1. Chapter 11
Exercise Solutions
EX11.1
vE = −VBE ( on ) ⇒ vE = −0.7 V
I C1 = I C 2 = 0.5 mA
vC1 = vC 2 = 10 − ( 0.5 )(10 ) ⇒ vC1 = vC 2 = 5 V
EX11.2
iC 2 1
= = 0.99
IQ ⎛ vd ⎞
1 + exp ⎜ ⎟
⎝ VT ⎠
⎛v ⎞ 1
1 + exp ⎜ d ⎟ =
⎝ VT ⎠ 0.99
⎛v ⎞ 1
exp ⎜ d ⎟= −1
⎝ VT ⎠ 0.99
⎡ 1 ⎤
vd = VT ln ⎢ − 1⎥ ⇒ vd = −119.5 mV
⎣ 0.99 ⎦
EX11.3
a. v1 = v2 = 0 ⇒ vE = 0.7 V
ΔVRC = ( 0.25 )( 8 ) = 2 V ⇒ vC1 = vC 2 = −3 V ⇒ vEC1 = 3.7 V
b. v1 = v2 = 2.5 V ⇒ vE = 3.2 V ⇒ vEC1 = 6.2 V
c. v1 = v2 = −2.5 V ⇒ vE = −1.8 V ⇒ vEC1 = 1.2 V
EX11.4
Let I Q = 1 mA, then I CQ1 = I CQ 2 = 0.5 mA
0.5
g m1 = g m 2 = = 19.23 mA / V
0.026
v 1
At vC 2 , Ad = c 2 = g m RC 2
vd 2
1
So, 150 = (19.23) RC 2 ⇒ RC 2 = 15.6 k Ω
2
v 1
At vC1 , Ad = c1 = − g m RC1
vd 2
1
So, −100 = − (19.23) RC1 ⇒ RC1 = 10.4 k Ω
2
If V + = +10 V and V − = −10 V , dc biasing is OK.
EX11.5
⎛ I Q RC ⎞ ⎡ ( 0.8 )(12 ) ⎤
−⎜ ⎟ −⎢ ⎥
Acm = ⎝ 2VT ⎠ = ⎣ 2 ( 0.026 ) ⎦ = −0.0594
(1 + β ) IQ RO 1+
(101)( 0.8 )(100 )
1+
VT β ( 0.026 )(100 )
vo = Acm vcm = ( −0.0594 )(1sin ω t )( mV ) = −59.4sin ω t ( μV )
EX11.6

2. (a) vd = 40 μV , vcm = 0
vo = ( 50 )( 40 ) ⇒ 2.0 mV
(b) vd = 40 μV vcm = 200 μV
Assuming Acm > 0
50
60 = 20 log10
Acm
Acm = 0.05
vo = ( 50 )( 40 ) + ( 0.05 )( 200 )
vo = 2.010 mV
EX11.7
I Q RC
a. Diff. Gain Ad =
4VT
For v1 = v2 = 5 V ⇒ Minimum collector voltage
IQ
vC 2 = 5 V ⇒ ⋅ RC = 15 − 5 = 10 V or I Q RC = 20 V for max. Ad
2
20
Then Ad = ⇒ Ad ( max ) = 192
2 ( 0.026 )
b. If I Q = 0.5 mA, RC = 40 kΩ
⎛ I Q RC ⎞
−⎜ ⎟
Acm = ⎝ 2VT ⎠
⎡ (1 + β ) I Q R0 ⎤
⎢1 + ⎥
⎣ VT β ⎦
Then
⎛ 20 ⎞
−⎜
⎜ 2 ( 0.026 ) ⎟
⎟
Acm = ⎝ ⎠ ⇒ Acm = −0.199
⎡ ( 201)( 0.5 )(100 ) ⎤
⎢1 +
⎣ ( 0.026 )( 200 ) ⎥
⎦
⎛ 192 ⎞
and C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 59.7 dB
⎝ 0.199 ⎠
EX11.8
Rid = 2 ⎡ rπ + (1 + β ) RE ⎤
⎣ ⎦
β VT (100 )( 0.026 )
rπ = = = 10.4 K
I CQ 0.25
Rid = 2 ⎡10.4 + (101)( 0.5 ) ⎤ = 122 K
⎣ ⎦
EX11.9
g m RC
Ad =
2 (1 + g m RE )
( 9.62 )(10 )
10 =
2 ⎡1 + ( 9.62 ) RE ⎤
⎣ ⎦
1 + ( 9.62 ) RE = 4.81 ⇒ RE = 0.396 K
Rid = 2 ⎡ rπ + (1 + β ) RE ⎤ = 2 ⎡10.4 + (101)( 0.396 ) ⎤
⎣ ⎦ ⎣ ⎦
Rid = 100.8 K

3. EX11.10
10 − VGS 4
= K n 3 (VGS 4 − VTN )
2
I1 =
R1
10 − VGS 4 = ( 0.1)( 80 )(VGS 4 − 0.8 )
2
10 − VGS 4 = 8 (VGS 4 − 1.6VGS 4 + 0.64 )
2
2
8VGS 4 − 11.8VGS 4 − 4.88 = 0
(11.8) + 4 ( 8)( 4.88)
2
11.8 ±
VGS 4 = = 1.81 V
2 (8)
10 − 1.81
I1 = I Q = = 0.102 mA
80
0.102
I D1 = ID2 = = 0.0512 mA
2
= K n1 (VGS 1 − VTN )
2
0.0512 = 0.050 (VGS 1 − 0.8 ) ⇒ VGS 1 = 1.81 V
2
v01 = v02 = 5 − ( 0.0512 )( 40 ) = 2.95 V
Max vcm : VDS 1 ( sat ) = VGS1 − VTN
= 1.81 − 0.8 = 1.01 V
vcm ( max ) = v01 − VDS 1 ( sat ) + VGS 1
= 2.95 − 1.01 + 1.81
vcm ( max ) = 3.75 V
Min vcm : VDS 4 ( sat ) = VGS 4 − VTN
= 1.81 − 0.8 = 1.01 V
vcm ( min ) = VGS 1 + VDS 4 ( sat ) − 5
= 1.81 + 1.01 − 5
vcm ( min ) = −2.18 V
−2.18 ≤ vcm ≤ 3.75 V
EX11.11
K n IQ (1)( 2 )
g f ( max ) = = ⇒ g f ( max ) = 1 mA / V
2 ( 2)
Ad = g f RD = (1)( 5 ) ⇒ Ad = 5
EX11.12
iD1 1 Kn ⎛ K ⎞ 2
= + ⋅ vd 1 − ⎜ n ⎟ vd
IQ 2 2IQ ⎜ 2 IQ ⎟
⎝ ⎠
Using the parameters in Example 11.11, K n = 0.5 mA / V 2 , I Q = 1 mA, then
iD1 1 0.5 ⎛ 0.5 ⎞ 2
= 0.90 = + ⋅ vd 1 − ⎜
⎜ 2 (1) ⎟ vd
⎟
IQ 2 2 (1) ⎝ ⎠
By trial and error, vd = 0.894 V
EX11.13
a.

4. IQ 0.5
gm = = = 9.615 mA/V
2VT 2 ( 0.026 )
VA 2 125
r02 = = = 500 kΩ
I C 2 0.25
VA 4 85
r04 = = = 340 kΩ
I C 4 0.25
Ad = g m ( r02 r04 ) = ( 9.615 ) ( 500 340 ) ⇒ Ad = 1946
b. (
Ad = g m r02 r04 RL )
Ad = ( 9.615 ) ⎡500 340 100 ⎤ ⇒ Ad = 644
⎣ ⎦
β VT (150 )( 0.026 )
c. rπ = = = 15.6 kΩ
I CQ 0.25
Rid = 2rπ ⇒ Rid = 31.2 kΩ
d. R0 = r02 r04 = 500 340 ⇒ R0 = 202 kΩ
EX11.14
80 80
I REF = (10 )(VGS 4 − 0.5 ) = ( 0.33)(VGS 5 − 0.5 )
2 2
2 2
VGS 4 + 2VGS 5 = 6
1
VGS 5 = ( 6 − VGS 4 )
2
10 1
(VGS 4 − 0.5) = ( 6 − VGS 4 ) − 0.5
0.33 2
6.0VGS 4 = 5.252 ⇒ VGS 4 = 0.8754 V
80
I REF = (10 )( 0.8754 − 0.5 ) = 53.37 μ A
2
2
Also I Q = I REF = 53.37 μ A
⎛ 80 ⎞
g m = 2kn I Q = 2 ⎜ ⎟ (10 )( 53.37 ) ⇒ 0.2066 mA/V
⎝ 2⎠
1 1
ro1 = ro8 = = = 1873.7 K
λ ID ⎛ 0.05337 ⎞
( 0.02 ) ⎜ ⎟
⎝ 2 ⎠
Ad = g m ( ro1 ro8 ) = ( 0.2066 )(1873.7 1873.7 )
Ad = 194
EX11.15
Ad = g m ( ro 2 RO )
400 = g m ( 500 101000 ) ⇒ g m = 0.8 mA / V
g m = 2 K n I DQ
⎛ 0.08 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞
0.8 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 40
⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠ 2
EX11.16

5. I e 6 = (1 + β ) I b 6 = I b 7
I c 7 = β I b 7 = β (1 + β ) I b 6
Ic7
= β (1 + β ) = (100 )(101) = 1.01× 104
Ib6
EX11.17
r + 300
ROA = π A
1+ β
⎛ β ⎞
I CA = ⎜ ⎟ I EA
⎝1+ β ⎠
⎛ β ⎞⎛ 1 ⎞
=⎜ ⎟⎜ ⎟ (1) ⇒ 9.803 μ A
⎝1+ β ⎠⎝ 1 + β ⎠
(100 )( 0.026 )
rπ A = = 265.2 K
0.009803
265.2 + 300
ROA = = 5.596 K
101
r + ROA
RO = π B 10
1+ β
(100 )( 0.026 )
rπ B = = 2.626 K
0.99
2.626 + 5.596
RO = 10
101
RO = 81.4 104 = 80.7 Ω
EX11.18
10 − 0.7 − ( −10 )
I1 = = 0.6 ⇒ R1 = 32.2 K
R1
⎛I ⎞
I Q R2 = VT ln ⎜ 1 ⎟ .
⎜I ⎟
⎝ Q ⎠
( 0.2 ) R2 = ( 0.026 ) ln ⎛
0.6 ⎞
⎜ ⎟ ⇒ R2 = 143 Ω
⎝ 0.2 ⎠
I R 6 = I1 ⇒ R3 = 0
10 = I C1 RC + VCE1 − 0.7
10.7 = ( 0.1) RC + 4 ⇒ RC = 67 K
vo 2 = −0.7 + 4 = 3.3 V
vE 4 = 3.3 − 1.4 = 1.9 V
vE 4 1.9
I R4 = ⇒ R4 = ⇒ R4 = 3.17 K
R4 0.6
vC 3 = vO 2 − 1.4 + vCE 4
= 3.3 − 1.4 + 3 = 4.9 V
10 − 4.9
I R 5 = I R 4 = 0.6 = ⇒ R5 = 8.5 K
R5
4.2 − 0.7
vE 5 = 4.9 − 0.7 = 4.2 V ⇒ R6 = = 5.83 K
0.6
0 − ( −10 )
R7 = ⇒ R7 = 2 K
5

6. EX11.19
Ri 2 = rπ 3 + (1 + β ) rπ 4
(100 )( 0.026 )
rπ 4 = = 4.333 K
0.6
(100 ) ( 0.026 )
2
β 2VT
rπ 3 ≈ = = 433.3 K
I R4 0.6
Ri 2 = 433.3 + (101)( 4.333) ⇒ Ri 2 = 871 K
Ri 3 = rπ 5 + (1 + β ⎡ R6 + rπ 6 + (1 + β ) R7 ⎤
⎣ ⎦
(100 )( 0.026 )
rπ 5 = = 4.333 K
0.6
(100 )( 0.026 )
rπ 6 = = 0.52 K
5
Ri 3 = 4.333 + (101) ⎡5.83 + 0.52 + (101)( 2 ) ⎤
⎣ ⎦
Ri 3 = 21.0 MΩ
gm 0.1
Ad 1 =
2
( RC Ri 2 ) g m = 0.026 = 3.846 mA/V
3.846
Ad 1 =
2
( 67 871) = 119.64
⎛I ⎞ 0.6
A2 = ⎜ R 4 ⎟ ( R5 Ri 3 ) = (8.5 21000 ) = 98.037
⎝ 2VT ⎠ 2 ( 0.026 )
A = Ad 1 ⋅ A2 = (119.64 )( 98.037 ) = 11, 729
EX11.20
1 1
fz = =
2π RO CO 2π (10 × 106 )( 0.2 × 10−12 )
f z = 79.6 kHz
1
fp =
2π Req CO
Req = 51.98 Ω from Example 11.20
1
fp =
2π ( 51.98 ) ( 0.2 × 10−12 )
f p = 15.3 GHz
TYU11.1
Vd = V1 − V2
= 2 + 0.005sin ω t − ( 0.5 − 0.005sin ω t ) ⇒ Vd = 1.5 + 0.010sin ω t ( V )
V1 + V2
Vcm =
2
2 + 0.005sin ω t + 0.5 − 0.005sin ω t
= ⇒ Vcm = 1.25 V
2
TYU11.2
For v1 = v2 = +4 V ⇒ Minimum vC1 = vC 2 = 4 V
IQ
I C1 = I C 2 = = 1 mA
2
10 − 4
RC = ⇒ RC = 6 kΩ
1

7. TYU11.3
From Equation (11.41)
g R
CMRR = m O
⎛ ΔRC ⎞
⎜ ⎟
⎝ RC ⎠
For CMRR |dB = 75 dB ⇒ CMRR = 5623.4
( 3.86 )(100 )
Then 5623.4 = ⋅ (10 )
ΔRC
Or ΔRC = 0.686 K
TYU11.4
From Equation (11.49)
1 + 2 RO g m
CMRR =
⎛ Δg ⎞
2⎜ m ⎟
⎝ gm ⎠
For CMRR |dB = 90 dB ⇒ CMRR = 31622.8
⎛ 1 + 2 (100 )( 3.86 ) ⎞
Then 31622.8 = ⎜ ⎟ ( 3.86 )
⎝ 2Δg m ⎠
Δg m 0.0472
Or Δg m = 0.0472 mA/V or = = 0.0122 ⇒ 1.22%
gm 3.86
TYU11.5
For v1 = v2 = 5 V ⇒ min vC1 = vC 2 = 5 V
So I C1 RC = 10 − 5 = 0.25 RC ⇒ RC = 20 kΩ
I Q RC ( 0.5 )( 20 )
Ad = = ⇒ Ad = 96.2 Let I Q = 0.5 mA
4VT 4 ( 0.026 )
96.2
C M RRdB = 95 db ⇒ C M RR = 5.62 × 104 ⇒ Acm = ⇒ Acm = 1.71× 10−3
5.62 × 104
⎛ I Q RC ⎞
⎜ ⎟
Acm = ⎝ 2VT ⎠ = 1.71× 10−3
⎡ (1 + β ) I Q R0 ⎤
⎢1 + ⎥
⎣ VT β ⎦
⎡ ( 0.5 )( 20 ) ⎤
⎢ ⎥
⎣ 2 ( 0.026 ) ⎦ = 1.71× 10−3
⎡ ( 201)( 0.5 ) R0 ⎤
⎢1 + ⎥
⎣ ( 0.026 )( 200 ) ⎦
1 + 19.33R0 = 1.125 × 105 ⇒ R0 = 5.82 × 103 kΩ
= 5.82 MΩ
We have R0 = r04 ⎡1 + g m 4 ( R2 rπ 4 ) ⎤
⎣ ⎦
VA 125
r04 = = = 250 kΩ
I Q 0.5
IQ 0.5
gm4 = = = 19.23 mA/V
VT 0.026
β VT ( 200 )( 0.026 )
rπ 2 = = = 10.4 kΩ
IQ 0.5

8. 5.820 = 250 ⎡1 + g m 4 ( R2 rπ 4 ) ⎤
⎣ ⎦
19.23 ( R2 rπ 2 ) = 22.28
(R 2 rπ 2 ) = 1.159 kΩ
R2 (10.4 )
= 1.159
R2 + 10.4
R2 (10.4 − 1.16 ) = (1.16 )(10.4 ) ⇒ R2 = 1.30 kΩ
Let I1 = 1 mA
⎛I ⎞
I Q R2 − I1 R3 = VT ln ⎜ 1 ⎟
⎜I ⎟
⎝ Q ⎠
( 0.5)(1.304 ) − (1) R3 = ( 0.026 ) ln ⎛
1 ⎞
⎜ ⎟ ⇒ R3 = 0.634 kΩ
⎝ 0.5 ⎠
If VBE ( Q3 ) ≅ 0.7 V
10 − 0.7 − ( −10 )
R1 + R3 = = 19.3 ⇒ R1 ≅ 18.7 kΩ
1
TYU11.6
a.
v0 = Ad vd + Acm vcm
vd = v1 − v2 = 0.505sin ω t − 0.495sin ω t
= 0.01sin ω t
v +v 0.505sin ω t + 0.495sin ω t
vcm = 1 2 =
2 2
= 0.50sin ω t
v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5sin ω t ) ⇒ v0 = 0.85sin ω t ( V )
b.
vd = v1 − v2
= 0.5 + 0.005sin ω t − ( 0.5 − 0.005sin ω t )
= 0.01sin ω t
v +v
vcm = 1 2
2
0.5 + 0.005sin ω t + 0.5 − 0.005sin ω t
=
2
= 0.5
v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5 ) ⇒ v0 = 0.25 + 0.6sin ω t ( V )
TYU11.7
IQ / 2 1
a. I B1 = I B 2 = = ⇒ I B1 = I B 2 = 6.62 μ A
(1 + β ) 151
β VT (150 )( 0.026 )
b. rπ = = = 3.9 kΩ
I CQ 1
Rid = 2rπ = 2 ( 3.9 ) = 7.8 kΩ
Vd 10sin ω t ( mV )
Ib = = ⇒ I b = 1.28sin ω t ( μ A )
Rid 7.8 kΩ
c. Ricm ≅ 2 (1 + β ) R0 = 2 (151)( 50 ) ⇒ 15.1 MΩ
Vcm 3sin ω t
Ib = = ⇒ I b = 0.199sin ω t ( μ A )
Ricm 15.1 MΩ

9. TYU11.8
Ad = g f RD
8 = g f ( 4 ) ⇒ g f ( max ) = 2 mA / V
Kn IQ
g f ( max ) =
2
⎛ 4⎞
( 2)
2
= K n ⎜ ⎟ ⇒ K n = 2 mA / V 2
⎝ 2⎠
TYU11.9
From Example 11-10, I Q = 0.587 mA
Kn2 IQ ( 0.1)( 0.587 )
Ad = ⋅ RD = ⋅ (16) ⇒ Ad = 2.74
2 2
1 1
For M 4 , R0 = = ⇒ R0 = 85.2 kΩ
λ4 I Q ( 0.02)( 0.587 )
g m = 2 K n (VGS 2 − VTN ) = 2 ( 0.1)( 2.71 − 1)
= 0.342 mA/V
− g m RD − ( 0.342)(16)
Acm = = ⇒ Acm = −0.0923
1 + 2 g m R0 1 + 2 ( 0.342)(85.2)
⎛ 2.74 ⎞
C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 29.4 dB
⎝ 0.0923 ⎠
TYU11.10
1
CMRR = ⎡1 + 2 2 K n I Q ⋅ Ro ⎤
2⎣ ⎦
CMRRdB = 60 dB ⇒ C M RR = 1000
1
1000 = ⎡1 + 2 2 ( 0.1)( 0.2 ) ⋅ R0 ⎤
2⎣ ⎦
2000 = 1 + 0.4 R0 ⇒ R0 ≅ 5 MΩ
TYU11.11
Ro = ro 4 + ro 2 (1 + g m 4 ro 4 )
Assume I REF = I O = 100 μ A and λ = 0.01 V −1
1 1
ro 2 = ro 4 = = ⇒1 MΩ
λ I D ( 0.01)( 0.1)
Let K n ( all devices ) = 0.1 mA / V 2
Then g m 4 = 2 K n I D = 2 ( 0.1)( 0.1) = 0.2 mA / V
Ro = 1000 + 1000 (1 + ( 0.2 )(1000 ) ) ⇒ 202 M Ω
ID 0.05
Now VGS1 = VGS 2 = + VTN = + 1 = 1.707 V
Kn 0.1
VDS1 ( sat ) = VGS 1 − VTN = 1.707 − 1 = 0.707 V
So vo1 ( min ) = +4 − VGS 1 + VDS 1 ( sat ) = 4 − 1.707 + 0.707
vo1 ( min ) = 3 V = 10 − I D RD = 10 − ( 0.05 ) RD ⇒ RD = 140 kΩ
For a one-sided output, the differential gain is:

10. 1
Ad = g m1 RD where g m1 = 2 K n I D
2
=2 ( 0.1)( 0.05) = 0.1414 mA / V
1
Ad = ( 0.1414 )(140 ) ⇒ Ad = 9.90
2
The common-mode gain is:
2 K n I Q ⋅ RD 2 ( 0.1)( 0.1) ⋅ (140 )
Acm = = ⇒ Acm = 0.0003465
1 + 2 2 K n I Q ⋅ Ro 1 + 2 2 ( 0.1)( 0.1) ⋅ ( 202000 )
Ad
Then CMRRdB = 20 log10 ⇒ CMRRdB = 89.1 dB
Acm
TYU11.12
IQ 0.5
a. I B5 = = ⇒ 15.3 nA
β (1 + β ) (180 )(181)
So I 0 = 15.3 nA
b. For a balanced condition
VEC 4 = VEC 3 = VEB 3 + VEB 5 ⇒ VEC 4 = 1.4 V
VCE 2 = VC 2 − VE 2 = (10 − 1.4 ) − ( −0.7 ) ⇒ VCE 2 = 9.3 V
TYU11.13
Ad = 2 g f ( r02 r04 )
IQ 0.2
gf = = = 1.923 mA/V
4VT 4 ( 0.026 )
VA 2 120
r02 = = = 1200 kΩ
I C 2 0.1
VA 4 80
r04 = = = 800 kΩ
I C 4 0.1
Ad = 2 (1.923) (1200 800 ) ⇒ Ad = 1846
TYU11.14
P = ( I Q + I REF ) ( 5 − ( −5 ) )
10 = ( 0.1 + I REF )(10 ) ⇒ I REF = 0.9 mA
5 − 0.7 − ( −5 ) 9.3
R1 = = ⇒ R1 = 10.3 k Ω
I REF 0.9
⎛I ⎞
I Q RE = VT ln ⎜ REF ⎟
⎜ I ⎟
⎝ Q ⎠
0.026 ⎛ 0.9 ⎞
RE = ln ⎜ ⎟ ⇒ RE = 0.571 kΩ
0.1 ⎝ 0.1 ⎠
VA 2 120
ro 2 = = ⇒ 2.4 M Ω
I C 2 0.05
VA 4 80
ro 4 = = ⇒ 1.6 M Ω
I C 4 0.05
0.05
gm = = 1.923 mA / V
0.026
( ) (
Ad = g m ro 2 ro 4 RL = (1.923) 2400 1600 90 ⇒ Ad = 158 )

11. TYU11.15
a.
R0 = r02 r04
120
r02 = = 1.2 MΩ
0.1
80
r04 = = 0.8 MΩ
0.1
R0 = 1.2 0.8 ⇒ R0 = 0.48 MΩ
b.
Ad ( open circuit ) = 2 g f ( r02 r04 )
(
Ad ( with load ) = 2 g f r02 r04 RL )
1
For Ad ( with load ) = Ad ( open circuit ) ⇒ RL = ( r02 r04 ) ⇒ RL = 0.48 MΩ
2
TYU11.16
2Kn 1
Ad = 2 ⋅
I Q ( λ2 + λ4 )
2 ( 0.1) 1
=2 ⋅ ⇒ Ad = 113
0.1 ( 0.01 + 0.015)
TYU11.17
75
For the MOSFET, I D = 25 − = 24.26 μ A
101
g m1 = 2 K n I D = 2 ( 20 )( 24.26 ) = 44.05 μ A/V
For the Bipolar, I E = 100 − 25 = 75 μ A
⎛ 100 ⎞
IC = ⎜ ⎟ ( 75 ) = 74.26 μ A
⎝ 101 ⎠
(100 )( 0.026 )
rπ 2 = = 35.0 K
0.07426
0.07426
gm2 = = 2.856 mA/V
0.026
g (1 + g m 2 rπ ) ( 44.05 ) ⎡1 + ( 2.856 )( 35 ) ⎤
⎣ ⎦
g m = m1
C
=
1 + g m1rπ 1 + ( 0.04405 )( 35 )
g m = 1.75 mA/V
C
TYU11.18
From Figure 11.43
80
r04 = = 160 kΩ
0.5
R0 ≅ β r04 = (150 )(160 ) kΩ ⇒ R0 = 24 MΩ
From Figure 11.44
1 1
r06 = = ⇒ r06 = 160 kΩ
λ I D ( 0.0125 )( 0.5 )
0.5 = 0.5 (VGS − 1) ⇒ VGS = 2 V
2
g m 6 = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2 − 1) = 1 mA / V
r04 = 160 kΩ
R0 = ( g m 6 )( r06 )( β r04 ) = (1)(160 )(150 )(160 ) ⇒ R0 = 3.840 MΩ

12. TYU11.19
From Equation (11.126)
2 (1 + β ) β VT 2 (121)(120 )( 0.026 )
Ri = = ⇒ Ri = 1.51 MΩ
IQ 0.5
β VT (120 )( 0.026 )
rπ 11 = = = 6.24 kΩ
IQ 0.5
′
RE = rπ 11 R3 = 6.24 0.1 = 0.0984 kΩ
IQ 0.5
g m11 = = = 19.23 mA/V
VT 0.026
VA 120
r011 = = = 240 kΩ
I Q 0.5
Then RC11 = r011 (1 + g m11 RE )
′
= 240 ⎡1 + (19.23)( 0.0984 ) ⎤
⎣ ⎦
= 694 kΩ
β VT (120 )( 0.026 )
rπ 8 = = = 1.56 kΩ
IC 8 2
Rb8 = rπ 8 + (1 + β ) R4 = 1.56 + (121)( 5 )
= 607 kΩ
Then RL 7 = RC11 Rb8 = 694 607 = 324 kΩ
⎛ IQ ⎞ ⎡ 0.5 ⎤
Then Av = ⎜ ⎟ RL 7 = ⎢ ⎥ ( 324 ) ⇒ Av = 3115
⎝ 2VT ⎠ ⎢ 2 ( 0.026 ) ⎥
⎣ ⎦
⎛r +Z ⎞
R0 = R4 || ⎜ π 8 ⎟
⎝ 1+ β ⎠
Z = RC11 RC 7
V 120
RC 7 = A = = 240 kΩ
I Q 0.5
Z = 694 240 = 178 kΩ
⎛ 1.56 + 178 ⎞
R0 = 5 || ⎜ ⎟ = 5 1.48 ⇒ R0 = 1.14 kΩ
⎝ 121 ⎠
TYU11.20
⎛ IQ ⎞
Av = ⎜ ⎟ RL 7
⎝ 2VT ⎠
⎛ 0.5 ⎞
⎜ 2 ( 0.026 ) ⎟ RL 7 ⇒ RL 7 = 104 kΩ
103 = ⎜ ⎟
⎝ ⎠