Neamen2eSMchap004

Neamen2eSMchap004 - Eloctro ‘c i uitAnal is and Desi n...

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Unformatted text preview: Eloctro ‘c i uitAnal is and Desi n 2"“ edition Chapter 4 Exercise Solutions 54.1 Icq 0.25 = —-- = --— = 9.6 A V 9"" VT 0.025 g, £"'_.2_r.n__/_ _ V7.6 _ {0.026)(120) __ a . n... Ito .— 0.25 =>r,,—1...:alcfl VA 150 f = = —- = :1 Ice 025 => rg 600 k9 E42 V V T ro=I—c3‘—=>fcq=—é-— 5 m ‘ 200 1:9 =5 [cg = 0.375 111A E43 V53 - V55[on) _ 0.92 — 0.? Re ' 100 => ng :1 0.0032 mA 1“. = (150](0.0022) = 0.33 mA I59: 0.33 . a. gm=-I%=D—02—G-=gm=117mfif\ V13 _ (03026)“.50) _ Q r" ' Icq _ 0.33 =’ —-——" '11'3 k VA 200 fn—E—mfirg—Gofikn b- 1’0 = -9muw(quRc}. Us = (fi)vs A» = 39- = —gm(J—) (rolchl 95 r: + R3 =-. —(12.7)(.1T;—1:81T0)(505|115) = -(12.7){o.1u55){14.54) =5 A1, = -19.6 134.4 _ Ice Fm - VT V55 - ‘f-EBUJI'H 1.145 - 0.1.0 I“ [so = ...__-.-_..—.—-- _—_- ___.__....— Rs :0 =9- Iaq = 0.0039 in). I“ = [90}(0.0039) =7 [cg = 0.801m:\ 0.801 = —— = _ V' 9m [1026 =5 in 30 3 mA/ Solutions Manual fig __ [0.0026){90} = e = 2.92 m r" Ice a L— L; 120 = —- = --— =1'D r” Ice 0.301 =’ -———'“ ° = -[30.3} (150“233) = —{ao.s)[o.055}(2.451 =:- 11., = -4.17 E45 Rm = 250||75 = 57.7 m . ' R2 . _ 75 \ - f - VT” =(R1-9- R?)(D) - (75+250)[D:‘ — L104 l.154—U.T = = 3.;3 A [5" 57.7 + (121mm; ‘1 [cg = 0.418 m.-\ (monome) - = —-—-—— = 7.46 kg a’ r" 0.413 0.413 = -—- = .08 V 9,... 0-026 1'5 mr'kf Va = —gmv..Rc R5,, = r, +(1+ fijRE = 7.46 +(l21)(0.6) = 30.1 m R, "R1 = zsqps = 57.7 in R,||R2||R,.b = 57.7ll801 = 33.54 m ,_ RtllkzllRm ‘ _ 3354 _ V‘ “[gufiugfl, + Rs V‘ " [3154+05) V‘ v; = (0.935% V§=V‘[1+[l:fi]RE] Then A, = (0.985)(-S.39) = ~82? - ‘u. R.» = r; +(1+ 3HRE) = 7.45 + (121mm) 9 FILE = 50.1kfl E4.6 As a first approximation. - .& Resulting gain is always smaller than this vaiuc. The affect of R5 is very small. RC S [—=10 c R E Elcctronic Circuit Analysis and Design, 2"d edition Solutions Manual 55]c(Rc+R.-:)+ch S:05(RC+RE)+25 So that RE=O.4S4kfl and RC=454kQ 05 11m = as = 0.0051101 Rm =(0.1)(1«- LURE = (0.1)(101}(0.454) = 459 km - R 1 l VT” =[Rl :R: J‘VCC =3: Rm 'VCC =E(459X5) 23 or V ---- TH — R] V”, = IRQRT" +Vag(an)+(l + mlmRE 23 -R— : (0.005)(459) +0.7 +(101){0.005)(0.454) which yields. RI = 24.] kg and R1 = 5.67 kg 154.? d1: analysis 1:: ., :I II E“ 33 DJ 11 E E DI 11 _ 12 — 0.1 ..10.2 _ 1.1 ‘ 12.75 +1101)(0.5) ' 63.25 ICQ =1.“ mA = 0.0174 :0 analysis V]; = hfgIb{Rc"RL) I - b _ hue + (l + h}¢)RE RL} A” = h.. +11 + 11,.)115 For Icq =13 mA h;.(:na.x) = 110 h!.{minj = 70 h..{mu} = 2 kn h..(min] = 1.11:9 410141121 _ _ 2 +{1111(0.5} “ 40141121 _ 4.54 1.1 +(71)(0.5) " A.(ma.x) = 2.54 A,(min) = 54.8 First approximation. A = ---—R—9 which predicts a E low value. Set 55- = 9 . Now RE vac a [camc +11E)+vEm 75 =(0.5)(912£ + RS) + 3.75 50 111£ = 0.625 Id). and RC =S.62 kn Rm = (0.1)(1 +1310. =(D.l)(101)(0.625) = 6.311111 1 1 v”, =E- Rm Va, = E(6.31)(75} V“. = (l +fi).'wRE +Vfl(an)+ ImRTH +11," {,9 = %_= 0.006 M T5 = (101)[0.006)(0.62S) + 0.7 + (0.006)(6.31) i —- 6.3! 7.5 + RI 1 11 1 Then Rl = 7.4-0 K1 and R1 = 42.8 I'd). E43 dc analysis 10 -— 0.7 Emma—01 Icq = 0.439 mA. 159 = 0.443- TIL-K V713 (UflQEHIOGJ In = = 0.00439 mA rw=E=W=SBZkQ g... = $3 = =1s.ss mA/V ra=%=%=228k9 1a) V.=-g..V,(r.ch) V: = vlr=[—1?B||r;].vl Raur, + R5 R5"; = 100115.92 = 559 to Then V, = [flJ-V, = 0.9131”, 559 + 05 Then A. = —(161.7)(0.918)= —143 1:- Rm = RBII‘F: = ? Rm 2- 5.59 kn R0 = Rcllfo = 10§|228 => R9 = 9.58 kn 54.10 = .__:§§c__ = E = _ {1] A r, +(1+ ,8)RE {095$ RE] (0'95 0.4 or A. = —4.75 Assume r: = 1.2. kfl from Example 4.5. Then —fi(2} _ 1.2 + (1 + ,6)(0.4) - 4'75 which yields 13 = 76 Te tYour ndcrstandin Cha tcr 4: Exercise Solutions E4.” dc analysis: Vm = 0.12," = mug. = [0 1:52 3 w 5 — 0.7 ,3“ = — = I” m + (126](5j Isa = 0.84 KIA. 3-3 _ (a.oza)[125) r =————-—-—-——=3.BTkQ ' Icq 0.34 L3, 1; = — = —— = 233 m '0 I” 0'34 I _ 12 — 0.7 — 10.2 _ 1.1 5" ’ 11.75 + (121mm) " 73.25 a' v“ z _’“”'(’°lchlIR”' "” = "5 Jag = 0.0150 I = 1.80. I =1.52 An = -9n{ru|1Rc-|1R:.] = —(32.3|(233]|2.31|5) 9" 5° , = v = 3.39 A. = 432.3;(135; : .4... = —50.4 “‘3 b. For Alc- =1.B = mm; = (1.8){1.SS] = 3.29 5- R0 = YDIIRC =5 & = 2.33 kn For Ausc- = —3.29 =:- Aug-E = 3.89 - 3.39 = 0.6 =:~ Mu. symmetrical swing 54‘” = 2 x (3.29) = .g'fi v pcak-to—pcak I 54.14 5.96:. dc load line: EC; =(1u +10) - mm: + Ra} oars V55 = '20 - Ic(10 + R5] 9“ = —icRc = -ic(10) .23“ 3- 9.: load line: - '121 , ‘ chq = 5 - (0.-L18|{a.6} — IT [OAIBMDfi] \ = 3 - 2.34 — 0.253 chq = 2.41 an M'cz variation (2.41— 0.512 = 3.32 x' pcak-to-pcak 134.13 Q7 veal 3.: . A _ a. dc load fine: fl r— tee Vac = Vcc - IERE — IcRc Vs: E12 -1c(Rz + Rc)=12 — 1.:[451 5"“ = V65? " 0-7 = “35(10): ICON) II: load lint: 50 Vcsq - 0-7 = red”) “I: 1*- "iCIRfl + W6 have u“ = —ic(o.5 + mm = -—ic{1.33) Vega = 20 - 1mm + Rs) Icq(10)+o.1 = 20-fcq(10+ R51 (1) Elgtronic Circuit Analysis and Design. 2'“1 edition Solutions Manual — 0.7 ‘ . [0 a I“ = £30039 3) m E416 15" =100+(101)Rs 100+(1011R5 R: . . k a. PM: = (12}— VrH = EMTHHIE) RT” = “11]” +3312; =[0.1][1'21IRE From (1) ICQIIG+10+REJ=20_D‘T =i2.IR5=12.1kQ . 1 Substitute (2) 1,,_07_VTH 11.3-fi—(12.1J{12I I = _—___-_._ = 2 (100 9‘3 baa-FEE): [9.3 5“ Rm +{1 +3113; 1-2.1+{121)(1) 10° + [101mg 1 16 I 1'6 0 01333 a 930(2n+ R5) = 19.3[100+[101)RE] 5" ' ‘ é 3" ' 120 ' ' m' 1 13. 500 + 930R; =1930 + 1949,3123 11.3 — 12—11432: 16.670 =1019.3R5 =3 RE =16.35 kn 00”“ = 133.: So Ill—{145.2} = 11.3 — (o.o1333){133.1] 1 [co = ...JMl— =3 [ca =o.531mA =3 R~=15.2-1 1:9 100 + (101)c16.35) R--——-—R 1_ “R .3. Va“ = 20 — (0.5311(10 + 16.35) = V339 = 6.0 v 311+}: =12-1= m Aves = Vcsa - 0-7 = 5 - 0-7 = 5-3 (12.1}(1523) = (13.24 —12.1JR2 Mu symmetrical swing = R2 = 55'... kn 2 x(5.3]= 1195‘! peak-:o-pcajc vac: =12 -[16]MI-(I.611H1 :3 v5“, = 3.99 ‘v' b- “0 =§mvrfflcilRL|= —gm(RC'HRLI=—Ue: E4.l5 :C = gay, 2 —g,.,.y_: . I“ = —5—'—°'—7—- = 0.90672 or - v“ = ic(RcHRL) ‘ 10 + {125)(3) I = 0.84 EDA. I = [1.547 A Go sq m 1-H \E'Ch: (“A (ism. VA 200 \ Icq 0.84 l a Vase = m - (0.34)(2.3) - (D.B47){5] ' = m -1.932 — 4.235 'cgq = 3.83 'v' 3.17 IL. Van Inn-J l-‘we Want Aic =1.6 — 0.1 = 1.5 ‘-""" m" = 3.99 -o.5 = 3.49 as" A?“ = E9. = 2.327 kn = Rcum AI: 1. RC!“ 2 ‘RL _ 2.327 m.- + R; 4 + R; -'2. =-. .32? R =5.'6 RD 33-; m [43 327)RL (‘1)(1 )=:- L a dc load ling E417 Va; 2 10 - Ic(7.3j Rrfi == R1113:- = 251150 =16.T 1:!) ac londlim: a: , SD _ . _ VTH =( )VCC = (- )[DJ 9“ = ‘lclflcufifJ = —:c(2.3||5) = —ic[1,5a) 31+ R: 00 + 25 (neglecting m] VT],- = 3.31 V . I _ V35 - Vaglon! _ 3.33 — 0.70 b' A“ = "‘5‘ 9° ‘ a” +u +5135 ‘ 16.? +021)“; = Aug; =(0.B¢)[I.SB) =13] v 163 Vcsh'nin} = 3.33 —1.33 = -..5 v ’ 137.7 a I” = 0.0191 [cg = 2.29 mA Vagrant) = 1.53 +1.33 = 5.15 V 50 mu: symmetrical swing = 2 x (1.33) = “55 V pubic-peak Test Your Qndcrstanding {ghagtcr 4: Exercise Solutigns .4" = m = a A \v' 9'"‘ VT 0.025 a '1'“ I v7.3 (omennm fr: __=____.._.._._1.35kn Icq 2.29 VA 100 = _ = _ = "' n r“ Ice 2.29 I k V: = [ Riki}: VJ lRI IIR‘ZHRH: + RS 1r”) = (1.36} +(121)(1fl43.7] => Rib =120 m and RIHRI = [6.7 142 Then R,||R1||Ri,, = 16.7"]20: 14.7 kfl Then v;= (——14'7 )V, = (0.9mm 143+05 Now Va = [£+gnvf)(REll’-I) = Vn[ l + r” r! r! V _ V; _ (0.9mm 1+[H fi]RE“ro 1+[1+ ‘3)RE ra r, r, So 1+ 13 0 967 R v, _ ( )[ r, ] ‘flr" _ (0.967)(1+ f3)RE“ra r‘ +(L +-,3)R£“rn R5“; = 1543.7 = 0.973 129 Then _ (O.967)(12l)(0.973) - = 0.956 136+(12!)(0.978) : in— b. Rm = r’ + (J + 3]RE]|rD = 1.35 + (121101973, =& Rag=120 1:9 (6) ‘ 7 “r, + R. "5121le 135+1m 5 = = .7 " RA!“ 1+ ,{3 3 12: which yields R, = 151:2 54.13 Vase =5V=>Isq =%=2.5mA in = = 0.0243 mA Vm = haan + Vazfiou) + (1 + 5115.335 fin- = Rmmr. + (1 + was] r. = 1.05 m RT" -203 Rn: + 203 => RT” = 95.5 kn 65 = RrHHZDS = i(95.6)(1o;=(00243)[95-5)+ 0.7 + 2.5(2) = 3.07 - _ 118R: _ . R; —118kfl. 1184-3: —9:I.E R2 = 504 M7 RF" =65k§2 v’=[ R*" ]-v, =[ -65 ]-V, = 0.99214 ' R.“ + R, 63+05 Then 3 (0.992)(1+fi)RE = (o.992}(1m)(2) 2 r, +(1+ 13)}?! L05+(101)(2) a=ww Neglecting RS . A, = 0.995 R R R . 5. R” = = gum 10! or R" =15.2§l . 1.05 Neglectlng R5. R5: —=: R.: =10.3$1 102 —-—— 1-34.19 . _ R: _ 1’13 — (31+Rz)(103 5 ,3 =100_ VA =125 V. Vsfion) = 0.7 V Icq = {1.75 mA 125 Thcnro=-_.—_=167kfl DAD ma- [momma] rs = -— - —'#.7— = Ice 04.) Vra - Vasfol'l} - {-5) En: +(1 + 3'le fiisc 1:... = 3.47 119 Iaq= [cq Electronic Circuit Analyais and Desigg, ’2“d edition Solutions Manual 2. = r" + (I + flJIREiIRLllrnl I“ = (Jfll—Ju +fim REIIFa-i-RL RIIIR'J ) 1°‘(R_.1|Rz+2.' ’5 In A] - I5 REH'N ) ( Rdan ) = —--—'— 1 __.—-- (REHTO +121. ( +3) RN32 + Z‘_ Assume that REHI’O = R5 HALE: = (U-UU + 3H: = (0.1)(101135 =10.1Rg Assume RL : I E2 Then -_ R: A: :2 la -—(R£ +I)(1G1)x ( 10135 ) 10.13; + 3.4T+{IDL][R5;]'.11¢Q] what: thLflLlirn z HENRI. = RElll *9 3 5 figmnumgflgx - 334-1 1 1.0le . 11-4.- 10135+3 l 1+RE 15 _ (101)(10.1)R§ _- R; + 1 1 __#__._.—_— " (1+ Ralpmflg + 3.4.7] + 10mg 1+ Rs [101){104112’3 - —-—_.—_'_——I‘—"_"_—_ 15 _ 10.13; + 3.47 + 10.13; + 3.4.712: +1013; (101}{10.1)R’£ = W 15 unfit:z + 114.5711; + 3.47 (101100.113: -_- 15[m.m§.; + 114.5731; + 3.47j 1020.111“ = 151.53%; + 17135512; + 52.05 363.631- - 17155512; - 52.05 = 0 16.733; - 33.01:; -1: u 33 a: #03)“ + «16.7; R; = __.___———-- 2(16.T} Mmusc+sign=>RE =2.Bkn Then RIIIR: = 1.0.1.35 = 20.3 kn. . _ R: _ . TH - (R1 D 1 R1111 —- —5 Rl(R1+ HQJUB) = Ril'uognm) — 5 {cg = 0.75 l—tzomuo) — a — 0.7 + 5 - (we) flz_————-——— ‘ 20.: +(mnm 1.57 = 3—(292) - 0.7 =. n! = 35.2 m R: R1 R1 .1 . 35.232 - = .0.2 = _ R. = ‘26. 16'! R. + R: 33.1 + H: 5 J————° E420 a. =10“. V5.5(Olfi = 0.7. Icq = 1.15 mA <- L. :I: II fix :1! u U ’3 3 l p. v” = fi—xtflfllfizfllfl) — a I _ v” —o.7-—{-5| an ‘ Rrg+(1+J)H£ chq =10.-!sqfis=é 6 5 - JEQRE = 5 =3' [sq = -R—E- : RE 5126 = ‘sz kn = R: =4.75 kn R‘rfi = (0.1)“ + “R: = IOJRE Then 1 - — 101 U - ‘ -U.T+ I _ 1—59.: R( )REU ) 3 3 3"" 1m lUJRE-I-{IOURE —1-{101}(4.TG) —u.7 00125 = L———-—— => 3.: 65.8 m (111.1)[4.76] 3:32 - = . R = 10.1 4. 6 =4B.1kfl RI”: (m) z ( if :1 (553m, -_- (43.1)(653) + (43.1w; (55.5 — 43.1w; =(4E.1J[65.5) =5- R; = 178.3 kn Y 11 tandin Cha ter 4: Exercise So ut'ons IQ 125 f0 — I” - m -100 m 54.21 av 00 . = M ___ :- = (1 M0026} = “a m Fen-3 130 Ice 1.20 I _ lu _ {17 g. 5 16 A g...|r'.=g...(i.r.1= 31.. 9° ‘ 100+ (1311110) "’ “ Z. = 1'11- '1'“. + 1CD = 0.35: 111A = 2.03 + (101)[4.?5{|1i|100] From Figure 4.21 = 2.03 + (101)[0.326||100} 3 < 11.. < a m Le: 11.: = 0 = 200+ (101){o.a191 93 < 5,, <170 Z.=B4.B kn 3<h0g<15 {.15 Assume RL=lkfl h..=1kn 1%,. =131 1'0 ——- (—ML)11 +311. 1 Rsllru + Rx. h. = 12 :15 = I— = 33.3 [:9 It = ( R1HR2 )IE 0. RJIIRa +2. I. R;- = R}, =10 kn 10 ( REIII'a ) ( R1 R2 ) A = — = _ 1+ 3 . ‘ 1.,- Hall’s +3; { ) £41112 + Z. R... = 11.1 +11+ h;.|(Rz EL figflra = 4.76n100 z 4.54 '3‘ A “( L54 ) 101 ( 4&1 =4+(135){101|10||83.3) ‘ ' 4.54+1 ( J 4311 +313) => Rm: 641110 =(ggg)mu(4s.1) __ hum “4 132.9 A" "354.33113... x =- A! = 30.0 1 (l +h1.)(R£ RI. ) _ Fir ‘- _ l c Ru—-—1+IHIIR5 ll'.,_------ml 4.16ll100 7_ h..+(1+h;:](RsRLl hoe) ” _ IOOIEGM (1353(10|}10{|s:1.31 => no _ 20.4 n A" ‘ (10 +100||6u 4 + (135)110I119H83‘31 {L _ 55.3 ) fl: " 10 +865 64] =- A, = 0.091 1 Rs -- h R: ._.. ___£_ — A' _ I (1+hu}(fia+flw) as -— + R; ho: _. Mymmfl) _ 1011513 +10 100 + 641 ‘1’ I0 9.» .-1a= 0.59 ’° = "=- = (H 53‘b(R£||R:-llro) = .l. M 1-: R" R5 ll ha, l+ hr]: 1. = — ’8 = 10 33.3 fl=093 0.0970 _ _ 1 + ,8 13:: — 1c 5 (MRI-urn) =1 £9 = m n . 101 =‘C(m (“WWW h- 35 = 1 1‘9. W— =ic(0.520) A _( 35.5 a 4 _MB_,' If as": =1.2s um. —- Av" = 1.035 v " " 1 + 86.5 541 —--' " ‘ ' Msfimumsymmcu'iulswinginomzvoluge Ro=10fla.33“[%%1-93] is = ZhPCI = V = 5.93l|0.03696 :- g = 35.3 n cc ic ircuit Anal sis and csi editi n S ' anual E422 Ra = 1.96 33—3 = 1.96||0.0204 = 35.7 n vm=25v. R,,,=2sm I 1 1 a 1 8 r,[m1n)= 2.93 RR 5 - 0.7 — ..3 . - = —= . 7 . 2.93 I" ' 25 +1101112) 227 a no 93 “H R0 = 1.96 fil- =1.93;|0.0290 = 28.6 0 I“ = 0.793 mA = 23.6 < RD < 33.7 n _ 1.39 = 0.193 = ‘_ v —‘—‘——. 9“ ‘ V1- 0.026 30 ° WV E423 V-rfl (0.0020(1001 = _ = —— = 3. 3 . '" Icq 0.793 23 m 3.21 V1. 125 _ "° = T}: ’ 0.793 " 158 m a. REIIRLIIrD = 2||0.5||1sa = 0.411133 2 0.4 A _ (1+3 (Rs Rnllru) " ‘ 1-. +(1+fi)(RsiIRL||’°} _ [101 0.4) _M 5 ' 3.23 + (101110.41 3 m b. Rib: ’r+(1+13)(RE”-RLHT0) 8.3 = 3.28 + (101x04; :1- Kg = 43.? kn r.- 3.23 R”: 1+3“ REW‘ 101 9“ :Rg= 32.0 Q. c. Iflmu] - Rs(min)=1.9 kn Rflmin} = 47.5 H? R1Lmax) = 52.5 kn fig-3:24.910!) ._ R,\._(47.3,_n v” ‘ (3. +102)“: ‘ _100 )m — 3'3'5 5 — 0..- — 2.373 __ 1.925 24.9+{1011(1.9) ‘ 210.3 [cg = 0.35% mA [sq-'- H;(mu)= 2.11:9 Rfimu] a: 52.5 m R; (min) = 47.5 k1"! RrH=2¢L9 kfl . _ & -_.. .. War — ( mo)(a] — ..6.5 I“ = mm[ 3 -D.T - 2.625 ] _ (100111.373) 243+ (101}(2.1) In; = 0.70? mA “(maxi = (“19326) = 3.03 m 237 For V351; = 2.5 V' 5 - 2.5 5 — 2.5 I“ = 12—: = "5'3" = 5 “A .rcq = (%)(51 = 4.93 mA =. 15.; = 0.0559 mA 1",.- = -10rn ., = = _t75g_°9§261 = 0.395 m ru=I:—;=:%=15.1kfl Fmvw = 9m[‘Ib’s) = ‘35 I. = (—R—EH’L) x {1 +311. Rsllfo + RI. RIIIR? ) I = __._......._. I " (mun, +2... ’ In - ( Rallfo ) ( R1IIR: ) = --- = ——-— l —------— A’ Is Rsfilru+RL ‘ +5] R1132“;a a. RE=RL=OJSm Ru. = ra + (1 + SHREIIRLIII'H 2 0.395 + {73){0.5||0;3|115.2] = 0396+ (715110.246; =' Rlb = Rallro = 031115.: = 0.434 m _ __ 0.434 _ R1IIR: ‘4’ ’ 1° ‘ (0.434 +0.3)E’6)(R11ER3 + 19.1) R. a: a: -_ _fl_ 1 3'38(R1ijflz+19.1) 0.267510, “.12, +191): 12.11122 RxIIR, = 5.933 m . R: . 1 . . __....._..._ L = —-- 1'- ‘TH (Rt +R:) cc RIIRxIIRz) -c Kiri-(6.9731(5) our data ndin 5 - Vsston} - Fm Rn; +f1 +3112: _ 5 - 0.: _ 1:7,, 0 6° 5.975 +(75uo51 fag: 2.95 = 4.3 -— Vm = V“, =1.“ = gamsfiuam 1. =5 R] = 26.0 kfl R182 __69.l.a_ 2532 R1+R2 I _ 6. TSUS + R2} = 2611: = R; = 9.53 kn b. For R3 = 4R; = 403.5) =9- Rg = 2 kg 5 - 2.5 2 -- Ina = 0.0164 mA = (T5}(0.026] Isa = =1.25 mA -- Icq = 1.23 mA “I 1.23 =1..':9 75 ,0 _ m _ 50.5 m z... = r, + {1 + mtflsllflnllro! = 1.59 +{75}[2|i0.51|60.9] =? Z“, = 31.3 kn Rzllrn = 2|I60.9 = 1.94 kg _ 1.94 RIIIR} ) _ ‘4’ ’ (1.94 + 0.5)(75)(RHIR5 +313 ' 10 R. pi: ) MIR, +315 0.155(31IIR1 + 31.3} = Rallfiz mun, .—. 6.33 1:9 10 = SD.4( 4.3 — v” 5.33 + (763(2) 1 . 1 VTH = 1.70 = Eflfllll321VCC = R—I'16‘33H51 Than 13:; = 0.0154 = 3,3, _ _ (15.5332 R; +32 ‘6'” ‘15.6+R: s.33{13.s + £2} = (15.533, =5- R; =9.6 kn E424 U— .T 1. 159 = i—fio— = 0.93 mA 3 ' 1013 = [co = 0.921 mA Vgcq =10 +10 - IcQRc - 13:3,: vim = 20 - (0.9213(5) - (0.931(103 =5- Vscq = 6.1 V Cha tcr4: Exc 'se olut'ons 5V1- _ (momma) _ _. _ = 2.82 m bl r? — Icq 0.921 I; 0.921 . = = — = 35.42 A V 9'" v:- 0.026 m l :0 = ng. and Vu- = 1’5 v: , 1 ) "= - m If: + "" “ Rah.+’ "5(Rflhn 9 4_E_ 9M5 2 %wflm . I — ii y ( 1 +9 ) 1+9m{REHrn-1 5 Rsnr. "‘ _ (35.42}(10 muzsz kn) ‘ 1+ (35.42)(m knuzm km :9- AE =D.937 u = = Univ-«RC = ngS'RC “E A, .-= ngc = {35.4'Z|(5} => A: = 177.] c, Vac-Q = 6.1V :9 Vgcq = V; — 1"; Vt,- = V; — VECQ = 0.7 - 5-1: -5.II V uc = Vc +1aRc For use = 0.5 =- z-c = +0.3 + 0.2 = —5.4 +10 RC . 0.2+ 5.4 to = =5 Current limited iufipax) = 0.921 =- uoipeak) = (0.9211(5) = 1.51 =5 WM-m-Pm = 1.12 mA E425 _ V55 -" VBE(°“) _ I. Inc - —"'—35 H1 4.5m; ‘ 100+[101H101 a 13¢ = 8.38 uh. [sq = 0.333 IDA fi—LE- = ——-—(1°G]{e'ms) =- r. = 3.10 150 r.— Icq {1.835 Icq 0.333 . = — = — m = . 3 , V gm VT 0.026 =- g 32 2 mil rn=VA=-—-x ==rrn=¢¢ Icq 0.335 Electronic Circuit Analysis and Design, 2“‘1 edition Solutions Manual “Rena wul} =[32.23)(0.909){u.0297}=.-1., = 0.370 ,_ r, _ .10_ _ AIL-1+3 _ #101 —0.0301kn RE ___ I¢_(RE+R. L-‘I' C. R. = = lD|l0.0307 = R. 7:30.79 & =R§=10kn E4215 5= IaRa+Vasfoni+I£Rs I _ 5-0.7 __ 4.3 5 " Ra+t101gRs ' 8944101112; I (100nm; C a He 4- (101m; 5 =IcRc+Vcs +1535 -5 ch = 10 - Ic (R4: + Rs) 100‘ cc anslysis 7-" ‘91-nVIIR-CHRL’ V5 = —V1,— E-R5=_Vl(1+ T1! rw fl I . V =- V °' (fwd-Ra) " Vo _ fur Va 3 ~=fi=..+a.(fic“m For Lg =1 134%. n: #=W=16kfl c (mo 1) I, = O = A 3 2.6 + Ra 33 41°23“) — 2.5 :- RE =24 m [C = l (1001(4.3) = 2.4 +(101mE (months) _“ = —-l——-— = R5 (m) = R; 4.23 kn E42? 8.. RT}; = TONE = 5.326 kn 5 , I. VTH = (a +TO)(ID] - D — -4.21\ 3—4.21410 _ 0.090 5.325 + [125){021 " 30.725 :13. = 2.93 ,uA. rm. = 0.355 m.-\ Ia: = S—Vcl __IC +(Vc: -U.T}—{-5II 5 1 (1+ ans 6 — ch 1:; 4.3 ' 2 0.355 ' —— + . 5 *(125)(1.5) (125mm 3 v v. 1- 0.366 —- 4 -c—' - " m = 5 T (116}{15I 0.511: = Vc1(0.2053) = V“ = 2.9:? V 15; = 0.359 mA r5. = [0.3691(013 — 5 =9 v5. = ——-L9".’6 v = Vca: = Vc: — V51 = 2.97? — [—4.925] =9 V95: =7.90 V 1 _[Vb1-0.7)-(—5)_5+'3.93-U.T BO? '— - — = 4.85 m.-"L 13 Ice: == (1+3)I5m => Ice: = 4.81mA V53 .1 Vc: - 0.7 = 2.93 — 0.7 = 2.23 Vc-gq; = 5 — V5: 2 5 - 2.23 = chq: = 3.32 V _ fivr _ glzsnoms) = 8 as m r“ _ - fem 13.355 3V 125 0.026 m = r = w =o.575 kn Ira: 4.81 qui 0.356 9...: — VT — 0.026 _ 14 1 InA/V [:21 = 4.31 = 9m: 3 VT “0.026 185 mAfV Te t You dcrstandin Run-1 = For: + {1 +3135: = 8.38 + (126}(0.2} '= 341.08 kn Rm: = hr: +(1+F3HRE:EIRL) = 0.970 +(126}(1.5[|10]=165 kn Va = (l + 3)102(R£2"RL] RC1 In: " (361+ Rub?)(_gnflb,fl) V5 vfl'l = I ‘ r'rl . r’ RC1 )(‘Emlr'V' v = 1 3 11‘- R A ( + H 5'1” L’kRCI + R”: RI“ A _ 0 u - V‘s _ _(126)(125)U 5|11"3( a l ) _ ' 5 +105 34135 = _{126}(125)(1-3°J(‘5—)( 1 170 34.00 A" :1 -}7.|- c. Rl = RINK-AIR.“ = (5.53)|I[34.1) =» R. = ILTG kfl an = 0—0:?) H = (“m-613:5) u = 004501110 => B; = ‘31 n 54.28 00 a. Arc-q: = (1—0100 111A} => Icq: = 0.990 mA Jam = I“: = -1— = 15.3. = 00099 m-\ 1 + a 101 15:31 0.0099 I3“ = 1+0 101 a 19m = 0.000090 1113.. Ice: = 0.0099 mA V31. = deems = 40000093010} = —0.00099 v -..-. 0 v5, = -0.7 V. v5; = —l.4 v I: = Ice: + fem = 0.990 + 0.0093 I; 51mA==>Vo=5—(1}(4)=1V Vcsq: = 1 - ("1.4) = 2.4 =5 Vcsq: =14 V. Icq: = 0.990 mm V559, = 1 — (—0-7) = 1.7 = Vc-am =1: v. 1cm = 0.0099 mA Cha :0: 4: xercise Solution V I b. r, = B—T- mdgm = if;- Ico T For Q], L (100140.020) _ = —---———‘ =‘.> . = -53 kg m 0.0098 F ‘ 0.0090 _ n gm} 7: 0.026 gm] — 0 3H min/X For Q; : (10000020) 0 = "'—"'— :r = ..33 k0 1'" 0.990 S” r 2 0.99 , = —— =9 = 33.]. 111A \- 9” 0.025 9"” K 1’01 = 1'02 = 35 «F: VI! = "(le V01 +§m2VIRJRC V5 2 +1012 '1}; . V02 = (H +9mzvn;}1'=2 = (1+ 3)L‘;Lr_._.3 rm ,1 r"; Than . _ (1—1-3 -. Pa = —{gm=Lrl Vall] .RC 1'_-] J'Il'l ('32 .1.; 1+;1+3)k— r,” v.- vfi- - r 14-(1+J)(—'-=-3 7—1; V _ 7:2 0—- 9ml+§m?{]+3) F'- RC -.! +_V='_ :+(1+33(:£) ."1 [9m +9m2i1+3)(rfl)]fic ,1” - __—__4__. ,. Electronic Circuit Analysis and Dcsigg, 2“d edition Solutions Manual 0.377 + (38.08|(101)(2'626) J {4} ., - A” = _ -6a.J 1+(101)(2'526\] 265.3; 153.8 . L. = — _ , = —T . a 1.999: =' —-————{ 6 9 d. R. = r... +l11-t331'n2 R. = 255.3 + {NH H1526] =? R. = 531 54.29 V5. - (—10) (V51 —0 7) — (—10) l. I = —-—-—-——+_______ 5‘ 20 [1 +5)t10) - 13133 - VES(°“J = V51 50 (1+'3)IBI =10~J§|R5—0.7 20 10——£J_T—151R3—0.T (IOIHIUJ 20 l 20 “OUIBI +Ian[fi) via: -—-—(101)[10) _ 0-2 846 20 + (101]{10} [102:]151 = 0.465 + 0.00851 [5: = {1.00464 mA =- V3; = —{].09251 =§ V31 = -‘0.Tg3 V =9 v3; = —l.493 V [Cl EOAE‘I mA. 1;: = 0.459 mI'L {1:10—11:93 20 9'13: =15: ‘11 Ia: = 0.00855 ruA =- [c1 = 0.865 mA 10 - 1.493 or IS: = T = 9.8515 fez =n.842 mA = 0.46035 mA Ic = Ic; + In = 0.464 + 0.842 = 1.306 [113. Va =10 - (1.3051(2) = 7.39 v Vase: = 7.39 — (—1.493) =9 V6502 = 5-55 V. Ice: = 0.542 m.-\ Vtsq: = 7.39 q (-0.793) =' Vcsq: = 3.18 V. {em = 0.464 mA b_ ICQ 1-,,3— gm=_.._ [60' VI FOI' Q1 2 100 0.0251 _ hr; = =5 1'“ =O.GD kn __0.46-4=_ _1_fl AV 9"” " 0.025 9'“ ’ " m I (100){0.025) =5 1'": = 3.09 kn I"; = 0.342 0.842 m = —— =3 . .‘ \-' 9 2 0-026 =>gm2 24m—\,’ 1'01 = "02 = DC c. Ra = l's-z +[1+ = 3.09 + (101K113) = 1013.1 Ru = rm +[1+ HHRmIIRn] = 5.60+ {101)[20||1013.1] CL 3.; = 1.936 M0 Note that Emlvlrl = {SIM and Pm: V111 = 13le 50 V = —{5Ibl + 3102134: Vs RE: ) M £11 an (R5! + RI! + ‘5)!“- 50 Va = "[Ib] +102“?ch = _{I.,. +( R“ )[1 + 3):..}Mc R5; + Ra .‘ . = = —[1 + (FER-flu +3)] 3R? = _[1+ (fiymn A” = 41.395 54.30 a. dc analysis 3 =100, Vggbon) = 0.7 V, VA = :10 Want: log: = 0.5 mA, chl .1 Van = It V E: +£24.33 =100kn Reflecting base currents: Test Your ndcrstandin I? = -—-- = 0. I1 .190 12 IRA V51 = IcmRs = (0.5)(o.5) = 0.25 v V151 = Vcsm + V51 = 4 + (1.25 _—. 1.25 50 Va: = Va: + Vasq: = 4.25 + 4 = 5.25 \r' Vac — Va 12 — 3.25 ' R =—-———-=——_ =- c Icq 0.5 => R; 1.5 kfl 1:3. 2 v5; + 0.7 = 0.25 + 0.7 = 0.95 v R3 R5 V _. ' _ __ '1 E” ‘ (R1 +RI+RJ){12) = 0'9“ ‘ 1011‘” => R: = 7.92 kn V32 = ch+ 0.7 = 4.25 + 0.7 a 4.95 . R2 + R: j V = _ a? (31+R:+R3 (12} R: + 7.92 .9‘ = —— =4 :1 mo (12] _ (4.951(100) R” ‘ 12 R1=100 — R2 - R3 = 100 — 33.3 — 7.92 - T.92 => R; = 33.3- kn =1 R; = 58.5 110 b. For both Q1 and Q: r _ fl: __ (100](o.o25) 1' — Icq - 0.5 =- r,“ = hr; = 5.2 H? Ice 0.5 9'“ = W = 0.026 = 51m = 9m: = 19.23 mA/V I'm =r92 =30 yo = —gm2 ng(RcIiRL) V 2 . 'L + 9m: VI! = 9m! Vin rs? . 1+ 3 v.,(__) 71-2 [+13 So Au=i=—QMZ[RC”RL}'Q Vs "" 1 3 3 =— m R, ‘ g 2( 1:11:41”) 100 A, = —(—— {19.23)(7.5||2) = A, = —30.1 101, Cha tor 4: xcrcise Solution [34.3 I a. dc analysis 1'3 = 80. V'ngon) = 0.7. VA : x. 2.32 - 0.T [.62 24.2 + {51mm = 54.1 ng = 0.0250 mA. ch = 2.00 mA Isa: Power dissipated in Rc = I};ch =E1.013{2) :9 PC = 8.0 m“. Powcr dissipalnd in R! = D. P; = o VCE = Vcc - 11: [RC +321 =12 — 2[2 + [11.5)] v” = 5.99 V Pr = IBVBE + 11: 1:5: = {0-0359H0-7) + (ENG-99} => P: = 1-1.0 mW m avr _ (BU)(U.O26] _ b. r.P _ ——ICQ _. 2.0 2.» n. _ 1.04 It!) For 12,- 18c05ul mV From the text. power dissipation in the transistor Pr = VcaoIcq - (V—JEYMCIIRL) = [6.99)(2 x10") 1 2 _ (ID—4:321?) {2 x1031|2 x10‘) 1’:- = [14: - 0.96) mu: => P: = 13.0 mW From new SPCA = ;§-(RCHRL‘JVP cos wi Pawn: dissipated in R; __ We!2 ._ 1 2 .1, [3. PL " RI- lrms_ rfl{Rc”RL] x RL x 2 an ‘ 3 1 0.018 I = [1.n4{1‘°)] " 2x103 x( ‘2 j =- P‘ = 0.479 mW Rc = 2 kn also so P;- = 3.0 + 0.479 =- P; = 8.43 mW Electronic Circuit Analxgis gel Design, 2'” edition Solutions Manual 84.32 {3 = '00. = 0.7 V. VA = o: ,_ R“, = RIIIR; = 10||53.a = 3.43 kn VT" = (E???) (5) g (10 $053.51)”) V13 = 0.7837 0.7337 - 0.7 IBQ — -— CLUUQQS mA Icq = 0.993 mA Vcsq = Vcc - Ichc 3.5 = 5 — (0.993JRC =? R = 2.52 kn b. Fewer in RC = PR = Ii-Rc = {o.993)’(2.52} =- P5 = 2.48 mW Powu in Q ; Pg 2 Ichcsq = (0.993)(2.5) =5 Pq = 2.48 mW I c. if = 0-993 cos u! 1 ac power: ; x (0.993}2 x 1% =1.24 mW in Ra- 1.34 --———— = 0.2§ 2.48 + 2.48 El ct nic ircuit Anal ' and Desi 2'“1 edition Ointions Manual Chapter 4 Problem Solutions 4-5 £351 ___ (120)(0.025) A. r, = 5.4 = Ice Icq 4'1 =5 Icq = 0.575 111?. _ r60 _ 2 _ 1 1 . 3. y.-».--- W VCEQ=§VCC=§(3)=2.5V r” = 1:11:- = {130121-026} z, r” =13“ m ch = Vcc — 1553c = 2.5 = 5.0 - (0.57MB;- ‘m =5 3.: = 5.55 59 VA 150 __.._—...— r°=E=T=Df§=75kn n 5 155 = = “‘13,? = 0.00432 mA b. ym='—=-gm=19.2mA/v ~- 0-025 _ _" — V35- = IBQRB + iaaion] 130 0.026 r... = =:' r, = 9.36 kn = [0.00433}(25) -!- 0.70 r; = 3i? =9 r9 = 350 = J'L'—“-'V = 0-821 .3 _ 303- __ (120]{0026} _‘ 42 I I b Tr — —-ICQ — #0578 - 4D kfl gm=%$200=oggeafcq=5.2mlk T3 ' =fl=m =222mAv 1.“ = E3; = W :5. r, = 0.525 m 9'" 0-; 0.025 f Icq 5.2 """'—“"_ VA 100 - V,‘ 200 Tn=——=—-—-,8=k3kfl ra=-I-—=fi=w9=335m ca 0-5' cc: - ' 1.! 4 3 yo = ~—gm(rg|§Rc )VI. VI = V5 ' 1'" \ BUDHRC) _ .7 All = _ m R = -—"""‘_— (a) Im=-2-L=O.0052mA g (Tu-FREIIHDII C} rn+R3 250 __ {120][17311433] _ (120)(4.22) [c =(120)(0.0052)=0.624m -- — -W - --—§r giflifi g_=24m,uv =’i~_=;15_-7_ 0.025 —-—-- _ (120)(0.025) _ m 4'6 . 1 _ ' r' v 0.624 a 2.: 5- VECQ = Ever: = a ‘v 1; =06 Vgcq =10 — Ichc =:' 3 = 10 — (0.5)Rc r s = (b) A,=-g.Rc r =_(24)(4)[ J: = Mk3. r'+R’ “250 I —£53—-E'-5-—0005 A.=—lsa 3” B ' we ' _h__ v0 Vasbn) + Iaqfia = Vaa = {0.70) + (0.005)(50] ('3 v‘ ' A‘ ' -133 a v5: = 0.95 V vs =—D.4265inlOOt V b 5.. — £g_gfig=..=mmm 4.4 T ' IC = fiV-r = (100)(0.025) = 5 2 m gm = V? I S ICQ S IDA 1‘" 1:0 :5 f'_._‘_—— 1.08 1.32 _ .4 co — m < —— . < ,,. . V = -—- = —- = = 00-2559 -3025:m_-9_§_5£§.m—M— r° co {Ls m—m r: - iVT: r-(mu) = W 2 2.59 kfl an ‘ c A.=--&-=-—u—°w°-1=>Au=—15.1 r,r + Rs 5.2 + 50 r.(min) = W .—. 1.53 m Elec nic Circuit Anal sis and Desi 2"" edition lutions Manual 4-7 0 35 R R1||r I. I: = 0.35 mA. I: = —‘— = 0.00347 mA c =_ ' ‘ R R 101 ( M‘ g'[R, Riflr, +R, C“ L) V; = 43123 = —(0.00a47}(10) R1|IR2IIG = fill-5.1.57 =0‘598k-Q =0 VE = "0.0307 v Av =_(los)[ 0.693 Jung): V; a V5 — V33(Dn} =5 V5 = --0.T35 V . A” =—45.3 b_ Vc = Vch + V5 = 3.5 — 0.735 = 2.77 V 4-9 I. Ice 2 I59 C ' 1 + a E 101 ' ‘ ‘3 ‘3‘ W _vc 5—2.7? = ID-Icn(1.2+0.2} = — — R = 0.43 m R": 1c 0.347 a 4-— Icq = 3.57 mA 3.5? (chk =-g mRJq R flr) I” - 15° _ 0.0233 M " Rallrfiks C ' Rallfia = Rn. = (0.1}(1 +5135 034? 100 =[0.1){101)(0.2)= 3.02 m =-_—- =13.3mA/V . = = g“ 0.025 r“ 0.347 238 m _(100)(D.026) _ 1 ’x-W-J-Wm L}H=R—1-Rry-[IU)—5 Rallr. = 14117.49 =4-28 *9 Vm = 130 Rm 4:10.500) + (1 +0)ng&5 — a 4.23 1 . _ - =_ — —(3.02)[10]- 3 ——[0.0233}(3.02) + 0.! A“ ( {4.2mm} 3" )=’ l A.=-31.7 +(1s:](0.0238}(0.2)-5 1 _ _ Railr. ) —(30.2) =1.49 =0 1;:l = 20.3 m a. A. _ g..( Ran,” + RS (RcHrul R. Rallr.r = 10||7.49 = 0.20 m 4.28 A... = —(13.3) [6.43288) =- Au = -'I'4.9 4.3 RTE = Rzllflz = 6|]1.5 = 1.2 kfl 1.5 1'- )v _ (1.5+6 _ VT;- — V3; on! _ 1.0 - 0.7 ‘ Rm + (1+fi}fls " 1.2+[101)(0.1) = 0.0155 mA. fag = 2.80 31A. Isq = 2.31 Vase = V+ - Ichc - IEQRE ){5) = 1.0 V -_- 5 — (2.0)(1) — (2.01)(0.1} =§ Vega = 1.92 V r = (100)§0.0202 a b. . “a r; = 1.07 m 2.80 9m=magm=lflflmA£V. ro=ou 20.332 _ 20.3 + R: — 3.02 =0 R; — 3.50 kfl b. r, = = L09 kg 3.54 3.57 = — = T V 3'" 0.020 13 “A, a. 0 v; 9 A _ —;SR<: ___ _ {150}[1.2) " ' r.+{1+5)fig 1.09+(1511[0.2p => .4... = -5.75 __—____ Chapter 4: Problem Solutions 4.IO R: 50 For Rc(min] = 1.9 kn and 35(mu)21.05 kn. = = = V " V” (:21 + 5:) °‘ (50 + “Jun 1° A“ = __....___._.2 1;“0351-2305 = —1.76 12_o7_m Sol.755].4.,\ 52.14 1 = --——‘—- = 0.0119 111A 5" 3.33 + (101m) 4_“ [ca = [.19 mA. fag = 1.20 [ILA l+fi vsco = 12 - (1.20m) - (1.193(2) (a) v“=[ £3 Jim)?! WWHCQR‘ Vgcq ‘—"— 3.42 V —'-_‘— 12=(%]rm(1)+5+1m(2) so that Ice = 1.99 mA 1.: 1,Q=%=a.0199m T R", = (o.1)(1+ mas =(o.1)(101)(1)=1o.1m 1 1 1/: RI =—-R-V =——l. m [mam RI m m Rl< 01x12) LP, Vcc=(l+fl]ImRE+Vfl(orz)+IEQRTH+Vm ' 12 =(101)(0.0199)(1)+(17 +(0.01'99)(10.1)+mi 521.2. 1;. 11;: RI which yields R,=I3.3!¢fl and R2=42kfl b. _ flag _ 4100)“) (b) A—m‘m= m = 4.95 Icq = 0.25 mA, I59 = 0.2525 IDA 13¢ = 0.0025 mA Iqus + Vaflon] + Iadfls + Rs] - 5 = 0 _ [lemmings] _ (o.oozs)(sn) + 0.. + (o.25253(o.1 +115) = a r"- -— — 2.13 RE =15.4 1:!) Vi) = gnVn-RC V L V" '; —..- —(o.0025)(50} — c..- = 41.325 v 5 = — rt "' ‘_' "IV? . ('1' +5 )RE Vc=Vcsq+Vs=3—D.825=2.175V =4,” Humane; Rc=1flfliaw r, 9.25 A = —_aR¢ = -(lon}c2] A _ 43¢ r,+(1+B)R: 2.15+(1013(1) "" r,+[l+3)R_= 9 Av = 434 r _ (momma) _ I“ m " ‘ 0.25 — ' _ I ‘ 41003113) _, C. Appromuon: Assume. r. docs nor vary significantly. A" = 10.4 + {WIND-H 5 w R; =2kfli3%= 2.11:9 or 1.91:9 12' =RB"{r'+U+5]R§] R; =1kn :e 5% = 1.05 m or 0495 m = 5°“[”" + “mm” For fig-(max) = 2.1 m m 35mm) 1" = “"2” =’ w = -(100 2.1a _ 1H 2.13 +(m1)(o.95) ‘ "" I! Elccgggnic Qircuit Analgsig and Designl 2“‘1 edition Solutigns Manual 4.13 t. 9 = IEQRS + ngfon} + {BQRS 0.7- 73.; = 0.70 mA. 73.; = 711—" = 0.00920 mA [cg = 0.74]. mA 9 = [0.75)Rs + 0.7 + (0.00925)(2) => RE = 11.0 m b. V; = 9 — (0.753(11) = 0.75 v V;- = V; — Vscq = 0.75 - 7 = -G.25 V _ Vc — [—9) _ 9 —5.25 RC 1m " 0.741 =5 RC = 3.71M) Tsr c. A. = -y... (H + Rs)(RCEIRI-H'°) r, = GEM = 131 kg 0.741 an 713 — —- 103 A. - '80 (3 71||10|1103) " " 201+ 2 ‘ A... = —43.9 d. R.=R5+r,=2+2.51=&-R.=4.31k9 4.14 (a) Vcc E [CARE +RE)+VCm 9 = 7642.2 + 2)+ 3.75 So the: 7m =125mA 1.25 =—-——=48.l MIA/V (b) g" 0026 r = (120)(0.026) 1.25 100 “Was—sow =250kfl Assume. circuit is to be designed to be bias stable. Rm = MR= = (0.1)(1 «7- 33m = (0.1)(121)(2) = 24.2 n 4.15 4.l6 R, = -4s.1(24.2u25)(sq[22§1] = -48.1(2.27](0.681) 01' R_=1;&=-745m=—745V7m - . 0 a, 150 = 0.00 mA. Isa = D7:- = 0-0121 mA I“. = 0.700 mA . 0.3 _ $75 = 1.9ng =7 R5 — [L012] 2? RB — 24.3 kfl 'c — ("5) 5 — 3 q - _ = = _. 4 kn R: [CO 0 738 => R; a 0.788 b. 9... — — 30.3 mA/V __ (65){D.026) _ Tr - — 75 To = = 95.3 kn . Rcllfa = mVu, V: = — 5 ‘° (Hanna +00)” “ ' Rcllro c: a 19.: _ m[.___..__ I “'5 g RCIITO +37. = 430.3% 3.541953 ) 2.a4||95.?. +4 G; = —11.6 mAg’V 4 " D" = 0.00047 ’5‘? = "‘s «7000(5) Jag = 0.547 InA a. aogh,.5120, 1057....52005 2.585005%- b?! high Sm nu: Rn: = RnIIRs = 5H1 = 0.833 H2 4.17 High-gain 5 v (5+1) ‘ = 0.033 + 3.7 Law-gain (5:I)Vs = 0.033 + 2.45 For 1, = 0.1030113 Ia- -—-- 0.2533175 1 ho: _lfl= h“ =1001l2 = 1.96 For RC1IR‘=Eol1—01‘l|* hu=2o=afifi [4|f4=50||2=1.9-2m |.41.,|rm = (120)(fl.1338]{1.96) = 43.2 041.4“in = (50)(0.2530)(1.92) = 39.0 39.0 5 H.111: 43.2 b. 11.: 1251111.. = 5|:3.7 = 2.13 m or R. = 5112.45 =1.54 m 1.64 < R. < 2.13 1:51 1 1 z _. g = ....._ l 11,. c 0.010 1 orRo — 'l — Sflllé = 3.70 m 3.70 < Ru < 3.85 kn 14 = 100m = 3.35 119 Assume an npn lmnsistar with 15 = 100 and VA=oo_ Lg: 05 I‘M-fi-50 Biasat Im=1mfi1 andch R5 =15! For a bias stab]: circuit Rm = (0.1)(1+ 03R; = (0.1}(101)(1) = 10.1 £59 I 1 101 V” =E- Rm -Vcc =E(10.1)(10)=E- Imz-l-olazomm Cha ter 4: Problem Solutions V” = 1391?,” +V,£[an)+(l +fljinE %“ ; (0.01)(10.1)+0.7+(10:](0.01){1) l which yields I"?I =55.8kfl and R1 = 123 kfl New z (100)(0.026) 1 gm L: 38.46mA/V = 0.026 V. = ~gnV.Rc where V]: AW .Vlz .v‘ R.HR2|er+R5 10110.6“ ' 01' v, = 0.67411, Then AV 2 = 40.674)!“ RC = —[_0.674)(38.46}Rc = ~50 which yiclds Rc = [.93 M1 With this Re. the dc bias is OK. 4.13 6 — 0.7 _ =—.—=0.0159 A 3' I” 10+(101}[3J m Ice =1.'59 mA. [59 = 1.711113 chq = {15 + 5) — {1.50}(0.a) — (1.713(3) r =2.6.{fl fl' 3:3,, = 5.33 v 1.59 = __ z - V b' 9"“ 0.020 g g———-——--"‘ an m“ ,fl=flw=r,=1.a4m.ro=x _ ‘43131211311 321130 (c) A* “ r, +(1 +1312. 'R,||R,, + 5:, Rib = r, +(l+fi)R£ =154+(101){3)= 3045151 R,an =10||3045 = 9.68 112 Then _ —{100)(6.s|163)_( 9‘53 ] "154+(101)(3) 9.68+05 2' A, = —1.06 . _ Rc , Ia — (RC+RL)1-5M) .‘b 2., (Am—\ i... Ra+rr+11+ 3135) ' __ RC 1.; A. ~— (31(RC+RL)(RE+H+(I +5185) _—_ —(100}(E§%) => A. =-. —1.59 Electronic Circuit nal sis and Desi 2"d edition (d:- Rh = R5 + MR, 2 05+10||3045 = 10.2 m -fi(RC“R-L) RIIIR-l r, +(l+fl)R£ 0243* + R, _-(I00)(63163) ' 154+(101)(3) Av =—|.01 A. = same as (c) => A. = -l.59 0:) A0: 9.68 :3» 9.68 + l 4.19 Let Vx=25v P=(I‘+Ic)vcc=>0.12=(!,,+rc)(25)=¢. IR+IC=4SM1 Lat [R =am. Iva-=40!“ £0-25 R1+R25 =3125kfl Solutions Manual 40 [no =TO—O=O.4}.M Let R5 = 2 1:9 . For a bias stable circuit Rm = (0.1)(1 +130}?E = (0.1)(101)(2) = 20.2 m 1 Vru =1? Rm 'Vcc = [BQRTH +Vu(0")+(l+fl)laaR£ 1 . 5202125) = (0.0004)(202) + 0.7 +(101)(U.0004)(2) which yields R1 = 64 £9 and R2 = 29.5 Id! 1-; = = 65 K1 Neglect Rs 0.04 _ fl. = 'fi-Rc [a - V: _ r: +(1+IB)RE —100R —10 A: R =26.7kfl With this RC, dc biasing is OK. 4.21 Need a voltage gain of g = 20. Assume a Sign inversion from a. common-emitter is not important. Use the configuration for Figure 4.28. Need an input resistance of fixlfl'3 R" _ 0.2mms R, =Rm||Rr bet Rm=som. Ru, =50m Rib = r: +{I 4153))?E a [l+fi)R5 = 25003 = 25 1:!) R 50 For =100. R =—“-'—=-—=0.495m ’3 "2 1+1; 101 Let RE=05m,V,_.C=10V, ICQ=02mA Then rm = $37)- = 0.002 m Vm = {new +V,E(on) +(l+fi)ImRE EL Rm 4/“ = jig—(sown) = (0.002)(50)+ 0.7 +(101)(0.002)(05) which yields R1 = 555 kg and R1 255%) Now I IA“: 80 —_6R,_. _ (100)(0.026) r‘+(l+ R; r" 0.2" =l3kfl _ ‘(100]Rc _ 13+(I01)(05) =5 Rc=123kfl [Notez 1'ch = (0.2)(12.7) = 254 V . So dc biasing is OK] Cha ter 4: Problem Solu ions 4.22 -HRC n+(1 +B)Rs First Ippmximttlm: 5:30. Ay= RC ., z—=10=>R,=1UR (A) RE c 5 Se: RC = 1213.5 V55- : Vcc — fc-(Rc + RE]=10 - Ic(13RE} For Vac = é-Vcc == 5 5 =10 — Ic[l3Rs) For Ic = 0.7 mA I; = 0.709, [a = 0.00375 mfi. =§ R; = 0.55 kn —- RC = 6.6 kn Bins 3:11:15:- R.1|R: = R“; = {n.l.)(l + am; = (0.1)(31)(o.55) z 1.46 kn m =‘(o‘709)(o.35) + 0.7 + Ec.ocs75)(4.46] 1 + E—(Memn) 8.87 = l—(ms) =. R = 5.03 m R; -'—— 5.033! = _ = 9. 5-03 + R: 4 4'5 =§ R: 3 1 kfl 1o 10 R. + R: 5.03 + 39.4 m a", + 0,325 ; 0.925 mA from Vac sourcc. New r, = = 2.97 kn _ mum) ’ 2.97 + (mums) IA»! = 11.1 4.23 -S'V' H = 120 Let Ice = 0.35 mm In = 0.35; mm. In = 0.00292 mA Miler—21:9. Foergq=4V=> 10 = 4 + (0.35)R¢ + (n.353)(2) _ {120}(0.026) RC = 15.11“), 1'" - T ='- 3.91 _ -5(RciIRL) _ _(120){15»11l10) — r.r _ 3.91 A... = -81.0 A» For bias subl: circuit mun: = RTH =(n.1)(1 + ME = (o.1)(121](2) :24: k9 v“, =( R“ )(10) -5= mewlnys R1 + R: R1 VTH' = 13911373 + Vazion) + (l + fillsofis - 5 femum) — 5 = (0302921942) + 0.? 1 + {121){o.oozgz)(2} - 5 Til—c242) = 1.477. R] = 154 m I 15”” = 24.2 =:. R2 = 23.4 kn 164 + R: 10 — = o. 2, . .05 = ‘ 164+?“ 05 035+0 2 gigzmfi 4.24 From Prob. 4-10: . R: so 9 = 12 = = V TH (R1 +R2){ J (30+10)(12) 10 12—D.T- 10 I” = 5.33 + (101m; [:9 = 1.19 mA. {:9 =1.20 mA Vac: = 12 -(1.19)(2)—[120)(1)= 3.42 V = 00119 mA Cha ter 4: Problem Solutions -HRC r.- + [1 + MR; First upprmdmntion: 6:30. Au= RC I,.m—=[Il R =IOR (A) RE l=> c 5 Sal RC: =1235 V5; 2 Vac - Ic[Rc+ £3} = 10 -Ic£13flg} 1 Fur Vac = EVCC = 5 5 = 10 - Icfllfisj F0: I: = 0.7 mA I; = 0.709, 13 = 0.00375 mA :5 Rs =0.55 kfl -— Rc = 5.6 109 Bias stable. => mull: == Rn: = (0.1)[1 1-H)Rs = (0.1){01}(0.55) = 4.06 m 10 =(0.709)10.55) + 0.? + (0.00075)(4.40) 1 . + EHABKIO) 0.07 = “14445): a, = 3.03 m 34 5.033; = . 6 R = 3 _ I . 3.03 + R! 4 i = g 4 10 10 ———- = -—— = 0.225 81 + R: 5.00 + 39.4 "m 0.7 + 0.225 § 0.925 31A from vac swan. (80)(6.5) 2.97 +{81}{0.55)= “'1 lAv] = 4.23 .3: 120 Let Icq = 3.35 111A, [5,; = 0.353 mA Isa = 0.00292 mA Letflz =1“). Foergg =4V=> :0 = 4 +(0.35)Rc + (0.1533(2) = {120)(0.026] RC = 13.1 it“, r,r 0.35 = 331“; A _ —fi{RcuRL} ___ _(120}(15.l|103 " _ r. 0.91 Au = -BI.D For hi1: stable circuit: RINK: = first = (0-1)“ +B)Rz = [0.1)m1)m = 24.2 m R: 1 lo — =—- -1C' —5 mun) } 5 R; 12"” ) VTH = IaoRrH + Vastuu) + (1 + flflsofis - 5 VTH = -f:—(24.2)(10}— s = {0.00292)(24.2) + 0.7 I + (1211(0.00292)(2) — 5 i(2¢2}=1.479, x. =104m Rm 761%: = 24.2 =9» W Its-1:02.81 = 9-052. 0.35 +0.05: = 0402 m 4.24 From Prob. 4.10 RT”. = Rdle = IDHSO = 3.33 kn . R: 50 . V = = = 7” (a.+0.)“2’ (50-010)“?j 10" I -12—°‘7-m —00119 A "— 8.33+(101)(1} ‘ ' m Icq = 1.19 m. In; =1.20 mA Vgcq =12 — (1.19)[2)-(120)[1) z 0.42 v Elcctronic Circuit Ana! ' and Desi n 2'“1 edition olutions Manual 4.27 {sq = 0.30 mA, IQ; = 0.792 mm fag = 0.003 1113 V; = 0.? + (D.OOB}(10] = 0.78 V Va = Ichc — 5 = (0.792)(4) — 5 = —1.03 V Vscq = 0.73 — (—1.33) = 2.61V Load line: Assume Vs remains consrant a: 2 0.78 V (.17 I Ru. 1: 1:. For 1 S 92c S 11 Avg: =11 — 3.42 = 2.53 =- Output voltagc swing = 5.16 \r' (Pubic-peak} 4.25 5 - 0.7 = = .00 I“ 50 +[101)(0.1 + 12.9) a 315 “A [co = 0.315 M. Isq = 0.319 13!. 1-3! $.79 chq = (5 + 5) - (0315)“) — {0.319){13) Vcsq = 3.96 V Aic = -1 ' "ea 2 m Coilector cm'rcnt swing = 0.752 - 0.03 = 0.712 EUR lfivgzl = {0.712)(2) = L42 V Output swing determiqu by current. 6.115" My: output swing = .84 V gut-mica: . . . 2.84 Swmg in :n cut-rent = T— .-. 0.71 mA peak-m-paak 4.28 6 - 0.7 . 1 = — = _ m: = -376.“ Inc 10+ (mm) o 0169 m For Me = 0.015 - 0.05 = 0.265 160 =1.69 m. In = 1.11;” a 1.6ch] = 1.62 Vcaq = (15 + 5) - {1-5953} - (1-71)(3) visa-(min) = 3.96 — 1.62 = 2.34 V554. = 5.38 V Output signal swing dam-mined by emu-rant: Mu output swing = 3.24 V Beak-m 4.26 100 For Rc=6kfl. veg-[m 035x6)=2.92v 0.35 v5 = 4,93, -V,£(on)= -m(10)~0.7 = —0.735V Thcn Va -_- vc -v5 = 2.92—(—0.735) = 3.66 V :38 «11- Ava: = Aic 41¢ =5 (45- 3.66) = 0146) so that Aic = 0.14 m (a) Total Avg = 2(45 - 3.66) = 1.68 V peak-to-pcak (b) Total Aic = 2(0.14) = 02.8 ml pcak-to-pcak Chagtcr 4: Problem Solutions 1 A1,: = —6—:Ay« For alum-111m symmen'iul swing For'vtdmin) = 1 V. AI!“ = 5.33 - l = 4.33 V Air: = Icq - 0.25 . .3 =1» |A1c1= %:l§- = 0.634 11111 Aves = cho q 0-5 . .‘ . . 1 Output smug unused by voltage. and [L‘sch - “45 m - Iflvcai Avg. = Mu. swing in output volug: = 1175 v pcak-lo-pcak Icq —- 0.25 = M 1m 0.04:3 A19 = 311:“; = m, = 0.342 111A Vcsq = 5 — [coil-11 (pukmfim) 0.545(Icq — 0.25) = [5 _ 1:9(1111 — 0.5 429 [0.545+1.1)Icq =5 —o.5+0.135 fcq = 2.32 nut, Ian 2 0.015? mA CAC- ‘ecuef “we Rn; = R1llflz = [G-1](1+H)R£ =(o.1)(131)(o.11=1.31kn . 1 Viv-H = 11—: - RTH - V+ = IaqR-rg + V35[Dn) +[1+ flflaq RE Rinsms} = {0.0157){1.31) + 0.1 1 ‘45 + (131110.01571[n.1) Ga 1 —-(9.05)=1.01=' R: = 8.96 kn R1 —‘-'“"‘—— 3.9632 Avc5(mu) = VCEQ " 1' Nc(max) = [C2 3.96 + R: Aves = 151.140.6875) . $0 = 1.31 :15 R2 = 2.27 kn 4.31 v - 1 = I (0.5375) and cm :0 1' =0.647 mA. V =10- 0.647 9 =4IIBV Vcan = Vcc - 1690?: + RE) =1. VCEQ = 9 _ {544.2} m cm ( )( J Then NC = [CE = 0.647 M 9 ' Ica(4-2)' 1 = I «106875) So Ava. = 11.14414) = (0.647)(2) = 1.294 V so Ian = 1‘64 “A and Van = 2'1 1V Voltage swing is well within the voltage 1 specification. Then Vm = fl' Rm 'Vcc “1“ Rm =(0-1)(l+ AB)RE Ava = 20.294) = 259 v peak-Eo-pcak R", = (o.1)(151)(2) = 30.2 m V =1 R111+V (on)+ 1+ 1' R m I: BE ( m m E 3» RTE = RxllR: = 101|ID = 5 kfl IM=E=Q0109M 4.32 R: 10 VTK=( )[15)—-9=( )(153‘-9=0 -;—(30.2)(9] =(0.0109)(3o.2)+o.7+(151)(0.0109)(2} 0 310 f 31‘ 9) 1" +1“ I _ — .1 — — 3 which yields R,=52.9m and 31:53.119 “‘3' 5+1181)(o.51 0'0369‘” Ian = 15.5 mA. ng 215.7 m 4.30 ch = 1a — (15.7)(n.5) =1 ch = 10.11.! dc load 11m: Electronic Circuit Anal ' csi 2""cdition Solutions Manual 1:. r. = “so 0'0” = 0.30 kn 15.6 _ (1+fiXREHR.) I R. MR... 4. A. _ '1 “3+ fiXREIIRL) RI RIM. + R: E... = 1.4.9 +(121)(2uz) => R... = 122 kg Rf, = r +(1 +13)(R5|[RL) = 0.30+(181)(05]|0.3) *— .I' A = 0.434 11.. = rr + (1 +15'HREHRL) or R..,=34.2kfl flue. RzflR, 149+51|5 R1flR2MRb=51|342=436m RFR‘ 1+3 = 121 =° _ (181)(05fi03) 4.36 J: Av=0806 Ro=325§2 A‘-03+(131)(o_1|o.3) 435+1 ——.-—'— 434 d. R.» = rr + (1 +5}(RsIIRL) R“, .—. 0.30 + (1s1)(u.135) => 3.! = 34.3 m _ r,+R,R1|1R,_ 0.3g]: R”'R‘u—1+IT“O {81: R,=6.l80. 4.33 5- RTE = RIHR: = IDHIO = 5 m R: = — O = —5 V R VTH I} V1,.fir _—_ [Hafiz-H + Vasfian) + (1 + 331.30 RE — 19 I I Rfllfi: _—5—o.‘r—§—102gum“11A "' 5(R1||R:+R.s) I" " s + (121)(2) ' Ice = 2.09 mA, Isa = 2.11 mA chq = 10 — (2.09)(1)— {2.11}(2] =9- Vcsg = 3.69 V Rea. = r: + (1 + BXREHRL) Vcc =10 V. For Vcsg = 5 V S=10—(1;3)Icqfiz 5:30. Fmflsnfljm b. Ice = 9.38 mA, Isq =10 mA, ng = 9.12:! mA v.4: fad “my. rr=fi$£fl=fljll 5w:- -_‘- ' \ Mia 12.. = 0.211 + (31)(o.5||o.5) =- W 1-07 = :3— 5K ._ Io _ RI: RLHRI ) ‘4' " :5 ‘ {I “(as + a.) (mun. +3... 157 n: _ L Rang: ) a _ {813(2) RIHR: + 20.46 - (Izaflu'ozs) 0.1975[Rllifiz +20A-EJ = R1HR2 c. r, -- = 1.49 kn Ring: =. 53415) Av: ([+fiXREHRL) RRIRiflRB ] r, +(1+ [3)(RE||R,_) Rlflkgufiz.6 + R, R“ = r, +(1+ flXREflRL) = L49+(121)(2||2) Rfl=1225kn. Rlfiazika=$225=4som A”: (121W) [480 J: 1.49 +(121)(:42)' 4so+s V-ra = IanrH + Vss(on) + (1 + ,3}I3qflg flit-(ammo) = (3.123534) + 0.7 + (m)(u.5) :- R! = 7.97 kn 7.9m; ._...— = '.04 R =13." n 7.9? + R; a i 4—w' k —_ Chapter 4: Problem Solutions (b) :0' = 0211+(01)(0sz)= 32.6 m - 05 5.04 _ A, = (31)[05+21§m) - (3|)(0.2)(0.134) 3‘. = 2.x? 4.35 3.. = RWHR”, where R... = r, +(I 4.0))?ls 5‘35 =035m 12m = 35. [m = r _ (l20)[0.026) ‘ ' 0.75 12,, = 4.15+(12l)(2) = 246 m Then R. = 120 = Rm|1245 =0 Rm = 235 m I” = g; = 0.00525 m V," =1'mRmr +Vfl[m1)+(l+fl)!,,mRIE T;- R”, are: = %(235)(5) = (0.00625)[23 s) l 1 =4.16kfl +0.7 + (121}(0.00625)(2) which yields R. = 319 m and R1 = 392 m 4.36 I» LuRs=2éfldecgq=§Vcc=12V =Isc=g—i—=O.5A Icq =0.493 A. fag = 6.53 [ILA _ (75)(0.026) ‘ 0.493 =3.969 Tr Io = (1+fl)h(-—R—E—') _ Rn: It _ 15(111'3 + Hub) Rut = r‘l + __n__ Rs 1178 A'_ fs-‘l+aj(R£+RL)(RTH+R-b) .. 24 RTE B = “(24 +3) (Rm +450) RrH 0'1“ ‘ Rm + 400 :5 HIE = 74.9 G {Minimum valuc) dc analysis: I v = _.... . V TH RI RrH cc = Jana-Rn; + Vsflnn) + 159195 flémsuza) = (0.00653){74.9) + 0.70 1 + (0.5)(24) = 12.75 13633 _—_ _ _.._.....__=7 _ = " R} 1360 136+Rz 49:?R2 1510 b. Vac th ling. —l \ 5mm: 1 If”? _. _¢ 4.?73 .11- ,L .1'1' , l Alc = —‘EAV:I For .00: = 0.493 a mm! = (0.493(6) => Mm swing in output voltage for this design = 5.92 V Euk-wfl I'.Ir 3.96 = = — = 0.0521 24 c. a. 1+flflflg 76 “24 1| =.'> E! = 52 Inn 4.37 3. R13 = Rgllfiz = 60||40 = 24 kn R: _ 40 _ _ , V” " (R. +R2)VGC‘ (40+60)(°} ‘2‘” 5-9.7—2 =—= .0 13¢ 24 +(51)(3) O 130 111A leg 9- 0.65" ERA, [sq = 0.553 ml. Vscq = 5 - Iagfls = 5 - (0.663H3] =- Vcsg = 3.01 V Electronic ircuit nal is and Desi 2nd edition "53 Isa-A liq-q \ Sure- -l __-—- Q‘f __ {SCHUJJZE} _ __ 80 _ .. —-——-o‘sso -— 2 kn. I'D —- - 123 kn Define R1: REIIRLHTO = swim = 1.69 m __ (1+B)RL _ gs1}{1.69) A» “ - Mann-691 =3- AI: = 0.97? C. f..- A. = (1 + 3)1.( REll": ) Rallfu + R1. RTE It = {3(RTH + Ru.) 3., _-. r, + (1 + 5131: 2 + [51)(1.59} = 33.1 Rsllrg = Sllra = 3Hl£3 = 2.93 2.93 24 ‘4‘ — (“(2.93 + 4) (24 + 33.2) a A. = «1.61 d. R.¢=f,+[l+§)R£IIRLliru =2+{51](1.59} :- Rw = 38.2 kn go 3 I R; = 3 = 0.0392“: Rn =38.7n e. Assume. variations in :-.r and re have negligible. cfl'octs R.=eo:5%— R1=63kfl. R. =57kn R;=4D¢5% R:=42kfl. R2=33m R; = 3 d: 5% as = 3.15 m. R; = 2.35 m R; = 4: 5% fig. = 4.2 1:11. R; = 3.8 Inf! “1- A' a {1 + 3(112111: IR.) (Eff-12133.6) th = r9 +{1-i- flXREHRLHru} Era-(max) = 25.2 kn. Rq-Hfmin) = 22.8 kn 3....(max) = 915 1:9, R.b(min) = 84.0 kn Rflmax). Edwin). Rug. = 85.6 kg Jig-[min], Rdmu). R“. = 137.4 kn Hz(ma.x)|lrn = 3.07 kn Rg(m.in)||ro = 2.79 169 For Rgfinixfl, Human}. RrHEmin) 2.79 22.3 A' “(51)(2.79 + 4.2) (22.8 + 37.4) 3 Ai = 4.31 Solutions Manual For Rflmu}. Rptmin). Rrg[ma.x) 3.07 25.2 ’1'“ (5”(3117 + 3.5) (25.2 + 33.5) s. A. = 5.05 4.38 The output of the emitter follower is +- 9.. ‘2’» “I— For v0 to be within 5% for a range of RL. we have M = (095)M RL(m1n) + R“ RL(max)+ Ru 4 = (0.9 5) 10 which yields 4 + R" I0 + R. R, = 0.364 k9. + R R We have R, = [G—ILHEI#]‘RE|¢ The first term dominates bet R1E|R2flRs a Rs , then R.Er'+R‘=:0.364=r'+4 l+_B 1+1? 01' 0354: rt +_4_=_-BV.L_+_i_ 1+5 1+}! 1690443) 1-1-13 omei Ice 1+}? 4 . . 4 The factor—15mm: rangcof—=0.044 to l+fl 91 4 V, ~——- = 0.0305. We can set Rn a 0.32 = —— 131 Ice Or I"...2 = 0.08125 mA. To take into account other factors. set [m = 0.15 m, 0.15 ’30 =m=000136mfi For Vega 551’. set Rz=fi=333kfl Cha ter 4: Probiem SolutiOns Dcsign a bias stable circuit. vm'=[ R1 )(10)—5=Ril(Rm)(1o)—s R. + R1 Rm = (0.1)(1+ MR5 = (0.1}(1 11)(33.3)= 370 m V,“ = ImRm +V,E(on)+(l +fl)ImR£ - 5 So 7:1—(37o)(10)—5 =(0.00135)(370)+ 0.7 +(111)(0.00136)(33.3)— 5 which yields R, = 594 Id"). and R2 = 981 kfl Now A. = (1+fiXRzllRJ Emu-Rm J r, +(1+ fixREHRL) Rm||R$ + RS Ri=r‘+(i+fi)(R£||RL) and QJVT . [ca ELL—‘90! RLI=4mJ r: = 15.6 m. R“, = 340.6 m _ (9903-314) . 370||3405 - 15.6+(91)(33.3||4) 370||3405+4 =° A,=0.9332 Flfi=90, 11"L = 10112 11., =71s.4 m _ (91]{33.3{l10) 370117154 _ 15.6+(91]1(33.31|10)'fiol_[?15».—4+4=> A,=0.9625 Fifi-1130. 10:41:12 13.222515205490101 _ (131)(33.31|4) 3701490 _ 225+(131}(33.3n4)' 37011490+4 = 1L = 0.9360 510: 130. R. =10m Rib = 1030 H2 _ (131](333110) svglloso ‘ 225+(131)(33.3||10) ' 370E1030+4 '1' A, = 0.9645 Now ua{min) = [1&(minfi - v3 = 3.735100: vo(max) =1A,(rnax]- v5 = 3.86 sin ax 93.1.7.3591, Va 4.39 L 873 = RllaRz = = 24 1:11 60 V = a T" (“+40%”) 5 V ,6 = 75 6 — 0.7 2‘! + (TENS) Icq = 0.954 IILA Isa = = 0.0131 111A 5:150 s — 0.7 = ——-—-----—— = 0.00630 111A {3“ 24 +(151)(5) {co = 1.02 [ILA 3 = 75 __ {7s)(0.025) _ 1 2; En 0.984 1‘.r g = 150 r, = 3.52 kn 3 r- 75 Rm 2 r, +{1+ 13‘1EREilRL)= 55-3 kn E = 150 Rm = 130 RR Av = (1+ 131113102.) _ RlIRzflRm r‘ +{1+ ,6)(RE||RL) RIHRZHR”, + RS For ,0 = 75. filming“, = 4ofiso||653 = 175 111 = (76)(01333) I 175 = 1.98 + (75)[0.33 3) 175 + 4 Av = 0.789 For 13 = 150 . 1015112102,,h = 4o||60i|130 = 20.3 m = (151)[o.833) _ 20.3 2 3.32 +(151)(0.333) 20.3 + 4 10:01:11 _ Rs RTE ) 3‘11“-(1+fi}(RE+RL)(RTH+R¢ ‘ s r 24 - .. .= .0 A"(76)(5+1) 24+65.3)=’—A h 3:150 s 24 = _ Ar = 19.5 A‘ “513(5) (24 + 130) g —— 19. b. c:ng is the same. as part (a) (b)For R, =50: 13: 75:». 11v =0754 13: 150:: A, =0.779 4.40 (a) rm=%=0.00617m v, = 139R, =(0.00511)(10) = v, =0.0517 v vs =V, +0.7: V; =0.7617V Eectronic ircu't al ' and Desi 2"“ edition Solutions Manual (b) 1m =(05 %J=o.494m (d) R R , I 44.7 I 0494 V’ = =1}; = (01303)“' gl=fl='_=, gm=19mIV Raflflm+Rs lql44'7+2 v, 0.026 ——-— Then - r‘ = £1, = (soxome) = r: = 42: m A, = (MOSSKOSOG) :9 A = 0.728 [m 0.494 ___ A; = 14.8_ (Unchanged) 5 =3. =fl g ,d = 304 ,4; Ice 0.494 ———— 4'41 v, = (1+ mmt , _¢_ 5 . rAr +(£+fi)RL R 50 A” = (1 +13) L r,+{1+fi)RL For fi=100. RL=05ILQ 104-44.? r = (100)(o.026) :51 m ' 05 _ (101x05) =————-—-—--——-=U.9066 The" Mm“) 5.2+(10i)(05} For 3:180, RL=SOOkQ 180 0.026 For R5=0 r,=%——l=9.36kfl -_ V_x (181](500) Va - [rt +gnVI}RLllro) Then =W = so that V _ «Vn 4.42 ‘- l+fi 1.. I = 1 mA, Vcsq = Vcc - Isa-R3 [ r kRLlrt) sq Now 5: ID -l(1)(RE-)='> R5251!“ V, +V‘. =V‘ Lag -—- 'i—fi'l— = or 10=I39Ra +Va£(°n)+IEQRE fi=V.—V'=V.+_l__fl_ve.__ m =(o.0099)Rs+0.7+(1)(5) + - [ r )(RLEM : RE .. 434 km S We find . h- I”. A. =£ = (1+fi)(RLflr;) _ (s 1)(o.s|304) V, r‘ +(l+fi)(RLIr,) 421+(s 11011304) “- R (81X05) ‘ :— = 0. 421+(31x05) = fl - Rib =5 +(I+fi)(RLI]I;)E421+(81K05)=44.7 m r _ glooflmozs} _ 2 63 kn R r ' " 0.99 - ' I, =[ 3 )1, and I. =[ ' )(H-p‘)! yo {1 +13)R; g101)(5) Rl'i'Rl-b r.+RL b Thcn Pg 4 I R r =’ "*’ = (1.995 = 0.995 . =4: 1+ ' ° — - 4. I, ( fi)(R'+R*In+RL] au._4.BZVpeakto-E§atbm «3,. 5(81)[ '° )0): A,- = :43 Ru. = fp+ C1 +5933 = 503 kn HEiIRu. = 434M503 = 234- kn 331 Ru, 234:!5 234 II “5 — .- “5 °= Rana-4+0; " = 234 + 0.7 234.7 v. = 0.997% :9. — “n :0 «403V cak-to cal: "5 " 0.997 5 _ ' p 'P c. a... = r, + 0 +4MREI1RL) Re. = 2.63 + (101mm; = 86.3 kfl RaIIR.. = 4341|as.3 = 12.3 In!) 72.3 V5 — D's — 0.99V5 — (U.99)(4.G3) u. .-. 3.99 V peak-m-pul: _ (1+ I?)le [31.] we ‘ r.+(1+fl){RsllR:.} "’” _ (101)(0.333) ‘ 2.63+(101)(0.833)(3‘99) yo = 3.87 V 4.43 PAW = iflrmflRL =1=1‘i(mu)(12) so idnm): 0289 A = iL(peak) = Jfloiggg) £,_[peak) = 0.409 A wiped) = idpeak} R,_ = (0.409)(12) = 4.91 V Need a gain of 13—1 = 0932 With R, = I0 Kl, we will not be ablc to meet this voltage gain requirement. Need to insert a buffer or an op-amp voltage follower (see Chapter 9) between RJ and Ca . Set 1E:z =05 A. ch =-;(12—(_12))=s v 24 = 159R, +1629 = (0.50% +3: RE = 32 Q 50 Let 13:50. 1m = E(05) =049 A _ 3v; _ (sz0026) _ " Ice ' 0.49 " 2‘65 9 RI, = r, +(1+fl)(R£l|RL)= 2.65 +(51)(32[\12} R, =-. 443 n 0.49 I =—=0.0C'93A= . Cha ter 4: Problem Solutions 24 R, 4—H1 So that R, + R1 = 245 (1 R1 RI+R Let 1.1—: 510}, =9BmA Vm a {24)-12 = 130R”; +Vnsla”) I “mica—12 [2341324) = Wm: +(05)(32) Now Rl = 245— R2 50 we obtain 4x10” R: +£1.0882R2 —16.7 = 0 which yields R2 = 175$} and R‘ =7DQ 4.44 (a) R”, = mull2 = 25.6||!O.4 = 7.40 m vm=[ R1 ](Vcc)=[—fl-—](18)=52V Rl + R1 104+ 25.6 V”, = ImRm +V.£(on)+ (i + ,B)[8QRE 5.2-0.7 I =——-——=0.0117m.4 “‘2 7.40+(126)(3) Then Ice = 1.46 mA and 1m 2 1.47 mA Vase = cc -ICQRC “159R: ch = 18—(l.46){4)— (L47)(3) =9 vac = 7.75 V (b) r = (125)(0.026) = 2.23 kn 1.46 1.46 = —--——— = 56.2 MAI V 3" 0.026 6‘. tie—.4 EH 6 3:5 R5 R5 fi (2'- R = Guam = 223.+?.40=0_0764m ' I+fl 126 _ 4404) 1, _ 4104133 “0410.04 “(wows-0764' ‘ or l" =—(0.974)f, v. = 4.04440 {fijwcllRI-J Elcctrgnic Circuit Analysis and Design, 2"" edition Solutions Manual R =E"-=i.93kfl=l.93Vlm4 .II I (c) 11; = [,(RSHREIIR') = {_(100||3"0.0764) = 110.0744} 1 v 1 = * °r ‘ 0.07 which yields = %(0.0744) = 1.93 I' V =-—‘3-=15.9 or A, V I 4.45 5(chR1) r, + R,nR, Lat R,]|R1 = 50 m, [m = 05m V”, = {,QRm +V,£(an)+(l+fl)1mRE 05 (100)(o.025} 1 =—=o.osm, =—-—=. .9 100 0 r, 05 5 2 m i - .123,H vac = (so)(12) = (0.005)(50} + 0.7 .1. R: R. +(101)(0.005}(05) which yields 1RI =500kfl and R1=55.6K2 100 1 12 “1 x 21 1 5.2 +50 (a) 11,: . 12512111. [3:100 = 10.7. Design criterion is meL (b) [m =05mA. [m =0505mA 11,:an = 12-(05)(12)-(0505)(05) = 11cm = 5.75 v 05 = R I =..—_= A A, 34ch L} g_ 0.026 1923mm A, =(1923)(12512)=> A, :15 4.46 E,(peak) = 25 ,uA, v,(peak) = 5 mV Sowcnccd R _—=-——=2 1 3: " 1‘ 25x11)" x0 2m I' From Problem 4.44 V 5 R IR 4-: — R R —‘ E 1: (LNG) 6“ L RsliRs+R.] Lat R¢=4kfl.RL=5kfl.RE=2kQ Now fi=120, so we haw: 120 R IR R ER 2: — m; :22 —5..5_ [121)(4fl5’[ R,|R,+R,] ({Mxfim] R5 R5 RxflliiE + R‘ RSHRE = so“: =1.92 1.11. so that R! = 0.192 m Assume Vow = 6 V Then = 0.909 vac a 1mm, +11,,,.)+vcm 12 = 152(4'+2]+6=: {m = 1m ,fimflum R, = 5+3?" =>0J92=m 1+1? 121 which yields Rm = 201 kfl Now Um = 139R,” +v,,E(on}+ {£91115 1 121 I =—=0.00833m.4.1 = — 1=1.oosm.A ‘9 120 E“ (120 ) 1 Va: = _‘ Rm 'Vcc = R, $20-06) = (0.00833}[20_1) +0.7 + (1.008)(2) which yields R‘l = 34.9 1:9 and R2 = 47.4 kg 4.47 a. Emitter current [sq = Icc = 0.5 nut 11.5 I = —- = 1104.95 A 3" 101 U m V5: = IEQRE = (0.5}[U => VE = 0.5 V V; = V; + V5£{0n) z 0.5 + 0.7 => VE = 1.20 V W: = V; +IBQR£ = 1.20 + [0.00495)(lfl'0) =3 W: = 1.7 V _ (100)(0.026) ' - (100)(0.004951 _ (100110304951 " 0.025 b.r IJ Skfl =5. gm = 19.0 mA/V _- Cha ter 4: Problem Solutio 5 V0 = ‘FMVV(R31IRL} Rs v (r + as)“ - HM _ "Iv-n J‘ _ 3' + n. + 115nm; ° 31-: . _ _— VS 1 1 (RE + RC) VIP m ‘1' "" + 1' *J-n— [9 r» RSIIRE] 0-5110; l )1 ( V [19“] I 1 ] 1 +0.05 5.25 0.05m 0.05m v,[4o_1g) = 40V; =5 V, = ——(D.4976)V5 v0 = (19](0.49TG}V5(100||1) AV = 9.16 C. ' ‘3»‘4 K. e 2'6. ‘4' "av- Vx Vx IX‘E+:_QMVIH VIII—“VAC R.- = 0.844100526 = R. = 49.5 n 4.48 20 —0.7 10 5. Ian = = 1.93 mA Ica =1.91 mA Vch = Vcc + Vssfon} - IcRc = 2s + 0.7 _ [1.91)(0.5) =. VECQ =13.3 v 4.49 Neglect efl'ect of ha. From Prohlam 4-16. assume 2.45 g 15.. 5 3.? kn 0050,= 3120 V0 = (MJbNRclIRf-J 1+ h}. . £3 + R0: I; V; I = , I = " (1 +0“) 5 Rs +RE||RM __ hf: RE ) A” (1+0c)‘flc"fifl(g£+me * ("F—1*") £5 + RsIIRu High gain deviccz Me = 3.7 kn. )1“ =120 3.7 I. — IE- — 0.0305 k9 R5111!“ = lfl||fl.0306 = 0.0305 120 10 1 A” _ (fi){6'5"5)(10 + D.0306)(1+ 0.0305) :5 A” = 2.71} Low gain device: J'I.e = 2.45 109, h}: = 80 Ric = Lg? = 0.03025 1013 Rani!“ = 10||0.03025 2 0.0302 00 10 1 A” " (E)(5'5|15)( 10 +0.03025) (1 + 0.0302) =>A..=2.70 50A,: constant 2.70 < A, < 2.71 c. R. = Baillie. We found 0.0302 < R.‘ < 0.0305 kn Ncglecdng ho“ Ra = R; = 6.5 m a. Small-signal voltage gain Av = mafia-HRH =* 25 = 9m(Rc|11! For Vgcq = 3 V => V: = --V5cq + V£3(on} = —3 + 0.7 => V;- = —2.3 "cc —- IcaR-c + W: = D 5 —2.3 2.? Re R—c =Icq= =[Co Forrfcq = 1 mA. fig = 2.7 kn 1 = -——— = .' V 9'“ 0.020 35 a “A! A, = [33.5)(2.Tl[1) = m Design trim-inn SlfiJfifl-d Ind V559 lldSflOd. Electmfljc Circuit Analysis and Design, 2“d edition Solutions Manual 101 I; :1: — 1.0]. mA V53 = [533 + V3503!) 5—0.? = . 1'“ =1» RE 426 kn 3R3: 5V1- (IGOHOJZE) r, = —— = —-—-——---— Icq 1 9 r: z 2.6 11.0, g... = 38.5 mAKV, r! = no b. 4.50 R 20 I. Vrm = (m)vac 2 (20+ 50)[10] =? V731 = 2.0 V RTHI = RiIIR: = 201130 =16 kn 2 -0.7 = = 0.0111 A h" 16+(101)(l) m Ln = 1.11 mA :9 - g =9 - 42 7 va “‘1 ’ 0.025 ——————--—---——9"'1 - ' r,“ = W =9 r, = 2.34 kn 1.11 —’—-—-— l' _ m a l' a m 01 - Ln _°L-__ WH:=( R‘ )Vcc=( 15 )(10}=1.50 V R3 + R! 15 + 35 Erma = Ralllh = ISIIES = 12.75 k0 1.50 —0.70 In: — — 0.0126 111A 1:: = 1.26 ILA 1.26 . = 9m: —- => gm: — 43.5 mA/V m = (muffins) = a; = 2.06 km r23 = on b. A»! = -9011 R61 = —(42.T)(2} a A,“ = -354 ' An: = —9na{Rc:llRLl = -(¢3-5)l*ll4l =:~ Ak, = -97 1:. Input msiscanc: 01' 2nd stage 11.; = muffin“: = 151|85!|2.06 = 12.75fl2flé = R1: = 1.77 510 A2. = ‘9’"1‘RCIHR1‘1) = —(42.71(2n1.77) AI,JL = —40.1 OVGI'IJI gain: A... = (—10.1)(—9'r) :0- A... = 31190 If we hld A... - A”: = (-55.‘)(—97) = 5111 Loading efl‘cct reduces wean gain 4.5L R: 12.7 . V = --—-—-——-- V = —-—--— 1‘2 ‘ 1"“ (R,+R,) cc (12.7+57.3)( ) =- Vryl = 1.905 V R-rm = R. "R: =12.TH57.3 =10.58 kn 1.905 -0.70 = —-—-----—— = 0.0047" A I31 10.53 +(121)(2) 'm Ic1= 0.572 111.4. 0.572 '22 HIV = :1 m = . 9"" (1025 M...— r“ =.-.- 2b r91 = 5.4-5 kn 0.312 or. r“ = 0.572 : ———'°‘ = x R. . ( 45 j V = —-'-— = 13 N“ (Ra-FR. CC 45-1-15 (1 =3- VTH3 = 9.0 1V. R”: = mum =15i115 =11.25 kn 9.0 - 0.70 = —---—-— = 3340' A I“ 11:25+{121)(1.6] “‘1 Ic: = 4.86 11121 4.35 = — m = - V 9"” 0.025 3—9 2 13' ""V 120 0.026 r” = 3 rfl = 0.642 kfl 1-22 = as: b. 131 = 0-5-77 mA. vczq.=12 — {0.572)(10) — (057mm 3 Vcsql = 5.13 V 15:: = $.90 VCEQ} = — 3 1’5qu = V 31 \P-au. lch‘ H.‘ \ Suere= -r lqif‘r.?1. a.” ' v Ind: 5'33 LL Ga. V“ flu-11"": \\€HF‘ ‘- I L!- 4.84- 7.“. - r8. Cha ter 4: Problem Solutions HI: = RJIIRIIIRm RIB ‘3 fr: + (1 + ISNREIHRLJ = 0.542 + [121)(1.6||0.25) R”. = 25.3 12.; = 15||45||2a.a Ra = 7.92 m a, A“ = —gm1(Rc1IlRa) = *{22M1UIIT-92J =: A” = —97.2 A z (1+ MIKE: E31.) W M2 +{1+ fiJ(RsalIRLJ _ (121101.210) _ ‘ 0.642+{121)(0.216] ‘ Ova-all gain =1—97.2Ho.9753= -943 0.975 d. 12.,- = mnmnm = 67.a|;12.7[|5.45 => .12.: = 3.51 kn 1": + Rs = R ' R4 1+ 3 1 53 whae .- = 0311mm“ = =- R5 = 5.29 0.642 +5.29 1:. _ ——12—1— 1.6 => 0.049]|1.6 : = 47.5 D eAi-- “'1 --'_"w A‘-36 ' c “ 0.210 m “‘ “3 ‘ " maul = (4.561(0210) = 1.05 V Max. output voltage swing = 2.10 V Beak-mink 4.52 _, 13 In {E 7244):: Ic,=69.9m21 69.9 I = --— = 0.699 91 100 "'A :3 In = [:5 1.4+01599) => 1:. = 2.08 mA (b) V, = Vt, +1!” +Vo V 11) V. =[fi+$+gflVfl](0-OSJ rt! 2 [100)(0026) "’ 59.9 69.9 = --— = 2533 xv 3'“! 0.026 m = 0.0372 k0 2 . 0.05 + 0.0372 + 683%] 05) 413.91% =12”; 2.08 2.08 g... wag—39M”, v,,[;+30]=vfl[—L+ 1 J 1.25 05 0.0322 V," (30.3) % V,,(28.88) = [%)(23.88) or (2) v_, =V; (0.00261) Then v V =1! 0.00261 ' v =V1. 93 3 '( U 0( ) V or = —" = 0.990 '4‘ V (:1 R15 = r11+(1+/3)[R:] Eiectron' ircuit Anal i and Desi n 2"" edition Solutions Manual 0 1 N I=—'-V +——— W ‘ 0.05 "[3" 0.051%] So that l s V‘ +(0 633)V 2638+——'— ‘ 0.05 ' ‘ 0.0310037: which yields E] Rf, = v—‘ = 053 Q [I ? 4.53 W: Va 1 1 a. Rn; = MIR; = 335nm = 91.0 m I: =—-+ = ‘1 —+— 05 r,, 05 rd v ( R) )V TH = C: L: V. V5, = I: + gnzvfl R1 + a, 0.0 0.05 1.25 = —— (10) =2.71: v [($11 ) 125+a35 Lflf -1; [Ling VTH=ImRTH+Va£1+Vh32+Izafim 0.05 ‘ ‘2 0.05 "1 1 1 1m =11+a11m=11+3121m V 1%). Wcfind4=R1=l 2717_1w 1' 1m: —'———'2—- => 151: 0.120 11:51 91.0 + (101) (1) .__.._‘ [cl =12.a 0A IC: = .3121 = 3(1-i- 3115: = (1001(101)(G-113 #Ai IQ: = 1.29 “1,41. I132 = 1.31 mA [RC = fcz + Ic1=1-'39 +0.01'23 =1-31m:\. if]:- = 10 — Inc-RC =10 - (1.3l)[2.2) = 7.12 V Vcfig = 7.12 - 1.31 = 5.31V ch: = Vcs: - V35: = 5.5}. — 0.7 Vcs; = 5.“ V Summary. 1:1 =12.3 “A. la: = 1.29 mA VC£1= 5.11V Vat: = 5.31V 0.0133 To find R“: 0. gm = 0.026 = 0.492 mA/V V 1.29 1 =-—-‘—— V - = __ = . V t i 1‘ 0.05 g" ’2 o. ,1 9'“ 0.026 49 6 “1‘” VII (2) Vn = T+gnlvll 03315-2) ll. 1 = 1;,[E + soyonsfiomn) or vx2 = (1.72% (3) V,I +1!“ +VI = 0 =:- V“ +(].':'2)'I./,l + V' = 0 so that V“ = {0.368}; and v“ = (1.72){—(0.368)V,] =-(0.633)V‘ Cha {or 4: Problem Solutions = - l' m l' V _ % (gully! +9 IVILRC Ilzgnzvl1+fi+ I v“ +gllIVl'I Vs = Va: + Va. V-n = Vs - an n: rod 1/, VI -V V VI: = + levr1)r1r: A +gu|lvfl = A 1-" Ar:fl rxllirx2 V” = V“ (1 + 5),": (3) V1: _=VA =—Vxl '"1 Then from (2) 100 0.026 m = LOLIWJ = 203 m L=VA[L+3MI + I J (100)“).026) J'Zul . rnl rxlurxl I": = T- = 2.02 kn v v V ' (1)1;=§..2VA+—L+—"’"5"EMIVA Cr: rrrl run Vo = -[9m1Vs'+ (9m: - 9m:)Vn2]Rc '1” r” ’1" Solving for V from Equation (2) and substitutin V" = (vs—V,2)(1+fi)('£) A g 7.“ into Equation (1)1 we find I 1 V. 1 1 531)]=V *+..+I +( +fl)(m 3(1+B} R _fl_ r“ a. all,“ n " I 1 1 i 1 I LIEU-+3) ‘ — —+ + 4...... + W: = - 9m1Vs + (gm: - gm!) ' —‘——L:'l:- Re '22 '2. gm ’nflrxz F... Gun?“ g": 1+{1+fl)(;;) Fol-{3:10am=100v.1c,=13m=:m A V; FHI=FNI=$=IUO£ZQ I.- = V; (1oo)(o.026} (49.5 — 0.492)(101}(% ’1": ’11 = :16 m =_ [a.492)+——T 2.2 I .__ = =-—-——=33.46mA/V 1 203) gull gm! I A... = -552 The“ c. m. = Rallflallfias J—+3s.4s+ 1 R _ 100 2.g|2.5 m. = m +{1 +5)m ' — 1 1 l 1 1 — ——+38.46+ +— +38.46 = 203 + (101){2.nz} = 407 m 100 100 male 100 2.6][2.6 12,, = 91.1140? = 74.4 m = R” or Q =Rg=22 kn R,=50.Dkfl Now 4.54 Imam!”m =0 - I 1 Replac: law by % £6: ICL EODlmA rd =-1-O-9=EDOkQ.raI= 10° = {0.00051 00 l =————-—-=38.46mAIV, = . gml 0.026 gml 03346rm‘i/V r‘z =W= 2.6 kiln" = 260 kfl Then R“ = 66.4 m Elccgmgig gircuit Analggig and Design, 2“d edition Solutions Manual 4.55 I. R1}; = Ranflg s 93.7“fi.3 =1 5.90 m VTH=( R: )Vcc R1 + R: 0.1 = _..————- = 0.7 V (6.3 +93.1)(12) 56 139 = ”_'%33 = 0.00949 mA Icq = 0.949 111A chq :12 — => VcEq = 6.31 V Transistor Pa = Ichcaq = (D.949](6.31) =- P9 = 5.99 mVV ac; PR = If,ch = (0.049)={51 =¢ P5 = 5.40 111W 0.? ‘1‘? 6.11 "-4 100 0.949 ' 105 m Peal: Sign! current = 0.949 111A |Vg[ma.x)§ :1: (5.as)(0.949) = 5.39 v 1 V“ mu) 1 5.39)2 Pnc=5'*%r=5[£s—] =* PRC = 2.42 111W 4.55 (a) 10 = 1,011, +Vu(an)+(l+fl)ImRE I” = [0—0.7 100+(121)(20) I...2 =O.443mA. 15c, =0.447 m r0 = = 0.00369 m For RC: Pm =(0.443}‘(10)= PM: =1.96mW For RE: P" =(0.447)1(20)=o p” =40 mW (b) AI: = 0.443 m. Ave; = (o.443)(10) = 4.43 1* Then P.EC = 5(mc)’ RC = %(0.443)‘(10) Pm: = 0.981mW 4.57 10 —0.7 I = —-—-—— = . a“ 50 + (151}{10} D "0596 “A 19.; = 0.394 1113. lgq = 0.90 mA VECQ = 20 - [U.394}[5} - [0.90)(10) 3 V'Ecq = 6.53 V Pq .3. lcqucq = (DJBQQHGJS) 9 Pa = 5.34 mVV PM .—‘: IE;ch = (0.394)1[5) => Fag z 4.0 mW PM é 13.;ch =(u.90)*(10)=~ PH 2 8.1mW V“ [—3 Hind. _ 1 s —___.. \h‘" rm. D. 81W (.53 1° —-1 1.43 kn 'AV“ 131'..- = 0.594 21.0.1414: (0.894){l.43)= 1.23 V . 5 . A“, = (5 +3)A1c 0.639 111.31 A1: = in = %(U.639}:(2) a P5”; = 0.405 mw FT“; = g . (0.094 - 0.539135) =- Fag = 0.15:1 mW FEE =0 E: 5.04 — 0.403 — 0.163 =:» E = 5.27 mW Chapter 4: Problem §olution§ 4.53 459 10-070 “PM = — = 0.00330 = = ' I“ 10° + ("um") I“ t I” 100 + (101M103 0 mm "m [cg = 0.030 mA, In = 0.045 mA Vcaq = 20 — (0.0301(10) - [0.040M10) => V 3 16 v Icq = 0.533 mA, [En = 0.346 In}; egg = . 14530 = 20 - [0.333)(10} - (D.345)(10) => Vang = 3.15 V Pa 2 Ichcgq = (0.033)(3.15) =0 Pq = 2.6:. mW PRC ; I‘Cqflc = (0.330)’{10)== P5,; = 1.02 mw 3.15 In 100 1'0 — — 119 kn Neglecting has: currents: a. RL=1kQ —1 -1 Sm: 10||1||119 = 0.902 m . —1 A“: ' 0.902 m 'm“ _1 .05;- = 0.030 = :00“: —_~ (0.902)(0.aaa) Aic = a 909 m - An. = 0.750 v For 1 [o 756): At}: = 0.330 = my": .—. [0.9001(0330) = 0.702 v Fm=c ' =‘PPR =0255 m“' R 10 2 1 —‘——-—' -= _C_ ’= a; =0.102m A“ (Rc+RL)A‘C (104-1) ‘6 Tm, = %(0.702)1(1) => '15—,“ = 0.200 mW b. R; = 10 kn 1 Pm.— = 510.030 - 0.702)°(10)= F“ = 0.0209 raw —-1 -1 _ ._ . l =—=— ' =. —.2 -. =23: w For me = 0.000 =1Auul = [0.330){4.00) = 402 Max. slaving dew-mined by voltage — 1 [3.16 ' =- =. W PM 2 1° :75“ 0499100 ...
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Neamen2eSMchap004 - Eloctro ‘c i uitAnal is and Desi n...

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