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