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
Full
Exercise Solutions (Chapters 1 - 16)
Problem Solutions (Chapters 1- 16)
Exercise 1--6
Full file at https://buklibry.com/download/solutions-manual-microelectronic-circuits-6th-edition-by-sedra-smith/
Ex: 1.25
T= 50K
ni ""'· BT;ne ···1Jgl(2K11
""' 7.3 X IOJ\50)JI1 e-1.12112xU2/IIJ-5
9.6 x IO-:l9/cm3
T"'"' 350K
n; ""· BTm e. lit/l!IITI
"" 7.3 X C·-l.la/tl X $.42 X I() • X = 4.15 X 10 11 /cml
Ex: 1.26
N 0 = 10 17/cm 3
From Exercise 3.1 n1 at
T ,.,. 350 K = 4.15 X 1011 /cnl
It, ""' N v """ 10 11/cm 3 ni2 p,a!-N f)
(4.l5 X 10 11 ) 2 1011
"' 1.72 x 106 /cnl
Ex: 1.27
At 300 K, n1 = 1.5 X IOH1/cm 3
Pp = N"
Want electron concentration
I w = TIP "" .5 X lO "' 1.5 X 104 /cm) 106 .2
;. N A = Pp = !!!.. TIP
( 1.5 X 10 10 { 1.5 X 104
1.5 X 1016fcmJ
Ex:l.28
a. u,·driff = Here negative sign indicates that electrons move in a direction oppOSite toE
We use
'-'··drifT '" -!J.-.E
·"-' 1350 X --=---, 2 x Hr·4
•.
4
"" 6.75 X 10 cm/s "'""' 6.75 x 10 mls
b. 1imc tnken to cross 2 J.l.ffi
length '= ::: 30 ps 6.75 X Hf
c. In driff current density Jn in
' qnJt,.E
= L6 X 10" 19 X 1{) 1t, X 1350 X l V 2 x w-·l
1.08 X 104 A!cm2
d. Ddff current 1, = Aqnu,.-dritr
""' Aqn11n£
= 0.25 X 10·-B X 1.08 X 104
"" 27 11A
Note 0.25 11-m 2 = 0.25 x 10· 8 cm2
• _ dn(x)
Sx. 1.29/,. ··· qD,-dx
From Figure E 1 • 2 9
llu = "" I (JJ.m) 3 4 2 >'I\
Dn " 35 <-1n·/s 35 x (10) (f.tmr /s:
= 35 X
dn _ 105 -0 _ ·····
dx- ·-
J , qD dn(x) n • lit
= 1.6 X X 35 X 108 X 105 "" 56 X 10- 6 AI( 11-m )2
= 56 11 A/( 11-m) 2
For 1. = I mA ""1. X A
=-> .4 = I nv\ -· !OJ IJ:A ::::.18
Jn 56 )J.AI(t.l.tn) 2
Ex;l.30
Using equation 1 . 4 5
D. = v,. llo li-p
Dn = Jl,V,. = 1350X25.9X w->
$!; 35 cm 2/s -;!
Dp = )LpVT "' 480 X 25.9 X 10
a.: 12.4 cm2/s
J<;x: 1. 31
Equation 3 . 50
W=
https://buklibry.com/download/solutions-manual-microelectronic-circuits-6th-edition-by-sedra-smith/
LetR 1 = JQkfl
<k:gaifi=R2 /R 1 = = R2 = lOkfl
liS j6> _, !)0
0;2985we 1 ·· · w coC:R 1
C"" I = 5SnP
0.2895 X 2-tt 104 X: 104
(b)SeOOlld•Order l!ection with transfer (unction:
T(.v} = • . . . .· 2 s· + 0.461J-1w,. + where the numera«>r cQiiflkient was selected io yiclcl a de gilin.ofunity
Ex: t1 21
the KHN circuit in Fig;ll. 24Choosi.ng
C=1nF
R "'·-1-."" 4 1 9 "" 15.9 kfl la),;C l1rl0 X 10
UsingEq.li.62 and selecting R1 ·= 10 kU
R1 = Ri "" lO kO
Using Eq.l1.63 and setting R2 10 Hl
R3 "" R2 (2Q- I) ""' 10(2 x 2- I) = 30 k!l
High .freuency gain = K "" 2 - a"". 1.5 V I V
T!w function to the output ofthe Jirst inte!rarotis
sKI(C!!L_
2 w., 2 s+sQ+w.,
Tims the. centre-frequency gain
"' K J1 ""' KQ lJ/Z 1.5 X 2 '"' 3 VI V CRw., 11.22
Select R 1 ""' R 2 = R 3 "" R5 = 10 kfl
=> C ·"" 1 "" 2.43 nF
J0.4293 X 2nl04 X 104
C4 = C6 = C ,., 2.43nF
Q = ..}i)A2'93w,, = 1.4 => R6 = -'L = 14 kfl 0.4684wP to)0 C
(c) Sccond·Order Section with Transfer-function; '
T(s) "' , 0.988..Jw- 11 + s0.1789w 1 , + 0.9883w 211
The circuit is similar to that in (b) above but with R 1 "' R2 "" R3 ·"' R 5 "" 10 kH
"" 1.6 nF
Q ·""" = 556 O.lr89
Thus Rt. = Q t 111 0 C ""' 55.6 kH
Placing the three sections in cascade, i.e. connecting the output of the first-order section to the input of the second-order section in (b) and the output of of section (b) to the input of (c) results in the I)V.:ralltmnsfcr function in eq. Ill . 2 5
given C "" lnF RL = 10 kH
R _1_ "' I 31.83 kH w.,c 2n5 x x
R 1 "' IOk!l
R2 = 10 k!l ""} RJ "" R 1 (2Q- I)
= 10(10 -· I) '"' 90 kl!
811 w 2 ''" 1,) 2 R11 ·'-" ""· 25.6 k!l
R 1 " n 5
OC gain"" x::. "' (2- b):: '" 3
Rf. '"' 3 X to '" 16.7 kfl 2- 115
Full
https://buklibry.com/download/solutions-manual-microelectronic-circuits-6th-edition-by-sedra-smith/
I = (toll4-0.7 20) + 20 "" 0.124 rnA
2.48 v
https://buklibry.com/download/solutions-manual-microelectronic-circuits-6th-edition-by-sedra-smith/
11.8 rrt (dB)
Numemtor is given by
a, (s- i.i)(s1 + ( 10l)2)(i + (3 X
f (6 X 103)2)
""a 1s(l + to")(.t2 + 9 X 106 )(i + 36 x 106 )
Degree of Numerator m "' 7
Pegree of Denominator N
Given that there is one zero at "' :
N-M:::J=t>NzS
:. T(s) * . 1 6 s + b1 s + b6 .• + + b 0
a 1 s{i + 106 )(i + 9 X l06 )(i + 36 X 106 )
From circuit : drawn from V00 rail = 21B
= cmrent rerumen to Vss nul
:.Power= (V 1JD+ Vs 5 )X2111 =t>
I mW = (1.65 + 1.65) X 2 In
.·. / 0 "' 1 mW = 151.5 11-A 4 X 1.65 V
'"'·· 18 ! 1.1 "' 126.3 JLA 11.9
<D
L--"""'--.---o +
V; + Vo
:.V 1(2i + 2s + I) (a) -+(b)
V0 (2.Y3 +2s+1)1 "' V 0 +2sV 1
Eq. (b)
V0 (4.$4 .+ :f1{4 + 4) + .l(2 + 4 + 2) + $(2 + 2) +'I)
=V0 + 2sV1
V0 (s) A 2s
V ( ) ""' T (s) 4 J 1 II 4• + 8s' + 8.f· + 4s
r( ) 0.5 .$ = 2 s +·2s +2s+ I
Poles are given by:
s' + 2sl + 2s + I ·"' 0
(s + I )(i + s + I) = 0
.:. Poles are s = - I nnd s
11.10
A_= I dB, A =20 dB, w/w, "' 1.3
using: A< ...,) = , o log [, t t 1(::f']
The easiest way to solve the circuit is to me nodal at nodes (I), {2), 0)
At nooe C\l 'if c.o 0
Va + V 0 + Vi O
I I! s 2s
:V, = V 0 (2.1 1 + + I) Eq. (a)
At node 12) "il = 0 ()
log(toAm;, 'w-1) = log(r2("'•YN.)
(dp
A flO ) logl(IO '"'" -I) UJ - 21og(• ,l<>>.r)-N = 11.3 N 12
The actual vnlue of stopband altenuation qm be cakulaled using the intt•ger value of N :
A{w,) = JOiog(l + IV= 12 _tv!'. = 27.35 dB actual anenuation
If the stopband specs are to he met exactly we need lo lind A_,.
t'J. 10 I ("'·I,,,,..{'
A,_= 20 N .12 =" OJl\24
l(llng( I + 1· 1 0.73 <JB
16.34
Note that the output is high if no word is selected. Thus, logically, high must correspond 10 logic 0 (and no transistor, as noted in the
Correspondingly, the words stored in are 0100,0000. 1000, 1001,0101,0001.01 10, and
16,35
Note lhat a IOta! of 14NMOS and 4PMOS are used.
16.36
(a) Forthe PMOS, with V8 =2.5 V
L 0 = (90/3)10- 6(1211.2)[(5- 1)2.5- 2.5 2 /2)
= JO x w·•oo)[4(2.5l- 2.5' 121
= 2.0625 mA
Thus the average charging is 2.06 mA
Time for precharge 1=CVI/
whence
1 = I X 10- 12 (5 - 0) I (2.06 X 10- 3 ) = 2.42 ns
(b) For the word-line rise.
T = RC = 5 x 10.1 x 2 x 10-" = JO n•
Here, vw = 5( I - -•"">
Thus the rise time (I 0% to 90%) is essentially the time 1 to 90%, where
0.9(5) = 5(1- e-'' 10 )
e·o'IO = O.J
and 1 = -10 ln(O.Il = 23 ns