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chap008

# chap008 - Electronic Circuit Analysis and Design 2"d...

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Unformatted text preview: Electronic Circuit Analysis and Design, 2"d edition .._____301Uti0n5 Manual Chapter 8 Exercise Solutions E8.3 58.1 For V“ = 0, mm“) = '35 = 1.2 A'= Inrmu) For ID = o :9 Vgshnax} = 24 v Maximum power when Vns = Nashua] = 12 \'r and 2 In = fang”) = 0.5 A no la 3:: ‘10 re Mam-nun: power at center or load line E82 Pm, = (0.05)(10) =- P5,” = 0.5 w E84 Pawn: = I'D - was = (”[12] =12 wan: C. Tank = Tlmb '3‘ P ‘ emit-Inn: Tm = 25 + (12](4) :5 Tim = ?3°c b. Tau: = sink ‘5' P ‘ aCIII—illk Ln: = 73 + (121(1) 3 Tc": = ESIC I. Tdu = Tu." + P ' Eda—cue T4,, = as +{12](3} = T“. =121°C 5.8.5 - 6am.“ = ELET—jm = %—33 = 3.5'c;w So 15 = 30 «- :i-RL a R; = 22.5 (1 Pa = Tm“ - Tm“, a W "m 9mm“ + 9w.-.“ + 9.“ “m = 200 — 25 b. Vac =15 v, Is...“ =21; 3-5 ”-5 +2 =5 Palm‘s = 29-2 W Va; .1. I5 - IcRL 0=15-ZRL=>R;=T.5Q => Maxim Pom =(1}(7.5] = 7.5 w = 25 + {29-21(0-5 + 2) => Tm. = 93°C Tun = Tomb + Pﬂmumcue—Ink + glut—mb) Electronic ggircuit Analygis and Design, 2'” edition Solutions Manual 58.6 E33 8. P9 = chq -Icq = (T.5)[T.5} Pg=56.3mW _1 v; 195.52“ ._ b —2 R1. '2' 1 #P‘—21.1mw E=(15)(7.5}:?E=113mw ngﬂzgﬂaq=lﬁ7y¢ 5 113 153.: 56.3 — 21.1: 35.2 mw ID——4 I. for.- = alga-:50 111A 0.1 b. 10 =5 ups-(min) = i — 2.? = 1.3 V 0., = — (i)(60)(0.050) =-. 4.7 v 50 muimm‘n swing is damned by drain-to-suurc: voltage. C. E: Van - Ipq =[10)(50|= 500 mW FE 31.25 — :- *'= =W’———"=”% 58.8 __1 vs a. EFL—EEL- => V, = \fmﬂbiz \/2(a}(25) = V; = 20 v 20 =>Vcc=D—E=#V£E=25V => Y5“ = 31.25 mw 58.9 F rm 4(3) :FZ=74W «VP 1r 2!) ..._--—.— =528 ‘1 " 4Vcc 4 25"" % - 1 v: (41’ KR; 43!. _ Hill _ ﬂ. _ 7 _ E‘ «(0.1) 4(01) “63' 40 => 19—0: 23.? mW d. ﬁ=%=§.%:n=52.3% 138.10 a w=IIo+UGSn-Y:—E “Ii =1 divas“ dug dl’o dvcsﬂ _dvﬂ_ dip" dv dim dva I? dvcs" _ 1H1 1 din“ JR; 2 Jim Al 0, = 0, in... = 0.050 A d”CHE" 1 So . = ———- —- —— = .-. dlbn J02 2 0.050 So inn =1}. + in; Fora small clung: in #0 — An. = Aim. — [—mm So Aim" = 1A”, 2 01’ dis" 1 di; 1 l l 1 a... - 2 a... E'E“§'ﬁ-°-°“ d n Then :55 = {moms} = 0.125 14:: Then 21 = + 0.125 =1.125 dye. ‘00 1 mdA, _ — = — . = . du: 1.125 = A 0 339 b. Fﬂw=5V.iL=0.25A=Ign.“359:0 Test Your Understanding Chagtcr 8: Exercise Solutions dye dip“ duo dvﬂ _ 1 1 1 1 1 1005.. = 2.24 0.0 = . '2 duo ( H 5) 011 d ..'.’J. = 1+ 0.112 =1.112 dun —d"°— ] ”4-0399 ”_dy1_1'112 L EB.” 11. Ice E %(2;:':) =VRLf= %=Bmﬁ RTH=RIHRI Var-H = (iii-11%.)“:C = ~Rl—l-RIH - Vcc Let Rm = (1 +6312...- =E76){O.1] = 7.5 1:11 1 R—-[7.5}(12J-n.7 = ._.J______ 0.107 7.5+ 7-5 —1‘ ‘ (91-2) = 2.33 =1» R. = 39.11:!) R. .._..__ 39.112; 391+ R2 ‘ 7‘5 =’ {39-1 - ”HR: — (7.6)(39.1) => R; = 913 [:9 b- PT= 5 (0-9Ica1’m. = §uosusnms1 = ”F"; = 33.9 mW E= Vccfcq =(12}(3} = 96 law '155’: 5— E=9&-33.9 =>E=sm mW PE 35.9 ﬂ—E—g—G-=g—4U.5% 158.12. a. R. = r.+(1+5)R‘5 and R's = a’RL =(1o]’(a}= 300 n R. = 1.5 kn = Rrgljﬂa I = —- = —--- = 23. A 0 112121 {10]2{8) a m _ (100](0ﬂ261 _ s——22'5 —0.116 [In R. = 0.116 + (101mm) = 50.9 kn Era-{3‘33} .' = 0.9 = —---— 1 a RTHJIS RTH + (50.9} 2, (30.9 - 1.511?” = (1.5)1303} => RT}? = 1.53 kn . R . 1 1'TH=( it JV‘cc=—-RTH'Vc-c RI+R2 R] _I—q— 22—“5- ’5') leg—- ,3 *100 —E|...-5 [TL-'1 Fry-0.7 ”FT =~ Til—(1.53103) =40.125}(1.531 +0.7 1 26.4.82 _ 26.4 + R: — {26.4 4.5313, = (1.53}(2s.4) 1.5-3 =3- R; 21.52 kn a. a; = (1.9ch =(n.91{1s]= 16.2 V 1'; = 0.91” = {o-s}{22.5] = 20.25 M :‘g = «is =(1u)(20.251= I: = 203 mA ‘15:: %(1.62J(o.2oa; => T; = 0.164 W 1-38.13 :. IC 2 [sq exp (\$2) =:r V35 2 Vrln (Ii) 1' In: 5 x 10“3 V35- — {00263111. (w) —- 0.5225 V :3” v.91 —_- V512 = 0.5225 1.... = {D = 159 up (0.5225) 0.026 = 5 KID-'13 exp (0'62”) 0.026 IEiII =12.5 mA Bloctronic Circuit Analysis and Design, 2."d edition Solutions Manual . 2 b. Vo—ZV, “em—.257mh ls: approzdmaﬁon: ac. ; 26.? mA. is, = 0.444 In)”. 26.7 x 10-3 2 «10-47| ID = 12.5 - 0.444 = 12.056 111A V35 = (0.026)1n( ) = 0.5551 12.056 'x 11]":| V9 = (0.026)1n( 5 x 10-” ) = 0.6216 2V0 = 1.243 v V53 = 2V}; - V35 = 0-5759 0.5769 0.026 ic, = 2 KID-.13 exp( ) = 0.366 mA 2nd approximation: (3,, = 0,, + is, = 20.? + 0.305 =‘-. 27.0 mm =1};u i5. = (2%){2'L6} =0 1:... = 27.1mA is, = 0.452 mA In =12.5 — 0.452 = ID =12.DS mA 271 x 10'3 2 x 10'” =9- VEE" = 0.6664 V V35... = (0.026) In ( 12.05 x 10-1 VD = {0.025)1n( 5 3‘10"” ) = 0.6215 "u" Evan = 1:143 V V55 = 1.243 - 0.6654 => 1;”: = 0.5766 v . _ O."66 . 1cP=2xlD ”exp( 3' )=>1¢ =0.855mA 0.026 _L___._ . . . 10 C. bar-IGV, IL: __=133m.-\ 0.015 is” — i1, = 133 MA =¥ iEI“ =13lmA fan = 2.15 111A =5 ID =12.5 - 3.15 alg =10.3 mA —3 V; = (0.02510: (”“3 " 1° 21/35 =1.335 V {D.026)1n( 131 0" v... ___*1_) 2 x 10"” :5 Vggn =0.70?4 ‘v' V53, = 1.235 - 0.7011- =5 V33, = 0.5275 V 11.5276 ‘1 0.026 J a ice = 0.130 m I'cp = 2 x 10'" exp[ 138.14 a. w=o=ﬂo.vaa=ﬂ.7v 11 —O.T 11-3 R1 — 6.2—5: IR]. —45.2 IRA 13.1 = If transistors are matchcd. than is: =i53 . . . I‘m Ina-Is1+lsa-Isa+l+ﬂ i --1 1+ 1 .1 1+ 1) m - E1 1 ‘5 El 41 1'51: 1450—2231'H=I'Ez =44.1mA . . in 44.1 Isi=im= =— b. FOfIJI=5V=\$v°=5V ill—1%? Ig=0.525A . ~ . 0.6 5 . :2; = 0.525 A, la; = 4—: => 13; = 15.2 111A _, . . 12-5.? V33 =5.n \4 =9-Lm = -—0-—25—-=25.2 [11A is; = 25.2 - 15.2 => I}; = 10.!) mA =:~ i3; = E =02“ mA _.__.il_._.._._.___ us..=5—CI.?=4.3V. In: 2 4.3 — (-12) _ 0.25 5.1.2 mA= '52 5'1 ' fag = _a_ =1.59 131.“. + i; =13: —i31=1.59 — 0.244 =- i; = 1.35 mi in 525 C. A:=—-=——=&AE=463 i; 1.35 From Equnticn (3.54) = (1 +010 =(41)(2so) = A’ 2R; 2(a) El. Test Your ndcrstandin Cha tcr 8: Exercise Solutions 58.15 IE = [£3 + In + [cs = {£3 +13; +£5.13: = {an + ﬂ-tfm + 135(1 + 3.1134 I5 = (1+ ﬁsh's: +ﬂu‘331m +35“ 4-31-1531'33 Emandﬁsmlugqmcn IE 5 5334.551'33 50 mar composiu cum-.11: gain is 13 —-:- 13334.35 Electronic Circuit Analysis and Design, 2“" edition ________—_Solutions Manual Chapter 8 Problem Solutions 13 = 1:32 = % 3 20.3 m 24—0.? 3.1 ‘ RE: 208 =>Rg=l.12kn I -R 1.57 7.2 b IAul=ngL =—C‘LT " =—-—( 0.025 )— 52 Vanna) = 12 v a V;- = L31“) = % =W 8.3 a. For mudmum powur delivorod to the load. so: [3. . V yam = LC 5 2 1:.- SH var: = 25 V = Vcspm} _ Vcc _ 25 “um I‘“ ' "RT " 01 [Cm = 2-50 mJ‘L < ram“: w as a V Icq = ——25 3 112.5 =12: mA Mamnmnpowuaxvps=ﬂ=4ov ‘ v” 2 Pq(ma.x) = Ice . —— = (0.125)(12.5) I PT 25 2 n=V—=E=U.625A ”5 z 1.56 W < Pun“ 80 -40 D 625 125 [sq = ﬁb- =1.25 KIA '1 VD°=5°V a 25’“ R 194m B _ 1.25 a;_ ' 1 2 1 3 l h. PLUM“) = E- ‘ [cg ' R]. = 3(0.125) (100) =1» mum) = 0.731 w .15" .ﬁ:_ 8.4 Maﬁmumpaworaths= V2” =25V £6," _ Pr _ 25 _ .35” ID‘ V55 ‘25 ‘l‘i‘ R; = “In =>RQ=2sn 3.2 Vac , l. Pq(ma.x) = Icq ~ —-2— V‘; Vpgmn 2Pq{:na.x) 2(20) 5 So! =——-—-=—==1.67A ca Vcc 24 Point (b): Maximum power dclivcrcd to load. Vac - (Vcclz) ‘24 - 12 Point (a): Will obtain maximum si a! current R _-. _...._ = ._. 7, an " Icq 1.67 3 M output. Point (o): Will obtain maximum signal voltage output. Electronic gircuit Analysis yd Dcsigg, 2“‘1 edition Solutions Manual 8.5 ¢ ‘1‘; = 3mg; L V: g,_ =2JKJW2 = 2,!(02x12s) =1 Aw lVol = (1)(2o)(o.5) = 10 v 1 VJ 1 gm)” — . = __ _ = .. = _ w L 2 RI. 2 20 ﬁ 32¢. E: 31.25 - 2.5 2» E: 23.75 w [5. V5; = 5 v. I; = 0.25(5 -4)3 = 0.2m, V135 = 37.5 V, P = 9.375 W vac = 5 v. ID = 0.2m - 4)“ =1.0 A. VD5=3OV. P_=30W Vac = 7 v. ID = 0.250: _ 4.)“ = 2.25 A, V“ = 17.5 v, P = 39.3?5 w var,- = a v. ID = a.25[2(5 - 0v“ - V35} _ 40 — V95- - —1-0— => VD: = 2-92 (13) P0 = POM -(S£0pe)(Tj -25) I =3.71A,£=19,§W .7 D At PD=0,1rJ.M=gg-+2s=5 ij=145°c Vaa = 9 v. ID = o.25[2[9 — 4W“ — v35] —“—— _ I10 - V95 __ T. —Tm _ T :- VDS — 1.33 V (C) P0,“: —_-——-";m ID = 331A. W or M C. ch. at Vac = 7 V. P = 39.375 W > P5, = [45*25 35 W m 94...“. = 60 => 6m,w=2°CIW 8.6 8.8 I Set VD: = “3—D = 25 V . a 2 PD used = Thu“: - I‘Mh 109 = 50 - 25 = 1.25 A adev-nu 20 2 or In = Kn(Vc:' m) 0 3 Tim" —- Tub 1.25 dn-“u PDJAHQ ﬁ+4=VG5=6JV ' = M = 2.5“C/w V ( R: v, 50 Gs— WR1+R:) on Tnun Let R R =10!) kn 1‘ +R 2 T‘ey - Tm}: = PD(9d.¢v—¢u| + HGm-I-ﬂlb} 5-5 = (T526) (501' =>£zi3kﬂ 150 — 25 = Pp(2.5 + aim-mu) =5 125 = P9 2-5 +0 “Hun w ..__._L_......= b- Pa = Iaansq = (1.2mm =- Pp_ = 31.25 w G. [Dunn = 1:90 =5 [DJ-n: = 2-5 A Vnsmu = Von #VDslmz = 50 V PD...“ 2: 31.25 W Qhaptcr 8: Problem Solutions 3.9 P0 = In - Vns = (41(5) = 20 W Th? - numb = PD(9d.u‘—cur + acne-ink + alni—tmb) T... — 25 = 20(1.75 + 0.3 + 3) = 111 b_ E = %(|‘L[ma.x]):RL =%(9-5)2{1} -_-,. P_:.= 43.02 mW 75;: 1q(v+ _ 1") + m0 - V‘) = . _ =2 :T§:.=136°C 95(20}+93(103="P'§ 94mW I" 43.02 Tdev —Tcua = PD ' axial—cue = (20M1-75) = 35 ﬂ = :2 = ““294 3' w Tcu.=Td..—35=135~35:M ' a 8.14 Tc." —- u’nl: = P5 - 9w.-.“ =[20)(0.8}=15 C . a. Iq(min) =‘ mm”) = i9- :- I (min) —- mu m. ng=Tcm—-16=101—16=>ng=55 C RL 0.1 -°————‘——-— 8'10 R=°_‘£:(‘_m=,g=11m 100 TI" - Tuna = PDwdev—eue + acne—amt.) b. P = I - V = = . 20° — 25 = 25(3 +9¢u._.mb) Q1 .9 c5: UUU){12)=> Pg: 12 W => ﬂea.-.“ = 4’C/W P(source) = 21,902] = 2._LV_V 8.11 — 1 V1 {10]2 c, p = —.—f~=———=o.sw 6...-..“ = Thug. --T...... = 175 - 25 = 10.01“, L 2 R1, 2:100) D.nnd 1-5 F5- =1.2 +2.4 = 3.6 W PD = Thu-m — Tani: T3— 0 5 edge—cu. + stun—Ink + sunk-cum ﬂ = _—i '-'—' 3— => 11 = 139% _175—25=P _10w 5 ‘ ' 10+1+4 ‘9'— 3.15 3.12 if: v; _ (W)2 1': E L - R], _ R; 9—5 -- 1 (W)1 1 (V‘)2 p=_.— —. V'--—-V+ E=Vcc L: s '2 Rx. +2 RI. ‘ V i V I 50E= Rx. -" cc- :7 _. .2____ _ - n =5 — 0'7 Vac-IQ 'l a a q=¥=>q=100\$5 s B. 13 8.16 I.. N63166: base currcnu. (a) VD: 2v” (sat) =Va; “Vm =VG: whim—x} :1 V+ — Vcshu) = 10 - 0.2 = 9.3 V V05 2 lC-«Va(max] 311d ID = 1L = K,(Va:)z ' = — 3—9. — ﬂ — V (11.1”) 2 u.(ma.x] In — RI. — 1 =:- Iq ._ 9-8 In). —°R—= KJVG‘) R_n—n.7-g-10| _ ‘ - 9.3 :9 W— V _ Vo(ma.x) 6: RI. '1‘... igﬂmu} = 21° =iaxlfmu) = 19.6 mA 50 ist(mjn) = 0 IG~V0(max}= lib-Ii] = M I'Limu) = IQ = 9.3 mA RL'Kn (5X04) mum} = —I.-_. = -9.a mA [10 _ mmun’ = Warn] 2 mo —'2ovafmu) + Imam) = ”(\$51 8.” 8.18 Elcctronic Circuit Analysis and Dcsigg, 2‘“1 edition Solutigns Manual vﬂmu) — 20.5%(mu)+100 = o 20.5: {20.53—4100 Vuimul= 2} ( ) :Voimax)=BV s u,=;-=1-: =15mA O ' 1-6 V = 15-: -—-—=2V 5’ K, .4 =>ut=10V 2 —;= 202'“ =10.2 mm v0 = iLRL and EL =5.1 = my“ — or EL=K,(vm) Then = K.R,_(v d_uo d—y: din-1 For v1 =10 2 and vcs=v,—v 4-(91 Va} + 40;: — no) dPn_ V wu=8V=:~— .17: inc =>—=0.BB9 Jul tit-a Atw=0. w=0=—=0 A191=L Vu=0.5=- 11': due JP: f-vof 01' v0 =2(v, — =2. 2(.-;- y.)(1 — 1—“) 4(10 - a) 1+4(1o --3) —=0.567 —J 3- V22 = Vr]n(1:)=(o.026}1n(5 x 10 V33 Vas=—=05987V 2 =7 VEE =1.1973 V Pr; = I'c - yo; = [5)(10) => Pq = 50 mW b. v°=-BV 1' --B='.~ L—DJ i: = ’30 an); ice 2-. 80 m 5 x 10—” J ic 30 K “I":1 U53 =Vrin(—I-:-)= (0. 026)ln(5—10_1'3" U35 2 0.6705 V v w = :9 — I133 + W = 0.5937 — 0.6703 —- 3 =5 vi = 4.072 v ”BE: = V3.9 -— use = 1.1973 - 0.6703 = 0.5265 V , _ 0.5265 Icn=Isexp(L:J)=5x10 Lacxp<ﬂ€l26) T 0 :5 IQ" = 0.311mA Pl. =61RL = [so)1(o.1} => PL = 540 mW Pa. = ic. » um,- =(o.311){1u — (—3)) =- Pq" = 5.60 In“? Pg, = ic, - ”EC 7-. (ED)(2) =:. Pg, =15!) mW 0‘. —:|‘|"2=l'v‘:c::.'.-.=25V=E 2 2 =>VEE=5.GV P" = (0.5)(10] =- Pn = PP = 3 mW (b) V55 = VG? "Vm =5 VD! =VGS “-2 VD; =10—va(max} and _fi£ +Vnu _v_q__(max)+ RL K. Hm) (2)0) so 10—w(mu}=,[email protected]+2_2= {Mtgluq so uohnur.) = s v . . 3 IDH=IL=T¢i2q=i§=BM ’8 V55: E+2=>Vcs=4v Thmvr=vo+V65—1:£=8+l—2.5 =3-u[=9.5V v usap=w — (w- %) =a-(9.5—2.5) vsap=lV=>Mpcutoﬁﬁigﬂ ___._.____ Qhaptcr 8: Problem Solutions 19;, = tilt; = (3)2(1) = F; = 54 mW PM“ = in" - Ups =(B)(10 — 3) =>- PMS! =16 111W PM: = in, 445.9 => P“: = 0 3.20 a. yo=2qv=>u=2§=>£§=ltﬂ=sa 3 tgn=ﬁ=>lsn=73.'2mA For ip- _ 25 min =:- i3; = 25 + 73.2 = 98."! mA 3 “95' — VT!!! (1—1;) — (0.026)Ln (m) = 0.7004 V 30—[24+n.7) 5. 3 98.2 = -—-—-— = Thu: . R1 => R; 98.2 =- RI 2 53.97 0 25 x 10" , VD = [0.025) in (m) = 0.3759 V V” = 291;. - v” = 2(o.s759) — 0.7004 = 0.4514 V V5.9 =9 3': = 0.203 [111%. b. Neglecting has: currcm 30 - 0.6 30 — 0.6 In: R1 = 53-97 =322545 111A. .5 Va =(0.0'.1£E)l1:.(a 213;) = 0.555 v Approximation for in is any. Diodumdnnsistorsmhed: E5: 1‘: = '4‘ m (b) 1"! +Vm¢ +Van" Van 3 ”0 For no large, EL =l'l- —- MIG/on H—L l,.,.,+l»’ =;—-—‘2— +Vm. van: ELK-III _ _ 0.4— :p =I5exp (VT—j = (6 x10 ‘2) exp ( 00,2164 to D .3 + N + N I r""-. 5 + Q ll 5‘ For 9,, small, I EV,+Eﬂ-Vm—- ”(g-[Hi- V” ] 2 It; 2 {mm ! v0 Hi. \$.; 2 It; ImRL v I = +_‘£_V .. A "’ 2 m X, New ﬂa— 1 -o95 JV, 1+1 [4.9.1 2 x" IWRL I So 1- J‘im—'—=—'-1=00526 2 K. 1993 095 I For R; = 0.1m. than ,fK I - De = 0.01052 Dr KJW=951 chanwritc g_=2 KJDQ =190mAIV This is the required transconductance for the output transistor. This implies a very large transistor. Electronic i uit Anal 5i and Desi 2'“ 8.23 Av -- ‘HI'IIRL So — 12 = -9n(2) =- g". =6 mA/V = I%.- '1' 1a.; :3 (exams) : Ice = 0.155 am But for mazdmum symmeu-ical swing. set 5mA=>|Aul>12 Maximum pews: 1-0 the loart 1 _ V5,: = (10]2 Etna) = 5 RI. 2(2) => Esmax] = 25 mW 15; = Vcc -Icq =(10)[5] = .50 mw SD 17 = 50% -1c_n-i_ .8 _ 130 ' R. = RT}; = 5 kn Vru = 1303-”: + Vazfon)+(1+ 19)];qu Isa 0.0273 InA SetR5=209 v“, = (momma) + 0.7 + (131)(0.021a){0.020} V”; = 0.957 v 1 EH=E'RTH'VCC 1 0.967 = — = R.(6)(1°)=’R' 02.0 m R; = 6.64 in La: Era =10 kn VTH = IsaR-ry + Veg +(1+ ﬂung-Rs = (0.153(10) + 0.7 +(101)(o.15){0.1} 1 l VT =3.715=—-R -V =—- H R: TH cc R1 (IUMIS) HI = 40.4 RR R: = 13.3 H? cdition S .25 8.26 1, max) = 112.5 mW Solutions Manual R: RI + R: 1.55 )V“ = (1.55 + 0.73)“), = 5.50 v Rm = R; HR; = o.73||1.55 = 0.496 m VTH=( _ Vm — V53 _ 6.80 —-0.70 ° ’ It” + (1 + an; “ 0.496 + (26)(0.02) Isa = 5.0 mA. Ice = 150 mA Is A, = -9mR';. md R1=a’RJ. = (332(3) = 72 n .— “Ia—q — Jig... => 5 TT A/V 9'" " V1— ' 0.026 ' A, = —(5.m(72) = _415 {Vol‘ = [2191- V: = (4151(00171 = 7.05 v 16:32:13.“? 3 — 1 V5 £2.35}: — _ . P =--—= P =34 W 1.. 2 RI. 2(3) =3- ; am E = 1:0 WE: = (0.15J{10) = 1.5 W T; 0.34.5 ﬂ—ﬁ— 1.5 ﬁll—23% a. Assuminzﬂm maﬁmmn powcris lacing delivered. than a 36 V3Ipeak)=36V=>V3=-Z-=9V 9 =>V SEHV =5.36V ""' J5 h 36 b. 17;: 2=Vg=25.5V R1, 2 _ c. Secondary I"... V... .. ”6 :- Irma-0314.1. P. Ip=o°iH=>I==ﬂﬁmA d. 15;: Ice -Vcc = (0.15)(36) = 5.4 W ——2—=a- -37% ”—5.4 "- 8.27 a E . ”e = (— +§mVI)R£ = Vt(r—+9m) E v.=V.r+v‘=>V,,=v,-—y' y: = (u. — 9.](1j—3) R's Pf 14-3.}? "_«=_='L__E__..____{1+I3}R'E._.._”_= 1"! 1+lj-B'R‘s Fr+(1+ﬁ)ﬂ': ”E In : whereR.‘ =(——‘-) R; 1‘1: \$.11: 1 gnaw}; v.- a.) .f-+(1+3)R's E=Icq Vcc For n = 50% : 1_ a 2 E = 0'5 = "1 ICQRL aaIchL E 100 ‘ Vcc 2Vcc so a, = Vcc Vac a: Vac Isa .12"; = (0.1)(501 ‘ L5. c. Rox'_*= 3V1- =49{0.0‘262 1+3 (1+5ch (50)(n.1) => & =0.255 ﬂ 8.23 hat :P bIcm olution I. With 1 10:1 transfamer mic. we need 1 mm gain of a through me transistor. . . . R . I¢=(1+3}'blﬂd1e= (ﬁt)!- summed i_,_ _ mph 1'. ‘3 ‘“+m(m|lﬂz+a.-.) where R.» = 1-,, +(1+ 3)R1=(1+ 3J3}. =(101)(U.3}= 30.3 Then a .—L (101)(—m——) Rul|R2+so.s RJIIR: __= .7 = - RINK: +ao.s 0° 92 O'RIIIR: 6951M Se: .2VCC_ , _Vcc_12_ - 31:0 _R‘=’IC°— 'x. —M-—1amA 13:: = Elﬁn-=045M Vra = IsoRrH + V35 1 . . 3—1 - Rn; - 1’6: = IaquH + V51: Riumsxm = (o.15){5.95) + 0.7 1 =- R; = 47.9 kn then R; = 8.13 kn b. I. = 0.9Icq =13.5 mA = if => [1, =135 [113. '15? %{o.135)’{a} :9 'PT = 72.9 mw E = Vchcq = (12)[15) = 'Ff =1ao mw 'l= =1ﬂ CPI]?! Eiectzonic gircuit Analysig and Degigg, 2"d edition Solutiong Manual 8.29 8.30 I. V}: = VIE; F; V; = \/2(8)[1) = 5.66 V = put output voltage V 5.65 If: R—:= T=O.Toah=pcakontpurcum:m Sat V = 0.9Vcc = BVP l0 minimize distortion Thena = LNG-3628 :9- g = 2,§§ b. New 1 I 1 0.709 Icq=—(—‘”) ( )=>ICQ=D.275A_ 0.9 = ﬁ 2.56 Then Pa = Vchq = (15)“).275) ﬂ. =- P2 = 4.95 W Power raring of mammot- I. Need a current gain of 8 Enough 1h: transismr. :- 2:9:(1+5)(iﬂ5‘—) I.‘ Rlllﬁ: + Rib when Rab =(1+ ﬂ)(0.9) .—. 90.9 m L = (—31% 101 RIIIR: + 90.9 or Rxllﬂz = 7.82 m ) = 0.0792 50: 2Vcc - 12 2IC'Q —- 0.9 m => [ca = a? — 13.3 IDA 13.3 13¢ = I'm-s 0.133 113A Then Rilfmzmz) = [0.1331(7.12}+ 0.7 :31 R] = 53.9 1:9 and R3 = 9.15 kn II. b. I.=(0.9)Icq=12 mA=—=-IL=12O 111A FE: «é-(onfm a T" = 57.5 mW PE= Vac-[cg =(12)(13.3} = T; =159.a mW F? 57.5 17—?snmﬁ'ﬂ—36J?!) 8.31 1. All transistors u: matched. l’nlI1-J‘L=ia:1-i-isa=(liners-l"Ii 15 61 - 1 . , 3 — — = . 3 (60+Gﬂjlcglg ZQOmA b. Pmm=6V,mR§=2ODQ. 5 io=m=lOJA=3ﬂmA§isa . 30 153— ﬁ—OAQZM is; = 3 - 0.492 = 2.508 In). 151 = ﬁg :15, = 41.111111 s1 ——-—-——-—- 3 iszg3mA=r£m=ﬁ=>49JSpA if = is: Him .2 49.18-41.11=>11 =8.07 11A Currant gain 30 x 10'3 a ‘1': a 01 x 10-5 a ———A‘ = 3'72 x 10 is; 30 x 10" V53: - VT 111(E) —(0-925)1“(5 x 10—19) V353 = [1.6453 V _ 131 _ 2.505 x10“) 1/“, ._ Vrm ( Is) ._[0.026)1n ( 5 x 10-19 V”, = 0.5907 v w = no ‘+ V353 — 1cm = s + 0.5453 — 0.5907 w = 6.0546 V Voltage gain an 6 = — = ., = 0.989 A" 9: 6.0646 ”—A Cha r 8: ProbEcm Solutions 8.32 8.33 ' . a. l I. For =1A.I =— 1 '° 3= so =‘ 3° In" R.- = 5m. +(1+ﬁ)[mn(r.a +£1 man]; We. can then write I’m a 7.2 mA and in 272 mA 10—- V 10 — m. V - 531 = .2[ W. + HE!) _ 20 M r' = (60)(u 026] = 0‘21? kn R1 R] 7.2 If. for simplicity. we assume'Vg-m = V553 = 0.7 V. 30 then i 10 _ v” 2120...... R. = 501.217 +(ﬁl)[2[|{ﬂ.217 + (61}{D.1])]]- —— = + 40 H: RL 1 = \$0.21? +61[2|16.J2]} If we “sum: ”0.314: = 4 V, than or R; = 46.4 kn 9—3 = ﬂ + 4.0 R15: hmhy'hcldsﬂ 1: R2 = 32.5 ('2 3.34 |.. b. For .9: = U, 9.3 15; — 32.5 I — 0.236 .JL 2 1' Since [534 = 101'5'13» then IE! = 1&5 = 2.863. c. W: can write I R. f,“ 510:1; r3+ 'll 1-1-51 2 1 +5: V 50 0.0 U Now 1-,; = 53—7' = 3—)I—2—5) = 0.4545 n I re: 2.56 ' _ ,63V-1- _ (120)(0.026| "” ' In “ H.286 “ [0‘91 9 So 1 0.4545 + 32.5 M \$9.32 Ea = - 2 51 327 a “12m 91 = 32-5IIO.G902 = 0.0900 V“ _ Then V‘—-V _ b. I. = K,(V,a +vn)’ =Tm 1 0.4545 0.0900 I R.“ = 5{+} or Q = 0.00534 9 5 = 10(Vsa - 2)2 =- V50 2 2-707 V 0- .‘D- 521—-3-‘-—‘=>Rl=32=1-45kﬂ RI 8.35 E c. R; = 100 n For a sinusoidal output signal: i5'_l [Flt—1E}: #- 5-2 R; *ZIOJ.‘ 1-125111“! ~£__—_ (i1 Inna—RLS 0.1:}: a=50mA _.1—o—(4 236+s) I 1—4.5 =1~Im=o.5231111\ 0.5 V5.51: 1! __._l___023+ 2:2.229v W“ — 5 + «\$236 — 2.229 a w- - 7.007 V (W - V56} -( 10) 1. 45 17.007 - V 14.61/52; — 57.41/11; + 41.4 = o 57.4 i 1/614)” - 4(14.5)(41.4) 204.6) I5: = = 10075:; - 2)! Vs: = v56, = 2.93 v I111 = 10(2.98— 2)2 :11 Im = 9.60 111A Vas =v: — Van =7-2.9s=4.02v V551 = 5 “4.02 = 0.9M!» I12, 3 n Form =0 IQ = 1:: +Ic2 +15: 511 {c3 ={1+13n)162 Im=1c1=( a, )Im'é‘gﬂ 1+5; 13.1 E = _L 1:: ﬁn(1+3p)181 Ian :1 [1+ 5n)_ﬂn(1—:%)Im p [Q = “- + an}ﬂn(1 f-FH >151 p +a.(1—§%P)rz1 +I111 ={51)(50)(%)Is1+(50)(L0)I£1+1151 IQ = 2313.11.31.51. +45.45Ig| 4- Is; lcctronic Circuit Anal si and Desi n 2“‘1 1111111011 Solutions Manual IE1 = L592 #A =9 [(3 =1.534 EA c:=(50)(:—lﬂ)(l. 692) => J's: = 76. 9 ,uA 11:3: (51)(50)(: —?-.)(l 592) =- fg; = 3.92 mA Ecuuse of 1-.” and Z. aeglcc: disc: of r11. Then neglecting ﬁn. 702. and run. WE- ﬁnd V Ix = ymavrl +gm2Vr: + grnLV-u + 1"”:2 Now '11-: = r,” VX. V11: 5 gmxvszl'n: 7:1. + z and VHS = (9m: le + ﬂmIV-rz)frr.1 = [ya-1 V1" + §m2(§m1 Vn fn-zﬂfwa Va: ‘—' ( r“ )iﬂml + gmlgml‘rf‘llrrl ' VX 7-,; + Z v.1- _ £—~J—ﬂ1:f;5;"= v and M1»: = 91M (“:1 z)r1a-Vx = (ff::2z)vx Then +1.,ﬂz V +1312 Then R0 = V_x m + z Ix = 1+ .31 +£1.52 + (H: +3152”: 1',” = SJ—IO 0.026) = 0.169 M“ 1.534 3 = 25 kn Cha ter 8: Problem Solutions 8.36 Thou Ra = 169 + 25 1+{10} + (10){50) + {10 + (10)(so}](50} 194 no = 25.911 =o.oons kn orRo = 7.4611 n. Neglect base currents. V33 = 2V9 = ZVT ln (Iai“) Is , - -: = 2(0.026) In (°1—“L) 3 V53 21.231V V351 + V55: = V33 I51 = [5: + [c2 13 [ :1 = _L a: ca (1+Bpjfm __ _ I3 [an — Eula: — ﬁn(Iﬁ:)-IE3 _ 15 [£1 — IE3 +ﬂn(l-:£B:)IE: I31 =Im[1 +5n(1—:l§;)] (I:B~>m=(—~+ﬁ raw __.,. )1 5: +5; V35: = V? ln [Ia—1. Vsaa= V1" ln £53 15 Is 31 20 (1.91)ch = (2—0)Ic3[1+(100)(§~ 1)] =I 3-]- 100:1010'! c: 20 + . a ea {c} = 190.05Ic3 VT 1n (l00.05fc3)+ VT In (Jr—cg) :-.. V33 I: Is I v, In (ﬁgs—IE) = v“ 5 100.0513” _ V35 7r) 1c: = —-[-—,—‘~_m 05 m(--V§B) = m Then In = 0.5247 BIA 8.37 Now Ic| =10!) 05Ic3 = 50 mA = IQ] Ica= (100)(: Law. 5247) = 49. 97 mA— - IE: 10 b. :10 V =:- ' n:— =0. 0 = ”0 HE: 100 1 A {Cl _ 10E) :31 NE 1 mA 4 x 19—“ V33 7-: 2(0.026)ln (w) = L269 V 0.1 V251 = (0.026)ln 10-” = 0.7134 V53; 2 1.269 — 0.7184 = 0.5506 V _ 9.5595 _ 1c: =10 ”exp(ﬂ‘026 ) 20.15: mA 2 2 —--_ Ha _ [10) _ PL—RL—IUO :PL—IW Pg: = ic1 415-3: = (0.1)(12 -10]\$ qu = 0.2 “I Pg; 2 iC'J - Vac: = (0.15?)(10 — [0.7 -12]} =- an = 3.34 mW ic; =[100Higﬁ = (100)“).157] =15.T mA Pg: 2 it: - ”CE? = (15.7}(10 - [-121) = P9, = 0.345 w 2 x 10"” => FEE =1.71195 V V551 + V133: + V531. = V83 —3 a. V“ = 3(0.026}1n (lo—KL) ‘- 1'rlli(£I---)+‘V-g-1n(Ic—_3)-i'-V::-11'1(I-E--_J) = V33 I5 I; I] , wm[“]=ns 99!; .1 (2221} 1C; =SHISVCXP 1].}- / 1.74195 = 20 2 "2 ’ ( N “1° ) °‘p( 0.026 ) Ic: = 0.20 A. Icl :10 mA. 1c: 2 0.5 mA 10 x 19") 1'33; = (0.026)“: (W :9» V25: = 0.58065 V Electronic Circuit Analysis and Desigg. 2nd edition Solutions Manual 0.2 v35: = (01126)]! (W) :5 Vin = 0.6585 V 0.5 x 10-3 V53; = (0.025)!“ ( 2 X 10—12) => V; E! = 0.50276 V __ 1 %:_l VB} ”- P‘*1°“’-2'g.-2 a => V3(ma.x) = 20 V R]; 20 :P‘=20W 2H ioImu) = ‘53- = -—1 A in + ic; +1.5: = —io[ma.x) = 1 A . i3. 1+,3p) £c_‘.(1+ﬁ)__ “n+5" (3, +5" -——-—E -1 . _1__ 1+5, i. 14.3)}: = ""[Hﬂa ( a. )+{an ( a. 1 ..[1+;.-(§)+(.%-2)‘] Ec5[1+ 0.06 + 0.0035] = 1 :19: = H.940 A ig, = 0.0564 A ii: = 3.38 mA ic: = 2.82 111.3! 2.52 x m-3 _ V38: -— (D.026}1n (m) — 0.047? V Van + V332 = 1.74195 — 0.5477 = 1.1942 1:2 1c: v _ = . Tln(5nfs)+vrln(!s) 11942 Icz = '45-. - 15:1,”!!! (101:2?) = v’zoums) ma 16": = 84.2 m3 ...
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chap008 - Electronic Circuit Analysis and Design 2"d...

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