)
15.8
R 1 R 2  100k 820k  89.1k
vth  v s
89.1k
 0.899v s | R th  89.1k 10k  8.99k
89.1k  10k
A V1  0.899

o1 R C1 R B 2 R I2

R th  r 1   o1  1R E1
| R I2  r 2   o2  1R E2  1.97k 1011.6k   164 k
R 3 R 4  43k 160k  33.9k | A V1  0.899
R 5 R 6  1. 2M 910k  518k | A V2  
AV3 
4

g m3 R E3 R L


1  g m3 R E 3 R L


10018k 33.9k 164k
8.99k  7.81k  1012k 

 o2 R C2 R G 3

r 2   o 2 1R E1
1.93mS 3. 0k 250 
1  1.93mS 3.0k 250

 4.51
100 4. 7k 518k 
1.97k  1011. 6k
 2.84
 0. 308 | A V  4.512.84 0.308   3. 95
5
15.9 Note that the dc equivalent circuits are identical for Q1 and Q2.
180k
V EQ 
15V  5.63V | R EQ  180k 300k  113 k
180k  300k
5.63  0.7 V
IB 
 2.31A | IC  100I B1  232A | I E  101I B1  234A
113 10120  k
V CE  15  2x10 4 I E  2x10 4 IC  5.71V
r 
100 0. 025V 
70  5.71V
 10.8k | r o 
 326k - Neglected
232A
232A
vth  v s
113k
 0.983v s | R th  113k 2k  1.97k
113k  2k
 o1 R I1 r  2 
10017k 10.8k
v1
 0. 983
 0.983
 3.02v s
vs
R th  r 1   o1  1R 5
1. 97k  10.8k  1012k
vo
vo
 g m2 R L | R L  100k 20k  16. 7k |
 40 232A 16.7k  154 v1
v1
v1
Av 
vo
 3.02 154   465 | R OUT  20k r o 2 20k
vs
R IN  R B1 r 1   o1 1R 5   113k 10.8k  1012k  73.8k
15.10 The ac equivalent circuit from Problem 15.9 becomes:




 o R I1 r 2   o2  1R 6
100 17k 10. 8k  10120k
v1
 0. 983
 0.983
 0.816
vs
R th  r 1   o1  1R 5
1. 97k  10.8k  10120k
100 16.7k
vo
 o2 R L


 0.822
v1
r 2   o2  1R 6
10. 8k  10120k
Av 
vo
 0.816 0.822   0.671 | The voltage gain is completely lost. | R OUT  20k
vs


R IN  113k r 1   o1  1R 5  113k 10.8k  10120k  107k
15.11
6
*Problem 15.11 - Multistage Amplifier
VCC 11 0 DC 15
VS 1 0 AC 1
RS 1 2 2K
*For output resistance
*VS 1 0 AC 0
*VO 10 0 AC 1
C1 2 3 10U
R1 3 0 180K
R2 11 3 300K
Q1 6 3 4 NBJT
RE1 4 5 2K
RE2 5 0 18K
C2 5 0 10U
RC1 11 6 20K
C3 6 7 10U
R3 7 0 180K
R4 11 7 300K
Q2 9 7 8 NBJT
RC2 11 9 20K
RE3 8 0 20K
C4 8 0 10U
C5 9 10 10U
RL 10 0 100K
.OP
.AC LIN 1 5KHZ 5KHZ
.MODEL NMOSFET NMOS VTO=1 KP=.001 LAMBDA=0.02
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PRINT AC VM(3) VP(3) IM(VS) IP(VS) VM(10) VP(10) IM(C5) IP(C5)
.END
VM(3)
1
A V  VM(10)  454 | R IN 
 74.7 k | R OUT 
 18.8 k
IM(VS)
IM(C5)
Results:
15.12
M1: Assume saturation: I DS 
0.05
V GS  2 2 and V GS  1800I D
2
0.05
V GS  2 2 or 45V 2GS  181V GS  180  0
2
 2. 22V, 1. 80V | V GS  1. 80 V and I DS = 1 mA
V GS  1800
V GS
V DS = 20  15000 0.001  1800(0.001) = 3.2 V > V GS  V TN
0.05
M2: Assume saturation: I DS 
V GS  2 2 and V GS  2500I D
2
0.05
V GS  2500
V GS  2 2 or 62.5V 2GS  251V GS  250  0
2
V GS  1.83 V and I DS = 0.723 mA
V DS = 20  2500 0.723mA  =18. 2 V > V GS  V TN


g m1  2 0. 050.001  10.0mS | g m2  2 0.05  7.23x10 4  8. 50mS
R IN  1800
1
1
 1800 100  94.7 | R OUT  2500
 2500 118  113
g m1
g m2
A V1  g m1 15k 1M  0. 01S14.8k  148
AV2  
g m2 2.5k 10k
1  g m2 2.5k 10k

8.5x10 3 2.5k 10k
1  8.5x10 3 2.5k 10k
 0.944
A V  A V1 A V 2 = +140
15.13 (a) RD1 = 750 
7
I DS1 does not depend upon RD1 and is unchanged.
V D1  15  750 I DS1  I B 2   15  750IDS1  15  750 0.005   11. 3V
V DS1  11. 3  1.00  10.3 V
15  0.7  11.3
 1.88 mA | V EC 2  11.3.7  1.87mA 4.7k  3.21V
1600
 12. 6A  I D1
IE 2 
IB 2
IE 3 
4700 I C 2  I B3   0. 7
3300

4700 IC 2  0.7 4.7k1.88mA   0.7

 2.47mA
3300
3300
I E3
 30. 9A  IC 2 | V CE 3  15  3300I E 3  6.86V
81
Q  Pt1: 5mA, 10.3V  | Q  Pt 2 : 1.88mA , 3.21V  | Q  Pt 3 : 2.47mA , 6.86 V 
IB 3 
(b) RD1 = 910 
V D1  15  910I D1  15  910 0. 005   10.5V | Q  Pt1 : 5mA , 9.45 V 
15  0.7  10.5
 2. 38 mA | V EC 2  10.5.7  2.36mA 4.7k  0.108 V
1600
Q2 is saturated! - The circuit will no longer operate as an amplifier.
IE 2 
V C 2  0.7 10. 5  0.7  0.1  0.7

 3.15mA
3300
3300
Q  Pt 3 : 3.15mA , 4.60V 
IE 3 
15.14
I C1  80
0.7  9 V
 325A | V EQ 2  9  9100I C1  6.04 V | R EQ 2  9.1k
100  8124  k
I C 2  80
9  0.7  6.04 V
 184A | VC1  9  9.1kI C1  I B2   6.06V
9.1  8112  k
V E1  9  IE1 24k  1.10V | V CE1  6.06  1.10   7.16V
V C2  9  I C2 43k  1.09V | V E2  V C1  0.7  6.76V | V EC 2  7. 85V
Q1: 325A,7.16 V 
Q 2 : 184A, 7.85V 
80
 6.15k
13.0mS
80

 10. 9k
7. 36mS
g m1  40325A   13.0mS | r 1 
g m2  40184A  7.36mS | r  2
100 k
 0.999v s | R th  100k 100  99.9
100 100k
vo  g m2 R C 2 v 2   7.36mS 43k v 2  317v2
vth  v s
v2   v th
Av 
15.15
8

80 R C1 r  2
R th  r 1
  0.999v
s
809.1k 10.9k 
99.9  6.15k
vo
 31763.5  2.01 x 10 4 or 86.1 dB
vs
 63.5v s
*Problem 15.15 - Two-stage Amplifier
VCC 8 0 DC 9
VEE 9 0 DC -9
VS 1 0 AC 1
RS 1 2 100
*For output resistance
*VS 1 0 AC 0
*IO 0 6 AC 1
C1 2 3 10U
RB 3 0 100K
Q1 5 3 4 NBJT
RE1 4 9 24K
C2 4 9 10U
RC1 8 5 9.1K
Q2 6 5 7 PBJT
RE2 8 7 12K
C3 7 8 10U
RC2 6 9 43K
*C4 6 10 10U - Not needed
.OP
.AC LIN 1 2.5KHZ 2.5KHZ
.MODEL NBJT NPN IS=1E-16 BF=80
.MODEL PBJT PNP IS=1E-16 BF=80
.PRINT AC VM(3) VP(3) IM(VS) IP(VS) VM(6) VP(6)
.END
VM(3)
Results: A V  VM(6)  1.86 x 104 | R IN 
 6.04 k | R OUT  VM(6)  43.0 k
IM(VS)
15.16
Assuming that M1 remains in saturation, then its drain current does not change
when the circuit is modified, and using the results from Prob. 15.12: I DS1 = 1 mA
and VDS1 = +5 V.
Assume that M2 is saturated:
0.05
2
I DS2 
V GS2  2 | V GS2  5  2500I D2 | Solving these
2
two simultaneous equations gives VGS2  1.67V and I DS2  2. 67mA .
15.17 dc equivalent circuit:
We assume saturation for J1 and forward-active region operation for Q2.
2
V EQ 2
2
 V

 18000I DS1 
I DS1  I DSS 1  GS1  | I DS1  0.005 1 
  I DS1  50A
V P 

1


3  0.7
V
 15  IDS1 240k  3.00V | R EQ 2  240k | IC 2  100
 215 A
240  1018. 2 k
V CE 2  15  8200I E 2  13. 2 V | Checking V DS1 : V D1  15  50  2.15 A240k  2.48 V
V DS1  2.48  50A 18k  1.58 V | V GS1  V P   50A 18k  1  0.1 V  M1 is saturated.
ac equivalent circuit:
9
1000.025 V 
g m1 
2
1
A V1 
v1
10
10
 6
g m1R L1   6
g R D1 r  2   o2  1R L 2
3
vs
10  10
10  10 3
5mA 50A  1.00mS | r 2 
6
6
215A

 11. 6k




R L 2  R E 2 R L 8. 2k 1k = 891 | A V1  1.00mS  240k 11.6k  101891  71. 4
AV2 
vo
1010.891k 

 0.886 | A V  71. 40.866   63.2 | R IN  1 M
v1
11.6k  1010. 891k
R OUT  R E 2
R th2  r 2
 o 2 1
 8.2k
240 k  11.6k
 8.2k 2.49k  1. 91 k
101
Note: R OUT and A V would be lower if ro1 were also included.
b  A V1 
6
v1
g m1R L1
10
 6
3
vs
1
 g m1R 5
10  10
A V1  0. 999
AV2 

 3.75
1.00mS  240k 11.6k  1010.891k
1  1.00mS 18k
vo
1010.891k

 0.886 | A V  3.75 0.866   3.25
v1
11.6k  1010. 891k
15.18
*Problem 15.18 - Two-stage Amplifier
VCC 8 0 DC 15
VS 1 0 AC 1
RS 1 2 1K
*For output resistance
*VS 1 0 AC 0
*VO 7 0 AC 1
C1 2 3 33U
RG 3 0 1MEG
J1 5 3 4 NJFET
RS1 4 0 18K
C2 4 0 33U
RD 8 5 240K
Q2 8 5 6 NBJT
RE 6 0 8.2K
C3 6 7 33U
RL 7 0 1K
.OP
.AC LIN 1 1KHZ 1KHZ
.MODEL NBJT NPN IS=1E-16 BF=100
.MODEL NJFET NJF BETA=0.005 VTO=-1
.PRINT AC VM(3) VP(3) IM(VS) IP(VS) VM(7) VP(7) IM(C3) IP(C3)
.END
VM(3)
1
Results: A V  VM(7)  63.1 | R IN 
 1. 00 M | R OUT 
 1. 91 k
IM(VS)
IM(C3)
15.19 Using the dc equivalent circuit:
10
240k
15V  10.0V | R TH  240k 120k  80.0k
240k  120k
15  10  0.7 V

 2.97 A | I c1  75I B1  223 A
80  76 18  k
V TH 
I B1
V E1  15 18000 I E1  10.9V | V C1  36000 IC1  8. 03 V | V EC 1  10.9  8.03  2.87 V
15  V SG2  8. 03
0. 004
| I SD2 
V SG2  4 2  V SG2  3.01 V, I SD2  1. 96 mA
5.1k
2
 15  5.1k1.96mA   5.00 V | V SG  V P  3.01  4  0. 99V | M 2 is saturated.
I SD2 
V SD2
Using the ac equivalent circuit:
r 1 
75 0.025V 
223A



 8.41k | g m 2  2 4 x10 3 1.96 x10 3  3.96mS
A v1 
v1
 o1R I
7536000 
 0.988
 0. 988
 284
vs
R th  r 1
988  8410
Av2 
vo
g m 2R L
3.96mS 0.836k  0.768


v1 1  g m2 R L 1  3. 96mS 0.836k
A v  0.768284   218 | R IN  R B r 1  7.61 k | R OUT  5100
1
 241 
g m2
(b) If the bypass capacitor is removed from the 18 k resistor, then
A v1 
v1
 o1R I
75 36000 
 0.988
 0. 988
 1.94
vs
R th  r 1   o1 118k
988  8410  76 18k
A v  0.7681.94   1.49 | Note that the input resistance will increase to
R IN  80k 8410  76 18k  75.6k
15.20
*Problem 15.20 - Two-stage Amplifier
VCC 8 0 DC 15
VS 1 0 AC 1
RS 1 2 1K
*For output resistance
*VS 1 0 AC 0
*VO 7 0 AC 1
C1 2 3 20U
R1 3 0 240K
R2 8 3 120K
Q1 4 3 5 PBJT
RE1 8 5 18K
C2 8 5 20U
RC 4 0 36K
M2 0 4 6 6 PFET
RS2 8 6 5.1K
C3 6 7 20U
RL 7 0 1K
.OP
11
.AC LIN 1 2KHZ 2KHZ
.MODEL PBJT PNP IS=1E-16 BF=75
.MODEL PFET PMOS KP=.004 VTO=4
.PRINT AC VM(3) VP(3) IM(VS) IP(VS) VM(7) VP(7) IM(C3) IP(C3)
.END
VM(3)
1
Results: A V  VM(7)  210 | R IN 
 7.92 k | R OUT 
 240 
IM(VS)
IM(C3)
15.21
a For  F  , I B1  0  I B2 | V BE1  0.7 V  V BE 2 | V C1  V BE 2  300kI B1  V BE1  1.40V
3  1. 4 V
0.7V
 4. 44A | I C1  4. 44 A | I C2  I E 2  IB 2 
 0  23.3 A
360 k
30k
Q1: 4. 44A ,1.40 V  | Q 2 : 23.3A, 2.30V 
I C1  I B 2 
b  Writing an equation for the current through the300 k resistor:
3V  360kI C1  I B 2   0.7V  0.7V
I E2  I B1 30k  0.7V
I B1 
and also I B1 
300k
300k
Solving these two simultaneous equations withI C1  75I B1 , I C 2  75I B 2 yields I B1  54.5nA,
I C1  4. 08A, I B 2  315nA, I C 2  23.6A | Q1: 4. 08A,1. 42V  | Q2 : 23.6A,2.28 V 
15.22
I C 2   F I E1   F
F  1
i
I C1 =  F  1I C1 | r 1 =  o  1r 2 | y11  1
F
v1


v1  i1 r 1  o1  1r 2 = 2r 1 | y 11 
0
v 1 0
i b2 
v2
v
v
r 2
v2
1
 2 | i1   2

| y12  
r o1 1  g m1r 2  2r o1
2r o1 r 1  r  2
2r o1  o1  1
2 o1r o1
y 21 
i2
v1
y 22 
i c1 
y 22 
15.23
12
1
i
| y12  1
2r 1
v2
v2  0
i2
v2

v 2 0
iC 2
v1

v 2 0
| i2 
v1 0
 o1 1r  2
g
g m2 = m2
r 1   o1  1r  2
2
v2
v
 1
v
  o 2i e1  i c1  2   o 2 o1
i c1  i c1  2   o2 i c1
ro 2
ro 2
 o1
r o2
v2
v2
v2
v2









2r
 R
 r
 o2 r  2
o1
r o1 1  o1 E  r o1 1  o 2 2  r o1 1 

 r 1  R E 
 r 1  r 2 


2
r
 2 
  o2
1

1
1
3
 o2 


r o2 2r o1 r o2 2r o2 2r o2
y11 
i1
v1

v 2 0
1
i
| y 12  1
r 1
v2
 0 | y 21 
v1 0
i2
v1
 g m1
v 2 0
g m2
g m2  g o 2
 g o2  g 2  g o1
1
f 2
 g m1
 g m1
1
1
1
1


F2  o2  f 1
1
I C 2   F IC1  IC1  g m2  g m1 | g o2  g o1 | y 21
y12 
i2
v2

v1  0
1

 R 
r o2 1  o2 E 
 r  2  R E 

1


 r
r o 2 1  o 2 o1 
 r 2  r o1 

1
r o2  o2  1
15.24
y11 
y12 
15.25
i1
v1
i2
v2
 0 | y12 
v 2 0

v1  0
i1
v2
 0 | y 21 
v1 0
i2
v1
 g m1 1 = g m1
v 2 0
1
1

| I DS2  I DS1
r o2 1  g m2 r o1   f 2r o1
I C 2   F2 I C1  g m2 =  o2 g m1
'
gm 
i2
v1
 g m1  o2  1  g m2
v2  0
r '  r 1 
r 'o  r o 2
 o2  1
 g m2  g m
 o2
 o1

 o1  o2   o1r 2   or  |  'o  'o1  o 2 1  'o1 o2
g m1 g m2
r o1  r  2
 o2  1
 ro 2
 o2 r o 2  r  2
 o2  1
 r o2 r o 2 

ro
r
|  'f  g m o  f
2
2
2
15.26
5
A V  G mR o  o2 gm1 o 2r o2   gm2 o2 r o2   o 2 f2  100 4075   3 x 10 !
15.27
Assume IC 2   F2 IC1  g m2 =  o 2 g m1 | I 'C = I C 2 = I C | I 'B  IB1 
g 'm   o2 g m1   o 2
IC 2
 F1 F2
g m2
 g m2  g m | r ' = r 1 =  or  |  'o  g m  or    2o
o 2
r 'o  r o 2 | 'f  g m r o  f
15.28
For forward - active region operation, we require VCB  0. V CB  V GS  I B R B 
V CB   V GS 
I EE
I
R B | For the JFET , V GS  0  V CB  EE R B . So the BJT will be
 F 1
F  1
in the forward active region for as long as IEE  I DSS  1mA or 0  I EE  1 mA
15.29
Dc equivalent circuit (vS = 0):
13
Q1: I C1   F1 IB1  100
1.5V  0. 7V
 8.52 A
200k  10191k


V CE1  V C1  V E1  0.7  1.5V  91kI E1   1.42 V | Q - point: 8.52A,1. 42V 
Q2 : I E 2 
1.5V  0.7V
100
 I C1  17. 0  8.52  8.48 A | I C2   F 2I E 2 
8.48 A = 8.40 A
47k
101


V EC 2  V E 2  V C 2  0.7  1.5  8.40A 1.5x10
 v v 
v c1
A V   o  c1  |
vs
v c1  v s 
5
 0. 940V | Q - point: 8. 40A ,0.940 V 
Ac equivalent circuit



1 
1
 g m1 47k
  0. 954
  40 8.52A 47k
g m2 
40 8. 40A 


vo
 g m2R L  40 8. 40A  1.5x10 5  50.4
vc1


|
A V  48.1
The two stage common-emitter/common-base cascade is usually called a cascode amplifier.
15.30
1 F 12  V BE 1 100 12  0.7

 20. 7 A | V C  12  3. 3x105 I C  5.17 V
a I C   F I E 
5
2  F 1
R EE
2 101 2.7x10
V CE  V C  0.7 V  5.87V | Q  Point  20.7A, 5.87V 
b  A dd  g m R C  40 20.7A 330k  273
o V T
1000.025 V 
R ID  2r   2
c A cc  
A dd  
R IC 
15.31
a I C   F I E 
2
IC
20.7A
 243 k | R OD  2R C  660 k
o R C
100330k

 0. 604
r    o  12R EE
122k  2101270k
gm RC
137
 137 | A cd  A cc | CMRR =
 227
2
0.604
r    o 12R EE
2

122k  2 101270k
 27.3 M
2
1 F 12  V BE 1 100 12  0.7

 20. 7 A
2  F 1 R EE
2 101 2.7x105
V C1  V C2  12  3.9x10 5 I C  3.93V | V CE  V C  0.7V   4.63V
Q  Point  20. 7A, 4.63V  | r  
A cc  
 oR C
r    o  12R EE

100 0.025 V 
 121k
20.7 A
100 390k
121k  101540k
 0.714 | v ic 
v C1  v C2  3.93  A cc v ic  3.93  0.714 5   0.360 V
14
5.000  5.000
 5.00 V
2
b  I C   F I E 
1  F 5V  V BE  12V  1 100 17 V  0.7V

 29.9 A
2 F  1
R EE
2 101 2.7x10 5
V C1  V C2  12  3.9x10 5 I C  0.339 V | Part (a ) is in error by 0.021 V
c The common - mode signal voltage applied to the base- emitter junction is
V be  V IC
r
121k
5
 11.1 mV  5mV.
r    o 12R EE
121k  101540k
A common - mode input voltage of 5 volts exceeds the small- signal limit.
15.32
a I E 
1 1. 5  0.7 V
60
 5.33A | I C   F I E 
I E  5.25A
2 75x10 3 
61
V CE  1.5  10 IC  0.7   1. 68V | Q - Pt: 5.25A ,1.68V 
60
 40IC  0.210mS | r  
 286k | A dd  g m R C  0.210mS 100k  21.0
gm
5
b g m
A cc  
 oR C
60100k

 0.636
r    o  12R EE
286k  61150k
For differential output: CMRR =
21.0

0
21.0
2
For single - ended output: CMRR =
 16.5, a paltry 24. 4 dB!
0.636
R ID  2r   572k | R IC 
r    o 12R EE
R OD  2R C  200 k | R OC
2
RC

 50k
2

286  61150 
k  4.72 M
2
15.33
*Problem 15.33
VCC 2 0 DC 12
VEE 1 0 DC -12
VIC 8 0 DC 0
VID1 4 8 AC 0.5
VID2 6 8 AC -0.5
RC1 2 3 330K
RC2 2 7 330K
Q1 3 4 5 NBJT
Q2 7 6 5 NBJT
REE 5 1 270K
.MODEL NBJT NPN BF=100 VA=60 IS=1FA
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7)
.TF V(7) VIC
.END
1
Results: A dd  VM(3,7)  241 | R ID 
 269 k | A cc = -0.602 | R IC  23. 2 M
IM(VID1)
15.34 (a)
15
IE 
1.5V  0.7V

2 4.7x10 4 

 8. 51A | I C   F I E 
100
I  8.43A
101 E
V CE  1.5  10 IC  0.7   1. 36V | Q - point: 8. 43A,1.36V 
100
 40I C  0. 337mS | r  
 297k | A dd  g m R C  0. 337mS 100k  33.7
gm
5
gm
A cc  
 oR C
100 100k

 1.02
r    o  12R EE
297k  10194k
For differential output: CMRR =
33. 7

0
33.7
2
For single - ended output: CMRR =
 16.5, a paltry 24.4 dB!
1. 02
R ID  2r   594 k | R IC 
r    o  12R EE
R OD  2R C  200 k | R OC 
2

297  10194 
2
k  4. 90 M
RC
 50 k
2
15.35 We should first check the feasibility of the design using the Rule-of-Thumb estimates
similar to those developed in Chapter 13 (Eq. 13.59):
The required A dd  200 46 db . Assuming we drop half the power supply voltage across R C :
V CC
 20 V CC  240. Thus, a gain of 200 appears feasible.
2
 V
100 0. 025V 
 2r   1M  r   500k | I C  o T 
 5. 00 A
r
500k
A dd  g mR C  40IC R L  40
R ID
IE 
I C 101
V  V BE 12  0.7V

I C  5.05A | R EE  EE

 1.12M
 F 100
2I E
25.05A 
A dd   g m R C  200 46dB  | R C 
200
200

 1.00M
gm
40 5x10 6


Checking the collector voltage: V C  12  990k5A   7V | Picking the closest 5% values
from the table in the Appendix: R EE  1.1 M and R C  1 M are the final design values.
These values give IC  5.09A and A dd  204 46. 2dB 
15.36 We should first check the feasibility of the design using the Rule-of-Thumb estimates
similar to those developed in Chapter 13 (Eq. 13.59):
The required Add  794 58 db. (This sounds pretty large - a significant fraction
of the BJT amplification factor  f .) Even assuming we choose to drop all of the positive
power supply voltage across R C (which provides no common - mode input range):
A dd  g m R C  40IC R L  40V CC  409   360.
Thus, a gain of 794 is not feasible with this topology!
16
15.37
I EE 100 400A

 198A | V CE  12  3. 9x10 4 IC  0.7   4.98V
2
101
2
100
Q - point: 198A, 4.98 V  | g m  40I C  7. 92mS | r  
 12.6k
gm
IC  FIE  F
A dd   g m R C  7.92mS 39k  309
A cc  
 oR C
100 39k

 0.0965
r    o  12R EE
12. 6k  101400k 
For differential output: CMRR =
309

0
309
2
For single - ended output: CMRR =
 1600 or
0.0965
64.1 dB
r    o  12R EE
12.6k  101400k 

k  20.2 M
2
2
(Note that this value is approaching the  o r o limit and hence is not really correct. )
R ID  2r   25.2 k | R IC 
R OD  2R C  78.0 k | R OC 
15.38
RC
 19. 5 k
2
I EE
75 400A
4

 197A | V C1  V C2  12  3.9x10 I C  4.32V
2
76
2
 4.32  0.7   5. 02V | Q - point: 197A,5. 02V 
IC  FIE  F
V CE
g m  40I C  7.88mS | r  
A cc  
75
 9.52k | A dd  g m R C  7.88mS 39k  307
gm
 oR C
75 39k

 0. 0962
r    o  12R EE
9. 52k  76400k
2.005  1.995
 2.00 V
2
v
0.01V
v C1  V C1  Add id  Acc v ic  4.32V  307
 0. 0962 2V   2.593 V
2
2
v
0. 01V
v C2  V C 2  A dd id  A cc v ic  4. 32V  307
 0.0962 2V   5.663 V
2
2
v OD  2.593  5.663  3.07 V
v id  2.005  1.995  0. 01V | v ic 
V CB  V C1  A cc V IC  V IC  0 | V IC 
15.39
R ID  2r  
4.32
 3.94 V
1  0.0962
2100 0.025V 
2 o V T
I
101
 IC
 1.00A | I EE  2 C  2
1A   2. 02 A
IC
5M
F
100
5
CMRR  g mR EE  10  R EE 
10 5
 2.5 G !
401.00A
17
15.40
I EE 100 20A
a I C   F I E   F

 9. 90A | V C 2  10  9.1x10 5 I C  0. 991V
2
101 2
g m  40I C  0. 396mS | A dd  g m R C  0.396mS 910k  360 | R EE =   A cc  0
v ID
 Acc v IC | For v s  0: v C 2  V C 2  0.991
2
For v s  2mV: v C2  0.991V  360 0.001V   00.001V   1. 35 V
v C2  V C 2  A dd
b  v s 
15.41
0.991V
 5.51 mV
180
I EE 120 200A

 99.2A | V O  10  1.10x10 5 IC  1.09 V
2
121
2
 40I C  3. 97mS | A dd  g m R C  3. 97mS 110k  437 | R EE =   A cc  0
a I C   F I E   F
gm
v id
=0
2
0.001V
For v s  1mV: V O  1.09 V and v o =  437 
= 0.219 V
2
V
1. 09
 5.00 mV
b  For v CB  0, v s  AO  2
437
 dd
2
For v s  0: V O  1. 09 V and v o = A dd
15.42
*Problem 15.42
VCC 2 0 DC 12
VEE 1 0 DC -12
V1 3 7 AC 1
V2 5 7 AC 0
VIC 7 0 DC 0
RC 2 6 110K
Q1 2 3 4 NBJT
Q2 6 5 4 NBJT
IEE 4 1 DC 200U
.MODEL NBJT NPN VA=60V BF=120
.OP
.AC LIN 1 1KHz 1KHZ
.PRINT AC IM(V1) IP(V1) VM(6) VP(6)
.TF V(6) VIC
.END
1
Results: A dd  VM(6)  193 | R ID 
 82.0 k | A cc = +0.0123 | R IC  45.8 M
IM(V1)
15.43
18
IC  FIE 


150 15  0.7
 47.4A | V EC  0.7  15  2x10 5 I C  6. 22V
151 2150k 
Q - points: 47.4A,6.22 V  | g m  40I C  1. 90mS | r  
150
 79. 0k
gm
A dd  g m R C  1.90mS 200k  380
A cc  
150200k
 oR C

 0.661
r    o  12R EE
79. 0k  151300k
R ID  2r   158k | R IC 
r    o 12R EE
2

79.0k  151300k
For a differential output: A dm  A dd  380 | A cm
For a single - ended output: A dm 
CMRR =
 22.7 M
2
 0 | CMRR = 
A dd
 190 | A cm  A cc  0.661
2
190
 287 or 49.2dB
0.661
15.44
IC  FIE 
100 10  0.7
 10.7A | V C1  VC 2  10  5. 6x105 I C  4.01V
101 2430k
V EC  0.7  4.01  4.71V | g m  40I C  0. 428mS | r  
100
 234k
gm
A dd   g m R C  0.428mS 560k  240
A cc  
100 560k
 oR C

 0.643
r    o  12R EE
234k  101860k
1  0.99
 0.995 V
2
v
0. 01V
v C1  V C1  Add id  Acc v ic  4. 01V  240
 0.643 0.995V   5.850 V
2
2
v
0.01V
v C2  V C 2  A dd id  A cc v ic  4.01V  240
 0.643 0. 995V   3. 450 V
2
2
5. 850  3. 450
v OD  5.850  3.450   2. 40 V | Note: A dd v id  2.40V and v OC 
 4.65
2
Also note: v OC  V C  A cc v ic  4.01  0.643 0.995V   4.65V
v id  1  0.99  0.01V | v ic 
15.45
*Problem 15.45
VCC 2 0 DC 10
VEE 1 0 DC -10
V1 4 8 AC 1
V2 6 8 AC 0
VIC 8 0 DC 0
RC1 5 1 560K
RC2 7 1 560K
Q1 5 4 3 PBJT
Q2 7 6 3 PBJT
REE 2 3 430K
.MODEL PBJT PNP VA=60V BF=100
.OP
19
.AC LIN 1 5KHz 5KHZ
.PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7)
.TF V(7) VIC
.END
1
A dd  VM(5,7)  213 | R ID 
 511 k
IM(V1)
Results:
213
A cc = 0.642 | R IC  37.5 M | CMRR =
 332  50.4dB
0.642
15.46


IEE
80 10A

 4.94A | V EC  0.7  3  3.9x10 5 IC  1.77V
2
81 2
80
Q - points: 4.94A,1.77 V  | g m  40I C  0.198mS | r  
 404 k
gm
IC  F
A dd  g m R C  0.198mS 390k   77. 2
A cc  
 oR C
80390k

 0.0385
r    o  12R EE
404k  8110M
R ID  2r   808k | R IC 
r    o  12R EE

808k  8110M 
 405 M
2
2
 or o
Note that RIC is similar to
so that RIC = 405 M will not be fully acheived.
2
 r
80
80
For example, if V A  80 V, o o 
 648 M
2
2 4. 94A
For a differential output: A dm  A dd  77.2 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
A dd
 38. 6 | A cm  A cc  0.661
2
38.6
 1000 or 60.0dB | V BC  0 requires VIC  V C  1. 07V and
0. 0385
Without detailed knowledge of EIE , we can only estimate that VIC should not exceed
V IC + 0.7  V CC  0.7V which allows 0.7V for biasing I EE  1.07 V  V IC  1.6V.
15.47
IEE 120 1mA

 496A | V C1  V C2  22  1.5x10 4 IC  14.6 V
2
121 2
120
Checking V EC  0.7  14.6   15. 3V | g m  40I C  19. 8mS | r  
 6.06k
gm
IC  F
A dd   g m R C  19.8mS 15k  297
A cc  
12015k
 oR C

 0.0149
r    o  12R EE
6.06k  1211M
v id  0.01  0  0.01V | v ic 
20
0. 01  0
 0. 005V
2
v id
0. 01V
 Acc v ic  14. 6V  297
 0.0149 0.005 V  16.09 V
2
2
v
0.01V
v C2  V C 2  A dd id  A cc v ic  14.6V  297
 0.0149 0. 005V   13.12 V
2
2
v OD  16. 09  13.12   2.97 V | Note: Add v id  2.97 V
v C1  V C1  Add
16.09  13.12
 14.6 and
2
 V C  A cc v ic  14.60  0. 01490.005 V   14. 6V
Also note: v OC 
v OC
15.48
IC  FIE 
A dd 
v od
R 
R 


| v od  v c1  v c2  i c1 R 
  i c2 R 



v id
2 
2 
vod  gm
A cd 
100 15V  0.7V
100
 70.8A | g m  40I C  2. 83mS | r  
 35. 3k
101 2100k 
gm
v id
2
R  R   gm v id R  R   g R v | A  g R  283

 


m
id
dd
m

2  
2 
2 
vod

R 

R 
| vod  v c1  v c2  i c1 R 
 i c 2 R 



v ic
2 
2 
For common - mode input, i c1  i c 2 
o
v ic
r   o  12R EE
vod  
o

 oR
R  
R 
vic R 
 R 
  v od  
v ic
r    o  12R EE
2  
2 
r    o  12R EE


A cd  
R
 oR
100100k
 0. 01
 . 00494
R r    o  12R EE
35.3k  101200k
CMRR 
15.49
283
 57300 or 95.2 dB
0. 00494
*Problem 15.49
VCC 2 0 DC 15
VEE 1 0 DC -15
V1 4 8 AC 0.5
V2 6 8 AC -0.5
VIC 8 0 DC 0
RC1 2 5 100.5K
RC2 2 7 99.5K
Q1 5 4 3 NBJT
Q2 7 6 3 NBJT
REE 3 1 100K
.MODEL NBJT NPN BF=100
.OP
.AC LIN 1 100 100
.PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7)
.TF V(5,7) VIC
.END
Results: A dd  VM(5,7)  274 | A cd = 0.00494 | CMRR  55500 or 94.9 dB
15.50
21
For a differential - mode input:
g m 
g m 
v

 v id

vod    id  v e g m 
 ve g m 
R  
R   g m R v id  g m Rv e
 2

2 
 2

2 


g m
vod  g m R v id 
ve  | At the emitter node:
gm


 vid

g m
 v

g m

 v e 
g m 
 g  
   id  ve 
g m 
 g  
 GEE v e  0
 2

  2


2
2
g

R
g
1
1
m
o EE
m
ve 
v id 
vid  v id | v od   g m R vid | A dd  g m R  300
2 g m r   o  12R EE
4 gm
For a common - mode input:
g m 
g m 
g m
vod   v ic  v e 
g m 
R  v ic  v e 
g m 
R  g m R v ic  v e   
g R vic  ve 




2
2
gm m
At the emitter node:
v ic
ve 
g m
g m
 v e 
 g  gm 
 g  
g m 
  GEE v e  0

2
2

 o  12R EE
r    o  12R EE
A cd 
v ic | v ic  v e 
r
v ic
r    o  12R EE
vod
g m
 oR
g m
g mR
g m
g mR



 0.00499
v ic
g m r    o  12R EE
g m 1  2g m R EE
g m 1  2g m R EE
1
g 
CMRR  2g m R EE  m   60000 or 95. 6 dB
 g m 
22
The MATLAB m-file listed below 'FET Bias' can be used to help find the drain currents in the
FET circuits in Problems 15.51 - 15. Use fzero('FET Bias',0) to find ids.
function f=bias(ids)
kn=4e-4; vto=1; gamma=0.0;
rss=62e3; vss=15;
vsb=2*ids*rss;
vtn=vto+gamma*(sqrt(vsb+0.6)-sqrt(0.6));
f=vss-vtn-sqrt(2*ids/kn)-vsb;
15.51
This solution made use of the m-file above. The solution to Problem 15.52 gives an
example of direct hand calculation.
2I DS
2IDS
V SS  V GS  2I DSSR SS | V GS  V TN 
| V SS  2I DSSR SS  V TN 
Kn
Kn


15  2I DSS 62x10 3  1 
2I DS
 I DS  107A | V GS  V TN  0.731V
4x10 4
V DS  15  62kI DS  V GS   10.1V  0.731V - Saturated | Q - pt: 107A,10.1V 
gm 
2107A 
2I DS

 0.293mS | A dd   g m R D   0.293mS 62k  18.2
V GS  V TN
0.731V
A cc  
gm R D
0.293mS 62k

 0. 487
1  2g m R SS
1  20.293mS 62k
For a differential output: A dm  Add  18.2 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
15.52
2IS 
A dd
 9.10 | A cm  Acc  0. 487
2
9.10
 18.7 | CMRR db  25.4 dB | R ID   | R IC  
0.487
12  V GS
K
12  VGS
A
2
 2 n V GS  V TN  
and for K n  400 2 and V TN  1V
220k
2
220k
V


2
2
12  V GS  88 V GS  2V GS  1 or 88 V GS  175 V GS  76  0 and V GS  1.348 V
ID  IS 
1 12  1.35 
5

  24.2A. V D  12  3. 3x10 I D   4. 01V
2  220k 
V DS  4.01   V GS   5.36V
gm 

2 24.2x10
2I D

V GS  V TN
0.348
A cc  
6
> V GS  V TN .
  1. 39mS |
Q - Point = 24.2A,5.36 V 
Add  g m R D  1.39ms 330k  45. 9
1. 39ms 330k
gm R D

 0.738
1  2g m R SS 1  21.39ms 220k
For a differential output: A dm  Add  45.9 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
A dd
 23.0 | A cm  A cc  0.738
2
23.0
 31. 2 | CMRR db  29.8 dB | R ID   | R IC  
0.738
23
15.53
*Problem 15.53
VCC 2 0 DC 12
VEE 1 0 DC -12
VIC 8 0 DC 0
VID1 4 8 AC 0.5
VID2 6 8 AC -0.5
RD1 2 3 330K
RD2 2 7 330K
M1 3 4 5 5 NFET
M2 7 6 5 5 NFET
REE 5 1 220K
.MODEL NFET NMOS KP=400U VTO=1
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7)
.TF V(7) VIC
.END
Results: A dd  VM(3,7)  45.9 | A cc = 0.738 | CMRR  31.1 | RID   | RIC  
15.54
R OD  2R D  5k  R D  2.5k | Selecting closest 5% value: R D  2.4k
20
A dd   g m R D  10 20  10 | g m 
4.17x10 

2 25x10 
10
 4.17mS  2K n I DS
2400
3 2
I DS
R SS 
15.55
24
3
 348A | V GS  V TN 
2I DS
Kn
 1.16V
V SS  V GS
5  1.16

 5.52k | Selecting closest 5% value: R SS  5.6k
2R SS
2348A 
This solution made use of the m-file above Problem 15.51.
V SS  V GS  2I DSSR SS | V GS  V TN 
V TN  V TO  
V
SB

2I DS
| V SS  2I DSSR SS  V TN 
Kn
 0.6  0. 6  V TO  
 2I
DSSR SS
Solving iteratively with R SS  62k | K n  400
I DS  91. 3A | V GS  V TN  0.676 V | V TN
 0.6  0.6

2IDS
Kn
A
| V TO  1V |  = 0.75 V yields
2
V
 3.01V | V GS  3.69V
V DS  15  62kI DS  V GS   12.9V  0.676 V - Saturated | Q - pt: 91.3A ,12.9V 
gm 
2I DS
291.3A

 0.270mS | A dd  g m R D   0.270mS 62k  16.7
V GS  V TN
0.676V
A cc  
g mR D
0.270mS 62k

 0. 486 assuming  = 0
1  2g m 1  R SS
1  20.270mS 62k 
For a differential output: A dm  Add  16.7 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
15.56
15.57
A dd
 8.35 | A cm  A cc  0. 486
2
8.35
 17. 2 | CMRR db  24.7 dB | R ID   | R IC  
0.486
*Problem 15.56
VCC 2 0 DC 15
VEE 1 0 DC -15
VIC 8 0 DC 0
VID1 4 8 AC 0.5
VID2 6 8 AC -0.5
RD1 2 3 62K
RD2 2 7 62K
M1 3 4 5 1 NFET
M2 7 6 5 1 NFET
REE 5 1 62K
.MODEL NFET NMOS KP=400U VTO=1 PHI=0.6 GAMMA=0.75
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(VID1) IP(VID1) VM(3,7) VP(3,7)
.TF V(7) VIC
.END
Results: A dd  VM(3,7)  16.8 | A cc = 0.439 | CMRR  25. 6 | R ID   | RIC  
This solution made use of the m-file above Problem 15.51.
25
V SS  V GS  2I DSSR SS | V GS  V TN 
V TN  V TO  
V
SB

2I DS
| V SS  2I DSSR SS  V TN 
Kn
 0.6  0. 6  V TO  
 2I
DSSR SS

A
Solving iteratively with R SS  220k | K n  400
I DS  20.3A | V GS  V TN
 0.6  0.6
2IDS
Kn
| V TO  1V |  = 0.75 V yields
2
V
 0.319 V | V TN  2.74V | V GS  3. 05V
V DS  12  330kIDS   V GS   8.35 V  0. 319V - Saturated | Q - pt: 20. 3A,8.35 V 
gm 
2I DS
220.3A 

 0.127mS | A dd   g m R D   0.127mS 330k  41.9
V GS  V TN
0.319 V
A cc  
g mR D
0.127mS 330k

 0.737 assuming  = 0
1  2g m 1  R SS
1  20.127mS 220k 
For a differential output: A dm  Add  41.9 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
A dd
 21. 0 | A cm  A cc  0.737
2
21.0
 28. 4 | CMRR db  29.1 dB | R ID   | R IC  
0.737
15.58
I DS 
I SS
 150A | V GS  V TN 
2
2I DS
=1 
Kn

2 1.5x10 4
4x10 4
 = 1.866 V | V
GS
 V TN  0.866 V
V DS  15  75kI DS  V GS   5. 62V  0.866V - Saturated | Q - pt : 150A,5. 62V 
gm 
2150A 
2I DS

 0.346mS | A dd   g m R D   0. 346mS 75k  26. 0
V GS  V TN
0. 866V
A cc  
gm R D
0.346mS 75k

 0.232
1  2g m R SS
1  20.346mS 160k
For a differential output: A dm  Add  26.0 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
15.59
26
A dd
 13.0 | A cm  A cc  0. 232
2
13.0
 56. 0 | CMRR db  35.0 dB | R ID   | R IC  
0.232
I DS 
I SS
 20A | V GS  V TN 
2
2I DS
=1
Kn

2 2x10
4x10
5
4
 =1. 316V | V
GS
 V TN  0.316 V
V DS  9  300kIDS   V GS   4. 32V  0.316 V - Saturated | Q - pt : 20A, 4.32V 
gm 
2I DS
220A 

 0.127mS | A dd  g m R D   0.127mS 300k  38. 0
V GS  V TN 0.316 V
A cc  
gm R D
0.127mS300k

 0.120
1  2g m R SS
1  20.127mS 1.25M
For a differential output: A dm  Add  38.0 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
A dd
 19.0 | A cm  A cc  0.120
2
19.0
 158 | CMRR db  44.0 dB | R ID   | R IC  
0.120
15.60
I DS =
I SS
=150A | V GS  V TN 
2
V TN  V TO  
V
SB

2I DS
= V TN 
Kn

2 1.5x10 4
4x10
 15  V
0.6  V
 0.6  0. 6  V TO  

V GS  0.866  1  0.75 15  V GS  0.6 
4
= V
TN
 0. 866V

GS
 0.6  0.6
GS
 3.86V | V TN  2.99 V
V DS  15  75kI DS  V GS   7. 61V  0.866 V - Saturated | Q - pt: 150A ,7.61V 
gm 
2150A 
2I DS

 0.346mS | A dd   g m R D   0. 346mS 75k   26. 0
V GS  V TN
0. 866V
A cc  
g mR D
0.346mS 75k

 0.233 assuming  = 0
1  2g m 1  R SS
1  20.346mS 160k
For a differential output: A dm  Add  26.0 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
A dd
 13.0 | A cm  A cc  0. 233
2
13.0
 55.8 | CMRR db  34.9 dB | R ID   | R IC  
0.233
15.61
I DS
I
= SS = 20A | V GS  V TN 
2
V TN  V TO  
V
SB
2I DS
= V TN 
Kn

 9V
GS
 0.6 
4x10
 9 V
0.6  V
 0.6  0. 6  1  0.75
V GS  0.316  1  0.75

2 2x10 5
GS
GS
4
= V
TN
 0.6  0.6
 0. 316V

 2.71V | V TN  2.39 V
27
V DS  9  300kIDS   V GS   5.71V  0.316 V - Saturated | Q - pt: 20A,5.71V 
220A 
2I DS

 0.127mS | A dd  g m R D   0.127mS 300k  38.1
V GS  V TN 0.316 V
gm 
A cc  
g mR D
0.127mS300k  0.120 assuming  = 0

1  2g m 1  R SS
1  20.127mS 1.25M
For a differential output: A dm  Add  38.1 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
A dd
 19.0 | A cm  A cc  0.120
2
19.0
 158 | CMRR db  44.0 dB | R ID   | R IC  
0.120
15.62
30
A dd   g m R D  10 20  31.6 | g m R D  31. 6 
2I DS R D
V GS  V TN
Maximum common - mode range requires minimum IDS R D  minimum V GS  V TN
Choosing V GS  V TN  0.25V to insure strong inversion operation,
I DS R D 
0. 2531.6 
 3. 95V | 0. 25V =
2
I SS  2I DS  312 A | R D 
15.63
I SD 
2
2I DS
0.25  0.005 
 I DS 
 156A
Kn
2
3. 95V
 25.3k  27k, the nearest 5% value.
156A
K
18  V SG
1 18  V SG 
2
A

  n V SG  V TP  
and for K n  200 2 and V TP  1V
2  56k 
2
112k
V
18  V SG  11. 2V SG  1  V SG  2.19V | V SG  V TP =1.19V | I SD  142A
2
V SD  V SG  91k ISD  18  7.27V  1.19V  Saturated| Q - Point = 142A ,7.27V 



g m  2 2x10 4 1. 42x10 4  0. 238mS | Add  g m R D  0.238mS 91k   21.7
A cc  
0.238mS 91k 
gm R D

 0.785
1  2g m R SS 1  20.238mS 56k
For a differential output: A dm  Add  21.7 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
15.64
28
A dd
 10. 9 | A cm  A cc  0.785
2
10.9
 13.9 | CMRR db  22. 9 dB | R ID   | R IC  
0.785
*Problem 15.64
VCC 2 0 DC 18
VEE 1 0 DC -18
VIC 8 0 DC 0
V1 4 8 AC 0.5
V2 6 8 AC -0.5
RD1 5 1 91K
RD2 7 1 91K
M1 5 4 3 3 PFET
M2 7 6 3 3 PFET
REE 2 3 56K
.MODEL PFET PMOS KP=200U VTO=-1
.OP
.AC LIN 1 3KHZ 3KHZ
.PRINT AC IM(V1) IP(V1) VM(5,7) VP(5,7)
.TF V(7) VIC
.END
Results: A dd  VM(5,7)  21.6 | A cc = 0.783 | CMRR  13.8 | R ID   | R IC  
15.65
I SD =
ISS
= 20A | V SG  V TP 
2
V TP  V TO  
V
BS
2I SD
=  V TP 
Kp


2 2x10 5
2x10 4
 = V

TP
 0.6  0.6  1  0.6 10  V SG  0. 6  0. 6

 0.447 V


V SG  0.447  1  0.6 10.6  V SG  0.6  V SG  2. 67V | V TP  2.23V
V SD  V SG  10  300k I SG   6.67 V  0. 447V - Saturated | Q - pt: 20A,6. 67V 

g m  2 2x10
A cc  
5
2x10   89.4S |
4
A dd  g m R D   89.4S 300k  26.8
g mR D
89.4S300k

 0.119 assuming  = 0
1  2g m 1  R SS
1  289.4S 1. 25M 
For a differential output: A dm  Add  26.8 | A cm  0 | CMRR = 
For a single - ended output: A dm 
CMRR =
A dd
 13. 4 | A cm  A cc  0.119
2
13.4
 113 | CMRR db  41.0 dB | R ID   | R IC  
0.119
15.66
Note: V SS and V DD should be 12V and V P  2V
ISS
= 10A | V O  12  820kI SD  3.80 V | For v S  0, v O  V O  3.80 V
2
 I

 10A

 V P  SD  1  2 
 1  1. 8V | V SG  V P  0.2V | V SD  0. 2V for pinchoff
 I DSS

 1mA

a I SD =
V SG
So V D  2 for pinchoff.
gm 
1
2V
21mA 10A   70.7S | A dd  g m R D   70.7S 820k  58.0
A cc  0 for R SS and r o   | v O  V O 
A dd
58.0
v id  3.80 
0.02  0  1.22V
2
2
v1
 0.2 V SG  V P   0.2 0.2   40.0mV | v1  80.0 mV from small - signal limit
2
A
Also v O  V O  dd v id  3.80  29. 0v id  2 V for pinchoff  v id  62.1mV | v1  62.1mV
2
29
15.67
150 22  0.7
 52.6 A | V CE  22  200kI C  0. 7  12.2 V
151 402k
1500.025 V 
Q  Po int  52.6A,12.2V  for both transistors | r  
 71. 3k
52.6A
a I C   F I E 
b  A cc  
A dd  
o R C
150200k

 0. 494
r   o  1R 1  2R EE 
71. 3k  151402k
150200k
o R C

 80. 4
r    o 1R1
71. 3k  1512k


R ID  2 r    o  1R 1  271.3k  1512k  747k
15.68
*Problem 15.68
VCC 2 0 DC 22
VEE 1 0 DC -22
VIC 10 0 DC 0
V1 4 10 AC 0.5
V2 8 10 AC -0.5
RC1 2 5 200K
RC2 2 9 200K
Q1 5 4 3 NBJT
Q2 9 8 7 NBJT
RE1 3 6 2K
RE2 7 6 2K
REE 6 1 200K
.MODEL NBJT NPN BF=150
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(V1) IP(V1) VM(5,9) VP(5,9)
.TF V(9) VIC
.END
Results:
30
A dd  VM(5,9)  79.9 | A cc = 0.494 | R ID 
1
 751 k
IM(V1)
15.69 (a)
b  IC   F I E 
100
100A   99.0 A
101
V CE  20  10 I C  0.7   10.8 V
5
Q  Po int  99. 0A,10. 8V  for both transistors | r  
100 0. 025V 
99. 0 A
 25.3k
'
A cc  
 oR L
100k

 0.165
r    o  1R EE
25. 3k 101600k
A dd  
 oR L
| R L  100k 500k 83. 3k | R 5  600k 2.5k 2. 49k
r    o 1R 5
A dd  
100 83. 3k
 30.1
25.3k  1012. 49k


R ID  2 r    o  1R 5  225. 3k 1012.49k  554k
15.70
*Problem 15.70
VCC 2 0 DC 20
VEE 1 0 DC -20
VIC 9 0 DC 0
V1 4 9 AC 0.5
V2 7 9 AC -0.5
RC1 2 5 100K
RC2 2 8 100K
RL 5 8 1MEG
Q1 5 4 3 NBJT
Q2 8 7 6 NBJT
REE 3 6 5K
IEE1 3 1 67.8U
RE1 3 1 600K
IEE2 6 1 67.8U
RE2 6 1 600K
.MODEL NBJT NPN BF=100
.OP
.AC LIN 1 1KHZ 1KHZ
.PRINT AC IM(V1) IP(V1) VM(5,8) VP(5,8)
.TF V(8) VIC
.END
Results: A dd  VM(5,8)  30.0 | A cc = 0.165 | R ID 
1
 555 k
IM(V1)
15.71 (a)
Note: Use RD = 75 k
31
b  IC   F
2
I EE 100
V 

100A  99.0 A | 99.0A = 200A 
1  GS   V GS = 1.19 V
2
101

4 
V CE   V GS  1.19 V | V DS  15  7.5x10 I C  V CE  0.7   7. 09 V
4
BJT Q  Points  99.0A,1.19V  | JFET Q  Points  99. 0A,7.09 V 
r 
100 0. 025V 
99.0 A
 25.3k | A cc   
10075k
 oR L

 0.0619
r    o  12R EE 
25.3k  1011.2M
A dd   g m R D  40 99. 0 A75k  297 | R ID  2r   50. 5k
15.72
Note: The sources of M1 and M2 should be connectd in Fig. P15.72.
2I DS1
4x10 4
 1 2
 1.894 V | V GS1  V TN  0.894 V
Kn
10 3
I DS1  I 2  400A | V GS1  V TN 
I SD3  I1  I DS1  500A  400A  100A | V DS1 = V SG3  1  2
10 4
4x10
4
 1.710V
V SG3  V TP  0.710 V | V SD3  V S1  V SG3  6  30kI SD3   1.89  1.71  6  3  2. 82V
Both M1 and M 3 are saturated. Q - points: M1 : 400A,1.71V  M 3 : 100A, 2.82V 



A dd   g m R D   2 4x10 4 10 3 30k  26. 8 | For r o  , A cc  0 | R ID  
32
15.73
I1 100 50A

 24.8 A | V CE2  12  V EB 3   V BE2   12 V
2 101 2
12 V
For V O  0  V EC 3  12 V | I C 3 
 500 A
24k
0.7V
Q - points: 24.8 A,12V  24.8 A,12V  500 A ,12V  | R C1 
 28. 2k
24.8A
a I C1   F
b  R C2 
r o2 
V EB 3
0.7V
1000.025 V 

 35.4k | r 2 
 101k
I C 2  I B3 24. 8A  5A
24.8A
60  12
100 0. 025V 
60  12
 2. 90M | r 3 
 5k | r o3 
 144k
24.8A
500A
500A
g m2
r o 2 R C2 r  3 g m3 r o 3 R 
2
4024.8A 
A dm 
2. 90M 35.4k 5k40 500A 144 k 24k  893
2
R ID  2r 2  202 k | c  R O = r o3 R  20. 6 k
A dm 
d R IC 
 o  1r o2
2

1012.90M
2
 147 M | e  v 2 is the non- inverting input
15.74 Note that the parameters of the transistors and values of R C have been carefully
adjusted to permit open-loop operation and achieve VO = 0.
*Problem 15.74 - Two Stage Amplifier
VCC 1 0 DC 12
VEE 2 0 DC -12
RC1 1 5 28.2K
RC2 1 7 33.9K
Q1 5 4 3 NBJT
Q2 7 6 3 NBJT
I1 3 2 DC 50U
Q3 8 7 1 PBJT
R 8 2 24K
V1 4 10 AC 0.5
V2 6 10 AC -0.5
VIC 10 0 DC 0
.MODEL NBJT NPN BF=100 VA=60
.MODEL PBJT PNP BF=100 VA=60 IS=0.288F
.OPTIONS TNOM=17.2
.OP
.AC LIN 1 1KHZ 1KHZ
.TF V(8) VIC
.PRINT AC VM(8) VP(8) IM(V1) IP(V1)
.END
A dm  VM(8)  1030 | A cm = 6. 07 x 10 3 | CMRR dB = 105 dB
Results:
1
R ID 
 239 k | R O  20.6 k
IM(V1)
15.75
33
15V
 300A | V C 2  15  2400I E3  V EB 3  15  0.729  0.7  13.6V
50k
I
300A
200A  80 200A 
  F 
 

 98.8A | I B 3  C 3 
 3.75A
 2  81  2 
F 3
80
a For V O  0, IC 3 
I C1  I C 2
V CE1  V CE 2  13.6  0. 7  14.3V | V EC 3  15  2400 IE 3  V O  14.3V
Q - points: 98.8A,14.3V  98. 8A,14.3V  300A,14. 3V 
RC 
15 13. 6
V
98.8  3.75  A
b  A v1 
 15.1k r 3 

80 0.025 V 
0.3mA
v c2
g 
   m1  R C r 3   o 3 1R E
 2 
v id

 6. 67k

40 98.8A 
A v1   
 15.1k 6.67k  812.4k  27.7
2



Av2 
Av 

80 50k
vo
o 3 R L


 19.9
v c2
r 3   o 3 1R E
6. 67k  812.4k 
v c2 vo
 27.719.9  551
v id v c 2
R ID  2r 1  2
 o1 V T
800.025V 
70 14.3
2
 40.5k | r o 3 
 281k
I C1
98.8A
0.3mA

c R OUT  50k r o 3 1 

d R IC 
15.76
 o1  1r o1
2


o R E

80 2.4k

 50k 281k 1 
 49. 0k
R C  r 3  R E 
 15.1k  6.67k  2. 4k 
81 70  14.3 

  34. 6M | e  v 2 is the non- inverting (+) input.
2  98. 8A 
15V
 300A V EC 3  15  V O  15  0  15 V
50k
I
300A
200A  80 200A 
  F 
 

 98.8A I B 3  C 3 
 3.75A
 2  81  2 
 F3
80
a For V O  0, IC 3 
I C1  I C 2
V CE1  15  V EB 3  V BE1   15. 0V
Q - points: 98.8A,15.0V  98. 8A15. 0V  300A,15.0V 
RC 
0.7
V
800.025 V
70 V  15V
 7.37k | r  3 
 6.67k | r o 3 
 283k
0. 3mA
0.3mA
98.8  3.75  A
v
g 
g 
b  A v1  c2    m1  R C r 3 2r o1 g m 3 R r o3    m1  R C r 3 g m3 R 


 2 
v id
2

A v1 
40 98.8A 
2
R ID  2r 1  2
c R IC 
15.77
34



7.37k 6.67k 40300A 50k  3530
 o1 V T
800.025V 
2
 40.5k
I C1
98.8A
 o1  1r o1
2


81  70  15 

  34. 8M
2 98.8A 

For V O  0, IC 3 
15 V
I
300A
81
 300A | I B 3  C3 
 3.75A | I E3 =
I = 304A
50k
 F3
80
80 C3
0.7 V  I E3 R E 0.7V  304A R E
200A  80 200A 
I C1  I C 2   F 
 

 98.8A | R C 

 2  81  2 
I C1  I B 3
98.8A  3.75A
V C2  15  2400I E 3  V EB 3  15  0.729  0.7  13.6V
A Vth3 = 
AV2  
g m1
R C = 2098.8A R C = -1.976x10 -3 R C | R th = R C
2
R th


 oR L
 oR R
| R L  R r o3 1 
 | A V  A Vth3 A V 2
 r 3   o  1R E
 R th  r 3  R E 
15.78
R OUT  R r o3 |
10
3

1
R OUT

1
1
1
I C3
1 1
I C3 R





| V O  0V
R r o3 R V A  V EC 3 R R V A  V EC 3
1 
9 
9
100 0.025 V 
1 
  R  1.11 k | I C3 
 8.11mA | r  3 
 308
R  70  9 
R
8.11mA
AV
2000

 6.165
A V2
324
I R
0.7
 20 C 2 C  20
neglecting I B3
RC
RC
1
1
r 3
r 3
A V 2  g m3 R OUT  40 8.11mA 1k  324 | A V1 
A V1  
I R r
g m2
R C r  3   20 RC2 Cr  3
2
C
3
0. 7
0.7V
0.7V 8.11mA
 6.165  R C  391 | I C1 
 I B3 

 1.87mA
RC
391
391
100
1
308
Selecting the closest 5% values: R = 1.1 k , R C  390  , I 1  3.74 mA
20
15.79
a For V O  0, IC 3  I 2  300A
V C2  15  2400I E 3  V EB 3  15  0. 729  0. 7  13.6V
200A  80 200A 
I C1  I C 2   F 
 

 98.8A
 2  81  2 
I B3 
I C 3 300A

 3.75A
 F3
80
V CE1  V CE 2  13.6  0. 7  14.3V | V EC 3  15  2400 IE 3  V O  14.3V
Q - points: 98.8A,14.3V  98. 8A,14.3V  300A,14. 3V 
RC 
80 0.025 V 
15 13. 6
V
 15.1k r 3 
 6. 67k
98.8  3.75  A
0.3mA
35
b  A v1 

v c2
g 
   m1  R C r 3   o 3 1R E
v id
 2 


40 98.8A 
A v1   
 15.1k 6.67k  812.4k  27.8
2



Av2 



vo
o 3 R L
o R E
70  14.3

| R L  r o 3 1 
 281k
 | r o3 
v c2
r 3   o 3 1R E
300A
 R C  r 3  R E 

802.4k

80 2.51M 
R L  281k1 
 999
  2.51M | A v2  
6.67k  812.4k
 15.1k  6.67k  2. 4k 
Av 
80 0.025 V 
v c2 vo
 V
 27.8999   27800 | R ID  2r 1  2 o1 T  2
 40.5k
v id v c 2
I C1
98.8A
R O  R L  2.51 M
15.80
200A  100 200A 
I C1  I C 2   F 
 

 99.0A | V CE1  V CE 2  15  V EB 3  V BE1   15V
 2  101  2 
For V O  0, IC 3  I 2  300A
V EC 3  15  V O  15V
Q - points: 99.0A,15.0V  99. 0A,15. 0V  300A ,15.0V 
15.81
A V  A V1 A V 2 | A V1 = 
AV2  
 o R OUT


g m1
R C r  3   o3  1R E
2


 oR OUT
 oR E

r o3 1 

r 3   o  1R E
 R th  r 3  R E 
r  3   o  1R E
I
80 200A
I C1  I C 2   F 1 
 98. 8A | For V O  0, I C 3  I2 = 300A
2 81
2
I
300A
81
IB 3  C 3 
 3.75A | I E 3 =
IC 3 = 303.8A | V EC 3  15  I E3 R E
F 3
80
80
RC 
0.7V  I E3 R E 0.7V  303.8AR E

I C1  I B 3
98.8A  3.75A
r 3 
70  15  303. 8A R E
80 0.025 V 
 6. 67k | g m1  4098.8A   3.95mS | r o3 
300A
300A
15.82
36
I SD2 
2I SD2
500A
 250A | V S = V SG = V TP +
1
2
Kp

2 2.5x10 -4
5x10 -3
  1.32 V
V D  15  0.7  14. 3V | V SD  V S  V D  1.32  14.3  15.6V | Q - pt : 250A,15. 6V 
I C 3  500A | V CE3 = V C3 - V E3 = 0 - -15 = 15V | Q - pt: 500A ,15V 
V BE
0.7V

 2.87k
I D2  IB 3 250A  500A
80
RD 
g m2 
AV 
20.005 0.00025   1.58x10
3
80 0. 025V 
 4k
0.5mA
1.58mS
r 3 
2.87k 4k  1.30
2
S | r 3 
vd2 v o
g
= A V1 A V 2 | A V1   m 2 R D
v id v d2
2
75V  15 V

A V 2  g m 3 r o 3 R 2   400.5mA 
2M  0. 02180k 2M  3300
 0.5mA

A V  1.30 3300   4300 | R IN   | R OUT  r o 3 R 2  180k 2M  165k
v 2 is the non- inverting (+) input
15.83
AV 
vd2 v o
g
80 0. 025V 
= A V1 A V 2 | A V1   m 2 R D r  3  | I C3  100A | r 3 
 20k
v id v d2
2
100A
500A
 250A | g m2  20.005 0.00025   1.58x10 3 S
2
V BE
0.7V
1. 58mS


 2.81k | A V1  
2. 81k 20k  1. 95
100A
I DS2  I B 3
2
250A 
80
I DS2 
RD
75 V  5V

A V 2  g m 3 r o 3 R 2   40100A 
10M  0.02800k 10M  2960
 100A

A V  1.95 2960   5770
15.84 Note that the parameters of the transistors and values of R D have been carefully
adjusted to permit open-loop operation and achieve VO = 0.
*Problem 15.84
VCC 7 0 DC 15
VEE 8 0 DC -15
V1 1 9 AC 0.5
V2 3 9 AC -0.5
VIC 9 0 DC 0
I1 7 2 DC 493.2U
R1 7 2 2MEG
M1 4 1 2 2 PFET
M2 5 3 2 2 PFET
RD1 4 8 2.863K
RD2 5 8 2.863K
Q3 6 5 8 NBJT
I2 7 6 DC 492.5U
37
R2 7 6 2MEG
.MODEL PFET PMOS KP=5M VTO=-1
.MODEL NBJT NPN BF=80 VA=75 IS=0.2881FA
.OP
.AC LIN 1 1000 1000
.TF V(6) VIC
.PRINT AC IM(V1) IP(V1) VM(6) VP(6)
.OPTIONS TNOM=17.2
.END
A dm  VM(6)  4630 | A cm = 1.46 | CMRR dB = 70.0 dB
Results:
R ID 
1
  | R O  164 k
IM(V1)
15.85
a I SD2 
500A
 250A | V S = V SG = V TP +
2
2I SD2
1
Kp

2 2.5x10 -4
5x10
-3
  1.32 V
V D  5  0.7  0.7  3.6V | V SD  V S  V D  1.32  3. 6  4.92 V | Q - pt: 250A ,4.92 V 
I C 3  IC 4  500A | I C 3   F IE 3   F  2IC 3  500A  I C 3 = 6.10A | I C 4 = 494A
For V O = 0, V CE4 = 5V and V CE3 = 5 - 0.7 = 4.30 V
Q - pts: 250A, 4.92V  250A, 4.92V  6.10A, 4.30V  494A ,5.00V 
V
 V BE 4
1.4 V
R D  BE 3

 5.60k | Based upon results for the Darlington
6.10A
I D2  IB 3
250A 
80
g
800.025 V 
circuit in (Eq . 15.53): A V1  m1 R D 2r 3 | r  3 
 328k
2
6.10A


g m1  2 0. 0050. 00025   1.58mS | A V1  
1.58mS
5.60k 656k  4.39
2
g m4 2
40 494A 2 75V  5V


 r o 4 R 2   
1M  9.88mS 108k 1M  963

2 3

2
3 494A

2
 4.39 963   4230 | R ID   | R OUT  r o 4 R 2  108k 1M  97.5 k
3
AV2  
AV
15.86
38
*Problem 15.86
*Vos (the dc value of V2) has been carefully adjusted to set Vo ≈ 0
VCC 8 0 DC 5
VEE 9 0 DC -5
V1 1 10 AC 0.5
V2 3 10 DC 1.21M AC -0.5
VIC 10 0 DC 0
I1 8 2 DC 496.3U
R1 8 2 1MEG
M1 4 1 2 2 PFET
M2 5 3 2 2 PFET
RD1 4 9 5.6K
RD2 5 9 5.6K
Q3 7 5 6 NBJT
Q4 7 6 9 NBJT
I2 8 7 DC 495U
R2 8 7 1MEG
.OP
.MODEL PFET PMOS KP=5M VTO=-1
.MODEL NBJT NPN BF=80 VA=75
.AC LIN 1 1000 1000
.TF V(7) VIC
.PRINT AC IM(V1) IP(V1) VM(7) VP(7)
.END
A dm  VM(7)  4080 | A cm = 2.58 | CMRR dB = 64.0 dB
Results:
R ID 
1
  | R O  96. 2 k
IM(V1)
15.87
I
100 100A
 49.5A | V EC 2  0.7V  15 V  0.7V   15. 0V
a I C1  I C 2   F 1 
2
101
2
100
For V O  0, IC 4   F I3 =
1.00mA  990A | V EC 4  0  15V   15.0V
101
1mA
I C 3 = I 2 + I B4  350A 
 360A | V CE3  V O  0.7V  15  14.3
101
Q  pts: 49.5A,15.0V  49.5A,15.0 V  360A ,14.3V  990A,15. 0V 
100 0. 025V 
0.7V
0.7V
b  R C 

 15.3k | r 3 
 6.94k
IC 2  I B3 49.5  3.60 A
360A
r o3 
AV 
50 14.3
g
g
 179k | A V  A V1 A V2 A V 3 = m1 R C r  3 g m3 r o3 1  m1 R C r 3  f 3
360A
2
2
40 49.5A 
c R O 
d R IC 
15.88
2
15.3k 6.94k 4064. 3  12200 | R ID  2r 1  2
100 0. 025V 
 101 k
49.5A
ro 3  r 4
100 0. 025V 
179k  2.53k
| r 4 
 2.53k | R O 
 1.80 k
 o4 1
990A
101
 o1  1r o1
2
| r o1 
50V  15V
1011. 31M
1. 31M | R IC 
 66.3 M e  v 2
49.5A
2
*Problem 15.88
*RC and Vos (see V2) have been carefully adjusted to set Vo ≈ 0
VCC 7 0 DC 15
VEE 8 0 DC -15
V1 1 9 DC 0.117M AC 0.5
V2 3 9 AC -0.5
VIC 9 0 DC 0
I1 7 2 DC 100U
Q1 4 1 2 PBJT
Q2 5 3 2 PBJT
RC1 4 8 15.8K
RC2 5 8 15.8K
Q3 6 5 8 NBJT
I2 7 6 DC 350U
Q4 8 6 10 PBJT
I3 7 10 DC 1M
.MODEL PBJT PNP BF=100 VA=50
.MODEL NBJT NPN BF=100 VA=50
.NODESET V(10)=0
.OP
39
.AC LIN 1 1000 1000
.TF V(10) VIC
.PRINT AC IM(V1) IP(V1) VM(10) VP(10)
.END
A dm  VM(10)  13800 | A cm = 0.0804 | CMRR dB =105 dB
Results:
R ID 
1
 133 k | R O  1.37 k
IM(V1)
15.89
a Working backwards from the output: V DS4  V DD  V O  12  0  12.0V | I DS4  I3  5.00mA
V GS4  V TN 
2IDS4
 0.75 
Kn
20.005 
0.005
I SD3  I2  2.00mA | V SG3   V TP 
 2.16V | V SD3  V DD  V GS4  12  2.16  9.84V
20.002 
2IDS4
 0.75 V 
Kn
0. 002
V D2  V DD  V SG3  12 V  2.16V  9.84 V | I DS1 = I DS2 

2 2.5x10 4
V GS2  0.75 V 
5x10
3
  1.07 V |
 2.16 V
I1
 250A
2
V DS1  V DS2  9.84 V  1. 07V   10.9V
Q  pts: 250A ,10.9V  250A,10. 9V  2.00mA, 9.84V  5.00mA ,12.0V 
b  A dm 

g m1
g m4 r o4
g
 f4
2.16V
R D g m3 r o 3 
 m1 R D f 3
| RD 
 8. 64k
2
1  g m 4 r o4
2
1 f 4
0.25mA
g m1  2 5x10
g m3 
g m4 
A dm 
15.90
40
3
1
 9.84
2.5x10 1 0. 0210.9   1.75mS | r o 3  0.015
 38. 3k
2mA
1
 12
3
0.02
2x10 1  0.015 9. 84  3.03mS | r o 4 
 12. 4k
5mA

4




25x10 5x10 1  0.02 12   7.87mS |
2 2x10
3
3
3
 f 3  g m3 r o3  116 |  f 4  97.6
1.75ms
97. 6
1
 868 | R ID   | R O 
 127
8.64k116 
2
1  97.6
g m4
*Problem 15.90
*The values of RD have been adjusted to bring the offset voltage to ≈ 0
VCC 8 0 DC 12
VEE 9 0 DC -12
V1 1 10 AC 1
V2 3 10 AC 1
VIC 10 0 DC 0
I1 2 9 DC 500U
M1 4 1 2 2 NFET
M2 5 3 2 2 NFET
RD1 8 4 8.28K
RD2 8 5 8.28K
M3 6 5 8 8 PFET
M4 8 6 7 7 NFET
I2 6 9 DC 2M
I3 7 9 DC 5M
.MODEL PFET PMOS KP=2M VTO=-0.75 LAMBDA=0.015
.MODEL NFET NMOS KP=5M VTO=0.75 LAMBDA=0.02
.OP
.AC LIN 1 1000 1000
.TF V(7) VIC
.PRINT AC VM(7) VP(7) IM(V1) IP(V1)
.END
A dm  VM(7)  802 | A cm  4.74x10 -7  0 | CMRR dB = 
Results:
1
30
R ID 
 10   | R O  126 
IM(V1)
15.91
a Working backwards from the output: V SD4  V O  V SS  0  5  5. 00V
I SD4  I 3  2. 00mA | V SG4  V TP 
V DS3  V O  V SG4  V SS  0  2.11  5  2.89V
I DS3  I 2  500A | V GS3  V TN 
2I DS4
 0.75V 
Kn

2 5x10
5x10
V D2   V SS  V GS3  5  1.15 V  3. 85V | I SD1 = I SD2 

2 3x10 4
20.002 
 2.11V
0.002
2I SD4
 0.7 
Kp
4
3
  1.15 V
I1
 300A
2

 1.25 V | V SD2 = V SD2  1.25  3. 85  5.10V
2x10 3
Q  pts: 300A ,5.10V  300A,5.10V  500A,2. 89V  2.00mA, 5. 00V 
V SG1  V SG2  0. 7V 
b  A dm 

g m1
g m4 r o4
g
 f4
1.15V
R D g m3 r o 3 
 m1 R D f 3
| RD 
 8.83k
2
1  g m 4 r o4
2
1 f 4
0.3mA
g m1  2 2x10
g m3 
g m4 
A dm 
15.92
3
3x10 1  0.015 5.10   1.14mS |
4



22x10 2x10 1  0.015 15   2.93mS |
r o3
2 5x10 3 5x10 4 1  0.022.89   2. 30mS | r o4
3
3
1
 2.89
0.02

 106k
0.5mA
1
 5. 00
 0. 015
 35.8k
2mA
f 3  g m 3r o3  244 |  f 4  105
1.14ms
105
1
 1220 | R ID   | R O 
 341
8.83k244 
2
1  105
g m4
Note: Use RE = 1k
41
a Working backwards from the output with V O = 0: V DS4  V CC  V O  5  0  5V
 I

 2mA

I DS4  I 3  2mA | V GS4   V P  DS4 1  5
 1  2. 76V | IC 3  I 2  500A
 I DSS4

 5mA

101
500A 1k  2.76 V   7.26 V
100
  V BE 2   5  0.505  0.7  0.7  4.50V
V EC 3  5  I E3 R E  V GS4  5V 
V CE 2  5  I E3 R E  V EB 3
I C1  I C 2   F
I1 100 200A
I R  V EB 3
0.505 V  0.7V

 99. 0A | R C  E3 E

 12.8k
2 101
2
I C2  I B 3
99. 0  5.00 A
V CE1  5  IC1R C   V BE 2   5  99.0A 12. 8k  0.7  4.43 V
Q  pts: 99.0A ,4. 43V  99.0A,4.50 V  500A, 7.26V  2.00mA ,5.00V 
b  Using current division at the collector of Q2 :
A dm 
r 3 

 g m 4 R L
g m1
RC
 o3 R E
 o3 r o3 1 

2 R C  r 3   o 3 1R E
 R C  r 3  R E 1  g m4 R L
100 0.025 V 
50V  7.26V
 5. 00k | r o 3 

500A
500A
g m4 
A dm 
2I DS4
2 2mA 
gm 4RL
1.79mS 5k

 1. 79mS |

 0.890
V GS4  V P
2.76  5
1  g m4 R L 1  1.79mS 5k
4099.0A 
2
12.8k100 
A dm  13900 | R ID  2r 1 = 2
42

12.8k  5.00k  1011k
100 0. 025V 
99.0A
115k1 


1011k
0.890
12.8k  5.00k  1k 
 50.5 k | R O 
1
 559 
gm 4
15.93
a Working backwards from the output with V O = 0: V DS4  V CC  V O  5  0  5V
I DS4  I 3  2mA | V GS4  V TN 
2I DS4
 0.7 
Kn
20. 002 
 1.59V | I C 3  I2  500A
0.005
V EC 3  5  V GS4  5  1.59  3.41V | VCE 2  5  V EB 3  V BE 2   5  0.7  0.7  5. 00V
I C1  I C 2   F
I1 100 200A
V EB 3
0.7V

 99. 0A | R C 

 7.45k
2 101
2
I C2  I B 3 99. 0  5.00 A
V CE1  5  IC1R C   V BE 2   5  99.0A 7. 45k  0.7  4. 96V
Q  pts: 99.0A ,4. 96V  99.0A,5.00V  500A, 3. 41V  2.00mA ,5.00 V 
b  Using current division at the collector of Q2 :
A dm 
r 3 
g m1
RC
g m4 R L
g
gm 4R L
 o3 r o 3
 m1 R C r 3  f 3
2 R C  r 3
1  g m 4R L
2
1  g m4 R L
100 0.025 V 
 5. 00k | g m4 
500A
A dm 
4099.0A 
2
R ID  2r 1 = 2
15.94
20.005 0.002   4.47mS
7. 45k 5. 00k40 50  3.41 1  4. 47mS 2k   11400
4.47mS 2k
100 0.025V 
1
 50.5 k | R O 
 224 
99.0A
g m4
*Problem 15.94
*The values of RC have been adjusted to set Vo ≈ 0.
VCC 8 0 DC 5
VEE 9 0 DC -5
VIC 10 0 DC 0
V1 1 10 AC 0.5
V2 3 10 AC -0.5
I1 2 9 DC 200U
Q1 4 1 2 NBJT
Q2 5 3 2 NBJT
RC1 8 4 8.00K
RC2 8 5 8.00K
Q3 6 5 8 PBJT
I2 6 9 DC 500U
M4 8 6 7 7 NFET
I3 7 9 DC 2M
RL 7 0 2K
.MODEL NBJT NPN BF=100 VA=50
.MODEL PBJT PNP BF=100 VA=50
.MODEl NFET NMOS KP=5M VTO=0.70
.OP
.AC LIN 1 2KHZ 2KHZ
.PRINT AC VM(7) VP(7) IM(V1) IP(V1)
.TF V(7) VIC
.END
A dm  VM(7)  11200 | A cm  0.0957 | CMRR dB =101 dB
Results:
R ID 
1
 56. 4 k   | R O  201 
IM(V1)
43
15.95
50 50A
I
I1 100 10A
 24.5 A
 4.95 A | I C 3  IC 4   F 2 

51 2
2
2 101 2
24.5 A 
V CE 2  V CC  IC 2  I B3 R C   V BE 2   3V  
300k  0.7   2.36 V
4.95A 
50 

50
250A   245A | V EC 5  3. 00 V | V C 4  0.7 V
For V O  0: IC 5   F I 3
51
24.5 A 
V C1  3  
300k  1.66V | V EC 3  V EC 4  1. 66  0.7  0.7  3.06V
4.95A 
50 

a I C1  I C 2   F
Q - pts: 4.95A ,2.36V  4.95A ,2.36 V  24.5A, 3.06V  24.5A,3. 06V  245A,3.00 V 
b  A dm  g m1 R C1 r  3 
r 3 
 
g m3
R C 2 r  5   o5  1R L
2

1R
r 5 o5 o 5 1LR L
500.025 V
500.025 V 
 5.10k
 51. 0k | r 5 
245A
24.5A
 o5  1R L
r 5   o5  1R L

512k 
 0. 952
5.10k  512k 
A dm  404.95A 300k 51.0k 
R ID  2r 1  2
40 24.5A
2
78k 5.10k  512k0.952  182
R  r  5 78  5.10
1000.025 V 
k  1.63 k

 1.01 M | R O = C 2
51
 o5  1
4.95A
c v A is the non- inverting input - v B is the inverting input
d A V  10V CC 10 V CC   30 2  900 | r  3  R C is substantailly reducing the gain
, R IN5 = 107k  R C 2 , and
Also, the input resistance of the emitter follower is low
is reducing the gain by an additional factor of almost 2.
15.96
I1 100 100A
I
50 200A
 49.5 A | I C3  I C 4   F 2 
 98.0 A
a I C1  I C 2   F 
2 101
2
2
51
2
98.0 A 

V CE 2  V CC  IC 2  I B3 R C   V BE 2   18V  49.5A 
120k  0.7   13. 0V

50 
50
For V O  0: IC 5   F I 3
750A   735A | V EC 4  18 V | V C 4  0.7V
51
98. 0 A 

V C1  18  49.5A 
120k  12.3V | V EC 3  V EC 4  12. 3  0.7  0.7   13. 7V

50 
Q - pts: 49.5A,13. 0V  49.5A ,13.0V  98. 0A,13.7V  98.0A,13.7V  735A ,18.0V 
b  A dm  g m1 R C1 r  3 
r 3 
500.025 V 
98. 0A
 o5  1R L
r 5   o5  1R L
44

 
g m3
R C 2 r  5   o5  1R L
2
 12.8k | r  5 
512k 
50 0. 025V 
1.70k  512k
735A
 0.984
 1 R
r o5 1LR
5
o5
L
 1.70k
A dm  4049.5A 120k 12.8k
R ID  2r 1  2
1000.025 V 
49.5A
40 98.0A
2
 101 k | R O =
170k 1.70k  512k0.984  2840
R C 2  r  5 170  1.70

k  3.37 k
 o5  1
51
c For positive V IC , V IC  V C1  12.3V | For negative VIC , the characteristics of I1 will
deterimine VIC . For the ideal current source, the negative limit of VIC is not defined.
d The actual voltage at the collector of Q4 would be
735A 

V C 4  18 V  I C 4  I B5 R C2  18V  98.0A 
170k  1.16V and V O = +1.86 V

50 
should be - 0.7V. The value of offset voltage required to bring the output back to zero is
V OS 
15.97
V O 1.86 V

 0.655 mV.
AV
2840
I1 100 70A

 34.7A | For V O = 0: I DS4  I C 3  1mA
2 101 2
 0.7  5  5.7V | V CE 2  0.7  5.0.7  5.0V
a I C1  I C 2   F
V CE1

V GS4  V P 1 


I DS4 
  5 1 
I DSS4 

1mA 
  2.76V | V CE 3   V GS4  2.76 V
5mA 
V DS4  V O   V EE  VGS4   0  5  2.76   2.24V
Q  pts: 34.7A ,5.70V  34.7A,5.00 V  1.00mA, 2.76V  1.00mA ,2.24 V 
b  A dm 
RC 
g m2
g
1000.025 V 
R C r 3 g m3 R O  m2 R C r  3 g m 3 R 2 f 4 r o 3  | r 3 
 72.1k

2
2
34.7A
V BE 3
0.7V
50V  2. 76V

 28.3 k | r o3 
 52.8k
I C 2  I B3 34.7 10.0A
1mA
2
50V  2. 24V 
 f 4  g m4 r o4   1mA 5mA 
  46.7
5


1mA
28. 3 k 72.1k400. 0015M 46.7 52.8k 9. 31 x 10
5
!
2
100 0.025V 
 2r 1 = 2
 144 k | R O  R 2 f 4 r o 3  5M 46.7 52.8k  1.65 M
34.7A
A dm 
R ID

40 34.7x10 6
15.98
I1 100 200A
100 12V

 99. 0A | For V O = 0: I C3   F IE 3 
 990A
2 101
2
101 12k
 12V  0V  12.0V | V CE 2  0. 7  0. 7  1.4V
a I C1  I C 2   F
V CE 3
RC 
12V  0.7 V
0.7 V

 104 k | V CE1  12  99.0A 104k  0.7  2.40V
IC 2  I B 3
99. 0  9. 90A
Q - points: 99.0A, 2,40V  99.0A,1.40V  990A,12 V 
45
b  A dm 
A dm 
g m2
 o 3 1R | r  100 0. 025V   2.53k
R C r 3   o3  1R
3
2
r 3   o 3 1R
990A
 
4099.0A 
2
R ID  2r 1 = 2
15.99

10112k
 189
104 k 2.53k  10112k2.53k
 10112k
100 0.025V 
R  r  3 104k  2.53k
 50.5 k | R O  C

 1.06 k
99.0A
 o 3 1
101
1000.025 V 
 16.7A For V O  0, V C1  0. 7V
300k
12  0. 7 11.3V
11. 3V
R  r 3 677k



 677k | R O  C

 6.7k
I C1  I B 3
IC1
16.7A
 o3  1
101
300k = 2r 1  I C1  2
RC
The RO specification cannot be meet if the RID specification is met and vice - versa.
Either R IDmust be reduced or R O must be increased, or both must be changed.
15.100
100 0.025V 
 5.00A | For V O  0, V C1  0.7V
1M
9  0.7
8.3V
8.3V
R  r 3 1.66M



 1. 66M | R O  C

 16.4k
I C1  I B 3
IC1
5.00A
 o3  1
101
1M = 2r 1  IC1  2
RC
The RO specification cannot be meet if the RID specification is met and vice - versa.
Either R IDmust be reduced or R O must be increased, or both must be changed.
15.101
a For V O  0, IC 6   F I E 6   F I 3 
I C5   F IE 5   F
 F4
100
5mA  4.95mA
101
IC 6
4.95mA
49.0A

 49.0A | I C4  IC 3  I 2  I B5  500A 
 500A
F 6
101
100
I C3
 I C3   F 3  2IC 3  500A  I C 3  9. 62A | I C4  500A  I C3  490A
 F3
I1 100 50A

 24.8A | V CE 6  18  0  18 V | V CE5  V CE 6  V BE 6  17.3V
2 101 2
 18  V BE 5  V BE 6  16.6V | V EC 3  V EC 4  V EB 4  15. 9V
I C1  I C 2   F
V EC 4
V CE1  V CE 2  18  V EB 4  V EB 3  V EB 2   17. 3V
Q  pts: 24.8A ,17.3V  24. 8A,17. 3V  9.62A ,15.9V  490A,16. 6V 
49.0A ,17.3V  4. 95mA,18.0V 
| RC 
1. 4V
1. 4V

 56. 9k
9.62 
I C2  I B 3 
24.8 
A

50 
b  Using the properties of the Darlington configuration (Eq . 15.54):
2

r o 2 and  'f  f 2
3
3
 o5  o6 R L
R IN5 

 2r 5   o5  o6 R L
Note that the correct expressions are: r 'o =
A dm 
46
g m2
g
2
R C R IN3 m 4 
 r
2
2 3 o 4


R IN3  2 o3 r  4  250
500.025 V 
 255k
490A
1000.025 V 
 100 100 2k  20.1M
4.95mA
4024.8A 
70 V 16.6V

20M

 177k | A dm 
56.9k 255k 3f 4 20.1M
490A
2
R IN5  2 o5 r 6   o5  o6 R L  2100 
R IN5 >> r o4
 f4
4070 11.6

3
RO 
3
R th5  2r 5
 o5  o6
 1155 | A dm  26500 88.5dB  | R ID  2r 1 = 2
a For V O  0, IC 6   F I E 6   F I 3 
 F4
24.8A
 202 k
100 0.025 V 
2
2
118k  2
r o 4  2r 5
3
49.0A
 3

 18.1 
o 5 o6
100 100 
15.102
I C5   F IE 5   F
100 0.025V 
100
5mA  4.95mA
101
IC 6
4.95mA
49.0A

 49.0A | I C4  IC 3  I 2  I B5  500A 
 500A
F 6
101
100
I C3
 I C3   F 3  2IC 3  500A  I C 3  9. 62A | I C4  500A  I C3  490A
 F3
I1 100 50A

 24.8A | V CE 6  22  0  22V | V CE 5  V CE 6  V BE6  21.3V
2 101 2
 22  V BE 5  V BE6  20. 6V | V EC 3  V EC 4  V EB 4  19.9V
I C1  I C 2   F
V EC 4
V CE1  V CE 2  22  V EB 4  V EB 3  V EB 2   21.3V
Q  pts: 24.8A ,21.3V  24.8A ,21.3V  9. 62A,19. 9V  490A,20. 6V 
1.4V
1.4 V

 56.9k
9.62 
I C 2  IB 3 
24. 8 
A

50 
b  Using the properties of the Darlington configuration (Eq . 15.54):
49.0A ,21. 3V  4.95mA ,22.0V 
| RC 
Note that the correct expressions are: r 'o =
2

r o 2 and  'f  f 2
3
3

g
g
2
 o 5 o6 R L
A dm  A V1 A V2 A V 3   m2 R C R IN3  m4  r o 4 R IN5 
 2
 2 3
2r 5  o 5 o6 R L 

R IN3  2 o3 r  4  250
500.025 V 
 255k
490A
R IN5  2 o5 r 6   o5  o6 R L  2100 
R IN5 >> r o4 
1000.025 V 
4.95mA
 100 100 2k  20.1M
4024.8A
70 V  20.6V

20M
 185k | A dm 
56.9k 255k  f 4

490A
2
3 20.1M
 f4
4070  20.6
1000.025 V 

 1208 | A dm  27700 88. 9dB  | R ID  2r 1 = 2
 202 k
3
3
24.8A
RO
2
100 0.025 V 
2
185k  2
r o 4  2r 5
R th5  2r 5
3
49.0A

 3

 22. 5 
 o5  o6
o 5 o6
100 100 
47
15.103
Since the transistor parameters are the same, V GS1  V SG2 
I SD2  IDS1 
2.2V
 1.1V
2
6x10 4
2
1.1  0.75   36.8 A
2
15.104
2.2V  V GS1  V SG2  V TN 
2IDS1
 V TP 
Kn

2
2.2  0.7  0.8  IDS1 
4 
 6x10
15.105
2
4x10
4
2I SD2
Kp

 |

where I SD2  I DS1
IDS1 
0.7
 ISD2  I DS1  29.7A
128.5
Since the values of IS and I E are the same, V BE1  V EB 2
1.35  V BE1  V EB 2  2V T ln
IC
1.35
| I C  10 15 exp
 196 A
IS
20.025 V 
15.106
1.35  V BE1  V EB 2  V T ln
IC 
IC
I
I 2C
+ V T ln C = V T ln
I S1
I S2
I S1I S2
1.35
5x10 10 exp 0.025
15
15
 438 A
15.107
15.108
*Problem 15.108
VCC 3 0 DC 10
VEE 5 0 DC -10
VBB 2 1 DC 1.3
VS 1 0 DC 0
Q1 3 2 4 NBJT
Q2 5 1 4 PBJT
RL 4 0 1K
.MODEL NBJT NPN IS=5FA BF=60
.MODEL PBJT PNP IS=1FA BF=50
.OP
.DC VS -10 +8.7 0.01
.PROBE
.END
15.109
Since the base currents are zeroF   , V BE1  V EB 2  250A 5k  1. 25V
1.25 V = V T ln
15.110
48
IC
I
I 2C
+ V T ln C = V T ln
| IC 
I S1
I S2
IS1I S2
25
 22.8 A
10 10 exp 0.1.025
15
16
V GS1  V SG2  0.5mA 4k  2. 00V | 2. 00V = V TN 

2
2.00  0.75  0.75  I DS1 
4 
 5x10
2
2x10
4

 |

2IDS1
 V TP 
Kn
I DS1 
2I SD2
| I SD2  I DS1
Kp
0.5
 I SD2  I DS1  9.38 A
163.3
15.111
5V
5V
IS 

 5.00mA | i S  I S  is
R L 1k
i max
 IS 
S
5V
5V
 I S  5.00mA | i min
 IS 
 I S  5.00mA
S
1k
1k
For I S = 5. 00mA, i max
 10. 0mA | i min
 0 | i DS  0.005 1  sin 2000t A
S
S
Power delivered from the supplies: P t   10 V i DS   10 VI S   0.052  sin 2000 t  W
P av 
1
T
T
 0.05 2  sin 2000tdt  100mW
0
2
 5  1
12.5mW
Signal power developed in RL : P ac   
 12.5mW |  = 100%
 12.5%
100mW
 2  1k
15.112
P ac 
1
T
5V 2 T 5V 2 T 


  10.0mW
5k 2 

 5k 2

P av 
1
T

5V T
5V T 
5V
 5V
 10.0mW

 5k 2
5k 2 

|
 =100%
49
15.113
P ac
T
P av
2
T
4
1
20

10itdt 

T 0
T
15.114
50
2
40t 


1 v t
4  T 
6400

dt

dt  3


T 0 R
T 0
R
T R
T
T
2
40
 it dt  T
0
T
4

0
T
4
t
0
2
dt 
100
3R
40t
1600
dt  2
TR
T R
*Problem 15.114(a) VBB = 0 V
VCC 3 0 DC 10
VEE 5 0 DC -10
VBB 2 1 DC 0
VS 1 0 DC 0 SIN(0 4 2000)
Q1 3 2 4 NBJT
Q2 5 1 4 PBJT
RL 4 0 2K
.MODEL NBJT NPN IS=5FA BF=60
.MODEL PBJT PNP IS=1FA BF=50
.OP
.TRAN 1U 2M
.FOUR 2000 V(4)
.PROBE
.END
*Problem 15.114(b) VBB = 1.3 V
VCC 3 0 DC 10
VEE 5 0 DC -10
VBB 2 1 DC 1.3
VS 1 0 DC 0 SIN(0 4 2000)
Q1 3 2 4 NBJT
Q2 5 1 4 PBJT
RL 4 0 2K
.MODEL NBJT NPN IS=5FA BF=60
.MODEL PBJT PNP IS=1FA BF=50
.OP
.TRAN 1U 2M
.FOUR 2000 V(4)
.PROBE
.END
100
50
3R
 tdt  R |   100% 50  66.7%
0
R
T
4
HARMONIC
NO
1
2
3
4
5
6
7
8
9
FREQUENCY
(HZ)
2.000E+03
4.000E+03
6.000E+03
8.000E+03
1.000E+04
1.200E+04
1.400E+04
1.600E+04
1.800E+04
FOURIER
NORMALIZED PHASE
COMPONENT COMPONENT (DEG)
NORMALIZED
PHASE (DEG)
3.056E+00 1.000E+00 -4.347E-01 0.000E+00
2.693E-02 8.811E-03 -1.300E+02 -1.296E+02
2.112E-01 6.910E-02 -1.744E+02 -1.740E+02
3.473E-02 1.136E-02 -1.550E+02 -1.545E+02
7.718E-02 2.525E-02 -1.678E+02 -1.674E+02
4.064E-02 1.330E-02 -1.679E+02 -1.675E+02
3.179E-02 1.040E-02 -1.580E+02 -1.576E+02
4.109E-02 1.345E-02 -1.736E+02 -1.731E+02
2.127E-02 6.960E-03 -1.568E+02 -1.564E+02
TOTAL HARMONIC DISTORTION = 7.831458E+00 PERCENT
HARMONIC
NO
1
2
3
4
5
6
7
8
9
FREQUENCY
(HZ)
2.000E+03
4.000E+03
6.000E+03
8.000E+03
1.000E+04
1.200E+04
1.400E+04
1.600E+04
1.800E+04
FOURIER
NORMALIZED PHASE
COMPONENT COMPONENT (DEG)
3.853E+00 1.000E+00 2.544E-01
1.221E-02 3.169E-03 6.765E+01
1.537E-02 3.990E-03 9.046E+01
1.504E-02 3.903E-03 5.520E+01
1.501E-02 3.897E-03 5.500E+01
1.531E-02 3.973E-03 4.231E+01
1.435E-02 3.726E-03 3.680E+01
1.467E-02 3.807E-03 2.823E+01
1.382E-02 3.587E-03 2.087E+01
NORMALIZED
PHASE (DEG)
0.000E+00
6.740E+01
9.020E+01
5.495E+01
5.475E+01
4.206E+01
3.654E+01
2.798E+01
2.062E+01
TOTAL HARMONIC DISTORTION = 1.064939E+00 PERCENT
15.115
V BE 2 0.7V

 70.0 mA .
R
10
0.07
 250i E  1000
 0.7  0.7  2500.07   19. 6 V
101
The current begins to limit at iE 
v S  1000i B  V BE1  V BE 2
15.116 Note: The current limiting will be much more dramatic if R 1 is increased to 10k.
*Problem 15.116
VCC 3 0 DC 50
VS 1 0 DC 1
R1 1 2 1K
Q1 3 2 4 NBJT
Q2 2 4 5 NBJT
R 4 5 10
RL 5 0 250
.MODEL NBJT NPN IS=1FA BF=100
.OP
.DC VS 1 50 .05
.PROBE
.END
15.117
51
For V BE 4 = 0.7V, V EB 5 = I 2 R B - V BE 4
V EB 5 = 1.2  0.7 = 0.5 V and Q5 is off .
V EQ  15  0.2  500A 50  14.8V
R EQ  50
I C 4  100
14.8  0.7 V  6. 98 mA
50 1012k
15.118
I 2 R G  V GS4  V SG5 | 0. 25mA 7k  V TN4 
1.75  0.75  0.75 
15.119
R OUT 
15.120
I C  100I B  100

2
I DS4 

 0. 005
2I DS4
 V TP5 
0. 005

2
  ISD5  I DS4  23.5 A
0.002 
r
1
o V T
1 VT
1 0.025 V



 25.0 m


1
100


1
I
100
I
100
10mA
n
o
o
C
E
1
2
9  0.7
V
 97.9 A | Looking back into the
200k 10182k 
transformer: R th 
1
n2
 r  
r
1 100 0.025V 

 253

 |
 o 1 101
97.9A
 o 1 
Desire to match the Thevenin equivalent resistance to LR:
vth 
 o  1n 2R L
r    o  1n 2 R L
1
253  10  n  5.03
n2
101253
v s  0.500v s | Using the ideal transformer
25. 6k  101253
1
1 vo
1 vo
 i1 R th  nv o | i 1  i 2 
| v th 
R th  nv o
n
n RL
n RL
vs 
relationships: v th
vo =
2I SD5
| I SD5  IDS4
0.002
v th
0.500 v s
| vo =
 0.0497 v s | v o = 0. 0497sin2000t
R th
253
n
5.03 
nR L
5.0310
2
0.0497  1
P o  
 0.124 mW


2  10
15.121
2M
 6V | R EQ  2M 2M  1M
2M  2M
6  0.7  12  V
100 0.025V 
 100I B  100
 22.8 A | r  
 110k
1M 101220k 
22. 8A
V EQ  12V
IO
V CE  12  I E 220k  12 
101
50  6.93 V
22. 8A 220k  6. 93V | r o 
 2.50M
100
22.8A


o R E

100220k

R OUT  r o 1 
  43.9 M
  110k1 
 1M  110k  220k 
 R th  r   R E 
15.122
52
The dc analysis is the same as Problem15.121. However, the bypass capacitor provides as
ac ground at the base of the transistor so that Rth = 0.


100 220k  
 oR E 
R OUT  r o 1 
  110k1 
  169 M
 r   R E 
 110k  220k 
15.123
A spread sheet will be used to assist in this design using F  o  100 & V A  70 V
The maximum current in the two bias resistors is0.2mA . To allow
some room for tolerances, choose I1  0.15mA . Neglecting the transistor base current,
12V
V
V
 80k | R 2  B R1  R 2   B 80k | R B  R 1 R 2
0.15mA
12
12

V B  0.7  or R  1 100V B  0. 7  R  | V  12  I R
I C  100

E
B 
CE
E E
R B  101R E
101 
IC

R1  R2 


70  V CE
 oR E
| R OUT  r o 1 

IC
 R B  r   R E 
Now, a spreadsheet MATLAB, MATHCAD, etc. can be used to explore the design space
ro 
with VB as the primary design variable.
VB
R2
R1
RB
RE
ro
Rout
0.500
0.600
0.700
0.800
0.900
1.000
1.100
1.200
1.300
1.400
1.500
1.600
1.700
1.800
1.900
2.000
3.33E+03
4.00E+03
4.67E+03
5.33E+03
6.00E+03
6.67E+03
7.33E+03
8.00E+03
8.67E+03
9.33E+03
1.00E+04
1.07E+04
1.13E+04
1.20E+04
1.27E+04
1.33E+04
7.67E+04
7.60E+04
7.53E+04
7.47E+04
7.40E+04
7.33E+04
7.27E+04
7.20E+04
7.13E+04
7.07E+04
7.00E+04
6.93E+04
6.87E+04
6.80E+04
6.73E+04
6.67E+04
3.19E+03
3.80E+03
4.39E+03
4.98E+03
5.55E+03
6.11E+03
6.66E+03
7.20E+03
7.73E+03
8.24E+03
8.75E+03
9.24E+03
9.73E+03
1.02E+04
1.07E+04
1.11E+04
-2.30E+02
-1.37E+02
-4.35E+01
4.97E+01
1.43E+02
2.37E+02
3.30E+02
4.24E+02
5.18E+02
6.11E+02
7.05E+02
8.00E+02
8.94E+02
9.88E+02
1.08E+03
1.18E+03
-1.74E+05
-8.00E+04
1.41E+04
1.08E+05
2.03E+05
2.97E+05
3.91E+05
4.86E+05
5.81E+05
6.76E+05
7.71E+05
8.66E+05
9.61E+05
1.06E+06
1.15E+06
1.25E+06
5.57E+05
9.73E+04
5.13E+03
1.80E+05
5.56E+05
1.09E+06
1.75E+06
2.52E+06
3.38E+06
4.31E+06
5.32E+06
6.38E+06
7.50E+06
8.68E+06
9.90E+06
1.12E+07
7.50E+04
6.80E+04
5.73E+03
1.02E+04
1.50E+02
1.00E+03
2.10E+05
1.07E+06
5.85E+05
8.86E+06
Two
possible
solutions
0.916
1.800
6.20E+03
1.20E+04
IO
1.04E-03
9.89E-04
The first solution is the lowest value of VB that was found to meet the output specification
using the nearest 5% values. The second is one in which the values were found to be very
53
close to existing standard 5% resistor values, but it uses twice the value of VB and has a
smaller output voltage compliance range.
15.124
330k
10V  3. 27V | R EQ  330k 680k  222k | Assume saturation:
330k  680k
5x10 4 
2
 3.27  30kI DS  3. 27  
V GS  1  V GS  1. 488V | I O = I DS  59.4 A
 2 
V EQ 
V GS
V DS  10  30kIDS  8.22 V | r o 

100  8.22 V
 1.82 M
59.4
A


3x10  1.82M1  0.254mS 30k  16.0 M
g m  2 5x10 4 59. 4x10 6 1  0.018.22  0.254 mS

R OUT  r o 1 g m
54
4
15.125
200k
 10V | R EQ  200k 100k  66. 7k
200k  100k
15  0.7 10
V
750.025 V 
 75IB  75
 96.7 A | r  
 19.4k
66.7k  76 43k 
96. 7A
V EQ  15V
IO
V EC  15  I E R E  15 
76
50  10.8 V
96.7 A 43k  10.8V | r o 
 629k
75
96.7A


o R E

75 43k

R OUT  r o 1 
  16.3 M
  629k1 
 66.7k  19. 4k  43k 
 R th  r   R E 
15.126
V EQ  9V
2M
 6 V | R EQ  2M 1M  667k | 9  105 I SD  V SG  6
2M  1M
V SG   V TP 

2I SD
2ISD 
5
| 10 ISD  3  0.75 
4   I O  I SD  20.2 A
Kp
7.5x10 

5
V SD  9  10 I SD  6.98V | r o 



100  6.98 V
 5.30 M
20. 2
A
g m  2 7.5x10 4 20. 2x10 6 1  0.016.98   0.180mS

  5.30M1  0.180mS10  101 M
R OUT  r o 1  g m R S   1  0.180mS 10 5
5
15.127



50V


1 2 2x10 4 1.75x10 4 R 5 
4 

1.75 x10
Note that including  in the gm expression will increase ROUT above this estimate.
Estimating R OUT  r o 1  g m R 5  
Hence neglecting it is a conservative simplification.

286k 1  2. 65x10
4

R E  2.5M  R 5  29.2k | Choose R 5  33k
V G  V DD  ID R 5  V SG  12 1.75x10
4

 1 


3.3x10 
4

2 1.75x10 4
2x10 4
  3.90V


R4
12V
12 = 3.90 | I2  25A | Assign I 2  20A | R 3  R 4 
 600k
R 3  R4
20A
R4 =
3. 90
R 3  R 4  =195k  R 4 = 200k | R 3  430k
12
15.128
55
V EQ  12V
68k
 8. 08V | R EQ  68k 33k  22. 2k | V B  8. 08  I B1  I B 2 R TH
68k  33k
 V B  0.7  12  V B  0.7  12 
V B  8. 08  

22.2k  V B  8.11V
126100k 
 126 20k

I C1   F I E1 
125 V B  0.7  12 
125  V B  0. 7  12

  158 A | I C2   F IE 2 

  31.7 A
126 
20k
126 
100k


V CE  0  8.11  0.7  8.87V | r o1 
r 1 
125 0.025 V
158 A
50  8.11V
158A
 19.8k | r 2 

125 0. 025V 
31.7A
 98.6k
R th1  22.2k 98. 6k  126 100k  22.2k


 oR E

12520k

R OUT1  r o1 1 
  15.2 M
  368k1 
 22. 2k 19. 8k  20k 
 R th  r 1  R E 
50  8.11V
r o2 
 1.83M | R th2  R TH r 1   o 120k
31.7A


R th2  22. 2k 19. 8k  126 20k  22.0k


o R E

125 100k 

R OUT2  r o 2 1 
  106 M
  1. 83M1 
R

r

R
22.
0k

98.6k

100k



th
2
E 
56

 368k | R th1  R EQ r 2  o  1100k
15.129
390k
V EQ  15V
 11. 9V | R EQ  390k 100k  79.6k
390k  100k
V  0.7  15 
I  I B1  I B 2  I B3
V B  11.9  0.7  E 4
R TH | I E 4  B
o  1
27k
I B1 
V B  1.4  15 
V  1.4  15 
V  1. 4  15
| IB2  B
| IB 3  B
126100k 
126 330k 
126 100 k 
I B1  I B 2  I B3  I E 4
Then: I E 4  126
11. 9  0.7  15 
 86. 9A | V E4  15  86.9A27k   12. 6V
79.6k  126 27k 
I C 3  IC1   F I E1 
I C 2   F I E2 
125 12.6  0.7  15 

  16.9 A | V CE1 = V CE 3 = 0  12.6  0.7 =13. 3V
126 
100k

125  12.6  0.7  15 

  5.11 A | V CE 2 = 0  12. 6  0. 7 = 13. 3V
126 
330k

16. 9A + 5.11A + 16.9A
 0. 311A  I E 4  Assumption is ok.
125
50  13. 3V
50  13. 3V

 3.75M | r o2 
 12.4M
16. 9A
5.11A
Checking: IB1  I B2  I B 3 
r o3  r o1
r 3  r 1 
125 0.025V 
125 0.025 V 
1250.025 V 
 185k | r 2 
 612k | r  4 
 36. 0k
16. 9 A
5.11 A
86.9 A
R IN3  R IN1  r 1   o  1100k   185k  126 100k   12.8M
R IN2  r  2   o  1330k   612k  126330k   42.3M
R O4  27k
R EQ  r  4
o  1
 27k
79.6k  36.0k
 887
126
R th1  R O 4 R IN2 R IN3  887 | R th2  R O4 R IN1 R IN3  887 | R th3  R O4 R IN1 R IN2  887


 oR E

125 100k 

R OUT 3  R OUT 1  r o1 1 
  168 M
  3.75M 1 
R

r

R
0.887k

185k
100k



th
1
E 



125 330k 

 oR E
R OUT 2  r o 2 1 
  12.4M 1 
  555 M
 R th  r 1  R E 
 0.887k  612k  330k 
57
15.130
20k
 4.07V | R EQ  20k 39k  13.2k
20k  39k
 1 V B  0. 7 1 V B  0.7 1 V B  0.7 
I B   


 | V B  4. 07V  13200I B  V B  3.95V
76 33k
76 16k
76 8.2k 
V EQ  12V
I C1 

 o 33k 

75 0  0.7V  3. 95V 

  97.2A | R OUT1  r o1 1 

76 
33k

 R th1  r 1  33k 

R th1  13.2k r 2   o 2  116k
R OUT1 
IC 2 
r 3   o3  18.2k 13.2k
60  8.75 
75 33k 

1 
 = 27. 4 M
97.2A  13.2k  19.3k  33k 


75 0  0.7V  3.95V 
 o 16k 

  201A | R OUT  r o2 1 

76 
16k

 R th2  r 2  16k 



R th2  13. 2k r 1   o1  133k r 3   o 3 18. 2k  13.2k
R OUT 2 
IC 3 
75 16k 

60  8. 75 
1 
 = 11.0 M
201A  13.2k  9. 33k  16k 

 o 8. 2k 

75 0  0.7V  3.95V 

  391A | R OUT  r o 3 1 

76 
8.2k

 R th2  r 2  8.2k 



R th3  13. 2k r 1   o1  133k r 2   o 2 116k  13. 2k
R OUT 3 
60  8. 75 
75 8. 2k 

1 
 = 4. 30 M
391A  13.2k  4.80k  8.2k 
15.131
V EQ  12V
2M
 6. 00V | R EQ  2M 2M  1. 00M | Assume saturation:
2M  2M
5
2I SD1
4
2.5x10
 6  V SG1  7.59 V
12  10 I SD1  V SG1  6 | V SG1  1 
I SD1  44.1 A, V SG1  1.59V, V SD1


50V  7.59V
1  100k 2250A 44.1A 1  0.02 7.69   22. 2 M
44.1A
2I SD2
5
12  4.7x10 ISD2  V SG2  6 | V SG2  1 
4
2.5x10
I SD2  10. 0 A, V SG2  1. 28 V, V SD2  6  V SG2  7. 28V
50 V  7. 28V
R OUT2  r o 2 1  g m2 R 2  
1  470k 2250A 10.0A1  0.027.28   210 M
10.0A
R OUT1  r o1 1  g m1R1  

15.132
58
*Problem 15.132
VCC 1 0 DC 12
R1 1 2 100K
R4 1 3 2MEG
R3 3 0 2MEG
R2 1 4 470K
M1 5 3 2 2 PFET
M2 6 3 4 4 PFET

VD1 5 0 DC 0
VD2 6 0 DC 0
.MODEL PFET PMOS VTO=-1 KP=250U LAMBDA=0.02
.OP
*.TF I(VD1) VD1
.TF I(VD2) VD2
.END
Results: IO1 = 44.4 A, ROUT1 = 22.1 M, IO1 = 10.1 A, ROUT1 = 209 M
15.133
For large A, I O 
V REF
5V

 100A
R
50k
For the small - signal model above,
vx  v s  i x  g m v r o | v =  Av s  v s   vs 1  A  | v s  i xR | Combining:
R OUT 
vx
50 V  10V
 R  r o 1  g m R 1 A  | r o 
 600k
ix
100A



g m  2 8x10 4 10 4 1  0.02 10   0.438mS


R OUT  50k  600k 1  0.438mS 50k 1  5x104
15.134
For large A, I O   F
 6.57x10
11
 !!
V REF 120 5V

 99.2 A
R
121 50k
For the small - signal model above,
vx  ve  i x  g m vr o | v =  Ave   ve  v e 1  A  | i x  Gve  g  1  A v e | Combining:
R OUT 
vx
1   f 1  A 
1


 r o 1   o  for g  1  A   G and f 1  A   1
ix
G  g  1  A g o
50V  10V
ro 
 605k | R OUT  605k121   73.2 M
99.2A
R OUT cannot exceed or o because of the loss of base current through r .
15.135 ROUT is limited to oro of the BJT. We need to increase the effective current gain of
the transistor which can be done by replacing Q1 with a Darlington configuration of two
transistors.
Now ROUT can approach the oro product of the Darlington which is R OUT 
2 2
 or o2
3
15.136
59
91k
 9.03 V | R EQ  91k 30k  22.6k
91k  30k
12  0.7  9. 03
V
 85I B 3  85
 9.34 A
22.6k  86240k 
V EQ  12V
IC 3
86
9.34A240k  0.7  9.03V
85
I
85 9.34A
I C1  I C 2   F C 3 
 4.62A | V EC 1  V EC 2  0.7  12  1.2M4. 62A   7.16 V
2
86
2
Q  po int s: 4. 62A,7. 62V  4.62A,7.62V  9. 34A, 9.03V 
V EC 3  12  I E R E  0.7  12 
r 3 
85 0.025 V 
9. 34A
 228k | r o3 
70  9.03 V
9.34A
 8.46M



85 240k 

o R E
R OUT3  r o 3 1 
  8.46M1 
  360M
R

r

R
22.6k

228k

240k



th
3
E 
g mR C
 204. 62A 1.2M   111 40.9dB 
2
 40 4. 62A 360M   6.65x10 4 96.5dB 
For a single - ended output, A V 
CMRR  g m1R OUT3
60
15.137
V EQ  15V
100k
 9.93V | R EQ  100k 51k  33.8k | Assume saturation:
100k  51k
9.93  15  7500I DS3  V GS3 | V GS3  1 
I DS1  I DS2 
2I DS3
4x10
2363A 
I DS3
 182A | V GS1  1 
2
2400A 
4
| I DS3  363 A, V GS3  2. 35V
 1.95V
V DS3  V GS1  7500IDS3  15   10.3V
V DS1  V DS2  15  36000 IDS1  V GS1  10. 4V | r o 3 

50V  10.3V
 166k
363A




R OUT3  r o 3 1 g m3 R S   166k 1  2 4x10 4 3. 63x10 4 1  0.0210.3 7.5k  903 k





A dd   g m R D   2 4x10 4 1. 82x10 4 1  0.02 10. 436k  0.419mS 36k  15.1
For a single - ended output, CMRR = g m1R OUT 3  0. 419mS 903k  378 51. 6 dB 
15.138
r o1
since the collector current of the
2
current source is twice that of the input transistors. For a single - ended output,
Assuming all devices are identical, R OUT =  o1
A dd  
g m1R C
RC
RC
g  r
 
| A cc  
=
| CMRR = m1 o1 o1  o1 f 1
r 
2

 o1r o1
2
2
2 o1 o1 

2 
Using our default paramters: CMRR  20 o1 V A1  2010070   140,000 103dB 
(Note that this analysis neglects the contribution of the output resistance r o of the input
pair.
If this resistance is included, a theoretical cancellation occurs and A cc = 0! Of
 r
course the output resistance expression R OUT  o o is not precise, but an improvement
2
over the CMRR expression above is possible.)
15.139
R OUT  r o 1  g m R S   F R S  g m r o R S  2K n I DS
V R S  I DS R S 
15.140
I1.5
DS R OUT
2K n


1
RS
I DS
 5x10   3.16 V
25x10 
0. 02 10 4
1.5
6
4
*Problem 15.140 - Fig. 15.72(a) - BJT Current Source Monte Carlo Analysis
*Generate a Voltage Source with 5% Tolerances
IEE 0 5 DC 1
REE 5 0 RTOL 15
EEE 1 0 5 0 -1
*
VO 4 0 AC 1
RE 1 2 RTOL 18.4K
R1 1 3 RTOL 113K
R2 3 0 RTOL 263K
61
Q1 4 3 2 NBJT
.OP
.DC VO 0 0 .01
.AC LIN 1 1000 1000
.PRINT AC IM(VO) IP(VO)
.MODEL NBJT NPN BF=150 VA=75
.MODEL RTOL RES (R=1 DEV 5%)
.MC 1000 DC I(VO) YMAX
*.MC 1000 AC IM(VO) YMAX
.END
Results - 3 limits: Io = 199 A ± 32.5 A, ROUT = 11.8 M ± 2.6 M
*Problem 15.140 - Fig. 15.72(b) - MOSFET Current Source
*Generate a Voltage Source with 5% Tolerance
IEE 0 5 DC 1
REE 5 0 RTOL 15
EEE 1 0 5 0 -1
*
VO 4 0 AC 1
RS 1 2 RTOL 18K
R3 1 3 RTOL 240K
R4 3 0 RTOL 510K
M1 4 3 2 2 NFET
.OP
.DC VO 0 0 .01
.AC LIN 1 1000 1000
.PRINT AC IM(VO) IP(VO)
.MODEL NFET NMOS KP=9.95M VTO=1 LAMBDA=0.01
.MODEL RTOL RES (R=1 DEV 5%)
.MC 1000 DC I(VO) YMAX
*.MC 1000 AC IM(VO) YMAX
.END
Results - 3 limits: Io = 201 A ± 34.7 A, ROUT = 21.7 M ± 3.6 M
62