Chapter 7 Solutions

Neamen (3rd Ed), Chapter 7 (Frequency Response) Problems

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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


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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


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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

=

∴ =

∴ = = =+


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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

=

∴ = = =


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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=


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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


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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

= = =


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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

−

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

= − ×

= −


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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

= = =+ ×


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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 −

=+

=+ ×

=


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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=

= ×=


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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


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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 + ×

=


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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

−

−

=+

×=

+ ×=


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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

−

= −

= × −=


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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

=

=

=


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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


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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

= = =


( ) ( ) ( )|| 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

⎡ ⎤⎡ ⎤= − = − = −⎢ ⎥⎢ ⎥+ +⎣ ⎦ ⎣ ⎦


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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= = =


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( ) ( )

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

⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥+⎝ ⎠ ⎝ ⎠⎢ ⎥ ⎢ ⎥= = =⎢ ⎥⎛ ⎞ ⎛ ⎞⎢ ⎥++ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎢ ⎥+ ⎝ ⎠⎣ ⎦⎝ ⎠⎣ ⎦


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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= = =


( ) ( ) ( )( ) ( )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

⎡ ⎤ ⎡ ⎤= = =⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

Chapter 7 Solutions

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Transcript of Chapter 7 Solutions