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Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 1/21
EERI 322 Problems : Neamen Chapter 7
7.2 SYSTEM TRANSFER FUNCTIONS Exercise 7.1.a For the circuit shown, answer the questions:
4PR kΩ=
SC =?
oV
iV
4SR kΩ=
Figure 7.2 i.) The corner frequency is 20f Hz= , determine the value of .SC From the corner frequency:
12
12
S
S
f
f
πτ
τπ
=
∴ =
The time constant is also given by
( )1 ( )
21 1 995
2 ( ) 2 (20)(8 )
S S P S
S P S
SS P
R R C
R R Cf
C nFf R R k
τ
π
π π
= +
∴ = +
∴ = = =+ Ω
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 2/21
ii.) Find the magnitude of the transfer function at 40f Hz= , 80Hz , and 200Hz Open-circuit time constant:
( ) (8 )(995 ) 7.96S S P SR R C k F msτ μ= + = Ω = Midband gain
4 0.58
P
P S
R kR R k
⎛ ⎞ Ω⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ Ω⎝ ⎠⎝ ⎠
The magnitude of the transfer function is
( )
( )( )
2
3
23
2( )1 2
2 (7.96 10 )0.51 2 (7.96 10 )
SP
S P S
fRT jfR R f
f
f
π τ
π τ
π
π
−
−
⎡ ⎤⎛ ⎞ ⎢ ⎥= ⎜ ⎟ ⎢ ⎥+⎝ ⎠ +⎣ ⎦⎡ ⎤
×⎢ ⎥= ⎢ ⎥+ ×⎢ ⎥⎣ ⎦
Evaluate at different frequencies:
( )( )
3
23
2 (40)(7.96 10 )( 40 0.5 0.4471 2 (40)(7.96 10 )
T j π
π
−
−
⎡ ⎤×⎢ ⎥= =⎢ ⎥
+ ×⎢ ⎥⎣ ⎦
( )( )
3
23
2 (80)(7.96 10 )( 80 0.5 0.4851 2 (80)(7.96 10 )
T j π
π
−
−
⎡ ⎤×⎢ ⎥= =⎢ ⎥
+ ×⎢ ⎥⎣ ⎦
( )( )
3
23
2 (200)(7.96 10 )( 200 0.5 0.4981 2 (200)(7.96 10 )
T j π
π
−
−
⎡ ⎤×⎢ ⎥= =⎢ ⎥
+ ×⎢ ⎥⎣ ⎦
Summary:
f 40 80 200 ( )T jf 0.447 0.485 0.498
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 3/21
Exercise 7.1.b For the circuit shown, if the corner frequency is 500f kHz= , determine the value of PC .
10PR kΩ=
oV
iV
10SR kΩ=
?PC =
Figure 7.3
From the corner frequency:
12
12
P
P
f
f
πτ
τπ
=
∴ =
The time constant is also given by
( || )1 ( || )
21 1 63.7
2 ( ) 2 (500 )(10 ||10 )
P S P S
S P S
SS P
R R C
R R Cf
C pFf R R kHz k k
τ
π
π π
=
∴ =
∴ = = =+
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 4/21
Exercise 7.2 For the circuit shown, the midband gain is 1dB− , and the corner frequencies are
100Lf Hz= and 1Hf MHz= .
?PR =
SC =?
oV
iV
1SR kΩ=
?PC =
Figure 7.10
a.) Determine PR , SC , and PC .
From the expression for the midband gain
1020 log 1
0.891
(1 0.891) 0.8918.2
P
P S
P
P S
P
P
RR R
RR R
RR k
⎛ ⎞= −⎜ ⎟+⎝ ⎠
∴ =+
∴ − =∴ = Ω
Lower cutoff-frequency 1
2 ( )1 1 173
2 ( ) 2 (100)(1 8.2 )
LS P S
SS P
fR R C
C nFf R R k k
π
π π
=+
∴ = = =+ +
Upper cutoff-frequency
6
12 ( || )
1 1 1792 ( || ) 2 (10 )(1 || 8.2 )
HS P P
PS P
fR R C
C pFf R R k k
π
π π
=
∴ = = =
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 5/21
b.) Determine the open-circuit and short-circuit time constants. Open-circuit time constant:
( )(1 8.2 )(173 )1.59
S S P SR R Ck k nFms
τ = += Ω+ Ω=
Closed-circuit time constant:
( || )(1 || 8.2 )(179 )160
P S P PR R Ck k pFns
τ == Ω Ω=
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 6/21
7.3 TRANSISTOR AMPLIFIERS WITH CIRCUIT CAPACITORS Exercise 7.3 For the following circuit:
iv
100SiR Ω= CC47μF
1R20kΩ
2R2.2kΩ
CR2kΩ
ER100Ω
CCV = 10V
OV
BE(on)
A
β= 200V =0.7V
V =∞
Figure 7.21
a.) Determine the expression for the time constant Sτ .
The small-signal equivalent circuit:
iv 1 2R || R= 1.98kΩ
rπ
OV
CR = 2kΩ
Vπ
bI
+
-
iR
mg Vπ
ER = 100Ω
100SiR Ω= CC 47μF=
iI
ibR
Small-signal equivalent
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 7/21
Open-circuit time constant: ( )S i Si CR R Cτ = +
b.) Determine the corner frequency and midband voltage gain.
First do a DC analysis 1 2
2
1 2
( )
|| 2.2 || 20 1.98
2.2 (10) 0.9912.2 20
0.991 0.7 13.2(1 ) 1.98 (201)(100)
(200)(13.2 ) 2.64
TH
TH CC
TH BE onBQ
TH E
CQ BQ
R R R k k k
RV V VR RV V
I AR R
I I A mA
μβ
β μ
= = Ω Ω = Ω
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠− −
= = =+ + +
= = =
Calculate small-signal parameters
(200)(0.026) 1.972.64
2.64 101.5 /0.026
T
CQ
CQm
T
Vr kI m
I mg mA VV
πβ
= = = Ω
= = =
Calculate resistances
1 2
(1 ) 1.97 (201)(100) 22.1|| 1.98|| 1.98 || 22.1 1.817
ib E
B
i B ib
R r R k kR R R kR R R k k k
π β= + + = + = Ω= = Ω= = Ω Ω = Ω
Calculate open-circuit time constant
( )(1.817 100 )(47 )90.1
S i Si CR R Ck F
ms
τμ
= += Ω+ Ω=
The corner frequency is given by
1 1 1.772 2 (90.1 )S
f Hzmsπτ π
= = =
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 8/21
Midband voltage gain
(max)( )
m C Bv
Si i B ib
g r R RAR R R R
π⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
But CQm
T
Ig
V= and T
CQ
VrIπβ
=
Therefore
( )( )3
(max)( )
(200)(2000) 1980(100 1817 (1980 22100)
208.65 82.22 10
17.2
C Bv
Si i B ib
R RAR R R Rβ
−
⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠
= − ×
= −
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 9/21
Exercise 7.4 The drain current is given as
2( )DQ n GS TNI K V V= − a.) Rearrange to determine GSV
2( )
( )
0.8 20.5
3.265
DQGS TN
n
DQGS TN
n
DQGS TN
n
IV V
K
IV V
K
IV V
K
V
= −
= −
= +
= +
=
The DC gate current is zero, therefore
3.265S GSQV V V= − = −
The source resistor ( 5) 5 1.735 2.17
0.8S S
DQ SS DQ
V VI R kR I− − +
= ⇒ = = = Ω
Given 0DQV =
The drain resistor is
5 6.250.8
DD DQD
DQ
V VR k
I−
= = = Ω
b.) The time constant is
3( ) (10 6.25) 10S D L C CR R C Cτ = + = + × ×
The corner frequency is 1 1
2 2 ( )S D L C
fR R Cπτ π
= =+
From this, the coupling capacitance can be calculated
3
1 1 4902 ( ) 2 (20)(16.25 10 )C
D L
C nFf R Rπ π
= = =+ ×
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 10/21
7.4 FREQUENCY RESPONSE: BIPOLAR TRANSISTOR Exercise 7.9 A bipolar transistor has parameters 0 150β = , 2C pFπ = , and 0.3C pFμ = , and is
biased at 0.5CQI mA= . Determine the beta cutoff frequency. Diffusion resistance
0
3
(150)(0.026)(0.5 10 )
7.8
T
CQ
VrI
k
πβ
−
=
=×
= Ω
Beta cutoff frequency
12
12 ( )
12 (7800)(2 0.3) 108.87
fr C C
MHz
βπ π μπ
π −
=+
=+ ×
=
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 11/21
Exercise 7.10 A BJT is biased at 1CI mA= , and its parameters are: 0 150β = , 4C pFπ = , and
0.5C pFμ = . Determine fβ and Tf . Diffusion resistance
0
3
(150)(0.026)(1 10 )
3.9
T
CQ
VrI
k
πβ
−
=
=×
= Ω
Beta cutoff frequency
12
12 ( )
12 (3900)(4 0.5) 109.07
fr C C
MHz
βπ π μπ
π −
=+
=+ ×
=
Cutoff frequency
0
6(150)(9.07 10 )1.36
Tf f
GHz
ββ=
= ×=
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 12/21
Exercise 7.11
DC Analysis Thevenin resistance
1 2|| 200 || 220 104.8THR R R k k k= = Ω Ω = Ω Thevenin voltage
2
1 2
220 (5) 2.62200 220TH CC
RV V VR R
⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠
Base current
2.62 0.7 9.316(1 ) 105 (101)(1)
TH BEBQ
TH E
V VI AR R
μβ
− −= = =
+ + +
Collector current
(100)(9.316 ) 931CQ BQI I A Aβ μ μ= = = Small-signal parameters Transconductance
931 35.83 /0.026
CQm
T
I Ag mA VV
μ= = =
Diffusion resistance
(100)(0.026) 2.79932
T
CQ
Vr kI Aπβ
μ= = = Ω
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 13/21
Miller capacitance [ ]
12 3
12 3
1 ( || )
(2 10 ) 1 (35.83 10 )(2200 || 4700)
(2 10 ) 1 (35.83 10 )(1498)
109
M m C LC C g R R
pF
μ
− −
− −
= +
⎡ ⎤= × + ×⎣ ⎦⎡ ⎤= × + ×⎣ ⎦
=
3dB frequency Base resistance
1 2|| || 100 || 200 || 220 51.17B SR r R R k k k k= = Ω Ω Ω = Ω Frequency
( )( )
( )( )
3
12
12 ||
12 51.17 || 2.79 10 109 10506
dBB
fR r C C
k kGHz
π π μπ
π −
=+
=Ω Ω + ×
=
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 14/21
7.5 FREQUENCY RESPONSE: THE FET Exercise 7.12 Small-signal parameters Transconductance
3 9
2
2 (0.2 10 )(400 10 )565.7 /
m n DQg K I
A Vμ
− −
=
= × ×
=
Unity-gain frequency (aka cutoff frequency)
6
12
2 ( )
565.7 102 (0.25 0.02) 10333
mT
gs gd
gfC C
MHz
π
π
−
−
=+
×=
+ ×=
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 15/21
Exercise 7.13 DC Analysis Gate voltage
( )
2
1 2
166 10166 234
4.15
G DDRV V
R R
V
⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞= ⎜ ⎟+⎝ ⎠
=
Drain current is given as
2( )D n GS TNI K V V= − KVL for input circuit 0G GS D SV V I R− − = Rearrange
G GSD
S
V VIR−
=
Combining the equations
2
2 2
3 2
2
2
2
( )
( 2 )
(0.5 10 )(500)( 4 4) 4.15
0.25 1 4.15
0.25 3.15
12.63.55
G GSn GS TN
S
n S GS TN GS TN G GS
GS GS GS
GS GS GS
GS
GS
GS
V VK V VR
K R V V V V V V
V V V
V V V
V
VV V
−
−− =
− + = −
× − + = −
− + = −
=
=
=
Transconductance
3
2 ( )
2(0.5 10 )(3.55 2)1.55 /
m n GS TNg K V V
mA V
−
= −
= × −=
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 16/21
Small-signal equivalent circuit
Miller capacitance
( )( )( ) ( )( )( )
1 ||
0.1 1 1.55 4 || 20
0.617
M gd m D LC C g R R
p m k k
pF
= +
= +
=
Input resistance
1 2||234 ||16697.1
GR R Rk kk
== Ω Ω= Ω
Time constant
( )( )( )( )
||
97.1 ||10 1.61714.7
P G i gs MR R C C
k k pns
τ = +
=
=
3dB frequency of small-signal voltage gain
( )
12
12 14.710.9
HP
f
nsMHz
πτ
π
=
=
=
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 17/21
7.6 HIGH-FREQUENCY RESPONSE OF TRANSISTOR CIRCUITS Exercise 7.14
DC Analysis Base voltage
0THV V= Input resistance
1 2||20 || 2010
THR R Rk kk
== Ω Ω= Ω
Base current
(1 )0 0.7 ( 5)
10 (126)(5 )6.72
TH BE EEBQ
TH E
V V VIR R
k kA
β
μ
− −=
+ +− − −
=+
=
Collector current
(125)(6.72 )840
CQ BQI I
AA
β
μμ
=
==
Small-signal parameters Diffusion resistance
(125)(0.026) 3.87
840T
CQ
Vr kIπβ
μ= = = Ω
Transconductance
840 32.3 /0.026
CQm
T
I Ag mA VV
μ= = =
Output resistance
200 238840
Ao
CQ
Vr kI μ
= = = Ω
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 18/21
High-frequency equivalent circuit
Output resistance
' || ||238 || 2.3 || 51.565
L o C LR r R Rk k kk
=== Ω
Miller Capacitance
( )( )( )
1 '
3 1 (32.3 )(1.565 )155
M m LC C g R
p m kpF
μ= +
= +
=
Input resistance
1 2||20 || 2010
BR R Rk kk
=== Ω
Equivalent resistance seen by capacitors
|| ||
1 ||10 || 3.87909 || 3.87736
eq S BR R R r
k k kk
π=
=== Ω
Time constant
( )
(736)(24 155 )131.7
P eq MR C C
p pns
πτ = +
= +=
Upper 3dB frequency
( )1 1 1.21
2 2 131.7HP
f MHznsπτ π
= = =
Midband voltage gain
( ) ( ) ( )|| 10 || 3.87' 32.3 (1.565 ) 37.2
|| 10 || 3.87 1B
v m LMB S
R r k kA g R m kR r R k k
π
π
⎡ ⎤⎡ ⎤= − = − = −⎢ ⎥⎢ ⎥+ +⎣ ⎦ ⎣ ⎦
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 19/21
Exercise 7.15
DC Analysis Base current
10 0.7 9.3 8.37(1 ) 100 (101)(10 )BQ
B E
I AR R k k
μβ
−= = =
+ + +
Collector current
(100)(8.37 ) 837CQ BQI I mA Aβ μ= = = Small-signal parameters Diffusion resistance
(100)(0.026) 3.10
837T
CQ
Vr kIπβ
μ= = = Ω
Transconductance
837 32.19 /0.026
CQm
T
Ig mA V
Vμ
= = =
For the input circuit, the time constant is
[ ]( )|| || 30.7 ||10 ||1 24 7131P E Sr R R C k k p psπ
π πτβ
⎡ ⎤⎛ ⎞= = =⎢ ⎥⎜ ⎟+⎝ ⎠⎣ ⎦
Upper 3dB frequency
1 1 223
2 2 (713 )HP
f MHzpπ
ππτ π= = =
For the output circuit, the time constant is [ ] [ ]( )|| 10 ||1 3 2.73P C LR R C k k p nsμ μτ = = = Upper 3dB frequency
1 1 58.42 2 (2.73 )H
P
f MHznπ
μπτ π= = =
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 20/21
Midband voltage gain
( ) ( )
3.1|| 10 ||1 101( || ) 32.22 (10 ||1 ) 0.87
3.110 || 1||1011
E
v m C LM
E S
r kR kA g R R m k k
kr kR R
π
π
β
β
⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥+⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥= = =⎢ ⎥⎛ ⎞ ⎛ ⎞⎢ ⎥++ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎢ ⎥+ ⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦
Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems
EERI322, 2008 (© CJVDM) 21/21
Exercise 7.16 DC Analysis Base voltage (Q1)
31
1 2 3
7.92( ) (12) 0.950258.8 33.3 7.92B
R kV V VR R R k k k
+⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ + + +⎝ ⎠⎝ ⎠
Neglect base currents
1 1 0.9502 0.7 500500
B BEC
E
V VI AR
μ− −= = =
Small-signal parameters Diffusion resistance
(100)(0.026) 5.2
500T
CQ
Vr kI mπβ
= = = Ω
Transconductance
500 19.23 /0.026
Cm
T
Ig mA VV
μ= = =
Miller capacitance
1 12 6MC C pFμ= = Base resistance 1 2 3|| 33.3 || 7.92 6.398BR R R k k k= = = Ω Time constant for input portion of circuit [ ][ ] [ ][ ]1 1 1|| || 1 || 6.398 || 5.2 24 6 22.24P S B MR R r C C k k k p p nsπ π πτ = + = + = 3dB frequency for input portion of circuit
1 1 7.152 2 (22.24 )H
P
f MHznsπ
ππτ π= = =
Time constant for output portion of circuit [ ] [ ]2|| 7.5 || 2 (3 ) 4.737P C LR R r C k k p nsμ π μτ = = = 3dB frequency for output portion of circuit
1 1 33.62 2 (4.737 )H
P
f MHznsμ
μπτ π= = =
Midband voltage gain
( ) ( ) ( )( ) ( )1 1
21 1
|| 6.4 || 5.2|| 19.23 7.5 || 2 22.5|| 6.4 || 5.2 1B
v m C LMB S
R r k kA g R R m k kR r R k k
π
π
⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦
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