hw5_solutions

hw5_solutions - a. VDS ! sat " # VGS $ VTN # 0 $ !...

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Unformatted text preview: a. VDS ! sat " # VGS $ VTN # 0 $ ! $2.5 " # 2.5 V VDS # 0.5 V % Biased in nonsaturation 2 I D # !1.1" & 2 ! 0 $ ( $2.5) " ! 0.5 " $ ! 0.5 " ' % I D # 2.48 mA ( ) i. VDS # 2.5 V % Biased in saturation I D # !1.1" ! 0 $ ! $2.5 " " % I D # 6.88 mA 2 ii. iii. b. % I D # 6.88 mA VDS = 5 V Same as (ii) VGS = 2 V VDS ! sat " # 2 $ ! $2.5" # 4.5 V VDS # 0.5 V % Nonsaturation Homework 5 Solutions i. ii. I D # !1.1" & 2( 2 $ ($2.5))(0.5) $ (0.5) 2 ' % I D # 4.68 mA ( ) ECE 2B Winter 2011 H.O. #18 VDS # 2.5 V % Nonsaturation I D # !1.1" & 2(2 $ ( $2.5))(2.5) $ (2.5) 2 ' % I D # 17.9 mA ( ) VDS # 5 V % Saturation 2 I D # !1.1" ! 2 $ ! $2.5" " % I D # 22.3 mA iii. ______________________________________________________________________________________ 3.5 (a) V DS !sat " # VGS $ VTN # 2.2 $ 0.4 # 1.8 V 2.2 # V DS * V DS !sat " # 1.8 % Saturation (b) V DS !sat " # VGS $ VTN # 1 $ 0.4 # 0.6 V V DS # $0.6 $ !$ 1" # 0.4 V + V DS !sat " # 0.6 V % Nonsaturation (c) VGS # 1 $ 1 # 0 % Cutoff ______________________________________________________________________________________ 3.6 MicaoelVctronic.s: $ 2rcu#t 0 % Cuis fand Design, 4th edition Chapter 3 ( r ) e SG # 2 2 Ci .2 i Analysto f By D. A. Neamen Problem Solutions (b) V SG # 2 V, V SD # 2 $ !$ 1" # 3 V ______________________________________________________________________________________ V SD !sat " # V SG , VTP # 2 , !$ 0.4" # 1.6 V So V SD # 3 * V SD !sat " # 1.6 % Saturation 3.2(c) V # 2 V, V # 2 $ 1 # 1 V 6 SG SD # R2 V SD %sat " # V SG $, VTP %##2 ,1!8 0.$ !101.% 3.6 V $ 4" # " 6 V VG ! & ' VDD & ' R1 + R 1. 8 ( 2 * So V SD )# 1 ( V2 *!sat " # ) 16 %3Nonsaturation SD A____e t_______ biased in s _ uration re____ ____ssum__ransistor _________at________gion _______________________________________________ V V + VGS 2 ID % S % G % K n !VGS + VTN " 3.7 RS RS 2 k3 + G 2 W / .5 " ! $ !VG 2 I3.6# Vn S0 % ! 0!VGS 2 "VTNS" + 0.8 " D 2 1 L .2 % V + 1.6V ( 0.64 GS GS 2 0.12 2 W / 4. $ $ 0 2 2W / 0VGS + 0.6VGS +-2096!% 1.2 "5 % 0 - # 5.79 .5 # 0 2 1L. 1L. 2 ______V____0.6_, _____" _( ______" ______%___4_____________________________________________ _ % __ _ ! 0.6 _ _ 4 ! 2.96 _ - V _ 2.0_6 V GS GS 2 VG + VGS 3.6 + 2.046 ID % % - I D % 0.777 mA 2 RS VDS % VDD + I D ! RD ( RS " VDS . VDS ! sat " % 10 + ! 0.777 " ! 4 ( 2 " - VDS % 5.34 V ______________________________________________________________________________________ Microelectronics: Circuit Analysis and Design, 4th edition Chapter 3 3.27 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 3.34 + 0.12 ( 2 (a) 0.35 # ) &!50 "!VGS % 0.4" $ VGS # 0.742 V *2' (b) V DS # 1.8 % !0.35"!2 " # 1.1 V V DS !sat " # VGS % VTN # 0.742 % 0.4 # 0.342 V V DS , V DS !sat " $ Saturation ___________% __V______sa_) __4 ___8__3.2 V _______________________________________________ (a) VGS _ 4 _ VDS ( _ t _ % _+ 0._ % _____ 3.35 If Sat I D % 0.25 ! 4 + 0.8 " % 2.56 2 V %( + DS 6 1.44 / VS # Non-G at ) 6 - 14 &!10 " % 5 # %2 V * ' 2 4 % I D RD ( VDS % K n RD 0 2 !VGSk+ VT " VDS + VDS 1 ( VDS 2 3 + . (+ W ( 2 VG # VGS - I D R S % 5 # VGS - ) n 2&) & R S !VGS % VTN " % 5 + &3 4 % ! 0.25 " !1" 0 2 ! 4 + 0.8 " VDS ) VDS * ( VDS * 2 '1 L ' 2 2 4 % 2.6VD + + 0.12DS 0.25V ( 2 5 % 2 # 2 GSS - ) V &!25"!0.5" VGS % 0.8VGS - 0.16 *S4 0.25VDS + 2.6VD2 ( ' % 0 ! " -0 6 7 +4 Or 0.75V 2.6 , .4V.GS6% 2.88 # 0 $ VGS # 1.71 V % 1.88 V VDS % + 0.122 ! 0.25 " ( 2 ID # ) &!25"!1.71 % 0.4 " # 2.58 mA 28 * 4 + 1.' 8 % 2.12 mA ID % 1 V DS # 10 % !2.58"!1.2 - 0.5" # 5.62 V ______________________________________________________________________________________ 2 GS 3.36 +W ( Let ) & # 20 for example, then *L' I C ! C;D9,=/ G;D ,,,,,,B""C, I B " # I B " L;LF ! /, CJ & K;K FJF ! K;LK,M RC ! C;+ FJG ( D9 ,,,,,,,,,,,, !C9 " 9I ! &6G;D%M G;HEK6 =/, RE I & ) FJF $;D9, " ' * ,,,,,,,,,,,,, , &J;G;EEFH ,J % " D & $ &C ! C;NO ,=/ IE ! ,,,,,,,,,,,, VC " 96;D9 ;HEK6 %$J % " K;6F I, - $G I C A C;NK , = A AAAAAAAAA! AAAAAA/AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, , CJ & K;K ! K;6D ,M RC ! 5.7 C;NK , + V BE ( ,,,,,K;6D ' R ' K;LK,M @ , C;D9 I I ' C;NK, / C ,,,,,,,,,%,I E " 'EoC&M8) =& , , )V & * ' B%$,B"4;,>9;CDE#F5,, TR E ! 6 " 9I ! 6;H G # , +V ( L & J ;D J;J) BE & , H IB ! ,,,,,,,,,, G;DG . FG - K " 9 . FG -F6!&M8) HH,=/ I C ! C;OO ,=/ & CJ " $ DO % $ 6;H % * VT ' -K I E ! C;ON,=/ + G;DG . FG ( & ,,,,,,N7&),, V BE " $G;GJL% ')) VCE ! CO & $C;OO % $ 6) &9C;FG -F66;H % % ON & " G;LFGE I, * $. % $ ' !, & O;OG;EEFG L ( VCE ! H;HO ,P & D;JN CO 6 ,,,,,,,,,, / " " " FFG , ! , F ! K;L,M RE F -6 & 9I - G;EEFG ,,,,,,,,,, I C " ,I E " $& ;J;D %$G;DG% " G;HEJD =/, L G EEFG ! J;JH66 I C ! C;LK,=/ IB ! CJ O DG I E " $ DG;% $ K;L % " # I B " H;JF ! /, ,,,,,,,,,, I B " F 0 / FFF I E ! C;LO,=/ ,,,,,,,,,, VCCE"! COI& $RCLK%96 %$& ;$C;LO %$$J%;L %K;6F I, V 9 - C C; " $ - G HEJD% K " AAAAAAAAAAAAA&AAKH AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, ! CO A D; AA & D;JOL , VCE ! C;OC,P , , , @%,C;OO ' I C ' C;LK,=/ C;OC ' VCE ' H;HO,P , , AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, , , .19 5 , E0F VCC ! I C RC " VCE , , H;9 & C;C , ,,,,,,, I C ! ! J;K9 =/, 6 , .V + , ,,,,,,, I C ! I S &Q8, BE ) , ,V ) - T* $ % !"#$%&'&#($%)"#*+,-"$#."(,/)0.1*;"K9 / CJ3K&*"4)5,6(7,&2"("%),,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,-708(&$,9, ' J *,0)2, & + ) ! J;DJNC P, , ,,,,,,, V ! V BB ! $J;JHO % '), :1,3;,/;,BE &0=&),,,,,,,,,,,,,,,,,,,,,,,,9,/,CJ ,&,CO,,,,),,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,>$%?'&=,@%'.("%)*, < , , ,,,, ,, * AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, , , , V " VCE F;9 " G;G , B?C I E ! CC ! ! E;D =/, RE F , , *-' * IE ' ( % ,,,,,,,, I C ! ( % + I E ! ( IG %#E;DE $ ! E;HIFJ =/, )& ) G, - & , * E;HIFJ . GE "J ' % ,,,,,,,, V BE ! #E;EFH $ ')( ( 9 . GE "GH % ! E;DFHI K, & ) ,,,,,,,, V BB ! V BE , I E R E ! E;DFHI , #E;D $#F $ ! F;GFD K, AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, , 5.20 !"#$%&'&#($%)"#*+,-"$#."(,/)0'1*"*,0)2,3&*"4)5,6(7,&2"("%),,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,-708(&$,9, B0C I C ! E 5,, VCE ! F K, :1,3;,/;,<&0=&),,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,>$%?'&=,@%'.("%)*, B?C I C ! AAA ! AAA $AA / AA ! E; AAAAAA AAAAAAAAAAA-I B AA#GFEA#F $AAAI C AAAAF6 =/, AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, , ,,,,,,, VCE ! F " #E;F6 $#6 $ ! G;E6 K, 5.24 G;6 " E;D B#C I E ! 9 ! V ! E;J9V /, E;G $ % 9 #= ! ,-) , HE ) F B 5,, I E " B B0C I C " 5,, I C " * * D % - ' & I E " * HD ' & I E , ' DE ' F +( + ( * GFE ,,,,,,, I C ! ( %#E;J9$ ! E;J6DG =/, 9 ! V B , HE ), V % 6;F ) ,,,I7&),, ) GFG &" * '* B ' ,, . V B " !K;DFJ L, ' * + HD ( 6 E # ,,,,,,, V ! DE" #E;J6DG$#+ $ " #F ;J9$(F $ ! "E;ELL K,,M,,<%(,8%**"?'&, F ! , DFJ ! $ ; ( % 9 ,,,,N$0)*"*(%$K";),@0(.E0G"%), ,,,,,,,, I E " " E;GKD =/, ,,,,,,, VCE ! E;F K, F I / VC I E;I , B,?,C,,, VCE !"EDE ! ,,C #DE$ !! E #F$K, ;D K ,, E CE F " E;I4 1 HE ,,,,,,,,IK " DE ! I E 2F % ,FD9)=/$,/ " DE ! I E #DK;MH $ , ! E* '#DE ; C! 6 3 + HD ( 0 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, ,,,I7&),, I E " E;JKEJ =/, , , 5.2,1,/)2,, V B " E;G % #E;JKEJ$#F$ ! 9 " !K;6FM L, AAAAAAAAAAAF AAAAD ,AAF $ AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, A " #E; AA E; AA ! E;DJJJ =/, , B0C I E ! G;9 5.25 *-' * GFE ' F ;+I ,,,,,,, I ! ( ;F ! E%M9 ! ( %#E;DJJJ$ ! E;DFDJ =/, B0C I C " ( G , - % E" E;)KGFG=/, 69 & E ) DE & ,,,,,,, V EC !E;M9! E;;I K, VE ! E G ,, I ! $#G9 " I 5 /, ! B,?,C,,, I B " -I B9E #GFE. I$B/ FC ! G;L =/,M,<%(,8%**"?'&, C ,,,,N,$0)*" (I $,")I*0(" $E;(K69,! E;EEF " E;K6K =/, "* % ! , . 0 "%) ,,,,,, I C E B ,,,,,,,, V EC ! E;F K, I C E;K6K ,,,,,,, - " F "" ;F " ME;JG , E ,,,,,,, I E !I B E;EEF;F =/, !G G;9 ME;JG HMGM ,,,,,,, 6C "! I E " "B ! G;F " E;EG9 !,G;GL9 =/, I I D % - MD;JG , , ,,,,,,, VC " #E;K6K $#DE $ ! F;F " !E;MM L, , ,C V EC " E;M9 ! #! E B,#,,,, N$0)*"*(%$,#.(%OO,;MM$ " D;GF L, , , I # E 5,, V #D ! F K, B,?,C,,,,- C"!ME;JG $EC;DE $ " MM;GF , AAAAAAAAAAAAAAAE;AAAAAAAAAAF !AAAAAAAAAAAAAAAAE;AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA, V ! AG F; AA V E , V ! AG ) 5,, I E " " #MH;GF$* E , ,,,,,,, I B " E * 9E ' 5,, . V E " E;MFGD L, ' 9E DE + ( , Problem 10.36 If the circuit shown in Fig. P10.36 is excited by the current waveform is (t ) shown in Fig. P10.35, determine i(t ) for t ≥ 0, given that R1 = 10 !, R2 = 5 !, and C = 0.02 F. i(t) is(t) R1 C R2 Figure P10.36: Circuit for Problems 10.36 to 10.38. Solution: The current waveform is given by is (t ) = In the s-domain, Is for t ≥ 0 is Is = 1 0.5s 1.5s2 + 16 +2 = . s s + 16 s(s2 + 16) 1.5 A 1 + 0.5 cos 4t A for t ≤ 0 for t ≥ 0. Since is (t ) was different from zero before t = 0, we need to account for the initial voltage across the capacitor. At t = 0− , the capacitor acts like an open circuit, in which case 1.5 × R1 R2 1.5 × 10 × 5 vC (0− ) = = = 5 V. R1 + R2 15 The s-domain circuit is shown in Fig. P10.36(a), where source transformation was applied to convert the current source into a voltage source. R1 V 1 sC v (0−) _C s I Figure P10.36(a) At node V, V − Is R1 V − vC(0− )/s V + + = 0, R1 1/sC R2 V s3 + 15s2 + 16s + 160 = R2 s(s + 15)(s2 + 16) = s3 + 15s2 + 16s + 160 . s(s + 15)(s + j4)(s − j4) ©2009 National Technology and Science Press which leads to I= All rights reserved. Do not reproduce or distribute. + IsR1 _ + R2 Partial fraction expansion gives I= with A1 = s I|s=0 = s3 + 15s2 + 16s + 160 (s + 15)(s2 + 16) = 0.67, s= 0 A1 A2 B B∗ + + + , s s + 15 s + j4 s − j4 A2 = (s + 15) I|s=−15 = B = (s + j4) I|s=− j4 = Hence, I= s3 + 15s2 + 16s + 160 s(s2 + 16) s3 + 15s2 + 16s + 160 s(s + 15)(s − j4) ◦ = 0.022, s=−15 = 0.161e j14.9 . s= − j 4 ◦ ◦ The corresponding time-domain current is 0.667 0.022 0.161e j14.9 0.161e− j14.9 + + + . s s + 15 s + j4 s − j4 (A). i(t ) = [0.667 + 0.022e−15t + 0.322 cos(4t − 14.9◦ )] u(t ) At t = 0, i(t ) = 1 A, as expected. Problem 10.46 For the circuit shown in Fig. P10.46, determine (a) H(s) = Vo /Vs and (b) h(t ), given that R1 = 1 k!, R2 = 4 k!, and C = 1 µ F. R2 _ R1 C Vs _ + + + Vo _ Figure P10.46: Op-amp circuit for Problem 10.46. Solution: (a) Application of KVL and KCL to the circuit under ideal op-amp conditions leads to Vo 1 + (R1 + R2 )Cs All rights reserved. Do not reproduce or distribute. H(s) = = Vs 1 + R1Cs = 5s + 1000 . s + 1000 ©2009 National Technology and Science Press Since both the numerator and denominator of H(s) have the same power of s, we need to apply long division. Hence, 5s + 1000 4000 = 5− . s + 1000 s + 1000 (b) f (t ) = 5" (t ) − 4000e−1000t u(t ). ...
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This note was uploaded on 04/19/2011 for the course ECE 2B taught by Professor York during the Spring '07 term at UCSB.

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