This document contains solutions to 10 exercises involving bipolar junction transistors and MOSFET circuits. The key points summarized are: 1) Exercise 10.1 calculates the reference and base currents in a BJT circuit. Exercise 10.2 calculates the reference current, zero bias current, and small signal output resistance. 2) Exercise 10.3 calculates the various currents in a multi-emitter BJT. Exercise 10.4 calculates the early effect resistance and voltage difference between two BJTs. 3) Exercise 10.5 again calculates reference current and zero bias current. Exercises 10.6 through 10.11 involve various MOSFET and BJT circuit calculations.

1. Chapter 10
Exercise Solutions
EX10.1
V + − VBE ( on ) 10 − 0.7
I REF = =
R1 15
I REF = 0.62 mA
I REF 0.62
I0 = =
2 2
1+ 1+
β 75
I 0 = 0.604 mA
EX10.2
V + − VBE ( on ) − V − 5 − 0.7 − ( −5 )
I REF = =
R1 12
I REF = 0.775 mA
I 0.775
I 0 = REF = = 0.7549 mA
2 2
1+ 1+
β 75
1 ΔV
ΔI 0 = ( 0.02 )( 0.7549 ) = 0.0151 mA and ΔI 0 = ΔVCE 2 ⇒ r0 = CE 2
r0 ΔI 0
4 V
r0 = = 265 kΩ = A ⇒ VA = ( 265 )( 0.7549 ) ⇒ VA ≅ 200 V
0.0151 I0
EX10.3
V + − 2VBE ( on ) 9 − 2 ( 0.7 )
I REF = =
R1 12
I REF = 0.6333 mA
I REF 0.6333
I0 = = = 0.6331 mA
2 2
1+ 1+
β (1 + β ) 75 ( 76 )
I 0 = 0.6331 mA = I C1
I0
I B1 = I B 2 = ⇒ I B1 = I B 2 = 8.44 μ A
β
I E 3 = I B1 + I B 2 ⇒ I E 3 = 16.88 μ A
I E3
I B3 = ⇒ I B 3 = 0.222 μ A
1+ β
EX10.4
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
V ⎛ I ⎞ 0.026 ⎛ 0.75 ⎞
RE = T ln ⎜ REF ⎟ = ln ⎜ ⎟ ⇒ RE = 3.54 kΩ
I 0 ⎝ I 0 ⎠ 0.025 ⎝ 0.025 ⎠
5 − 0.7
R1 = ⇒ R1 = 5.73 kΩ
0.75
VBE1 − VBE 2 = I 0 RE = ( 0.025 )( 3.54 ) ⇒ VBE1 − VBE 2 = 88.5 mV
EX10.5

2. 5 − 0.7 − ( −5 )
I REF = ⇒ I REF = 0.775 mA
12
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
⎛ 0.775 ⎞
I 0 ( 6 ) = ( 0.026 ) ln ⎜ ⎟ ⇒ I 0 ≅ 16.6 μ A
⎝ I0 ⎠
EX10.6
⎛I ⎞
I 0 RE = VT ln ⎜ REF ⎟
⎝ I0 ⎠
0.026 ⎛ 0.70 ⎞
RE = ln ⎜ ⎟ ⇒ RE = 3.465 kΩ
0.025 ⎝ 0.025 ⎠
I 0.025
gm2 = 0 = ⇒ g m 2 = 0.9615 mA/V
VT 0.026
β VT (150 )( 0.026 )
rπ 2 = = = 156 kΩ
I0 0.025
VA 100
r02 = = = 4000 kΩ
I 0 0.025
′
RE = RE || rπ 2 = 3.47 ||156 = 3.39 kΩ
R0 = r02 (1 + g m 2 RE ) = 4000 ⎡1 + ( 0.962 )( 3.39 ) ⎤
′ ⎣ ⎦
R0 = 17.04 MΩ
1 3
dI 0 = ⋅ dVC 2 = ⇒ dI 0 = 0.176 μ A
R0 17, 040
EX10.7
I REF = I R + I BR + I B1 + ... + I BN
I 01
I R = I 01 = I 02 = ... = I 0 N and I BR = I B1 = I B 2 = ... = I BN =
β
⎛I ⎞ ⎛ N +1⎞
I REF = I 01 + ( N + 1) ⎜ 01 ⎟ = I 01 ⎜ 1 + ⎟
⎝ β ⎠ ⎝ β ⎠
I REF
So I 01 = I 02 = ... = I 0 N =
N +1
1+
β
I 01 1
= 0.90 =
I REF N +1
1+
50
N +1 1
1+ =
50 0.9
⎛ 1 ⎞
N +1 = ⎜ − 1⎟ ( 50 )
⎝ 0.9 ⎠
⎛ 1 ⎞
N =⎜ − 1⎟ ( 50 ) − 1
⎝ 0.9 ⎠
N = 4.55 ⇒ N = 4
EX10.8

3. VDS ( sat ) = 1 V = VGS 2 − VTN = VGS 2 − 2 ⇒ VGS 2 = 3 V
⎛ μ C ⎞⎛W ⎞
I O = K n 2 (VGS 2 − VTN ) = ⎜ n ox ⎟ ⎜ ⎟ (VGS 2 − VTN )
2 2
⎝ 2 ⎠ ⎝ L ⎠2
⎛W ⎞ ⎛W ⎞
0.20 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 10
2
⎝ L ⎠2 ⎝ L ⎠2
⎛ μ C ⎞⎛W ⎞
I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 1 − VTN )
2
⎝ 2 ⎠ ⎝ L ⎠1
VGS 1 = VGS 2
⎛W ⎞ ⎛W ⎞
0.5 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 25
2
⎝ L ⎠1 ⎝ L ⎠1
VGS 3 = V + − VGS 1 = 10 − 3 = 7 V
⎛ μ C ⎞⎛W ⎞
I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 3 − VTN )
2
⎝ 2 ⎠ ⎝ L ⎠3
⎛W ⎞ ⎛W ⎞
0.5 = ( 0.020 ) ⎜ ⎟ ( 7 − 2 ) ⇒ ⎜ ⎟ = 1
2
⎝ L ⎠3 ⎝ L ⎠3
EX10.9
I REF = K n (VGS − VTN )
2
a.
0.020 = 0.080 (VGS − 1)
2
VGS = 1.5 V all transistors
b.
VG 4 = VGS 3 + VGS1 + V − = 1.5 + 1.5 − 5 = −2 V
VS 4 = VG 4 − VGS 4 = −2 − 1.5 = −3.5 V
VD 4 ( min ) = VS 4 + VDS 4 ( sat ) and VDS 4 ( sat ) = VGS 4 − VTN = 1.5 − 1 = 0.5 V
So VD 4 ( min ) = −3.5 + 0.5 ⇒ VD 4 ( min ) = −3.0 V
c.
R0 = r04 + r02 (1 + g m r04 )
1 1
r02 = r04 = = = 2500 kΩ
λ I 0 ( 0.02 )( 0.020 )
g m = 2 K n (VGS − VTN ) = 2 ( 0.080 )(1.5 − 1) ⇒ g m = 0.080 mA / V
R0 = 2500 + 2500 (1 + ( 0.080 )( 2500 ) ) ⇒ R0 = 505 MΩ
EX10.10
For Q2 : vDS ( min ) = VP = 2 V ⇒ VS ( min ) = vDS ( min ) − 5 = 2 − 5 ⇒ VS ( min ) = −3 V
I 0 = I DSS 2 (1 + λ vDS 2 ) = 0.5 (1 + ( 0.15 )( 2 ) ) ⇒ I 0 = 0.65 mA
2
⎛ v ⎞
I 0 = I DSS1 ⎜ 1 − GS 1 ⎟
⎝ VP1 ⎠
2
⎛ v ⎞
0.65 = 0.80 ⎜ 1 − GS1 ⎟
⎝ −2 ⎠
vGS 1
= 0.0986 ⇒ vGS 1 = −0.197 V
−2
vGS 1 = VI − VS − 0.197 = VI − ( −3) ⇒ VI ( min ) = −3.2 V

4. Vgs 2 = 0, Vgs1 = −VX
VX VX − V1
IX = + + g m1VX (1)
r02 r01
V1 V1 − VX
+ = g m1VX (2)
RD r01
⎛ 1 ⎞
VX ⎜ + g m1 ⎟
V1 = ⎝ r01 ⎠
1 1
+
RD r01
1 ⎛ 1 ⎞
⎜ + g m1 ⎟
IX 1 1 1 r01 ⎝ r01 ⎠
= = + + g m1 −
VX R0 r02 r01 1 1
+
RD r01
⎡ 1 ⎤
1 ⎛ 1 ⎞ ⎢ r01 ⎥
= + ⎜ + g m1 ⎟ ⎢1 − ⎥
r02 ⎝ r01 ⎠⎢ 1 1 ⎥
+
⎢ RD r01 ⎥
⎣ ⎦
⎛ 1 ⎞
1 ⎛ 1 ⎞ ⎜ RD ⎟
= + ⎜ + g m1 ⎟ ⎜ ⎟
r02 ⎝ r01 ⎠⎜ 1 + 1 ⎟
⎜R ⎟
⎝ D r01 ⎠

5. 1 1 ⎛ 1 ⎞
For RD r01 ⇒ ≅ + ⎜ + g m1 ⎟
R0 r02 ⎝ r01 ⎠
For Q1:
2I ⎛ VGS 1 ⎞ 2 ( 0.8 ) ⎛ −0.197 ⎞
g m1 = DSS 1 ⎜1 − ⎟= ⎜1 − ⎟
VP ⎝ VP ⎠ 2 ⎝ −2 ⎠
g m1 = 0.721 mA/V
1 1
r0 = = = 10.3 kΩ
λ I 0 ( 0.15 )( 0.65 )
1 1 1
= + + 0.721 = 0.915 ⇒ R0 = 1.09 kΩ
R0 10.3 10.3
EX10.11
⎛V ⎞
a. I REF = I S exp ⎜ EB 2 ⎟
⎝ VT ⎠
⎛I ⎞ ⎛ 0.5 × 10−3 ⎞
VEB 2 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ −12 ⎟ ⇒ VEB 2 = 0.521 V
⎝ IS ⎠ ⎝ 10 ⎠
5 − 0.521
b. R1 = ⇒ R1 = 8.96 kΩ
0.5
c. Combining Equations (10.79), (10.80), and (10.81), we find
⎛ VEC 2 ⎞
⎡ ⎜1 + ⎟
⎛ V ⎞⎤ ⎛ V ⎞
I S 0 ⎢ exp ⎜ I ⎟ ⎥ ⎜ 1 + CEo ⎟ = I REF × ⎝
VAP ⎠
⎣ ⎝ VT ⎠ ⎦ ⎝ VAN ⎠ ⎛ VEB 2 ⎞
⎜1 + ⎟
⎝ VAP ⎠
⎛ 2.5 ⎞
⎡ ⎛V ⎞ ⎤⎛ 2.5 ⎞ ⎜1 + ⎟
= ( 0.5 × 10−3 ) ⎝
100 ⎠
10−12 ⎢ exp ⎜ I ⎟ ⎥ ⎜ 1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 100 ⎠ ⎛ 0.521 ⎞
⎜1 + ⎟
⎝ 100 ⎠
⎛V ⎞ ⎛ VI ⎞
1.025 ×10 −12 exp ⎜ I ⎟ = 4.974 × 10 ⇒ VI = 0.521 V
−4 8
⎟ = 5.098 × 10 exp ⎜
⎝ VT ⎠ ⎝ VT ⎠
⎛ 1 ⎞ 1
−⎜ ⎟ −
d. Av = ⎝ VT ⎠ = 0.026 = −38.46 ⇒ A = −1923
1 1 1 1 0.01 + 0.01
v
+ +
VAN VAP 100 100
EX10.12
I CQ 0.8
gm = = = 30.77 mA/V
VT 0.026
VA 80
r0 = r02 = = = 100 kΩ
I CQ 0.8
a. V0 = − g mVπ 1 ( r0 || r02 ) , Vπ 1 = Vi
Av = − g m ( r0 || r02 ) = − ( 30.77 ) [100 ||100] ⇒ Av = −1538
b. Av = − g m ( r0 || r02 || RL )
1540
Av = − = −770 − 770 = − ( 30.77 )( 50 || RL ) ⇒ ( 50 || RL ) = 25 ⇒ RL = 50 kΩ
2
EX10.13
(a) Neglecting effect of λ and RL

6. I O = I REF = K n (VIQ − VTN )
2
0.40 = 0.25 (VIQ − 1)
2
Then VIQ = 2.265 V
1 1
b. r0 = r02 = = = 125 kΩ
λ I 0 ( 0.02 )( 0.4 )
g m = 2 K n (VIQ − VTN ) = 2 ( 0.25 )( 2.26 − 1) = 0.632 mA/V
Av = − g m ( r0 || r02 ) = − ( 0.632 )(125 ||125 ) ⇒ Av = −39.5
39.4
c. Av = − g m ( r0 || r02 || RL ) − = − ( 0.632 )( 62.5 || RL ) ⇒ 62.5 || RL = 31.25 ⇒ RL = 62.5 kΩ
2
TYU10.1
For I 0 = 0.75 mA
⎛ 2⎞ ⎛ 2 ⎞
I REF = I 0 ⎜ 1 + ⎟ = ( 0.75 ) ⎜1 + ⎟
⎝ β⎠ ⎝ 100 ⎠
I REF = 0.765 mA
V + − VBE ( on ) − V −
I REF =
R1
5 − 0.7 − ( −5 )
R1 =
0.765
R1 = 12.2 kΩ
TYU10.2
10 − ( 0.7 )( 2 )
I REF = = 0.717 mA
12
I 0 ≈ I REF = 0.717
V 100
r0 = A = ⇒ r0 = 139.5 kΩ
I 0 0.717
1 4
ΔI 0 = ΔVCE 2 = ⇒ ΔI 0 = 0.0287 mA
r0 139.5
TYU10.3
1 0.50
I 0 = I REF ⋅ = ⇒ I 0 = 0.4996 mA
⎛ 2 ⎞ ⎛ 2 ⎞
⎜1 +
⎜ β (1 + β ) ⎟ ⎜ 1 + 50 ( 51) ⎟
⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠
I0
I B3 = ⇒ I B 3 = 9.99 μ A
β
⎛1+ β ⎞
I E3 = ⎜ ⎟ I C 3 = I E 3 = 0.5096 mA
⎝ β ⎠
IE3 0.5096
IC 2 = = ⇒ I C 2 = 0.490 mA = I C1
⎛ 2⎞ ⎛ 2⎞
⎜ 1 + ⎟ ⎜ 1 + 50 ⎟
⎝ β⎠ ⎝ ⎠
IC 2
I B1 = I B 2 = ⇒ I B1 = I B 2 = 9.80 μ A
β
TYU10.4

7. ′
k n1 ⎛ W ⎞ ′
kn 3 ⎛ W ⎞
⎜ ⎟ (VGS 1 − VTN 1 ) = ⎜ ⎟ ( 5 − VGS 1 − VTN 3 )
2 2
I REF =
2 ⎝ L ⎠1 2 ⎝ L ⎠3
1
⎡⎛ 40 ⎞ ⎛ 17.3 ⎞ ⎤
2
⎢⎜ 38 ⎟ ⎜ 2.70 ⎟ ⎥ (VGS1 − 0.98 ) = ( 3.98 − VGS 1 ) ⇒ VGS 1 = 1.814 V
⎣⎝ ⎠ ⎝ ⎠⎦
⎛ 0.040 ⎞
⎟ (17.3)(1.814 − 0.98 )
2
I REF = ⎜
⎝ 2 ⎠
I REF = 0.241 mA
′
kn 2 ⎛ W ⎞
⎜ ⎟ (VGS 1 − VTN 2 )
2
IO =
2 ⎝ L ⎠2
⎛ 0.042 ⎞
⎟ ( 6.92 )(1.814 − 1.0 )
2
=⎜
⎝ 2 ⎠
I O = 0.0963 mA
TYU10.5
a. From Equation (10.52),
3 ⎛ 3 ⎞
⎜ 1− ⎟
12 × 10 + ⎜ 12 ⎟ × 1.8
VGS 1 = ( )
3 ⎜ 3 ⎟
1+ ⎜1+ ⎟
12 ⎝ 12 ⎠
⎛ 0.5 ⎞ ⎛ 1 − 0.5 ⎞
VGS 1 = ⎜ ⎟ (10 ) + ⎜ ⎟ × (1.8 )
⎝ 1 + 0.5 ⎠ ⎝ 1 + 0.5 ⎠
VGS 1 = 3.93 V also VDS1 = 3.93 V
I REF = (12 )( 0.020 ) [3.93 − 1.8] ⎡1 + ( 0.01)( 3.93) ⎤ ⇒ I REF = 1.13 mA
2
⎣ ⎦
(W / L )2 (1 + λVDS 2 )
b. I 0 = I REF × ×
(W / L )1 (1 + λVDS1 )
⎛ 6 ⎞ ⎡1 + ( 0.01)( 2 ) ⎤
I 0 = (1.13) × ⎜ ⎟ × ⎣ ⎦ ⇒ I = 0.555 mA
⎝ 12 ⎠ ⎡1 + ( 0.01)( 3.93) ⎤
0
⎣ ⎦
c. For VDS 2 = 6 V ⇒ I o = 0.576 mA
TYU10.6
K n1 (VGS 1 − VTN ) = K n3 (VGS 3 − VTN )
2 2
⎛ 0.10 ⎞
⎜ 0.25 ⎟ (VGS 3 − 2 )
VGS 1 − 2 = ⎜ ⎟
⎝ ⎠
VGS 1 − 2 = ( 0.6325 )(VGS 3 − 2 )
VGS 3 = 10 − VGS 1
VGS 1 − 2 = ( 0.6325 )(10 − VGS 1 ) − ( 0.6325 )( 2 )
1.6325VGS1 = 7.06 ⇒ VGS 1 = 4.325 V
I REF = K n1 (VGS 1 − VTN ) = ( 0.25 ) ( 4.325 − 2 ) ⇒ I REF = 1.35 mA
2 2
I O = 3K n 2 (VGS 1 − VTN ) = 3 ( 0.25 )( 4.325 − 2 )
2 2
I O = 3I REF ⇒ I O = 4.05 mA
TYU10.7

8. I REF = 0.20 = K n1 (VGS 1 − VTN ) = 0.15 (VGS 1 − 1) ⇒ VGS 1 = VGS 2 = 2.15 V
2 2
0.15
I O = K n 2 (VGS 2 − VTN ) = ( 2.15 − 1) ⇒ I O = 0.10 mA
2 2
2
I O = K n 3 (VGS 3 − VTN )
2
0.10 = 0.15 (VGS 3 − 1) ⇒ VGS 3 = 1.82 V
2
TYU10.8
All transistors are identical
⇒ I 0 = I REF = 250 μ A
I REF = K n (VGS − VTN )
2
0.25 = 0.20 (VGS − 1) ⇒ VGS = 2.12 V
2
TYU10.9
For Q1 : iD = I DSS 1 (1 + λ vDS1 )
2
⎛ v ⎞
For Q2 : iD = I DSS 2 ⎜ 1 − GS 2 ⎟ (1 + λ vDS 2 )
⎝ VP ⎠
vGS 2 = −vDS1 and vDS 2 = VDS − vDS 1
So
2
⎡ −v ⎤
I DSS 1 (1 + λ vDS1 ) = I DSS 2 ⎢1 − DS 1 ⎥ ⎡1 + λ (VDS − vDS1 ) ⎤
⎣ VP ⎦ ⎣ ⎦
I DSS 1 = I DSS 2
2
⎡ v ⎤
⎡1 + ( 0.1) vDS1 ⎤ = ⎢1 − DS1 ⎥ ⎡1 + ( 0.1) ( 3) − ( 0.1) vDS 1 ⎤
⎣ ⎦
⎣ 2 ⎦ ⎣ ⎦
1 + 0.1vDS 1 = (1 − vDS 1 + 0.25vDS 1 ) (1.3 − 0.1vDS 1 )
2
3 2
This becomes 0.025vDS 1 − 0.425vDS 1 + 1.5vDS 1 − 0.3 = 0
We find vDS 1 = 0.2127 V, vDS 2 = 2.787 V, vGS 2 = −0.2127 V
iD = I DSS 1 (1 + λ vDS 1 ) = 2 ⎡1 + ( 0.1)( 0.2127 ) ⎤
⎣ ⎦
iD = 2.04 mA
R0 = r02 + r01 (1 + g m 2 r02 )
2 I DSS ⎛ vGS 2 ⎞ 2 ( 2 ) ⎛ −0.2127 ⎞
gm = ⎜1 − ⎟= ⎜1 − ⎟
−VP ⎝ VP ⎠ 2 ⎝ −2 ⎠
g m = 1.787 mA/V
1 1
r02 = r04 = = = 5 kΩ
λ I DSS ( 0.1)( 2 )
R0 = 5 + 5 ⎡1 + (1.787 )( 5 ) ⎤ ⇒ R0 = 54.7 kΩ
⎣ ⎦
TYU10.10
⎛ 0.1× 10−3 ⎞
a. VEB 2 = ( 0.026 ) ln ⎜ −14 ⎟
⇒ VEB 2 = 0.557 V
⎝ 5 × 10 ⎠
5 − 0.557
b. R1 = ⇒ R1 = 44.4 kΩ
0.1
c.

9. ⎛ VEC 2 ⎞
⎡ ⎛V
⎞ ⎤ ⎛ VCE 0 ⎞ ⎜ 1+ V ⎟
I S 0 ⎢ exp ⎜ I
⎟⎥ ⎜ 1+ ⎟ = I REF × ⎜ AP ⎟
⎣ ⎝ VT
⎠⎦ ⎝ VAN ⎠ ⎜ 1 + VEB 2 ⎟
⎜ VAP ⎟
⎝ ⎠
⎛ 2.5 ⎞
⎡ ⎛ VI ⎞ ⎤ ⎛ 2.5 ⎞ ⎜ 1 + 100 ⎟
⎟ = ( 0.1× 10 ) ⎜ 0.557 ⎟
−14 −3
5 × 10 ⎢ exp ⎜ ⎟ ⎥ ⎜ 1 +
⎣ ⎝ VT ⎠ ⎦ ⎝ 100 ⎠ ⎜
⎜1+ ⎟
⎟
⎝ 100 ⎠
⎛V ⎞
( 5.125 ×10−14 ) exp ⎜ VI ⎟ = 1.019 ×10−4
⎝ T⎠
⎛V ⎞
⎟ = 1.988 × 10 ⇒ VI = 0.557 V
9
exp ⎜ I
⎝ VT ⎠
1
−
d. Av = 0.026 ⇒ Av = −1923
1 1
+
100 100
TYU10.11
I REF = K p1 (VSG + VTP )
2
a.
0.25 = 0.20 (VSG − 1) ⇒ VSG = 2.12 V
2
b. From Equation (10.89)
⎡1 + λP (V + − VSG ) ⎤ K (V − V )2
VDSO = Vo = ⎣ ⎦− n I TN
λn + λP I REF ( λn + λP )
1 + ( 0.015 )(10 − 2.12 ) ( 0.2 )(VI − 1)
2
5= −
0.030 0.25 ( 0.030 )
0.15 = 1.12 − 0.8 (VI − 1) ⇒ VI = 2.10 V
2
−2 K n (VI − VTN )
c. Av =
I REF ( λn + λP )
2 ( 0.2 )( 2.10 − 1.0 )
Av = − ⇒ Av = −58.7
0.25 ( 0.030 )
TYU10.12
I REF = K p1 (VSG + VTP )
2
(a)
80 = 50 (VSG − 1) ⇒ VSG = 2.26 V
2
⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2
(b) VDSo = Vo = ⎣ ⎦− n I TN
λn + λ p I REF ( λn + λ p )
⎡1 + ( 0.015 )(10 − 2.26 ) ⎤ ( 50 )(VI − 1)
2
5= ⎣ ⎦−
0.030 (80 )( 0.030 )
20.83 (VI − 1) = 32.2 ⇒ VI = 2.243 V
2
−2 K n (VI − VTN ) −2 ( 50 )( 2.243 − 1)
(c) Av = = ⇒ Av = −51.8
I REF ( λn + λ p ) (80 )( 0.030 )
TYU10.13
a.

10. IC 0 0.5
gm = = ⇒ g m = 19.2 mA/V
VT 0.026
VAN 120
r0 = = ⇒ r0 = 240 kΩ
I CQ 0.5
VAP 80
r02 = = ⇒ r02 = 160 kΩ
I CQ 0.5
b. Av = − g m ( r0 || r02 || RL ) = − (19.2 ) [ 240 ||160 || 50] ⇒ Av = −631
TYU10.14
1
I C = 1mA, g m = = 38.46 mA/V
0.026
(100 )( 0.026 )
rπ 1 = rπ 2 = = 2.6 K
1
80
rO1 = rO 2 = = 80 K
1
120
rO = = 120 K
1
1
RO1 = 2.6 80 = 0.0257 K
38.46
For R1 = 9.3 K
R ′ = R1( RO1 + RE ) = 9.3 ( 0.0257 + 1) = 0.924 K
RE = 1 [ 2.6 + 0.924] = 0.779 K
′′
RO 2 = 80 ⎡1 + ( 38.46 )( 0.779 ) ⎤ = 2476.7 K
⎣ ⎦
Av = − g m ( rO || RO 2 ) = − ( 38.46 )(120 || 2476.7 ) = − ( 38.46 )(114.5 )
Av = −4404
For RL = 100 K
Av = −38.46 ⎡114.5 100 ⎤ = −2053
⎣ ⎦
For RL = 10 K
Av = −38.46 [114.5 ||10] = −354
TYU10.15
M 1 and M 2 identical ⇒ I o = I REF
a.
I O = K n (VI − VYN )
2
0.25 = 0.2 (VI − 1)
2
VI = 2.12 V
g m = 2 K n (VI − VTN ) = 2 ( 0.2 )( 2.12 − 1) ⇒ g m = 0.447 mA/V
1 1
r0 n = = ⇒ r0 n = 400 kΩ
λn I 0 ( 0.01)( 0.25 )
1 1
r0 p = = ⇒ r0 p = 200 kΩ
λ p I 0 ( 0.02 )( 0.25 )
b. Av = − g m ( r0 || r02 || RL )
Av = − ( 0.447 ) [ 400 || 200 ||100] ⇒ Av = −25.5