Ideal Bode plot for Prob. 16-27.
16-28.
Given:
C
= 50 pF
A
v
= 200,000
Solution:
C
in
=
C
(
A
v
+ 1)
(Eq. 16-26)
C
in
= 50 pF(200,000 + 1)
C
in
= 10
μ
F
Answer:
The Miller input capacitance is 10
μ
F.

1-81
16-29.
Given:
C
= 100 pF
A
v
= 150,000
R
L
= 10 k
Ω
R
in
= 1 k
Ω
Solution:
C
in
=
C
(
A
v
+ 1)
(Eq. 16-26)
C
in
= 100 pF(150,000 + 1)
C
in
= 15 μF
f
2
= 1/(2
π
RC
)
f
2
= 1/[2
π
(1 k
Ω
)(15 μF)]
f
2
= 11 Hz
A
v(dB)
= 20 log A
v
(Eq. 16-9)
A
v(dB)
= 20 log(150,000)
A
v(dB)
= 104 dB
Answer:
See figure below.
Ideal Bode plot for Prob. 16-29.
16-30.
Given:
T
R
= 10 μs
Solution:
f
2
= 0.35/
T
R
(Eq. 16-29)
f
2
= 0.35/10 μs
f
2
= 35 kHz
Answer:
The upper cutoff frequency is 35 kHz.
16-31.
Given:
T
R
= 0.25 μs
Solution:
f
2
= 0.35/
T
R
(Eq. 16-29)
f
2
= 0.35/0.25 μs
f
2
= 1.4 MHz
Answer:
The bandwidth is 1.4 MHz.
16-32.
Given:
f
2
= 100 kHz
Solution:
f
2
= 0.35/
T
R
(Eq. 16-29)
T
R
= 0.35/
f
2
T
R
= 0.35/100 kHz
T
R
= 3.5 μs
Answer:
The risetime is 3.5 μs.
16-33.
Given:
C
= 1 μF
R
= 50
Ω
Solution:
R
in
=
Z
in(Stage)
in
1
2
||
||
β
e
R
R
R
r
′
=
R
in
= 717
Ω
f
C
1
= 1/(2
π
R
in
C
)
f
C
1
= 1/[2
π
(717
Ω
)(1 μF)]
f
C
1
= 222 Hz
Answer:
The lower cutoff frequency for the base
coupling circuit is 222 Hz.
16-34.
Given:
C
= 4.7 μF
R
C
= 3.6 k
Ω
R
L
= 10 k
Ω
Solution:
R
=
R
C
+
R
L
R
= 36 k
Ω
+ 10 k
Ω
R
= 13.6 k
Ω
f
C
1
= 1/(2
π
RC
)
f
C
1
= 1/[2
π
(13.6 k
Ω
)(4.7 μF)]
f
C
1
= 2.49 Hz
Answer:
The lower cutoff frequency for the collector
coupling circuit is 2.49 Hz.
16-35.
Given:
C
= 25 μF
R
C
= 3.6 k
Ω
R
L
= 10 k
Ω
Solution:
f
C
1
= 1/(2
π
Z
out
C
)
f
C
1
= 1/[2
π
(22.4
Ω
)(25 μF)]
f
C
1
= 284 Hz
Answer:
The lower cutoff frequency for the emitter
bypass circuit is 284 Hz.
16-36.
Given:
2 pF
c
C
′ =
10 pF
e
C
′ =
5 pF
Stray
C
′
=
R
1
= 10 k
Ω
R
2
= 2.2 k
Ω
R
C
= 3.6 k
Ω
R
L
= 10 k
Ω
R
G
= 50 k
Ω
β
= 200
Solution:
r
g
=
R
G
||
R
1
||
R
2
r
g
= 50
Ω
||10 k
Ω
||2.2 k
Ω
r
g
= 48
Ω
C
in(m)
= 236 pF
in(m)
246 pF
e
C
C
C
′
=
+
=
Base
f
2
= 1/(2
π
r
g
C
)
f
2
= 1/[2
π
(48
Ω
)(246 pF)]
f
2
= 13.5 MHz
Collector
Stray
7 pF
e
C
C
C
′
=
+
=
C
out(m)
= 2 pF
R
=
R
C
+
R
L
R
= 3.6 k
Ω
+ 10 k
Ω
R
= 13.6 k
Ω
f
2
= 1/(2
π
RC
)
f
2
= 1/[2
π
(13.6
Ω
)(1.3 pF)]
f
2
= 8.59 MHz
Answer:
The high cutoff frequency for the base is 13.5
MHz and the collector is 8.59 MHz.

1-82
16-37.
Given:
g
m
= 16.5 mS
C
iSS
= 30 pF
C
oSS
= 20 pF
C
rSS
= 5 pF
Solution:
C
gd
=
C
rSS
= 5 pF
C
gS
=
C
iSS
–
C
rSS
C
gS
= 30 pF – 5 pF
C
gS
= 25 pF
C
dS
=
C
oSS
–
C
rSS
C
dS
= 20 pF – 5 pF
C
dS
= 15 pF
Answer: C
gS
= 25 pF,
C
gd
=
C
rSS
= 5 pF,
C
dS
= 15 pF.
16-38.
Given:
R
1
= 2 M
Ω
R
2
= 1 M
Ω
R
D
= 1 k
Ω
R
L
= 10 k
Ω
R
G
= 50
Ω
C
in
= 0.01 μF
C
out
= 1 μF
Solution:
R
in
=
Z
in(Stage)
R
in
=
R
1
||
R
2
R
in
= 667 k
Ω
f
C
1
= 1/(2
π
R
in
C
)
f
C
1
= 1/[2
π
(667 k
Ω
)(0.01 μF)]
f
C
1
= 23.9 Hz or 14.5 Hz
Answer:
The dominant low cutoff frequency is 23.9 Hz.
16-39.
Given:
R
1
= 2 M
Ω
R
2
= 1 M
Ω
R
D
= 1 k
Ω
R
L
= 10 k
Ω
R
G
= 50 k
Ω
C
gd
= 5 pF
(from Prob. 16-37.)
C
gS
= 25 pF
(from Prob. 16-37.)
C
dS
= 015 pF
(from Prob. 16-37.)
Solution:
A
v
= g
m
r
d
A
v
= (16.5 mS)(1 k
Ω
||10 k
Ω
)
A
v
= 15
C
in(m)
=
C
gd
(A
v
+ 1)
(Eq. 16-40)
C
in(m)
= 5 pF(15 + 1)
C
in(m)
= 80 pF
C
=
C
gS
+
C
in(m)
C
= 25 pF + 80 pF
C
= 105 pF
R
=
R
G
||
R
1
||
R
2
R
= 50
Ω
||2 M
Ω
||1 M
Ω
R
= 50
Ω
Gate
f
2
= 1/(2
π
RC
)
f
2
= 1/[2
π
(50
Ω
)(105 pF)]
f
2
= 30.3 MHz
Collector
C
out(M)
=
C
gd
[(
A
v
+ 1)/
A
v
]
(Eq. 16-41)
C
out(M)
= 5 pF[(15 + 1)/15]
C
out(M)
= 5.3 pF
C
=
C
dS
+
C
out(m)
C
= 15 pF + 5.3 pF
Drain
f
2
= 1/(2
π
RC
)
f
2
= 1/[2
π
(909
Ω
)(20.3 pF)]
f
2
= 8.61 MHz
Answer:
The high frequency cutoff for the gate is 30.3
MHz and the drain is 8.61 MHz.

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- Resistor, Electrical resistance, kΩ