1. Chapter 7
Exercise Solutions
EX7.1
(a)
(i)
RS = RP = 4 kΩ
1 1
ω= =
rS ( RS + RP ) CS
1 1
CS = =
2π f ( RS + RP ) 2π ( 20 )( 4 + 4 ) × 10 3
CS = 0.995 μ F
(ii)
⎛ RP ⎞ ω rS
T ( jω ) = ⎜ ⎟
⎝ RS + RP ⎠ 1 + ω 2 rS2
rS = ( RS + RP ) CS = 7.96 × 10 −3
RP 4
= = 0.5
RS + RP 4 + 4
f = 40 Hz
( 0.5)( 2π )( 40 ) ( 7.96 × 10 −3 )
T ( jω ) =
( )
2
1 + ⎡ 2π ( 40 ) 7.96 × 10 −3 ⎤
⎣ ⎦
T ( jω ) = 0.447
f = 80 Hz
( 0.5)( 2π )(80 ) ( 7.96 × 10 −3 )
T ( jω ) =
( )
2
1 + ⎡ 2π ( 80 ) 7.96 × 10 −3 ⎤
⎣ ⎦
T ( jω ) = 0.485
f = 200 Hz
( 0.5 )( 2π )( 200 ) ( 7.96 × 10 −3 )
T ( jω ) =
( )
2
1 + ⎡ 2π ( 200 ) 7.96 × 10 −3 ⎤
⎣ ⎦
T ( jω ) = 0.498
(b)
1 1
ω= =
rP ( RS RP ) CP
1
CP =
2π f ( RS RP )
1
=
(
2π 500 × 10 3
) (10 10 ) × 10 3
CP = 63.7 pF
EX7.2
a.

2. ⎛ RP ⎞ RP RP
20 log10 ⎜ ⎟ = −1 ⇒ = 0.891 = ⇒ (1 − 0.891) RP = 0.891 ⇒ RP = 8.20 kΩ
⎝ RP + RS ⎠ RP + RS RP + 1
1 1
fL = ⇒ CS =
2π ( RS + RP ) CS 2π (100 )(1 + 8.20 ) × 10 3
CS = 0.173 μ F
1 1
fH = ⇒ CP =
2π ( RS RP ) CP 2π 10( ) (1 || 8.20 ) × 10
6 3
CP = 179 pF
b.
rS = ( RS + RP ) CS
( )(
rS = 1 × 10 3 + 8.20 × 10 3 0.173 × 10 −6 )
rS = 1.59 ms open-circuit time-constant
rP = ( RS RP ) CP
rP = (1 8.20 ) × 10 3 179 × 10 −12( )
rP = 0.160 μ s short-circuit time-constant
EX7.3
a. rS = ( Ri + RS i ) CC
b.
1
f =
2π rS
RTH = R1 R2 = 2.2 20 = 1.98 kΩ
⎛ R2 ⎞ ⎛ 2.2 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 0.991 V
⎝ R1 + R2 ⎠ ⎝ 2.2 + 20 ⎠
VTH − VBE ( on ) 0.991 − 0.7
I BQ = = = 0.0132 mA
RTH + (1 + β ) RE 1.98 + ( 201)( 0.1)
I CQ = 2.636 mA
β VT ( 200 )( 0.026 )
rπ = = = 1.97 kΩ
I CQ 2.636
ICQ 2.636
gm = = = 101.4 mA/V
VT 0.026
Rib = τ π + (1 + β ) RE = 1.97 + ( 201)( 0.1) = 22.1 kΩ
RB = R1 R2 = 1.98 kΩ
Ri = RB Rib = 1.98 22.1 = 1.817 kΩ
rS = ( Ri + RSi ) CC
= (1.817 + 0.1) ×10 3 ( )( 47 × 10 )−6
= 90.1 ms
1
f = ⇒ f = 1.77 Hz
(
2π 90.1 × 10 −3 )
Midband Gain

3. − β RC Ri
Aν = ⋅
rπ + (1 + β ) RE Ri + RSi
− ( 200 )( 2 ) 1.82
= ⋅
1.97 + ( 201)( 0.1) 1.82 + 0.1
Aν = −17.2
EX7.4
I DQ = K n (VGS − VTN )
2
a.
0.8
+ 2 = VGS ⇒ VGS = 3.265 V
0.5
VS − ( −5 )
VS = −3.265 ⇒ I DQ =
RS
−3.265 + 5
RS = ⇒ RS = 2.17 kΩ
0.8
5
VD = 0 ⇒ RD = ⇒ RD = 6.25 kΩ
0.8
b.
rS = ( RD + RL ) CC = (10 + 6.25 ) × 10 3 × CC
1 1
f = ⇒ CC =
2π rS (
2π f 16.25 × 10 3 )
1
CC = ⇒ CC = 0.49 μ F
(
2π ( 20 ) 16.25 × 10 3 )
EX7.5
rS = ( RL + Ro ) CC 2
1 1
f = ⇒ CC 2 =
2π rS 2π f ( RL + Ro )
⎧ rπ + ( RS RB ) ⎫
⎪ ⎪
R0 = RE r0 ⎨ ⎬
⎪
⎩ 1+ β ⎪
⎭
From Example 7-5, R0 = 35.5 Ω
1
CC 2 =
2π (10 ) ⎡10 × 10 3 + 35.5⎤
⎣ ⎦
CC 2 = 1.59 μ F
EX7.6
I DQ = K P (VSG + VTP )
2
a.
1
− ( −2 ) = VSG ⇒ VSG = 3.41 V
0.5
VS = 3.41
5 − 3.41
RS = ⇒ RS = 1.59 kΩ
1
5
For VSDG = VSGQ ⇒ VD = 0 ⇒ RD = ⇒ RD = 5 kΩ
1

4. b. rP = ( RD RL ) CL
1 1
f = ⇒ CL =
2π rP 2π f ( RD RL )
1
CL = ⇒ CL = 47.7 pF
( )
2π 10 6 ( 5 10 ) × 10 3
EX7.7
(a)
RTH = 5 K
VTH = −3.7527
−3.7527 − 0.7 − ( −5) 0.54726
I BQ = =
5 + (101)( 0.5) 55.5
= 0.00986
I CQ = 0.986 mA
gm = 37.925 rπ = 2.637 K
5 − Vo Vo
0.986 = − = 1 − Vo ( 0.4 )
5 5
Vo = 0.035
(b)
Rib
RS ϭ 0.1 k⍀
V0
ϩ
Vi ϩ RTH ϭ RC ͉͉RL ϭ
Ϫ V␲ r␲
5 k⍀ gmV␲ 2.5 k⍀
Ϫ
RE ϭ
0.5 k⍀
Vo = − gmVπ ( RC RL )
Rib = rπ + (1 + β ) RE = 2.64 + (101)( 0.5) = 53.14 K
Vb Vb Vb
Vπ = = =
⎛1+ β ⎞ ⎛ 101 ⎞ 14.885
1+ ⎜ ⎟ RE 1 + ⎜ 2.637 ⎟ ( 0.5)
⎝ rπ ⎠ ⎝ ⎠
⎛ RTH Rib ⎞ ⎛ 5 53.14 ⎞
Vb = ⎜ ⎟ Vi = ⎜ ⎟ Vi
⎝ RTH Rib + RS ⎠ ⎝ 5 53.14 + 0.1 ⎠
⎛ 4.57 ⎞
=⎜ ⎟ Vi = 0.9786
⎝ 4.57 + 0.1 ⎠
(37.925)( 2.5)
Av = − ( 0.9786 ) = Av = −6.23
14.885
(c)
EX7.8
a.

5. 0 − 0.7 − ( −10 )
I BQ = = 0.0230 mA
0.5 + (101)( 4 )
I CQ = 2.30 mA
β VT (100 )( 0.026 )
rπ = = = 1.13 kΩ
I CQ 2.30
I CQ 2.30
gm = = = 88.46 mA / V
VT 0.026
RE ( RS + rπ ) CE
rB =
RS + rπ + (1 + β ) RE
=
( 4 × 10 ) ( 0.5 + 1.13) C
3
E
0.5 + 1.13 + (101)( 4 )
1 1
rB = = = 0.7958 ms
2π f B 2π ( 200 )
0.796 × 10 −3
rB = 16.07CE ⇒ CE = ⇒ CE = 49.5 μ F
16.07
b.
rA = RE CE = 4 × 10 3 ( )( 49.5 × 10 ) ⇒ r
−6
A = 0.198 s
1 1
fA = = ⇒ f A = 0.804 H Z
2π rA 2π ( 0.198 )
EX7.9
β 0VT (150 )( 0.026 )
rπ = = = 7.8 kΩ
I CQ 0.5
1
fβ =
2π rπ ( Cπ + Cμ )
1
= ⇒ f β = 8.87 MH Z
(
2π 7.8 × 10 3
) ( 2 + 0.3) × 10 −12
EX7.10
β 0VT (150)(0.026)
rπ = = ⇒ rπ = 3.9 kΩ
I CQ 1
1
fβ =
2π rπ ( Cπ + Cμ )
1
= ⇒ f β = 9.07 MHz
(
2π 3.9 × 10 3
) ( 4 + 0.5) (10 )
−12
fT = β 0 f β = (150 )( 9.07 ) ⇒ fT = 1.36 GHz
EX7.11
RTH = R1 R2 = 200 220 = 104.8 kΩ
⎛ R2 ⎞ ⎛ 220 ⎞
VTH = ⎜ ⎟ VCC = ⎜ ⎟ (5) = 2.619 V
⎝ R1 + R2 ⎠ ⎝ 200 + 220 ⎠
2.62 − 0.7
I BQ = = 0.009316 mA
105 + (101)(1)
I CQ = 0.9316 mA

6. I CQ 0.9316
gm = = ⇒ gm = 35.83 mA/V
VT 0.026
β VT (100)(0.026)
rπ = = ⇒ rπ = 2.79 kΩ
I CQ 0.932
a.
C M = Cμ ⎡1 + gm ( RC RL ) ⎤
⎣ ⎦
C M = ( 2 ) ⎡1 + ( 35.83 )( 2.2 4.7 ) ⎤ ⇒ C M = 109 pF
⎣ ⎦
b.
RB = rS R1 R2 = 100 200 220 = 51.17 kΩ
1
f3 dB =
2π ( RB rπ ) (Cπ + Cμ )
1
= ⇒ f3dB = 0.506 MHz
2π [ 51.17 2.79] × 10 3 × (10 + 109 ) × 10 −12
EX7.12
gm
fT =
2π ( Cgs + Cgd )
gm = 2 K n I DQ
=2 ( 0.2 )( 0.4 )
= 0.5657 mA/V
0.5657 × 10 −3
fT = ⇒ fT = 333 MHz
2π ( 0.25 + 0.02 ) × 10 −12
EX7.13
dc analysis
⎛ R2 ⎞ ⎛ 166 ⎞
VG = ⎜ ⎟ VDD = ⎜
R1 + R2 ⎠ ⎟ (10 ) = 4.15 V
⎝ ⎝ 166 + 234 ⎠
V
I D = S and VS = VG − VGS
RS
VG − VGS
K n (VGS − VTN ) =
2
RS
( 0.5)( 0.5) (VGS − 4VGS + 4 ) = 4.15 − VGS
2
0.25VGS − 3.15 = 0 ⇒ VGS = 3.55 V
2
gm = 2K n (VGS − VTN ) = 2 ( 0.5 )( 3.55 − 2 )
= 1.55 mA/V
Small-signal equivalent circuit.
Ri ϭ 10 k⍀
V0
ϩ Cgd
Cgs
RG ϭ
Vi ϩ Vgs RD RL
Ϫ R1͉͉R2
gmVgs
Ϫ
a. C M = Cgd (1 + gm ( RD RL ) )
C M = ( 0.1) ⎡1 + (1.55) ( 4 20 )⎤ ⇒ C M = 0.617 pF
⎣ ⎦
b.

7. 1
fH =
2πτ P
τ P = ( RG Ri ) ( Cgs + C M )
RG = R1 R2 = 234 166 = 97.1 kΩ
RG Ri = 97.1 10 = 9.07 kΩ
( )
rP = 9.07 × 10 3 (1 + 0.617 ) × 10 −12 = 14.7 ns
1
fH = ⇒ f H = 10.9 MHz
(
2π 14.7 × 10 −9 )
EX7.14
dc analysis
VTH = 0, RTH = 10 kΩ
0 − 0.7 − ( −5 )
I BQ = = 0.00672 mA
10 + (126 )( 5 )
I CQ = 0.840 mA
β VT (125)( 0.026 )
rπ = = = 3.87 kΩ
I CQ 0.840
I CQ 0.840
gm = = = 32.3 mA/V
VT 0.026
VA 200
r0 = = = 238 kΩ
I CQ 0.84
High-frequency equivalent circuit
C␮
RS ϭ 1 k⍀
V0
ϩ
ϩ RB ϭ
Vi V r␲ r0 RC RL
Ϫ R1͉͉R2 ␲ C␲ gmV␲
Ϫ
a.
Miller Capacitance
C M = Cμ (1 + gm RL )
′
′
RL = r0 RC RL
′
RL = 238 2.3 5 = 1.565 kΩ
C M = ( 3 ) ⎡1 + ( 32.3 )(1.57 ) ⎤ ⇒ Cμ = 155 pF
⎣ ⎦
b.
Req = RS RB rπ = RS R1 R2 rπ
Req = 1 20 20 3.87 = 0.736 kΩ
rP = Re q ( Cπ + C M )
( )
= 0.736 × 10 3 ( 24 + 155 ) × 10 −12
−7
= 1.314 × 10
1
fH = ⇒ f H = 1.21 MHz
(
2π 1.314 × 10 −7 )
c.

8. ⎡ RB τ π ⎤
( Av )M ′
= − gm RL ⎢ ⎥
⎢ RB τ π + RS ⎥
⎣ ⎦
⎡ 10 3.87 ⎤
( Av )M = − ( 32.3 )(1.565 ) ⎢ ⎥ ⇒ ( Av ) M = −37.2
⎣ 10 3.87 + 1 ⎦
EX7.15
The dc analysis
10 − 0.7
I BQ = = 0.00838 mA
100 + (101)(10 )
I CQ = 0.838 mA
β VT (100 )( 0.026 )
rπ = = = 3.10 kΩ
I CQ 0.838
I CQ
gm = = 32.22 mA/V
VT
For the input
⎡⎛ r ⎞ ⎤
rPπ = ⎢⎜ π ⎟ RE RS ⎥ Cπ
⎣⎝ 1 + β ⎠ ⎦
⎡ 3.10 ⎤
=⎢ 10 1⎥ × 10 3 × 24 × 10 −12
⎣ 101 ⎦
= 7.13 × 10 −10 s
1 1
f Hπ = = ⇒ f Hπ = 223 MHz
(
2π rPπ 2π 7.13 × 10 −10 )
For the output
rP μ = [ RC RL ] Cμ = (10 1) × 10 3 × 3 × 10 −12
= 2.73 × 10 −9
1 1
fHμ = = ⇒ f H μ = 58.4 MHz
(
2π rP μ 2π 2.73 × 10 −9 )
⎡ ⎛ rπ ⎞ ⎤
⎢ RE ⎜ ⎟ ⎥
RL ) ⎢ ⎝1+ β ⎠ ⎥
( Aν )M = gm ( RC
⎢ ⎥
⎛ rπ ⎞
⎢ RE ⎜ ⎟ + RS ⎥
⎢
⎣ ⎝1+ β ⎠ ⎥
⎦
⎡ ⎛ 3.1 ⎞ ⎤
⎢ 10 ⎜ 101 ⎟ ⎥
= ( 32.22 )(10 1) ⎢ ⎝ ⎠ ⎥⇒ A
( ν )M = 0.870
⎢ ⎛ 3.1 ⎞ ⎥
⎢ 10 ⎜ 101 ⎟ + 1 ⎥
⎝ ⎠ ⎦
⎣
EX7.16
⎛ R3 ⎞ ⎛ 7.92 ⎞
VB1 = ⎜ ⎟ (12 ) = ⎜ (12 ) = 0.9502 V
⎝ R1 + R2 + R3 ⎠ ⎝ 58.8 + 33.3 + 7.92 ⎟
⎠
Neglecting lose currents
0.9502 − 0.7
IC = = 0.50 mA
0.5
β VT (100 )( 0.026 )
rπ = = = 5.2 K
IC 0.5
IC 0.5
gm = = = 19.23 mA/V
VT 0.026
From Eq (7.127(a)),

9. τ Pπ = [ RS RB1 rπ ][Cπ 1 + C M 1 ]
RB1 = R2 R3 = 33.3 7.92 = 6.398 K
C M 1 = 2Cμ 1 = 6 pF
Then
τ Pπ = [1 6.398 5.2] × 10 × [24 + 6] × 10 ⇒ 22.24 ns
3 −12
1 1
f Hπ = = ⇒ 7.15 MHz
2πτ Pπ 2π ( 22.24 × 10 −9 )
From Eq (7.128(a))
τ P μ = [ RC RL ] C μ 2 = ( 7.5 2 ) × 10 × 3 × 10 ⇒ 4.737 ns
3 −12
1 1
fHμ = = = 33.6 MHz
2πτ Pμ 2π ( 4.737 × 10 −9 )
From Eq. 7.133
⎡ RB1 rπ 1 ⎤
Av = gm 2 ( RC RC ) ⎢ ⎥
⎣ RB1 rπ 1 + RS ⎦
M
⎡ 6.40 5.2 ⎤
= (19.23)( 7.5 2 ) ⎢ ⎥
⎣ 6.40 5.2 + 1 ⎦
⎡ 2.869 ⎤
= (19.23)(1.579 ) ⎢
⎣ 2.869 + 1 ⎥
⎦
Av M
= 22.5
TYU7.1
a.
V0 = − ( gmVπ ) RL
rπ
Vπ = × Vi
1
rπ + + RS
sCC
V0 ( s ) − gm rπ RL
T (s) = =
Vi ( s ) rπ + RS + (1 / sCC )
− gm rπ RL ( sCC )
=
1 + s ( rπ + RS ) CC
gm rπ = β
− β RL ⎛ s ( rπ + RS ) CC ⎞
T (s) = ×⎜ ⎟
rπ + RS ⎜ 1 + s (r + R ) C ⎟
⎝ π S C ⎠
Then τ = ( rπ + RS ) CC
b.
1
f3− dB =
2π ( rπ + RS ) CC
1
f3− dB = ⇒ f3 dB = 53.1 Hz
2π ⎡ 2 × 10 + 1 × 10 3 ⎤ ⎡10 −6 ⎤
⎣
3
⎦⎣ ⎦
rg R ( 2 )( 50 )( 4 )
T ( jω ) max = π m L =
rπ + RS 2 +1
T ( jω ) max = 133
c.

10. ͉T( j␻)͉
133
53.1 Hz f
TYU7.2
a.
⎛ 1 ⎞
V0 = − g mVπ ⎜ RL ⎟
⎝ sCL ⎠
⎛ r ⎞
Vπ = ⎜ π ⎟ × Vi
⎝ rπ + RS ⎠
⎛ 1 ⎞
V0 ( s ) rπ ⎜ RL × sC ⎟
T (s) = = − gm ⎜ L ⎟
Vi ( s ) rπ + RS ⎜ 1 ⎟
⎜ RL + sC ⎟
⎝ L ⎠
− β RL ⎛ 1 ⎞
T (s) = ×⎜ ⎟
rπ + RS ⎝ 1 + sRL CL ⎠
Then τ = RL CL
b.
1 1
f3− dB = = ⇒ f3dB = 3.18 MH Z
( )(
2π RL CL 2π 5 × 10 10 × 10 −12
3
)
gm rπ RL ( 75)(1.5)( 5)
T ( jω ) = =
rπ + RS 1.5 + 0.5
T ( jω ) max = 281
c.
281
͉T( j␻)͉
f 3.18 MHz
TYU7.3
a. Open-circuit time constant ( CL → open )
rS = ( RS + rπ ) CC
( )
= ( 0.25 + 2 ) × 10 3 2 × 10 −6 = 4.5 ms
Short-circuit time constant ( CC → short )
( )(
rP = RL CL = 4 × 10 3 50 × 10 −12 )
rP = 0.2 μ s
b. Midband gain

11. ⎛ rπ ⎞
V0 = − gmVπ RL , Vπ = ⎜ ⎟ Vi
⎝ τ π + RS ⎠
V −g r R
Av = 0 = m π L
Vi rπ + RS
− ( 65 )( 2 )( 4 )
=
2 + 0.25
Av = −231
c.
1 1
fL = = ⇒ f L = 35.4 Hz
(
2π rS 2π 4.5 × 10 −3 )
1 1
fH = = ⇒ f H = 0.796 MHz
(
2π rP 2π 0.2 × 10 −6 )
TYU7.4 Computer Analysis
TYU7.5 Computer Analysis
TYU7.6
βV (100 )( 0.026 )
rπ = 0 T = = 10.4 kΩ
I CQ 0.25
1
fβ =
2π rπ ( Cπ + C μ )
1 1
Cπ + Cμ = =
( )(
2π f β rπ 2π 11.5 × 10 6 10.4 × 10 3 )
Cπ + Cμ = 1.33 pF
Cμ = 0.1 pF ⇒ Cπ = 1.23 pF
TYU7.7
β0
h fe =
1 + j ( f / fβ )
f β = 5 MH Z , β 0 = 100
At f = 50 MH Z
100
h fe = ⇒ h fe = 9.95
2
⎛ 50 ⎞
1+ ⎜ ⎟
⎝ 5 ⎠
⎛ 50 ⎞
Phase = − tan −1 ⎜ ⎟ ⇒ Phase = −84.3°
⎝ 5 ⎠
TYU7.8
f 500
fβ = T = ⇒ f β = 4.17 MHz
β 0 120
1
fβ =
2π rπ (Cπ + C μ )
1 1
Cπ + Cμ = =
2π f β rπ 2π (4.167 × 10 6 )(5 × 10 3 )
Cπ + Cμ = 7.639 pF
Cμ = 0.2 pF ⇒ Cπ = 7.44 pF

12. TYU7.9
(a)
gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 3 − 1) ⇒ gm = 1.6 mA/V
′
gm = 80% of gm = 1.28 mA/V
gm
′
gm =
1 + gm rS
gm
1 + gm rS =
′
gm
⎛ gm
1 ⎞ 1 ⎛ 1.6 ⎞
rS = ⎜ − 1⎟ = ⎜ − 1⎟
⎝
gm ′
gm ⎠ 1.6 ⎝ 1.28 ⎠
rS = 0.156 kΩ ⇒ rS = 156 ohms
(b)
gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 5 − 1) ⇒ gm = 3.2 mA/V
gm 3.2
′
gm = = = 2.134
1 + gm rS 1 + ( 3.2 )( 0.156 )
Δgm 3.2 − 2.134
= ⇒ A 33.3% reduction
gm 3.2
TYU7.10
gm
fT =
2π ( CgsT + C gdT )
gm
=
2π ( Cgs + Cgsp + Cgdp )
gm
Cgs = − Cgsp − C gdp
2π fT
0.5 × 10 −3
= − ( 0.01 + 0.01) × 10 −12 ⇒ Cgs = 0.139 pF
(
2π 500 × 10 6
)
TYU7.11
gm
fT =
2π (Cgs + Cgsp + Cgdp )
Cgsp = Cgdp
gm 1× 10−3
2Cgsp = − C gs = − 0.4 × 10−12
2π fT 2π (350 × 106 )
2Cgsp = 0.0547 pF ⇒ Cgsp = Cgdp ≅ 0.0274 pF
TYU7.12
dc analysis
⎛ 50 ⎞
VG = ⎜ ⎟ (10 ) − 5 = −2.5
⎝ 50 + 150 ⎠
VS − ( −5 )
VS = VG − VGS . I D =
RS
VG − VGS + 5
K n (VGS − VTN ) =
2
RS

13. (1)( 2 ) ⎡VGS − 1.6VGS + 0.64 ⎤ = −2.5 − VGS + 5
⎣
2
⎦
2VGS − 2.2VGS − 1.22 = 0
2
( 2.2 ) + 4 ( 2 )(1.22 )
2
2.2 ±
VGS = ⇒ VGS = 1.505 V
2 (2)
gm = 2 K n (VGS − VTN ) = 2 (1)(1.505 − 0.8 )
= 1.41 mA/V
Equivalent circuit
Ri Cgd
V0
ϩ
ϩ RG ϭ
Vi Vgs
Ϫ R1͉͉R2 Cgs
RD
gmVgs
Ϫ
(a) C M = Cgd (1 + gm RD ) = ( 0.2 ) ⎡1 + (1.42 )( 5 ) ⎤ ⇒ C M = 1.61 pF
⎣ ⎦
(b)
τ P = ( Ri RG ) ( Cgs + CM )
rP = [ 20 50 150 ] × 10 3 × ( 2 + 1.61) × 10 −12
( )( )
= 13 × 10 3 3.62 × 10 −12 = 4.71 × 10 −8 s
1 1
fH = = ⇒ f H = 3.38 MHz
(
2π rP 2π 4.71 × 10 −8 )
⎛ RG ⎞
c. ( Av )M = − gm RD ⎜ ⎟
⎝ RG + RS ⎠
⎛ 37.5 ⎞
( Av )M = − (1.41)( 5 ) ⎜ ⎟ ⇒ ( Av ) M = −4.60
⎝ 37.5 + 20 ⎠
TYU7.13 Computer Analysis