1. Chapter 14
Exercise Solutions
EX14.1
−50
a. ACL = ⇒ ACL = −49.949
⎡ ⎛ 1 ⎞ ⎤
⎢1 + ⎜ 5 × 104 ⎟ ( 51) ⎥
⎣ ⎝ ⎠ ⎦
dACL 51 dA
b. = 10 × ⇒ CL = 0.0102%
ACL 5 × 104 ACL
−50
ACL = ⇒ ACL = −49.943
⎡ 51 ⎤
⎢1 + 4.5 × 104 ⎥
⎣ ⎦
EX14.2
a. For R0 = 0
= + (1 + 104 ) = 0.1 + 103
1 1 1
Ri f 10 10
⇒ Ri f = 10−3 kΩ = 1 Ω
b. For R0 = 10 kΩ
1 1 1 ⎡1 + 104 + 1 ⎤ 104
= + ×⎢ ⎥ ≅ 0.1 +
Ri f 10 10 ⎣ 1 + 1 + 1 ⎦ 3 (10 )
Rif = 3 × 10−3 kΩ ⇒ Ri f = 3 Ω
EX14.3
⎛ 40 ⎞
40 (1 + 104 ) + 99 ⎜ 1 + ⎟
Ri f = ⎝ 1 ⎠
99
1+
1
4 × 10 + 4.059 × 103
5
≅
100
Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ
EX14.4
R
1 + 2 = 100
R1
1 1 ⎡ 105 ⎤
a. = ⎢ ⎥ = 10 ⇒ R0 f = 0.1 Ω
R0 f 100 ⎣100 ⎦
1 1 ⎡ 105 ⎤
b. = ⎢ ⎥ = 10
2
R0 f 10 ⎣100 ⎦
R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω
EX14.5
From Equation (14.43)

2. ACL 0
ACL ( f ) =
f
1+ j ⋅
f PD ( A0 /ACL 0 )
25 25
= =
f f
1+ j ⋅ 1+ j ⋅
( 50 ) (104 / 25) 2 × 104
a. f = 2 kHz
v0
= 25 ⇒ v0 ( peak ) = 1.25 mV
vI
b. f = 20 kHz
v0 1
= ⋅ 25 ⇒ v0 ( peak ) = 0.884 mV
vI 2
c. f = 100 kHz
v0 25 25
= = = 4.90
vI 1 + (100 / 20 ) 5.099
2
⇒ v0 = 0.245 mV
EX14.6
Full-scale response = 1× 5 = 5 V
5
t= ⇒ t = 2.5 μ s
2
EX14.7
SR 0.63 × 106
a. FPBW = =
2π V0 ( max ) 2π (1)
FPBW = 1.0 × 105 ⇒ FPBW = 100 kHz
0.63 × 106
b. FPBW = = 1.0 × 104
2π (10 )
⇒ FPBW = 10 kHz
EX14.8
⎛I ⎞ ⎛ 1.85 × 10−14 ⎞
V0 S = VT ln ⎜ S 2 ⎟ = (0.026) ln ⎜ −14 ⎟
⎝ I S1 ⎠ ⎝ 2 × 10 ⎠
⇒ V0 S = 2.03 mV
EX14.9
We need
iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V
By Equation (14.60(a))
⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC1 = I S 1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜ 1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 3 ⎢ exp ⎜ EB 3 ⎟ ⎥ ⎜1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
By Equation (14.60(b))

3. ⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC 2 = I S 2 ⎢ exp ⎜ BE 2 ⎟ ⎥ ⎜ 1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 4 ⎢ exp ⎜ EB 4 ⎟ ⎥ ⎜ 1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
I S1 = I S 2, take the ratio:
⎛v −v ⎞ I
exp ⎜ BE1 BE 2 ⎟ = S 3
⎝ VT ⎠ IS 4
⎛I ⎞
vBE1 − vBE 2 = V0 S = VT ln ⎜ S 3 ⎟
⎝ IS 4 ⎠
= 0.026 ⋅ ln (1.05 )
⇒ V0 S = 1.27 m V
EX14.10
1 I Q ⎛ ΔK n ⎞
VOS = ⋅ ⋅⎜ ⎟
2 2Kn ⎝ Kn ⎠
1 150 ⎛ ΔK n ⎞
0.020 = ⋅ ⋅
2 2 ( 50 ) ⎜ 50 ⎟
⎝ ⎠
⇒ ΔK n = 1.63μ A / V 2
ΔK n 1.63
⇒ = ⇒ 3.26%
Kn 50
EX14.11
⎛ R5 ⎞ +
Want ⎜ ⎟V = 5 m V
⎝ R5 + R4 ⎠
R5
R5 R4 so × V + = 0.005
R4
( 0.005)(100 )
R5 = = 0.05 kΩ
10
⇒ R5 = 50 Ω
EX14.12
R1′ = 25 1 = 0.9615 kΩ
′
R2 = 75 1 = 0.9868 kΩ
For I Q = 100 μ A ⇒ iC1 = iC 2 = 50 μ A
From Equation (14.75)
⎛ 50 × 10−6 ⎞
( 0.026 ) ln ⎜ −14 ⎟ + ( 0.050 )( 0.9615 )
⎝ 10 ⎠
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 )
⎝ IS 4 ⎠
0.58065 + 0.048075
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934
⎝ IS 4 ⎠

4. ⎛i ⎞
ln ⎜ C 2 ⎟ = 22.284
⎝ IS 4 ⎠
50 × 10−6
= 4.7625 × 109
IS 4
I S 4 ≅ 1.05 × 10−14 A
EX14.13
From Equation (14.79)
⎛ R ⎞
v0 = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟
⎝ R1 ⎠
For v0 = 0
⎛ 100 ⎞
0 = (1.1× 10−6 ) (100 kΩ ) − (1.0 × 10−6 ) R3 ⎜ 1 + ⎟
⎝ 10 ⎠
R3 (11) = (1.1)(100 kΩ ) ⇒ R3 = 10 kΩ
TYU14.1
v1CM ( max ) = V + − VSD1 ( sat ) − VSG1
v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1
We have:
I REF = 100 μ A, kn = 80 μ A / V 2 , k ′ = 40 μ A / V 2 ,
′ p
⎛W ⎞
⎜ ⎟ = 25
⎝L ⎠
For M1 :
⎛ 40 ⎞
I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP )
2
⎝ 2 ⎠
So 50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V
2
VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V
Then
vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V
For M 4 :
⎛ 80 ⎞
I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN )
2
⎝ 2⎠
So 100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V
2
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184
So
V − − 0.184 ≤ vCM ≤ V + − 1.13 V
TYU14.2
vo ( max ) = V + − VSD 8 ( sat ) − VSG10
vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat )
Now

5. 50
VSG 8 = VSG10 = + 0.5 = 0.816 V
( 40/ 2 )( 25)
VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V
So vo ( max ) = V + − 0.316 − 0.816 = V + − 1.13
Also
50
VGS 6 = + 0.5 = 0.724 V
(80/ 2 )( 25)
100
VGS 4 = + 0.5 = 0.816 V
(80/ 2 )( 25)
VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
So
vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54
Then
V − + 0.54 ≤ vo ≤ V + − 1.13 V
TYU14.3
500
ACL ( ideal ) = −= −25
20
Within 0.1% ⇒ −25 + ( 0.001)( 25 ) ⇒ ACL = −24.975
−25
−24.975 =
⎡ 26 ⎤
⎢1 + A ⎥
⎣ 0L ⎦
26 −25
= − 1 = 0.0010
A0 L −24.975
A0 L = 25.974
TYU14.4
ACL ( ∞ )
a. ACL =
⎡ A (∞) ⎤
1 + ⎢ CL ⎥
⎣ A0 L ⎦
R 495
ACL ( ∞ ) = 1 + 2 = 1 + = 100
R1 5
100
ACL = ⇒ ACL = 99.90
100
1+ 5
10
ACL ( ∞ ) = 100
dACL 100
b. = 10 × 5 = 0.01%
ACL 10
ACL = 99.90 − ( 0.0001)( 99.90 )
⇒ ACL = 99.89
TYU14.5

6. ACL ( ∞ ) − ACL ACL 1
= 1− = 1−
ACL ( ∞ ) ACL ( ∞ ) A (∞)
1 + CL
A0 L
ACL ( ∞ ) ACL ( ∞ )
1+ −1
A0 L A0 L
So 0.001 = =
A (∞) A (∞)
1 + CL 1 + CL
A0 L A0 L
ACL ( ∞ )
0.001 = 0.999 ⋅
A0 L
0.001 0.001
ACL ( ∞ ) = ⋅ A0 L = ⋅ (104 ) ⇒ ACL ( ∞ ) = 10.010
0.999 0.999
or
ACL = (1 − 0.001)(10.010 ) ⇒ ACL = 10.0
TYU14.6
ii ⎛ Rif ⎞
=⎜ ⎟
I1 ⎝ Ri ⎠
iI 0.1
a. = = 1× 10−5
i1 104
iI 10
b. = = 1× 10−3
i1 104
TYU14.7
Voltage follower R2 = 0, R1 = ∞
Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 )
≅ 5 × 106 kΩ ⇒ Rif = 5000 MΩ
TYU14.8
f3dB =
fT
=
(105 ) (10 ) ⇒ 20 kHz
ACL 0 50
SR
f max = f 3dB =
2π V0 ( max )
SR 0.8 × 106
V0 ( max ) = =
2π f3dB 2π ( 20 × 103 )
⇒ V0 ( max ) = 6.37 V
TYU14.9
a. v0 = I B1 R3 = (10−6 )( 200 × 103 )
⇒ v0 = 0.20 V
b. R4 = R1 R2 R3 = 100 50 200
⇒ R4 = 28.6 kΩ