chap002

chap002 - Electrgnjc Circuit Analysis and Design 2'“...

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Unformatted text preview: Electrgnjc Circuit Analysis and Design, 2'“ ﬁitign Solutions Man_£al Chapter 2 Exercise Solutions E2.l 24 - 12 — 0.6 0.10 b. Vﬂ{max) = 24 +12 = 36 V C. a. ip(pea.k) = = 114 mA v: = '24sin u! = 12.6 12.5 T1- = 31.7” wt] = sin—1 ( By symme-u'y. wt; = 130 — 31.7 = 1483" .3 — . ‘70 = x 100% => 32.1% 5.2.2 v. # T'sin{21r60t] V V, = _"'_ {R VH1 75 E23 v.=1205in(2360t]. V. :05. R = 2.5 kn Full-wave rectiﬁer Tumsmn0112 =v 052:2. W = 120 — 0.7 =1z9.3 v 'lr’r =119.3 —- 100 =19.3 V V... _ 119.3 _ arm; Era—unis >< 1051(193; C = 2.06 x10'5 = 20.6 x 111-“ =:. C = we HF SOC: [32.4 v. = 50 sin(2:r60t), ‘11. = 0.7, R =10 k9 Full-wave rectiﬁer v... _ {50 -1.4) 2fRV. " mama x 104m) C: 2.025 >-<10"5 =2n.25 x 10—5 => 6' =20: 14F C: E25 Using Eq. (2-10) _ 21. _ 2E4) _ a- 2.. uﬁt —. V” — v~--—_II,5 — 3.3“ 10 S Vps 5 14V, V3 :16 20 5 R}, S 100 5.6 5.6 Iﬁmu) :2 ﬁ- = 0.28 A, Iﬂmin] = 1—0-0— : 0.056 A Electronic Circuit Analysis and Design, 2"“ edition Solutions Manual fauna} VLimin) = V1.(nnminal} + I;(min}rz __ Was-(max - Vz I; max - . - vpstmn’ _ oﬁvz __ “JV-Pﬁmu) = + Efﬁziminiiilj) _ |Eps§ming—Vz[!;(min) [aimin)= 13°“ —0.:so Vrsimin) —- 0.9V: - 0.1Vps(max] _ o 0585 A _ [14 — 51519501410 * 5.6](56) . . _ ' , _ " 10 — (0.9)(5.6) - (0.1}(14) Malina—6:21ttfgassssiu'“= 1688 _ 2352 - 246.4 Va Res = = 0.143 =91-L3‘P'g " 3.55 ‘ Izhmur) = 591.5 mA - Powurimin) = Iz(max) - Vz = (U.5915)[5.6) Power = 3.31 W Vpsimu! — V2 14 — 5.6 132.10 R‘ = Izimu}+ Idmin) = 0.5915 +0355 __ 3.4; ‘ 0.6475 R; z 139 152.7 I: = Kai..1L R, For V,,(min) and Idmax), then 11—9 I: (min) = (Minimum Zener current is zero.) For V,s(max) and 14min), then -0.1=O 116-94330“ £2.11 Izimax) = The characteristic of the minimum Zener current being one-tenth of the maximum value is violated. The proper cirwit operation is questionable. 132.8 fz = -——-—V”(m:)- V2 — A,_(I'rw.x) 50 10-9 0.0153 -1L(max) 0r IL(max) = 35.4 mi E19 % chulation = “M — “(mini Vdnominal} Vdnominal) = 5.6 VAL-nu) = P},(nominal) + Iz(ma.x)rz = 5.5 + {0‘5915)[1.5} = 5.457 st u Unde andin Cha ter 2: Exercise Solutions v, =o.7v E1” Foruy<5. D2011 =>VB=—5V=>V3=4.3V Vu=--0-5V. 1131:0- Dlturusonwhcnvr=15=>vg =l.8V 1m=f= wjjm=r=q37mA Aug 1 R2 1 "’ '7' =- -—--——-=— For H; > ..a, A“ 3 => R1 + R: 3 E2 I6 \$R|=gﬂg ' (a) 132.12 For 1’, = D, vﬁmax) = —2 15" Now, Ava = 8 V , so that va(min) = -10V 1-32.13 A: #5 goes negative. D turns on and yo = +5 V‘ As us goes posme D turns oﬂ‘. 35 Output. 5 sqwe wave oscillating between +5 and +35 volts. E114 o W =4l4’ I: 1 11.4 9.; = 0.5895 mA Set I = 19‘ v; = 4.4 - 0.6 — (ﬁ.5895)(0.5) m = 3.505 "U. la H Summary: D 5 m 5 3.5. yo = 4.4 For In > 3.5. D2 tum: of! and when u: a 9.4. 9¢=10 Electronic Circuit Analysis and Design, 2"“ edition Solutions Manual E118 I. In. 2 vie-‘1’ A 6.4 in" I,” = (o.a}(1.s x10"”)[ “ mus 3‘10"”) (“'5‘ I35 =12.8 111A b. We have Bo = {12.8H1) =12.8 volts, The diode mus! be rovers: biasad so that Vps > 12.8 V0115. 13.2.19 The equivalent circuit is +5»; .I. + :szh'iv r‘ 2.559. \‘EMY :I R ﬂ-IV 5~1.7-0.2 I——-—r!—+R—-—15 mA 5-l.T—U.'2 3.1 Tf+R-——15——ﬁ—-G.207kﬂ =2D7ﬂ R=207-15=>£=|§29 Electr nic Circuit Anal sis and Desi n 2"d edition Problem Solutions 2.1 13:: Rs 1% V,=0.6V,rf=209 FDTVI=1ﬂV, M33( R R+rf ( 1 1+0.02 Chapter 2 2.3 )(10 — 0.6} )(9.4) 2.4 olutions Manual (:1) v,(max) = 1% = 40V (b) PIV = |v,(max)| = 40 V (c) :f —r I L l l I =--- rdr=—-— 40 ' xdx va(avg) give” 231'; 5m 40 g 40 40 =§[-cosx]o =E;[—(—l-1]]=— 01' va(avg)== 12.7 V (:1) 50% yo = vs - 2V1. => ps(ma.x} = vufmaac) + 2V, 3. For Va{ma.x) = 25 V =# uﬂmu} z 25 + 2(0.7} = 26.4 V N. 150 N1 N; " 26.4 a F; _ 6‘06 b. For w[ma.x) = 100 V : y5(max] = 101.4 ‘v' N,_ 160 N, _ _ N; ' 101.4 a T3 - “'3 From part (a) pm = 2v5(max) — V}, = 2(26.4) —0.7 or PW =52.lV or. from part (b) PIV = 2(101.4]-0.? or PI V = 202.1 V Electronic Circuit Anal sis and Desi n 2“Cl edition Solutions Manual 2.5 a. when) = 24 V : :rﬂmu) = 24 + 2(O.T) uﬂmu) = 25.1} V __ 25.4 _ _ _ F vsfrms) — ﬂ : palms) _— 1:.96 \ . _ VM __ VM 1" b' ‘ :2ch =’ C ‘ Hm: '24 _ C=W=C=266IﬁF . v"l ‘ C. iDJn-u: = TEE—(1 +21? g—E- . 24 24 'DJn-I — +2FV mm. = 5.03 A 2.6 (:1) vs (max) = 24 + 0.7 = 2447 V v max vsirms)= ’= vscmamsv V V 24 V zangu-za-C: M =--————....._ M ' 12c m»: (603(150)(05) 01' C=5333yF (c) For the half-wave rectiﬁer K, V 24 24 I =— [+4 “ =— 1+4 — “m [ E 150[ “42(03] Or iD_M=10.OA 2.7 2.8 4. (i #7? 'L): ‘1: 5 _ Vé:[2|/ US A era—u 7h :11 \ vs (r) = 24 sin at Now intavguﬁifow We have for x, S at 5 .1:2 , 24sinx—12.7 ID = ———-—-——— R To ﬁnd I. and 3:2 , 245ian 212.7 :61 =0558 rad x2 = :r—OSSS = 2.584 rad Then . 1 " 245inx-12.7 'a(m’g) = 2 = Of 2:21.83—LOQE: R R -——-—--—— Fraction of time diode is conducting = x2 -x1 “00% = 2584-0558Kwo% 21: 211' or Frac ' n = 32.2% Power rating 1' x . 2 Em 2R1?”=ﬁjigdt=£J[MSInx—12.7:l (ix Ta 2:“ R N94)1 sin1 x — 2(12.7)(24) sin x +(12.7)2 ]dx [Mfg — “14hr —2(12.7)(24](—cas.r): +{12.7]2x‘::| For R: 1.19 9 . then PM =17.9 w Cha ter 2: Problem Solutions 2.9 2.11 R=£=lsoﬂ Foru.>0, (V,=n) 0.1 vs(max)=vo(max)+V, =15+D.7 Or vﬁmax): i5.7V Then 9 RH U. vsfm) =§= 11.1V '- no R New L LE: £=los _ N2 11.1 N2 .0 = (_R_=IIm_) RaIIRL + R; V=Vy =C_ V” __ 15 RR 2 156 ' 2ch ZIRV, 2(60)(150)(o_4) 1" L - 1:25-13 — - m or an .—. "I. = 0A3“ 1:. PW = 2v, (mag—VF = 2(1 5.7)_0.7 or PIV = 30.7 V 2.10 For u.- > D V; = 6.3 V R.- =129. V: = 6.8 6.3 —4.8 a. I: = 1—2 = 125mA IL=I1-Iz=l25—Iz 25<IL<120mA=>40<RL<1929 b. Pz = Isz = {100)(48} =# P; = 430 mW P; = ILVu =(120)[4.B)= P: = 576 mW Vq=0 2.13 Village“: RL+R1 =Pi RL ‘ 1 *- II= =H'r_=4ﬂrpﬁ M) = g: -y[ RL-l-Rl 2 I 10 L—J—BE=>I; 226.3111A 'U. fz=If—IL=¢>Iz=18.TmA 19 Electronic ircuit Anal sis and Desi n 2"“ edition Solutions Manual b. Pzimnx) = 400 mW => 1mm) = “10% = 4a m NW 6 6 . I =-—=D.012A, I ' =——=. =. 1.5mm) = I: — mm“) = 45 — 40 Arm) 500 Jun“) 1000 0006 A . 10 Now =# I = 5 A = — nfmm} :11 RI. R _ Vps(min)_vz =- R; = 2 m " 12(min)+IL(max) For R‘ = 175:) c" 15 6 it _-. 5?.1mA I; = 26.3 1:: 30.3 mA 230=W= 12(min)=0‘030 A z{ma.xl)o— 40 m.'\ a ILLn'unJ = al.1—— 40 = 1r.1 mA The“ R; = m a R; = 535:: [2(max)=0.l+0.02=0.12 A and 2.14 R = VPs(max)“Vz a. From Eq. (2-23) 12(max)+IL[“ﬁ") Mm“) _ 500[2o -10]~ 50[15 —- 10] or ( ) ’ 15 — (as-mm: —(n.n{201 230:»—-——V” “m ‘5 v =413v 5000 450 D.12+D.006 2’ ‘ 4 Iz(ma.x} = 1.1875 A D 2.17 130nm) = 0.11375 A Using Figure 2.17 . 2-21b a. Vp5=30£25%=>15SVpsSZSV 20-10 R. = =3 R. = 3.039 For Vps(min) '- I=I II+I 2' 20:25 A 1;. P2 = (1.1375){10) => Pg =11.9 w ’ mm" “mm 3+ “1 P; = I;(mu)% =(u.5){1o)=> P; = 5 w R, = 152751” => R. = more _._.....__ I _ 2.15 b. For Vjﬁsltmu) (a) As approximation, assume {2(max) and [2(min) => IILmax) = 25}; m =- Mini-x} = 75 “LA are the same as In problem 2-14. For Idan 2 D a Izmu) = 75 “1A %(max) = Whom) + Iz(ma.x)rz Vzu = Vz - Iz'rz =10 — (0.025)(5} = 9.875 V = 20 + (n_453}(2) = 20.906 Va(ma.x) = 9.875 +(D.01'5)[5J = 10.25 Wﬂmin) = Vo[nom} + Iz[min)rz Wmin} = 9-375 + (0305)“) = 9-90 = 20 + (n.0453)(2) = 20.0905 M 20.905 — 20.0906 ._ Mi ._ b. 95 Rag = T x 100% c. % Reg — Vanuatu) x100% => 5’6 Reg — 3.5% =- 96 Reg = 4.08% 2.13 2.16 From Equation (221(3)) . V (min}—-V 24—16 _ VL(max)-V,_(mln) Rf =_rs‘__*___g,__,=__ 93ch -— WXIOO% [2(m1n)+ [L(max) 4D+400 or = VLKWm)+ I‘zAEImU‘)!’z *(Vdmm) + 12(mink) 12,. =18.2 Q VL("°’") Also = [12(max)—rz(min)](a) = m V, ziz c = Va 6 ZIRC 313V, So RaR,+r,=132+2=2o.2n Iz(ma.x)—Iz(min}=0.l A Then 24 C = 2(60)(1)(202) :> C=9901ﬂF Cha ter 2: Problem Solutions -———-———.__._____.___.—E.____.___.____ 2.19 V: = Vza +1211 Vz(nnm) = B V 8 = Vza + (u.1)(o.5) => Vzu = 7.95 v 1‘ = Vs-(mu) - Vz(nom) _ 12-8 I R; _ 3 Potlg=02A\$Iz=1JS3A = 1.333 A For I}; =1A\$Iz=0333A Vﬂmax) = Vzo + Iz(ma.x}rz = 1.95 + (1.133)(c-.5) = 3.5155 VL{min) = Vzn + 12(min)?z _= 7.95 + (o.333]{o.5) = 5.1165 AV; = 0.4 V (JV; 0.4 %R=g- Vofmm) — —5-=>%ch_5.0% =—-—VM :3: = V” K. C ZJRC 2;}in R=1§+n=3+05=35§1 Then 8 C = 2(60)(35)(0.8} 2.20 (a) For 405 u, so. both diodes are conducting: v0 = 0 For 0 S v, S 3, Zencr not in breakdown, so f, =: 0 . v0 = 0 => C: 0.023817 For v, >3 (b) For v, < 0 , both diodes forward biased . 0 - v, _,I = ID .15“ v, =—10v,iI =4,“ For v, >3, fl = V333 . At v, = Low, 20.35mA 2.21 (a) t K [K 2K V: +l'5 V1=%x15=5V=>fDrw55.7, rig—2w _ , 15—V V w... 2 1=T1,y°=V1-|-B,T 111—91) 15- vu—0‘7)__1Iu-3-7 1 2 1 V! 15.7 0.7 (1 1 1 _ ._.... a: - - — z 2.‘ 1+ 2 4-1 lr't: 1+2+1) "OI 9) 1 5.55 = 119(15):? #9 = 5-3-14; + 3442 E + ((3)51, = D for 05v, \$5.7 Then ID— v, — [it + 3.42) v, —-v0 _ 2.5 l 1 or . _ 0.5u, — 3.42 {D 1 Forv,=15. in=558mA Elgtmnic Circuit Analysis and Design, 2'” edition Solutions Manual ' . 30—10.? For Inf—30V. 1.. 100+“) 9o =i(1ﬁ)+10.7 = 12.5 'v' = 9.175 A 5;? .1.) v‘) ’5' f‘ u. 2.22 11.5 __ um 20 (a) For Duff, v0= 56 20)-1o= 3.33v n Then for v, S 3.33+O.7 =4‘03V =9 v0 = 333"" For v, > 4.03 . v9 = VI -0.'?: do For v, =10, v:JI =93 2.24 + 5 - ‘UI "‘4'! I no R=E£K \$1, = 0.5 v w = 15 sin .41 (b) For v, \$4.03V , in :0 For v, >4.03, fp+ﬂ=m 10 20 Which yields in @551” 41605 For vIr =10, in =0395mA 2.25 a V... = 0 2.23 can-l4 ha ter 2' Problem Solutions b V... = D 218 ¢ U; -—1 U4 10 uh L VI "‘ __L a. ForV,=0=>l/'.=2.TV h. For?’.,=ﬂ.7V=»V,=2.0V VT = 0.6 2.29 C -—l 1.: H.4- I’I "’ T." V 5 J“ 226 O . . nc posmblc example IS shown. 230 L will tend to block the transient_ signals For circuit in Figure P242703) Dz will limit the voltage to +14 V and 43.7 V. (i) For V, = +3 V Power ratings depends on number of pulses per second and duration of pulse. 2.27 V, = 0 I. 1" 4o lectronic Circuit nal sis anch i 2 ““ edition Soluti ns Manual 2.31 For Figure P2.27(a) 2.33 2.34 1.0-0.6 I” IDI—m:IQ:—D.94MA IEZE‘O v. = [51(95): v9 =3.93 v 5—0.5 b. In: - m=1m —D.-HmA IQ: =0 113 = 11:11:95) : Vg = 4.18 V c. Sum as (5) Q - 5 10 = 2 (0.5) +0.5 + I(9.5) :5 I = 0.954 mA :4, = {(9.5) =- Ivll = 9.15 v [m = ID; ='-;-: [m =IE; =0.+8‘2 mA 0.. I=Im=fm=0 Vg=lﬂ b. 10 5: M95} + 0.5 + 1(05) =5 I=Im =D.94 mA 1Q1=U I4; =10 - {(9.5) =5 V9 = 1.07 v c. 10 = I(9.5) + 0.5 + I(0.5) + 5 :5 {=1}; =0.44n1A 1m =0 V. = 10 — I(9.5) a 1-3 = 5.32 V d. 10 = I(9.5) + 0.5 + \$10.5) :5 z: 9,2“ :95 [51:15.1 = é=>IQ1 =IQ2=U.432111A 'v0 = 10 —J’{9.5) =5- V2 = 0.042 v L M=VE=OQDLDLDLOII Vq=IL4V Iii—1.4 = =0. I 9.5 . -D.6 [DI 2132:340—sﬂIQ] =I23 =7.6mA IQ; = In]. +Ipz — I = 2(7.6) - 0.589 =- 12; =14.6 mA h. V1=W=5V DimquomDaoE 10 = [(9.5] +0.6 + %(0.5] +5 =5 w f01=Ipr=é=>Ipl=IE3=DJEGmA IQ; =0 W. =10 — I(9.5)=10 — (0.451)[9.5) :5 V9 = 5.12 v C. K=5V.V:=UD1OE.D2,D30[1 v9 = 5.4 v 1-.— m i“ =5 =0.' 9 9.0 {m = =5- [21 = 7.5 mA 0..) [pt—=3 ID: mA (1. V1=5V,V:=2VD1OE,D2,D:OE V9 = 4.4 V 10 — 5.4 = = . 0 I 9.5 - 4.4 -0.5 —-2 = —— 1 =30 A I“ 0.5 : -D3-'-—m in; In: = In: -I=3.E -|].539=> IQ; =10}. mm 2.35 (a) DI MID! alelon So 192:0 lO—O.6-(-O.6)_ 10 NOWK=--0.5V, I'm: RI+R1 \$539 In, =125mA Ifl =10~0.5—(125)(2)=> 9; =55 V ,8, .wﬂm ID, = I,“ - [m = 2.2- 1.25 =. ID, = 0.95 m (b) DII on, D, 011,13, oﬁ' Sofm=0 V1-4... __ R1 6 or _ [0.20.833MA I..=—4'4'('5)=£=o.94m gm. 10 ID: = I,u -1m = 0.94 —0.333 :5 ID, = 0.107 mA V, = .7an, -5 = (o.94)(s)—5=> VI = —0.3 V a ter '_ 1c lutions (c) All diodes are on P] =4.4V, V1 =_o.5V [0—0. —4.4 IA =05mA= RS = R, =10m I I," zols+05=lmA 2.1%: R2 R1 = Skﬂ -O.6—(-S) In2=1-5’"A=T=> 33:293‘9 vo=uz for —4.65<w<4.65 2.36 2.38 For v, small, both diodes off L 0.5 v0 =[ols+s)v, = 00909"I When vI — v0 = 0.6 . DI turns on. So we have v, — 0.09091), = 0.6 :3 v, = 0.66, v0 = 0.05 “W For B, on v, —0.6-va_ 1:, -v0 v0 . . -————-—-—-——+—--——=—— which yields 5 5 05 v _ 2v, -0.6 9 12 Whenzva = D2 turns on. Then 1:, = 5 HE, R: = 10 k9 o.6=-—V'1'2—':>v,=3_91/ mmm on =v2=o Now for v, > 3.9 rm = 10;“ — Elli-El = 1.35 — 1.0 m+vl'v° =i+h22£ In: =D.BEmA S S 0.5 0.5 Which yields 1:. R: = 10 1:11, B; = 5 kn. D: 0E. D; an VO=M;FOI' v,=l0::-vo=l.15V min 22 1o—o.7—§-1D1 Va = IR: — 10 => V! = —3.57 V +‘OV 2.39 15-f%+o.7) _ fawn-FEE 10 20 2D 15 0.7 0.7 1 l 1 £0 ﬁ‘ﬁ'ﬁ~%(ﬁ+ﬁ+ﬁ)-“(3€) 143:6.975‘1' -\ov For In > a. when D: um: 03' 10 —0.7 I _ 20 W = IUD k9) = 4.65 V = 0.465 mm Elggggggic Cirggit Analysis and Dcsigg, 2"d edition Solutigng Manual 2.40 - V 1., 10K VI ° V I04: VI _. V1. IDKID 10K :— a: 5.. V1=15V,V:=IDV Diodcaﬁ V.=7.5V.V.=5V=>Vu=-2.5V Inna b. V1=10V,V5=15V Diodeon Va-Vb_Vb VI VI’VL 1o “'10 33+ 10 15 -15_ L 1 1 1 10+1u'v" 1o+1o)+“(1o+ﬁ) 1 1 "D'5(ﬁ+ E) 2.62 antic) => V..=s.55 v 15—6.55 6.55 In -- l—o-l—oafn —0.19mA =K=W—M V2 =0.6 V 2.41 93:0, D10EDaon 10-15 I: 15 = 0.5 mA u.=1o—(o.5)(5)=>uo=7.5vrorn<w<7.5_ Fww=30V. D: oﬂ’.va=10V Dem-minch whmV,=10 I_Pf-2.5 25 V, =10=I(ID]+2.5% I=0.75 mA 9: = (0.75)(25) + 2.5 = 21.25 2.42 I. V = V = CI h. V = 4.4 V Vi] = 3.3 V C. V9] = 4.4 V, VI: = 3.3 V [F Logic “1" It:ch degrade: as itgocs draugh additional logic 311:8. 2.43 I. Vm=Vgg=5V b. mam 21221-2; c-MM Logic “0" ligml degrades u 1: goes Ihmugh Iddiﬁnnal logic gates. 2.44 2.45 2.47 (V: ANDVz) OR (V; AND V1} 111 - 1.5 — 0.2 —_- -——--—-—- = 2 = 0.012 I R+ w 1 mA 3.3 g — = 691."ﬂ R+ 1” 0.012 ' = .7 1o - 1-1 _ v, I‘ u 75 a VR=IV. I=0.8mA V155 3 1+ (0.3){2} var.- = 2.6 V 2.48 In; =1111¢A 0.6 x 10" = (1)(1.6 x 10‘“) (1017}A A = 3.75 x 10“ c1112 ...
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chap002 - Electrgnjc Circuit Analysis and Design 2'“...

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