Chapter 14 Problem Solutions 14.1

80

(max) 4.5 (max) 56.25 mV

od

i

o i

vA

vv v

= = −

= ⇒ =

So 56.25(max) 39.77 mV2i rmsv = =

14.2

(a)

24.5 0.028125 mA1604.5 4.5 mA1L

i

i

= =

= =

Output Circuit 4.528 mA= 4.5 0.05625 V80

oi i

vv v

A−= − = ⇒ = −

(b) 4.515 mA

(min) 300

oo

L L

L

viR R

R

≈ = =

⇒ = Ω

14.3 (1) 2 Vov = (2) 2 12.5 mVv = (3) 42 10OLA = × (4) 1 8 Vv μ= (5) 1000OLA = 14.4 From Eq. (14.4)

2 1

2

1

2 1

23

1

2

1 23 3

1

2

1

2

1

/11 1

/1511 1

2 10

15115 1

2 10 2 10

15.0075 (0.9925)

15.12

CL

oL

R RAR

A RR R

RR

RR R

RRR

RR

−=⎛ ⎞

+ +⎜ ⎟⎝ ⎠

−− =

⎛ ⎞+ +⎜ ⎟× ⎝ ⎠

⎛ ⎞⎜ ⎟

⎛ ⎞ ⎝ ⎠+ + =⎜ ⎟× ×⎝ ⎠

=

=

14.5

1 01 10 0 1

1 2

01

0

01

1 2 1 2

and

so that

1 1 1

IL

i

L

I

i

v vv v v v A vR R R

vv

A

vv vR R R R R

−− = + = −

= −

⎛ ⎞+ = + +⎜ ⎟

⎝ ⎠

So

01 2 0 1 2

1 1 1 1 1I

L i

v vR R A R R R

⎡ ⎤⎛ ⎞= − + + +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

Then 0 1

2 0 1 2

(1/ )1 1 1 1 1

CLI

L i

v R Av

R A R R R

−= =⎡ ⎤⎛ ⎞

+ + +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

From Equation (14.20) for LR = ∞ and 0 0R =

0

2

(1 )1 1 11

L

if i

AR R R

+= + ⋅

a. For 1 kiR = Ω

3

3

(1/ 20)1 1 1 1 1

100 20 100 1100.05

[0.01 1.06 10 ]

CLA

−

−=⎡ ⎤⎛ ⎞+ + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

−=+ ×

or

3

4.52

1 1 1 10 90.8 1 100

CL

ifif

A

RR

⇒ = −

+= + ⇒ = Ω

b. For 10 kiR = Ω

3

4

(1/ 20)1 1 1 1 1

100 20 100 10100.05

[0.01 1.6 10 ]

CLA

−

−=⎡ ⎤⎛ ⎞+ + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

−=+ ×

or 4.92CLA⇒ = −

31 1 1 10 98.9 10 100 if

if

RR

+= + ⇒ = Ω

c. For 100 kiR = Ω

3

5

(1/ 20)1 1 1 1 1

100 20 100 100100.05

[0.01 7 10 ]

CLA

−

−=⎡ ⎤⎛ ⎞+ + +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

−=+ ×

or

3

4.965

1 1 1 10 99.8 100 100

CL

ifif

A

RR

⇒ = −

+= + ⇒ = Ω

14.6

2

1

2

1

1

11 1

oCL

i

OL

RRvA

v RA R

⎛ ⎞+⎜ ⎟

⎝ ⎠= =⎡ ⎤⎛ ⎞

+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

For the ideal:

2

1

0.101 500.002

RR

⎛ ⎞+ = =⎜ ⎟

⎝ ⎠

( ) (0.10)(1 0.001) 0.0999ov actual = − = So

0.0999 50 49.9510.002 1 (50)OLA

= =+

which yields 1000OLA =

14.7 From Equation (14.18)

211

1

2

1

1 1 1

OL

oovf

L o

AR Rv

Av

R R R

⎛ ⎞− −⎜ ⎟⎝ ⎠= =

⎛ ⎞+ +⎜ ⎟

⎝ ⎠

Or 3

3

1 1 1

31 1

5 10 11 100 (4.99999 10 )

1 1 1 1.1110 1 100

4.504495 10

o

o

v v v

v v

⎛ ⎞×− −⎜ ⎟ − ×⎝ ⎠= ⋅ = ⋅⎛ ⎞+ +⎜ ⎟⎝ ⎠

= − × ⋅

Now 11

1 1 1

iv vi Kv R v

−= ≡

Then 1 1 1iv v KR v− =

which yields

11 1ivv

KR=

+

Now, from Equation (14.20) 3

3

11 5 101 1 101 110 100 1

10 1005.0011 10(0.1) (0.01) 45.15495

1.11

K

⎡ ⎤+ × +⎢ ⎥= + ⎢ ⎥

⎢ ⎥+ +⎢ ⎥⎣ ⎦

⎡ ⎤×= + =⎢ ⎥⎣ ⎦

Then

( )( )1 45.15495 10 1 452.5495i iv v

v = =+

We find 3

1 4.504495 10452.5495

io

vv ⎡ ⎤= − × ⎢ ⎥⎣ ⎦

Or 1

1 9.9536ovf

i

vAv

= = −

For the second stage, LR = ∞

3

32 1 1

3

1 1 11

1

5 10 11 100

4.950485 101 11 100

1 1 1 5 10 49.61485110 100 1

100

1 (49.61485)(10) 1 497.1485

o

o o o

v v v

K

v v vvKR

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

⎡ ⎤⎢ ⎥+ ×≡ + =⎢ ⎥⎢ ⎥+⎢ ⎥⎣ ⎦

′ = = =+ +

Then 3

2

1

4.950485 10 9.95776497.1485

o

o

vv

− ×= = −

So 2 ( 9.9536)( 9.95776) 99.12o

vf vfi

vA Av

= = − − ⇒ =

14.8 a.

1 01 1

3 1 2

0I

i

v vv v vR R R R

−− + + =+

(1)

01

3 1 2 2 3

1 1 1 I

i i

v vvR R R R R R R⎡ ⎤

+ + = +⎢ ⎥+ +⎣ ⎦

0 0 0 0 1

0 2

0L d

L

v v A v v vR R R

− −+ + = (2)

or

010

0 2 2 0

1 1 1 L d

L

A vvvR R R R R⎡ ⎤

+ + = +⎢ ⎥⎣ ⎦

1

3

Id i

i

v vv RR R

⎛ ⎞−= ⋅⎜ ⎟+⎝ ⎠

(3)

So substituting numbers: 0

11 1

10 20 10 40 40 10 20Iv vv 1⎡ ⎤+ + = +⎢ ⎥+ +⎣ ⎦

(1)

or 1 0[0.15833] [0.025] [0.03333]Iv v v= +

41

0(10 )1 1 1

1 0.5 40 40 0.5dvvv ⎡ ⎤+ + = +⎢ ⎥⎣ ⎦

(2)

or [ ] [ ] ( )4

0 13.025 0.025 2 10 dv v v= + ×

( )1120 0.6667

10 20I

d Iv vv v v−⎛ ⎞= ⋅ = −⎜ ⎟+⎝ ⎠

(3)

So [ ] [ ] ( )( )( )4

0 1 13.025 0.025 2 10 0.6667 Iv v v v= + × − (2) or

[ ] 4 40 13.025 1.333 10 1.333 10Iv v v= × − ×

From (1): ( ) ( )1 0 0.1579 0.2105Iv v v= +

Then [ ] ( ) ( )4 4

0 0

3 40

3.025 1.333 10 1.333 10 0.1579 0.2105

2.1078 10 1.0524 10I I

I

v v v v

v v

= × − × +⎡ ⎤⎣ ⎦⎡ ⎤ ⎡ ⎤× = ×⎣ ⎦ ⎣ ⎦

or 0 4.993CLI

vAv

= =

To find :ifR Use Equation (14.27)

( )3

1 2

1

0.5 0.511 40

101 1 0.5 0.5 0.5110 40 1 40 40(40)

(1.5125) {(0.125)(1.5125) 0.0003125} 25

I

d

I d

i

vv

i v v

⎛ ⎞+ +⎜ ⎟⎝ ⎠

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

= − −

or (1.5125) {0.18875} 25I I di v v= −

Now (20)d I i Iv i R i= = and 1 (20)I Iv v i= −

So (1.5125) [ (20)] [0.18875] 25 (20)[505.3] (0.18875)

I I I I

I I

i v i ii v

= − ⋅ −=

or

2677 kI

I

vi

= Ω

Now 10 2677 2.687 Mif ifR R= + ⇒ = Ω

To determine 0 :fR Using Equation (14.36)

30

20 0

1

1 1 1 10400.5 11

10 20

L

f

i

ARR R

R R

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= ⋅ = ⋅

′ ⎢ ⎥ ⎢ ⎥++⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

or 0 3.5 fR′ = Ω

Then 0 1 k 3.5 fR = Ω Ω

0 3.49 fR⇒ = Ω

b. Using Equation (14.16)

3

5( 10) (0.05)%10

CL CL

CL CL

dA dAA A

⎛ ⎞= − ⇒ = −⎜ ⎟⎝ ⎠

14.9

0 0 0

0

0L d I

i

v A v v vR R

− −+ = and 0d Iv v v= −

So 0 0 0

00 0

0 00

0 0 0

4 4

0

4 40

( ) 0

1 1 1

1 (10 ) 1 (10 )0.2 0.2 100 100 0.2

[5.000501 10 ] [5.000001 10 ]

L II

i i

L LI

i i

I

I

v A v vv vR R R R

A Av vR R R R R

v v

v v

− ⋅ − + − =

⎡ ⎤ ⎡ ⎤+ + = +⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤1+ + = +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

× = ×

So 0 0.9999CLI

vAv

= =

b. Set 0Iv =

0 0 00 0

0

00 0

0 0

and

1 1

L dd

i

L

i

v A v vi v vR R

Ai vR R R

−= + = −

⎡ ⎤= + +⎢ ⎥

⎣ ⎦

Then 0

0 0 0

1 1 1L

f i

AR R R R

= + +

or 4

0

1 1 (10 ) 10.2 0.2 100fR

= + +

which yields 0 0.02 fR ≅ Ω

14.10

1 01 1 2 1

01 21

20 10 401 1 1

20 10 40 20 10 40

I I

I I

v vv v v v

vv v v

−− −+ =

⎡ ⎤+ + = + +⎢ ⎥⎣ ⎦

and 0 0 1Lv A v= − so that 01

0L

vvA

= −

Then

1 2 0 3

20

0 1 2

0 0

0 0

1 1 7(0.05) (0.10)40 402 10

[2.50875 10 ]1.993 3.986

2 1.993 0.352

I I

I I

v v v

vv v v

v v %v v

−

⎧ ⎫⎛ ⎞+ = − + ⋅⎨ ⎬⎜ ⎟× ⎝ ⎠⎩ ⎭= − ×

⇒ = − −

Δ Δ−= ⇒ =

14.11

2 2 240 4 0.8

40 10 5Bv v v v⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ (1)

01

10 40AA v vv v −−

=

01 1 110 40 10 40A

vv v ⎛ ⎞+ = +⎜ ⎟⎝ ⎠

1 0(0.1) (0.025) (0.125)Av v v+ = (2)

0 0 0 ( )L d L B Av A v A v v= = − (3) or

0 0 2

02

0

02

0

[0.8 ]

0.8

0.8

L A

AL

AL

v A v vv v vA

vv vA

= −

− = −

⇒ = −

Then

01 0 2

0

1 2 0 3

20

0

2 1

(0.1) (0.025) (0.125) 0.8

0.125(0.1) (0.1) 0.02510

[2.5125 10 ]

3.9801

0.0199 0.49754

L

d

d

d

vv v v

A

v v v

vvA

v v

A%

A

−

⎡ ⎤+ = −⎢ ⎥

⎣ ⎦⎡ ⎤− = − +⎢ ⎥⎣ ⎦

= − ×

⇒ = =−

Δ⇒ = ⇒

14.12 a. Considering the second op-amp and Equation (14.20), we have

2

1 1 1 1 100 1010.10110 0.1 (0.1)(11)1

0.1ifR

⎡ ⎤⎢ ⎥+= + ⋅ = +⎢ ⎥⎢ ⎥+⎢ ⎥⎣ ⎦

So 2 0.0109 kifR = Ω The effective load on the first op-amp is then

1 20.1 0.1109 kL ifR R= + = Ω Again using Equation (14.20), we have

11 1001 1 1 110.0170.1109 0.101 110 1 11.0171

0.1109 1ifR

+ += + ⋅ = +

+ +

so that 99.1 ifR = Ω

b. To determine 0 :fR For the first op-amp, we can write, using Equation (14.36)

0

20 1 0

1

1 1 1 100401 11

1||10||

L

f

i

ARR R

R R

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥= ⋅ = ⋅⎢ ⎥ ⎢ ⎥++⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

which yields 0 1 0.021 kfR = Ω For the second op-amp, then

0

20 0

1 0 1

1 1

1( ) ||

1 1000.101 1

(0.121) ||10

L

f

f i

ARR R

R R R

⎡ ⎤⎢ ⎥⎢ ⎥= ⋅⎢ ⎥+⎢ ⎥+⎣ ⎦

⎡ ⎤⎢ ⎥⎢ ⎥= ⋅⎢ ⎥+⎢ ⎥⎣ ⎦

or 0 18.4 fR = Ω

c. To find the gain, consider the second op-amp.

01 2 2 2 02( )0.1 0.1

d d d

i

v v v v vR

− − − −+ = (1)

01 022

1 1 10.1 0.1 10 0.1 0.1dv vv ⎛ ⎞+ + + = −⎜ ⎟

⎝ ⎠

or 01 2 02(10) (20.1) (10)dv v v+ = −

02 0 2 02 2

0

( ) 00.1

L d dv A v v vR

− − −+ = (2)

02 022

02 2

100 1 01 1 0.1 0.1

(11) (90) 0

d

d

v vv

v v

⎛ ⎞− − + =⎜ ⎟⎝ ⎠

− =

or 2 02 (0.1222)dv v=

Then Equation (1) becomes 01 02 02(10) (0.1222)(20.1) (10)v v v+ = −

or 01 02 (1.246)v v= −

Now consider the first op-amp.

1 1 1 01( )1 1

I d d d

i

v v v v vR

− − − −+ = (1)

1 011 1 1(1) (1)1 10 1I dv v v⎛ ⎞+ + + = −⎜ ⎟⎝ ⎠

or 1 01(1) (2.1) (1)I dv v v+ = −

01 01 0 1 01 1

0

( ) 00.1109 1

L d dv v A v v vR

− − −+ + = (2)

01 1

01 1

1 1 1 100 1 00.1109 1 1 1 1

(11.017) (99) 0

d

d

v v

v v

⎛ ⎞ ⎛ ⎞+ + − − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− =

or 1 01(0.1113)dv v=

Then Equation (1) becomes

01 01(1) (0.1113)(2.1)Iv v v+ = − or 01(1.234)Iv v= − We had 01 02 (1.246)v v= − So 02 (1.246)(1.234)Iv v=

or 02 0.650I

vv

=

d. Ideal 02 1I

vv

=

So ratio of actual to ideal 0.650.= 14.13 (a) For the op-amp. 6

0 3 10L dBA f⋅ = 6

3 4

10 50 Hz2 10dBf = =

×

For the closed-loop amplifier. 6

310 40 kHz25dBf = =

(b) Open-loop amplifier. 4 4

2

33

44

3 2

43

3 2

2 10 2 10| |1 1

2 100.25 1.94 101 (0.25)

2 105 3.92 101 (5)

dBdB

dB

dB

A Afj f

f f

f f A

f f A−

× ×= ⇒ =⎛ ⎞+ + ⎜ ⎟⎝ ⎠

×= ⇒ = = ×+

×= ⇒ = = ×+

Closed-loop amplifier

3 2

3 2

250.25 24.251 (0.25)255 4.90

1 (5)

dB

dB

f f A

f f A−

= ⇒ = =+

= ⇒ = =+

14.14 The open loop gain can be written as

00

6

( )1 1

5 10

L

PD

AA ff fj j

f

=⎛ ⎞⎛ ⎞+ ⋅ + ⋅⎜ ⎟⎜ ⎟×⎝ ⎠⎝ ⎠

where 50 2 10 .A = ×

The closed-loop response is 0

01L

CLL

AAAβ

=+

At low frequency, 5

5

2 101001 (2 10 )β

×=+ ×

So that 39.995 10 .β −= × Assuming the second pole is the same for both the open-loop and closed-loop, then

1 16tan tan

5 10PD

f ff

φ − −⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

For a phase margin of 80 ,° 100 .φ = − ° So

16100 90 tan

5 10f− ⎛ ⎞− = − − ⎜ ⎟×⎝ ⎠

or 58.816 10 Hzf = ×

Then 0

5

2 25 5

6

12 10

8.816 10 8.816 101 15 10

L

PD

A

f

=×=

⎛ ⎞ ⎛ ⎞× ×+ +⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

or 5

58.816 10 1.9696 10PDf× ≅ ×

or 4.48 HzPDf =

14.15 (a) 1st stage

3 3(10) 1 100 dB dBf MHz f kHz− −= ⇒ = 2nd stage

3 3(50) 1 20 dB dBf MHz f kHz− −= ⇒ = Bandwidth of overall system 20 kHz≅ (b) If each stage has the same gain, so

2 500 22.36K K= ⇒ = Then bandwidth of each stage

3 3(22.36) 1 44.7 dB dBf MHz f kHz− −= ⇒ = 14.16

3

4

32 24

3 3

1

5 10200 40 101 1

o

dB

odB

dB dB

AAfj

fAA f Hz

ff f

−

−

− −

=+

×= ⇒ = ⇒ =⎛ ⎞ ⎛ ⎞

+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Then 4(5 10 )(40) 2 T Tf f MHz= × ⇒ =

14.17

4 6

63 3

2

33

(5 10 ) 10 20

(25) 10 40

25

1 140 10

PD PD

dB dB

vov v

dB

f f Hz

f f kHz

AA A

f fjf

− −

−

× = ⇒ =

⇒ =

= ⇒ =⎛ ⎞+ + ⎜ ⎟×⎝ ⎠

At 30.5 20 dBf f kHz−= =

2

25 22.361 (0.5)

vA = =+

At 32 80 dBf f kHz−= =

2

25 11.181 (2)

vA = =+

14.18

3 6(20 10 ) 10 50vf vfMAX MAXA A× ⋅ = ⇒ =

14.19 From Equation (14.55),

6

0

10 102 2 (10)P

SRF P BWVπ π

×= =

or max 159 kHzF P BW f= =

14.20 a. Using Equation (14.55),

6

0 3

8 102 (250 10 )PVπ

×=×

or 0 5.09 VPV =

b.

Period 63

1 1 4 10 s250 10

Tf

−= = = ××

One-fourth period 1 sμ=

0

0

Slope 8 V/ s1

8 V

P

P

VSR

sV

μμ

= = =

⇒ =

14.21 For input (a), maximum output is 5 V.

1 V/μsS R = so

For input (b), maximum output is 2 V.

For input (c), maximum output is 0.5 V so the output is

14.22 For input (a), 01max 3 V.v =

Then 02 max

3(3) 9 Vv = =

For input (b), 01max 1.5 V.v =

Then ( )02 max

3 1.5 4.5Vv = =

14.23

6

3

20 , 0.8 /.8 10 6.37

2 2 (20 10 )

MAX

po poMAX

f kHz SR V sSRV V Vf

μ

π π

= =

0 ×= = ⇒ =×

14.24

11 1 exp ,BE

ST

VI IV

⎛ ⎞= ⎜ ⎟

⎝ ⎠ 2

2 2 exp BES

T

VI IV

⎛ ⎞= ⎜ ⎟

⎝ ⎠

Want 1 2 ,I I= so

14 1

1

2 14 2

1 2

5 10 (1 )exp1

5 10 (1 )exp

(1 ) exp(1 )

BE

T

BE

T

BE BE

T

VxVI

I VxV

V Vxx V

−

−

⎛ ⎞× + ⎜ ⎟

⎝ ⎠= =⎛ ⎞

× − ⎜ ⎟⎝ ⎠

⎛ ⎞−+= ⎜ ⎟− ⎝ ⎠

Or

2 11 exp exp1

0.0025exp 1.100.026

OSBE BE

T T

VV Vxx V V

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

Now 1 (1 )(1.10)x x+ = − ⇒

0.0476 4.76%x = ⇒ 14.25 From Equation (14.62),

1 2

3

44

1 1

1 1

CE CE

AN S AN

EB ECS

AP AP

v vV I Vv vIV V

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

For 2 0.6 V,CEv = then 4 5 V.ECv = We have 1 5 VCEv = so

3

4

5 0.61 180 800.6 51 180 80

S

S

II

⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟= ⋅⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

or 2

32

4

(1.0625) 1.112(1.0075)

S

S

II

= =

So ( )( )14

3 10 1.112SI −= or

143 1.112 10 ASI −= ×

14.26

By superposition:

2

1

2

1

( ) 50

( ) 1 51

o i i i

o os os os

Rv v v vR

Rv v v vR

= − ⋅ = −

⎛ ⎞= + ⋅ =⎜ ⎟⎝ ⎠

So ( ) ( ) 50 51o o i o os i osv v v v v v v= + = − +

For 20 iv mV= and 2.5 osv mV+ 50(0.02) 51(0.0025) 0.8725 ov V= − + = −

For 20 iv mV= and 2.5 osv mV= − 50(0.02) 51( 0.0025) 1.1275 ov V= − + − = −

So 1.1275 0.8725 ov V− ≤ ≤ −

14.27

( )[ ]( )

50 50 2.5mV sin mV

0.125 0.25sin vo i

o

v v t

v t

ω

ω

= − = − ± +⎡ ⎤⎣ ⎦= ± −

14.28

38

4

0.5 10 5 10 10

I A−

−×= = ×

Also

0

1 io

odV II C V Idt tdt C C

= ⇒ = = ⋅∫

Then 8

36

5 105 10 10 10

t t s−

−

×= ⇒ =×

14.29

a. 01100| | 10 110

v ⎛ ⎞= +⎜ ⎟⎝ ⎠

or 01| | 110 mVv =

Then

02 0150| | | | (5) 10 1 (110)(5) (10)(6)10

v v ⎛ ⎞= + + = +⎜ ⎟⎝ ⎠

or 02| | 610 mVv =

14.30

0v due to Iv

01(0.5) 1 0.9545 V

1.1v ⎛ ⎞= + =⎜ ⎟

⎝ ⎠

Wiper arm at 10 V,V + = (using superposition)

1 51

1 5 4

|| 0.0909(10) (10)|| 0.0909 10

0.090

R RvR R R

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

Then 011 (0.090) 0.0901

v ⎛ ⎞= − = −⎜ ⎟⎝ ⎠

Wiper arm in center, 1 0v = and 02 0v = Wiper arm at 10 V,V − = − 1 0.090v = − So

03 0.090v = Finally, total output 0 :v (from superposition) Wiper arm at ,V +

0 0.8645 Vv = Wiper arm in center,

0 0.9545 Vv =

Wiper arm at ,V −

0 1.0445 Vv = 14.31 a. 1 2 0.5 || 25 0.490 kR R′ ′= = = Ω or

1 2 490 R R′ ′= = Ω b. From Equation (14.75),

6

114

6

214

125 10(0.026) ln (0.125)2 10

125 10(0.026) ln (0.125)2.2 10

R

R

−

−

−

−

⎛ ⎞× ′+⎜ ⎟×⎝ ⎠⎛ ⎞× ′= +⎜ ⎟×⎝ ⎠

1 2

2 1

0.586452 (0.125) 0.583974 (0.125)0.002478 (0.125)( )

R RR R′ ′+ = +′ ′= −

So 2 1 0.0198 k 19.8 R R′ ′− = Ω ⇒ Ω Then

2 1

2 1

(1 )0.0198

(1 )(0.5)(1 )(50) (0.5)(50) 0.0198

(0.5) (1 )(50) (0.5) (50)25(1 ) 25 0.0198

50.5 50 0.5 50(0.5 50 )(25 25 ) (25 )(50.5 50 ) 0.0198

(50.5 50 )(0.5 50 )

x x

x x

R x R R RR x R R xR

x xx x

x xx xx x x x

x x

− ×− =

+ − +− − =

+ − +− − =

− ++ − − − =

− +

{ } { }{ } { }

2 2 2

2

2

2

2

25 0.5 0.5 50 50 50.5 50 0.0198 25.25 2525 25 2500

25 0.5 0.0198 25.25 2500 2500

0.5 0.019998 1.98 1.981.98 2.98 0.48 0

2.98 (2.98) 4(1.98)(0.48)2(1.98)

x x x x x x x x

x x x

x x xx x

x

− + − − + = + − −

− = + −

− = + −− + =

± −=

So 0.183x =

and 1 0.817x− = 14.32

1 1

2 2

||15 0.5 ||15 0.4839 k|| 35 0.5 || 35 0.4930 k

R RR R

′ = = = Ω′ = = = Ω

From Equation (14.75),

1 21 1 2 2

3 4

12 2 1 1

2

1 1 12 2

2 2 2

1 12

2 2

(0.026) ln (0.026) ln

(0.026) ln

(0.026) ln 1

(0.026) ln (0.4930) 1 (0.9815)

C CC C

S S

CC C

C

C CC

C C

C CC

C C

i ii R i RI I

ii R i R

i

i i Ri Ri i R

i iii i

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

′⎛ ⎞ ⎡ ⎤′= − ⋅⎜ ⎟ ⎢ ⎥′⎝ ⎠ ⎣ ⎦

⎛ ⎞= −⎜ ⎟

⎝ ⎠

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦

By trial and error:

1 252 ACi μ= and 2 248 ACi μ= or

1

2

1.0155C

C

ii

=

14.33 From Eq. (14.79), we have

21 2 2 3

1

1 2

1

1 A 2 A

o B B

B B

Rv I R I RR

I Iμ μ

⎛ ⎞= − +⎜ ⎟

⎝ ⎠= =

Setting 0,ov = we have

( )( ) ( )

( )( )

6 3 63

3

3 36

2000 10 200 10 2 10 120

200 10 9.09 K2 10 11

R

R R

− −

−

−

⎛ ⎞= × − × +⎜ ⎟⎝ ⎠

×= ⇒ =×

14.34

2

1

1

1 80

6.329 K

RRR

+ =

=

( )1 2

5sin mVf I

B

V v tI I I

ω= =

+ =

(a) 26 3

0 0(10 )(500 10 ) 0.50 V

I X B

O o

V V I IV v−

= ⇒ = == × ⇒ =

(b)

( ) ( )( )[ ]( )

1 2

02 2

1 2 1 1 2

6 3

1 1 1 1

80 5sin mV 10 500 10

0.5 0.4sin v

o XXB

X B o I B

o

o

v VV IR R

vV I v R v I RR R R R R

v t

v t

ω

ω

−

−+ =

⎛ ⎞ ⎛ ⎞+ + = ⇒ = + +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= + ×⎡ ⎤⎣ ⎦= +

14.35 a.

For 2 1 A,BI μ= then ( )( )6 4

0 10 10v −= −

or 0 0.010 Vv = −

b. If a 10 kΩ resistor is included in the feedback loop

Now 0 2 1(10) (10) 0B Bv I I= − + = Circuit is compensated if 1 2 .B BI I= 14.36 From Equation (14.83), we have

0 2 0Sv R I= where 2 40 kR = Ω and 0 3 A.SI μ= Then

( )( )3 60 40 10 3 10v −= × ×

or 0 0.12 Vv =

14.37 a. Assume all bias currents are in the same direction and into each op-amp.

( ) ( )( )6 501 1 01100 k 10 10 0.1 VBv I v−= Ω = ⇒ =

Then ( ) ( )

( )( ) ( )( )02 01 1

6 4

5 50 k

0.1 5 10 5 10

0.5 0.05

Bv v I−

= − + Ω

= − + ×

= − +

or 02 0.45 Vv = −

b. Connect 3 10 ||100 9.09 kR = = Ω resistor to noninverting terminal of first op-amp, and

3 10 || 50 8.33 kR = = Ω resistor to noninverting terminal of second op-amp.

14.38 a. For a constant current through a capacitor.

0 0

1 t

v I dtC

= ∫

or 6

0 06

0.1 10 (0.1)10

v t v t−

−

×= ⋅ ⇒ =

b. At 10 s,t = 0 1 Vv = c. Then

124

0 06

100 10 (10 )10

v t v t−

−−

×= ⋅ ⇒ =

At 10 s,t = 0 1 mVv = 14.39 a. Assume all bias currents are into the op-amp.

( ) ( ) ( )6 301 1 50 k 10 10 50 10Bv I −= Ω = × ×

or 01 02 0.5 Vv v= =

( )( ) ( )( )6 303 011 10 10 20 10v v −= − + × ×

or 03 0.3 Vv = −

b. 10 || 50 8.33 kA AR R= ⇒ = Ω

20 || 20 10 kB BR R= ⇒ = Ω c. Assume the worst case offset current, that is, 0 1 2S B BI I I= − or 0 2 1.S B BI I I= − From Equation (14.83),

( )( )3 601 2 0 50 10 2 10Sv R I −= = × ×

or 01 02 0.1 Vv v= =

( )( )( ) ( ) ( )

03 01 0 2

6 3

1

1 0.1 2 10 20 10Sv v I R

−

= − −

= − − × ×

or 03 0.14 Vv = −

14.40 a. Using Equation (14.79), Circuit (a),

( )( ) ( )( )6 3 6 30

500.8 10 50 10 0.8 10 25 10 150

v − − ⎛ ⎞= × × − × × +⎜ ⎟⎝ ⎠

or 0 0v =

Circuit (b),

( )( ) ( )( )6 3 6 30

2

500.8 10 50 10 0.8 10 10 150

4 10 1.6

v − −

−

⎛ ⎞= × × − × +⎜ ⎟⎝ ⎠

= × −

or 0 1.56 Vv = −

b. Assume 1 0.7 ABI μ= and 2 0.9 A,BI μ= then using Equation (14.79): Circuit (a),

( )( ) ( )( )6 3 6 30

500.7 10 50 10 0.9 10 25 10 150

0.035 0.045

v − − ⎛ ⎞= × × − × × +⎜ ⎟⎝ ⎠

= −

or 0 0.010 Vv = − Circuit (b),

( )( ) ( )( )6 3 6 60

500.7 10 50 10 0.9 10 10 150

0.035 1.8

v − − ⎛ ⎞= × × − × +⎜ ⎟⎝ ⎠

= −

or 0 1.765 Vv = − 14.41 a. If 0,R =

0,max 0

3 6 3

0,max

0,max

1001 (100 k )10

(11)(10 10 ) (2 10 )(100 10 )0.110 0.20

0.310 V

S Bv V I

vv

− −

⎛ ⎞= + + Ω⎜ ⎟⎝ ⎠

= × + × ×= +⇒ =

b.

10.1 ||100 9.17 kR R= = Ω = 14.42

a. 2

(15) 0.010 Vi

i

RR R

⎛ ⎞=⎜ ⎟+⎝ ⎠

2

2

15 0.00066671515(1 0.0006667) 0.0006667

RR

=+

− =

Then 2 22.48 MR = Ω

b. 1 1|| 15 ||10 6 ki FR R R R= = ⇒ = Ω 14.43 a. Assume the offset voltage polarities are such as to produce the worst case values, but the bias currents are in the same direction. Use superposition:

Offset voltages

01 01

02

02

100| | 1 (10) 110 mV | |10

50| | (5)(110) 1 (10)10

| | 610 mV

v v

v

v

⎛ ⎞= + = =⎜ ⎟⎝ ⎠

⎛ ⎞= + +⎜ ⎟⎝ ⎠

⇒ =

Bias Currents: 6 3

01 (100 k ) (2 10 )(100 10 ) 0.2 VBv I −= Ω = × × = Then

6 302 ( 5)(0.2) (2 10 )(50 10 ) 0.9 Vv −= − + × × = −

Worst case: 01v is positive and 02v is negative, then

01 0.31 Vv = and 02 1.51 Vv = − b. Compensation network:

If we want

20 mV and 10 V

8.33 (10) 0.0208.33

B

B C

C

R V VR R

R

+ +⎛ ⎞= =⎜ ⎟+⎝ ⎠

⎛ ⎞=⎜ ⎟+⎝ ⎠

or 4.15 MCR ≅ Ω 14.44 Assume bias currents are in same direction, but assume polarity of offset voltages are such as to produce the worst case output. a. Let 1 5.5 A,BI μ= 2 4.5 ABI μ= Bias Current Effects:

01 1 02

03 1 01 03

(50 k ) 0.275 V 0.275 V(20 k ) 0.165 V

B

B

v I vv I v v

= Ω = ⇒ == Ω − ⇒ = −

Offset Voltage Effects:

01 02

03 01 03

50(5) 1 30 mV 30 mV10

205 1 40 mV20

v v

v v v

⎛ ⎞= + = ⇒ =⎜ ⎟⎝ ⎠

⎛ ⎞= − − + ⇒ = −⎜ ⎟⎝ ⎠

Total Effect: 01 0.305 Vv = and 02 0.305 Vv = 03 0.205 Vv = − 14.45 For circuit (a), effect of bias current:

3 90 (50 10 )(100 10 ) 5 mVv −= × × ⇒

Effect of offset voltage

050(2) 1 4 mV50

v ⎛ ⎞= + =⎜ ⎟⎝ ⎠

So net output voltage is 0 9 mVv = For circuit (b), effect of bias current: Let 2 550 nA,BI = 1 450 nA,BI = then from Equation (14.79),

9 3 9 60

2

50(450 10 )(50 10 ) (550 10 )(10 ) 150

2.25 10 1.1

v − −

−

⎛ ⎞= × × − × +⎜ ⎟⎝ ⎠

= × −

or 0 1.0775 Vv = −

If the offset voltage is negative, then 0 ( 2)(2) 4 mVv = − = −

So the net output voltage is 0 1.0815 Vv = −

14.46 a. At 25 C,T = ° 0 2 mVSV = so the output voltage for each circuit is

0 4 mVv = b. For 50 C,T = ° the offset voltage for is

0 2 mV (0.0067)(25) 2.1675 mVSV = + = so the output voltage for each circuit is

0 4.335 mVv = 14.47 a. At 25 C,T = ° 0 1 mV,SV = then

01 0150(1) 1 6 mV10

v v⎛ ⎞= + ⇒ =⎜ ⎟⎝ ⎠

and

02 01

02

60 601 (1) 120 20

6(4) (1)(4) 28 mV

v v

v

⎛ ⎞ ⎛ ⎞= + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + ⇒ =

b. At 50 C,T = ° 0 1 (0.0033)(25) 1.0825 mV,SV = + = then

01 01(1.0825)(6) 6.495 mVv v= ⇒ = and

02 (6.495)(4) (1.0825)(4)v = + or

02 30.31 mVv = 14.48

0

0

25 C; 500 nA, 200 nA50 C, 500 nA (8 nA / C)(25 C) 700 nA

200 nA (2 nA / C)(25 C) 250 nA

B S

B

S

I II

I

° = =° = + ° ° =

= + ° ° =

a. Circuit (a): For ,BI bias current cancellation, 0 0v = Circuit (b): For ,BI Equation (14.79),

9 3 9 60

0

50(500 10 )(50 10 ) (500 10 )(10 ) 150

0.025 1.00 0.975 V

v

v

− − ⎛ ⎞= × × − × +⎜ ⎟⎝ ⎠

= − ⇒ = −

b. Due to offset bias currents. Circuit (a):

9 30 0(200 10 )(50 10 ) 0.010 Vv v−= × × ⇒ =

Circuit (b): 2

1

Let 600 nA400 nA

B

B

II

==

Then 9 3 9 6

0

0

50(400 10 )(50 10 ) (600 10 )(10 ) 150

0.020 1.20 1.18 V

v

v

− − ⎛ ⎞= × × − × +⎜ ⎟⎝ ⎠

= − ⇒ = −

c. Circuit (a): Due to ,BI 0v = Circuit (b): Due to ,BI

9 3 9 60

0

50(700 10 )(50 10 ) (700 10 )(10 ) 150

0.035 1.40 1.365 V

v

v

− − ⎛ ⎞= × × − × +⎜ ⎟⎝ ⎠

= − ⇒ = −

Circuit (a): Due to 0 ,SI 9 3

0 0(250 10 )(50 10 ) 0.0125 Vv v−= × × ⇒ =

Circuit (b): Due to 0 ,SI

2

1

Let 825 nA575 nA

B

B

II

==

Then 9 3 9 6

0

0

50(575 10 )(50 10 ) (825 10 )(10 ) 150

0.02875 1.65 1.62 V

v

v

− − ⎛ ⎞= × × − × +⎜ ⎟⎝ ⎠

= − ⇒ = −

14.49

0

0

25 C; 2 A, 0.2 A50 C, 2 A (0.020 A / C)(25 C 2.5 A

0.2 A (0.005 A / C)(25 C) 0.325 A

B S

B

S

I II )

I

μ μμ μ μ

μ μ μ

° = =° = + ° ° =

= + ° ° =

a. Due to :BI (Assume bias currents into op-amp). 6 3

01

01

(50 k ) (2 10 )(50 10 )0.10 V

Bv Iv

−= Ω = × ×⇒ =

02 01

6 3 6 3

60 601 (60 k ) (50 k ) 120 20

(0.1)(4) (2 10 )(60 10 ) (2 10 )(60 10 )4

B Bv v I I

− −

⎛ ⎞ ⎛ ⎞= + + Ω − Ω +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + × × − × ×

or 02 0.12 Vv =

b. Due to 0 :SI

1

1

2

1st op-amp. Let 2.1 A2nd op-amp. Let 2.1 A

1.9 A

B

B

B

III

μμμ

===

6 301 1

01

(50 k ) (2.1 10 )(50 10 )0.105 V

Bv Iv

−= Ω = × ×⇒ =

02 01 1 2

6 3 6 3

60 601 (60 k ) (50 k ) 120 20

(0.105)(4) (2.1 10 )(60 10 ) (1.9 10 )(50 10 )(4)

B Bv v I I

− −

⎛ ⎞ ⎛ ⎞= + + Ω − Ω +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + × × − × ×

or 02 0.166 Vv =

c. Due to :BI

6 301 01

01 02

6 3 6 3

(2.5 10 )(50 10 ) 0.125 V

60 601 (60 k ) (50 k ) 120 20

(0.125)(4) (2.5 10 )(60 10 ) (2.5 10 )(50 10 (4)

B B

v v

v v I I

−

−

= × × ⇒ =

⎛ ⎞ ⎛ ⎞= + + Ω − Ω +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + × × − × ×

or 02 0.15 Vv =

Due to 0 :SI

1

2

Let 2.625 A2.3375 A

B

B

II

μμ

==

6 301 1

01

(50 k ) (2.6625 10 )(50 10 )1.133 V

Bv Iv

−= Ω = × ×⇒ =

02 01 1 2

6 3 6 3

60 601 (60 k ) (50 k ) 120 20

(0.133)(4) (2.6625 10 )(60 10 ) (2.3375 10 )(50 10 )(4)

B Bv v I I

− −

⎛ ⎞ ⎛ ⎞= + + Ω − Ω +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= + × × − × ×

or 02 0.224 Vv =

14.50

41

3 4B I

Rv vR R

⎛ ⎞= ⎜ ⎟+⎝ ⎠

and 20 2

1

( ) 1I BRv v vR

⎛ ⎞= +⎜ ⎟

⎝ ⎠

or

4 20 2 2

3 4 1

( ) 1I IR Rv v v

R R R⎛ ⎞⎛ ⎞

= +⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠

or 1.Iv

20 1

1

( )I IRv v vR

= − ⋅

Then

4 2 20 2 1

3 4 1 1

1 I IR R Rv v v

R R R R⎛ ⎞⎛ ⎞

= + − ⋅⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠

we can write 2 2d

I cmVv v= + and 1 2

dI cm

Vv V= − ⋅ Then

4 2 20

3 4 1 1

12 2d d

cm cmV VR R Rv V V

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

Common-mode gain

0 4 2 2

3 4 1 1

1cmcm

v R R RAV R R R R

⎛ ⎞⎛ ⎞= = + −⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠

Differential mode gain

0 4 2 2

3 4 1 1

1 12d

d

v R R RAV R R R R

⎡ ⎤⎛ ⎞⎛ ⎞= = + +⎢ ⎥⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠⎣ ⎦

Then

4 2 2

3 4 1 1

4 2 2

3 4 1 1

4 2 2

3 1 14

3

4 2 2

3 1 14

3

1 12

1

1 1 12

1

1 11

d

cm

AC M R RA

R R RR R R R

R R RR R R R

R R RR R RR

RC M R R

R R RR R RR

R

=

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

+ −⎢ ⎥⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠⎣ ⎦⎡ ⎤⎢ ⎥

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

⎛ ⎞⋅ ⋅ + −⎜ ⎟⎛ ⎞ ⎝ ⎠+⎜ ⎟⎝ ⎠

Minimum CMRR ⇒ Maximum denominator 4

3

maximum RR

⇒ and 2

1

minimum .RR

Then

4

3

2

1

(1.02)(50) 5.204(0.98)(10)(0.98)(50) 4.804(1.02)(10)

RRRR

= =

= =

Then 1 5.204 (5.804) (4.804)2 6.204

5.204 (5.804) (4.804)6.204

1 (9.6725)2(0.06447)

CMRR

⎡ ⎤⋅ +⎢ ⎥⎣ ⎦=⎡ ⎤⋅ −⎢ ⎥⎣ ⎦

⋅=

1075.0 20log (75.0)37.5 dB

dB

dB

CMRR CMRRCMRR

= ⇒ =⇒ =

14.51 Use the results of Problem 14.50:

Let 4

3

1 50 (1 2 )(5)1 10

R x xR x

+ ⎛ ⎞= ⋅ ≈ +⎜ ⎟− ⎝ ⎠

Let 2

1

1 50 (1 2 )(5)1 10

R x xR x

− ⎛ ⎞= ⋅ ≈ −⎜ ⎟+ ⎝ ⎠

Then

( ) ( ) ( )( )

( ) ( ) ( )( )

( )

2 2

2 2

22

1 2 51 6 10 1 2 52 6 10

1 2 56 10 1 2 5

6 10

1 30 10 100 30 10 100230 10 100 30 10 100

1 [60 200 ] 30 100220 20

xx x

xCMRR

xx x

x

x x x x

x x x x

x xx x

+⎡ ⎤⋅ − + −⎢ ⎥+⎣ ⎦=

+⎡ ⎤⋅ − − −⎢ ⎥+⎣ ⎦

⎡ ⎤+ − + − −⎣ ⎦=⎡ ⎤+ − − − −⎣ ⎦

⋅ − −= =

a. For 90 dB 31,623dBCMRR CMRR= ⇒ = x will be small, neglect the 2x term. Then 3020 0.0000474 0.00474%

31,623x x= ⇒ = =

b. For 60 dB 1000.dBCMRR CMRR= ⇒ = Then 3020 0.0015 0.15%

1000x x= ⇒ = =