Chapter 12 Problem Solutions 12.1 (a)

5

5

5 5

15 10120

1 (5 10 )120 (120)(5 10 ) 5 10

0.008331

fAAAβ

ββ

β

=+

×=+ ×

+ × = ×=

(b) 3

3

3 3

5 101201 (5 10 )

120 (120)(5 10 ) 5 100.008133

ββ

β

×=+ ×

+ × = ×=

12.2 (a) 0.151f

AAA

ββ

= =+

T Aβ= (i) T = ∞ (ii) 4 380 dB 10 1.5 10A A T= ⇒ = ⇒ = × (iii) 15T =

(i) 1 6.667fAβ

= =

(ii) 6.662fA = (iii) 6.25fA = (b) (i) T = ∞ (ii) 32.5 10T = × (iii) 25T =

(i) 1 4.00fAβ

= =

(ii) 3.9984fA = (iii) 3.846fA = 12.3 (a)

1 1251f

AAAβ β

= ≅ =+

0.0080β = (b)

[ ][ ]

(125)(0.9975) 124.6875

124.68751 (0.008)

124.6875 1 (0.008) A124.6875 1 0.9975

49,875

fA

AA

AA

A

= =

=+

+ =

= −=

12.4

(a) 4

4

3

101001 (10 )

9.9 10β

β −

=+

= ×

(b)

4

100 ( 0.10) 0.001 0.10%10

f f

f

f

f

dA A dAA A A

dAA

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= − = − ⇒ = −⎜ ⎟⎝ ⎠

12.5

4

4

4

3

0.001 ( 0.10)5 10

500

5 105001 (5 10 )

1.98 10

f f

f

f

f

dA A dAA A A

A

A

ββ −

⎛ ⎞= ⎜ ⎟⎝ ⎠

− = −×

=

×=+ ×

= ×

12.6 (a) For Fig. P12.6(a)

1

1 1 1

200 101 200 1 10

o o

i o

v vv vβ β

= =+ +

1 1

200 10 401 200 1 10fA

β β⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠

21 1 1 150 (1 200 )(1 10 ) 1 210 2000β β β β= + + = + +

21 12000 210 49 0β β+ − =

1

1

210 44100 4(2000)(49)2(2000)

0.1126

β

β

− ± +=

=

For Fig P12.6(b)

2

2

(200)(10)401 (200)(10)0.0245

ββ

=+

=

Fig. P12.6(a) (b)

1 180180 10 (8.4634)(4.704)

1 (180)(0.1126) 1 (10)(0.1126)39.81

39.81 40 0.475%40

f

f

f

f

A

A

AdAA

=

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟+ +⎝ ⎠⎝ ⎠=

−= ⇒ −

Fig. P12.6(b) (180)(10) 39.91

1 (180)(10)(0.0245)39.91 40 0.225%

40

f

f

f

A

dAA

= =+

−= ⇒ −

(c) Fig. P12.6(b) is a better feedback circuit. 12.7 (a) ( 10)( 15)( 20) 3000O

O S

V V VV V V

ε ε

ε β= − − − = −= +

So 3000( )O O SV V Vβ= − + We find

30001 3000

Ovf

S

VAV β

−= =+

For 3000120 0.0081 3000vfA β

β−= − = ⇒ =+

(b) Now ( 9)( 13.5)( 18) 2187OV V Vε ε= − − − = − Then

2187 2187 118.241 2187 1 2187(0.008)vfA

β− −= = = −+ +

120 118.24% change 100 1.47% change120

−= × ⇒

12.8

5(10 )(4) (50) 8B Bf f kHz= ⇒ = 12.9 (a) 5

3 3(50) (10 )(4) 8dB dBf f kHz− −= ⇒ =

(b) 53 3(10) (10 )(4) 40dB dBf f kHz− −= ⇒ =

12.10

30(50)(20 10 ) 5A× = so 5

0 2 10A = × 12.11

Low freq. 0

01fAAA β

=+

5000100 0.00981 (5000)

ββ

= ⇒ =+

Freq. response

1 2

1 2

1 2

1 2 1 2

1 2 1 2

5000

1 1

(5000)(0.0098)1 11 1

5000

1 1 49

5000

1 49

5000

50

f

f fj jf fAA

Af fj jf f

f fj jf f

f f jf jfj jf f f f

f f jf jfj jf f f f

β

⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= =

+ +⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=⎛ ⎞⎛ ⎞+ + + ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Also 0 100

1 1 1

ff

A B A B A B

AA

f f f f f fj j j j j jf f f f f f

= =⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

So

1 2 1 2

100 10011 1

50 50 50A B A B

f f f f f f jf jfj j j j j jf f f f f f f f

=⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + + + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠

Then

1 2

1 1 1 150 50A Bf f f f

+ = +

and 1 2

1 150A Bf f f f

=

1 10f = and 2 2000f = 1 1 1 1 0.002 0.000010 0.002010

50(10) 50(2000)A Bf f+ = + = + =

and

6

1 1 1(50)(10)(2000) 10

B

A B A

ff f f

= ⇒ =

Then 6

1 0.00201010

B

B

ff

+ =

6 2 3

6 2 3

10 1 2.01 1010 2.01 10 1 0

B B

B B

f ff f

− −

− −

+ = +

− × + =

3 6 6

6

3 4

6

2.01 10 4.0401 10 4(10 )(1)2(10 )

2.01 10 2.0025 102(10 )

B

B

f

f

− − −

−

− −

−

× ± × −=

× ± ×=

+ sign 31.105 10 HzBf = × + sign 29.05 10 HzAf = ×

12.12 (a) Fig. P12.6(a)

1

1

1

1 1

3

1

200

110

200 1 (10)(0.1126)1 (0.1126)1

200 (4.704)1 22.52

940.73 940.73 123.5223.52 1

(23.52)40 (23.

1(23.52)

f

dB

fjf

A

fjf

fjf

f fj jf f

fjf

f

−

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

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

= = ⋅+ +

= =+

52)(100) 2.352 kHz⇒

Fig P12.6(b)

1

1

1

3

1

(200)(10)

12000

(0.0245)(200)(10)1 1 491

2000 1 (50)(100) 5 KHz50 1

(50)

f

dB

fjfA

fjf fjf

ffj

f

−

+= =

+ + ++

= ⋅ = ⇒+

(b) Overall feedback ⇒ wider bandwidth. 12.13

0 1 2 1

00

0

1000(100) (1) (100)(10) (1)(1) 10001

i n

i n

v A A v A vSv v vN

= +

= + = + ⇒ = =

12.14 (a)

(b)

Circuit (b) – less distortion 12.15 (a) Low input R ⇒ Shunt input Low output R ⇒ Shunt output Or a Shunt-Shunt circuit (b) High input R ⇒ Series input High output R ⇒ Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit 12.16 (a) 4 5(max) (1 ) 10(1 10 ) (max) 10i i iR R T R k= + = + ⇒ ≅ Ω

34

10(min) 101 1 10

ii

RR k

T−= = ≅ Ω

+ +

Or (min) 1iR = Ω

(b) 4 4(max) (1 ) 1(1 10 ) (max) 10o o oR R T R k= + = + ⇒ ≅ Ω

44

1(min) 101 1 10

oo

RR k

T−= = ≅ Ω

+ +

Or (min) 0.1oR = Ω 12.17

Overall Transconductance Amplifier, og

i

iAv

= Series output = current signal and Shunt input = current

signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown.

12.18

1

1 2

||||

ix

i

R RV V

R R Rε⎛ ⎞

= − ⋅⎜ ⎟+⎝ ⎠

12.19

3

50 48 2mV

5 2.5 10 V/V0.002

0.048 0.0096 V/V5

5 100 V/V0.05

b

b

i f

ov

f

o

orf

i

V V V

VA

VVVVAV

ε

ε

β

= − = − =

= = = ×

= = =

= = =

12.20

2 2

1 1

1 20 19vfR RAR R

⎛ ⎞≈ + = ⇒ =⎜ ⎟⎝ ⎠

d S iv i R=

0

1 2

( )S d s ds

v v v v viR R− − −

= + (1)

0 0 0

0 2

( ) 0L d s dv A v v v vR R

− − −+ = (2)

00

0 2 0 2

( )1 1 L d S dA v v vv

R R R R⎛ ⎞ −

+ = +⎜ ⎟⎝ ⎠

0

0 20

0 2

( )

1 1

L d S dA v v vR R

v

R R

−+=

⎛ ⎞+⎜ ⎟

⎝ ⎠

From (1):

0

2 0 2

1 2

0 2

0

0 22

2 21 2 1 2

0 0

( )1

1 1

111 1 1 1

1 1

L d S d

S d S dS

L

S S d

d S i

A v v vR R Rv v v v

iR R

R R

AR RRi v v

R RR R R RR R

v i R

⎡ ⎤−⋅ +⎢ ⎥− − ⎣ ⎦= + −⎛ ⎞

+⎜ ⎟⎝ ⎠⎛ ⎞⎛ ⎞ −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟= + − − + +⎜ ⎟⎜ ⎟+ +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠=

02 2

1 2 0 0 2 1 2 0 2

2 2

0 0

02 2 2

0 1 1 0 0 0 1 1 0 0

0 2

1 1 1 1 1 11 11

1 1

1 1 1 1 1 11

Li

S S

LS i S

S

AR RRR R R R R R R R R

i vR RR R

AR R Ri R vR R R R R R R R R R

i R R

⎧ ⎫⎡ ⎤⎛ ⎞ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎛ ⎞+ + + − + + −⎪ ⎪⎢ ⎥⎜ ⎟⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟⎪ ⎪⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎝ ⎠⎢ ⎥+ =⎨ ⎬ ⎢ ⎥⎪ ⎪+ +⎢ ⎥⎪ ⎪ ⎢ ⎥⎣ ⎦⎩ ⎭⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪+ + + ⋅ + + = + ⋅ +⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

+ 0 02 20

1 1 1 1

1 1 (1)i L SR RR RR A vR R R R

⎧ ⎫⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞⎪ ⎪+ + + + = + +⎨ ⎬⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪⎣ ⎦ ⎣ ⎦⎩ ⎭

Let 2 190 kR ,= Ω 1 10 kR = Ω

5

7

0.1 0.10.1 190 100 20 10 2010 10

(1.000219 10 ) (20.01)

S S

S S

i v

i v

⎧ ⎫⎡ ⎤ ⎡ ⎤+ + ⋅ + + = +⎨ ⎬⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎩ ⎭× =

55 10 k 500 MSif if

S

vR Ri

= ≅ × Ω ⇒ ≅ Ω

Output Resistance

0

0 2 1

1

1 2

0 1

0 0 0 1 2 2 1

1

||||

||||1 1 1

( || ) ||

|| 10 ||100 9.09

X L d XX

i

id X

i

L iX

X f i i

i

V A v VIR R R R

R Rv V

R R RA R RI

V R R R R R R R R R

R R

−= +

+−

= ⋅+

⋅= = + +

+ +

= =

5

0

4

50 0

1 1 10 9.09 10.1 0.1 9.09 190 190 9.09

10 4.566 10 0.005022.19 10 k 0.0219

f

f f

R

R R−

⎛ ⎞= + ⋅ +⎜ ⎟+ +⎝ ⎠

= + × += × Ω ⇒ = Ω

12.21 a.

0

1 2

( )S d S dv v v v vR R− − −

= and 0d

vvA

=

0

1 2 2 1 2

0 0

2 1 2

0 2

1 2 2 1

2

10

2

1

1 1

1 1

1 1 11 1

1

11 1

S Sd

S

S

v v v vR R R R R

v vR A R R

v RvR R R A R

RRv

v RA R

⎛ ⎞+ = + +⎜ ⎟

⎝ ⎠⎛ ⎞

= + +⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + +⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦⎛ ⎞

+⎜ ⎟⎝ ⎠=⎛ ⎞

+ +⎜ ⎟⎝ ⎠

which can be written as 0

2

1

1 / 1vf

S

v AAv RA

R

= =⎡ ⎤⎛ ⎞

+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

b. 2

1

1

1 RR

β =+

c. 5

5

10201 (10 )β

=+

So

5

5

10 120 0.0499910

β β−

= ⇒ =

Then 2 2

1 1

1 11 1 19.0040.04999

R RR Rβ

= − = − ⇒ =

d. 49 10A → × 4

4

9 10 19.999561 (9 10 )(0.04999)fA ×= =

+ ×

434.444 10 2.222 10 % 0.005%

20f f

f f

A AA A

−−Δ Δ− ×= = − × ⇒ = −

12.22

[ ]

1000 90.9 A/A1 1 (1000)(0.01)

1 90.91 1 (0.01)(1000)

(1 ) 10 1 (0.01)(1000) 110 k

iif

i i

iif if

i i

of o i i of

AAA

RR R

AR R A R

β

ββ

= = =+ +

= = ⇒ = Ω+ +

= + = + ⇒ = Ω

12.23

50 47.5 2.5 A

5 2000 A/A0.0025

0.0475 0.0095 A/A5

5 100 A/A0.05

i fb

oi

fbi

o

oif

i

I I II

AIIII

AI

ε

ε

μ

β

= − = − =

= = =

= = =

= = =

12.24 a.

Assume that 1V is at virtual ground.

0 fb FV I R= − Now

00 0

3 3

fb Ffb

fb S

I RVI I IR R

I I Iε

= + = −

= −

and 0

0 ii

II A IAε= =

so 0

fb Si

II IA

= −

From above

03

00

3

03 3

1

1

11 1 1

Ffb

FS

i

F FS

i

RI IR

I RI IA R

R RI IR A R

⎛ ⎞+ =⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞

− + =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎡ ⎤⎛ ⎞ ⎛ ⎞+ = + +⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦

or

30

3

3

1

11 1

11

F

ifS F

i

iif

i

F

RRI

AI R

A R

A AARR

⎛ ⎞+⎜ ⎟

⎝ ⎠= =⎡ ⎤⎛ ⎞

+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

= =+⎛ ⎞

+⎜ ⎟⎝ ⎠

b.

3

1

1i

FRR

β =⎛ ⎞

+⎜ ⎟⎝ ⎠

c. 5

5

10251 (10 ) iβ

=+

so

5

5

10 125 0.0399910i iβ β

−= ⇒ =

so 3 3

1 11 1 24.00.03999

F F

i

R RR Rβ

= − = − ⇒ =

d. 5 5 410 (0.15)(10 ) 8.5 10iA = − = ×

so 4

4

8.5 10 24.99891 (8.5 10 )(0.03999)ifA ×= =

+ ×

so 3

5 31.10 10 4.41 10 4.41 10 %25

if

if

AA

−− −Δ ×= − = − × ⇒ − ×

12.24 b.

( )0 1 2m gs m LV g V g V Rπ= + (1)

0i gsV V V= + (2)

1 12

2 2

10 0FEm

E E

V V h Vg V Vr R r R

ππ π

π π

⎛ ⎞++ + = ⇒ + =⎜ ⎟

⎝ ⎠ (3)

11

1

0m gsD

V V Vg Vr R

π π

π

−= + =

or

1 1 11

D m gsD

V VV R g V

r Rπ π

π

⎡ ⎤= + −⎢ ⎥

⎣ ⎦ (4)

Then

11

2 1

1 0FE Dm gs

E D

V Vh RV g Vr R r R

π ππ

π π

⎛ ⎞ ⎡ ⎤++ + − =⎜ ⎟ ⎢ ⎥

⎝ ⎠ ⎣ ⎦ (3)

1 11

2 2 2

1 1FE D Dm gs

E E E

h R RV g Vr R r R Rππ π

⎧ ⎫⎛ ⎞ ⎛ ⎞+⎪ ⎪+ + =⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪ ⎝ ⎠⎝ ⎠⎩ ⎭

Let

11

2

1 Dm gs

eq E

RV g VR Rπ

⎛ ⎞⋅ = ⎜ ⎟

⎝ ⎠

so 11

2

Dm eq gs

E

RV g R VRπ

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Then

10 1 1 2

2

Dm gs m m eq gs L

E

RV g V g g R V RR

⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦ (1)

so

( )10 1 2 0

2

1 Dm L m eq i

E

RV g R g R V VR

⎡ ⎤⎛ ⎞= + −⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

so

11 2

20

11 2

2

1

1 1

Dm L m eq

Ev

i Dm L m eq

E

Rg R g RRVA

V Rg R g RR

⎡ ⎤⎛ ⎞+⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦= =⎡ ⎤⎛ ⎞

+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

c. Set 0iV =

1 2X

X m gs mL

VI g V g VRπ+ + =

gs XV V= − From part (b), we have

1 11 1

2 2

D Dm eq gs m eq X

E E

R RV g R V g R VR Rπ

⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Then

11 2

0 2

1 1X Dm m eq

X L E

I Rg g RV R R R

⎛ ⎞= = + ⎜ ⎟

⎝ ⎠

or

01 1

1 22

1 1L

m Dm m eq

E

R Rg Rg g R

R

=⎛ ⎞⎜ ⎟⎝ ⎠

12.25

1,S fb iI I I V I Rε ε= + =

00

3fb

VI IR

= + and 0 1 fb FV V I R= −

00 i

i

II A I IAε ε= ⇒ =

Now

13

1

3 3

1 ( )

1

fb i fb F

Ffb i

fb S e

I A I V I RR

R VI A IR R

I I I

ε

ε

= + −

⎡ ⎤+ = +⎢ ⎥

⎣ ⎦= −

1

3 3

1

3 3 3

1

( ) 1R

1 1

FS e i e

F FS e i

i

R VI I A IR

R R VI I AR R R

VIRε

⎡ ⎤− + = +⎢ ⎥

⎣ ⎦⎡ ⎤⎡ ⎤ ⎛ ⎞

+ = + + +⎢ ⎥⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎣ ⎦

=

13 3 3

31

3 3

1 11 1

1

1 11

F FS i

i

F

ifS F

ii

R RI V AR R R R

RRVR

I R AR R R

⎧ ⎫⎡ ⎤⎡ ⎤ ⎛ ⎞⎪ ⎪+ = ⋅ + + +⎨ ⎢ ⎥ ⎬⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

⎛ ⎞+⎜ ⎟

⎝ ⎠= =⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪⋅ + + +⎨ ⎬⎢ ⎥⎜ ⎟

⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

The 31/ R term in the denominator will be negligible. Using the results of Problem 12.15:

5

4

251 (25) 102

5 10 k 0.5

if

if if

R

R R−

=⎧ ⎫⎡ ⎤+⎨ ⎬⎣ ⎦⎩ ⎭

≅ × Ω ⇒ = Ω

Output Resistance (Let 0)LZ =

3

X XX i

F i

X

F i

V VI A IR R R

VIR R

ε

ε

= + ++

=+

so

0 3 3

11 1 , 24iX F

X f F i

AI RV R R R R R

+= = + =

+

Let 240 k ,FR = Ω 3 10 R k= Ω 51 1 10 1

10 240 2ofR+= ++

so 30 0 05

240 2 2.42 10 k or 2.42 1 10 1

F if f f

i

R RR R RA

−+ +≈ = ⇒ ≈ × Ω ≈ Ω+ +

12.26

[ ] [ ][ ] [ ]

5 0.2 A/V1 1 (4.8)(5)

1 (10) 1 (4.8)(5) 250 k

1 (10) 1 (4.8)(5) 250 k

gfz

if i z

of o z

AgA Ag

R R Ag

R R Ag

βββ

= = =+ +

= + = + = Ω

= + = + = Ω

12.27

0

0

40 38 2.0 mV

8mA 4 A/V2mV

38mV 4.75 V/A8mA8mA 0.2 A/V40mV

i fb

og

fbz

gfi

V V V

IA

VVII

A V

ε

ε

β

= − = − =

= = =

= = =

= = =

12.28

0(1 ) SFE

EFE E

V VhI Ih R

ε−+= ⋅ =

Also 0 ( )FE gI h A Vε= so 0

FE g

IV

h Aε =

Then 0

0

0

0

0 0

1

1 1

(1 ) 1

11 (1 ) 1 ( )

SFE

FE E FE g E

SFE

FE FE g E E

g FE E S

FE g E E

FE g E FE g

S E g FE E S FE g E

V Ih Ih R h A R

Vh Ih h A R R

A h R VIh A R R

h A R h AI IV R A h R V h A R

+⋅ = −

⎡ ⎤++ =⎢ ⎥

⎢ ⎥⎣ ⎦⎡ ⎤+ +

=⎢ ⎥⎢ ⎥⎣ ⎦

⎡ ⎤= ⋅ ⇒ ≈⎢ ⎥

+ + +⎢ ⎥⎣ ⎦

b. z EB R=

c. 5

5

5 10101 (5 10 ) zβ

×=+ ×

5

5

5 10 110 0.099998 k5 10z z ERβ β

× −= ⇒ = = Ω

×

d. If 55.5 10gA → × then 5

5

5.5 10 10.00001821 (5.5 10 )(0.099998)gfA ×= =

+ ×

541.82 10 1.82 10 %

10gf

gf

AA

−−Δ ×= ⇒ ×

12.29

(1 ) ,E FE gI h A Vε= + E SE

VI IR

ε= − and ,S iV I Rε = S S S iV V V V I Rε ε= − = −

Now ( ) 11 ( )FE g S i S S i SE

h A I R V I R IR

+ = ⋅ − −

( ) i1 1 SFE g i S

E E

VRh A R IR R

⎡ ⎤+ + + =⎢ ⎥

⎣ ⎦

( )1 1S iif E FE g i

S E

V RR R h A R

I R⎡ ⎤

= = + + +⎢ ⎥⎣ ⎦

From Problem 12.16: ( ) 51 5 10 mSFE g FE gh A h A+ ≈ = ×

0.1 kER ≈ Ω

so 5 20(0.1) (5 10 )(20) 10.1ifR ⎡ ⎤= × + +⎢ ⎥⎣ ⎦

or 610 kifR = Ω

gV A Vr

πε

π

=

0

( )XX m

V VI g VR

επ

− −= + (1)

( )( || )X g E iV I A V R Rε ε= − + (2)

or 1 ( || ) ( || )g i X E iV A R R I R Rε ε⎡ ⎤+ = −⎣ ⎦ Now:

0 0

XX m g

VVI g A r VR R

επ ε= + + (1)

0 0

( || )11 ( || )

X E i XX m g

g E i

I R R VI g A rR A R R Rπ

⎡ ⎤⎛ ⎞ −= + +⎢ ⎥⎜ ⎟ +⎢ ⎥⎝ ⎠ ⎣ ⎦

0

00

( || )111 ( || )

Xf

X

E im g

g E i

VRI

R RR g A rR A R Rπ

=

⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪= + + ⎢ ⎥⎨ ⎬⎜ ⎟ +⎢ ⎥⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

55 10 mSm g FE gg r A h Aπ = = ×

Let 100FEh = so 35 10 mSgA = × || 0.1 || 20 0.1kE iR R = ≈ Ω

Then

50 3

1 0.150 1 5 1050 1 (5 10 )(0.1)fR

⎧ ⎫⎡ ⎤⎪ ⎪⎛ ⎞= + × +⎨ ⎬⎜ ⎟ ⎢ ⎥+ ×⎝ ⎠⎪ ⎪⎣ ⎦⎩ ⎭

or 0 5.04 MfR = Ω

12.30

5 0.2 V/μA1 1 (4.8)(5)

1 40 1 1 (4.8)(5)

1 40 1 1 (4.8)(5)

zzf

g z

iif if

g z

oof of

g z

AAA

RR RA

RR RA

β

β

β

= = =+ +

= = ⇒ = Ω+ +

= = ⇒ = Ω+ +

12.31

40 38 2 μA

8 V42 μA

38 μA4.75 8 V8 V0.240 μA

i fb

oz

fbg

o

ozf

i

I I I

VA

IIVV

AI

ε

ε

β

= − = − =

= = =

= = =

= = =

12.32 a.

Assuming 1V is at virtual ground

0( ) fb FV I R− = − and 00( ) z

z

VV A I IAε ε− = − ⇒ =

fb SI I Iε= −

So 00 ( )S F S F F

z

VV I I R I R R

Aε⎛ ⎞

= − = − ⎜ ⎟⎝ ⎠

0 1 FS F

z

RV I RA

⎡ ⎤+ =⎢ ⎥

⎣ ⎦

so 0

1

F z Fzf

S z FF

z

V R A RAI A RR

A

= = =+⎡ ⎤

+⎢ ⎥⎣ ⎦

or 111

z zzf

z gz

F

A AAA

AR

β= =

+⎛ ⎞+ ⎜ ⎟

⎝ ⎠

b. 1g

FRβ =

c. 6

46

5 105 101 (5 10 ) gβ

×× =+ ×

6

45

6

5 10 15 10 1.98 10

5 10g gβ β −

× −×= ⇒ = ×

×

1 50.5 kF Fg

R Rβ

= ⇒ = Ω

d. 6 6(0.9)(5 10 ) 4.5 10zA = × = ×

64

6 5

4.5 10 4.994 101 (4.5 10 )(1.98 10 )zfA −

×= = ×+ × ×

34

55.4939 1.11 10 0.111%5 10

zf

zf

AA

−Δ= − = − × ⇒ −

×

12.33

1 0 0,i z xV I R V A I V A Iε ε ε= − = − ⇒ =

fb SI I Iε= − and 0 1 fb FV V I R− = −

1

1 11

1

1

( )

1

1

z S F

z S F Fi i

z FS F

i i

Fif

S z F

i i

A I V I I R

V VA V I R RR R

A RI R VR R

V RRI A R

R R

ε ε− = − −

⎛ ⎞ ⎛ ⎞− = − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎡ ⎤

= + +⎢ ⎥⎣ ⎦

= =⎡ ⎤

+ +⎢ ⎥⎣ ⎦

Or, using the results from problem 12.18.

[ ]

3

6 3

3 3

3

50.5 105 10 50.5 101

10 10 10 10

50.5 10 99.79 1 500 5.05

if

if

R

R

×=⎡ ⎤× ×+ +⎢ ⎥× ×⎣ ⎦

×= ⇒ = Ω+ +

12.34 Assume 0.2 CQI mA=

Then (100)(0.026) 13 0.2

r kπ = = Ω

(1) 1 2 2 1 2

1 1 1i A A o i oAA

i i i

V V V V V VV VR R R R R R R R

⎛ ⎞− −= + ⇒ + = + +⎜ ⎟

⎝ ⎠

Now 1 1 1 (12)

10 10 10 1 10i o

A i o AV V V V V V⎛ ⎞+ = + + ⇒ + =⎜ ⎟

⎝ ⎠

or 1 ( )12A i oV V V= +

(2) 2

(1 )v o o AFE

o

A V V V Vh

R r Rε

π

⎛ ⎞− −+ =⎜ ⎟+⎝ ⎠

where i AV V Vε = − Then

0 2

( )(1 )v i A o o A

FEA V V V V V

hR r Rπ

⎛ ⎞− − −+ =⎜ ⎟+⎝ ⎠

we find

2 2

(1 ) (1 ) (1 )v i FE o FE o v A FE A

o o o

A V h V h V A V h VR r R r R R r Rπ π π

+ + +− − = −

+ + +

Then 3 3(5 10 )(101) (101) (5 10 )(101) 114 14 10 14 10

i o oA

V V VV

⎛ ⎞× ×− − = −⎜ ⎟⎝ ⎠

Rearranging terms, we find

10.97

1 1( ) ( 10.97 ) 0.997512 12

ovf

i

i i iif i

i i Ai A

i

A i o i i i

VAV

V V VR RI V VV V

R

V V V V V V

= =

⎛ ⎞= = = ⎜ ⎟−⎛ ⎞− ⎝ ⎠

⎜ ⎟⎝ ⎠

= + = + =

Then 1 (10 ) 4

1 0.9975if ifR k R M⎛ ⎞= Ω ⇒ = Ω⎜ ⎟−⎝ ⎠

To find the output resistance:

2 1

( )(1 )||

v FE xx

o i

A V h VIR r R R Rε

π

++ =

+ +

1

1 2

||||

ix

i

R RV V

R R Rε⎛ ⎞

= − ⋅⎜ ⎟+⎝ ⎠

Now 1 || 1 ||10 0.909iR R = =

Then 0.0833 xV Vε = −

Now

{ }

3

3

(5 10 )(0.0833) 1 1(101)1 13 10 0.909

3.012 10 0.0917

x x

x

I V

V

⎧ ⎫⎡ ⎤× +⎪ ⎪= +⎨ ⎬⎢ ⎥+ +⎪ ⎪⎣ ⎦⎩ ⎭

= × +

Or 43.32 10 0.332x

of ofx

V R k RI

−= = × Ω ⇒ = Ω

12.35 a. Neglecting base currents

2 2

1 0

0.5 mA, 12 (0.5)(22.6) 0.7 V0.5 mA 0

C C

C

I VI v

= = − == ⇒ =

Then 3 2 mACI =

b. 1 21

(100)(0.026) 5.2 k0.5

FE T

C

h Vr rIπ π

⋅= = = = Ω

1 2

3

3

0.5 19.23 mA / V0.026

(100)(0.026) 1.3 k2

2 76.92 mA / V0.026

m m

m

g g

r

g

π

= = =

= = Ω

= =

( )

1 21 1 1 2

1 1

1 2 1 1 21

11 2 2

1

0

1( ) 0 1

( )

m m

m

i S b

V Vg V g Vr r

V V g V Vr

VV R r V V

r

π ππ π

π π

π π π ππ

ππ π

π

+ + + =

⎛ ⎞+ + = ⇒ = −⎜ ⎟

⎝ ⎠

= + − +

or 1 2 21

1 Si b

RV V V V

rπ ππ

⎛ ⎞= + − +⎜ ⎟

⎝ ⎠

But 2 1V Vπ π= − so

1 21

2 Si b

RV V V

rππ

⎛ ⎞= + +⎜ ⎟

⎝ ⎠ (2)

02 02 01 2

3

0mC

V V Vg VR rπ

π

−+ + = (3)

3 0 0 23 3

3 2

3 02 0

bm

L

V V V Vg V

r R RV V V

ππ

π

π

−+ = +

= −

so

202 0 0

3 2 2

1 1 1( ) bFE

L

VhV V Vr R R Rπ

⎛ ⎞ ⎛ ⎞+− = + −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ (4)

2 0 2 2

2 1 1

0b bV V V VR R r

π

π

−+ + = (5)

Substitute numbers into (2), (3), (4) and (5):

2 212

5.2i bV V Vπ⎛ ⎞= − + +⎜ ⎟⎝ ⎠

2 2(2.192)i bV V Vπ= − + (2)

02 2 01 1 1(19.23) 0

22.6 1.3 1.3V V Vπ

⎛ ⎞ ⎛ ⎞+ + − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

02 2 0(0.8135) (19.23) (0.7692) 0V V Vπ+ − = (3)

02 0 2101 101 1 1 11.3 1.3 4 50 50bV V V⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

02 0 2(77.69) (77.96) (0.02)bV V V= − (4)

2 0 21 1 1 1 0

50 10 50 5.2bV V Vπ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 0 2(0.120) (0.020) (0.1923) 0bV V Vπ− + = (5)

From (2): 2 2 (2.192).b iV V Vπ= + Substitute in (4) and (5) to obtain:

02 0 2(77.69) (77.96) [ (2.192)](0.02)iV V V Vπ= − + (4 )′

2 0 2[ (2.192)](0.120) (0.020) (0.1923) 0iV V V Vπ π+ − + = (5 )′ So we now have the following three equations:

02 2 0(0.8135) (19.23) (0.7692) 0V V Vπ+ − = (3)

02 (77.69)V

0 2(77.96) (0.02) (0.04384)iV V Vπ= − − (4 )′

2 0(0.120) (0.4553) (0.020) 0iV V Vπ+ − = (5 )′ From (3): 02 0 2(0.9455) (23.64).V V Vπ= − Substitute for 02V in (4′) to obtain:

2 0 2(77.69)[ (0.9455) (23.64)] (77.96) (0.02) (0.04384)o iV V V V Vπ π− = − − or

0 20 (4.504) (0.02) (1836.5)iV V Vπ= − + Next, solve (5 )′ for 2 :Vπ

2 0(0.120) (0.4553) (0.020) 0iV V Vπ+ − =

2 0 (0.04393) (0.2636)iV V Vπ = − Finally,

0 0

0

0 (4.504) (0.02) (1836.5)[ (0.04393) (0.2636)]0 (85.18) (484.12)

i i

i

V V V VV V

= − + −= −

So 0 484.12 5.68

85.18vf vfi

VA AV

= = ⇒ =

12.36 a. 1 2|| 400 || 75 63.2 kTHR R R= = = Ω

2

1 2

1

1

75 (10) 1.579 V75 400

1.579 0.7 0.007106 mA63.2 (121)(0.5)0.853 mA

TH CC

BQ

CQ

RV VR R

I

I

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠−= =

+=

1

2

2

3

10 (0.853)(8.8) 2.49 V2.49 0.7 0.497 mA

3.610 (0.497)(13) 3.54 V3.54 0.7 2.03 mA

1.4

C

C

C

C

V

I

V

I

= − =−≈ =

= − =−≈ =

Then

1

1

2

2

3

3

(120)(0.026) 3.66 k0.8530.853 32.81 mA / V0.026

(120)(0.026) 6.28 k0.4970.497 19.12 mA / V0.026

(120)(0.026) 1.54 k2.032.03 78.08 mA / V

0.026

m

m

m

r

g

r

g

r

g

π

π

π

= = Ω

= =

= = Ω

= =

= = Ω

= =

b.

1 1 1 1i iV V V V V Vπ ε ε π= + ⇒ = − (1)

1 1 1 01 1

1 1m

E F

V V V Vg Vr R R

π ε επ

π

−+ = + (2)

2 1 1 1 2( )( || )m CV g V R rπ π π= − (3)

3 0 32 2

2 3

0mC

V V Vg VR r

π ππ

π

++ + = (4)

3 0 0 13 3

3 3m

E F

V V V Vg Vr R R

π επ

π

−+ = + (5)

Substitute numbers in (2), (3), (4) and (5): 0

1 11 1 132.81 ( )

3.66 0.5 10 10iVV V Vπ π

⎛ ⎞ ⎛ ⎞+ = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 1 0(35.18) (2.10) (0.10)iV V Vπ = − (2)

2 1(32.81) (88 || 6.28)V Vπ π= − or 2 1(120.2)V Vπ π= − (3)

3 0 32(19.12) 0

13 13 1.54V V V

V π ππ + + + =

or 2 3 0(19.12) (0.7263) (0.07692) 0V V Vπ π+ + = (4)

13 0

1 1 178.081.54 1.4 10 10

iV VV V ππ

−⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 3 0 1(78.73) (0.8143) (0.10) (0.10)iV V V Vπ π= − + (5)

Now substituting 2 1(120.2)V Vπ π= − in (4):

1 3 0(19.12)[ (120.2)] (0.7263) (0.07692) 0V V Vπ π− + + =

or 1 3 0(2298.2) (0.7263) (0.07692) 0V V Vπ π− + + =

Then 3 1 0(3164.3) (0.1059)V V Vπ π= −

Substituting 3 1 0(3164.3) (0.1059)V V Vπ π= − in (5):

1 0 0 1(78.73)[ (3164.3) (0.1059)] (0.8143) (0.10) (0.10)iV V V V Vπ π− = − + or 5

1 0(2.49 10 ) (9.152) (0.10)iV V Vπ × − = − Then

5 71 0 (3.674 10 ) (4.014 10 )iV V Vπ

− −= × − × Now substituting 5

1 0 (3.674 10 )V Vπ−= ×

7(4.014 10 )iV −− × in (2): 5 7

0(35.18)[ (3.674) 10 ) (4.014 10 )]iV V− −× − ×

or 0

0

(2.10) (0.10)(0.1013) (2.10)

i

i

V VV V

= −=

So 0 20.7i

VV

=

c. iif

i

VRI

= and 1 1i RB bI I I= +

11

iRB

B

VIR

=

11

1b

VIr

π

π

=

Now 5 7

14

1

(20.7 )(3.674 10 ) (4.014 10 )

(7.60 10 )i i

i

V V V

V Vπ

π

− −

−

= × − ×

= ×

Then

4

4

(7.60 10 )63.2 3.66

10.01582 2.077 10

iif

i i

VRV V −

−

=×+

=+ ×

or 62.4 kifR = Ω

d. To determine 0 :fR Equation (1) is modified to 1 1 0eV Vπ + = ( 0)iV = Equation (5) is modified to:

3 0 1(78.73) (0.8143) (0.10)XV I V Vπ π+ = + (5) Now

1 0(35.18) (0.10)V Vπ = − (2)

2 1(120.2)V Vπ π= − (3)

2 3 0(19.12) (0.7263) (0.07692) 0V V Vπ π+ + = (4) Now

1 0 (0.002843)V Vπ = − so

2 0

2 0

( )(0.002843)(120.2)(0.3417)

V VV V

π

π

= − −=

Then 0 3 0(0.3417)(19.12) (0.7263) (0.07692) 0V V Vπ+ + + =

or 3 0 (9.101)V Vπ = − (4) So then

0 (9.101)(78.73) XV I− +

0 0(0.8143) (0.10)( )(0.002843)V V= + − or

0 (717.3)XI V= (5) or

00 00.00139 k 1.39 f f

X

VR RI

= = Ω ⇒ = Ω

12.37 (a)

(1) 1 11

i A A OAm

E F

V V V VVg Vr R Rππ

− −+ = +

(2) 1 11 2

0B Bm

C

V Vg VR rπ

π

+ + =

(3) 2 22 3

0C C om

C

V V Vg VR rπ

π

−+ + =

(4) 3 33

C O O Am

F

V V V Vg Vr Rππ

− −+ =

1 2 3i A B C OV V V V V V V Vπ π π= − = = −

(1) ( ) 11

1 A OAi A m

E F

V VVV V gr R Rπ

⎛ ⎞ −− + = +⎜ ⎟

⎝ ⎠

(2) 11 2

1 1 ( ) 0B m i AC

V g V VR rπ

⎛ ⎞+ + − =⎜ ⎟

⎝ ⎠

(3) 22 3 3

1 1 0OC m B

C

VV g V

R r rπ π

⎛ ⎞+ + − =⎜ ⎟

⎝ ⎠

(4) ( ) 33

1 O AC O m

F

V VV V g

r Rπ

⎛ ⎞ −− + =⎜ ⎟

⎝ ⎠

1 1(100)(0.026) 14.30.182 K 550 mA/V

14.3 0.026mr gπ = = = =

2 2(100)(0.026) 4.620.563 K 178 mA/V

4.62 0.026mr gπ = = = =

3 3(100)(0.026) 4.470.582 K 172 mA/V

4.47 0.026mr gπ = = = =

(1) ( ) 1 5500.182 0.05 1.2

A OAi A

V VVV V −⎛ ⎞− + = +⎜ ⎟⎝ ⎠

(2) 1 1 (550)( ) 00.3 0.563B i AV V V⎛ ⎞+ + − =⎜ ⎟

⎝ ⎠

(3) 1 1 178 00.65 0.582 0.582

OC B

VV V⎛ ⎞+ + − =⎜ ⎟⎝ ⎠

(4) ( ) 11720.582 1.2

O AC O

V VV V −⎛ ⎞− + =⎜ ⎟⎝ ⎠

(1) ( )(555.5) (20) ( )(0.8333)i A A A OV V V V V− = + − (2) (5.109) 550( ) 0B i AV V V+ − = (3) (3.257) 178 (1.718) 0C B OV V V+ − = (4) ( )(173.7) ( )(0.8333)C O O AV V V V− = − (1) (555.5) (0.8333) (576.3)i O AV V V+ = (2) (5.109) 550 (550) 0B i AV V V+ − = (3) (3.257) 178 (1.718) 0C B OV V V+ − = (4) (173.7) (0.8333) (174.5)C A OV V V+ = From (2) (107.7) (107.7)B A iV V V= − From (4) (1.0046) (0.004797)C O AV V V= − Substitute into (3)

[ ](3.257) (1.0046) (0.004797)O AV V− (178)[ (107.7) (107.7)] (1.718) 0A i oV V V+ − − = (3.272) (0.01562) (19170.6) (19170.6) (1.718) 0O A A i OV V V V V− + − − = (19170.6) (19170.6) (1.554)A i OV V V= −

(1.00) (0.00008106)A i OV V V= − Substitute into (1)

[ ](555.5) (0.8333) (576.3) (1.00) (0.00008106)(576.3) (0.0467)

i o i o

i o

V V V VV V

+ = −= −

(0.880) (20.8)

23.6

o i

ovf

i

V VV AV

=

= =

Ideal 1.2 0.05 25.0

0.05F E

vfE

R RAR+ += = =

(b) iif

i

VRI

= and 1

1 1

i Ai

V V VIr r

π

π π

−= =

We have (1.00) (0.00008106)(1.00) (23.6) (0.00008106)(0.99809)

A i o

i i

A i

V V VV V

V V

= −= −=

Then (1 0.99809)

(0.01051)0.182

ii i

VI V

−= =

95.1 K(0.01051)

iif if

i

VR RV

= ⇒ =

To find ,ofR set 0iV =

33 3

3

3

33

1( )( )

x AX m

F

C X

X AX C X m

F

V V VI g V

r RV V V

V VI V V gr R

ππ

π

π

π

−+ + =

= −−

+ − + =

For 0,iV = we have (1.0046) (0.004797)

(576.3) (0.8333)(0.001446)(1.0046) (0.001446)(0.004797)(1.0046)

C X A

A X

A X

C X X

C X

V V VV VV VV V VV V

= −=

== −=

(1 0.004797)1(1.0046 1.0) 1720.582 1.2

(0.7991) (0.8293)(0.03024)

33.1 K

XX X

X X X

X X

Xof

X

VI V

I V VI V

VRI

−⎛ ⎞+ − + =⎜ ⎟⎝ ⎠

+ ==

= =

12.38

1 1 8 KD ER R= = ( )GG GS D C LV V I I R= + +

10 10 ( 0.7)8 8

D DD C

V VI I− − += =

( )DGG TN D C L

n

IV V I I Rk

= + + +

( )

2

2

10 10 9.34.5 1 (1.8) (1.8)8(0.4) 8 8

836 10 8 18 1.8 16.74 1.83.2

36 4.472 10 42.74 3.6

3.6 6.74 4.472 10

12.96 48.528 45.4276 19.999(10 )12.96 28.528 154.6 0

D D D

D D D

D D

D D

D D D

D D

V V V

V V V

V V

V V

V V VV V

− − −⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − + + − + −

= − + −

− = −

− + = −

− − =

28.528 813.847 4(12.96)(154.6)2(12.96)

4.726 V

D

D

V

V

± +=

=

10 4.726 0.6593 mA8

9.3 4.7268

0.5718 mA

D

C

I

I

−= =

−=

=

0.6593 1 2.284 V0.4GSV = + =

4.5 2.284 2.216VOV = − = 4.726 2.216 2.51VDS D OV V V= − = − =

( )( )sat 1.28VDSV = Also

( ) (0.6593 0.5718)(1.8) 2.216 O D C LV I I R V= + = + = OK

2 (4.726 0.7) 2.216 3.21 ECQV V= + − = (b)

(1) i gs oV V V= +

(2) 11 2

Am gs

D

VV g VR r

π

π

+ =

(3) 22 2

0Bm

E

VV g VR r

ππ

π

+ + =

B A gs i oV V V V V Vπ = − = −

(2) 11 2

( )A B Am i o

D

V V Vg V VR rπ

−+ − =

(3) 22 2

( ) 0B B Am B A

E

V V V g V VR rπ

−+ + − =

(4) 1 2

1 2( ) ( )

Om gs m

L

Om i o m B A

L

Vg V g VR

Vg V V g V VR

π+ =

− + − =

1 1

22

T

2

2 2 (0.4)(0.6593) 1.027 mA/V0.5718 21.99 mA/V0.026

(100)(0.026) 4.547 K0.5718

m n D

Cm

g K IIgV

rπ

= = =

= = =

= =

(2) (1.027)( )8 4.547

A B Ai o

V V VV V −+ − =

(3) (21.99)( ) 08 4.547B B A

B AV V V V V−+ + − =

(2) 0.125 1.027 1.027 0.2199 0.2199A i o B AV V V V V+ − = − (3) 0.125 0.2199 0.2199 21.99 21.99 0B B A B AV V V V V+ − + − = (2) 0.3449 0.2199 1.027 1.027A B i oV V V V− = − + (3) 22.21 22.335A BV V=

1.0056A BV V= Then (0.3449)(1.0056) 0.2199 1.027 1.027B B i oV V V V− = − + 0.1269 1.027 1.027B i oV V V= − +

8.093 8.093B i oV V V= − + 8.138 8.138A i oV V V= − +

Then

(4) [ ](1.027)( ) (21.99) 8.093 8.093 8.138 8.1381.8

oi o i o i o

VV V V V V V− + − + + − =

1.027 1.027 177.965 177.965 178.955 178.955 0.5556i o i o i o oV V V V V V V− − + + − = 2.017 2.5726 0i oV V− =

0.784 AVo

i

VV

= =

Set 0iV =

2 1

2 1( ) ( )

XX m m gs

L

XX m B A m X

L

VI g V g VR

VI g V V g VR

π+ + =

+ − + − =

8.0938.138

B X

A X

V VV V

==

[ ](21.99) 8.093 8.138 1.027( ) (0.5556)2.572

X X X X X

X X

I V V V VI V

+ − + − ==

0.389 KXof

X

V RI

= =

12.39 a. 2 2 (0.5)(0.5) 1 /m n DQg K I mA V= = =

0 ( )m gs SV g V R=

0i gsV V V= + so 0gs iV V V= − Then

0 0( )m S iV g R V V= − 1(2) 0.667

1 1 (1)(2)m S

v vm S

g RA Ag R

= = ⇒ =+ +

To determine 0 fR

XX m gs

S

VI g VR

+ = and gs XV V= −

0

1 1Xm

X f S

I gV R R

= = +

so 0 01 1 2 0.667 k

1f S fm

R R Rg

= = ⇒ = Ω

b. For 20.8 /nK mA V=

2 (0.8)(0.5) 1.265 mA / Vmg = = (1.265)(2) 0.7167

1 (1.265)(2)vA = =+

0.7167 0.667 7.45% increase0.667

f

f

AA

Δ −= ⇒

0

0

0

0

1 || 2 0.7905 || 21.2650.5666 k

0.5666 0.667 15.05% decrease0.667

f

f

f

f

R

RR

R

= =

= Ω

Δ −= ⇒

12.40 dc analysis:

1

1

150 || 47 35.8 k47 (25) 5.96 V

47 150

TH

TH

R ,

V

= = Ω

⎛ ⎞= =⎜ ⎟+⎝ ⎠

2

2

33 || 47 19.4 k33 (25) 10.3 V

33 47

TH

TH

R ,

V

= = Ω

⎛ ⎞= =⎜ ⎟+⎝ ⎠

1

1

5.96 0.7 0.0187 mA35.8 (51)(4.8)(50)(0.0187) 0.935 mA

B

C

I

I

−= =+

= =

2

2

10.3 0.7 0.03705 mA19.4 (51)(4.7)(50)(0.03705) 1.85 mA

B

C

I

I

−= =+

= =

1

2

(50)(0.026) 1.39 k ;0.935

(50)(0.026) 0.703 k1.85

r

r

π

π

= = Ω

= = Ω

1

2

0.935 35.96 mA / V0.0261.85 71.15 mA / V0.026

m

m

g

g

= =

= =

1S eV V Vπ= + (1)

1 01 1

1 1

e em

F

V V V Vg Vr R R

ππ

π

−+ = + (2)

2 2 21 1

1 2 2

0mC B

V V Vg VR R r

π π ππ

π

+ + + = (3)

0 02 2

2

0em

C F

V V Vg VR Rπ

−+ + = (4)

Substitute numerical values in (2), (3) and (4): 1e SV V Vπ= − (1)

11 1 0

1 1 1(35.96) ( )1.39 0.1 4.7 4.7SV V V V Vπ

π π⎛ ⎞ ⎛ ⎞+ = − + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 1 0(46.89) (10.213) (0.2128)SV V Vπ = − (2)

1 21 1 1(35.96) 0

10 19.4 0.703V Vπ π

⎛ ⎞+ + + =⎜ ⎟⎝ ⎠

or 1 2(35.96) (1.574) 0V Vπ π+ = (3)

2 0 11 1 1(71.15) ( ) 0

4.7 4.7 4.7SV V V Vπ π⎛ ⎞ ⎛ ⎞+ + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 2 0 1(71.15) (0.4255) (0.2128) (0.2128) 0SV V V Vπ π+ − + = (4)

From (3): 2 1(22.85)V Vπ π= − Then substitute in (4):

1 0 1(71.15) (22.85) (0.4255) (0.2128) (0.2128) 0SV V V Vπ π− + − + = or

1 0(1625.6) (0.4255) (0.2128) 0SV V Vπ− + − = From (2): 1 0(0.2178) (0.004538)SV V Vπ = − Then

0 0(1625.6)[ (0.2178) (0.004538)] (0.4255) (0.2128) 0S SV V V V− − + − = or 0(354.3) (7.802) 0SV V− + = Finally

0 45.4S

VV

⇒ =

12.41 For example, use a z-stage amplifier. Each stage shown in Fig. 12.29. 12.42

For 3 33

:2n

nk WM K

L′ ⎛ ⎞= ⋅⎜ ⎟⎝ ⎠

Let 3

25WL

⎛ ⎞ =⎜ ⎟⎝ ⎠

Then 23

0.080 (25) 1 /2nK mA V⎛ ⎞= =⎜ ⎟

⎝ ⎠

Want 0ov = for 0,iv = so that 2

3 2 30.4 1 ( )D Q GS TNI I V V= = = ⋅ −

Then 30.4 2 2.63 1GSV V= + =

For 3 3 22.63 12GS G D DV V V I R= ⇒ = − Or 2.63 12 (0.1) 93.7 D DR R k= − ⇒ = Ω

3 3 32 2 (1)(0.4) 1.26 /m n Dg K I mA V= = =

( ) ( )1

1 2

10 0.0769120 10A o o o

RV V V VR R

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

(Small amount of feedback) (1) 1 2i gs gs AV V V V= − +

(2) 1 2 1 20m gs m gs gs gsg V g V V V+ = ⇒ = − Then

2 212 ( )2i gs A gs A iV V V V V V= − + ⇒ = −

2 0.0384 0.5gs o iV V V= − (3) 2 [0.03846 0.5 ]B m gs D m D o iV g V R g R V V= − = − −

(4) 3gs B oV V V= − and [ ]3 3 1 2|| ( )o m gs LV g V R R R= + So

[ ]3 1 2|| ( ) ( )o m L B oV g R R R V V= + − Then

( ) ( )3 1 2|| 0.03846 0.5o m L m D o i oV g R R R g R V V V= + − − −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦ Or

[ ][ ] [ ][ ]3 1 2 3 1 21 || ( ) (0.03846) 1 || ( ) 0.5o m L m D m L m D iV g R R R g R g R R R g R V⎡ ⎤+ + + = + ⋅⎣ ⎦ Now

1 2|| ( ) 4 ||130 3.88LR R R+ = = So

[ ]1 (1.26)(3.88) (93.7)(0.03846) 1 (1.26)(3.88)(0.5) (93.7)o m m iV g g V⎡ ⎤+ + =⎣ ⎦ Rearranging terms, we find

229 10 1.11 /5.89 17.6

o mm

i m

V g g mA VV g

= = ⇒ =+

We have

1 2

0.0802 1.11 2 (0.1) 772 2n

m Dk W W W Wg I

L L L L′⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

12.43 Assuming an ideal op-amp, then from Equation (12.58)

0 1

2

201 1000.2S

I RI R

= + = =

Then 1

2

99RR

=

For example, set 2 5 kR = Ω and 1 495 kR = Ω 12.44

(a) 1 1

1

2

2

100 (0.2) 0.198 1 101

10 (0.198)(40) 2.08 2.08 0.7 1.38

1100 (1.38) 1.37 101

FEC E

FE

C

E

C

hI I mAh

V V

I mA

I mA

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠= − =

−= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

For 1 :Q

1

1

(100)(0.026) 13.1 0.198

0.198 7.62 /0.026m

r k

g mA V

π = = Ω

= =

For 2:Q

2

2

(100)(0.026) 1.90 1.37

1.37 52.7 /0.026m

r k

g mA V

π = = Ω

= =

(b)

1 1 1

1

eS

S F

V V V VIR r R

π π π

π

−= + + (1)

2 21 1

1 2

0em

C

V V Vg VR r

π ππ

π

++ + = (2)

2 12 2

2

e em

E F

V V V Vg Vr R R

π ππ

π

−+ = + (3)

Substitute numerical values in (1), (2), and (3):

1

1

1 1 1 110 13.1 10 10

(0.2763) (0.10)

S e

S e

I V V

I V V

π

π

⎛ ⎞ ⎛ ⎞= + + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − (1)

1 2

1 2

1 1 1(7.62) 040 1.90 40

(7.62) (0.5513) (0.025) 0

e

e

V V V

V V V

π π

π π

⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+ + = (2)

2 1

2 1

1 1 1 152.71.90 1 10 10

(53.23) (1.10) (0.10)

e

e

V V V

V V V

π π

π π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − (3) From (3),

2 1(48.39) (0.0909)eV V Vπ π= + Substituting into (1),

[ ]1 2 1(0.2763) (0.10) (48.39) (0.0909)SI V V Vπ π π= − + or

1 2(0.2672) (4.839)SI V Vπ π= − (1′) and substituting into (2),

( ) ( ) ( )1 2 2 1(7.62) (0.5513) 0.025 48.39 0.0909 0V V V Vπ π π π+ + + =⎡ ⎤⎣ ⎦ or

1 2 1 2(7.622) (1.761) 0 (0.2310)V V V Vπ π π π+ = ⇒ = − (2′) Then substituting (2′) into (1′), we obtain

2 2(0.2672)( )(0.2310) (4.839)SI V Vπ π= − − or

2 (4.901)SI Vπ= − Now

22 2

2

2 22(52.7) (42.16)

2 0.5

CO m

C L

RI g V

R R

V V

π

π π

⎛ ⎞= − ⎜ ⎟+⎝ ⎠

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠

Then

(42.16)4.901

SO

II −⎛ ⎞= − ⎜ ⎟⎝ ⎠

or

8.60Oif

s

IAI

= =

(c) 1i

S

VRIπ= and ||i S ifR R R=

We had 1 2 (0.2310)V Vπ π= − and 2 (4.901)SI Vπ= −

so 1

1(4.901) (21.22)0.2310S

VI Vππ

−⎛ ⎞= − =⎜ ⎟⎝ ⎠

Then 1 1 0.04713

21.22iS

VRIπ= = =

Finally 10

0.04713 47.4 10

ifif

if

RR

R= ⇒ = Ω

+

12.45 (a)

(1) 1 1 2

1 1

ei

FS B

V V VIRR R r

π π

π

−= +

(2) 1 21 1

1 2 2

0Cm

C B

V Vg V

R R rπ

ππ

+ + =

(3) 2 2 2 12 2

2 2

e em

E F

V V V Vg Vr R R

π ππ

π

−+ = +

(4) 22 2

2

( ) Co m

C L

RI g V

R Rπ⎛ ⎞

= − ⎜ ⎟+⎝ ⎠

Now

(1)′ 1 2

1 1

ei

FS B F

V VIRR R r R

π

π

= −

So 2 11 1

Fe i F

S B F

RV V I RR R r R π

π

⎛ ⎞= ⎜ ⎟ −⎜ ⎟⎝ ⎠

Now, from (2) 2 2 2

1 11 2 2

0em

c B

V V Vg V

R R rπ π

ππ

++ + =

(2)′ 2 21 1

2 1 21 2 2

1 0em

C BC B

V Vg V

r R RR R rπ

ππ π

⎛ ⎞+ + + =⎜ ⎟

⎝ ⎠

Also

(3)′ 12 2 2

2 2

1 1 1m e

F E F

Vg V V

r R R Rπ

ππ

⎛ ⎞ ⎛ ⎞+ + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

And

(4)′ 22

2 2

O C L

m C

I R RV

g Rπ⎛ ⎞⎛ ⎞+

= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Substitute (1)′ into (2)′ and (3)′

(2)″ 21 1 1

2 1 21 2 2 1 1

1 1 0Fm i F

C BC B S B F

V Rg V V I Rr R RR R r R R r R

ππ π

π π π

⎡ ⎤⎛ ⎞ ⎢ ⎥+ + + − − =⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

(3)″ 12 2 1

2 2 1 1

1 1 1 Fm i F

F E F S B F

V Rg V V I Rr R R R R R r R

ππ π

π π

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

Solve for 1Vπ from (2)″ and substitute into (3)″. Also use Equation (4)′. (b) 1 1 2 20 80 16 KBR R R= = =

120 (10) 2 V

100THV ⎛ ⎞= =⎜ ⎟⎝ ⎠

1

1

2 0.7 0.0111 mA16 (101)(1)1.11 mA

BQ

CQ

I

I

−= =+

=

2

2

15 85 12.75K

15 (10) 1.5V100

TH

TH

R

V

= =

⎛ ⎞= =⎜ ⎟⎝ ⎠

2

2

1.5 0.7 0.01265 mA12.75 (101)(0.5)1.265 mA

BQ

CQ

I

I

−= =+

=

1

2

1.11 42.69 mA/V0.0261.265 48.65 mA/V0.026

m

m

g

g

= =

= =

1

2

(100)(0.026) 2.34K1.11

(100)(0.026) 2.06K1.265

r

r

π

π

= =

= =

Now 1 2 2 12.75 1.729KC BR R = =

1 1 1 1 16 2.34 10 1.695KS B F B FR R r R R r Rπ π≅ = =

1 2 2 1.729 2.06 0.940 KC BR R rπ = = Now

(2)″ 21 1

1 2

1 1 1042.69 (10)2.06 0.940 1.729 1.695

46.587 1.064 5.784 0

i

i

VV V I

V V I

ππ π

π π

⎛ ⎞ ⎡ ⎤+ + + ⋅ −⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦+ − =

(3)″ 12 1

2 1

101 1 1 10 (10)2.06 10 0.5 10 1.695

49.03 12.29 21

i

i

VV V I

V V I

ππ π

π π

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

From (2)″ 1 2(0.1242) (0.02284)iV I Vπ π= − Then (3)″ [ ]2 249.03 12.29 (0.1242) (0.02284) 21i iV I V Iπ π= − −

249.31 19.47 iV Iπ = −

From (4)′ 24 4 (0.0411)

48.65 4o

oIV Iπ

+⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Then [ ](49.31) (0.0411) 19.47o iI I− = −

9.61oi

i

I AI

= =

12.46 a.

1

13.5 || 38.3 9.98 k13.5 (10) 2.606 V

13.5 38.3(120)(2.606 0.7) 1.75 mA

9.98 (121)(1)

TH

TH

C

R

V

I

= = Ω

⎛ ⎞= =⎜ ⎟+⎝ ⎠−= =

+

( )( )1

2

10 1.75 3 4.75 V4.75 0.7 0.50 mA

8.1

C

C

V

I

= − =−≈ =

1

1

(120)(0.026) 1.78 k1.751.75 67.31 mA / V0.026m

r

g

π = = Ω

= =

2

2

(120)(0.026) 6.24 k0.500.50 19.23 mA / V0.026m

r

g

π = = Ω

= =

b.

1 1 1 2

1||S e

S B F

V V V V VR R r R

π π π

π

− −= + (1)

2 2 21 1

1 2

0em

C

V V Vg VR r

π ππ

π

++ + = (2)

2 2 2 12 2

2 2m

E F

V V V Vg Vr R R

π ε ε ππ

π

−+ = + (3)

and

0 2 2 2( )m CV g V Rπ= − (4) Substitute numerical values in (1), (2), and (3)

1 21

1 2

1 10.6 0.6 9.98 ||1.78 1.2 1.2

(1.67) (4.011) (0.8333)

S e

S e

V V VV

V V V

ππ

π

⎡ ⎤= + + −⎢ ⎥⎣ ⎦= − (1)

21 2

1 1(67.31) 03 6.24 3

eVV Vπ π⎛ ⎞+ + + =⎜ ⎟⎝ ⎠

or 1 2 2(67.31) (0.4936) (0.3333) 0eV V Vπ π+ + = (2)

2 2 21

1 19.236.24 8.1 1.2 1.2

e eV V VV ππ⎛ ⎞+ = + −⎜ ⎟⎝ ⎠

or 2 2 1(19.39) (0.9568) (0.8333)eV V Vπ π= − (3)

From (1) 2 1(4.813) (2.00)e SV V Vπ= −

Then 1 2 1(67.31) (0.4936) (0.3333)[ (4.813) (2.00)] 0SV V V Vπ π π+ + − =

or 1 2(68.91) (0.4936) (0.6666) 0SV V Vπ π+ − = (2′)

and [ ]2 1 1(19.39) (0.9568) (4.813) (2.00) (0.8333)SV V V Vπ π π= − −

or 2 1(19.39) (3.772) (1.914)SV V Vπ π= − (3′)

We find 1 2(0.009673) (0.007163)SV V Vπ π= −

Then 2 2(19.39) (3.772)[ (0.009673) (0.007163)] (1.914)S SV V V Vπ π= − −

2 (19.42) ( 1.878)SV Vπ = − or 2 (0.09670)SV Vπ = − so that

0 (19.23)(4)( )(0.09670)SV V= − − Then

0 7.44S

VV

=

12.47

Using the circuit from Problem 12.32, we have 1if

S

VRIπ= .

Where 1SS

S

V VIR

π−=

From Problem 12.32 1 2(0.009673) (0.007163)

(0.009673) (0.007163)( )(0.09670)(0.01037)

S

S S

S

V V VV VV

π π= −= − −=

So (0.01037) (0.6) 0.00629 k

(0.01037)S

ifS S

VRV V

⋅= = Ω

−

or 6.29 ifR = Ω

12.48

1

1

1.4 ||17.9 1.298 k1.4 (10) 0.7254 V

1.4 17.90.7254 0.7 0.0196 mA

1.298(50)(0.0196) 0.98 mA

TH

TH

B

C

R

V

I

I

= = Ω

⎛ ⎞= =⎜ ⎟+⎝ ⎠−= =

= =

Neglecting dc base currents, 2

2

2

10 (0.98)(7) 3.14 V3.14 0.7 3.25 mA0.25 0.550 (3.25) 3.19 mA51

B

E

C

V

I

I

= − =−= =+

⎛ ⎞= =⎜ ⎟⎝ ⎠

1

1

(50)(0.026) 1.33 k0.980.98 37.7 mA / V

0.026m

r

g

π = = Ω

= =

2

2

(50)(0.026) 0.408 k3.193.19 123 mA / V0.026m

r

g

π = = Ω

= =

1 1 1

1 2|| ||SF

V V VIR R r R

π π

π 1

−= + (1)

2 2 21 1

2 1

0em

C

V V Vg Vr R

π ππ

π

++ + = (2)

2 2 12 2

2 1

em

E

V V Vg Vr R

ππ

π

−+ = (3)

2 1 1 11

1 2

e

E E F

V V V VVR R R

π π− −= + (4)

Enter numerical values in (1), (2), (3) and (4): 1 1 1

17.9 ||1.4 ||1.33 5SV V VI π π −

= +

or

1 1(1.722) (0.20)SI V Vπ= − (1)

2 2 21(37.7) 0

0.408 7eV V V

V π ππ

++ + =

or 1 2 2(37.7) (2.594) (0.1429) 0eV V Vπ π+ + = (2)

2 2 12(123)

0.408 0.25eV V V

Vππ

−+ =

or 2 2 1(125.5) (4) (4)eV V Vπ = − (3)

2 1 1 11

0.25 0.50 5eV V V VV π− −

= +

or 2 1 1(4) (6.20) (0.20)eV V Vπ= − (4)

From (4): 2 1 1(1.55) (0.05)eV V Vπ= −

Then substituting in (3): 2 1 1 1(125.5) (4)[ (1.55) (0.05)] (4)V V V Vπ π= − −

or 2 1 1(125.5) (2.20) (0.20)V V Vπ π= − (3′)

and substituting in (2): ( ) ( ) ( ) ( ) ( )1 2 1 137.7 2.594 0.1429 1.55 0.05 0V V V Vπ π π+ + − =⎡ ⎤⎣ ⎦

or 1 2 1(37.69) (2.594) (0.2215) 0V V Vπ π+ + =

Now 1 1 2(170.16) (11.71)V V Vπ π= − −

Then substituting in (1): 1 1 2(1.722) (0.20)[ (170.16) (11.71)]SI V V Vπ π π= − − −

or 1 2(35.75) (2.342)SI V Vπ π= +

and substituting in (3′): 2 1 2 1(125.5) (2.20)[ (170.16) (11.71)] (0.20)V V V Vπ π π π= − − −

or 2 1

1 2

(151.3) (374.55)(0.4040)

V VV V

π π

π π

= −= −

so that

Then 2 2

2

(35.75)[ (0.4040)] (2.342)(12.10)

S

S

I V VI V

π π

π

= − += −

20 2 2

2

2 2

( )

2.2(123) (64.43)2.2 2

Cm

C L

RI g V

R R

V V

π

π π

⎛ ⎞= − ⎜ ⎟+⎝ ⎠

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠

or 2 0(0.01552)V Iπ = − Then

0 01 5.33(0.01552)(12.10)S S

I II I

= ⇒ =

12.49

For example, use the circuit shown in Figure P12.30 12.50

1 2 3

1 2

3

6.24 k , 3.12 k , 1.56 k19.23mA / V 38.46 mA / V,76.92mA / V

m m

m

r r rg , gg

π π π= Ω = Ω = Ω= ==

1 1S eV V Vπ= + (1)

1 1 1 31 1

1 1

e e em

E F

V V V Vg Vr R R

ππ

π

−+ = + (2)

2 1 1 1 2( || )m CV g V R rπ π π= − (3)

3 3 32 2

2 3

0em

C

V V Vg VR r

π ππ

π

++ + = (4)

3 3 3 13 3

3 2

e e em

E F

V V V Vg Vr R R

ππ

π

−+ = + (5)

Enter numerical values in (2)-(5): 1

1 1 31 1 1(19.23)

6.24 0.1 0.8 0.8e eV V V Vπ

π⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 1 1 3(19.39) (11.25) (1.25)e eV V Vπ = − (2)

2 1 1(19.23) (5 || 3.12) (36.94)V V Vπ π π= − = − (3)

2 3 31 1 1(38.46) 02 1.56 2eV V Vπ π

⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 2 3 3(38.46) (1.141) (0.5) 0eV V Vπ π+ + = (4)

3 3 31 1 1 176.92

1.56 0.1 0.8 0.8e eV V Vπ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

or 3 3 1(77.56) (11.25) (1.25)e eV V Vπ = − (5)

From (1) 1 1S eV V Vπ = − Then

1 1 3( )(19.39) (11.25) (1.25)S e e eV V V V− = − or 1 3(19.39) (30.64) (1.25)S e eV V V= − (2′)

2 1(36.94) (36.94)S eV V Vπ = − + (3′)

1 3 3(38.46)[ (36.94) (36.94)] (1.141) (0.5) 0S e eV V V Vπ− + + + = (4′) From (5): 3 3 1(6.894) (0.1111)e eV V Vπ= + Then

1 3 1(19.39) (30.64) (1.25)[ (6.894) (0.1111)]S e eV V V Vπ= − +

or 1 3(19.39) (30.50) (8.6175)S eV V Vπ= − (2″)

and 1 3 3 1(1420.7) (1420.7) (1.141) (0.5)[ (6.894) (0.1111)] 0S e eV V V V Vπ π− + + + + =

or 1 3(1420.7) (1420.76) (4.588) 0S eV V Vπ− + + = (4″)

From (2″): 1 3(0.6357) (0.2825)e SV V Vπ= +

Then substituting in (4″): 3 3

3

(1420.7) (1420.76)[ (0.6357) (0.2825)] (4.588) 0(517.5) (405.95) 0

S S

S

V V V VV V

π π

π

− + + + =− + =

Now 0 3 3 376.92mI g V Vπ π= = or 3 0 (0.0130)V Iπ =

Then 0(517.5) (0.0130)(405.95) 0SV I− + = or

0 98.06 mA / VS

IV

=

12.52

1 2

1 2

3

3

(100)(0.026) 5.2 k0.5

0.5 19.23 mA / V0.026

(100)(0.026) 1.3 k2

2 76.92 mA / V0.026

m m

m

r r

g g

r

g

π π

π

= = = Ω

= = =

= = Ω

= =

1 21 1 2 2

1 2

0m mV Vg V g Vr r

π ππ π

π π

+ + + = (1)

Since 1 2r rπ π= and 1 2 ,m mg g= then 1 2V Vπ π= −

1 2 3 2 32S e eV V V V V Vπ π π= − + = − + (2)

3 3 32 2

3 2

0em

C

V V Vg Vr R

π ππ

π

++ + = (3)

3 3 23 3

3 2

em

F

V V Vg Vr R r

π ππ

π π

+ = + (4)

30 3 3

3

( )Cm

C L

RI g V

R R π⎛ ⎞

= −⎜ ⎟+⎝ ⎠ (5)

From (2): 3 22e SV V Vπ= +

(19.23) 3 32 2

1 ( 2 ) 01.3 18.6 18.6 SV V

V V Vπ ππ π+ + + + =

or 2 3(19.23) (0.8230) (0.05376) 0SV V Vπ π+ + = (3′)

23 2

1 176.92 ( 2 )1.3 10 5.2S

VV V V ππ π⎛ ⎞ ⎛ ⎞+ = + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 3 2(77.69) (0.3923) (0.1) SV V Vπ π= + (4′)

0 3 32 (76.92) (51.28)

2 1I V Vπ π

⎛ ⎞= − = −⎜ ⎟+⎝ ⎠ (5′)

From (3′): 2 3(0.04255) (0.002780) SV V Vπ π= − −

Then 3 3

3

(77.69) (0.3923)[ (0.04255) (0.002780) ] (0.1)(77.71) (0.0989)

S S

S

V V V VV V

π π

π

= − − +=

or 3 (0.001273) SV Vπ =

so that 0 (51.28)(0.001273) SI V= −

or 0 (0.0653) mA/VS

IV

= −

12.52

3

2

4 5

5 10

1 110 10

Af fj j

×=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1 14 5Phase tan 2 tan

10 10f fφ − −⎛ ⎞ ⎛ ⎞= = − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

By trial and error, when 51.095 10 , 180f Hz φ= × ≅ ° For 1T = at 51.095 10 ,f Hz= ×

( ) ( )( )( )

3 33

2 2

4 5

5 10 5 101 1 4.84 10

10.996 2.1991 1

10 10f f

β ββ −

× ×= ⇒ = ⇒ = ×

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

12.53 Use the basic circuit shown in Figure 12.27.

For the ideal case

0 1

i E

IV R

=

we want 30 10 A/V 1 mA/V

i

IV

−= =

Set 1 kER = Ω Since the op-amp has a finite gain, finite input resistance, and finite output resistance, the closed-loop gain is slightly less than the ideal. ER will need to be slightly decreased to increase the gain. 12.54 dc analysis

(on) 05 0.7 0.0343

100 (51)(0.5)(50)(0.0343) 1.71 mA

E E EB B B CC

B

C

I R V I R V

I

I

+ + + =−= =

+= =

Then (50)(0.026) 0.760 k1.71

rπ = = Ω

1.71 65.77 mA/V0.026mg = =

a.

To determine :ifR

0 ( ) 0||S

B F

V V VIR r R

π π

π

− −+ + = (1)

0 0 ( )m

C F

V V Vg VR R

ππ

− −= + (2)

Now from (2):

0

0

1 1(65.77)10 1 10

(65.67) (1.10)

VV V

V V

ππ

π

⎛ ⎞− = +⎜ ⎟⎝ ⎠

=

or 0 (59.7)V Vπ=

and from (1): 0 0

100 0.760 10 10(0.8543) (0.1)(59.7) 0

(6.824)

S

S

S

V V VI

I V VI V

π π

π π

π

+ + + =

+ + == −

Now ( ) 147 if if

S

VR RI

π−= ⇒ = Ω

To determine 0 :fR

X XX m

C F B

V VI g VR R R r π

π

= + −+

(3)

( )( )

( )B

XB F

R rV V

R r Rπ

ππ

⎛ ⎞−= ⎜ ⎟⎜ ⎟+⎝ ⎠

(4)

Now (100 0.760)

( ) (0.07014)(100 0.760) 10 X XV V Vπ

⎛ ⎞−= = −⎜ ⎟⎜ ⎟+⎝ ⎠

so 1 1 (65.77)(0.07014)1 10.754X XI V ⎛ ⎞= + +⎜ ⎟⎝ ⎠

0 0 175 Xf f

X

VR RI

= ⇒ = Ω

b. From part (a), we find

6.824SI

Vπ = −

then

0 (59.7)6.824

SIV −⎛ ⎞= ⎜ ⎟⎝ ⎠

or

0 8.75 kS

VV

= − Ω

c. If capacitance is finite, a phase shift will be introduced. 12.55 dc analysis: GS DSV V=

2

2

2

( )

10 (0.20)(8)( 2)

10 1.6( 4 4)

DD GSD n GS TN

D

GS GS

GS GS GS

V VI K V V

RV V

V V V

−= = −

− = −

− = − +

2

2

1.6 5.4 3.6 0

5.4 (5.4) 4(1.6)(3.6)3.95 V

2(1.6)10 3.95 0.756 mA

82 2 (0.2)(0.756) 0.778 mA/V

GS GS

GS

D

m n D m

V V

V

I

g K I g

− − =

± += =

−= =

= = ⇒ =

a.

0

0

0

1 1

gs S gs

S F

Sgs

S F S F

V V V VR R

V VV

R R R R

− −+ =

⎛ ⎞+ = +⎜ ⎟

⎝ ⎠ (1)

00

0

0

1 1 1

gsm gs

D F

gs mD F F

V VV g VR R

V V gR R R

−+ + =

⎛ ⎞ ⎛ ⎞+ = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (2)

So from (1): 01 1

10 100 10 100S

gsV VV ⎛ ⎞+ = +⎜ ⎟

⎝ ⎠

or 0

0

(0.11) (0.10) (0.010)(0.909) (0.0909)

gs S

gs S

V V VV V V

= +

= +

Then from (2):

0

0

0

0

1 1 1 0.7788 100 100

(0.135) ( 0.768)( 0.768)[ (0.909) (0.0909)]

(0.2048) (0.6981)

gs

gs

S

S

V V

V VV V

V V

⎛ ⎞ ⎛ ⎞+ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −

= − += −

so 0 3.41vS

VAV

= = −

b. We have 0(0.909) (0.0909)

(0.909) (0.0909) (3.41 )0.599

gs S

S S

S

V V VV V

V

= +

= + −=

Now 0 0 ( 3.41 )

0.599S S

zfS gsS S S

S

V V V RAV VI V V

R

−= = =

− −

or ( 3.41)(10) 85.0 V/ma

0.401zf zfA A−= ⇒ = −

c. 0.401 0.599 (10) 14.9 k0.401

gs

gs S Sif if

S S S

VV V VR RI R V

= = = ⋅ ⇒ = Ω

d.

1010 100

(0.0909)

X XX m gs

D S F

Sgs X X

S F

X

V VI g VR R R

RV V VR R

V

= + ++

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

00

1 1(0.778)(0.0909)8 10 100

1 0.2048 4.88 k

X X

Xf

X f

I V

I RV R

⎡ ⎤= + +⎢ ⎥+⎣ ⎦

= = ⇒ = Ω

12.56

As 0 100, 1010

Fm

S S

V RgV R

− −→ ∞ = = = −

To be within 10% of ideal, 0 10(0.9) 9S

VV

= − = −

From Problem 12.41, we had 0(0.909) (0.0909)

(0.909) ( 9 )(0.0909)0.0909

gs S

S S

S

V V VV V

V

= +

= + −=

Also from Problem 12.41, we had 0 (0.135) (0.010 )gs mV V g= −

or ( 9 )(0.135) (0.0909) (0.010 )

1.215 0.000909 0.0909S S m

m

V V gg

− = −− = −

or 13.36 mA/Vmg =

12.57 dc analysis

24 ||150 20.7 k24 (12) 1.655 V

24 150

TH

TH

R

V

= = Ω

⎛ ⎞= =⎜ ⎟+⎝ ⎠

1.655 0.7 0.00556 mA20.7 (151)(1)BQI −= =

+

so 0.834 mACQI =

(150)(0.026) 4.68 k0.834

0.834 32.08 mA / V0.026m

r

g

π = = Ω

= =

0

||S

S B F

V V V V VR R r R

π π π

π

− −= + (1)

0 0 0mC F

V V Vg VR R

ππ

−+ + = (2)

From (1):

01 1 15 5 20.7 || 4.68S

F F

V VV

R Rπ⎡ ⎤

= + + −⎢ ⎥⎣ ⎦

or

01(0.20) 0.4620SF F

VV V

R Rπ⎛ ⎞

= + −⎜ ⎟⎝ ⎠

From (2):

01 1 132.08 0

6F F

V VR Rπ

⎛ ⎞ ⎛ ⎞− + + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

so

010.1667

132.08

F

F

VR

V

R

π

⎛ ⎞− +⎜ ⎟

⎝ ⎠=⎛ ⎞

−⎜ ⎟⎝ ⎠

(2)

Then

00

10.16671(0.20) 0.4620

132.08

FS

F F

F

VR V

VR R

R

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

Neglect the FR in the denominator term. Now

0 005 (0.20)

5SS

V VV VV

= − ⇒ = − = −

( )00 0

2

2

0.1667 1(0.20)(0.20) (0.4620 1)

32.08

1.283 (0.4620 1)(0.1667 1) 32.081.206 32.71 1 0

FF F

F

F F F F

F F

V RV R R V

R

R R R RR R

− +⎡ ⎤− = + −⎢ ⎥

⎣ ⎦− = − + + −

− − =

232.71 (32.71) 4(1.206)(1)2(1.206)FR

± +=

so that 27.2 kFR = Ω

12.58 dc analysis

4 ||15 3.16 k4 12 2.526 V

4 15

TH B

TH

R R

V

= = Ω =

⎛ ⎞= =⎜ ⎟+⎝ ⎠

2.526 0.7 0.002513.16 (181)(4)0.452 mA

BQ

CQ

I

I

−= =+

=

(180)(0.026) 10.4 k0.452

0.452 17.4 mA/V0.026m

r

g

π = = Ω

= =

1 1 1 0

||i

S B F

V V V V VR R r R

π π π

π

− −= + (1)

21 0

|| ||mC B

Vg VR R r

ππ

π

+ = (2)

32 0

|| ||mC B

Vg VR R r

ππ

π

+ = (3)

0 0 0 13 0m

C L F

V V V Vg VR R R

ππ

−+ + + = (4)

Now || || 8 || 3.16 ||10.4 1.86 k|| 3.16 ||10.4 2.42 k

C B

B

R R rR r

π

π

= = Ω= = Ω

Now substituting in (2): 2

1(17.4) 01.86V

V ππ + = or 2 1(32.36)V Vπ π= −

and substituting in (3): 3

2

31

(17.4) 01.86

(17.4)[ (32.36) ] 01.86

VV

VV

ππ

ππ

+ =

− + =

or 3 1(1047.3)V Vπ π= Substitute numerical values in (1):

01

1 1 110 10 2.42

i

F F

V VV

R Rπ⎛ ⎞

= + + −⎜ ⎟⎝ ⎠

or

01

1(0.10) 0.513iF F

VV V

R Rπ⎛ ⎞

= + −⎜ ⎟⎝ ⎠

Substitute numerical values in (4):

11 0

41 0

0

14

1 1 1(17.4)(1047.3) 08 4

1 11.822 10 0.375 0

10.375

11.822 10

F F

F F

F

F

VV V

R R

V VR R

VR

V

R

ππ

π

π

⎛ ⎞+ + + − =⎜ ⎟

⎝ ⎠⎛ ⎞ ⎛ ⎞

× − + + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞− +⎜ ⎟

⎝ ⎠=× −

so that

00

4

10.3751(0.10) 0.513

11.822 10

Fi

F F

F

VR V

VR R

R

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

We have 0 80i

VV

= − or 0

80iV

V = −

4

10.375(0.10) 1 10.513

180 1.822 10

F

F F

F

RR R

R

⎡ ⎤⎛ ⎞− +⎢ ⎥⎜ ⎟⎛ ⎞ ⎝ ⎠⎢ ⎥− = + −⎜ ⎟ ⎢ ⎥⎝ ⎠ × −⎢ ⎥

⎢ ⎥⎣ ⎦

Neglect that 1/ FR term in the denominator.

4

2 4

(0.513 1)(0.375 1)(0.00125 ) 11.822 10

22.775 (0.513 1)(0.375 1) 1.822 10

F FF

F

F F F F

R RRR

R R R R

+ +− = − −

×

= + + + ×

We find 2 422.58 1.822 10 1 0F FR R− × − =

4 4 21.822 10 (1.822 10 ) 4(22.58)(1)2(22.58)FR

× ± × +=

or 0.807 MFR = Ω

12.59

a. 1( )S d d

S F

V V V VR R

− − − −=

or

11 1 0Sd

S F S F

V VVR R R R

⎛ ⎞+ + + =⎜ ⎟

⎝ ⎠ (1)

0 1 11

1 2

( )d

F

V V V VVR R R− − −

= +

or

01

1 1 2

1 1 1 d

F F

V VV

R R R R R⎛ ⎞

= + + +⎜ ⎟⎝ ⎠

(2)

and 0 0L dV A V= or 0

0d

L

VVA

=

Substitute numerical values in (1) and (2): 0 14

1 1 05 10 5 1010

SV V V⎛ ⎞⋅ + + + =⎜ ⎟⎝ ⎠

or 4

0 1(0.3 10 ) (0.20) (0.10) 0SV V V−× + + = (1)

0 01 4

1 1 1 150 50 10 10 1010V VV ⎛ ⎞ ⎛ ⎞= + + + ⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

or 5

0 1(0.02 10 ) (0.22)V V−− = (2)

Then 5

1 00.02 10

0.22V V

−⎛ ⎞−= ⎜ ⎟⎝ ⎠

and 5

40 0

4 50

0.02 10(0.3 10 ) (0.20) (0.10) 00.22

0.3 10 0.4545 10 0.00909 (0.20) 0

S

S

V V V

V V

−−

− −

⎡ ⎤⎛ ⎞−× + + =⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

⎡ ⎤× − × + + =⎣ ⎦

Then 0 03

0.20 21.949.115 10S S

V VV V−

−= ⇒ = −×

b. ( )

d d Sif

S d S d

S

V V RRV V V V

R

− − ⋅= =

− − +

Now 04

0

21.9410

Sd

L

V VVA

= = −

Then 4

4

(21.94 10 )(5)1 21.94 10ifR

−

−

×=− ×

or 21.099 10 k 10.99 if ifR R−= × Ω ⇒ = Ω

c. Because of the 0L dA V source,

0 0fR =

12.60 For example, use the circuit shown in Figure 12.41 12.61 Break the loop

tI Iε=

Now 0 0

1

0i tF S i

V VA I

R R R R+ + =

+

0Sr

S i F S i

R VI

R R R R R⎛ ⎞

= ⋅⎜ ⎟+ +⎝ ⎠

or 0 ( )S ir F S i

S

R RV I R R R

R⎛ ⎞+

= ⋅ +⎜ ⎟⎝ ⎠

Then

1

1 1 ( ) 0S ii t r F S i

F S i S

R RA I I R R R

R R R R R⎛ ⎞ ⎡ ⎤⎛ ⎞+

+ + × + =⎜ ⎟ ⎢ ⎥⎜ ⎟⎜ ⎟+ ⎝ ⎠⎣ ⎦⎝ ⎠

1

1 1 ( )

ir

t S iF S i

F S i S

AIT TI R R R R R

R R R R R

= − ⇒ =⎡ ⎤ ⎛ ⎞++ +⎢ ⎥ ⎜ ⎟+ ⎝ ⎠⎣ ⎦

12.62

1 1 1 01 1

1 1m

E F

V V V Vg Vr R R

π ε επ

π

−+ = + (1)

1 1 1 1 1 21 2

0 ( )( )rm r m C

C

Vg V V g V R rR rπ π π

π

+ = ⇒ = − (2)

2 tV Vπ = so that

3 0 32

2 3

0m tC

V V Vg VR r

π π

π

++ + = (3)

3 0 0 13 3

3 3m

E F

V V V Vg Vr R R

π επ

π

−+ = + (4)

From (4):

10 3 3

3 3

1 1 1m

E F F

VV V g

R R r Rε

ππ

⎛ ⎞ ⎛ ⎞+ = + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

But 1 1V Vε π= −

so

13 3

30

3

1

1 1

mF

E F

VV gr R

V

R R

ππ

π

⎛ ⎞+ −⎜ ⎟

⎝ ⎠=⎛ ⎞

+⎜ ⎟⎝ ⎠

Then

13 3

31 1

1 1

3

11 1 1

1 1

mF

mE F

FE F

VV gr R

V gr R R

RR R

ππ

ππ

π

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

⎛ ⎞⎝ ⎠⎝ ⎠⎣ ⎦ ⋅ +⎜ ⎟⎝ ⎠

(1′)

and

13 3

32 3

2 32

3

11 1 0

1 1

mF

m tC

CE F

VV gr R

g V VR r

RR R

ππ

ππ

π

⎛ ⎞+ −⎜ ⎟⎛ ⎞ ⎝ ⎠+ + + =⎜ ⎟

⎛ ⎞⎝ ⎠ ⋅ +⎜ ⎟⎝ ⎠

(3′)

From (3′), solve for 3Vπ and substitute into (1′). Then from (1′), solve for 1Vπ and substitute into (2). Then

.r

t

VTV

= −

12.63

1

0r er r

S F

V VV VR r Rπ

−+ + = (1)

2 21

1 2

0em t

C

V V Vg VR r

π π

π

++ + = (2)

22 2

2

e e rm

E F

V V V Vg Vr R R

ππ

π

−+ = + (3)

Using the parameters from Problem 12.29, we obtain 1 1 1 0

10 15.8 10 10e

rVV ⎛ ⎞+ + − =⎜ ⎟

⎝ ⎠

or (0.2633) (0.10)r eV V= (1)

21 1(7.62) 040 2.28 40

et

VV Vπ⎛ ⎞+ + + =⎜ ⎟⎝ ⎠

or 2(7.62) (0.4636) (0.025) 0t eV V Vπ+ + = (2)

21 1 152.7

2.28 1 10 10r

eVV Vπ

⎛ ⎞ ⎛ ⎞+ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 2 (53.14) (1.10) (0.10)e rV V Vπ = −

Then 2 (0.0207) (0.001882)e rV V Vπ = − (3)

Substituting in (2): (7.62) (0.4636)[ (0.0207) (0.001882)] (0.025) 0t e r eV V V V+ − + =

or (7.62) (0.03460) (0.0008725) 0t e rV V V+ − =

From (1) (2.633)e rV V= Then

(7.62) (2.633)(0.03460) (0.0008725) 0(7.62) (0.09023) 0

t r r

t r

V V VV V

+ − =+ =

or 84.45r

t

VV

= −

Now

84.45r

t

VT TV

= − ⇒ =

12.64

tV Vπ = −

0 0m

C F B

V Vg V

R R R rππ

= ++

(1)

and

0B

rB F

R rV V

R r Rπ

π

⎛ ⎞= ⎜ ⎟⎜ ⎟+⎝ ⎠

(2)

Now

01 1(65.77)1 10 100 0.760

V Vπ

⎛ ⎞= +⎜ ⎟⎜ ⎟+⎝ ⎠

or 0(65.77) (1.0930)V Vπ = and

0 00.754 (0.07011)

10 0.754rV V V⎛ ⎞= =⎜ ⎟+⎝ ⎠

so 0 (14.26) rV V= Then (65.77)( ) (14.26) (1.0930)t rV V− =

4.22r

t

VV

= − so that 4.22T =

12.64 Want 1 12f MHz= for a phase margin of 45°

( ) ( ) 40 80 0 10dBT dB T= ⇒ = Then

( ) ( )

6

0

1 112 10PD

TT f

f fj jf

=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟×⎝ ⎠⎝ ⎠

Set 1f f= and 1T = So

4

26

10112 101 2

PD

T

f

= =⎛ ⎞×+ ⋅⎜ ⎟⎝ ⎠

which yields 6 412 10 10 1.70

2 PDPD

f kHzf× = ⇒ =

12.65

a. 1 12 4tan 2 tan

5 10 10f fφ − −⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

or 1 1 4180 180

1802 4180 tan 2 tan 1.05 10 Hz5 10 10

f f f− −⎛ ⎞ ⎛ ⎞− = − − ⇒ ≈ ×⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

b. 5

180 2 24 4

2 4

(10 )( ) 11.05 10 1.05 101 1

5 10 10

T f β= =⎡ ⎤⎛ ⎞ ⎛ ⎞× ×+ +⎢ ⎥⎜ ⎟ ⎜ ⎟× ⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

5(10 )1(21.02)(2.105)

β= or

44.42 10β −= × 12.65

( )

40 0

0

0

80dB 10

01f

A AAA

Aβ

= ⇒ =

=+

or ( )4

4

105 0.21 10

ββ

= ⇒ ≈+

Then ( ) 400 0.2 10T Aβ= = ×

Inserting a dominate pole

1 1 16 7tan tan tan

10 10PD

f f ff

φ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

If we want a phase margin of 45°, then 1 1

6 7135 90 tan tan10 10

f f− −⎛ ⎞ ⎛ ⎞− ° ≈ − ° − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

By trial and error, 0.845MHzf ≈ Then

4

2 226

0.2 10 110.8450.845 10 0.845 11 1 101PD

T

f

×= = ×⎛ ⎞ ⎛ ⎞× ⎛ ⎞ ++ + ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

( )( )6 40.845 10 0.2 10

1.309 1.0036PDf× ×≈

so 555HzPDf = 12.66

(a) 3

2

4 5

(10 )

1 110 10

vT Af fj j

ββ= =⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1 14 52 tan tan

10 10f fφ − −= − −

Set 180φ = − ° By trial and error, 44.58 10 Hzf = × (b) Set 1T = at 44.58 10 Hzf = ×

3

22 24 4

4 5

3

(10 )14.58 10 4.58 101 1

10 10

(10 )1 0.02417(21.976)(1.10)

β

β β

=⎛ ⎞⎛ ⎞ ⎛ ⎞× ×⎜ ⎟+ +⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= ⇒ =

(c) 3

3

10 39.71 (10 )(0.02417)ovfA = =

+

(d) Wait 1T < at 44.58 10 Hz,f = × so system is stable for smaller values of .β 12.67

1 1 14 4 5tan tan tan

10 5 10 10f f fφ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

At 48.1 10 Hz,f = × 180.28φ = − ° Determine ( )T f at this frequency.

3

2 2 24 4 4

4 4 5

3

1 1 1(10 )8.1 10 8.1 10 8.1 101 1 1

10 5 10 10

(10 )(8.161)(1.904)(1.287)

T β

β

= × × ×⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

a. For 0.005β = ( ) 0.250 1 StableT f = < ⇒

b. For 0.05β = ( ) 2.50 1 UnstableT f = > ⇒

12.68 (b)Phase margin 80 100φ= °⇒ = − °

1 13 4100 2 tan tan

10 5 10f fφ − −⎛ ⎞ ⎛ ⎞= − = − −⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

By trial and error, 31.16 10 f Hz= ×

Then 3

22 23 3

3 4

34

(5 10 )11.16 10 1.16 101 1

10 5 10

(5 10 ) 4.7 10(2.35)(1.00)

T β

β β −

×= =⎛ ⎞⎛ ⎞ ⎛ ⎞× ×⎜ ⎟+ ⋅ +⎜ ⎟ ⎜ ⎟⎜ ⎟ ×⎝ ⎠ ⎝ ⎠⎝ ⎠

×= ⇒ = ×

12.69

c. For 0.005,β =

( ) 1(0 dB)T f = at 42.10 10 Hzf ≈ × Then

4 4 41 1 1

3 4 5

2.10 10 2.10 10 2.10 10tan tan tan10 10 10

87.27 64.54 11.86

φ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − − −

or 163.7φ = −

System is stable. Phase margin 16.3= ° For 0.05,β =

( ) 1 (0 dB)T f = at 46.44 10 Hzf ≈ ×

Then 4 4 4

1 1 13 4 5

6.44 10 6.44 10 6.44 10tan tan tan10 10 10

89.11 81.17 32.78

φ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − − −

or 203.1 System is unstable.φ = − °⇒

12.70

5

4 5 5

(10 )

1 1 15 10 10 5 10

T Af f fj j j

ββ= =⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟× ×⎝ ⎠⎝ ⎠⎝ ⎠

Phase Margin 60 120φ= °⇒ = − ° So

1 1 14 5 5120 tan tan tan

5 10 10 5 10f f f− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

By trial and error, at 510 ,f Hz= 120φ ≅ − ° Then

( )( )( )( )

5

2 2 25 5 5

4 5 5

55

(10 )110 10 101 1 1

5 10 10 5 10

101 3.22 10

2.236 1.414 1.02

T β

ββ −

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ ⋅ + ⋅ +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= ⇒ = ×

12.71

(a) 5

35

10100 9.99 101 (10 )

ββ

−= ⇒ = ×+

3 5

2 2

3 5

2 2

3 5

5

(9.99 10 )(10 )1

1 110 10

9991

1 110 10

3.08 10 Hz

vT Af f

f f

f

β−×= = =

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ×

Phase 1 1

3 5

5 51 1

3 5

tan tan10 103.08 10 3.08 10tan tan

10 1089.81 72.01161.8

f fφ

φ

− −

− −

= − −

× ×= − −

= − −= −

Stable (b) Phase Margin 180 161.8 18.2= − = ° 12.72

(a) 1 1 14 6 7tan tan tan

5 10 10 5 10f f fφ − − −= − − −

× ×

For 180φ = − ° By trial and error, 6

180 7.25 10 Hzf = ×

(b) 5

2 2 26 6 6

4 6 7

4

(0.10)(10 )

7.25 10 7.25 10 7.25 101 1 15 10 10 5 10

0(145)(7.319)(1.01)9.33

T

T

=⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

1=

=

System is unstable 4

2 2 2

4 6 7

7

101

1 1 15 10 10 5 10

2.14 10 Hz

Tf f f

f

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= ×

7 7 71 1 1

4 6 7

2.14 10 2.14 10 2.14 10tan tan tan5 10 10 5 10

89.87 87.32 23.17200.4

φ

φ

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠= − − −= − °

(c) For 67.25 10 Hzf = × 5

2 2 26 6 6

4 6 7

(0.0010)(10 )

7.25 10 7.25 10 7.25 101 1 15 10 10 5 10

100(145)(7.319)(1.01)0.0933

T

T

=⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

=

System is stable.

2 2 2

4 6 7

6

1001

1 1 15 10 10 5 10

2.13 10 Hz

Tf f f

f

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= ×

6 6 6

1 1 14 6 7

2.13 10 2.13 10 2.13 10tan tan tan5 10 10 5 10

88.66 64.85 2.44155.9

φ

φ

− − −× × ×= − − −× ×

= − − −= − °

12.73

(a) 1 1180 1803 5

5180

180 tan 2 tan5 10 10

1.05 10 Hz

f f

f

φ − −= − = − −×

= ×

(b) 3

2 25 5

3 5

0180

(0.0045)(2 10 )

1.05 10 1.05 101 15 10 10

9(21.02)(2.1025)

0.204

T

T f

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

=

= =

System is stable

2 2

3 5

4

91

1 15 10 10

3.88 10 Hz

Tf f

f

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

= ×

4 41 1

3 5

3.88 10 3.88 10tan 2 tan5 10 10

82.66 42.41125.1

φ

φ

− −× ×= − −×

= − −= − °

(c) 3

2 25 5

3 5

(0.15)(2 10 )

1.05 10 1.05 101 15 10 10

300(21.02)(2.1025)6.79

T

T

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

=

=

System is unstable

2 2

3 5

5

3001

1 15 10 10

2.33 10 Hz

Tf f

f

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

= ×

5 51 1

3 5

2.33 10 2.33 10tan 2 tan5 10 10

88.77 133.54222.3

φ

φ

− −× ×= − −×

= − −= − °

12.74 Phase Margin 45 135φ= °⇒ = − °

1 1 1 13 4 5 6

135

tan tan tan tan10 10 10 10

f f f fφ

− − − −

= − °

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

At 410 Hz,f = 135.6φ = − °

3

2 24 4

3 4

2 24 4

5 6

3

11 1(10 )10 101 110 10

1 1

10 101 110 10

(10 )1(10.05)(1.414)(1.005)(1.00)

T

β

β

=

= × × ×⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

× ×⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

or 0.01428β =

12.75

3

6 6

1 1500011 300 10

1 1

1 12 10 25 10

PD

Tff jj

f

f fj j

= × ×⎛ ⎞ ⎛ ⎞++ ⎜ ⎟⎜ ⎟ ×⎝ ⎠⎝ ⎠

× ×⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠

Phase Margin 45 135φ= °⇒ = − ° at 300 kHzf = 3 3

1 13

300 10 300 10135 tan tan 0 0300 10

90 45PDf

− −⎛ ⎞ ⎛ ⎞× ×− ° = − − − −⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠= − ° − °

Now

23

2 23

3

1@ 300 kHz50001

300 101 2 1 1

300 10 500012

300 10 2 84.8 Hz5000

PD

PD

PD PD

T f

T

f

f

f f

= =

= ≈⎛ ⎞×+ ⋅ ⋅ ⋅⎜ ⎟⎝ ⎠

⎛ ⎞× ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

×≈ ⇒ =

12.76 (a) At 610 Hz,f =

6 61 1

4 6

10 10tan 2 tan10 10

89.4 90 180

φ − −= − −

= − − ≈ − °

3

2 26 6

4 6

3

10

10 101 110 10

10 5(100)(2)

T =⎡ ⎤⎛ ⎞ ⎛ ⎞

+ +⎢ ⎥⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= =

1T > at 6180 10 Hz,f f= = System is unstable.

(b) 3

2

4 6

10

1 1 110 10PD

Tf f fj j j

f

=⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

Phase margin 45 135φ= °⇒ = −

1 1 14 6135 tan tan 2 tan

10 10PD

f f ff

φ − − −= − = − − −

410 Hzf ≅ 3

2 2 24 4 4

4 6

3

4

4

3

10110 10 101 1 1

10 10

10110 (1.414)(1.0)

(10 )(1.414)10

14.14 Hz

PD

PD

PD

PD

T

f

f

f

f

= =⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ + +⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎣ ⎦

≅⎛ ⎞⎜ ⎟⎝ ⎠

=

=

12.77

(a) 1 1 1180 180 1804 4 5180 tan tan tan

10 5 10 10f f fφ − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

4180 8.06 10 Hzf ≅ ×

(b) 2 2 24 4 4

4 4 5

500

8.06 10 8.06 10 8.06 101 1 110 5 10 10

500(8.122)(1.897)(1.284)25.3

T

T

=⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=

=

(c)

4 4 5

500

1 1 1 110 5 10 10PD

Tf f f fj j j j

f

=⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞+ + + +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟×⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠

Phase Margin 60 120φ= °⇒ = − °

1 1 1 14 4 5120 tan tan tan tan

10 5 10 10PD

f f f ff

− − − −− = − − − −×

Assume 1tan 90PD

ff

− ≅ °

Then 34.2 10 Hzf ≅ ×

2 2 2 23 3 3 3

4 4 5

23

3

50014.2 10 4.2 10 4.2 10 4.2 101 1 1 1

10 5 10 10

50014.2 101 (1.085)(1.004)(1.0)

4.2 10 500(1.0846)(1.0035)(1.0)

9.14 Hz

PD

PD

PD

PD

T

f

f

ff

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × × ×+ + + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

=⎛ ⎞×+ ⎜ ⎟⎝ ⎠

× ≅

= 12.78

(a) 4

4

1050 0.01991 (10 )

ββ

= ⇒ =+

4

5

(0.0199)(10 )

1 110PD

Tf fj j

f

=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

Phase margin 45 135φ= °⇒ = − °

1 15135 tan tan

10PD

f ff

− −− = − −

510 Hzf = 4

2 25 5

5

5 4

(0.0199)(10 )110 101 1

10

10 (0.0199)(10 )1.414

711 Hz

PD

PD

PD

T

f

ff

= =⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

=

(b) 4

4

1020 0.04991 (10 )

ββ

= ⇒ =+

4

5

4

2 2

5

5

(0.0499)(10 )

1 1711 10

(0.0499)(10 )1

1 1711 10

1.76 10 Hz

Tf fj j

Tf f

f

=⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= =⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ×

5 51 1

5

1.76 10 1.76 10tan tan711 10

89.77 60.40150.2

φ

φ

− −⎛ ⎞ ⎛ ⎞× ×= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − −= −

Phase Margin 180 150.2 29.8= − = °

12.79 (a) 5100 dB 10O OA A= ⇒ =

5

5

1020 0.049991 (10 )

ββ

= ⇒ =+

5

6 7

(0.04999)(10 )

1 1 110 10PD

Tf f fj j j

f

=⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

Phase Margin 45 135φ= °⇒ = − °

1 1 16 7135 tan tan tan

10 10PD

f f ff

− − −− = − − −

610 Hzf ≈ 5

2 2 26 6 6

6 7

5

26

6 5

(0.04999)(10 )110 10 101 1 1

10 10

(0.04999)(10 )1101 (1.414)(1.005)

10 (0.04999)(10 )(1.414)(1.005)2.84 Hz

PD

PD

PD

PD

T

f

f

ff

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞

+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

=⎛ ⎞

+ ⎜ ⎟⎝ ⎠

=

=

(b) 5

5

105 0.199991 (10 )

ββ

= ⇒ =+

5

2 2 2

6 7

6

(0.19999)(10 )1

1 1 1284 10 10

2.25 10 Hz

Tf f f

f

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= ×

6 6 61 1 1

6 7

2.25 10 2.25 10 2.25 10tan tan tan284 10 10

89.99 66.04 12.68168.7

φ

φ

− − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − − −= −

Phase Margin 180 168.7 11.3= − = ° 12.80 a. 5(0) 100 dB (0) 10T T= ⇒ =

5

6 6

10( )1 1 1

10 5 10 10 10

T ff f fj j j

=⎛ ⎞⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟⎜ ⎟× ×⎝ ⎠⎝ ⎠⎝ ⎠

1T = =

5

2 2 2

6 6

1 1 110

1 1 110 5 10 10 10f f f

= × × ×⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠ ⎝ ⎠

By trial and error 0.976 MHzf =

61 1 10.976 10 0.976 0.976tan tan tan

10 5 1090 11.05 5.574 106.6

φ − − −⎛ ⎞× ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= − ° − ° − ° = − °

Phase Margin 180 106.6 73.4= ° − ° = °

b. 11

PF

fC

′ ∝ so 1

10 7520Pf

=′

or 1 2.67 HzPf ′ =

Now 1T = =

5

2 2

6

1 110

1 12.67 5 10

f f= × ×

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠

2

6

1

110 10

f×

⎛ ⎞+ ⎜ ⎟×⎝ ⎠

By trial and error 52.66 10 Hzf ≈ ×

then 5

1 1 12.66 10 0.266 0.266tan tan tan2.67 5 10

90 3.045 1.524 94.57

φ − − −⎛ ⎞× ⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= − ° − ° − ° = − °

Phase Margin 180 94.57 85.4= ° − ° = ° 12.81

(a) 31

2dBfπτ− = where ( )1 2o i iR R Cτ =

3 12 7(500 1000) 10 2 10 6.67 10 sτ− −= × × × ⇒ = × Then

3 37

1 239 2 (6.67 10 )dB dBf f kHzπ− −−= ⇒ =

×

(b) For

10 Hz,PDf = 1 1 0.0159 2 2 (10)PD

sf

τπ π

= = =

Then ( ) ( )1 2o i i MR R C Cτ = +

( ) ( )30.0159 500 1000 10 i MC C= × × + or ( ) 8 124.77 10 2 10 0.0477 i M M MC C x C C Fμ− −+ = × = + ⇒ = 12.82 Assuming a phase margin of 45 .°

1 16 6135 90 tan tan

2 10 25 10f f− −⎛ ⎞ ⎛ ⎞− ° ≈ − ° − −⎜ ⎟ ⎜ ⎟× ×⎝ ⎠ ⎝ ⎠

By trial and error, 1.74 MHzf ≈ Then

26

2 2

115000

1.74 101

1 1

1.74 1.741 12 25

PD

T

f

=

= ×⎛ ⎞×+ ⎜ ⎟⎝ ⎠

× ×⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 61.74 10 5000

(1.325)(1.0024)PDf× ≈

so 462 HzPDf = 12.83

5

5

1020 0.049991 (10 )

ββ

= ⇒ =+

Phase Margin 60 120φ= ° ⇒ = − °

1 1 16 7120 tan tan tan

10 5 10PD

f f ff

− − −− = − − −×

Assume 1tan 90PD

ff

− = °

1 56tan 30 5.77 10 Hz

10f f− ≅ ° ⇒ = ×

5

2 2 25 5 5

6 7

3

25

(0.04999)(10 )15.77 10 5.77 10 5.77 101 1 1

10 5 10

4.999 1015.77 101 (1.155)(1.0)

PD

PD

T

f

f

= =⎛ ⎞ ⎛ ⎞ ⎛ ⎞× × ×+ + +⎜ ⎟ ⎜ ⎟ ⎜ ⎟×⎝ ⎠ ⎝ ⎠⎝ ⎠

×=⎛ ⎞×+ ⎜ ⎟⎝ ⎠

5 35.77 10 4.999 10(1.155)(1.0)PDf

× ×

133.3 HzPDf =