This document contains solutions to problems involving BJT circuit analysis. Key points include: 1) Problem 5.1 analyzes two common emitter amplifier circuits and calculates currents and voltages. 2) Problem 5.2 calculates alpha and collector current values for two common emitter circuits with different beta values. 3) The remaining problems involve calculating various parameter values like currents, voltages, and gains for additional common emitter, common base, and common collector BJT circuits. Circuit analysis methods like Kirchhoff's laws and transistor biasing equations are applied. 4) Graphs and calculations of operating points are presented for some circuits. Effects of component value variations are considered, such as the impact on collector

1. Chapter 5
Problem Solutions
5.1
iC 510
(a) β= = ⇒ β = 85
iB 6
β 85
α= = ⇒ α = 0.9884
1 + β 86
iE = (1 + β ) iB = ( 86 )( 6 ) ⇒ iE = 516 μ A
2.65
(b) β= ⇒ β = 53
0.050
53
α = ⇒ α = 0.9815
54
iE = (1 + β ) iB = ( 54 )( 0.050 ) ⇒ iE = 2.70 mA
5.2
β 110
(a) For β = 110: α = = = 0.99099
1+ β 111
180
For β = 180: α = = 0.99448
181
0.99099 ≤ α ≤ 0.99448
(b) I C = β I B = 110 ( 50 μ A ) ⇒ I C = 5.50 mA
or I C = 180 ( 50 μ A ) ⇒ I C = 9.00 mA
so 5.50 ≤ I C ≤ 9.0 mA
5.3
1.12
(a) iB = ⇒ 9.33 μ A
120
⎛ 121 ⎞
iE = (1.12 ) ⎜ ⎟ = 1.13 mA
⎝ 120 ⎠
120
α = = 0.9917
121
50
(b) iB = = 2.5 mA
20
⎛ 21 ⎞
iE = ⎜ ⎟ ( 50 ) = 52.5 mA
⎝ 20 ⎠
20
α = = 0.9524
21
5.4
(a)
α
α β=
1−α
0.9 9
0.95 19
0.98 49
0.99 99
0.995 199
0.999 999
(b)

2. β
β α=
1+ β
20 0.9524
50 0.9804
100 0.9901
150 0.9934
220 0.9955
400 0.9975
5.5
1.2
(a) IB = ⇒ 14.8 μ A
81
⎛ 80 ⎞
I C = (1.2 ) ⎜ ⎟ = 1.185 mA
⎝ 81 ⎠
80
α= = 0.9877
81
VC = 5 − (1.185 )( 2 ) = 2.63 V
0.80
(b) IB = ⇒ 9.88 μ A
81
⎛ 80 ⎞
I C = ( 0.80 ) ⎜ ⎟ = 0.790 mA
⎝ 81 ⎠
80
α= = 0.9877
81
VC = 5 − ( 0.790 )( 2 ) = 3.42 V
(c) Yes, VC > VB so B-C junction is reverse biased in both areas.
5.6
5
For VC = 0, I C = = 2.5 mA
2
IC 2.5
IE = = ⇒ I E = 2.546 mA
α 0.982
5.7
0.75
(a) IB = ⇒ 12.3 μ A
61
⎛ 60 ⎞
IC = ( 0.75 ) ⎜ ⎟ = 0.738 mA
⎝ 61 ⎠
60
α = = 0.9836
61
VC = I C RC − 10 = ( 0.738 )( 5 ) − 10
VC = −6.31 V
1.5
(b) IB = ⇒ 24.6 μ A
61
⎛ 60 ⎞
I C = (1.5 ) ⎜ ⎟ = 1.475 mA
⎝ 61 ⎠
⎛ 60 ⎞
α = ⎜ ⎟ = 0.9836
⎝ 61 ⎠
VC = (1.475 )( 5 ) − 10 ⇒ VC = −2.625 V
(c) Yes, VC < 0 in both cases so that B-C junction is reverse biased.

3. 5.8
VC − ( −10 ) 10 − 1.2
IC = = = 1.76 mA
RC 5
IC 1.76
IE = = ⇒ I E = 1.774 mA
α 0.992
5.9
I C = I S eVRE / VT
⎛ 0.685 ⎞
= 10−13 exp ⎜ ⎟ ⇒ I C = 27.67 mA
⎝ 0.026 ⎠
⎛1+ β ⎞ ⎛ 91 ⎞
IE = ⎜ ⎟ I C = ⎜ ⎟ ( 27.67 )
⎝ β ⎠ ⎝ 90 ⎠
I D = 27.98 mA
I 27.67
IB = C = ⇒ I B = 0.307 mA
β 90
5.10
Device 1: iE = I Eo1evEB / VT ⇒ 0.5 × 10−3 = I Eo1e0.650 / 0.026
So that
I EO1 = 6.94 × 10−15 A
Device 2: 12.2 × 10−3 = I Eo 2 e0.650 / 0.026
Or
I Eo 2 = 1.69 × 10−13 A
I Eo 2 1.69 × 10−13
Ratio of areas = = ⇒ Ratio = 24.4
I Eo1 6.94 × 10−15
5.11
VA 250
(a) ro = = ⇒ ro = 250 k Ω
IC 1
VA 250
(b) ro = = ⇒ ro = 2.50 M Ω
IC 0.1
5.12
BVC B 0 60
BVC E 0 = =
3 β 3
100
BVC E 0 = 12.9 V
5.13
BVC B 0
BVC E 0 =
3 β
220 220
56 = ⇒3β = = 3.93
3 β 56
β = 60.6
5.14

4. BVC B 0
BVC E 0 =
3 β
BVC B 0 = ( BVC E 0 ) 3 β = ( 50 ) 3 50
BVC B 0 = 184 V
5.15
−0.7 − ( −10 )
(a) IE = = 1.86 mA
5
⎛ 75 ⎞
I C = (1.86 ) ⎜ ⎟ = 1.836 mA
⎝ 76 ⎠
VC = −0.7 + 4 = 3.3 V
10 − 3.3
RC = ⇒ RC = 3.65 K
1.836
0.5
(b) IB = = 0.00658 mA
76
VB = I B RB = ( 0.00658 )( 25 ) ⇒ VB = 0.164 V
⎛ 75 ⎞
I C = ( 0.5 ) ⎜ ⎟ = 0.493 mA
⎝ 76 ⎠
−1 − ( −5 )
RC = ⇒ RC = 8.11 K
0.493
I
(c) O = E (10 ) + 0.7 + I E ( 4 ) − 8
76
7.3 = I E ( 4 + 0.132 ) ⇒ I E = 1.767 mA
⎛ 75 ⎞
I C = (1.767 ) ⎜ ⎟ = 1.744 mA
⎝ 76 ⎠
VCE = 8 − (1.744 )( 4 ) − ⎡(1.767 )( 4 ) − 8⎤
⎣ ⎦
= 16 − 6.972 − 7.068 ⇒ VCE = 1.96 V
⎛I ⎞
(d) 5 = I E (10 ) + ⎜ E ⎟ ( 20 ) + 0.7 + I E ( 2 ) = I E (10 + 0.263 + 2 ) + 0.7
⎝ 76 ⎠
I E = 0.3506 mA ⇒ I B = 4.61 μ A VC = 5 − ( 0.3506 )(10 )
VC = 1.49 V
5.16
For Fig. 5.15 (a) RE = 5 + 5% = 5.25 K
−0.7 − ( −10 )
IE = = 1.77 mA
5.25
I C = 1.75 mA
10 − 3.3
RC = = 3.83 K
1.75
RE = 5 − 5% = 4.75 K
−0.7 − ( −10 )
IE = = 1.96 mA
4.75
I C = 1.93 mA
10 − 3.3
RC = = 3.47 K
1.93
So 1.75 ≤ I C ≤ 1.93 mA 3.47 ≤ RC ≤ 3.83 K
For Fig. 5.15(c) RE = 4 + 5% = 4.2 K

5. 8 − 0.7
IB = = 0.0222 mA I C = 1.66 mA
10 + ( 76 )( 4.2 )
I E = 1.69 mA
VCE = 16 − (1.66 )( 4 ) − (1.69 )( 4.2 )
= 16 − 6.64 − 7.098 ⇒ VCE = 2.26 V
RE = 4 − 5% = 3.8 K
8 − 0.7
IB = = 0.0244 I C = 1.83 mA
10 + ( 76 )( 3.8 )
I E = 1.86 mA
VCE = 16 − (1.83)( 4 ) − (1.86 )( 3.8 )
= 16 − 7.32 − 7.068
VCE = 1.61 V
So 1.66 ≤ I C ≤ 1.83 mA 1.61 ≤ VCE ≤ 2.26 V
5.17
VBB − VEB 2.5 − 0.7
RB = = ⇒ RB = 120 k Ω
IB 0.015
I CQ = ( 70 )(15μ A ) ⇒ 1.05 mA
VCC − VECQ 5 − 2.5
RC = = ⇒ RC = 2.38 k Ω
I CQ 1.05
5.18
(a)
−VB − ( −1)
VB = − I B RB ⇒ I B = =
RB 500
I B = 2.0 μ A
VE = −1 − 0.7 = −1.7 V
VE − ( −3) −1.7 + 3
IE = = = 0.2708 mA
RE 4.8
IE 0.2708
= (1 + β ) = = 135.4 ⇒ β = 134.4
IB 0.002
β
α= ⇒ α = 0.9926
1+ β
I C = β I B ⇒ I C = 0.269 mA
VCE = 3 − VE = 3 − ( −1.7 ) ⇒ VCE = 4.7 V
(b)
5−4
IE = ⇒ I E = 0.5 mA
2
4 = 0.7 + I B RB + ( I B + I C ) RC − 5
I B + IC = I E
I B + IC = I E
4 = 0.7 + I B (100 ) + ( 0.5 )( 8 ) − 5
IE 0.5
I B = 0.043 ⇒ = (1 + β ) = = 11.63
IB 0.043
β
β = 10.63, α = ⇒ α = 0.9140
1+ β
5.19

6. VB − 0.7 − ( −5 )
VB + 4.3
IE = =
3 3
⎛ 50 ⎞ ⎛ VB + 4.3 ⎞ ⎛ 50 ⎞
IC = ⎜ ⎟ I E = ⎜ ⎜ ⎟
⎝ 51 ⎠ ⎝ 3 ⎟ ⎝ 51 ⎠
⎠
⎛ V + 4.3 ⎞ ⎛ 50 ⎞
VC = 5 − I C RC = 5 − ⎜ B ⎜ ⎟ (10 )
⎝ 3 ⎟ ⎝ 51 ⎠
⎠
Now VB = VC , so VB [1 + 3.27 ] = 5 − 14.1 = −9.05
VB = −2.12 V
−2.12 + 4.3
IE = ⇒ I E = 0.727 mA
3
5.20
10 − VE 10 − 2
IE = = ⇒ I E = 0.80 mA
10 10
VB = VE − 0.7 = 2 − 0.7 = 1.3 V
VB 1.3
IB = = ⇒ I B = 0.026 mA
RB 50
I C = I E − I B = 0.80 − 0.026 ⇒ I C = 0.774 mA
I C 0.774
β= = ⇒ β = 29.77
I B 0.026
β 29.77
α= = ⇒ α = 0.9675
1+ β 30.77
VEC = VE − VC = VE − ( I C RC − 10 )
= 2 − ⎡( 0.774 )(10 ) − 10 ⎤
⎣ ⎦
VEC = 4.26 V
Load line developed assuming the VB voltage can change and the RB resistor is removed.
1 mA
Q-point
0.774
IC
4.26 20 VEC
5.21
5 − 0.7
IB = ⇒ 17.2 μ A
250
I C = (120 )( 0.0172 ) = 2.064 mA
VC = ( 2.064 )(1.5 ) − 5 = −1.90 V
VEC = 5 − ( −1.90 ) ⇒ VEC = 6.90 V

7. IC (mA)
6.67
Q-point
2.06
6.9 10 V
EC (V)
5.22
⎛ 50 ⎞
I C = ⎜ ⎟ (1) = 0.98 mA
⎝ 51 ⎠
VC = I C RC − 9 = ( 0.98 )( 4.7 ) − 9 or VC = −4.39 V
1
IB = = 0.0196 mA
51
VE = I B RB + VEB ( on ) = ( 0.0196 )( 50 ) + 0.7 or VE = 1.68 V
5.23
⎛ 50 ⎞ 0.5
I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA, I B = = 0.0098 mA
⎝ 51 ⎠ 51
VE = I B RB + VEB ( on ) = ( 0.0098 )( 50 ) + 0.7 or VE = 1.19 V
VC = I C RC − 9 = ( 0.49 )( 4.7 ) − 9 = −6.70 V
Then VEC = VE − VC = 1.19 − ( −6.7 ) == 7.89 V
PQ = I CVEC + I BVEB = ( 0.49 )( 7.89 ) + ( 0.0098 )( 0.7 ) or PQ = 3.87 mW
Power Dissipated = PS = I Q ( 9 − VE ) = ( 0.5 )( 9 − 1.19 )
Or PS = 3.91 mW
5.24
I
I E1 = I E 2 = ⇒ I E1 = I E 2 = 0.5 mA
2
I C1 = I C 2 ≈ 0.5 mA
VC1 = VC 2 = 5 − ( 0.5 )( 4 ) ⇒ VC1 = VC 2 = 3 V
5.25
2 − 0.7 1.3
(a) RE = 0 I B = =
RB RB
⎛ 1.3 ⎞ 5 − 2
I C = ( 80 ) ⎜ ⎟ = = 0.8 ⇒ RC = 3.75 K
⎝ RB ⎠ RC
RB = 130 K
0.8 ⎛ 81 ⎞
(b) RE = 1 K IB = = 0.010 mA I E = 0.8 ⎜ ⎟ = 0.81 mA
80 ⎝ 80 ⎠
2 = ( 0.010 )( RB ) + 0.7 + ( 0.81)(1) ⇒ RB = 49 K
5 = ( 0.8 ) RC + 2 + ( 0.81)(1) ⇒ RC = 2.74 K
2 − 0.7
(c) For part (a) IB = = 0.01 mA
130

8. I C = (120 )( 0.01) ⇒ I C = 1.20 mA
VCE = 5 − (1.2 )( 3.75 ) ⇒ VCE = 0.5 V
For part (b) 2 = I B ( 49 ) + 0.7 + (121) I B (1)
I B = 0.00765 mA, I E = 0.925 mA, I C = 0.918 mA
VCE = 5 − ( 0.918 )( 2.74 ) − ( 0.925 )(1) ⇒ VCE = 1.56 V
Including RE result in smaller changes in Q-point values.
5.26
VCC − VBE ( on )
a. I BQ =
RB
I CQ 2
I BQ = == 0.0333 mA
β 60
24 − 0.7
RB = ⇒ RB = 699 kΩ
0.0333
VCC − VCEQ 24 − 12
I CQ = ⇒ RC = ⇒ RC = 6 kΩ
RC 2
VCC − VBE ( on ) 24 − 0.7
b. I BQ = =
RB 699
= 0.0333 mA ( Unchanged )
I CQ = β I BQ = (100 )( 0.0333) ⇒ I CQ = 3.33 mA
VCEQ = VCC − I CQ RC = 24 − ( 3.33)( 6 ) ⇒ VCEQ = 4.02 V
(c) VCE = VCC − I C RC = 24 − I C ( 6 )
IC (mA)
4
Q-pt (␤ ϭ 100)
3.33
Q-pt (␤ ϭ 60)
2
4.02 12 24 VCE
5.27
a. VB = 0 ⇒ Cutoff ⇒ I E = 0, VC = 6 V
1 − 0.7
b. VB = 1 V, I E = ⇒ I E = 0.3 mA
1
I C ≈ I E ⇒ VC = 6 − ( 0.3)(10 ) ⇒ VC = 3 V
c. VB = 2 V. Assume active-mode
2 − 0.7
IE = = I E = 1.3 mA ≈ I C
1
VC = 6 − (1.3)(10 ) = −7 V!

9. Transistor in saturation
2 − 0.7
IE = ⇒ I E = 1.3 mA
1
VE = 1.3 V, VCE ( sat ) = 0.2 V
VC = VE + VCE ( sat ) = 1.3 + 0.2 ⇒ VC = 1.5 V
5.28
a. VBB = 0.
⎛ RL ⎞ ⎛ 10 ⎞
Cutoff V0 = ⎜ ⎟ VCC = ⎜ ⎟ ( 5)
⎝ RC + RL ⎠ ⎝ 10 + 5 ⎠
V0 = 3.33 V
b. VBB = 1 V
1 − 0.7
IB = ⇒ 6 μA
50
I C = β I B = ( 75 )( 6 ) ⇒ I C = 0.45 mA
5 − V0 V
= IC + 0
5 10
⎛1 1 ⎞
1 − 0.45 = V0 ⎜ + ⎟ ⇒ V0 = 1.83 V
⎝ 5 10 ⎠
c. Transistor in saturation V0 = VCE ( sat ) = 0.2 V
5.29
(a) β = 100
⎛ 100 ⎞
(i) I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.0990 mA
⎝ 101 ⎠
VO = 5 − ( 0.099 )( 5 ) ⇒ VO = 4.505 V
⎛ 100 ⎞
(ii) I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.495 mA
⎝ 101 ⎠
VO = 5 − ( 0.495 )( 5 ) ⇒ VO = 2.525 V
(iii) I Q = 2 mA Transistor is in saturation
VO = −VBE ( sat ) + VCE ( sat ) = −0.7 + 0.2 ⇒ VO = −0.5 V
(b) β = 150
⎛ 150 ⎞
(i) I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.09934 mA
⎝ 151 ⎠
VO = 5 − ( 0.09934 )( 5 ) ⇒ VO = 4.503 V
4.503 − 4.505
% change = × 100% = −0.044%
4.503
⎛ 150 ⎞
(ii) I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.4967 mA
⎝ 151 ⎠
VO = 5 − ( 0.4967 )( 5 ) ⇒ VO = 2.517 V
2.517 − 2.525
% change = × 100% = −0.32%
2.525
(iii) I Q = 2 mA Transistor in saturation
Vo = −8.5 V No change
5.30

10. 5 − 0.5
VCB = 0.5 V ⇒ VO = 0.5 V , I C = = 0.90 mA
5
⎛ 101 ⎞
IQ = ⎜ ⎟ ( 0.90 ) ⇒ I Q = 0.909 mA
⎝ 100 ⎠
5.31
For I Q = 0, then PQ = 0
⎛ 50 ⎞
For I Q = 0.5 mA, I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA
⎝ 51 ⎠
0.5
IB = = 0.0098 mA, VB = 0.490 V , VE = 1.19 V
51
VC = ( 0.49 )( 4.7 ) − 9 = −6.70 V ⇒ VEC = 7.89 V
P ≅ I CVEC = ( 0.49 )( 7.89 ) ⇒ P = 3.87 mW
For I Q = 1.0 mA, Using the same calculations as above, we find P = 5.95 mW
For I Q = 1.5 mA, P = 6.26 mW
For I Q = 2 mA, P = 4.80 mW
For I Q = 2.5 mA, P = 1.57 mW
For I Q = 3 mA, Transistor is in saturation.
0.7 + I B ( 50 ) = 0.2 + I C ( 4.7 ) − 9
I E = IQ = I B + IC ⇒ I B = 3 − IC
Then, 0.7 + ( 3 − I C ) ( 50 ) = 0.2 + I C ( 4.7 ) − 9
Which yields I C = 2.916 mA and I B = 0.084 mA
P = I BVEB + I CVEC = ( 0.084 )( 0.7 ) + ( 2.916 )( 0.2 ) or P = 0.642 mW
5.32
VEE − VEB ( on ) 9 − 0.7
IE = = ⇒ I E = 2.075 mA
RE 4
I C = α I E = ( 0.9920 ) ( 2.075 ) ⇒ I C = 2.06 mA
VBC + I C RC = VCC
VBC = 9 − ( 2.06 ) ( 2.2 ) ⇒ VBC = 4.47 V
5.33
VCC − VCEQ 12 − 6
I CQ = = = 2.73 mA
RC 2.2
I CQ 2.73
I BQ = = ⇒ I BQ = 0.091 mA
β 30
0.7 − ( −12 )
IR2 = = 0.127 mA
100
I R1 = I R 2 + I BQ = 0.127 + 0.091 = 0.218 mA
V1 = I R1 R1 + 0.7 = ( 0.218 )(15 ) + 0.7 ⇒ V1 = 3.97 V
5.34

11. For VCE = 4.5
5 − 4.5
I CQ = = 0.5 mA
1
0.5
I BQ = = 0.02 mA
25
0.7 − ( −5 )
I R2 = = 0.057 mA
100
I R1 = I R 2 + I BQ = 0.057 + 0.02 = 0.077 mA
V1 = I R1 R1 + VBE ( on ) = ( 0.077 )(15 ) + 0.7 = 1.86 V
For VCE = 1.0
5 −1
I CQ = = 4 mA
1
4
I BQ = = 0.16 mA
25
I R2 = 0.057 mA
I R1 = I R 2 + I BQ = 0.057 + 0.16 = 0.217 mA
V1 = ( 0.217 )(15 ) + 0.7 ⇒ 3.96 V
So 1.86 ≤ V1 ≤ 3.96 V
IC
5
4
Range of
Q-pt values
0.5
0 1 4.5 5
5.35
5 − 2.5
(a) RC = =5K
0.5
0.5
IB = = 0.00417 mA
120
5 − 0.7
RB = = 1032 K
0.00417
IC (mA)
1.0
Q-point
0.5
2.5 5 V (V)
CE
(b) Choose RC = 5.1 K
RB = 1 MΩ

12. For RB = 1 MΩ + 10% = 1.1 M, RC = 5.1 k + 10% = 5.61 K
5 − 0.7
I BQ = = 3.91 μ A ⇒ I CQ = 0.469 mA
1.1
VCEQ = 2.37 V
RB = 1 MΩ + 10% = 1.1M, RC = 5.1 K − 10% = 4.59 K
I BQ = 3.91 μ A ⇒ I CQ = 0.469 mA
VCEQ = 2.85 V
RB = 1 MΩ − 10% = 0.90 MΩ RC = 5.1 k + 10% = 5.61 K
5 − 0.7
I BQ = = 4.78 μ A ⇒ I C = 0.573 mA
0.90
VCEQ = 1.78 V
RB = 1 MΩ − 10% = 0.90 MΩ RC = 5.1 k − 10% = 4.59 K
I BQ = 4.78 μ A ⇒ I C = 0.573 mA
VCEQ = 2.37 V
IC (mA)
1.09
0.891
0.573
0.469
1.78 2.37 2.85 5 V (V)
CE
5.36
VE 2 = 5 − VBE 2 VE1 = 5 − VBE1
VO = VE 2 − VE1 = ( 5 − VBE 2 ) − ( 5 − VBE1 )
VO = VBE1 − VBE 2
⎛I ⎞
We have VBE1 = VE ln ⎜ E1 ⎟
⎝ I EO ⎠
⎛I ⎞
VBE 2 = VT ln ⎜ E 2 ⎟
⎝ I EO ⎠
⎡ ⎛I ⎞ ⎛I ⎞⎤
VO = VT ⎢ln ⎜ E1 ⎟ − ln ⎜ E 2 ⎟⎥
⎣ ⎝ I EO ⎠ ⎝ I EO ⎠⎦
⎛I ⎞ ⎛ 10 ⎞
VO = VT ln ⎜ E1 ⎟ = VT ln ⎜ I ⎟
⎝ IE2 ⎠ ⎝ I ⎠
kT
VO = ln (10 )
e
5.37
(a) RE = 0
VI − 0.7 (120 )( 4 )
IB = IC = β I B VO = 5 − I C ( 4 ) = 5 − (VI − 0.7 )
200 200
When VO = 0.2, 0.2 = 5 − 2.4 VI + 1.68 ⇒ VI = 2.7

13. VO(V)
5
0.2
0.7 2.7 5 VI (V)
(b) RE = 1 K
VI − 0.7 V − 0.7
IB = = I I C = β IB
200 + (121)(1) 321
(120 )( 4 )
VO = 5 − (VI − 0.7 )
321
When VO = 0.2 = 5 − 1.495VI + 1.047
VI = 3.91 V
VO(V)
5
0.2
0.7 3.91 5 VI (V)
5.38
For 4.3 ≤ VI ≤ 5 Q is cutoff I C = 0
VO = 0
If Q reaches saturation, VO = 4.8
4.8
IC = = 1.2 mA
4
1.2 5 − 0.7 − VI
IB = = 0.015 = ⇒ VI = 1.6
80 180
So VI ≤ 1.6, VO = 4.8

14. VO(V)
4.8
1.6 4.3 5 VI (V)
5.39
(a) For VI ≥ 4.3, Q is off and VO = 0
⎛ 101 ⎞
When transistor enters saturation, 5 = ⎜ ⎟ I C (1) + 0.2 + I C ( 4 ) ⇒ I C = 0.958 mA
⎝ 100 ⎠
VO = 3.832 V
I B = 0.00958 mA
⎛ 101 ⎞
5=⎜ ⎟ ( 0.958 )(1) + 0.7 + ( 0.00958 )(180 ) + VI
⎝ 100 ⎠
VI = 5 − 0.7 − 0.9676 − 1.7244 ⇒ VI = 1.61 V
For VI = 0, transistor in saturation
5 = I E (1) + 0.2 + I C ( 4 ) ⇒ 5 = I C (1) + I B (1) + 0.2 + I C ( 4 )
5 = I E (1) + 0.7 + I B (180 ) 5 = I C (1) + I B (1) + 0.7 + I B (180 )
I E = IC + I B
4.8 = 5 I C + I B (1)
4.3 = 1I C + 181I B
I B = 4.8 − 5 I C
4.3 = I C + (181)( 4.8 − 5 I C )
904 I C = 864.5
I C = 0.956 mA
VO = 3.825 V

15. VO(V)
3.832
3.825
1.61 4.3 5 VI (V)
5.40
RTH = R1 R2 = 33 10 = 7.67 kΩ
⎛ R2 ⎞ ⎛ 10 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ (18 ) = 4.186 V
⎝ R1 + R2 ⎠ ⎝ 10 + 33 ⎠
V − VBE ( on ) 4.186 − 0.7
I BQ = TH =
RTH + (1 + β ) RE 7.67 + ( 51)(1)
I BQ = 0.0594 mA
I CQ = β I BQ ⇒ I CQ = 2.97 mA
I EQ = 3.03 mA
VCEQ = VCC − I CQ RC − I EQ RE
= 18 − ( 2.97 )( 2.2 ) − ( 3.03)(1) ⇒ VCEQ = 8.44 V
5.41
I CQ = 1.2 mA, VCEQ = 9 V , RTH = 50 k Ω
1.2
Also I B = = 0.015 mA
80
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
⎛ R2 ⎞ 1 1
VTH = ⎜ ⎟ (VCC ) = ⋅ RTH ⋅ VCC = ( 50 )(18 )
⎝ R1 + R2 ⎠ R1 R1
1
Then ( 50 )(18) = ( 0.015)( 50 ) + 0.7 + ( 81)( 0.015)(1) or R1 = 338 k Ω.
R1
338R2
Then = 50 ⇒ R2 = 58.7 k Ω
338 + R2
⎛ 81 ⎞
I EQ = ⎜ ⎟ (1.2 ) = 1.215 mA
⎝ 80 ⎠
18 = I CQ RC + VCEQ + I EQ RE
18 = (1.2 ) RC + 9 + (1.215 )(1) ⇒ RC = 6.49 k Ω
5.42

16. RTH = R1 R2 = 20 15 = 8.57 k Ω
⎛ R2 ⎞ ⎛ 15 ⎞
VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 4.29 V
⎝ R1 + R2 ⎠ ⎝ 15 + 20 ⎠
I EQ
VCC = I EQ RE + VEB ( on ) + ⋅ RTH + VTH
1+ β
⎛ 8.57 ⎞
10 = I EQ (1) + 0.7 + I EQ ⎜ ⎟ + 4.29
⎝ 101 ⎠
10 − 0.7 − 4.29 5.01
Then I EQ = = ⇒ I EQ = 4.62 mA
8.57 1.085
1+
101
I EQ ⎛ 4.62 ⎞
VB = ⋅ RTH + VTH = ⎜ ⎟ ( 8.57 ) + 4.29 or VB = 4.68 V
1+ β ⎝ 101 ⎠
5.43
(a)
RTH = 42 58 = 24.36 K
⎛ 42 ⎞
VTH = ⎜ ⎟ ( 24 ) = 10.08 V
⎝ 100 ⎠
10.08 − 0.7 9.38
I BQ = = ⇒ 7.30 μ A
24.36 + (126 )(10 ) 1284.36
I CQ = 0.913 mA I EQ = 0.9202
VCEQ = 14.8 V
IC (mA)
2.38
Q-point
0.913
14.8 24 V
CE (V)
(b)
R1 + 5% = 60.9, R2 + 5% = 44.1 RTH = 25.58 K VTH = 10.08
10.08 − 0.7 9.38
I BQ = = ⇒ 7.30 μ A
25.58 + 126 (10 ) 1285.58
I CQ = 0.912 mA I EQ = 0.919
VCEQ = 14.81

17. R1 + 5% = 60.9, R2 − 5% = 39.90 RTH = 24.11 K VTH = 9.50
9.50 − 0.7 8.8
I BQ = = = 6.85 μ A
24.11 + (126 )(10 ) 1284.11
I CQ = 0.857 mA I EQ = 0.8635 mA
VCEQ = 15.37 V
R1 − 5% = 55.1 K R2 + 5% = 44.1 K RTH = 24.50 K VTH = 10.67 V
10.67 − 0.7 9.97
I BQ = = = 7.76 μ A
24.50 + 1260 1284.5
I CQ = 0.970 mA I EQ = 0.978 mA
VCEQ = 14.22 V
R1 − 5% = 55.1 K R2 − 5% = 39.90 RTH = 23.14 K VTH = 10.08
10.08 − 0.7 9.38
I BQ = = = 7.31 μ A
23.14 + 1260 1283.14
I CQ = 0.914 mA I EQ = 0.9211 mA
VCEQ = 14.79 V
So we have 0.857 ≤ I CQ ≤ 0.970 mA
14.22 ≤ VCEQ ≤ 15.37 V
5.44
a.
RTH = R1 R2 = 25 8 = 6.06 kΩ
⎛ R2 ⎞ ⎛ 8 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 24 )
⎝ R1 + R2 ⎠ ⎝ 8 + 25 ⎠
= 5.82 V
V − VBE (on) 5.82 − 0.7
I BQ = TH =
RTH + (1 + β ) RE 6.06 + ( 76 )(1)
I BQ = 0.0624 mA, I CQ = 4.68 mA
I EQ = 4.74
VCEQ = VCC − I CQ RC − I EQ RE
= 24 − ( 4.68 )( 3) − ( 4.74 )(1)
VCEQ = 5.22 V
b.
5.82 − 0.7
I BQ = ⇒ I BQ = 0.0326 mA
6.06 + (151)(1)
I CQ = 4.89 mA
I EQ = 4.92
VCEQ = 24 − ( 4.89 )( 3) − ( 4.92 )(1)
VCEQ = 4.41 V
5.45
(a)

18. I CQ ≅ I EQ = 0.4 mA
3 3
RC = ⇒ RC = 7.5 k Ω; RE = ⇒ RE = 7.5 k Ω
0.4 0.4
9
R1 + R2 ≅ = 112.5 k Ω
( 0.2 )( 0.4 )
⎛ R2 ⎞
VTH = ⎜ ⎟ (VCC ) = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
⎝ R1 + R2 ⎠
RR (112.5 − R2 ) R2 0.4
RTH = 1 2 = , I BQ = = 0.004 mA
R1 + R2 112.5 100
⎛ 9 ⎞ ⎡ (112.5 − R2 ) R2 ⎤
R2 ⎜ ⎟ = ( 0.004 ) ⎢ ⎥ + 0.7 + (101)( 0.004 )( 7.5 )
⎝ 112.5 ⎠ ⎣ 112.5 ⎦
We obtain R2 ( 0.08 ) = 0.004 R2 − 3.56 × 10−5 R2 + 3.73
2
From this quadratic, we find R2 = 48 k Ω ⇒ R1 = 64.5 k Ω
(b) Standard resistor values:
Set RE = RC = 7.5 k Ω and R1 = 62 k Ω, R2 = 47 k Ω
Now RTH = R1 R2 = 62 47 = 26.7 k Ω
⎛ R2 ⎞ ⎛ 47 ⎞
VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ ( 9 ) = 3.88 V
⎝ R1 + R2 ⎠ ⎝ 47 + 62 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
3.88 − 0.7
So I BQ = = 0.00406 mA
26.7 + (101)( 7.5 )
Then I CQ = 0.406 mA
VRC = VRE = ( 0.406 )( 7.5 ) = 3.05 V
5.46
(a)
RTH = R1 R2 = 12 2 = 1.714 K
⎛ R2 ⎞ ⎛ 2⎞
VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⇒ VTH = −3.571 V
⎝ R1 + R2 ⎠ ⎝ 14 ⎠
(b)
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5
−3.57 = I BQ (1.714 ) + 0.7 + (101) I BQ ( 0.5 ) − 5
5 − 0.7 − 3.571 0.729
I BQ = = ⇒ 13.96 μ A
1.714 + (101)( 0.5 ) 52.21
I CQ = 1.396 mA, I EQ = 1.410 mA
VCEQ = 10 − (1.396 )( 5 ) − (1.41)( 0.5 ) ⇒ VCEQ = 2.32 V
(d)

19. RE = 0.5 + 5% = 0.525 K RC = 5 + 5% = 5.25 K
0.729
I BQ = ⇒ 13.32 μ A
1.714 + (101)( 0.525 )
I CQ = 1.332 mA I EQ = 1.345 mA
VCEQ = 10 − (1.332 )( 5.25 ) − (1.345 )( 0.525 )
= 10 − 6.993 − 0.7061 ⇒ VCEQ = 2.30 V
RE = 0.5 + 5% = 0.525 K RC = 5 − 5% = 4.75 K
I CQ = 1.332 mA I EQ = 1.345 mA
VCEQ = 10 − (1.332 )( 4.75 ) − (1.345 )( 0.525 )
= 10 − 6.327 − 0.7061 ⇒ VCEQ = 2.97 V
RE = 0.5 − 5% = 0.475 K RC = 5 + 5% = 5.25 K
0.729
I BQ = ⇒ 14.67 μ A
1.714 + (101)( 0.475 )
I CQ = 1.467 mA I EQ = 1.482 mA
VCEQ = 10 − (1.467 )( 5.25 ) − (1.482 )( 0.475 )
= 10 − 7.70175 − 0.70395 ⇒ VCEQ = 1.59 V
RE = 0.5 − 5% = 0.475 K RC = 5 − 5% = 4.75 K
I CQ = 1.467 mA I EQ = 1.482 mA
VCEQ = 10 − (1.467 )( 4.75 ) − (1.482 )( 0.475 )
= 10 − 6.96825 − 0.70395 ⇒ VCEQ = 2.33 V
5.47
RTH = R1 R2 = 9 1 = 0.91 kΩ
⎛ R2 ⎞ ⎛ 1 ⎞
VTH = ⎜ ⎟ ( −12 ) = ⎜ ⎟ ( −12 ) = −1.2 V
⎝ R1 + R2 ⎠ ⎝ 1+ 9 ⎠
I EQ RE + VEB ( on ) + I BQ RTH + VTH = 0
−VTH − VEB (on) 1.2 − 0.7
I BQ = =
RTH + (1 + β ) RE 0.90 + ( 76 )( 0.1)
I BQ = 0.0588, I CQ = 4.41 mA
I EQ = 4.47 mA
Center of load line ⇒ VECQ = 6 V
I EQ RE + VECQ + I CQ RC − 12 = 0
( 4.47 ) ( 0.1) + 6 + ( 4.41) RC = 12 ⇒ RC = 1.26 kΩ
5.48
(a)
RTH = 36 68 = 23.5 K
⎛ 36 ⎞
VTH = ⎜ ⎟ (10 ) = 3.46 V
⎝ 36 + 68 ⎠
3.46 − 0.7
I BQ = = 0.00178 mA
23.5 + ( 51)( 30 )
I CQ = 0.0888 mA I EQ = 0.0906 mA
VCE = 10 − ( 0.0888 )( 42 ) − ( 0.0906 )( 30 )
= 10 − 3.73 − 2.72 ⇒ VCE = 3.55 V

20. (b)
R1 = 22.7, R2 = 12 K, RC = 14 K, RE = 10 K
RTH = 7.85 k VTH = 3.46
3.46 − 0.7
I BQ = = 0.00533 mA
7.85 + ( 51)(10 )
I CQ = 0.266 mA I EQ = 0.272 mA
VCE = 10 − ( 0.266 )(14 ) − ( 0.272 )(10 )
VCE = 3.56 V
5.49
(a)
⎛ R2 ⎞ ⎛ 68 ⎞
RTH = 36 68 = 23.5 K VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = 1.54 V
⎝ R1 + R2 ⎠ ⎝ 36 + 68 ⎠
5 = ( 51) I BQ ( 30 ) + 0.7 + I B ( 23.5 ) + 1.54
2.76
I BQ = = 1.78 μA ⇒ I CQ = 0.0888 mA
1553.5
I EQ = 0.0906 mA
VECQ = 10 − ( 0.0906 )( 30 ) − ( 0.0888 )( 42 )
= 10 − 2.718 − 3.7296 ⇒ VECQ = 3.55 V
(b)
RTH = 12 22.7 = 7.85 K
VTH = 1.54 RE = 10 K RC = 14 K
5 = ( 51) I BQ (10 ) + 0.7 + I B ( 7.85 ) + 1.54
2.76
I BQ = = 5.33 μ A I CQ = 0.266 mA
517.85
I EQ = 0.272 mA
VECQ = 10 − ( 0.272 )(10 ) − ( 0.266 )(14 )
= 10 − 2.72 − 3.724 ⇒ VECQ = 3.56 V
5.50
(a)
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.5 ) = 5.05 k Ω
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
R1
I CQ 0.8
I BQ = = = 0.008 mA
β 100
1
Then ( 5.05)(10 ) = ( 0.008)( 5.05) + 0.7 + (101)( 0.008)( 0.5)
R1
44.1R2
or R1 = 44.1 k Ω, = 5.05 ⇒ R2 = 5.70 k Ω
44.1 + R2
⎛ 101 ⎞
Now I EQ = ⎜ ⎟ ( 0.8 ) = 0.808 mA
⎝ 100 ⎠
VCC = I CQ RC + VCEQ + I EQ RE
10 = ( 0.8 ) RC + 5 + ( 0.808 )( 0.5 )
RC = 5.75 k Ω
(b) For 75 ≤ β ≤ 150

21. ⎛ R2 ⎞ ⎛ 5.7 ⎞
VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.145 V
⎝ R1 + R2 ⎠ ⎝ 5.7 + 44.1 ⎠
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
1.145 − 0.7
For β = 75, I BQ = = 0.0103 mA
5.05 + ( 76 )( 0.5 )
Then I CQ = ( 75 )( 0.0103) = 0.775 mA
1.145 − 0.7
For β = 150, I BQ = = 0.00552 mA
5.05 + (151)( 0.5 )
Then I CQ = 0.829 mA
ΔI CQ 0.829 − 0.775
% Change = = × 100% ⇒ % Change = 6.75%
I CQ 0.80
(c) For RE = 1 k Ω
RTH = ( 0.1)(101)(1) = 10.1 k Ω
1 1
VTH = ⋅ RTH ⋅ VCC = (10.1)(10 ) = ( 0.008 )(10.1) + 0.7 + (101)( 0.008 )(1)
R1 R1
which yields R1 = 63.6 k Ω
63.6 R2
And = 10.1 ⇒ R2 = 12.0 k Ω
63.6 + R2
⎛ R2 ⎞ ⎛ 12 ⎞
Now VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.587 V
⎝ R1 + R2 ⎠ ⎝ 12 + 63.6 ⎠
1.587 − 0.7
For β = 75, I BQ = = 0.0103 mA
10.1 + ( 76 )(1)
So I CQ = 0.773 mA
1.587 − 0.7
For β = 150, I BQ = = 0.00551 mA
10.1 + (151)(1)
Then I CQ = 0.826 mA
ΔI CQ 0.826 − 0.773
% Change = = × 100% ⇒ % Change = 6.63%
I CQ 0.8
5.51
VCC ≅ I CQ ( RC + RE ) + VCEQ
10 = ( 0.8 )( RC + RE ) + 5 ⇒ RC + RE = 6.25 k Ω
Let RE = 0.875 k Ω
Then, for bias stable RTH = ( 0.1)(121)( 0.875 ) = 10.6 k Ω
0.8
I BQ = = 0.00667 mA
120
1
(10.6 )(10 ) = ( 0.00667 )(10.6 ) + 0.7 + (121)( 0.00667 )( 0.875 )
R1
71.8R2
So R1 = 71.8 k Ω and = 10.6 ⇒ R2 = 12.4 k Ω
71.8 + R2
10
Then I R ≅ = 0.119 mA
71.8 + 12.4
This is close to the design specification.
5.52

22. I CQ ≈ I EQ ⇒ VCEQ = VCC − I CQ ( RC + RE )
6 = 12 − I CQ ( 2 + 0.2 )
I CQ = 2.73 mA, I BQ = 0.0218 mA
VCEQ = 6 V
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
⎛ R2 ⎞
VTH = ⎜ ⎟ (12 ) − 6, RTH = R 1 R2
⎝ R1 + R2 ⎠
Bias stable ⇒ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(126 )( 0.2 ) = 2.52 kΩ
⎛ 1⎞
VTH = ⎜ ⎟ ( RTH )(12 ) − 6
⎝ R1 ⎠
1
( 2.52 )(12 ) − 6 = ( 0.0218)( 2.52 ) + 0.7 + (126 )( 0.0218)( 0.2 ) − 6
R1
1
( 30.24 ) = 0.7549 + 0.5494
R1
23.2R 2
R1 = 23.2 kΩ, = 2.52
23.2 + R 2
R2 = 2.83 kΩ
5.53
⎛ 81 ⎞
a. I CQ = 1 mA. I EQ = ⎜ ⎟ (1) = 1.01 mA
⎝ 80 ⎠
VCEQ = 12 − (1)( 2 ) − (1.01)( 0.2 ) ⇒ VCEQ = 9.80 V
1
I BQ = = 0.0125 mA
80
RTH = + ( 0.1) (1 + β ) RE = ( 0.1)( 81)( 0.2 ) = 1.62 kΩ
⎛ R2 ⎞ 1 1
VTH = ⎜ ⎟ (12 ) − 6 = ( RTH )(12 ) − 6 = (19.44 ) − 6
⎝ R1 + R2 ⎠ R1 R1
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
1
(19.44 ) − 6 = ( 0.0125 )(1.62 ) + 0.7 + ( 81)( 0.0125)( 0.2 ) − 6
R1
1
(19.44 ) = 0.923
R1
21.1R2
R1 = 21.1 kΩ, = 1.62
21.1 + R2
R2 = 1.75 kΩ
b.
R1 = 22.2 kΩ or R1 = 20.0 kΩ
R2 = 1.84 kΩ or R2 = 1.66 kΩ
R2 ( max ) , R1 ( min )
RTH = (1.84 ) ( 20.0 ) = 1.685 kΩ
⎛ 1.84 ⎞
VTH = ⎜ ⎟ (12 ) − 6 = −4.99 V
⎝ 1.84 + 20.0 ⎠
−4.99 − 0.7 − ( −6 ) 0.31
I BQ = = = 0.0173 mA
1.685 + ( 81)( 0.2 ) 17.89
I CQ = 1.39 mA, I EQ = 1.40 mA

23. For max, RC ⇒ VCE = 12 − (1.39 )( 2 ) − (1.40 )( 0.2 )
VCE = 8.94 V
R2 ( min ) , R1 ( max )
RTH = (1.66 ) ( 22.2 ) = 1.545 kΩ
⎛ 1.66 ⎞
VTH = ⎜ ⎟ (12 ) − 6 = −5.165 V
⎝ 1.66 + 22.2 ⎠
−5.165 − 0.7 + 6 0.135
I BQ = = = 0.00761 mA ⇒ I CQ = 0.609 mA, I E = 0.616
1.545 + ( 81)( 0.20 ) 17.745
VCEQ = 12 − ( 0.609 )( 2 ) − ( 0.616 )( 0.2 )
VCEQ = V 10.7 V
So 0.609 ≤ I C ≤ 1.39 mA
8.94 ≤ VCEQ ≤ 10.7 V
5.54
VCEQ ≅ VCC − I CQ ( RC + RE )
5 = 12 − 3 ( RC + RE ) ⇒ RC + RE = 2.33 k Ω
Let RE = 0.333 k Ω and RC = 2 k Ω
Nominal value of β = 100
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.333) = 3.36 k Ω
3
I BQ = = 0.03 mA
100
1 1
VTH = ⋅ RTH ⋅ (12 ) − 6 = ( 3.36 )(12 ) − 6
R1 R1
Then VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
1
( 3.36 )(12 ) − 6 = ( 0.03)( 3.36 ) + 0.7 + (101)( 0.03)( 0.333) − 6
R1
which yields R1 = 22.3 k Ω and R2 = 3.96 k Ω
⎛ R2 ⎞ ⎛ 3.96 ⎞
Now VTH = ⎜ ⎟ (12 ) − 6 = ⎜ ⎟ (12 ) − 6 or VTH = −4.19 V
⎝ R1 + R2 ⎠ ⎝ 3.96 + 22.3 ⎠
For β = 75, VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6
VTH + 6 − 0.7 −4.19 + 6 − 0.7
I BQ = = = 0.0387 mA ⇒ I C = 2.90 mA
RTH + (1 + β ) RE 3.36 + ( 76 )( 0.333)
−4.19 + 6 − 0.7
For β = 150, I BQ = = 0.0207 mA
3.36 + (151)( 0.333)
Then I C = 3.10 mA
Specifications are met.
5.55
RTH = R1 R2 = 3 12 = 2.4 k Ω
⎛ R2 ⎞ ⎛ 12 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 20 ) = 16 V
⎝ R1 + R2 ⎠ ⎝ 12 + 3 ⎠
(a) For β = 75
20 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
20 − 0.7 − 16 = I BQ ⎡( 76 )( 2 ) + 2.4 ⎤
⎣ ⎦
So I BQ = 0.0214 mA, I CQ = 1.60 mA, I EQ = 1.62 E
VECQ = 20 − (1.6 )(1) − (1.62 )( 2 ) or VECQ = 15.16 V

24. (b) For β = 100, we find I BQ = 0.0161 mA, I CQ = 1.61 mA, VECQ = 15.13 V , I EQ = 1.63 mA
5.56
I CQ = 4.8 mA → I EQ = 4.84 mA
VCEQ = VCC − I CQ RC − I EQ RE
6 = 18 − ( 4.8 )( 2 ) − ( 4.84 ) RE ⇒ RE = 0.496 kΩ
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.496 ) = 6.0 kΩ
VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
I BQ = 0.040 mA
1 1
VTH = ⋅ RTH ⋅VCC = ( 6.0 )(18 )
R1 R1
1
( 6.0 )(18) = ( 0.04 )( 6.0 ) + 0.70 + (121)( 0.04 )( 0.496 )
R1
1
(108 ) = 3.34
R1
32.3 R2
R1 = 32.3 kΩ, = 6.0
32.3 + R2
R2 = 7.37 kΩ
5.57
For nominal β = 70
2
I BQ = = 0.0286 mA → I EQ = 2.03 mA
70
VCEQ = VCC − I CQ RC − I EQ RE
10 = 20 − ( 2 )( 4 ) − ( 2.03) RE ⇒ RE = 0.985 K
RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 71)( 0.985 ) = 6.99 K
VTH = I BQ RTH + VBE ( on ) + I EQ RE
1
⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE
R1
1
( 6.99 )( 20 ) = ( 0.0286 )( 6.99 ) + 0.70 + ( 2.03)( 0.985 )
R1
1
(139.8 ) = 2.90
R1
48.2 R2
R1 = 48.2 K, = 6.99
48.2 + R2
R2 = 8.18 K
Check: For β = 50
⎛ 8.18 ⎞
VTH = ⎜ ⎟ ( 20 ) = 2.90
⎝ 8.18 + 48.2 ⎠
V − VBE ( on ) 2.90 − 0.7
I BQ = TH = = 0.0384 mA
RTH + (1 + β ) RE 6.99 + ( 51)( 0.985 )
I CQ = 1.92 mA
For β = 90
2.90 − 0.7
I BQ = = 0.0228 mA
6.99 + ( 91)( 0.985 )
I CQ = 2.05 mA
Design criterion is satisfied.

25. 5.58
I CQ = 1 mA → I EQ = 1.017 mA
VCEQ = VCC − I CQ RC − I EQ RE
5 = 15 − (1)( 5 ) − (1.017 ) RE ⇒ RE = 4.92 kΩ
Bias stable: RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)( 4.92 ) = 30.0 kΩ
1
I BQ = = 0.0167 mA
60
1
VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE
R1
1
( 30.0 )(15 ) = ( 0.0167 )( 30.0 ) + 0.70 + (1.017 )( 4.92 )
R1
1
( 448.5) = 6.197
R1
72.5 R2
R1 = 72.5 kΩ, = 30.0
72.5 + R2
R2 = 51.2 kΩ
Check: For β = 45
⎛ 51.2 ⎞
VTH = ⎜ ⎟ (15 ) = 6.21V
⎝ 51.2 + 72.5 ⎠
V − VBE ( on ) 6.21 − 0.7
I BQ = TH = = 0.0215 mA
RTH + (1 + β ) RE 30 + ( 46 )( 4.92 )
ΔI C
I CQ = 0.967 mA, = 3.27%
IC
Check: For β = 75
6.21 − 0.7
I BQ = = 0.0136 mA
30.0 + ( 76 )( 4.92 )
ΔI C
I CQ = 1.023 mA, = 2.31%
IC
Design criterion is satisfied.
5.59
(a)
VCC ≅ I CQ ( RC + RE ) + VCEQ
3 = ( 0.1)( 5 RE + RE ) + 1.4 ⇒ RE = 2.67 k Ω
100
RC = 13.3 k Ω, I BQ = = 0.833 μ A
120
RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2.67 ) = 32.3 k Ω
1 1
VTH = ⋅ RTH ⋅ VCC = ( 32.3)( 3)
R1 R1
= I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE
= ( 0.000833)( 32.3) + 0.7 + (121)( 0.000833)( 2.67 )
which gives R1 = 97.3 k Ω, and R2 = 48.4 k Ω
(b)

26. 3 3
IR ≅ = ⇒ 20.6 μ A
R1 + R2 97.3 + 48.4
I CQ = 100 μ A
P = ( I CQ + I R )VCC = (100 + 20.6 )( 3)
or P = 362 μW
5.60
5 − VE 5
IE = = = 1.67 mA
RE 3
RTH = R1 || R2 = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 3) = 30.3 kΩ
⎛ R2 ⎞ 1
VTH = ⎜ ⎟ ( 4 ) − 2 = ⋅ RTH ⋅ ( 4 ) − 2
⎝ R1 + R2 ⎠ R1
I EQ
I BQ = = 0.0165 mA
1+ β
5 = I EQ RE + VEB ( on ) + I B RTH + VTH
1
5 = (1.67 )( 3) + 0.7 + ( 0.0165 )( 30.3) + ( 30.3)( 4 ) − 2
R1
1
0.80 = ( 30.3)( 4 ) ⇒ R1 = 152 kΩ
R1
152 R2
= 30.3 ⇒ R2 = 37.8 kΩ
152 + R2
5.61
a. RTH = R1 R2 = 10 20 ⇒ RTH = 6.67 kΩ
⎛ R2 ⎞ ⎛ 20 ⎞
VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⇒ VTH = 1.67 V
⎝ R1 + R2 ⎠ ⎝ 20 + 10 ⎠
b.
10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
10 − 0.7 − 1.67 7.63
I BQ = = ⇒ I BQ = 0.0593 mA
6.67 + ( 61)( 2 ) 128.7
I CQ = 3.56 mA, I EQ =3.62 mA
VE = 10 − I EQ RE = 10 − ( 3.62 )( 2 )
VE = 2.76 V
VC = I CQ RC − 10 = ( 3.56 )( 2.2 ) − 10
VC = −2.17 V
5.62
V + − V − ≅ I CQ ( RC + RE ) + VECQ
20 = ( 0.5 )( RC + RE ) + 8 ⇒ ( RC + RE ) = 24 k Ω
Let RE = 10 k Ω then RC = 14 k Ω
Let β = 60 from previous problem.
RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)(10 )
Or RTH = 61 k Ω

27. 0.5
I BQ = = 0.00833 mA
60
⎛ R2 ⎞ 1
VTH =⎜ ⎟ (10 ) − 5 = ⋅ RTH ⋅10 − 5
⎝ R1 + R2 ⎠ R1
Now 10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH
1
10 = ( 61)( 0.00833)(10 ) + 0.7 + ( 0.00833) ( 61) + ( 61) (10 ) − 5
R1
Then R1 = 70.0 k Ω and R2 = 474 k Ω
10 10
IR ≅ = ⇒ 18.4 μ A
R1 + R2 70 + 474
So the 40 μ A current limit is met.
5.63
a. RTH = R1 R2 = 35 20 ⇒ RTH = 12.7 kΩ
⎛ R2 ⎞ ⎛ 20 ⎞
VTH = ⎜ ⎟ (7) − 5 = ⎜ ⎟ ( 7 ) − 5 ⇒ VTH = −2.45 V
⎝ R1 + R2 ⎠ ⎝ 20 + 35 ⎠
b.
VTH − VBE ( on ) − ( −10 )
I BQ =
RTH + (1 + β ) RE
−2.45 − 0.7 + 10
= ⇒ I BQ = 0.136 mA
12.7 + ( 76 )( 0.5 )
I CQ = 10.2 mA, I EQ = 10.4 mA
VCEQ = 20 − I CQ RC − I EQ RE
= 20 − (10.2 )( 0.8 ) − (10.4 )( 0.5 )
VCEQ = 6.64 V
c.
R2 = 20 + 5% = 21 kΩ
R1 = 35 − 5% = 33.25 kΩ
RE = 0.5 − 5% = 0.475 kΩ
RTH = R1 R2 = 21 33.25 = 12.9 kΩ
⎛ R2 ⎞
VTH = ⎜ ⎟ (7) − 5
⎝ R1 + R2 ⎠
⎛ 21 ⎞
=⎜ ⎟ ( 7 ) − 5 = −2.29 V
⎝ 21 + 33.25 ⎠
−2.29 − 0.7 − ( −10 )
I BQ = = 0.143 mA
12.9 + ( 76 )( 0.475 )
I CQ = 10.7 mA, I EQ = 10.9 mA
For RC = 0.8 + 5% = 0.84 kΩ
VCEQ = 20 − (10.7 )( 0.84 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 5.83 V
For RC = 0.8 − 5% = 0.76 kΩ

28. VCEQ = 20 − (10.7 )( 0.76 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 6.69 V
R2 = 20 − 5% = 19 kΩ
R1 = 35 + 5% = 36.75 kΩ
RE = 0.5 + 5% = 0.525 kΩ
RTH = R1 R2 = 19 36.75 = 12.5 kΩ
⎛ 19 ⎞
VTH = ⎜ ⎟ ( 7 ) − 5 = −2.61 V
⎝ 19 + 36.75 ⎠
−2.61 − 0.7 − ( −10 )
I BQ = = 0.128 mA
12.5 + ( 76 )( 0.525 )
I CQ = 9.58 mA, I EQ = 9.70 mA
For RC = 0.84 kΩ
VCEQ = 20 − ( 9.58 )( 0.84 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 6.86 V
For RC = 0.76 kΩ
VCEQ = 20 − ( 9.58 )( 0.76 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 7.63 V
So 9.58 ≤ I CQ ≤ 10.7 mA and 5.83 ≤ VCEQ ≤ 7.63 V
5.64
a.
RTH = 500 kΩ 500 kΩ 70 kΩ = 250 kΩ 70 kΩ ⇒ RTH = 54.7 kΩ
5 − VTH 3 − VTH VTH − ( −5 )
+ =
500 500 70
5 3 5 ⎛ 1 1 1 ⎞
+ − = VTH ⎜ + + ⎟ − 0.0554 = VTH ( 0.0183)
500 500 70 ⎝ 500 500 70 ⎠
VTH = −3.03 V
b.
VTH − VBE ( on ) − ( −5 )
I BQ =
RTH + (1 + β ) RE
−3.03 − 0.7 + 5
=
54.7 + (101)( 5 )
I BQ = 0.00227 mA
I CQ = 0.227 mA, I EQ = 0.229
VCEQ = 20 − ( 0.227 )( 50 ) − ( 0.229 )( 5 )
VCEQ = 7.51 V
5.65
RTH = 30 || 60 || 20 ⇒ RTH = 10 kΩ
5 − VTH 5 − VTH VTH
+ =
30 60 20
⎛ 5 5 ⎞ ⎛ 1 1 1 ⎞
⎜ + ⎟ = VTH ⎜ + + ⎟
⎝ 30 60 ⎠ ⎝ 30 60 20 ⎠
VTH = 2.5 V
For β = 100

29. VTH − VBE ( on ) − ( −5 )
I BQ =
RTH + (1 + β ) RE
2.5 − 0.7 + 5
=
10 + (101)( 0.2 )
I BQ = 0.225 mA
I CQ = 22.5 mA, I EQ = 22.7 mA
VCEQ = 15 − ( 22.5 )( 0.5 ) − ( 22.7 ) ( 0.2 )
VCEQ = −0.79! In saturation ⇒ VCEQ = 0.2 V
VTH = I BQ RTH + VBE + I EQ RE − 5
2.5 + 5 − 0.7 = I BQ (10 ) + I EQ ( 0.2 )
6.8 = I BQ (10 ) + I EQ ( 0.2 )
14.8 = I CQ ( 0.5 ) + I EQ ( 0.2 )
Transistor in saturation,
I EQ = I BQ + I CQ
6.8 = I BQ (10 ) + I BQ ( 0.2 ) + I CQ ( 0.2 )
6.8 = I BQ (10.2 ) + I CQ ( 0.2 )
51× 14.8 = I BQ ( 0.2 ) + I CQ ( 0.7 )
754.8 = I BQ (10.2 ) + I CQ ( 35.7 )
748 = I CQ ( 35.5 )
I CQ = 21.1 mA
VCEQ = 0.2 V
5.66
I CQ = 50 μ A, I BQ = 0.625 μ A, I EQ = 50.6 μ A
(a)
1
RE = = 19.8 K
0.0506
5 = ( 0.050 ) RC + 5 + ( 0.0506 )(19.8 ) − 5
RC = 80 K
RTH = R1 R2 Design bias stable circuit.
RTH = ( 0.1)( 51)(19.8 ) = 101 K
⎛ R2 ⎞ 1
VTH = ⎜ ⎟ (10 ) − 5 = ⋅ RTH ⋅ (10 ) − 5
⎝ R1 + R2 ⎠ R1
1
So (101)(10 ) − 5 = I BQ (101) + 0.7 + ( 0.0506 )(19.8 ) − 5
R1
1
(1010 ) = 0.0631 + 0.7 + 1
R1
573 R2
R1 = 573 K = 101
573 + R2
R2 = 123 K
(b)

30. RTH = 101 K, VTH = −3.23 V
VTH = I BQ RTH + 0.7 + (121)(19.8 ) I BQ − 5
1.07 = I BQ (101 + 2395.8 ) ⇒ I BQ = 0.429 μ A
I CQ = 0.0514 mA, I EQ = 0.0519 mA
VCEQ = 10 − ( 0.0514 )( 80 ) − ( 0.0519 )(19.8 )
= 10 − 4.11 − 1.03 ⇒ VCEQ = 4.86 V
5.67
(a)
2
I EQ = 0.80 mA, RE = = 2.5 K
0.8
12 = I CQ RC + VCEQ + I EQ RE
12 = ( 0.8 ) RC + 7 + 2 ⇒ RC = 3.75 K
VTH = I BQ RTH + VBE + I EQ RE
For a bias stable circuit
RTH = ( 0.1)(12.1)( 2.5 ) = 30.25 K
1 1
⋅ RTH ⋅ VCC = ( 30.25 )(12 ) = ( 0.00667 )( 30.25 ) + 0.7 + 2
R1 R1
1
( 363) = 2.90 ⇒ R1 = 125 K
R1
125 R2
= 30.25 ⇒ R2 = 39.9 K
125 + R2
(b) Let RE = 2.4 K, RC = 3.9 K
R1 = 120 K R2 = 39 K
Then
RTH = R1 R2 = 120 39 = 29.4 K
⎛ 39 ⎞
VTH = ⎜ ⎟ (12 ) = 2.94 V
⎝ 120 + 39 ⎠
2.94 − 0.7 2.24
I BQ = = ⇒ 7.00 μ A
29.4 + (121)( 2.4 ) 319.8
I CQ = 0.841 mΑ I EQ = 0.848 mA
VCEQ = 12 − ( 0.841)( 3.9 ) − ( 0.848 )( 2.4 ) = 12 − 3.28 − 2.04
VCEQ = 6.68 V
5.68
(a) I CQ = 100 μ A, I BQ = 1.18 μ A, I EQ = 101 μ A
2
RE = ⇒ RE = 19.8 K
0.101
9 = ( 0.101)(19.8 ) + 6 + ( 0.1) RC − 9
RC = 100 K

31. Design a bias stable circuit.
BTH = ( 0.1)( 86 )(19.8 ) = 170 K
⎛ R2 ⎞ 1
VTH = ⎜ ⎟ (18 ) − 9 = (170 )(18 ) − 9
⎝ R1 + R2 ⎠ R1
1
9 = ( 0.101)(19.8 ) + 0.7 + ( 0.00118 )(170 ) + (170 )(18 ) − 9
R1
203R2
R1 = 203 K = 170
203 + R2
R2 = 1046 K
(b) β = 125
9 = (126 ) I BQ (19.8 ) + 0.7 + I BQ (170 ) + (15.07 − 9 )
2.23
I BQ = = 0.837 μ A I CQ = 0.1046 mA
2664.8
I EQ = 0.1054 mA
VECQ = 18 − ( 0.1046 )(100 ) − ( 0.1054 )(19.8 )
VECQ = 5.45 V
5.69
(a)
⎛ 51 ⎞ 3
I EQ ≈ ⎜ ⎟ ( 20 ) = 20.4 mA ⇒ RE = = 0.147 K
⎝ 50 ⎠ 20.4
6
VRC = 18 − 9 − 3 = 6 V RC = = 0.3 K
20
For a bias stable circuit.
RTH = ( 0.1)( 51)( 0.147 ) = 0.750 K
V + = I EQ RE + VEB + I BQ RTH + VTH
1
18 = 3 + 0.7 + ( 0.4 )( 0.75 ) + ( 0.75 )(18)
R1
1
14 = (13.5) ⇒ R1 = 0.964 K
R1
( 0.964 ) R2
= 0.75 ⇒ R2 = 3.38 K
0.964 + R2
(b) Let RE = 0.15 K, RC = 0.3 K
R1 = 1.0 K, R2 = 3.3 K
⎛ 3.38 ⎞
RTH = 1/13.3 = 0.767 K, VTH = ⎜ ⎟ (18 ) = 13.8 V
⎝ 1 + 3.38 ⎠
18 − 13.8 − 0.7 3.5
I BQ = = = 0.416 mA
0.767 + ( 51)( 0.15 ) 8.417
I CQ = 20.8 mA, I EQ = 21.2 mA
VECQ = 18 − ( 20.8 )( 0.3) − ( 21.2 )( 0.15 )
= 18 − 6.24 − 3.18
VECQ = 8.58 V
5.70

32. RTH = R1 R2 = 100 40 = 28.6 kΩ
⎛ R2 ⎞ ⎛ 40 ⎞
VTH = ⎜ ⎟ (10 ) = ⎜ ⎟ (10 ) = 2.86 V
⎝ R1 + R2 ⎠ ⎝ 40 + 100 ⎠
VTH − VBE ( on ) 2.86 − 0.7
I B1 = =
RTH + (1 + β ) RE1 28.6 + (121) (1)
I B1 = 0.0144 mA
I C1 = 1.73 mA, I E1 = 1.75 mA
10 − VB 2
= I C1 + I B 2
3
VB 2 − VBE ( on ) − ( −10 )
IE2 =
5
10 − VB 2 VB 2 − 0.7 + 10
= I C1 +
3 (121) ( 5 )
10 9.3 ⎛1 1 ⎞
− 1.73 − = VB 2 ⎜ +
⎜ 3 (121) ( 5 ) ⎟
⎟
3 605 ⎝ ⎠
1.588 = VB 2 ( 0.335 ) ⇒ VB 2 = 4.74 V
4.74 − 0.7 − ( −10 )
IE2 = ⇒ I E 2 = 2.808 mA
5
I B2 = 0.0232 mA
I C 2 = 2.785 mA
VCEQ1 = 4.74 − (1.75 ) (1) ⇒ VCEQ1 = 2.99 V
VCEQ 2 = 10 − ( 4.74 − 0.7 ) ⇒ VCEQ 2 = 5.96 V
5.71
VE1 = −0.7
−0.7 − ( −5 )
I R1 = = 0.215 mA
20
VE 2 = −0.7 − 0.7 = −1.4
−1.4 − ( −5 )
IE2 = ⇒ I E 2 = 3.6 mA
1
I B 2 = 0.0444 mA
I C 2 = 3.56 mA
I E1 = I R1 + I B 2 = 0.215 + 0.0444
I E1 = 0.259 mA
I B1 = 0.00320 mA
I C1 = 0.256 mA
5.72
Current through V − source = I E1 + I E 2 and I E1 = I E 2 = (1 + β ) I B1 = ( 51)( 8.26 ) μ A
So total current = 2 ( 51)( 8.26 ) μ A=843 μ A
P − = I ⋅ V − = ( 0.843)( 5 ) ⇒ P − = 4.22 mW
(From V − source)
From Example 5.19, I Q = 0.413 mA
⎛ 50 ⎞
So I C 0 = ⎜ ⎟ ( 0.413) = 0.405 mA
⎝ 51 ⎠

33. P + = I ⋅ V + = ( 0.405 )( 5 ) ⇒ P + = 2.03 mW
(From V + source)
5.73
RTH = R1 R2 = 50 100 = 33.3 kΩ
⎛ R2 ⎞
VTH = ⎜ ⎟ (10 ) − 5
⎝ R1 + R2 ⎠
⎛ 100 ⎞
=⎜ ⎟ (10 ) − 5 = 1.67 V
⎝ 100 + 50 ⎠
5 = I E1 RE1 + VEB ( on ) + I B1 RTH + VTH
⎛ 101 ⎞
I E1 = ⎜ ⎟ ( 0.8 ) = 0.808 mA
⎝ 100 ⎠
I B1 = 0.008 mA
5 = ( 0.808 ) RE1 + 0.7 + ( 0.008 )( 33.3) + 1.67
RE1 = 2.93 kΩ
VE1 = 5 − ( 0.808 ) ( 2.93) = 2.63 V
VC1 = VE1 − VECQ1 = 2.63 − 3.5 = −0.87 V
VE 2 = −0.87 − 0.70 = −1.57 V
−1.57 − ( −5 )
IE2 = = 0.808 ⇒ RE 2 = 4.25 kΩ
RE 2
VCEQ 2 = 4 ⇒ VC 2 = −1.57 + 4 = 2.43 V
5 − 2.43
RC 2 = ⇒ RC 2 = 3.21 kΩ
0.8
I RC1 = I C1 − I B 2 = 0.8 − 0.008 = 0.792 mA
−0.87 − ( −5 )
RC1 = ⇒ RC1 = 5.21 kΩ
0.792