Microelectronic circuits analysis and design 3rd edition rashid solutions manual

Microelectronic Circuits Analysis and Design 3rd Edition Rashid SOLUTIONS MANUAL Full download: https://testbanklive.com/download/microelectronic-circuitsanalysis-and-design-3rd-edition-rashid-solutions-manual/

Chapter 2 2.1 (a) vd = vO/Ao = 15 V/2 × 105 = 75 V (b) ii = vd/Ri = 75 V/2 × 106 = 37.5 × 10–12 A 2.2 vO = (v+ – v–) = AovS = 105 × 50 × 10–6 = 5 V 2.3 vO = – Aovd = – AvvS = – 105 × 10 × 10–6 = –1 V 2.4 vO = Ao (v+ – v–) = 2 × 105 (0 – 2 sin 377t) Output saturates at 14 V. 2.5 vO = Ao (v+ – v–) = 2 × 105 (75 + 25)10–6 = +20 V; at saturation, vO = +14 V 2.6 (a) vd = v1 – v2 = (120 – 80) V = 40 V 200 v v (b) vc = 1 2 = = 100 V 2 2 (c) 20 log CMMR = 90 log CMMR = 90/20 or | Ad/Ac | = 31,622 | Ac | = | Ad |/31,622 =

2 105 = 6.32 31, 622

(d) vO = Advd ± Acvc = 2 105 40 10–6 6.32 80 10–6 = 8 ± 50.56 10–6 V 2.7 (a) vd = v2 – v1 = 60 V – 100 V = – 40 V v v 60 100 (b) vc = 1 2 =

2

2

V = 80 V (c) 20 log CMRR = 90 dB log CMRR = 4.5 Ad Ac

= 3.1622 × 104 A

Ac =

d

3.1622 104

=

2 105 3.1622 104

= 6.3247

(d) vO = Ad vd + Ac vc = 2 × 105 (– 40 × 10–6) + 6.3247 × 80 × 10–6 = – 8 5.05976 × 10–4 = – 7.9994 V or – 8.00050 V 2.8 Choose 1 k < R1 < 10 M; let R1 = 10 k.

Af = 100 = 1 + RF/R1, RF = 900 k vO = vS (1 + RF/R1) = 100 × 10–3 × 100 = 10 V Problem 2.8 VS 1 0 RS 1 2 RF 4 3 R1 3 0 VCC 5 0 VEE 0 6 XA1 2 3 . SUBCKT OPAMP RI 1 2 RO 5 3 EA 5 4 . ENDS OPAMP . PROBE . TF V(4) VS . END

DC 1K 990K 10K 15V 15V 4 1 1E12 50 1

100MV

0 2

OPAMP 3 4

2

5E5

is

2.9 vd

iS = id =

Ri

, vd =

Rs +

vO

vs +

–

Ad

vd

Ri

v

o

Rout O

Ao Ri

, Ri >> Rs R

RfI vS R1

Substituting for vO = 1  

iS =

v

+ –

–

Then iS = id =

Ro

id

1Rf R1

Rf

i

vS, Rin =

Ao Ri

vS iS

=

Ao Ri

(i)

1 Rf R1

iO from the circuit v Av iO =

O

v o d

Ro

(ii) O

Ri Rf

Rf

iO Ro

R1

+ vd

+ vO –

Ao vd

Ri

–

Also – vd =

R1 vO R1 Rf

e

Assuming R i >> R 1

(iii)

From (ii) and (iii), 1

=

Rout

iO

=

1R A1 oR R1

Vo

f

1

Ro

R1 Rf

R o

Assuming (R1 + Rf) >>

(iv)

1 Ao R1 (R1 Rf ) R Rout =

o

1 Ao R1 (R1 Rf )

(a) From (i) 25 103 1012

A R o

Rin =

i

= 250 × 1012

= 1 990 k 10 k

1 Rf R1 Rout =

50 3

1 25 10 10

4

(10 k 990 k)

(b) Rin = 5 × 105 × 1012/100 = 5 × 1015  Rout =

50 5

1 5 10 10

4

1000 103

2.10 Let R1 = 10 k. Then Av = 10 = 1  

RIF  , RF = 90 k R1

vO = 500 mV × 10 = 5 V

= 0.01

=

50 1 250

= 0.2

2.11 0

< t < sms

vo

15

0

1.5

7.1

8

9.5

t(ms)

vS needed for vO = 15 V RFI   20  , vS = 3 V = vS 1  5 R1 

15 = vS 1  

3 vS = 2t, t =

= 1.5 ms

2

0 < t < 5 ms

10 80 t+ 3 3 3 10

3= t=

71 ms 10

v– =

vO R1 R1 RF

2.12 (i)

vO = Ao vd = Ad (vS – v–) vO Ad

v– = vS –

(ii)

From (i) and (ii) 1

R 1

vS = vO

Ao

R1 Rf

v Af =

(a) Af =

O

vS

A R R =

o

1

f

R1 RF Ao R1

5 105 40 103 40 103 5 k 105 104

= 3.999968

(b) vO = Af vS = 3.999968 vS RIF  30 k =1+ = 4, vO = 4 vS Error insignificant R1  10 k

(c) For Ao , Af = 1    2.13 vO = vS 1  

vO =

15 V

(iii)

RIF  R1 

= 5 vS

1 5 vO = 50 sin 2000 t = 15 V, = sin 1

, = 17.45° 50

vO

197.45° 342.55° 17.4°

162.5°

q

2.14 Assuming vd to be negligible, vO ≃ v+ = vS. Hence, vO = vS = 5 V iS = ii = vd/Ri and vd = vO/Ao iS = ii =

vO Ao Ri

≃

v

Sv

,

Ao Ri

= Rs + Ao Ri

iS = 10 k + 5 × 105 × 105 × 2 × 106 ≃ 1012

iO =

vOI AoIvdI Ro

vOI

(i)

Ri Rs – Ro vd

Ri

vo

+

Rs

vS

+ A v o d –

iS

Also vd =

e vOe Ri

(ii)

Ri Rs

(i) and (ii) give iO = 1

=

Rout

vO A v oRO (R i R i) Ro iO vO

=

sv

1A Ro (R Ri ) i Ro

O

Ri Rs s

1 Ri Rs

Ro << (R i+ R s) 1 Ao Ri (Ri Rs ) Hence Rout =

=

Ro 1 Ao Ri (Ri Rs ) 75

–5

1 5 105 2 106

(2 106 10 4 )

≃ 15 × 10 

– iO

Ri Ro

+

+ –

Rs

+ vO –

Aovd

2.15 (a) From Eq. (2.13) fbw = Ao fb = 2 × 105 × 10 = 2 MHz From Eq. (2.31) we get 1 R R F /1 Af (j) = 1 jf (1 RF R1 ) f bw RF = 50 k, R1 = 15 k 1 50 k 15 k Af (j) = 1 jf 1 50 k 15 k  2 106 4.333 = 1 j 100 103 (4.333) 2 106 =

4.333 = 4.235 –12.23° 1 j 0.2167

From Eq. (2.32) R fc = Ao fb = =

1 A f R1 RF o b

15k 15 k 50 k

× 2 × 105 × 10 Hz

= 461.54 kHz (b) 2 4 3 vS

RF R1

VS 2 0 AC 0.1V RF 4 3 50K R1 3 0 15K XA1 2 3 4 0 * Vi+Vi– Vo+ Vo– * Subcircuit definition for UA741-AC . SUBCKT UA741-AC 1 2 3 * Subcircuit name Vi+ Vi– Vo+ RI 1 2 2MEG RO 6 3 75 GB 4 5 1 2 0.5M R1 5 4 10K C1 5 4 1.5619UF EA 4 6 54 2E+5 D1 3 7 DMOD D2 4 7 DMOD . MODEL DMOD D (BV = 14V) . ENDS UA741-AC . AC DEC 10 100 HZ 1 MEGHZ . PRINT AC VM(4) . PROBE . END

4 Vo–

2.16 (a) R1 = RF = 15 k fbw = Ao fb = 2 × 105 × 10 = 2 MHz From Eq. (2.31) Af (j) =

1RF R1 1 jf 1 R R f F

=

=

1

11 1 jf 1 1 2 10

bw

=

6

2 1 j 100 103 10 6

2 = 1.998 –5.71° 1 j 0.1

(b) From Eq. (2.32) fc = Ao fb =

15 k × 2 × 105 × 10 = 1 MHz 15 k + 15 k

VS 2 0 AC 0.1V RF 4 3 15K R1 3 0 15K XA1 2 3 4 0 * Vi+Vi– Vo+ Vo– * Subcircuit definition for UA741-AC * SUBCKT UA741-AC 1 2 3 4 * Subcircuit name Vi+ Vi– Vo+ Vo– RI 1 2 2MEG RO 6 3 75 GB 4 5 1 2 0.5M R1 5 4 10K C1 5 4 1.5619UF EA 4 6 5 4 2E+5 D1 3 7 DMOD D2 4 7 DMOD . MODEL DMOD D (BV = 14V) . ENDS UA741-AC . AC DEC 10 100HZ 1MEGHZ . PRINT AC VM(4) . PROBE . END

2.17 From Eq. (2.35) vO = 1 

R

F

R

A

R va 

RB Ra

A

R vb 

Rb

A

vc

Rc

20 RA = Ra || Rb || Rc = vO = 1

40k 202

k

3

20(2)   3 20 3 20 

20(3)

20 k 3 20 = 3

2 2 1 =–3V 3 3

2.18 iS =

vS 50 mV = 10 A (R1 = Rs) = Rs 5 k 5V

vO = – A f v S ,

RF

50 mV

= 100 = Af

= 100, RF = 100 × 5 k = 500 k

R1 Rx = R1 || RF = 5 k || 500 k = 4.95 k

2.19 vS

if

iS –

Rs = R1

RF

Ro

vd

Ri

vO

+ Aovd Rx rf

Assume Ri >> Rx Rin =

where

vS = R1 + Ri || rf iS

(i)

vd if

rf = –

– vd = if RF + Ro if + Ao vd or

rf =

v

d

if rf =

,

v

d

(1 + Ao) = RF + Ro

if

RF R o 1 Ao

(ii)

The expression for Rout is the same as in Problem 2.14. (a) Rin = 5 k + 1012 ||

500 103 50 1 25 103

≃ 5 k + 1012 || 20 ≃ 5020  Ro 1 Ao Ri (Ri Rs )

Rout =

50

=

1 25 103 1012

(1012 5 k)

50 103 

25

≃ 2 × 10–3

From Eq. (2.45) RF R1 Af = 

11 R R A F

=

1

o

500k 5k 1 1 500 k 5 k 

= – 99.60 25 103

vO = – 50 × 10–3 × 99.6 = – 4.98 V

500 103 50 1 5 105

(b) Rin = 5 k + 1012 ||

= 5 k + 1012 || 1 ≃ 5001  Ro 1 Ao Ri (Ri Rs )

Rout =

=

50 5

12

1 5 10 10

(1012 5 k)

≃ 100 × 10–6 = 10–4 (c) From Eq. (2.45) RF R1 Af = 

11 R

R A E

=

1

500k 5k 1 1 500 k 5 k 

o

– 99.98 5 105

Problem 2.19 VIN R1 RF RX XA1

4 4 2 1 1

0 2 3 0 2

DC 10MV 5K 500K 4.95K 3 0 LF411-DC

. SUBCKT LF411-DC 1 RI 1 2 1E12 RO 5 3 50 EA 4 5 1 2 . ENDS LF411-DC . TF V(3) VIN . PROBE . END

2

3

4

25E3

2.20 Af =

eRF

R1

= , vO = AfvS = Saturation

vO = – 14 V 2.21 (a) From Eq. (2.46) x=

1

1

RF 

Ao 



R1 

=

1 2 105

1

50 k 

 10 k

= 3 × 10–5

From Eq. (2.45) Af =

e

RF =  R1 (1 x)

50 k 10 k (1 3 10 5 )

(b) vO = Af vS = – 4.9998 × 100 × 10–3 = – 0.4998 V

= – 4.9998

(c) From Eq. (2.47) Vo = x

RF = 3 × 10–5 × 5 = 15 × 10–5 R1 = 0.15 mV = 0.03%

2.22 (a) From Eq. (2.13) fbw = Ao fb = 2 × 105 × 10 = 2 MHz From Eq. (2.56) Af = 

RF R1 1 jf 1 R R f F

=

=

50 k 15k =

1

1 jf 1 150 k 15 k 

2 106

bw

3.333 1 j (100 103 ) (1 3.333) 2 106 3.333 = – 3.258 –12.19° = 1 j 0.216 R1 15k = 0.230 = R1 RF 15 k 50 k

fc = fbw = 0.23 × 2 × 106 = 460 kHz (b) Problem 2.22 VS 1 0 AC 0.1V R1 1 3 15K RF 3 4 50K RX 2 0 10K XA1 2 3 4 0 UA741-AC . SUBCKT UA741-AC 1 2 3 4 RI 1 2 2MEG RO 6 3 75 GB 4 5 1 2 0.5 R1 5 4 10K C4 5 4 1.5619UF EA 4 6 5 4 2E5 D1 3 7 DMOD D2 4 7 DMOD . MODEL DMOD D (BV = 14V) . ENDS UA741-AC . PRINT AC VM(4) . AC DEC 10 100 1MEGHZ . PROBE . END

40 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.23 R1 = RF = 15 k

(a)

fbw = Ao fb = 2 × 105 × 10 = 2 MHz From Eq. (2.56) Af = 

1

RF R1 =

1 jf 1 R R f F

1

1 j 100 103 (1 1) 2 106

bw

1 = – 0.99 – 5.71° 1 j 0.1 fc = fbw, = 0.5 =

fc = 0.5 × 2 × 106 = 1 MHz 2.24 (a)

R1

R2

R3

v1

vi

vO Li

iS

iF

Rx

i

Ignoring vd (virtual ground) iS =

i=

vi R1

= iF, v1 = – iF R2 =

v1 R = + v1 2 Rx R1 Rx

i1 = iF + i =

vi R vi 2 R1 Rx R1 R

Also

vO = v1 – i1R3 = – v

vi

=

R2

1

R1 

R3 R3  R2

R

1

2

2

R i

R1 vO

vi R R2 = vi 2 R1 R1

R1

3

vi

R1 Rx

 Rx

41

20 k 

© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

– 10 =

1 50 k 

20 k

20 k

20 k

Rx

42 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

10 =

2  2

20 k 

5

Rx

20 k R

, 25 = 2 

x

20 k Rx =

= 869.5 

23

SR 2Vm

(b) SR = 6 V/s, fs(max) = 6 106 fs(max) =

(c)

= 95 kHz

2 10

VS R1

5 5

0 3

AC 10MV 50K

R2

2

0

1K

R3

3

6

20K

RX R4 XA1

6 4 2

0 6 3

869.5 20K 4 0 UA741-AC

. SUBCKT UA741-AC1

2

3

RI

1

2

2MEG

RO

6

3

75

GB

4

5

1

R1 C1

5 5

4 4

10K 1.5619UF

EA D1 D2

4 3 4

6 7 7

5 4 DMOD DMOD

2

4

0.5 M

1E5

. MODEL DMOD D (BV = 14V) . ENDS UA741-AC . AC DEC 10 100 20MEGHZ . PRINT AC VM(4) VP(4) . PROBE . END 2.25 From Eq. (2.60) R vO =

F

R1 40k = 

v 1

R

R

Fv

R2 2

2

40 k

Fv

R3

3

40 k (3) 

43

(2)

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20 k

20 k

20 k



= – (4 – 6 – 4) = + 6 V

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2.26 R2

il

R1

iS v

R4

i

iF

R3

vi

+

v

d

S

v

–

iS = iF =

o

vS , v1 = – iF R2 = R1

R2 v1 = + vS R1 R4 R4

i=

R

v S

i1 = iF + i =

Also

= – vS

= –

vS

R2

2

R1

–R 3

= –

R2

1

R1 R4

vS R2 v R 2 – vS R3 + S R 3 R1 R1 R4 R1

R – R3 

R1

2

+ vS

R1

vO = v1 – i R3 =

vO

v R2 R1

S

1

R vS R1 R4 2

R1

1

R2

 R1

 R1 R4

R

3

R1  R2

3

R ,

Given

R4 

O

v

= vS

12

= – 240

0.05

Taking R2 = 6 R1 and R3 = 9 R4 240 = 6 1 9 

3

R = 6 10

3

R2 

R  R2

R3 = 30 R2 Let R2 = 10 k and R3 = 300 k. Then 9 300 k 100 R4 = =

45 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

3 k= 33.33 k R1 =

10  k 6

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2.27 From Eq. (2.67) RF R , 5 = (220 – 200) mV × F R1 R1

vO = (va – vb) RF = 250 R1

Let Ra = R1 = 2 k; then RF = 0.5 M = Rx 2.28 vO = – Ao (v+ – v–)

(i)

12

Ri = 10

Ignoring current through Ri v+ =

va Rx

and v– =

Ra Rx

RF

e

vb

R1

e

R1 RF

v

O

R1 RF

Substitution in (i) gives

vO =

eAoe vae

Rx eAoe RFevb eAoe R1evO

Ra Rx

R1 RF

R1 RF

For Ra = R1 and RF = Rx

vO =

eAoe vae

RF

A

R1 RF

RFevb

e

o

eAoe

R1 RF

R1evO R1 RF R v

R = – Ao

F

R1 RF

(va – vb) + Ao

1 O

(ii)

R1 RF

A R (R R ) or

vO =

o

F

1

F

(vb – va)

1 Ao R1 (R1 RF ) = (a) vO =

Ao RF (vb va) R1 (1 Ao ) RF

25 103 0.5 106 (200 220) 103 2 103 (1 25 103 ) 0.5 106 5 105 0.5 106 20 10 3

(b) vO =

2.29

2 103 (1 5 105 ) 0.5 106

=

250 106 2 10 50 106 0.5 106 3

≃ 5.05 V

≃ 5V

47

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(a) Ao , Eq. (ii) from the last problem reduces to R R | A f | = vO/vb – va = F , 200 = F R1 R1 Let R1 = 2.5 k. Then RF = 0.5 M.

48 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(b) For Ao = 5 ×105 A oR F (vb – va) R1 R1 Ao RF

vO = =

Af



5 105 0.5 106 20 10 3 2.5 103 2.5 103 5 105 0.5 106

=4V

= 200 × 20 × 10–3 = 4 V

Hardly any error.

2.30 From Problem 2.28 oAa vx R

vO =

oAF Rb

Ra Rx vO =

v

o 1 A b

R1 RF

Rv

R1 RF

5 105 15 103 20 103 2 103 20 103

5 105 50 103 5 103 5 103 50 103



5 105 5 103Ov 5 103 50 103 = 6.818 × 103 + 2.777 × 103 + 4545 × 103 vO vO ≃ – 0.2 V 2.31 (a) i6 v1

i4

R4

i5

i

R5

7

R6 R7

v2

R1

R3

+

vO

– R2

v –v

R 2

v+ =

v2,

i4 =

R1 R2

R4

3

R7

v –v ,

i5 =

–

1

R5

v –v

v i7 =

–

1

,

i6 =

3

O

R6

i4 = i5 v –v v1 – v– = – 3 R5 R4

(i) v v

v v

49

v

© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

i5 = i6 + i7,

–

R5

3

=

3

O

R6

3

R7

v v v3 v3 v 3 – – =– – – O R5 R6 R5 R6 R7

50 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

– v3

1 R5

v 1 v =– – – O  R7  R5 R6

1  R6

(ii)

From (i) v v v v1 – – = – – 3 R5 R5 R4 R4 v1

–v

R4

v1

1 

1 –

R4

R5

R5

R4 R5  

– v–

v3

= –

 R4 R5

R4

v3

= –

(iii)

R5

From (ii) R R R R – v3

5

6

6



R R 7

7

v = –

5

R5 R6 R7

v –

–

R5

O

(iv)

R6

From (iii) using (iv) v1 R4

Let

R4 R5

–v –

R4 R5

X=

R5 R6 R7

1 =

–

R5 R5 R6 R6 R7 R5 R7

v–

vO

– R5

R6

R5 R6 R7 R5 R6 R6 R7 R5 R7 R R v1 – v– 4 5 R4 R5 R4

= –

1 R5

R R

v X

–

– v–

– v–

4

1 5

+

R4 R5

4

R5

R R 5 

1

R4 R5

R5

vO R6

v– 1 – R5 R5 v =–

R5

R4

1

–

v = –

X

1

1

R5 

1

R4

v O

R5 1

–

R6 v O

R5

R6

1 v1 R4  v– =

R

vO

R6

5

1

515

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R4 R5  R4 R5 –

R

1 R5

Also (v+ – v–)Ao = vO. Substituting for v+ and v– 1 v R5  O R6 1 1 –   R5 R5 R4 R5

v R14 R2 v2 – R4 R5 R1 R2

1

v1 R2 R1 R2

v2 –

R4 R4 R5 R4 R5

–

1 R5

–

1 R5

vO Ao

=



vO 

R5 R6 R4 R5 X 1 –

R R

=

R 4

5

5



vO Ao R

5

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R2 v R1 R 2 2

v1

– R

R4 R5 1

1

–

X

R

R R 4



R2



4

R5  R4e R5

–



R

–

1 X

R

5

5

v6 /R + A

vO o

R 5

o

R 

5

1

1 R6



R5 



5

5

1 5

X X 1 

R4 R5

–

o

1

v1

R R R4  4 R R

2

R 5R

4

X

R R

4

1

X

 5

v –

R1 R2

vO =

R

5

5

1 R5

=

1 

A

R5

Assume Ao is large

R2

v –

R R R4  4 R R

2

R1 R2

4

VO =

v1

1

1 5

–

1 X

R

R

5

5

5



X R5 R4 R5 1 – R5  R4 R5

1 X

R5

Given R4 = R5 = R6 = R7 = 50 k R2

v – 2

R1 R2

R4 R5 R4

VO =

X=

v1 –

X 5R

X R4 R5 – X 50 k 50k 50k 50 k = 350 k 50 k  3 v 150 k 250 k

v1 –

100 k 50 k – 50 k 350 k  50 k 3100 k – 50 k 3

vO = 0.6 v – 0.6 v =

2

50 k 250 k

1

2

3 v2 – v1  2.32 (a)

53 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

v _

i F

1 vi

i1 i3 RF R1 R3 3 i2 R2 R4 4 vO 2

54 Š 2017 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

i1 =

vi – v– v , i3 = – , iF = i1 + i3 R1 R3

iF =

vi v– v– – – R1 R1 R3 v –v

Also

–

iF =

–

vO

=

RF

v

O

vi

–v

R1 i

=

+

R2

R

–

–

R3

1  R3

RF 

(i)

v

v

v –

i

, R4

R2

+

=

R2

R4

1

1

= v

–

v

R2 vi

v

R1

1

1

R

v –v i2 =

R1

1 –

–

i

=

RF

v

R4

2

R2 R4 vi = v +

R4 ,

v+ = vi

R4

R R 2

4

vO = Ao (v+ – v–)

Also

vO = v+ – v–, Ao

v– =

e

v– = v+ –

vO Ao

vie R4 v – O Ao R2 R4

(ii)

Substituting in (i) –

vO RF

=

vi R1

–

R4

RF

vO vi

i

R2 R4

vO

1

v

–

1

1

R1

vO 1 Ao

R1

R3

+ RF 

1

1 R3

RF 

vO 1 Ao

=

vi R1

1

R1

–

1  R3

R4 R2 R4

RF 

1

v i

R1

1

1  R3

RF 

For large Ao

55 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

vO vi

1 R – R R R 1

= –

= –

= –

1

R 41 R 11R  3 1 RF 1 Ao 1 R1 1 R3 1 RF  4

1 1R 4

2

F

– R2 R4 R 11 R 1 R3 1 R F 1 RF

RF

RF R4 1

R1

R2 R4

1 R1

R3

vO – 500k 500k 100 k 1 = vi 40 k 80 k + 100 k

1 RF

1 1 40 k 50 k

500 k

= – 12.5 + 277 k (20 10–6 + 20 10–6 + 2 10–6) = – 12.5 + 13.0 = 0.5

56 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Frequency response A(j) = vO vi

Ao , where fu = fb = Aofb 1 + jf f b

(j) = – = –

1 R 1 – R4 R 2 R 14R 1 R1 1 R 3 1 RF 1 Ao j 1 R1 1 R3 1 RF 

F

1 R1 – R4 R2 R41 R1 1 R3 1 RF 1 RF 1 R1 1 R3 1 RF 1 jf f b  1 R1

= – 1

1

FR

1 1

R2 R 4R

1 R1

R4

–

1 1  R3

Ao

RF  Ao

1

R

1 3

R

F

1 1 1   R1 R3 RF 

jf Ao

f

b

For large Ao, the second term in the denominator tends to zero. Therefore, vO vi

(j) = –

1

 1 R4 – R 2 R4 R 1 R 1 1R 3 1R F 1 RF 1 R1 1 R3 1 RF jf f bω  – RF R1 – R4 R2 R41 R1 1 R3 1 RF

=

1 + jf RF 1 R1 1 R3 1 RF 

= vO vi

f bω

500 k 40 k – 500 k 100 k 80 k 100 k 1 40 k 1 50 k 1 500 k  1 jf 500 k 106 1 40 k 1 50 k 1 500 k  12.5 –13.05

(j) =

0.55

=+

1 jf 23.5 10 –6

1 jf 23.5 10 –6

2.33 (a) i2

1 2

R2

R1

i1 vo1

3

v1

2

vo2

vO

R3

49 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

R4

v

(v(+)1 – v(–)1)Ao = vo1, v(+)1 = v1, v(–)1 = vo1

51 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

(v1 – vo1) =

vo1 =0 Ao

For large Ao, v1 = vo1, and similarly vo2 = v2.

Now i1 = i2; then i1 =

vo1 – v(–)3 R1

i2 =

v(–)3 – vo R2

vo1 – v(–)3 = R1 vo1

–

v(–)3

R1

–

v(–)3

v(–)3 – vO R2 =

R1

–

=

1

R2

– vo1

R2 =

R1 R2

(–)3



R

1

1

=

o1

vo1

R2  R1

v

vo 

R2

o

R1

v(–)3 =

1

, v

R2

v

2

vO

–

R1

R R v(–)3

vO

–

R2

v(–)3

R2

v(–)3

R2

R2 R1 v1 + vO R1 R2 R1 R2 R

v(+)3 = vo2

or

v(+)3 – v(–)3 =

4

R3 R4

[v(+)3 = – v(–)3] Ao = vO

,

vO Ao

Substituting for v(–)3 and v(+)3 v2

R4 R2 R v 1 – v1 – vO = O R3 R4 R1 R2 R1 R2 Ao

For large Ao R

R –

4

v2 =

R3 R4 R vO

1

R1 R2

R

2

R1 R2

R = – v1

2

R1 R2

1

v1 =

vO

R1 R2 R

+ v2

4

R3 R4 R R R 50

© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

vO = – v1

R2 + v2 R1

4

1

2

R1 R3 R4

For R1 = R3 = 10 k and R2 = R4 = 1 M, vO = – v1 

106 106  + v 2 10 4 10 4

10 k + 1 M 10 k + 1 M

51 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

= – v1 100 + v2 100 = (– 100) v1 – (– 100) v2 a1 = – 100, a2 = – 100 (b) Assume v2 = 0; then vo = a1v1. fmax =

SR 6 V / µs = = 95.5 kHz 2 vO 2 10

(c) Problem 2.33 VS1 2 VS2 5 R1 4 R3 7 R2 9 R4 8 XA1 2 XA2 5 XA3 8 . SUBCKT RI 1 RO 6 GB 4 R1 5 C1 5 EA 4 D1 3 D2 4 . MODEL . ENDS . AC DEC . PRINT . PROBE . END

0 AC 1MV 0 AC 2MV 9 10K 8 10K 10 1MEG 0 1MEG 4 4 0 LF411_AC 7 7 0 LF411_AC 9 10 0 LF411_AC LF411_AC 1 2 3 4 2 2E12 3 50 5 1 2 0.1M 4 10K 4 0.79UF 6 5 4 2E5 7 DMOD 7 DMOD DMOD D (BV = 14 V) LF411_AC 10 100 20MEGHZ AC VM(10)

52 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.34 +

v

R 1

R2 ix

R

i1 vo1

–

Rx

i

ix

3

vO

R3

ix

R4 2

vo2 R5

v2

Assume there is virtual short circuit at the input of op-amps 1 and 2. Then v(–)1 = v1, v(–)2 = v2 Current ix through Rx is given by ix =

v1 v Rx

2

vo1 = v1 + ix R2 = v1 +

vo2 = v2 – ix R3 = v2 –

1v

v2 Rx

R2

(i)

v1 v 2 R3 Rx

(ii)

53 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

v o1

i1 = iF, i1 =

v

v (–)3

, iF =

R1

v

( )3

O

RF

i1 = iF vo1 v(–)3 R1 vo1 v()3 R1

v(–)3 vO

=

RF v()3

=

R1

v(–)3

v(–)3

R1

RF

vO



RF

=

RF

vo1

vO

R1

RF

1 1 R R F = 1 R1 RF R1 RF 

v(–)3

R1 R F

v(–)3

=

R1 RF

v

vO

o1

R1

RF

R R 1

v(–)3 =

v

F

o1

R1 RF

=

R R 1

+

R1

v

F

O

R1 RF RF

RF R1 vo1 + vO R1 RF R1 RF

R v(+)3 =

5

[v(+)3 – v(–)3] Ao = vO

vo2,

R4 R5

R5 RF R1 vo2 – vo1 – R4 R5 R1 RF R1 RF Substituting for vo1 and vo2 from (i) and (ii)  R R R 5

3

v2 (v1 v2 )

R R 4

5

R



R1 vO R1 RF

=

R5 R4 R5

F

v1 (v1 v2 )

R R x  1 v2 –

F

R5

R

 R3

R4 R5 Rx

1

v +

vO Ao

vO =

R

R

2

1

R x 

v O

vO =

R 1

A

F

R5 R3  v2

R4 R5 Rx G iv

e n

R 2

54 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

= R

3,

R1 = R4 and RF = R5,

R1 vO = R1 RF

–

RF v2  R1 RF

RF RF v1 – R1 RF R1 RF RF R2 v1 + R1 RF Rx

RF R1 RF

RF  R2 RF v1 – v1 R1 RF R1 RF Rx =

RF 2RF (v2 – v1) – R R1 RF 1 RF

R2 v1 Rx

R2 RF v1 + Rx R1 RF

R2 Rx

v2

R2 v2 Rx

RF + R1 RF 2RF + R1 RF

R2 v 2 Rx  R2 v2 Rx

55 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

=

2RF  RF (v2 – v1) + R R 1 F R1 RF RF 2R  1 2 R1 RF  Rx

= (v2 – v1)

R1 vO

= (v – v ) 2

R1 RF

1

vO = (v2 – v1)

R2 (v2 – v1) Rx



RF R1 R

F

R

F

1

2R2

1 

Rx

2



2R

R1  Rx 2.35 From Eq. (2.73) vO

=

1

vs1 vs 2

2R RF  Rg R1

Take RF = 500 k, R1 = 10 k and R = 1 k. Then 21000 2000 vO × 50 = 1+ × 50 ≃ R Rg vs1 vs 2 g Rg can be varied from 100 to 200 .

2.36 Av =

vO vs2 vs1

FR

R1

For R1 = 1 k, RF = 750 k.

Problem 2.36 VS1 1 0 DC 1MV VS2 4 0 DC 2MV R1 2 5 1K RA 3 6 1K RX 6 0 750 RF 5 7 750K XA1 1 2 2 0 OPAMP XA2 4 4 3 0 OPAMP XA3 6 5 7 0 OPAMP 4 . SUBCKT OPAMP 1 2 3 RI 1 2 2E12 RO 5 3 50 56 EA 4 5 1 2 2E5 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. . ENDS OPAMP

. OP . END

57 Š 2017 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.37 t

1 v dt R2 CF 0 S

vO = Vco – For 0 t 1 ms vO = 0 –

1 10 5

t



(1) dt = – 105 (t) 0

For t = 1 ms, vO = – 10 × 10–3 = – 100 V. The amplifier goes in saturation at vO = – 14 V and Vco = – 14 V. For 1 t 2 ms 5

t1

vO = – 14 – 10

5

(1) dt

= – 14 + 105 (t – 1)terms

0

For t = 2 ms, vO = – 14 + 105 (10 –3). The amplifier goes in saturation at vO = + 14 V and Vco = 14 V. For 2 t 3 ms t2

(1) dt = 14 – 105 (t – 2)terms

vO = 14 – 105 0

At t = 3 ms vO = Vsat = – 14 V Vco = – 14 V For 3 t 4 t3

(–1) dt = – 14 + 105 (t – 3)

vO = – 14 – 105 0

At t = 4 ms vO = Vsat = + 14 V Vco = 14 V Waveform is shown below. v vO vS t (ms) 1

2

3

4

58 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2.38 vS = vS =

10 t 10 3

0 t 1 ms

10 t + 20 103

1 ms t 2 ms

For 0 t 1 ms, Vco = 0. t

1 vO = 0 –

1

10

F

R C 010

3

10 t 2

1 t dt = 

103 10 3 2

At t = 1 ms t2 Vo = – 5 V = – 107

2

2.39 R1 || RF = R1 for RF >> R1 CX = CF (1 + AO) = 0.01 F (1 + 5 × 105) 5000 F 1 CY = C F 1 ≃ C =F 0.01 F  Ao From Example 2.11 A(s) =

Ao 1 R1Cx s

3-dB frequency b

=

1

1

=

R1Cx

=

10 4 5000 10 6

1

rad/s

50

fb = 3.18 × 10–3 Hz 2.40 Since A = 10 at f =

1 = 0.16 Hz 2

fb = 10 × 0.16 = 1.6 Hz Let C1 = 1 F. Then i

=

R1 =

1 2 1.6 99.5 103 10 6 10

= 99.5 ms

100 k 6.25

59 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

F

=

1.6

= 6.25 s, RF =

1 µF

= 6.25 M

60 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Problem 2.40 VS 1 CF 3 R1 1 RF 3 RX 2 VCC 5 VEE 0 XA1 2 . Vi+ Vi–

0 0 3 4 0 0 6 3 +VCC

AC 1MV 1UF 100K 6.3MEG 100K 12V 12V 5 6 –VEE Vo

4

UA741

. LIB NOM.LIB . AC DEC 101 0.01HZ 10HZ . PROBE . END 2.41 (a) Given values

ABP 20 *

R1 

110 3

RL 

2010 3

fH 

1000

Using Eq. (2.86), dc gain (for an ideal op-amp and neglecting the effect of RL)

RF 

ABP R1

RF 2 10

4

From Eq. 2.87

1 CF  2 RF fH

CF 7.958 10 9

CF = 8 nF (b) For SPICE verification,

61 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

59 Š 2017 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Problem 2.41 VS 1 0

AC

1MV

R1 2 1 1k RF 3 2 20k CF 3 2 8nF Ri 2 0 2Meg E1 N002 0 2 0 -2E+5 Vin 1 0 AC 1mV 0 RL 3 0 20k Ro 3 N002 75 .ac dec 101 0.1 100kHz

. PROBE . END Plot vm(3)

Voltage magnitude (mV)

20

15

10

5

0

1m

10 m

100 m

1

10

100

1k

10 k

100 k

Frequency (Hz)

2.42 (a) From Eq. (2.108) fb =

1 1 , 104 = 2 R1C1 2 R1C1

For C1 = 0.1 F, R1 = 159 .

From Eq. (2.110), Af(max) =

eR F

R1

, RF = 20 × 159 

RF = 3180  (b) Problem 2.42 VS 1 R1 1 C1 7 RF 3

0 7 3 4

AC 0.1V 159 0.1UF 3180

58 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

RX 2 0 3180 VCC 5 0 15V VEE 0 6 15V XA1 2 3 5 6 . LIB NOM.LIB . AC DEC 10 100Hz 1MEGHZ . PROBE . END

4

UA741

2.43 (a)

d

= R1C1 = 2 × 103 × 10–8 = 20 s

(b) fb =

1 2 R1C1

1 2 2 10 10 8 3

R (c) Af(max) =

10 k

F

R1

≃ 7958 Hz

2k

=5

2.44 Using Eq. (2.110) APB =

RF R1

Assume R1 = 5 k and APB = 20.

RF = 5 × 20 = 100 k fL =

1 , 2 RC1C1

fH =

1 2 RFC1

1 2 5k 5k

= 6.4 nF

1 2 10 103 6.4 10 9

250Hz

C1 =

For SPICE verification: using op-amp model of Fig. 2.3(a) as

Problem 2.44 R1 3 2 5k RF 4 3 100k C1

59 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

2 1 6.4nF Ri 3 0 2Meg

61 Š 2017 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

E1 N002 0 3 0 -2E+5 Vin 1 0 AC 1mV 0 RL 4 0 20k Ro 4 N002 75 .ac dec 101 0.1 100kHz . PROBE . END Plot

2.45 (a) R1

i1

R2

i2

– vi

R4 vo1

+

R5

i4

i4 R6

– A2 +

vo

Assume ii = 0 0–v i1 =

v –(1)

R1

vo1 v– ( 2) R4

= i2 =

=

v –(1)

o1

(i)

R2

v– ( 2) vo R5

(ii)

60 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

From (i) –

–

–

R1 v(1) R1

 v–(1) 

v–(1) R1

v– (1) 

v(1)

R2

=

R2 v(1)

vo1 R2

–

R2

vo1

=–

R2



1 1  v o1  = R1 R 2  R2 R R

v– (1)

v– (1)

v(1) – vo1

=

1

v 2

R1 R 2 1R

R2

=

o1

R2

= vo1,

R1

v– (1) = vo1 .

R1 R1 R 2

 vo1 = Ao1(V+ (1) – v– (1)) = Ao v – v 

vo1 = Ao (vi) – Aovo1 . vo1 + Aovo1

vo1 = 1 Ao R1

e

1

R1 R 2 

= Aovi

R1 R 2 Ao R

o1

R1

R1 R1 R 2

R1

e

i

e

. vi

R 2

Now v+2 = vi 

vo1 – v–( 2) R4

=

v–(2) – vO R5

v v v vo1 – –(2) = –(2) – O R R R R4 4 5 5

61 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

vOO = Ao (V+ (2) – v– (2)),

v

= v+ (2) – v– (2)

Ao vO = vi – v– (2), Ao vo1 R4

+

vO R5

= v– (2)

v– (2) = vi –

vO Ao

R4 R 5  R 4 R5

62 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

vo1

+

R4

 vO = v –v O (R 4 R 5) i

R5

Ao

v1

.

vO (R 4 R 5)

+

R5

Ao

vO 

+

R1 R 2 R 4

1 Ao R1

vO

Ao 

1

R5

vO

4

= vi

= vi

R4 R5

R

Ao

Ao R 4 R 5

.

vi

R1 R 2 R 4

1 Ao R1

R

4

= vi

R 4 R5

–

R 4 R5

R  5 

R4 R5

– vO

R 4 R5

A 5

–

o

R 4 R5

R

R

R1 R 2 R4

1 R1 Ao



A

R R

o – 1R1 R1 R 2Ao R4 1 R5 R 4 R 5  Ao R 4 R 5

4

5

vO = vi

Now let Ao =

R4 R5

R5

R 4 R5

R

R 4 R5

4

5

Ao 1 jf / f b R

R 4

A

R R 5



4

vo = vi

o – 1jf /f b1R 1 R1 R 2.Ao 1jf /f b R 4 R 4 R 5 .1 jf / f b  1 R5 R 4 R5 5

Ao

R 4 R5

=

Ao –  R1 R 4 R5 1jf /f bR 4 R A R R 4 o 1 2   1 R R R R jf 5 4 5 4 .

R R A fb R5 R 4 R 5 Ao 4 5 o

R 4 R5 R 4 R5

Ao

– R 4

jf f

R 4

63



R 1 R 4 Ao  R1 R 2

© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

=

 R R

1 4

R5



5

1 R5

=

jf

4

5

.

fb

1 R1R 4 

R4 jf f b Ao R1 R 2 R 4R 5jf .

Ao f b R 4 R5

R 4 R5  – R 4 R5

R R

R 4 R 5 Ao

R 4 R 5 Ao

R 4 R5  R 4 R5 – =



b



1 jf

R4

R1R 4 

Ao f b R1 R 2 RI4 RI5  jf  1 .

Ao f b R 4 R5



64 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

R 4 R5  – R R 4

1 R1R 4 

5

1

=

R4 1 jf

R R 2

A f

o

b

RI4 RI5  jf  .

Ao f b R4

1

R1 R 2 RR 1 4

check o  R1 R 4 – R 2 R 5 R1 R 4 Further A f R o

fc1 =

b

1

= 90.9 103

R1 R 2

fc2 = Ao f b

R4

e

R 4 R5

=

106 50 50 500

=

106 50 550

6

R5

SR 2Vo

(b) fmax <

SR 6 = –6 2Vo 10 2 10

fmax =

= 95.49 kHz 2.46 i3

R3

R2 i2 vi

R1

3

R4

4

i1

i4

i5

vO 7

2 v(–)2 5

v –v (a) i1 =

i

(–) 1

R1

–v

v , i2 =

(–) 1

o1

R2

v ,

i3 =

–v

(–) 1

O

R3

i1 = i2 + i3 Substituting for i1, i2, and i3 vi

65

© 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

–

v(–)1

=

v(–)1 vo1 v(– vO )1 R1 R1

– R2

– R2

R3

R3

66 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

vi

v(–)1 =

v(–)1

R1

R1

R2

Also

v(–)1 o1v – – R3

Ov

(i)

R2

R3

i4 = i5 vo1 – v(–)2 R4 vo1 R4

–

v(–) 2

v(–)2 – vO

=

R5 v(–) 2

=

R4

–

R5

vO

(ii)

R5

Further (0 – v(–)1)Ao = vo1, (0 – v(–)2) Ao = vO or v v – v(–)1 = o1 , – v(–)2 = O Ao Ao or o1v

v(–)1 =

Ao 1 vi = v (–)1 

1

R1

R2

R1

=

vo1 1 Ao

v

O

, v(–)2 =

Ao 1 – R3

1

vo1 vO – R1 R3

R1

1 R2

vo1

– R3

1 1 = – vo1  R Ro  1

–

vO

R2

R3

1

1

1

v

O

–

5

R3 

R2

(iii) R2

R3

From (ii) vo1 R4

= v(–)2

1 R

1 4

R

–

vO R

5

=

vo1 = –

vO 1

1

Ao

R4

R4

1 vO 

–

vO

R5

R5 1

–

R4

Ao

67

vO Substituting in (iii) we have

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R4

R5 R5 vi

=

R4 1

R1 For large AD –

1

R5 

Ao R4

R4

v o

v

R5

 O

1

1

1

Ao R1

R2

1

1

 

R3 

R2

vO R3 vi

=

R1

R4 R5

=

vO

1

–

vo

R2

R3 R4 – R2 R5 R2 R3 R5

R4

=

R3

R2 R5

–

1 vO R3

vO

68 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

Substituting the values R2 R3 R5 vO 100k 150k 160k = = 20 k 150 k 20 k – 100 k 160 k vi R1 R3 R4 – R2 R5  = – 9.23 (b) Problem 2.46 VS 1 0 AC 0.1V R1 1 3 20K R2 3 4 100K R3 3 7 150K R4 4 6 20K R5 6 7 160K RX1 2 0 10K RX2 5 0 10K XA1 2 3 4 0 UA741-AC XA2 5 6 7 0 UA741-AC . SUBCKT UA741_AC 1 2 3 4 RI 1 2 2MEG RO 6 3 75 GB 4 5 1 2 0.1M R1 5 4 10K C1 5 4 1.59UF EA 4 6 5 4 2E5 D1 3 7 DMOD D2 4 7 DMOD . MODEL DMOD D (BV = 14V) . AC DEC 10 100 20MEG . PRINT AC VM/7 . PROBE . END 2.47 (a)

R1

i1

i2

R2 i5

R5 RF

3

i4 1

2 vi

R4 6

4

2

vO

7

5

R 3

v2

R6

0–v i1 = i2, –

v(–)1

=

v(–)1

–

vo1

,

R1

v(–)1 –

–v

v (–)1

i1 =

,

v(–)1

–

i2 =

(–)1

o1

R2

vo1

69 © 2017 Cengage Learning®. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

R1

R2

R2

R1

R2

R2

67 Š 2017 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

1

v(–)1

or

v 1 = o1 R2 R2 



R

1

v =

R1 R 2 

v(–)1

(i)

o1

or

v(–)1 =

R1 R2  R2

vo1 – v(–)2

v(–) 2

o1

–v

v

(–)2

R4

,

i5 =

(–) 2

O

R5

v(–) 2 v v – = o1 – O R5 R4 R5

R4 – v(–)2 

vo1

v(–) 2 – vO R5

= 

R4

i4 =

1

R1 R2

v –v i4 = i5,

R

1

1

R4

R5 

vo1

=

4

v 5

vO

R4

R R v(–)2

–

=

R4 R5

R5 v

o1

R4

O

(ii)

R5

Also

(v1 – v(–)1) Ao = vo1

(iii)

And

(v2 – v(–)2) Ao = vO

(iv)

From (iii) R1

v –v 1

o1

R1 R2 

v1 Ao –

vo1 1

A = v  o

o1

v o1A oR1 = vo1 R1 R2

o A1

R

= v1 Ao

R1 R2  vo1 = v1

Ao 1+ A R R R o

1

1

2

From (ii) R R v(–)2

4

5

A = v1

v

1

o

O

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1 + Ao R1 R1 R2

R4 R5

R R 5

e

From (iv) v2 –

R R 4

4

1 + Ao R1 R1 R2

R4

R5 vO e v R4 R5 1 Ao R1 R1 R2 

R5



Ao v1

o

1

R5 v2 –

R R 4

5



e 1

1

2

4

o

O

O

vO =

A o

vO 1

= A 2

R5

vO

5

R R 1

5

R4 v R4 R5

Ao v1 1 A R

v

R4 R5

R4 – R R

R R

1 A R

5

R R

1

o 1

R4 R5

=

R5

A v

4

v(–)2 =

R4

R4 vO

R R o

4

5

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1 vO 



4R

R4 R5 

Ao

4

4

R R

1A R 5

o

v1

1 + Ao R1 R1 R2

v1 1

–

1A R R R o

oA

5

R4 R5

v2

vO =

R

= v2 –

1

1

2

R5 R4 R5



1A R o

R R 4

4

5

For large Ao v2 vO =

– R R

A R

R 4

o

5

v1 R

1



R5

2

4

Given R1 = R5 = 10 k and R2 = R4 = 500 k k; then R R 4

vO =

R4

= –

R

R R v2 –

5

R1 R2 R5 R1

1

2

R1 v

R4 

1

5

v1

R4 R4 R5 

– –

R4

v2

= a1 v1 – b1 v2 where R R a1 =

1

2

5

R1 a1 =

R R

R ,

b1 =

4

5

R4

R4

10 k + 500 k 10 k = – 1.02 10 k 500 k R R

b1 =

4

5

R4

500 k + 10 k =

500 k

= – 1.02

vO = (– 1.02) v1 – (– 1.02) v2 Note b1 is same as a2.

SR SR (b) fmax = = 2 10 V 2 . Vout

=

6 V / µs 2 10 V

= 95.5 kHz

2.48 Same as Problem 2.47.

2.49 A1 = 5, A2 = 7, A3 = 3, B1 = 2, B2 = 1, B3 = 6 A = 5 + 7 + 3 = 15, B = 2 + 1 + 6 = 9, C = A – B – 1 = 15 – 9 – 1 = 5 R

68R

R

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C > 0, Rx = , Ry =

F

= C

F

C

F

5

For RA = 20 k, RF = MRA = 15 × 20 k = 300 k.

R 300 k 300 k = 60 k = 60 k, R a = e F = Ry = A1 5 5 1300

R Rb =

F

A2

=

7

k = 42.86 k

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Rc =

300 k = 100 k 3 R

R1 =

F

B1

300 =

R R2 =

F

B2

2

k = 150 k

300 =

1

R k = 300 k, R3 =

300 k

F

B3

=

6

= 50 k

From Eq. (2.117) RB = R1 || R2 || R3 || Ry = (150 || 300 || 50 || 60) k = 33.33 k RB || RF = 33.33 || 300 = 30 k 2.50 A1 = 5, A2 = 9, A3 = 3, B1 = 8, B2 = 2, B3 = 6 A = 5 + 9 + 3 = 17, B = 8 + 2 + 6 = 16, C = A – B – 1 = 17 – 16 – 1 = 0 Rx = , Ry = ; M = 17 Assume RA = 20 k; then RF = MRA = 340 k.

Ra =

Rc =

R2 =

eRF

A1 eRF

A3 eRF

B2

=

R 340 340 = 37.77 k k = 68 k, R b = e F = 5 9 A2

=

340 3

= 113.33 k, R 1 =

= 340/2 = 170 k, R 3 =

eRF

eRF

B3

B1

=

=

340 = 42.5 k 8

340 = 56.66 k 6

From Eq. (2.117) RB = (R1 || R2 || R3 || Ry) = R1 || R2 || R3 =

340 340  k = 21.25 k || 170 || 8 6

RB || RF = (21.25 || 340) k = 20 k 2.51 Referring to Fig. (2.34) and Eq. (2.119)

vo (5 va 7 vb 3 vc ) (2 v1 v2 6 v3 ) vo A1 va A2 vb A3 vc B1v1 B2 v2 B3 v3 where

A A1 A2 A3 5 7 3 15 B B1 B2 A3 2 1 6 9 C A B 1 15 9 1 5 Since C 0, R

,R

RF RF 70

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x

y

C

5

For RA 20K

RF MRA , where M is the largest value of A . Thus in this problem, M 15

71 Š 2017 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

RF MRA 15 x 20K 300 K  R 300K Ry F 60 K  C 5 R 300K Ra F 60 K  A1 5 R 300K Rb F 42.86 K  A2 7 R 300K Rc F 100 K  A3 3 R 300K R1 F 150 K  B1 2 R 300K R2 F 300 K  B2 1 R 300K R3 F 50 K  B3 6 Rx  Check: From Eq. (2.116)

RA Ra || Rb || Rc 60 K || 42.86 K ||100 K 20 K  From Eq. (2.117)

|| R || R 150 K || 300 K || 50 K || 60 K

R R || R B

1

2

3

y

150

K 7

From Eq. (2.118)

150K

R || R B

F

||300 K 20 K

7

Therefore, RA (RB || RF ) is satisfied.

2.52 Given

vo (5 va 9 vb 3 vc ) (8 v1 2 v2 6 v3 ) Rmin 20 K  vo A1va A2 vb A3 vc B1v1 B2 v2 B3 v3 where

A A1 A2 A3 5 9 3 17 B B1 B2 B3 8 2 6 16 C A B a 17 16 1 0 Since C 0 , Rx 0,

RF NRmin

Ry 0

and N 9

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RF 9 x 20 K 180K

71 Š 2017 Cengage LearningŽ. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

RF 180K 36 K  A1 5 R 180K Rb F 20 K  A2 9 R 180K Rc F 60 K  A3 3 R 180K R1 F 22.5 K  B1 8 Ra

RF 180K 90 K  B2 2 R 180K R3 F 30 K  B3 6 R2

From Eq. (2.116)

RA Ra || Rb || Rc 36 K || 20K || 60 K 10.58 K 

From Eq. (2.117)

RB R1 || R2 || R3 22.5 K || 90 K || 30 K 11.25 K 

From Eq. (2.118)

RB || RF 11.25 K || 80 K 10.58 K 

Therefore,

RA (RB || RF ) is satisfied 2.53 Closed loop gain A f 

RF R1



1 x106 1x104

From Eq. (2.120)

vo (min) v Af

s

vo (max) Af

13 13 v s 100 100 Maximum input signal = 130 mV 2.54

1 x106  100 10K  13 Peak input signal = 130 mV 100 (b) Undistorted output voltage = 12V (a) Closed loop gain A f

Limiting Op-amp current = 18 mA

100

Allowable load current iL

Minimum load resistance R L

vo 12 18 RF 1K  v 12

= 18

o

RL

18 12 6 mA

2K 6 mA

2.55

SR 10

V s

(a) From Eq. (2.126)

SR 6.286 vs f SR 10 f 1.59 x108 Hz 6 6.286 vs 1 x10 x 6.286 x10 x103 (b) From Eq. (2.128)

SR 2 xVm

f s (max)

10 159.2KHz 106 x 2 x10

2.56 From Eq. (2.123)

tr 0.3 s tr 2.2  tr 136.36 x109 s 2.2 From Eq. (2.125)

f max

1 2

1 1.16 MHz 2 x136.36 x10 9

2.57

SR 10

V , VS 10V s

(a) From Eq. (2.128) 6 VS 10x1x10 10 6 s SR 10 1 1 f max 1.59 x105 KHz 6 2 2 x10

(b) From Eq. (2.128)

f

s (max)

SR 2V m

10 2 x10 s

 2

106

159 KHz

2.58

SR 10

V , s

f s 250 KHz

(a) From Eq. (2.127)

Vm

SR 2f

10 6.37V 1 x10 x 2 x250 x103 6

s

(b) From Eq. (2.127)

SR 2V

f s (max)

10 x106 159.1 KHz 2 x10

m

2.59 (a) From Eq. (2.130)

voo (1

RF

)v R1

io

vio 6 mV (1

voo

50K ) 6 mV 15 K

voo 26 mV (b)

voo vio 6 mV 2.60

vio 6 mV voo 12V Referring to Fig. (2.40) and using the expression given below

v oo

vio

v io

t v io

CR

12 6 mV (1

t

(1

t

)

CR

)

CR t ≃ 2 x103 CR t 2 x103 x 0.1 x106 x10 x103 2 s 2.61 (a) If IB is the biasing current Vob = RFIB = 50 103 500 10–9 = 25 mV (b) Offset minimizing resistance 2.62

Rx = R1 || R F (15 || 50) k = 11.538 k  RIF  Voi 1 =  R1  Using Eq. (2.138)

RI  Rx Iio = 1  F  R1 

Rx Iio



=1 



RF (R1 || RF) Iio = 1 50 (15 || 50)k Iio 15 R1 

65 15 50 15 65

=

103 200 10–9 = 10–2 V

2.63 (a) Voi = RF Iio = 150 103 300 10–9 = 45 mV (b) Vob = 150 103 500 10–9 = 75 mV 2.64 (a) R1 || R F = (15 || 50) k = 11.54 k = Rx Vob = 0 Voi = RF Iio = 50 103 300 10–9 = 15 mV (b) For Rx = 0, Eq. (2.2.134) gives

Vob = RF IB = 50 103 500 10–9 = 25 mV 2.65 (a) From Eq. (2.142) for vC(t = 0) +vof =

Vio I t – e io t + Vio – Rx IB1 CF R1 CF 6 10 3 10 7 15 103

12 =

=

t–

300 109 107

t + 6 10–3 – 10 103

6 t – 3t + 6 10–3 – 5 10–3 1.5

= 4t – 3t,

t ≃ 12 s

(b) Rx = 0, hardly makes any difference. The value remains the same.

2.66 From Fig.P2.66

is ≃ if

i f i1 io io i f i1 Since vo i1 R2

i i o

i i f

1

vo f

R2 Since vs is R1 ≃ if R1 v v io s o

500 10–9

R1

R2

vo i f RF i o

io

vs

RF vs vs RF vs

i f

R1 vs

(1

R1 Given

RF

R2

R1 R2

R1

RF

(1

R1

)

) R2 (1)

R2 RF 1M ,

R2 1K , io 500 mA,

vs 10 mV

From Eq. (1) above

103 K

10 mV 500 mA 

R1

(1 

1K

)

R1 ≃ 20  2.67 Power produced by the photodiode P = Dp a = 10–6 × 40 × 10–2 = 400 nW iS = PDi = 400 × 10–9 × 1 = 400 nA RF = | – vO/iS | =

1 107 = 2.5 M = 400 nA 4

2.68 For RF >> R2, VS =

eR1eR2

RF

IM

Let RF = 100 k; then R2 = 100 .

R1 =

V SRF = I M R2

200 103 100 103 100 106 10 2

2.69 zin = – R (R1/RF) Assume R1 = RF = 10 k and R = 15 k. Then zin = – 15 k 2.70

=2M

V

V

vO = – 50 ve + 5 = – 50 (vref – vS) + 5 = 50 vS – 50 vref + 5 Let R1 = 10 k and R2 = 10 k. RF = 50 10 = 500 k = R3.

–12V R3=RF RF 12V +12V

Rx=R2 vref R1

+

3

vO

vS

–

2.71 There is no unique solution. Hint: Use an op-amp comparator to compare the dc voltage with the sifted sine wave to produce a square-wave output and then pass the signal to a R-C circuit, this is to differentiate.

2.72 Same as Problem 2.70.

2.73 Ao = 1 + RF/R1 = 100 From Eq. (2.31) Af (j) =

1RF R1 1 jf 1 R R f F

1

bw

From Eq. (2.32) 1

fc = fbw 

, fbw = 4 106 Hz

1 RF R1 fbw = 4 106 Hz fc = 4 106

1 = 40 kHz 100

Let R1 = 2.5 k. Then RF = 99, RF = 2.5 × 99 k R1 = 247.5 k = 250 k Voltage swing:

We know fs =

SR = 2 vO

SR 2 11

= 25 103 Hz

SR = 25 103 2 Ã&#x2014; 11 = 1.73 V/s Problem 2.73 VS 1 RX 1 R2 3 R3 3 XA1 2 . SUBCKT RI 1 RO 6

0 AC 1MV 2 10K 0 2.5K 4 250K 3 40 LF411_AC LF411_AC 1 2 3 2 2E12 3 50

GB

5

4

R1 5 C1 5 EA 4 D1 3 D2 4 . MODEL . ENDS . AC DEC . PRINT . PROBE . END

1

2

0 .1M

4 10K 4 0.79UF 6 5 4 2E5 7 DMOD 7 DMOD DMOD D (BV = 14V) LF411_AC 10 100 20MEGHZ AC VM(4)

4

2.74 Use the circuit of Problem 2.33 to design the desired differential amplifier circuit with Af = 5 kV/V and Ri 500 k.

– v1

+

R1

R2

vo1 – +

– v2

+

R3 vo2

– R4

From Problem 2.33 R1 = R3 = 10 k, R2 = R4 = 1 M Av1 = – 100 v1 + 100 v2 = – 100 (v1 – v2) Now add one stage of noninverting op-amp of gain 50.

Af = Av1 Av2 = 100 50 = 5000 = 5 kV/V Problem 2.74 VS1 2 VS2 5 R1 4 R3 7 R2 9 R4 8 R6 12 R7 12 R8 10 XA1 2 XA2 5 XA3 8 XA4 11 . SUBCKT RI 1 RO 6 GB 4 R1 5 C1 5 EA 4 D1 3 D2 4 . MODEL . ENDS

0 AC 1MV 0 AC 2MV 9 10K 8 10K 10 1MEG 0 1MEG 0 1K 13 49K 11 1K 4 4 0 7 7 0 9 10 0 12 13 0 LF411_AC 1 2 2E12 3 50 5 1 2 4 10K 4 0.79UF 6 5 4 7 DMOD 7 DMOD DMOD D (BV LF411_AC

LF411_AC LF411_AC LF411_AC LF411_AC 4 2 3

0.1M

2E5

= 14V)

+ vO

. . . .

PRINT AC AC DEC 10 PROBE END

VM(13) 100 20MEGH2

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