Solution Manual of Microelectronic Circuits (6th Edition) - Adel S Sedra & Kenneth Carless Smith (1) PDF

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Oxford University Press, Inc., publishes works that further Oxford University's objective of excellence in research, scholarship, and education. Oxford New York Auckland Cape Town Dares Salaam Hong Kong Karachi Kuala Lumpur Madrid Melbourne Mexico City Nairobi New Delhi Shanghai Taipei Toronto With offices in Argentina Austria Brazil Chile Czech Republic France Greece Guatemala Hungary Italy Japan Poland Portugal Singapore South Korea Switzerland Thailand Turkey Ukraine Vietnam

Copyright © 20 II by Oxford University Press, Inc. Published by Oxford University Press, Inc. 198 Madison Avenue, New York, New York 10016 http://www.oup.com Oxford is a registered trademark of Oxford University Press All rights reserved. No part of this publication m11y be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission of Oxford University Press.

ISBN: 978-0-19-976570-6

Printing number: 9 8 7 6 S 4 3 2 I Printed in the United States of America on acid-free paper

Contents Exercise Solutions (Chapters 1 - 16) Problem Solutions (Chapters 1- 16)

Preface

This manual contains complete solutions for all exercises and end-of-chapter problems included in the book Microelectronic Circuits, International Sixth Edition, by Adel S. Sedra and Kenneth C. Smith. We are grateful to Mandana Amiri, Shahriar Mirabbasi, Roberto Rosales, Alok Berry, Norman Cox, John Wilson, Clark Kinnaird, Roger King, Marc Cahay, Kathleen Muhonen, Angela Rasmussen, Mike Green, John Davis, Dan Moore, and Bob Krueger, who assisted in the preparation of this manual. We also acknowledge the contribution of Ralph Duncan and Brian Silveira to previous editions of this manual. Communications concerning detected errors should be sent to the attention of the Engineering Editor, mail to Oxford University Press, 198 Madison Avenue, New York, New York, USA 10016 or e-mail to [email protected]. Needless to say, they would be greatly appreciated. A website for the book is available at www.oup.com/sedra-xse

lV

Exercise 1--1

Ex: 1.1 When output terminals arc open circuited For circuit a. n0 c = u,U) For circuit b. l'oc (,(t) X R, When output tenninals arc short-circuited .., CJrcmt . . a. i ,.. "' l'~{f) .·or For circuit b. f.,.

v 0 ·~ 10 mY X

If R, D = 0001

1•.1

2 V => D "" 0010

15V=>D=IIII (b) (i) +I V (ii) +2 V (iii} +4 V (iv) +8 V (e) The closest discrete value rep.rcsented by Dis 5 V; thus D ~" OIOL TI1c error is -0.2 V or -0.21 5.2 X 100 •·~· -·4%

I

F..x: 1.6 (al T

(c) T ""

""

I: ( 1 +

r:.J

l ~

iJ

_I +

25

l't.

10 II)'·' V "' IOp. V 101' + 10 --

X

.4

.2

;. N A = Pp = !!!..

=

I nv\ -· . !OJ IJ:A ::::.18 Jn 56 )J.AI(t.l.tn) 2

TIP

Ex;l.30

( 1.5 X 10 10{ 1.5 X 104

Using equation 1 . 4 5

1.5 X 1016fcmJ

D. llo

~ =

v,.

li-p

Dn = Jl,V,. = 1350X25.9X

Ex:l.28 a. u,·driff = -~J.-nE Here negative sign indicates that electrons move in a direction oppOSite toE We use

'-'··drifT

$!;

Dp

X

35 cm /s '

=

)LpVT "'

a.:

12.4 cm2/s

J> ND

Aqu:(. 1._!!,/!._ + Jl!c .. ) ,N v

Equationl.. 5U(,

L,N,~

1

., lO"' X 1.6 X I()'

x[

v w~-'---' --

Equation 1. 52 X J'

X

5

19 :>> Np Nn

I

:::W·_;;;'

I,( e

\'/\' .. 1 ·-

I)

lY ,l

,.,

•. r:quatJon

1

~

. .;- 3(. 1 <

)

-

.

tl > N1) . ;V, ,

Ex:

l. 36

::. A,1,v"w

/l

1.()~ X

X

•J

1().

1.6 X 10- IQ

1.66 >: 10

11

(_!_-- + ___1_,)(0.814- 0.605)

ern

!(}IX

10 1"

0.166 IJ.rll

r;:.~-~--~~-

A .j2~sqN 0 V 0

Kx: 1. 37 [ " " v~

In

.-~ampk

N ~.>

1. 2;>

'f.

:

10' /em· and

V1

1

••,."

~~

" ' "~------,~-~"""

t~'(~; • ;~)v,. + vi?J

Ex: 1.33

to''inn·'

In the ll·Tt';!ion nf thi' pn jm:tinn diode

n,, J' . ,

,\1 , II, 11,.

IO"'tnn'

ll-?

>-::

ll)'')'

10"'

(,J)l{

u~ing

Q1

)<

!()

Clll

equation 1- 53

' ·V V , Aq( -~~-"'-'·-~) W iv_, 1 /v "·

0.601\

Jllll

Exercise 1--8

"·' I0- 4 xL6XIQ-ItJ ( IO

18 . 16) XIO X6,(}8XI0-Scm

1018

+ 1016

Ex: 1.40 r::quation 1 . 7 4 L2 ::J!.

•"' 9.63 pC

D,.

I "" Is "" llqn/(..!!.JL.. + [)~ ) l.,.N[) L.N.,

Reverse Current

10···!4 X 1.6 X 10-l? X ( 1.5 X 10111 ) 2

x( 5

X

10 · 10-4 X 10 16

+

18 ) 10 X 10-~ X J0 1K

7.3 X }(}" 15 A

(5 X 10- 4/ 5 --: 25 ns Equation 1 . 81

cd ~ (;:) 1 In example 1.30 N,.. "'" l0 18/crn1 • N0

""

l0 16/cm 3

A'>suming NA >> Np '~'r::

-.fl ""

25 ns

:.CJ "" ( 25 X 10-~ ). O.J X 10-J 25.9 X 10 ~ ( 10

18

10 16)(

X

%.5pF !

+ 10!6 0.814

1018

= 3.2 pF Bqualion 1 • 71

Ci""

C;o

R 3.2

X

10- 12

~ = 1.12pF

Ex: 1. 39 C1 =

'

1Sl. "" dV

-!!...{-.r/) dV

d

=-{-.rXls(e dv d

= Trl~(e ~ dV.

VN

T-J)]

I'll'

r- I)

)

Exercise 2--1

Ex: 2.1

Therefore:

The rrrinimum number of terrrrinals required by a single op amp is five: two input terminals, one output terrrrinal, one terminal for positive power supply and one terminal for negative power supply. The minimum number of terminals required by a quad op amp is 14: each op amp requires two input terminals and one output terminal (accounting for 12 terminals for the four op amps). In addition, the four op amp can all share one terminal for positive power supply and one terminal for negative power supply.

Ex: 2.2

v2

=

-

V 1)

isA = J.LGmR. For G"' = 10 mAN and 100 we have:

J.L

104 YN Or equiva-

100 X 10 X 10 A lently 80dB

Ex: 2.4 The gain and input resistance of the inverting amplifier circuit shown in Figure 2.5 are

- R2 andR 1 respectively. Therefore, we have:

Equation are v3 = A(v2 V;d

V 3 = J.LGmR(V 2

That is the open-loop gain of the op amp

v 1,

-

V;cm

-

R•

v 1 );

R 1 = 100 kfl and

1 2(v 1 + v2 )

=

a)

= 0 - .1_ = -0.02 Y = -2 mY 103

R 2 = -10::=>R 2 RJ Thus: R 2 = 10 X 100 kfl

IOR 1

IMfl

+ 0.002 y

0- ( -0.002)

Ex: 2.5

2mY

R= IOkfl

-1 mY

!(-2mY+O) 2 3

b) -10 = lO (5- v 1)=> v 1 = 5.01 Y V;d

= v2

= 5- 5.01 = 0.01 Y = 10 mY

11 1

-

~(5.01

+ 5) = 5.005 y

.::5Y

1.1 From Table

c) 113 11;d

=

A(11 2 -11 1 )

=

v2

= 103 (0.998-1.002) =-4Y

~( 1.002 +

0.998) =

=

IY

I

V0

R,

d)

-3.6 = 103 [712- (-3.6)]

=> V;d

J2

=

= v2

-

=

I -(v 1

2

R; =

= -3.6036- (-3.6)

-0.0036 Y =

IO\v2 + 3.6)

-3.6036 y V1

=

+ Vo) ~

-3.6 mY

Ex: 2.3 From Figure E2.3 we have: V 3 = f.L V" and -

G"'V 1)R

=

G,R(V 2

-

V 1)

0

V; - Ri;

0- Ri; = - Ri; Ri; = -R=>R, i;

~'0 ~;" = o

-R

-IOkfl

v

__!

V;

thus R.I =

I 2[3.6 + (-3.6)]

-3.6 y

Vd = (G"'V 2

~ 0 I; ~ 0 , i.e., output is open circuit

The negative input terminal of the op amp, i.e., V; is a virtual ground, thus V; = 0

v 1 = 0.998 - 1.002 = -4 mY

-

R,

we have:

and V; is a virtual ground (V; = 0),

Q= •

0 => R.I = 0 fl

I;

Since we are assuming that the op amp in this transresistance amplifier is ideal. the op amp has zero output resistance and therefore the output resistance of this trans resistance amplifier is also zero. That is R0 = 0

n.

Exercise 2-2

R =Jb1cll

~v.

Rt 0:5 rnA

C.onnectlng.the sig~IUI source shown in Figure. e~:s to the inp!ll (If this. amplifier we have: V; js a virtual .gruund that js V1 = o. thi,~s the ~r" tent flowing through the to kll resistt)l' connected ·between V1 and ground is i.ero. Thercfui'e V0

V,-Rx0.5mA,O-IOK'X05mA

""

=

-sv

For.the cifuuit shown above we have:

v::: 15Hz

9.9 .k!! 9.9 kH ::: 10 Hl

With this value of R:~ the new value of the output de voltage (using equ:ttiossible amplitude that can be accommodated at the output without incurring SR distortion is: V o "" V o ,..,( fM ) '"' I0 5 fM

X

! "-" 5

2 V (peak)

Exercise 3-1

Ex: 3 .1

(c)

Refer ro fiig3. 3(a). for V 1 ;:;-: 0, the diode q:mdu V 0 ..~ V1 For V1 < 0, the diOde is cut-off, zero current flows thro1igh R:md V 0

""

0, The

res1dts is lhe tnmsfer characteristic in Fig E~ .1.

OAt

1~

Ex:3.2 see Figure). 3aand:3. 3b f)uriilg the positive half (>f the sinusoid, the diotic is Jnrward bia.~e...