chap010 - Electronic ercuit Analysis ang Design. 2'”...

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Unformatted text preview: Electronic ercuit Analysis ang Design. 2'” edition Solutions Manual Chapter 10 Exercise Solutions EIOJ I _ V+ — Vgflon _ 10 — 0.7 REF — RI —' [BEE = 0.62 IDA 135}- 0.62 1° = *‘0‘ = “"2" 1 + ['3' 1+ fi' In = 0.604 mA 510.2 For 10 .-= 0.75 mA 2 2 has]? — [0(1 + - {0.75)(l + 155E =O.765 mA V" — V55(on) n V- R: _ s -— 0-7 — (-5) R‘ ' 0.765 R: = 12.2 kn 510.3 V+—V33(On)—V- 5-—0.7-(-5) 1m - a. - —*17— Ala = {0.02){0.754) = 0.0151mA and [3.1.9 = LAT/cs; =3 1'; = Ach fa a.“ 4 VA 0 = 0.0151 = “5 m n n =0 VA = [255)(0.754) => V: ; 200 V EH14 V" - zvasfon} 9 -- 2(D.T) 1 = _— = _..__.._. R" R; 12 1555 = 0.0333 mA 10 = _“'E:;_ = m = 0.6331mA 1 m 1 + an +0) + 7506) 12 = 0.6331 111A = IQ: I0 131 =13: = jig-=5 I31 =13: = 3.44 0A 13.». = fax +45: = In =16.ss 0A 10: 1+3 Ln: = =5 Ia: = 0.222 HA E105 I“; z 1,031.33; =U.Tl.7 VA 100 To—-J:u-— m=>fu—139kn 1 4 = —- V = — 31° rafl‘ 9“ 139 = A! = 0.0288 mA => [0 = 0.4996 IDA 10 153=§=°'153_=9_-9.9"_A I3; = (1 E'Bjicg = 15; =U.5095 111A 153 0.5096 In = — 2 " ‘2 (1+ 3) (1+ 5-.) =;. 1a, = 0.490 mm = Ia; 1c: 151= 152 '-= T: 1m = [3: 39-50 “A 1310.7 LIE; 2 V1- 1n (IRE?) In R __ V_1-l (In?) _ 0.0251 (0.75 E ' la n I. ‘ 0.025 “ 0.025 :5 RE = 3.54 m R: = 5 -— 0.7 n 75 3 R: = 5.73 kn V331 - Van = 10125 = [0.025)(3.54) =;. 1/55, — Vgfl = 00.5 mv E103 IRE;- = w : IEEE = mA [012; = V-rll'l (IMF) I003) = (0.026}ln( 5 Io ; 15.5 11A Electronic Qircuit Analysig and Design, 2"" edition Solutions Manual 510.9 1 LEE = V71“ (1113:) In 0.026 0.70 R” " 0.025 I“ (0.025) a Rs ' 3‘" m In 0.025 __ gm, _ W _ m =:~ 9...; _ 0.962 mA/v m ___ 131/,- = g150)(o.nza) =1“ m 1., 0.025 V... we run — I—n' — 0.025 — 4000 R;- .—. hum = 3.4mm = 3.39 kn £0 = r.,(1+ gmzx'g) = 400110 + (o.9&2}(3.39)] no =17.o4s m 1 3 ‘u" ‘ E'Wc’ ‘ 17.1145 =1» .11. = 0.176 11A EIOJO IMP=IR+IBH+151+...+IEN IR=I01=I02=...=I¢N mdlflfl=I31=I53=HI=IaN=I§ Ins:- =Iu1 +[N+1)(£Ei) =1." (1+ fl 13 501 -.r - —_r — ’w‘ 61'- 0‘2—...— 0N-W 1+ .6 I111 _ _ 1 fur—o'go-T-Tfi +— 50 514-1 1 1+ so _fi N+1=(—15—1)(50) N= L—1(so)-1 0.9 N=4.55=>£!=§ E10.” 1. From EquationfleZ). 1- i 12 10+ 3 x(1.s) 1+ E ya“ = (11.3.5)(1°)+ x (1'8) V65: = 3.93 V also V951 = 3.93 V Iasr = (12)(o.nzu)[3.93 — 1.s]'[1 + (0.111 )(3.93}} a {BEE = 1.13 m}. b. In = IRE,- x by”) x —-———-—“ My”) [WILL (1+ AVDm) _ 1 KW 11 -11.13)x (12) [1111031113331] =# [g = 0.555 111.31 C. For V55; = 6 V => In = 0.576 mA EIOJZ 2 2 Knl(VGSI _VTN') = 1%an 'an) 0.10 . V1351 - 2 = (H ’(Vcsz - '3) V651 — 2 =(o.r:‘.:12}(v.-,-3a - 2) V653 = 10 —- V551 V631 — 2 = (0.532)(1o — V551) — {0.532)[2} 1.133sz51 = 7.056 = V531 = 4.32 v [m = 19.04;“ Wm): = ((125)1(432—2)2 :1 [m = 1.3511111 10 z 3Kn1[V651"Vm)z = “0-25X432'2): Io =3IREF => I0 =4.04mA E10.13 Voshar) 2 l V = V052 -VTN = Vast ’ 2 =9 Van = 3 V C W 10 = KHI(VGSI ‘Vm)1 = (Van ‘Vm)2 L 2 0.20 40.02%!) (3—2): = =10 L 1 L 2 #,C W ’m’ "‘ [fill—1%“ "‘91 ’1 V65! 3 V65: 05 = (0.020)?) (3— 2)2 =-.. [E] = 25 L I L I van =vuvafl =1o-3 = 7 v C W {REF 3 (V653 ‘Vm )1 1 05 = (0.020)[%l(7 - 2)2 =9 = 1 E10.” 3- [w = K110!“ " m): 0.020 = o.uao{vas - 1]“ V“ = 1.5 V I.“ maisnors Tes L: de tandin IL, V01. == VGS: +VG5'I +V' = 1.5 +1.5— 5 = -2V V5. = My. — V65“ = -2 -- 1.5 = —3.5 V Vpdmin) = V5. 4- Vpsdsax) and Vm(mr)=Vm —Vm = LS— l =05V SO Vndmin) :1 —3.5 + 0.5 =9- Vm(min) = -3.0 V c. Ho = P04 4- r0:(1 +§anIJ 1 1 "’2 .r" Mo (0.021(0020) 25 m g_ = my,” - v,” ) = 2(0.080)(15— 1) = g. = 0.030 m IV Ru = 2500 + 2500(1+(0.050)(2500)) = no = 505 M0 510.15 Ins: = 0-20: KII(VGSI ‘er = 0-1500“ " '32 =3 V05, =ch = 2.15‘.’ [a = Kn2(VG\51 ‘Vm)2 =9§113(9-‘v15‘1)2 =° ID = 0.10M 2 In = Kama” "Vm) 0.10:0.15(vm- 1)2 = v6,3 =1.32 v E10.16 510. All Emsismrs u: identical =- 10 = Insp- = 250 ,uA 2 1:5! "" K-(Vas "Vm) 0.25 = 0.200%; — 1)3 => ya: = 2.12 V 17 F0! Q: : 1135th = lel = 2 V = Vs(min) = #ns(min) - 5 = 2 - 5 => vspnin) = —3 v Io = 1935:“ +Auasq} = 0.5(] + (0.15)(2)) 3Y9 = 0.65 mA. 2 In = {8351(1- Us“) VP: u 1 0.6-5 = 030(1 - 6:1) “3" = 0.0986- => mm = 41.197 v “as: = V: - Vs - 0.197 = V; - (—3) =- ngminz = —3.2 v Cha ter 10: Excrciscs Iuti ns + 9M1 VX = 9m: VI (2) _.+_- Rn nu l 1 _(__+,-.) __3__;+_1_... _*o= an rm 9"“ ;+; RD PM L 1 ro = — -.— 1- ,0: + (rm +9911) T_+':1_' RD 1'01 L 1 =—+(—+gml) lfiol) '02 I'01 _+_ 3-D To: Electrgnic Circuit Analysis and Design, 2"" edition Solutions Manual FOI' RD < r51 ‘ 1 - 1 1 =3. -— = —— —- m Re 1'01 + (I'm +9 1) For Q; : g _ 21.9551 (1- V951) _ 203.8)(1- —0.197 "" 1vp1 V; 2 —2 - gm. 2 0.721 mAIV l 1 r” E — (o.15)(o.ss) ‘ 1” m 1 1 1 ._..=_.._ _ _7 =, Rn 10_3+10_3+021 0915 :9 For Q1: in = Ins:1[1+ 19135:) 2 For Q21 in = [5351(1 _ {1 + A9352} 905') = *VDSI 311d lms: = V05 - was: So [5551“ + Avon) "VD £1 VP a =I5551[1— ] {1+A(VDS‘VDEI)] Ins-51 = Lass: [l+[0.l)vn51] 2 = [1 _ "‘25‘] [1 + (0.1)(3) _ (0.1).:931} A 1 + 0.1.195, = (1 — van + 0.259%”)(13 - 011nm) This becomes flimsy-1.,1 — 0.42M”. + 1.53;”. — 0.3 = 0 We find 92:] = 0.212 V. v23; 2: 2.79 V, Vgfl = -0.212 V in = fps-5'3“ + Amps-g) = 3“ + [D.1)(0.212)} i2 = 2.04 mA Ra = ’01 + rn1{1+ 91112702) 9 = 2fnss(1_ Ira-5:) __ 2(3) (1- —0.213) —Vp V; 2 —’.’ l 1 1‘ = = —— = — = 02 PM “on (alum 5 kn Ra = s + 5{1+(1.79)(5)J =5 E = 54.3 I -.3 Vzm = V: In ( fl“) = (fl.026}ln lu-u __ s - 0.521 b- R1 T => 3; = 3.95 kn c. From Equation (10.72) VEC: 1+ V: Vcso _ ( VAF ) Isa[exp (VT)](1+ VAN) _ [35,:- x .— V 2.5 -12 i 10 [cxp(vr)] (1+ 100) 'f 2.5 (1+ _ —:s ._ (0.5 x10 )——-—(1 + 0521 100 v 1.03 x10.12 exp = 5.125 x 10" V:- V exp ..-.. «LETS x 10" VT =9 V! = 0.521 V d. _(_1_) 1 A” = V1- : {1ng = —38.46 1 + 1 __1__+_l_ 0.01+0.01 VAN Ku- 100 100 => Ax = -—1923 1510.20 0.1 x10" V = 0.0 -——--— a. “a ( 25w: ( 5 x 10‘“) => V55: = 0.557 V b. Rl = 5 —U.557 T =- 31:44.4 m V: cha =— rs«[=x»<w>1(~v,.~> V 1+ 5c: =1REFx _VA.L 1+V5m V4.9 Test Your Understanding Chagtcr 10: Exercise §olutions (5.125 x 10‘”) exp( V exp =1.988 x 109 = V: = 0.557 V E1021 2 " [REF = Kp-I(VSG +VTP) 0.25 = 0.20(Vsc — 1)2 =- ng = 2.12 V I]. From Equation [10.89] [MAW ‘Vsafl Knm Wm )1 Vm=v, =-—-————— 2.. +2., Imp.” +215] 5 = 1 +(n.o15)(10 - 2.12] _ (0.22m — 1)’ 0.030 u.2s(o.oao} 0.15 = 1.12 — 0.8(V: — 1)2 => V; = 2.10 V _ -2K.(V, Wm) “ A ' 1mm, +3.3 A = _2(o.2 (2.10 — 1.0) " a.25(o.ozu} =3- AI: = -58.7 E1022 (a) [W = ENG/55 +v,,}2 sozsow,G - I)’ :9 vw = 2.26V [ii-APUP _Vw] __ “90’: ‘14"): (b) VD," = v, = __ ’1:- +ln ‘FREF(’1- +3...) 5 = [1 + (0.015)(10 — 2.26)] _ (5mm — 1)’ 0.030 (su)(o.nao) 20.33414 --1)2 = 32.2 a V: = 2.24 v —2K‘(v, -V,.~) = —2(so)(2.24 ~:) (:3 ’1': (“AAA/1F) (Bo)(o.o30) «=4m A, = —g.,.rra||m} = -(30.B}[100|1100] =:» An- = «154:0 b. A» = -9m(ro||ra=||RL) — T70 = —(30.3)(5U“RL) =5 (SOIIRLJ = 25 =3~ RE = 5|] kn £10.24 Ica 6.5 ‘_ m = — = —— m = ‘ V 9 _ v,- o.025 5 -——’ 19 2 “A; VAN 120 ru=E=fi=>r9=240kn VA; 30 1' = — = a— = on Ice 0.5 =)' 1’9: 150 a. A. = ~gm(ruumnm = —(19.2)[24ou160n501 => AB = —6.'l-I Electronic Circuit Analysis and Dcsig, 2“ edition Solutions Manual E1025 (a) Neglccling effect of A and RL 2 la =1“: = .[Vm-Vm) 0.40 = 035(qu - 1)” Then ME = 2.26 V l 1 = = _ = — = 2 bl fo '0: Ala (0.02)[0.{) 1 5 kn g_ = 2K,(V,a — v...) = 2(0.25)(2.26 * 1) = 0.63 mAIV ' ,4” = _;m(r,||r.,2} = —{D.63)(125|I125) c. A» = —gm(ra||ro:HRL) - = -(0.63](62.5|\RL] =7 62.55134, = 31.25 a. R; = 62.5 10'} 510.26 M. ansz identical =- 10 = has! ‘- IO=KH(VI-Vm)z 0.25 = 0.2(Vr - 1): V: = 2.12 V g. = 2K_(v, 41m): 2(02)(2.12 —- 1) 3 gm .-. 0.448 mAEV I. l ran—EE—Wz—sifirnn—‘00kn 1 Apro ' (0.02)[0.25) ’flflfl‘fl '1 A» = -ym(foilfoallfi£-J A. = —(0.44s)[400||200||mo] =~ A... = “25.5 Electr n'c Circuit Ana] sis and esi 2nd edition Solutions Manual Chapter 10 Problem Solutions 10.1 I ‘I _0_2VT_V_ " ‘ _ ’ ‘ m";— 2V1 + 1:3: = V35: + [:33 R” (—2»; — v-) = v” +1.33; Ic=7;:{2V,-(2V.,+V+)( R1 )-—vfl} RI +3.2 b. V..=V55lJ:IdR1 =3; 1 1 , _, [C = E{2V—Y - 5(2i'1 + V )— V35} I —__V_ 01’ C— 2R c. Ic=2mA=:-—(2%m=fia=2.5kfl :1 -'2{0.T} w (—10) I1=Iz=2mA= R1+Rz =31+R2=L3 k0 gRI=R2=115m 10.2 bar 1 I“ = 2 = 2 1+ E 1+ 3'6 IQ: = Q = 0.952 mA 10.3 V’- on —V' (a) I”: “I: } a or I —- . — --E RI: 5 07 { 5):: Rl=58.6k§1 0.5 (b) R, Jim: 0-0.7-(_15)= Im- 05 R, = 28.6 kn Advantage: Requires smaller resistance. (c) For part (a): 29.3 mm") = (53.5)(0.9s) = 0526 MA . 29.3 t' = ———-— = 0.476 m “(m0 (58.6)(105) M0 = 0525— 0.475 = 0.05 MA = 5% 14.3 For part (b): “(maka = 0526 mfi. 14.3 (236)005) Mo = 0.03 m =9 13% Io(min) = = 0.476 m 10.4 2 2 S. Leer —In(1+§) -— (1+m) 01' 135:; = 2.04 mA 15—0.? 31 _ M4 => 10-7.01 kn VA Bi:I b. ra=?O-=?=40kfl Ma 1 1 _ AV” _ :0- :- AID .. (3)93) _ 0.2325 mA AIu 0.2325 Mu _ Io —— 2 =3- ~7— “11.5% 10.5 rm g) min 3.5) R1 '- 054 :R; =7.95kfl 10.6 a. 11.153:- = 5‘0'7 =[L239IILA 13 ' In = “a: => In = 0.230 mA 1+ 5 V4 _ 50 _ b. TD—T—m—zijkn I l = — . = — = l A Ah ’0 AVEC (2177) (1.3) 0 09599 in =0 I9 = 0.236 M 1 c. AI.) — (3.3) — 0.0152 mA :9- I! = 0.245 mm. Electronic Circuit Analysis and Design, 2"d edition Solutions Manual 10.7 _ __ 5-0.T-(—-5) a. 1115p — I — Rl 2) RI =93 kn b. In =ZIuF d» Ig =2mfi. 5 — . c. For Vac-1(mjn] = 0.7:. RC; = 2” 7 =>R§3 :2.” m 10.8 Io=flIc1 _ Ic: Io IMF=IcI+Im+Im—Im+—§-+F l n 1+n I =1 (1+—+—)=r (1+-———) ASAP :3 [B a c: 5 __ In 1+1! ‘ n(” 5 J or I, = JEEP—— (1+ 1+1!) ,3 l0.9 Using the maul: of Prohlnm 10-8. 2:: 21”: =1-I355=1.06mA 14-5—6 5,0,7 111— 1.06 =>El —4.06kfl 10.10 Firs: approximation - BE area of Q2 is 3 times that of 9,. R. '=- 5'03 ‘HJ = 12,: 13.5 m 05 Second approximation - take. into account It v31?“ variation. For 9,: ’_cr = exp[3w:fia] [a v, or V3.51 " Vans: = VT 10(10‘] la: 0.7 ~15” = 0.026in[é) =9 VIE = 0.632 V for IC = 0.5 mA Then = 5~ 0.682 - (—5) R. 05 = Rl=lflfi4m Now -J In =_’c =—___..“‘1:7 => 15, =2.03x10'” A cx 23—; ¢XP[4] VT 0.02.6 For Q2‘ -1 In =—-L =fl—= 6.09xl0‘” A Van [0.682) exp ————- EXP VT 0.026 [0.“ I2 = 21. and [3 =3], (3) 11:1.0m.I,=I5mA (b) I, = 0.25m. f, = 0.75 me! (C) I, =0.167 mA. 1'2 =0.333mA 10.12 I.. In=Ic1fltdIREp=Ic1+I33=ch+1—I:3_fl __ VBE _ Hal V33 Isa w15': +Im+ R2 -— —E + R: 3101 V35 I = I -..—... __ REF cpl-“14“”-i-(1_{_-fi.}fi2 V33 ( '2 ) ~ —— = I 1 IR” (1 +5213: ° + an +3) I“? _ fl... I _ (1 '5‘ Huh ° _ 2 1 _...... ( + #0 +51) 2: I -(o‘ro)(1+ 2 )+___“-"' ' “F ‘ ' (so){51) (mun) {115;- = 0.700216 + 0.000354 {SEE = 0.7011 m = 1° ' H“) R: => R; =12.2T kn 10.13 I. he = 1:73 Ind-Ins? = {:3 + I35- : Ian +11:sfi Iss=hm+1m +Ia-z +...+Ia:¢ =(1+N)Igfi = (1+NJIc3 .6 1+N)Icn Than I = I S——-— as! 03+ 3“ +13) (1+ ) Bu +13) 6 b. Ins? — (0.5)[1 + tsunsn] — 0.5012 mA R1 = 5— 201.7 — —5) 0.5012 5 R: = 17.15 kn Cha tea-10: Pro lcm Solutions 10.14 IMF=I0(1+ 0013)) =‘“"”’ [1 + =. IEEE = 0.5004 ERA 5— .7 — _ =M=Rl=1119kfl R: 0.5004 v1?v V7: -?I/ Io=fzur-( P4! In = 0-8 IRA In, = (0.5)(1 + 1 2 1+ fl(2'+.8)) min) =5 13:: = 0.5024 m R: — _ 1a - 203.7) 0.5024 d» R; = 20.69 kn 2 [50)(51) 10.16 T‘hcmflysiaismctlymesmuinthcm. Wehave l Io = 1MP- (1+ -—2 ) 5(2+fi) 10.” In =2 IRA. I5: = % = 0.0257 mA 1:1 =1 mA. L9] = 71—5 = 0.0133 1111‘; 153 = 131 + I31 = 9.6133 443.0251r = 0.04 mA Isa 0.04 In: — l+fi — 76 — 0.000526 mA 135; =Ic1 +I33 => 1355 = 1.000525 z 1 11A R1 = 10-20.?) =E=>R1 =B.fikfl IRE? 1 —— 10.13 .1. We have. R” g #203 1'3; — In n.- IREF — 0.5 — 160 Then hzmafio=54hfifl 1 5 '1. A10 = E *AVC = =» A10 = 0.781 EA 10.19 V3; =Vg-ln (IMF) Is 103'"1 —1.5 0.7 = (D.fl?fi}lll I =' I: = 2.03- K 10 A 5' 2 x 10'3 AI 2 mA, V33 — (0.026)!n = 0.713 V R.=£‘%fl=fll =7.14kfl _ VT {33? _ 0.025 ( 2 j R" " Ia l ( Ia ) - 0.050 In 0.050 Electrgnic Circuit Analysis and Design, 2"" edition Soiutions Manual [0.20 10 - 0.7 = 0.465 m 20 I. Inst-‘2 Let 2":9 V33 ; Vrln (IRE?) 10'a 0.7 = {0.026) ln( ) = 15 = 2.03 x 10'” A Then 0.465 x 10" V55 = [0.026]I.n ) = 0.680 V Then I”,- .=. ng's-a—D- =» 15H = 0.465 mA b. RE: fim(lup) =0.025-Ln(0.466) In In 0.10 0.10 => RE = 4000 10.21 v*—v —v- 5-0.7-(4) a I =___.§£L_—__.=_____ U "5‘ R. I00 2’ 1:55:93!“ Ion. = v, :n[’—;§f-] =0 10(10): 0.0261n[m] 0 0 By trial and arm-r. [0 E 6.8 psi R1: = (111(1 +3—1RE) Now r,z =§=4 41 MO. 0.0068 =-—-——-=0.262m.A/V 3" 0.026 r z 100)(0.026} " 0.0053 80 R; = ,1||R£ = 382fll0 = 9.74 Id). Then . R, =4.4l[1+(0.262)(9.?4)] = R, :53 M: (d) V... —V... = Ion. = (0.0063X10} =. =382kfl V3.“ -V.EE = V 10.22 1 AI = — - AV 0 Re C R4: = f02(l + an: R's) VA 30 7‘02 In = m”. -- 4.6 Mn [0 0.0174 ‘_ (so)(0.025) 3 rue — ——o.0174 119.5 kn R's = 8.5“er = 7||I]9.5 '5 = 5.51 kn 3.. = (4.6)[1 + (0.659)(6.61)] =0 R4, = 24.9 M9 Now 1 Mo _. {5) :0 an. .. 0.20102. [0.23 Re = f02[1+ gnu-Rte) W119” R's = 3511?}: VA _ 75 _ fag 10 0.025 gm: — v; —- -- 0.952 DJAIV _ 5V,- _ (30}(0025) _ m _ In _ Mama _ 03.2 m _ Vr Im - w. 3-15. mg " Io h( In ) ' 0.025 In (0.025) = 3.54 kn '5 = 3.544|83.2 = 3.40 m Ru = 3[1+{0.952)(3.4)}=12.a M9 0: -L.Av -i--um A “'21., “"120‘“ ” 50 Ah: = 0.234 a 0-9362: In 25 10.24 Let R, =5kfl, Then 12-07— -12 [gar =m7'g—‘l: I,“ Now 10R. :19 1:1[Im']: [0 RE = 0'0261n[4'—66]=0 RE 5 l m 0.10 0.10 --—--—— Qhagter 10: Problem Solutions 10.25 10 — 0.7 - —10 40 V3; ; v10: (“‘5‘”) In; 2 = 0.1325 mA Is '10"3 Is 0.7 = (0.026}1n( ) =- 15 = 2.03 x 10'“ A New 0mfixm“ V55 = (0.026)1u( 2-03 x ID_15 ) .—.. 0.681 V Va E: = 0.631 V 80 _.: 10 - U.681—.[—10) Ins? — m :5 IEEE =0.4;83 mA I0 Inn?) = (0.026} ll]. InRs = V:- In (13”) By Will and error. => In Q 8.? yA V539 = Veg; - 103.5- =0.651 — (D.OOBT)(12) =5 V35: = 0.5765 V 10.26 Van: + IMF-Rm = V55: + I033: V351 -' Van = 103:: - IMF-Rs: For matched lransistnn V33; = V1111 (IMF) I: van Vrln (I???) = 30.33: - IMPREL Outpmrcsisunneloofiuginmthncoflncwofflgisincrmai 10.27 1.. 1 VI]. VII = _ m V. —— — :x m +9. 2+ m + RI (1) V V..- V} =.- V” + (-4"- + gmavn + '4) Rs: (2) ff; 7'0] V5: = Var! [1 + +9m1 + fun 1‘0: .. 10 - 0.7 1 = R“ 13.6 +5 — Lit—15° mm = 2.0 m m ' 0.5 75 To] = 53:13:51 0.5 = 0.5 mA , 9,,“ z __ = 10.23 mAfV 0.026 , 1 1 Vx _ V“ [1 + (1—6 +1923 + 150)(5)] = v,.L = 143001009) 1 1 V' Than I' = 1!“ 0.01009 [— 19.23 — —-x x "( ) 2.6+ +150 +115 I;r = 152(01900) + V§(0.0135) 9R3=E£=lfifi3kn 13c Vx -v. Ix = + ymavwz 1'0: V- = IxIRsaIKrn +33] V“, =.. _r‘.'.£__)v‘ 2 ('r2+R|'b Then Ix = V—" - I—xmmum + as); 1'02 1’0: 7r? I - 9m: (IX3[R£2||(712 + Rs: = Re: =D Ins; =Io an; =15 kn gm: =19.23 LEAP! Electronic Circuit Analysis and Dcsigg, '2‘.”I edition Solutions Manual Ix = — * —{5||(2.s + 3.55)] _c19.23)(fifi)(m[5m2.s + 3.53)] 1,; = Vx[0.00666] — Ix{o.a1asa) - “(2243) Ix[23.145) = Vx(0.00556) :- 30 = [Ex- = 3.45 MB _E..___.._—.— b. When R51 = RE: =0 - 53 are: = 150 kn 10.28 Assumc all transistors are snatched. 8.. 21,351 = V933 + IoRE I V33; = V11n( In V333 = VTID - Is 2v1-1n(l"") — V, In (51) = Iafls Is 5 b. Va;=0.7VIl1mA=§ID-l=fsexp(£i) 0.026 or I9 = 2.03 x 10'” A V5; It 0.]. ILA. 0.1 x 10" => V33 == (0.fl26)ln ) .-. 0.640 V Sim: Io = Imp. flan 0.51m VBE=IUR£='RE=T DIRE=SA kn 10.29 m ‘ M = 9.5 =2. R, = 18.6 m R: Inr= I. ., . Rs: = o °“5- (“‘3”) => Rm =10.17 kn { 5 _.._ = . ‘03) => Rm 2 438 kn V352 = 0.7 -— Inglis: = 0.7 —(D.01)(10.17) =9' Vzgz = 0.595 V V333 = 0.7 - («Rag = CLT - (B.03}(2.{38} #- VEE: = 0.527 V 10.30 (a) Van =Vsn I = V’—2VREI -V' “F Rl + R2 New 21,851+ Ins-3R2 2 Van + [0R5 01' [0R2 = 2V3£I ’Vass + Inn-R: We have iii [a Vm =Vr 1n and V”, =1}, In — I: I, ('3) La 31: R: and [o z [REF =5 van = Van 5V3: Then Va: = 10R; _IREFRZ = !0(R£‘R2) so V” —V‘ - 110M; -R,) [an =10 = 2 R: =V "V -Io & +10 3R2 R: Then [a = V -V 2R5 (c) Want I0 = 05m _s—(-s) _ So 122— 2‘05) =9 [as—10m _ 5—2(o.7)—(-5) _ 2R2—-—-——05 5:12:42 Then R,=R1=8.6k§2 10.3] 20 — 0.7 - 0.7 12 I” = 21'”;- = 3.1 IDA In: = I331: = 1.55 111.4. I0: = 3:331:- = 4.65 IDA 1- Inar =- =1.5$ mA 11. V55: = -In1-Rcz -(-10) = —(3.l){2) + 10 # VEE = 3.8 V Vac: =10 - 10234:: = 10 - [1.55)[3) =5 VEm = 5.35 V Vac: = 10 - Io; RC: = 10 - [£.65)(1) =§ V52; = 5.35 V _ ._.. _. Chagtcr 10: Problem Solutions l0.32 h' V” = i v I.. Is: Won 1 + (0.02114) - In = {200][ 1 => In = 203 nA his; ; 20 B I" = 2.325 link I +£0.02K1'89) 2.32 Now V55 -— 0.7 = (0.026) In C- VDS‘? = 5 v =v =v' =n.mv 55 5" 1+(n.nz)(5) I. = {200) => Io —:. 216 ,IJA Than 2115 approfimatim I”! E 20 491.722} = 132 mm 10.35 s I {05 IoL=2IREF=L54mA (21Hr =V + ‘5’ =1+ —=2V 61' ml KM 2 I9: = IRE;- =2.32 mA I..=3In;=s.9sm 10:55: ’w =K 1m. " K.. '1 K. b. A! the MEG Of umfion. Veg = V35- : 0.722 V a. Cl: 4‘64 =>Rgi=2JJkfl 0'5 _ m — 0.122 _ Mann) = {0.5)(o.95}(+) => roman) = 0.475 mA Re: - ——2'32 a Jig; - 4.0 kn S 0.; a - . RC: = 1 a 722 =- 85: r. 1.33 kn o 5‘95 0.475 3 In 5 0.525 m 10.33 2 Is: appmximalion (b) 10=Kn2[ 1125: +le_vmz] III Iflp=10-0.7=1M — a 6‘3” I(min)-(0')1;E£+l-105 =3- V3 = 0.7 V a assumed a - ‘0 0.5 ' V333 = Ins;- - Ran = (“(3) = 3 V =# Iflmin) = 0.451 EUR 2 Mum} = (9.5}[1/‘34’ + 1 — 0.95] V 0.5 HE! R = 3 1:“ ’J— => Io(ma.x) = 0.551 am So V1154 r Vu3—3VQRsa-E-= =55! =0.75 kn 045151550551 mA Im=lmA 1531:2111}; foa=4mA _ 1+ 0.02) 2) .- Io - (200) => In = 200 EA Electronic ircuit Anal si and Desi n 2ndcdition Solutions Manual [0.36 IOJT (a) mesat)=VG,—Vm or VG; = m(sa!)+Vm =0.2+0.8=l.0 k; W ID = ?(:](Vas ' Vm )2 so = 48(3LK](0.2)’ :3 [E] = 25 (b) V655 “Vm = 2W0; ' VTN) V6,, = 0.8+2(0.2) =3 VG” 2 L2 V (C) V,” (min) = 2V”, (m!) = 2(02) =9 Vm(min] = 0.4 V 10.33 (I) It =m+gnvflz Vnnuat) = 2V = VG” —VTm = Vm2 - 15: r. v = 35 v (2) 1 =1” 3.93.. G” i r 1 W 1 I, In =["-“-Cu)[—] (V031 'sz) V‘II =Vx 0 V3! = “VA 2 L 2 5° 25:) = (20)[%] (35 — 15)2 = (I) I, =V4—V4[i+s.] W 1 :9, r9 = (2) I; = —-*-+ gnV, = VA = r,[1x - 39;] a—-—‘——-—-— r. Then I W [REF = [iflncu - le) W ’e 100 = (mm?) (35-15)’ = 2 !.=Y‘--r. ‘+s. xii-V, 333/. W I; r. ra I = I I: =fl—Il-gnralz +gmvx +833”; r’ I Now vm=|o—v,;n =1o-35=55v _ _ 1 W - z I.[2 + 2.5] ~ V.[r_ + s. + 3.1;] so 100=(20)[I]3(65— 1.5) = Since g_ >> l = 02 r. L 3 -’.[2+ 34,15 K{g.)(1+ 3.x.) Then 10.39 3:. = R a .3315... a. Prom Eqmtion (10.10). I; ' s..(1+g.r.) 7 Usually, gflr. >>2 . so that 5 1 - i I is: 25 V = = 5 —""'— R-E__ 651 Van 1 5 ()+ 1+“: 8.. "' 2—5 25 M47 1 — 0.447 = _ __ 0.5 (1+u.447){5}+ (1+M4‘r)‘ I VGs-l = Van = 1.71 V (0.5) __ Chapter [0: Emblem Solutions [0 = [13){15)(1.74 — o.5]’{1+(n.02)(2)] = {415M104} => In = 0.432 mA c. In a {415)[1 +(0.02}(4}1 '4? I9 = 0.443 mA 10.40 I k' w . (a) In” = (V55: +Vrr)- I J W I = Bax-L‘le: +Vrr) But Vml = 3—1156, So 25(va _ 0’4): = 5(3 ‘ Vsa: " (3-4)2 which yields v35, =1.08V and Vm = 1.9211 I“, = 20(15)(1Jox?.—0.4)2 = :m. = 231 pa 1.. _ m - :2 _ o 6 I“, (vii/L)l 25 Then 10 = (0.6)(231) = 139 M (b) V5,:(mr) = 1.03—— 0.4 = 0.63 V V,: = 3- 0.68 = 2.32 = 10R then R=2—'33=9 R=16.7k£2 0.139 -—---—-———-— 10.41 V”, (mt) = 0.25 = Vm +Vr, = V55 — 0.4 = V”, = 0.65 v [a = (sz +vw )1 25 = 4T°[—‘:-l(0.65 —o.4)1 = [ Eifiiiwfi [W k ’ w {gar = '?’[I]J(Vms + Vim): I’m = 3— 0.65 = 235V Then 40 W 1 W 75=— -u 2. —, _ = 2[Lb 35 04) =: [L]: 0.986 HE H I! M Q Im=75m= hl :_-_-’ ll 0’: O 10.42 a. I", = mug, 44.”): 100 = moisz — 2)2 => Va: = J V FBI VDI. = *3 V, In a 100 [UK 13- R0 = 1'01 + T02(1+9m1'0i} 1 1 = = —- =: — = no '°‘ '°‘ Mo (0.025(u.1) 5 m g," = 2Kn(VGS -Vm)= 20100—2) = 0.2 mAIV R4; = 500 + 500[1+{0.2)(500)l R4; = 51 Mn I 5 __._.. v =_ ._., AIu Ru A Di 51 =3! Aura 0118 1115. 10.43 Vim = -Ix1'o2 Vs: = (1x - nggu)rm + J'an = (Ix + ymfxrn: )flu + Ixfuz Vs: = Ix[faz + (l + 9mfnz)fm] = -V}.s v —v v 1 Ix=§ngaa+J—-fi=i-Vsa(§m+-—-) 7G6 '06 rGG v Ix = J— — I): (a... + i)[roz + (1 +sm’02)’°*1 r08 1'0! 1;:{1 + (gm + L)[fu: + (1 +gmm}ml} = V_x 1'05 7'06 V2: T; = = rug +{1+ gmfogflrn: + “- +gmfoz)l'o¢] In 2 In» = 0.2 um = 03(sz - 1}3 Vos‘ = 2 V Electronic Circuit Analysis and DesignI 2'“1 edition Solutions Manual 8. = 2K.(Vas "Vm) = 2(0-2K3- 1] = 0-4 "=4 I V I“, = 150 = 50(3) (25—05)1 => = 0.75 rog=rg4=roq=-1—=-——1—-=250kfl L! L] AIa (fl.fl2)(0.2) k! W 2 IRE: = (V564 +Vrr) R, = 250 + [1 + (0.4)(250n 4 x {250+{1+(0,4)(250)]{250)} 150:2 31') (25—05)1 = [E] 1.88 30 = 2575750 1:!) L ‘ L ‘ = R.” = 2.53 x 10’ 9 10.46 I. As a first appruximation I044 k' W k: W IREF = 30 = 30(Vas1 - 1}: => Van = 2 V I a " ~ (V651 _VTN) {V651 -me Then V351 = = 4 V I 3 r The second appmma‘ lion k = (V55; +Vrrf 2 L 2 1 ‘ a so = swag. — 1) [1+ [0.0mm (1) 50(20)(Vas. -0:5)1 =50(5)(Van —05) 2 of % = (V051 _ I): a V651 =1“? (2) 50(20)(V651 -05) = 20(10:){Vm4 _05) . (3) VSGA +Vass +ch :6 Then From (1 ) 4025., -05)2 = (v6.3 - 0.5)2 = van = 2(an - 05) + 05 From (2) 504,... - 05}2 = (van - 0.5)2 =» V654 = JEWGII ‘05)+0'5 Then (3) becomes b. From I PSpice analysis. In = 7109 11A for Vm :- «Ewan —05) +05+2(an —05)+05+vc,, =6 - -1 V M To = 77-14 #A 5°? VD: = 3 V- m mm ‘ z 0.05 A . 7a- which yields v6,l =1.36V and ‘5 M" " °' 0065 van = 2.22 v , V“. = 2.42 v ’0 = Kn (V05: " Vm )2“ + AAVGEI) = so{1.902 —1)’[1+(0.02)(1.962)] 0' W The,“ 10.47 ' . 1: first m0 IRE" (Van ‘Vm )1 = 50(1?-0)(1.36—05)1 ‘ a. l app on‘ 1 7 1 or 1,... = ID = 0.740 m. In? = 3“ = 30033 '1} =* Vm = 2 V V6,, =14”2 =1.36V Vm(sat) =14”: —V,,, = 1.36-05 = I sz(mr)=o.sév Mammmmnm m 45 In.” = so = sow“. _ .1)‘[1 + (0.023(211 ' 0r V55. =1.9a v = v”: Vnszfiat} :05 V = Van ' Vm = Van ‘05 =9 (a = K (Van _Vm)3(1+ zvmz) Vcnzlv , Telvuygoodnppmximmm ID =50”; =—’5~—[-W-) (van --v,,,)2 I. = so “A 2 L , =50[1V_] (1.0.5)t =3. =4 [3. From a PSpioe analysis. In = 50.00 #A for VD: = L2 L2 —1Vanfifl:coulputrcsiamiqu=76.9Mn. Van = Van 31V =’ Than - fw=15°=£ E (V 'V )1 A! =i-Vm=_‘_=ooszm 2 L , a“ T” ° R4, 76.5 ' W 2 W = “(310-05) =9 [-5) =11 a: a change. of 0.065%. I Van +Vm. +Vsn = 5 2V6n =6~I=SV = V“: =25V Cha ter 10: Problem Solut' ns 10.43 I 10.50 (a) K“ =50(5)=250MIV2 1RH=W 2 L , .R Also _ 1 (W10 R - K 110' [1‘ (w/L)I ] its: =40(I)(ch '13): I 2 I 5 in: = '80“va ‘12): = [1— = (a.94)(0.534) 11"“ (0.25)(0.05) 50 Java," -1.2)= «Ema? — 1.2) R = 6.1 1 Id! which yields 5.325 (b) v" — V' = Imom) +12“, V”? ' 55%" _ 1'2“ L2 V503 (“I ) = V563 +Vrr The“ 2 1m = 50: 2‘15me 45): = Vm = 12 I V [0.040(vm — L2) ]-R = 24 WW, - 1.49(Vm —12)— 12 The“ which yields v,,,,(sa:) = 121-05: 0.71 v V” g 2'69 v and V5“ = 3'41 V Also Now 24 - 3.42 — 2.69 [m = so = 50(5)(Vw —0:3)1 =9 V6,, = 0.947 v has: = “200 = 1’ m = 395 M‘- Thcn I 39.5 n 9 (V‘—V')fl=0.71+0.947=[.66V "T‘ ' m I2 = (L25)(895) = I 12 m 1,: 03 895 =7L6 (c) Ia, =25=50(%] (0947-05)2 =9 =25 ' ( x ) m 5 5 L r‘ =4{395) =353 ya [02 =75=20(iy-] (121—05)2 :2 [E] =7.44 10-51 L a L .s We have VW=2.69V and vm=3.42v "’49 so 10 269 342 389 Iky=——-—---—n-—IR . =53 [JEF=I9.45‘UA I W 1 Then I»? =[Eflncu11']3{vcss 'Vm) I: =(G'2X19'45) = 3'39 m m = (20) (E) (“m _ I): I, = [1.25)(19.45) = 24.3 ,uA L , :, ={03)(19.45}= [55613 W W W _ = __ = __ = a .4 = :(L) (L) (L) 11.2 L 4(19 5) 77304 10.52 1 . [0| =[Eflncu {V653 ‘me For “as = 0' ‘D = IDES“ + has] I I. VD = -5 V. #105 = 5 IMF _ L 3 Or In: — I'D =(2)[1+[0.05](5)] L i =>i =2.SmA 0) «am 1 1151" 1 - ‘ 1:. V9 = U. 995 = 107 => =22Jl #_ in =(2)[1 + [0.05)(10)] =0- ' = 3 A m0) 000) Wm '”—”‘ L I- In: L ,— D.1 ' W C. Vp=5V,905=15V _...._2_ ip = (2)[1+(0.05)(15)] =5» '2 =15 111A Electronic Circuit Analfiis and Design. 2'“‘ edition Soiutigns Manual 10.54 . I VG!) = 1—— Iu Ipss( VP Va: 1 ._.(.- VP 3.. VG; = [0.293)(—4) = 4.17 v 'I'henla=%l.ndvs=-Vcs -Vcs _ _{—1.17} R: In 2 10.55 V I. IMF=I51EIP( or V33. = W In (“TR”) = (0.026)1n( I51 1 x 10" 5 x 10-u : Vim = 0.5568 5 — 0.5568 = — =9 1:. RI 1 c. From Equation (10.72) and letting Vega = V552 = 2.5 V 1.0208x10-12ex 3; =(m..)[L03125) V: 1.00696 Then v, = 0.0261n(1003613x10’) 80 v, = 05339 V -(1 VT (IIVAN) +(1/Vu) 1 d. A»: _ —38.46 L+L ‘ 0.00333+0.0125 120 30 A... = «1846 Av: 10.56 I. V“ = Vrln(1up) = [D.D26)ln( 0.5 x10") [5; 10"“ = Va: 2 0.5203 5 -- 0.5203 13- Rt=—3'5—=’§J:L95_m :. From Equlflm (10.72) Ippliu with slight modifications ch 1+ V V In”? 380 1+ 50° =;R£P. ...__V4.L V1- VAp 1+ V331 VAN -1'a V'an) ( 2-5) (5x10 VT 1+80 2.5 1+—-- .. - -J, I} _(0.3K10 ) 120 _ V550 _, 1.02033 5. 55 ’ 0 " = 0" D ' 1 2: x 1 exp( VT { ° "1 ) 1.00434 V353 = =' V; = 5 - 0.5334 => Vi = 4.452 V _ v = —g1,rVT) ‘ A (UKm)+unaa 1 Av = Q = —33.45 L + _1_ 0.00533 + 0.0125 120 30 AI: = —1345 10.57r Ignore. {W115}: = 5 specification 0. M. and M: matched. so we must have Vsn: = VSG = V563 = Vase = 2-5 V For MI, and M3: _ 1 w 1 [w ~[3p,C“IZ)I(Vm +V,,) (MAJ/m) 100 = 14%) (2.5 - 1H1 + (0.02)(2.5}] " (T), “'23 = g). = (g). FCI' Ma: Io = [épnCuI‘ZLJOEVG ' VTN)2(1+ LVN) 100 = 200;) (2 - 1)’[1 + (0.02){2.5}] W —— :9. 4.75 ’(£). Cha ter 10' P ham Sol ti ns 1 1 "' 'M-“P‘m-W-“m l W = {CI :2 — C _.... 1 gm no [21”: L]fla = 2 (0.02)(4.75)(a.1) g... = 0.195 mA/v A» = -am[ro.llro,) = —(u.195)(soo;|sao) g. AI = “48.75 l0.58 " Ina-r z Kps(vsc +Vn')! 100 = 100W“; - 1}“ =:. wig = 2 v b. Vpsn=Vpn=5v From Equafinn (10.87) 1000/; - 1)1 = 100D + {0.02)(s}] x [1 - (0-02)(2)]I1 - (MEHSJI (V: - 1)’ = (1.1){u.9s)(a.so) = 9.9504 =5 vI =1.575 v c- Alf = _9M[rfiullrflp) 1 1 11337 ‘ (n.uz)(u.1) 3,, = 2 Kurm = 2 (o.1)(o.1) = 0.2 an IV A... = _[o.2)(son||500) run = "up = = 500 m fir Ax = —50 10.59 a. Usingmruullsofprobluu 10.27. wefinddu: resis- mWhmMcouedm-ofagmbe = 1'02 1+—L “3+3” 1-0! 1": Hm [hum + Rm wmfiahmmefrommehueafazmvd QI. Wefound l l 1 —+§mI+—- =—+—ffl—_—._rfl— R: l E (i I ) I + r'l +§m1 + r01 b. A, = —,m(fg|[RL"Ro) 10.60 Cum: rcsism of Wilson source E 5'0: 2 Thu. Av = 'gmU'DII-RO) _ Var _ £_ fa; _ IREF -— — _ VAN _ [20 __ ra — I“? — 0'2 —600 kn I 0.2 gm = flair; = = 7.69 InAfV Av = 4.59 [600 = -7.59[ean||1s.oou] =9- Au = —4447 10.61 3" gm(Mu)=2'V KflIREF 9m(Mo) = 2‘/[u.25)cu.2) = W I l a = = __ n -_- 250 kg '° MIKE; gamma) ” b ,0, = 1 I =- r0: =15? Aplazr = momma) b. A... = -gm[ro,.|[ro,) = —[0.447](250H167} =‘* A... = —44.8 c. R; = 250"].51' = fon‘lfg, or R‘ = 100 kn 10.62 Since V“; = D. the cimuil becomes Elcctro ic Circuit Anal si and Desi n 2"“ edition Soluti 11 Manual 10.63 A» = ‘gmlt‘RullRU3) From the results of IFEI‘s: R92 :17” +".,2(1+3;’LI) From results of Problem 10.62 R0] = 5.1 + rn4(1+ 3;"...1) We find gm, 2 2 [0.05)[20}(0208) = 0556 M f 1’ l I ’ " r“ ‘ ' (momma) ' 625 m l l = = —— = ———— = 625 m r” r“ 21,: m (0.02)(0.oa) g; = 2 (0.02)(40)(o.03) = 0506 mA N Then R“, = 625+625[l+(0506)(625)] = 199 MD. 120, = 525+ 625[l +{0506)(625)] = 199 ML: Than A, = —(0555g199ooq|199ooo] =¢ A, = 46,317 r A: Rd =ru2+g;rfl3+‘2l] I or R.=nz+c.;(|+gf.r.z) r l 01 A = % = ‘3ml(roIIIRn) Now 3,,“ = 2,}(0.050)(20)(0.10] = 0.632 m r V r = ——-1— ‘— —L---- = kfl. " 21,1,”a _ (o.02)(u10} g; = 2 KP!” = 2 (onzo)(so)(o.1) = 0.30 mA/V l 1 a = —-—-—- =-—— = m "1 '3’ 2,1,,” (omoxm) Then 1:, = 500+soo[1+(o.3)(soo)] =- 201 m A, = —(0.632)(50q]201000) = A, = -3I5 ...
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