1. Chapter 15
Exercise Solutions
EX15.1
For the circuit shown in Figure 15.7
1
f 3dB =
2π RC
or
1 1
RC = = = 3.979 × 10−6
2π f3dB 2π ( 40 × 103 )
For R = 75 K
Then
C = 5.31×10−11 = 53.1 pF
We have C3 = 1.414C = 75.1 pF
C4 = 0.707C = 37.5 pF
EX15.2
1
fC =
CReq
or
1 1
C= =
f c Req (105 )( 20 × 106 )
C = 0.5 pF
EX15.3
C1 30
Low-frequency gain: T = − =− = −6
C2 5
fC C2 (100 × 10 )( 5 × 10 )
3 −12
f 3dB = = ⇒ f 3dB = 6.63 kHz
2π CF 2π (12 × 10−12 )
EX15.4
1
f0 =
2π 3RC
1 1
RC = = = 6.13 × 10−6
2π f 0 3 2π (15 × 10 ) 3
3
Let C = 0.001 μF = 1 nF
Then R = 6.13 kΩ so R2 = 8R = 49 kΩ
EX15.5
1 1
f0 = ⇒C =
2π RC 2π f 0 R
1
C= ⇒ C ≅ 0.02 μ F
2π ( 800 ) (104 )
R2 = 2 R1 = 2 (10 ) ⇒ R2 = 20 kΩ
EX15.6

2. ⎛ R1 ⎞
VTH = ⎜ ⎟ VH
⎝ R1 + R2 ⎠
⎛ R1 ⎞
2=⎜ ⎟ (12)
⎝ R1 + 20 ⎠
2 ( R1 + 20 ) = 12 R1
40 = 10 R1 ⇒ R1 = 4 kΩ
EX15.7
⎛ R1 ⎞
VTH − VTL = ⎜ ⎟ (VH − VL )
⎝ R1 + R2 ⎠
⎛ R1 ⎞
0.10 = ⎜ ⎟ (10 − [ −10])
⎝ R1 + R2 ⎠
R2 20 R
1+ = = 200 ⇒ 2 = 199
R1 0.10 R1
⎛ R2 ⎞
VS = ⎜ ⎟ VREF
⎝ R1 + R2 ⎠
⎛ R ⎞ ⎛ 1 ⎞
VREF = ⎜1 + 1 ⎟ VS = ⎜ 1 + ⎟ (1) ⇒ VREF = 1.005 V
⎝ R2 ⎠ ⎝ 199 ⎠
VH − VBE ( on ) − Vγ
I=
R + 0.1
10 − 0.7 − 0.7
R + 0.1 = = 43 kΩ
0.2
R = 42.9 kΩ
EX15.8
At t = 0− , let v0 = −5 so v X = −2.5. For t > 0
⎛ −t ⎞
v X = 10 + ( −2.5 − 10 ) exp ⎜ ⎟
⎝ rX ⎠
When v X = 5.0, output switches
⎛ t ⎞
5.0 = 10 − 12.5 exp ⎜ − 1 ⎟
⎝ rX ⎠
⎛ t ⎞ 10 − 5 5.0
exp ⎜ − 1 ⎟ = =
⎝ rX ⎠ 12.5 12.5
⎛ t ⎞ 12.5 ⎛ 12.5 ⎞
exp ⎜ + 1 ⎟ = ⇒ t1 = rX ⋅ ln ⎜ ⎟ ⇒ t1 = rX ( 0.916 )
⎝ rX ⎠ 5.0 ⎝ 5.0 ⎠
During the next part of the cycle
⎛ t ⎞
v X = −5 + ( 5 − [ −5]) exp ⎜ − ⎟
⎝ rX ⎠
When v X = −2.5, output switches

3. ⎛ t ⎞
−2.5 = −5 + 10 exp ⎜ − 2 ⎟
⎝ rX ⎠
⎛ t ⎞ 5 − 2.5 2.5
exp ⎜ − 2 ⎟ = =
⎝ rX ⎠ 10 10
⎛ t ⎞ 10 ⎛ 10 ⎞
exp ⎜ + 2 ⎟ = ⇒ t2 = rX ⋅ ln ⎜ ⎟ ⇒ t2 = rX (1.39 )
⎝ rX ⎠ 2.5 ⎝ 2.5 ⎠
1
Period = t1 + t2 = T = ⎡( 0.916 ) + (1.39 ) ⎤ rX = 2.31rX ⇒ Frequency =
⎣ ⎦ 2.31rX
rX = ( 50 × 103 )( 0.01× 10 −6 ) = 5 × 10 −4 s ⇒ f = 866 Hz
t1 ( 0.916 )
Duty cycle = × 100% = × 100% ⇒ Duty cycle = 39.7%
t1 + t2 ( 0.916 ) + (1.39 )
EX15.9
a.
rX = RX C X
⎛ R1 ⎞ ⎛ 10 ⎞
vY = ⎜ ⎟ v0 = ⎜ ⎟ (12 ) = 1.2 V
⎝ R1 + R2 ⎠ ⎝ 10 + 90 ⎠
R1
β= = 0.10
R1 + R2
⎡ 0.7 ⎤
⎡1 + Vγ /VP ⎤ ⎢ 1 + 12 ⎥
T = rX ln ⎢ ⎥ = rX ln ⎢ ⎥
⎣ 1− β ⎦ ⎢1 − (0.10) ⎥
⎢
⎣ ⎥
⎦
T = 50 × 10−6 = rX ln [1.18] = (0.162) rX
50 × 10−6
RX = ⇒ RX = 3.09 kΩ
(0.1× 10−6 )(0.162)
b. Recovery time
⎛ t ⎞
v X = VP + (−1.2 − VP ) exp ⎜ − ⎟
⎝ rX ⎠
When v X = Vγ , t = t2

4. ⎛ t ⎞
0.7 = 12 + ( −1.2 − 12 ) exp ⎜ − 2 ⎟
⎝ rX ⎠
⎛ t ⎞ 12 − 0.7
exp ⎜ − 2 ⎟ = = 0.856
⎝ rX ⎠ 13.2
⎛ 1 ⎞
t2 = rX ln ⎜ ⎟ = ( 0.155 ) rX
⎝ 0.856 ⎠
rX = ( 3.09 × 103 )( 0.1× 10−6 ) = 3.09 × 10−4 ⇒ t2 = 48.0 μ s
EX15.10
T = 1.1 RC
T = 75 × 10−6
Let C = 10 nF
Then
75 × 10−6
R= = 6.82 K
(1.1) (10 ×10−9 )
EX15.11
1 1
f = = ⇒ f = 802 Hz
0.693 ( RA + 2 RB ) C ( 0.693) ⎡ 20 + 2 ( 80 ) ⎤ × 103 × ( 0.01× 10−6 )
⎣ ⎦
R + RB 20 + 80
Duty cycle = A × 100% = × 100% ⇒ Duty cycle = 55.6%
RA + 2 RB 20 + 2 ( 80 )
EX15.12
1 VP2
a. P= ⋅
2 RL
VP = 2 RL P = 2 ( 8 )(1) ⇒ VP = 4 V
VP 4
IP = = ⇒ I P = 0.5 A
RL 8
b. VCE = 12 − 4 = 8 V
I C ≈ 0.5 A
So P = I C ⋅ VCE = ( 0.5 )( 8 ) ⇒ P = 4 W
EX15.13
VP = 2 RL PL = 2 ( 8 )(10 ) = 12.65 V
⎛ V ⎞
PS = VS ⎜ P ⎟
⎝ π RL ⎠
PS π R2 (10 ) π ( 8 )
VS = =
VP 12.65
VS = 19.9 V
EX15.14
dV0 dV dV
Line regulation = +
= 0 ⋅ Z+
dV dVZ dV
Now

5. dV0 ⎛ 10 ⎞
= ⎜1 + ⎟ = 2
dVZ ⎝ 10 ⎠
dVZ ⎛ rZ ⎞ 10
+
=⎜ ⎟= = 0.00227
dV ⎝ rZ + R1 ⎠ 10 + 4400
So Line regulation = ( 2 )( 0.00227 ) = 0.00454
0.454%
EX15.15
V1 V0 − V1 ⎛1 1⎞ V
= ⇒ V1 ⎜ + ⎟ = 0
10 10 ⎝ 10 10 ⎠ 10
⎛ 2⎞ V V
V1 ⎜ ⎟ = 0 ⇒ V0 = 2V1 ⇒ V1 = 0
⎝ 10 ⎠ 10 2
V0 − V1 V0 V0 − A0 L (VZ − V1 )
+ + =0
10 RL R0
V0 V0 V0 A0 LVZ V1 A0 LV1
+ + − = −
10 RL R0 R0 10 R0
V0 A V
= − 0L 0
2(10) 2 R0
V0 V 1000 ( 6.3) V0 (1000 ) V0
+ I0 + 0 − = −
10 0.5 0.5 20 2 ( 0.5 )
V0 [0.10 + 2.0 − 0.05 + 1000] + I 0 = 12, 600
V0 (1002.05) + I 0 = 12, 600
For I 0 = 1 mA ⇒ V0 = 12.5732
For I 0 = 100 mA ⇒ V0 = 12.4744
V0 ( NL ) − V0 ( FL )
Load reg = × 100%
V0 ( NL )
12.5732 − 12.4744
= × 100%
12.5732
Load reg = 0.786%
EX15.16
a.

6. VZ − 3VBE ( on )
IC 3 =
R1 + R2 + R3
5.6 − 3 ( 0.6 ) 3.8
IC 3 = = ⇒ I C 3 = 0.482 mA
3.9 + 3.4 + 0.576 7.88
⎛I ⎞
I C 4 R4 = VT ln ⎜ C 3 ⎟
⎝ IC 4 ⎠
⎛ 0.482 ⎞
I C 4 (0.1) = (0.026) ln ⎜ ⎟
⎝ IC 4 ⎠
By trial and error
I C 4 = 0.213 mA
VB 7 = 2(0.6) + (0.482)(3.9) ⇒ VB 7 = 3.08 V
b.
⎛ R13 ⎞
⎜ ⎟ V0 = VB 8 = VB 7
⎝ R13 + R12 ⎠
⎛ 2.23 ⎞
⎜ ⎟ (5) = 3.08
⎝ 2.23 + R12 ⎠
( 2.23)( 5 ) = ( 3.08 )( 2.23) + ( 3.08) R12
11.15 = 6.868 = 3.08R12 ⇒ R12 = 1.39 kΩ
TYU15.1
1
f 3dB =
2π RC
1 1
RC = = = 1.59 × 10−5
2π f3dB 2π (104 )
Let C = 0.01 μ F ⇒ R = 1.59 kΩ
Then
C1 = 0.03546 μ F
C2 = 0.01392 μ F
C3 = 0.002024 μ F
1 1
T = =
6 6
⎛ f ⎞ ⎛ 20 ⎞
1+ ⎜ ⎟ 1+ ⎜ ⎟
⎝ f3d ⎠ ⎝ 10 ⎠
T = 0.124 or T = −18.1 dB
TYU15.2
1 1
f 3dB = ⇒ RC =
2π RC 2π f3dB
1
RC = = 3.18 × 10−6
2π ( 50 × 103 )
Let C = 0.001 μ F = 1 nF ⇒ R = 3.18 kΩ
Then

7. R1 = 2.94 kΩ
R2 = 3.44 kΩ
R3 = 1.22 kΩ
R4 = 8.31 kΩ
1
T = 0.01 =
8
⎛ f ⎞
1 + ⎜ 3− dB ⎟
⎝ f ⎠
8 2
⎛ f ⎞ ⎛ 1 ⎞ 4
1 + ⎜ 3− dB ⎟ = ⎜ ⎟ = 10
⎝ f ⎠ ⎝ 0.01 ⎠
2
⎛ f3− dB ⎞ f 3dB
⎜ ⎟ ≅ 10 ⇒ f = ⇒ f ≅ 15.8 kHz
⎝ f ⎠ 10
TYU15.3
1
1-pole T = ⇒ −3.87 dB
2
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
2-pole T = ⇒ −4.88 dB
4
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
3-pole T = ⇒ −6.0 dB
6
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
1
4-pole T = ⇒ −7.24 dB
8
⎛ 12 ⎞
1+ ⎜ ⎟
⎝ 10 ⎠
TYU15.4
1
Req =
fC C
1 1
or f C C = = = 2 × 10−7
Req 5 × 106
If C = 10 pF ⇒ fC = 20 kHz
TYU15.5
1 1
f0 = = ⇒ f 0 ≅ 65 kHz
2π 6 RC 2π 6 (10 )(100 × 10−12 )
4
R2 = 29 R = 29 (104 ) ⇒ R2 = 290 kΩ
TYU15.6

8. 1 1
f0 = = ⇒ f 0 = 7.12 MHz
⎛ CC ⎞ ⎡ (10−9 ) 2 ⎤
2π L ⋅ ⎜ 1 2 ⎟ 2π (10 ) ⎢
−6
−9
⋅⎥
⎝ C1 + C2 ⎠ ⎣ 2 × 10 ⎦
C2
= gm R
C1
C2 1 1
gm = ⋅ = ⇒ g m = 0.25 mA / V
C1 R 4 × 103
We have
⎛ k ′ ⎞⎛ W ⎞
g m = 2 ⎜ ⎟⎜ ⎟ (VGS − VTh )
⎝ 2 ⎠⎝ L ⎠
k ′ ≅ 20 μ A / V 2 , VGS − VTh ≅ 1 V
W 0.25 × 10−3
So = = 12.5
L ( 20 × 10−6 ) (1)
and a value of W / L = 12.5 is certainly reasonable.
TYU15.7
⎛R ⎞
VTH = − ⎜ 1 ⎟ VL
⎝ R2 ⎠
⎛R ⎞ R
0.10 = − ⎜ 1 ⎟ ( −10 ) ⇒ 1 = 0.010
⎝ R2 ⎠ R2
Let R1 = 0.10 kΩ then R2 = 10 kΩ
TYU15.8
a.
⎛ R2 ⎞ ⎛ 10 ⎞
VS = ⎜ ⎟ VREF = ⎜ ⎟ ( 2)
⎝ R1 + R2 ⎠ ⎝ 1 + 10 ⎠
VS = 1.82 V
⎛ R1 ⎞ ⎛ 1 ⎞
VTH = VS + ⎜ ⎟ VH = 1.82 + ⎜ ⎟ (10 )
⎝ R1 + R2 ⎠ ⎝ 1 + 10 ⎠
VTH = 2.73 V
⎛ R1 ⎞ ⎛ 1 ⎞
VTL = VS + ⎜ ⎟ VL = 1.82 + ⎜ ⎟ ( −10 )
⎝ R1 + R2 ⎠ ⎝ 1 + 10 ⎠
VTL = 0.91 V
b.

9. TYU15.9
⎛ R ⎞
VS = ⎜ 1 + 1 ⎟ VREF
⎝ R2 ⎠
⎛R ⎞ ⎛R ⎞
VTH = VS − ⎜ 1 ⎟ VL and VTL = VS − ⎜ 1 ⎟ VH
⎝ R2 ⎠ ⎝ R2 ⎠
⎛R ⎞
Hysteresis Width = VTH − VTL = ⎜ 1 ⎟ (VH − VL )
⎝ R2 ⎠
⎛R ⎞ ⎛R ⎞
2.5 = ⎜ 1 ⎟ ( 5 − [ −5]) = 10 ⎜ 1 ⎟
⎝ R2 ⎠ ⎝ R2 ⎠
R
So 1 = 0.25
R2
Then
⎛ R ⎞
VS = −1 = ⎜ 1 + 1 ⎟ VREF = (1 + 0.25)VREF ⇒ VREF = −0.8 V
⎝ R2 ⎠
Then
VTH = −1 − ( 0.25 )( −5 ) ⇒ VTH = 0.25 V
VTL = −1 − ( 0.25 )( 5 ) ⇒ VTL = −2.25 V
TYU15.10
⎛ R1 ⎞ ⎛ 10 ⎞ 1
vX = ⎜ ⎟ v0 = ⎜ ⎟ v0 = v0
⎝ R1 + R2 ⎠ ⎝ 10 + 20 ⎠ 3
10
t = 0, vX = −
3
⎛ 10 ⎞ ⎛ t ⎞
v X = 10 + ⎜ − − 10 ⎟ exp ⎜ − ⎟
⎝ 3 ⎠ ⎝ rX ⎠
10
Output switches when v X =
3

10. 10 ⎛ t1 ⎞
= 10 − 13.33 exp ⎜− ⎟
3 ⎝ rX ⎠
⎛ t ⎞ 10 − 3.33 6.67
exp ⎜ − 1 ⎟ = =
⎝ rX ⎠ 13.33 13.33
⎛ t ⎞ 13.33
exp ⎜ + 1 ⎟ = ≅2
⎝ rX ⎠ 6.67
t1 = rX ln (2) = (0.693)rX
T = 2(0.693)rX
1
f =
2(0.693)rX
rX = RX C X = (104 )( 0.1×10 −6 ) = 1×10 −3 ⇒ f = 722 Hz ⇒ Duty cycle = 50%
TYU15.11
⎛ R1 ⎞ 20
β =⎜ ⎟= = 0.333
⎝ R1 + R2 ⎠ 20 + 40
rX = RX C X = (104 )( 0.01× 10−6 ) = 1× 10−4
⎡ 0.7 ⎤
⎛ 1 + Vγ / VP ⎞ ⎢ 1+ 8 ⎥
⎟ = (1× 10 ) ln ⎢ ⎥ ⇒ T = 48.9 μ s
−4
T = rX ln ⎜
⎝ 1− β ⎠ ⎢1 − 0.333 ⎥
⎢
⎣ ⎥
⎦
Recovery time

11. ⎛ R1 ⎞ ⎛ 20 ⎞
vY = ⎜ ⎟ v0 = ⎜ ⎟ (8) = 2.667 V
⎝ R1 + R2 ⎠ ⎝ 20 + 40 ⎠
⎛ t ⎞
0.7 = 8 + ( −2.667 − 8 ) exp ⎜ − 2 ⎟
⎝ rX ⎠
⎛ t ⎞ 8 − 0.7
exp ⎜ − 2 ⎟ = = 0.6844
⎝ rX ⎠ 10.66
⎛ 1 ⎞
t2 = rX ln ⎜ ⎟ ⇒ t2 = 37.9 μ s
⎝ 0.685 ⎠
TYU15.12
1
f =
( 0.693)( RA + RB ) C
1
RA + RB =
( 0.693) fC
Let C = 0.01 μ F, f = 1kHz
1
RA + RB = = 1.443 × 105
( 0.693) (103 )( 0.01×10−6 )
RA + RB
Duty cycle = 55 = × 100%
RA + 2 RB
55 =
(1.443 ×10 ) (100 )
5
(1.443 ×10 ) + R
5
B
RB =
(1.443 ×10 ) (100 − 55) ⇒ R
5
= 118 kΩ so RA = 26.2 kΩ
B
55
TYU15.13
v01 ⎛ R2 ⎞ ⎛ 30 ⎞
a. = ⎜ 1 + ⎟ = ⎜1 + ⎟ = 2.5
vI ⎝ R1 ⎠ ⎝ 20 ⎠
v02 R 50
=− 4 =− = −2.5
vI R3 20
1 VL2 1 [12 − (−12)]2
(b) P= ⋅ = ⋅ = 240 mW
2 RL 2 1.2
Or
P = 0.24 W
12
c. = V pi = 4.8 V
2.5