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chap006 - Electronic Circuit Analysis and DesignI 2nd...

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Unformatted text preview: Electronic Circuit Analysis and DesignI 2nd edition Solutions Manual Chapter 6 ExerCisa SOlutionS C. If. = 0.4 sin a)! = #4, = -[3.9|(0.4)sinut 56.1 lid, = -3.56 sing! g_ =2K—(VG£_V1N) and 'AI V951 = 6 «- 3.56 = 3.4-1 1 _ K {V )2 VGS1 = 2.39 + 0.4 = 3.39 D - n 63‘ TN 0-75 = 05W“ 433): =3 V53 = 2.025 V VG“ “ VTN = 3-29 “ 2 = 1-2? g =2(05)(2.025-0£) a g = [22 mA IV 50 V05! >V631‘Vm =5 13mm EGA- "- : [Mn (Va: " 1"1'.'tl)1].1 = [(0.01)(05)(2.015—o.s)’]4 = 5:133“) 7:12-IDQ{E}=’!iDQ=0-533 [ILA 109 = [90,56 ‘1‘”er a. Van = Van *15030 E61 0.33: = 2(1/55 - 1}1 =. 1/5; = 1.55 v gm=1'.it'u(‘.ffiJ 4’“) and ID=KI(VG$‘V7-N): _.__...._ ,1 . =2K v —v =22 1.65-1 =VcS-Vm = 7?— and b g" r[ 56 l TPD ( X ) :3 gm 2 '16 mA/V. rg = 3c r, =[RK.(VG, vmfl‘ 4152]" A, = ”—° = -9mfio = -(216)(G) r,=[(oals)(2)]‘= n=333m ”54.11555. 56.3- 136 5 _ = V — 2 I I' [be K11 G: TM) 109 = Kn(VLT-S- TN): =VGI-Vm = ?’Q~ 0.4 = 0.5(Vas — 2}’ => V2: = 2.39 v . VDSQ = VDD - IDQRD = 10 — (0.4)(10) g“ =2K.(Vm-Vm)=2K. 10$ :3» Vpsq = S V K b. g_ = Mania, — v”) = 2(05)(239 - 2) a g... = 0.39 mA/V 55.6 r°=[A!DQ]-l, A=0#ro=ou __ Y — fl— , = ."—°_. = —g...Rn = —(o.59)(1o) 2:124”! ”is 1’- 5 A, = —fl.9 Electronic Circuit Analysi§ and Design, 2'“ edition Solutions Manual 040 = ' :9 = 0.104 a”) '7 231035)“ "——— g_ = z (Ejm' = 2,}(05)(0.75) = 1.22 mA IV For (a). g... = 33 = (1.22)(0.153) = g_,, = 0.137 M IV For (1)). gm, = (1.22)(o.104)= g“, = 0.127 m IV 56.? 3- fog = Kn (V65 " m)! = (0‘25)(2 ‘03): => I52 = “.35 [11.4. Vpsq = Van - Ipqflp = 5 - (0.35H5} => 1352 = 3.2 V b. g" = 1K,(VG, — m) = 2(0.25)(2—03) => 9... =0.6 mA/V. re =ac c. A. = 1 = -§mRD = —[o.s)(5; 56.8 v. = 9,, = 0.1 sin wt i4 = y..u,, = (D.6}{O.]Jsinwt ii = 0.053inwt mA n, = (—3}(D.1}sin.ut = -0.Ssin .ul “min = IDQ +1} = 036+ 0.065inwt =ip EA. up: = VDSQ + 94. = 3.3 - 0.3sin‘r! = Va: 56.9 1"ng = 3 V and for; = 0.5 m.-"L aac=5013=39=4m I” = K,(vm -|v,,|)1 0.5 =10}; -1y2 :: v5.5 = 1.7.1 \’ => V5.3 = 5 - 1.71 = V95”- : 3.29 \- -4-u = “gm RD g.=2 Igloo =2 W05} 9... = 1.41 mA/\" A. = —(1.413(4J:.4. = -5.54 A, = a = —w.4 = _D.465in-..~:I = _5_“ Vi I". II'I => v. = 0.08168inu! 56.10 1 a. Vsa = 9 - Ians. [are = KP(V5'G 'lvfll) Vsc = 9 —(21(1.2)(vsa — 2)” = 9 - 2.4(V525 - 4Vsc + i) 2.4vs’a — 3.51/55} 0.6 = a 3.5 :L- \/[3.5)’ - 4(2.4}(o.5) 2(2.4) v55 = 3.51V. Ipq = 2(3.51 — 2)1 a log = 4.56 mA Vsoq = 9 + 9 — IDQU-z + 1) = 13 — [4.56)[2.2) : V559 = 7.97 V V55 = V], = ngsaRo A. = —ngo = -(6.04}(1) => A,: = -6.D4. H" , Lu ? =23 K =(2o)(25)=soomrv= '55: g+15=35V=>V5=-25V Au="§m-RD y... = 2(o.5)(2.5 — 1.5} = 1 mA/V ForA.=—4.0=~Ro=4kfl. VD = 5 - (0.5m) = a v = Vpsq = 3 — (—2.5) = 5.5 V E612 3.. mm Ra = Vcs = Vns => transistor biased in SEE. region In = K.(Va: ‘me = K-(Vns ‘Vm )1 Va: =Vw ‘Ian =vm— K.R.(vo. - m)‘ Test Your understanding Chapter 6: Egcrcise Solutiong V95- =15 — (0.151(1unvn; — 1.11)1 K = 1;, + g... V..Rs = V“ = v: _ 1+9 R- =15 — 1.50/35 - 16%;; + 3.24) W: __ —9' V RD In a — m 5H 1.51/3.S - 4-4V“ — 10.11 = o g_ = 215.0%; —v,,,) = 2(03](-1.35+2) V 1.4:h1/(43fl1 +(4)(1.5}{1u.11) D _ s _ = 1.01 mA/V 2(1-51 A 3 ya = —ngo _ —(1.043(T.331 =5 VDSq =4.45 v ” V. 1 +ng5 ' 1 +(1.o-1)(4) hm = (0.1539145 .15}2 a. 190 =1.os mA =:- A. = —1.38 6.14 b. Neglecdng client of RC,- : E a. 5: [MRS +Vm. and A» = -!m(RD"R1) Ina = K,(Vsa +1117): 3" = 219M: 4!“) = 2(0.15)(4.45-1.3) =, g z u 795 mAfV 0.3 = 0.50%; + 0.5)’ => V5,; = 11.455 v A. = -(o.795)(10[15) =:~ A. 2 4.55 5 = (0.5185 + 0.455 = R5 = 5.57 m Van =10 -— [chLe + RD} 1:. Kg =» establishes Va; = V05 =5- ésscntiafly no cfi'cct 3 =10 _ [O‘BHS‘fij t‘RD} on small-signal voltage gain. RD = W2: :5 R2 = 3.08 kn V V IDQ = 0.3[2 —- V55}: = -—55— = i 3.2(1 — {V513 + V5113) = VSG 3.21%; -13.avsa + 12.5 .—. n 13.3 t \/{13.s)' — 4(3.2)(12.8) ' = v = mv, = - "v.- R Vsa 26m) 0 9V 9{Rni1rol 9' ( ollfo) vs; = 1.35 v =9 19¢, = 0.3(2 — 1.35:1 A» = 7° = -9m[Rollro} = M g_ =2K,(v,a +11") =2(05)(o.455+os) = 1.21 mAJV b. Vase = Von - IDQKRD + RS} _ 1 _ 1 ro-m—mfi—SZJM} 6 =10 - (Hamlin + 4} A. = —(1.27}(3.oa|152.5) = A, = 4.12. _ 10 - (11333)“) - 6 _ RD — 0.338 => RQ —- 7.53 1:51 56.15 C. V0 = Univ-giro v; =v,,+v;,==.v,,=v.—La So 1'3 =9mrufl’. -V21! 1 _ 39._ gmrn _ (411591 ‘ " " 1". 1+ gmro 1+ [HIE-0} E cctrouic lrcuit Anal sis and Desi n 2“‘1 edition Snlutibns Manual v ' . 1" +§mv'; = l'_: Ind Lil = —“, h=gmu+—=Ro=f0”—=501|i :9 £9 E 0.25 kn WW! 35 =4 kflza- AN 3 9”("0| RS} 1+9mfrnllflsi (43(3‘TI = ' = .7 fl . I, = --—--— TDIIR; 30"“ J k s ‘i 1+[4H3-1‘] = A. = 0.937 15.6.16 V1259 = V00 - IaaRs 5=1 10—(1. 5)Rs=>RE=3.33kfl 1m =K_V{G, —v,,,) =5 15: (”(1113—05) Voszzmv =va_v5_vc..5. R: R: V = _ . 314-32) DD 400 ID So R; = 280.8 [(0. R1 = 1192 1:9 =§VG=T.02V=( Neglecting R5.- . A” = gmtflsflrn) 1+§m(R-5'|1To} 1.. = [2190)" = [(0.015)(1.5}]" = 44-4 kn Rsllro = 3.33II44.4 = 3.1 R10 3,, = Lila!” =2 (1X15) =25er {2.451(32) _ v- =Wfifu—DJB‘I 9—,." Rsllr "2—15 1333115“ =0. I1108||3.1 =2 R4: = 0.35 m 55.17 a . 9.3 __ V“ = (R. fay“ = (mu-9.3)“ =0.531\-‘ Im=K.-V(sa -|an) = ‘90”: VG‘IVWI} _ 5 -— 1':- .. R;- Thm 10.015102 — 0.521- 0.5)” = 5 _ 2;. 20’s - 1.151)1 = 5 - 1/: 2W3: ~ 2-752)”: +1.90?) = 5 _ 0;. 2V5“ — 4521',- — 1.19 z 0 1.52 2 Jun)” + 4(2)(:.19) 2(2) Vs=2.50V=rqu= gu=zm_ W: 0394120111! v. = 5—5'25=0.5mA A»: 3-K: . R1ER2 l+g_R, !«'.',HR,+R£r 4 . . = (039 15) ‘ 707u93 :3 5:11.770 1+(0594)(5) 70.7fl93+05 — Neglecting R .. A, =08” mzfisl 0—m=° 51" 0-594 = R5 = 0.915 m B618 (:1) gfl=2 Knlm=>2=2 K_(DB]=> K =l.25mAlV2 Kn =#"—C"vE=L=25 (0.020)?) 9, L L 50 E£=¢525 Jr...— 1 DQ=KIM Wm): I’D-8:1-25‘V05'2}: =:» 1‘9: = 2 B \' 1:. r0 2 ”mar: [(.)001).(0 s )1“ = 25 m ___ 9m[mi|RL‘1 ' 1+QM(FD1|RL) rDHRL = 1'35i|4= 3.55 _ (2113.30) __ A... — ‘—'"_1 + (21(3 00 =. .1. — 0.556 Rfl=gl— ro=-1125=>Ra~0.5kn 136.19 2 92 = p(VSG ”WI?” 3 =2(vs._-, - 2)2 =- V55 = 3.22 v 5 — V 5 —- 3.22 Ipq = £556 = 3 = R5 a» R: = 0.593 kn = wool“ .-. [(0.02)(a)}" = 10.7 m g_=2 Kg” =2 (2X3)=4.9mAIV Hmfi'nllfls} 1 +9...(rol1.Rs) To‘IRS' =15.T||0.593 = 0.573 kg (2.9)(0573) ‘ A = —— .1. = 073-. " 1+ (4.9)(0513) :——-‘—— ForRL=o¢. Av: If A. is reduced by 1.0% =- A. = 0.73? — 0.073? = 0.653 Stu-(rullRSIIRLI ,1, = -——-———-—-——-— l+!m(fa1lREIIRL) Test Your undergtanding Chapter 6: Exercise Solutions LG! TOIIRSIIRL = .1: (4.9): l + (1.9): t = 9.402 = 0.573"RJ_ 0.573RL EL 4- 0.573 :1 (0.57:1 - 0.002151 = {0.402)(0.573) 0.563 = = 0.663 = 4.9::[1— 0.663) = 0.102 = R: =1.35 kn 136.20 .,.-.- 3L=005m=gm=2asmAn ll’0 “In — = R R z 2. 4 = R .. r. all 1 DIM—4+1“) [4 — 2.133,; = {2.4)(4) =- R9 = 5 kn gfl=2 Kl‘rm 235=2JK_(05)= K. =4.09 m4.le 1 log: K.(Va5 ‘Vm) 0.5 = 4.09{Vas —1)"’ =1. V135 =1.35 v :Vsz—IJSVI i-‘b=5—(0.5){6)=2V We have V“ =3.35 >VG§ —V,,, = 1.35— 1 = 035V :0 Biased in the saturation region E621 Vi! = gravidflflllfil.) and V”. = V. A» = PM‘RDIIRL] 5— V =_;:£= K rvmt: "Iva-D s - 1:25 ={11{4)(V.=a - 0.01” a - V15 = 411/515 -1003“; + 0.04} 411-c-5.41'_-a—2.=14=0 s 4::- 1/15 41 + 1411-1112 44) [be V =——————-——__. 3G 2(4) 11.33 = 1.7] V I” = ° 11'”: 0.022 111A gfl=2JK Im=21“1)(0.822)=l.81m14lv A. =11.01](2t|4] =(10111(1.311= m __ = "" I91". ‘ "—131=4I|0-°°- 9W Rm = Rs 56.22 C W K". = __’”- .. [__] =(0.020)(30)= 1.15:1:21/1'2 2 L , K_ = '” C" £[%lm =(0—020}(1) 0.020 mmvz —- - =- —— => = -3.94 A“ Kn 0. 020 53—— The transition point is dctcrmined from Kn "can — vma = van _ V1741." I "F”(vas: vmn} n2 v55: —- 0.3 = [5 — 0.8-) — {8.94HVGSI — 0.5) (s - 0.51 + (3.941(00) +0.3 ”55‘ = 1 +3.04 pas, = 1.22 V For Q-point in middle of saturation rcgion VG: _ 1.22 — 0.5 _— 2 +0.33VG$=1.UIV =E—"C" K:_[W1K“ =.(0015)(2)= 0.030mmv2 -36 K.)=(36( 0.)03o —103 ”0111!1 1.08 =(0'015)(E]1 =1 (%]. = 72 The transition point is found from Vcsg—l={19-1]‘{5NPGSI-n 10—1+5+l =2.29V 1+5 ”GS: = For (2-point in nfiddl: of saturation region 2.29 - 1 Vac: 2 -.-1=>Vr-[c=1.615 \- E624 (a) Transition points: For 14,: v0, = v“, -|vm|=s— 1.2 = 313v For M1: K.1[{Vm )1“ + hm” = Kakvm )20 + ‘12P” - v“ D] zsqvi. +(0-011V3u] .. ”[02” 1+ (0 .(01) 5-) (0- 03%)] Icctronic Circuit Anal 5i and Desi 2nd edition Solutions Manual Hing“ +(o.01)v;‘]= l512 — 03:11am“M {0.00% mg,“ +0.00l4-4vm —0512 = 0 which yields um i 0.388 V Then middle. of saturation region 3.3 - 0.388 [log = + 0.358 :3 VDEQ! = 2_CI9‘I V Knlkvcsl ' VTND}2(1+ Alva )] = K..[(Vm.)‘(1 + 12%» - ”011] flout/cg; — 0.512(1+[u.01][2.094])‘_ = 25[(1.2]’(1+[o.01}[5 — 2.0941): m[(va,-, — 0.31%].02093‘; =1.m (v55, — 0.3}: = 0.145 :3- VEE] = 1.15 v b_ 100 = Kuhn”. —o.s)’(l+(o.01)(2.094))] JDQ =.-(0.25][(0.145)2(1.02094]] => IDQ = 37.0 HA 'Fgml = m = wmtrmnrm 3.1: 2K41(Vcsa "Va-no) = 2(o.25)(1.13 — 0.5) = 0.19 mA/V —n.19 c. A. A”: (G.037)(fl.01 +0.01) =’ fl 15.6.25 1 Ho = Rs: “ —' 9”: 9m: = 0.532 mA/V, R5; = 3 kn 1 - Boas“ m—a|11.a3=§2_1.32m 56.26 I.. fog: = 211124.. Vqug = ID V J'aq:-fl.—-:=10=ER$: =>R§3=5lcfl a lag: 3 Ka2(v551 ‘de 2 =1(V::52 — if =- V501 = 141V 9 Va; = 3.41 V [0 - 31%! “than an. = ——;-—— = am =13 m For VpJ-qi =10 V =- Vs; = 3.1] -10 = -6.39 V Then R51= w =, 11., = 1.71 m IDI = .1013: - I'm)1 2 = [(1,651 - 2): =9 V651 = 3.11 \' R: 65‘ (R1+R,)£ 3 100131 R: _ 1 121+ R, “' R1 Rm 3-“ = income) — (mm) 1 => RL = 556 kn 586R; _ _ _ _ - EST :2, * 2"” =° We - 200m: —(2001(a85} => R7 = 304 kfi' b‘ Sm=21aniIDQI =2'JU‘X2) =:' gm = 9...; = 2.63 mA/V From Exampl: 6-16 A _ -9Ml§m2RDl(-RFJHRL) k |+§maiflsallRLi REINRL = Si|4 = 2.22 m —{2.83)(2.83)(3.3)(2.22) A“ 2 1+ (2.53)(2.22) => 14., = —8.06 R... = —1-|| £5: = —1-||5 = 0.353115 9M2 2.53 =- R0 = 0.330 kn E627 "' IDQI= nI(Vo:I'Vm:)1 1= Lewes, — 2)“ => V051 = V652 = 2.91 \-' E5 = 10 kn =3 V51 = Iqus - 10 = {1)(10) -- IO =0 _ _ Ra Va; -23: _ (—R: +3“ EJUD) = (3%) [10} a R; = 145.5 kn Vase: = 3.5 =- V5: =15 V =- 3.5+ 2.91 :9- Vm = 6.41 Pa: = (m)(10) =s.41 R2 + R3 = 320.3 = R: +1i5.3 =9 R3=175 kn T113513:+Ra+fla=500=fl1+175+l‘£5.5 =5 R] = 179.5 Hi How Vs: = 3.5 =- VD: = Vs: +Vanz =3.5+3.5=TV Test u nde tandin 10—? 503.9: 1 #RQ=3|¢Q 11. From Examplc 6-18: A» = ‘91!“ RD g_l = affirm = 2 (1.2)(1) = 2.19 mA {V A. = —(2.19}(3) = A. = —6.57 156.28 From Example 6-13: 9... = 2.98 mAJV. I'o = 42.1119 MIR; = 420||1so =123 kn .“ 12.1132. . "" ‘ (£11111: +3.)“ / 126 " 123 + 23 )W = 0.36314 -ym"}.1rollfloilfla1 1.; = -12331111.3331(42.11|1.1||4} At: 42.37 :3 A -—3 -....— [42.1;|1.31) =.- 415711135) 3 [L E629 V3 = [1303.3 = [1.2}(2.7} = 1.24 V V1: = V: + Vosq = 3.24 +12 =1524 — :2 RD =20—121—H3 RE =3.97kn Va: 2 In - 1055(1 — W) 2 1.2: 4(1 - V—E’i) =3 515-1: 0.352 V.» Va,- :1: (3.4521141 = 4.333 1},- = 1.;— + 12;; = 3.24 —1.333 = 1.33 v . R: _ R: _ vs a (R1+Rz){201_ (5)123] — 1.33 =9 R; =47 kn. R; = 453 kn l 1 ° = 11119.; (3.0115102) =16? m _ 21055(1_ Vas‘ _ 2(_4)(1_ 1.356) 901 - (—VF} . — l' . VP 3 3 = L46 mA/Y A. = *9mEToIIRDEIRL} = 41451037133711.” =3 1-1... = —2.37 E630 _ R: ‘ V51 (121+ R1)(VDD} Ch ter 6: Exercise Solutions v ‘ v —v 1 1nm=1955(1-_g) =5(1__G‘_.i) VP 2 172 V511 51 2 = -— --— =6 —— 3(1 2 +2) 2 73 20—1! andlpq1= 51 4. 2 “1111-1120;“TI = ($4.15) V2 20 - 1/5,: 24({4 - 7.61151 + 57.73) = EVE-1 —132.4V51 +1335.24 6V§1 — 181.4%. +1366.24 = 3 131.4 i 1/(13131’ — 4(3)(1333.24} v = 51 2(6) V5; =14.2 V = V551 =11: —14.2 = 3 V > V; SO V5] = 15.0 9 V331 = 17.2 — 15 =1.2<Vp—Q 20-16 0111qu: :Ioo1=1mA V5901 = 30 - 19011351 + Rm} =20-(1113) ==v V5199: =12 V v3. = 133.113. =111<11= 4 v = Va: Ina: = 1033(1 - -—.i = 61/}. —— 731's, + 215 = o 1/ 73:1: 1/1711” —4(5}(2161 ’2 2(6) => V5: = 7.59 V or = 5.05 V For V5; = 5.03 V => V65; = 4 -- 5.08 = -1.03 transistor biascd 0N lcctronic Circuit nal is and Desi 2nd edition Solutions Manual [Baa-J Elf—E: [DQI =1.11mA b Von=2u-Vsa =20-5.05:V253 =1‘L9V I} h. V32 = PMIVIMHDI = “Emivu‘RDl V0 = gm2V¢.:(RnIIRL) V92 = VII: + H) =5 Vail = A— 1 + 9m2(R$21|RL) V0 ‘PMIRDI A..=—-' V. — 1+ 9m2(R32ll-RL) ll 2I953( V65) gnu 1- —."' IVPI 1'9- - Hi]. _ 1:3 _ v _ 2 {I 2 ) — 2 4 mad gm = 2(25 ( _, _) = 2.15 mA/V Than A, -[2 “(4) = —2 as = A” Va: 2 Vas_ . 2—3[l-v—r) =- VP —0.a V55 = [man—3.5} =- v5; = -1.75 -V 3 - (-10) A150 Ina: Re 2:131:22: 35:5.aam Rs 2fnss( V65) 2(3)( 1.75) 5!". = 1— -— = .— ]_ .— lvpi VP 3. 3.5 .—. 2.29 mA/V m = m = 50 m v. = V9. +ngsVy = V}. = WET A“ = 1/3 = ngS-"ru _ (2.29)[5.as|]50] V- 1+ymRero " 1+{2.29)[s.33||su] a .m— gmtasnm _ _ = 0.911— 0.186 = 0.745 1+9m(Rs\iRu (2.29){R5!RL) _ 7 1+(2.29)(RsI!RL) _ 0‘ 45 {2.29)(RslIRL)(1 - 0.745) = 0.745 =§ RSIIRL = 1.28 kn 5.38121: 9-1-23 5.55“!- R; => (5.33 — 1.28)R;_ = {1.38)(5.85} => R; = 1.54 kn Eectronic icuit Anal sis and Des’ n 2“d edition Soluti ns Manual Chapter 6 Problem Solutions #flcfli W ‘ (a) s..=2 Kai’s =2 [HZ—17:)!” 0.5: 2 (o.%020)(—)(0.5):E=12.5 W (b) 1., =[&§L‘XI)(VGS ’Vm}1 0.5 = (0.02)(12.5)(sz _ 0.0)2 :- Vi: = 2.21 V 6.2 J“ C... W ‘a’ “=2 “”2 [ ’2 II)“ (“129): = (10)(¥)(100} => 3:): = 0.025 (b) 1 =[#.2 C“I—W-)(vm +v,, 100- - (10)(0.525)(Vsc - 1.2): :- VEE = 4.2 V 6.E 6.3 ID = Kn(Vcs ‘ m):(1+lvm) 1m. 1+1VDs1 3.4 1+MID} ___,=——-—..——_ —— 1m 1+Ame =’ 3.0 ‘ 1+,\(.=.) 3.4[1+ 5A] = 3011+ 10).] 3.4 — 3.0 = Aux-10 - (3.4) - 5) =0. A = QflfiQS 1 1 '“ " 1'1; ' (0.0300)(3) =’ 515—1—949—3- 6.4 :D=1 1 6.5 A” = -gm(rnllflg} = -[I}(50“10) E? (L. = -3.33 E=W=~ID=Qfi3EmA 6‘6 v v 10 5 _ DD - Dsq ___ — ‘- R” ‘ Inc 0.5 For Vase = ‘1 V for ample. #(IC ('1 10:2- =““2—( 1% )(Wm‘ Hm 0.5:(0.03:](%)( 2-0.B} W 115 :—= . L b, g_=2 K‘Im =2K.[Vasq 42,.) 9,, = 2(0.030){11.5}(2 — 0.0} =5 5..., = 0.035 mMV [W = (0.030)[11.0)(2 - 0.0)2 = 0.501 mA 1 _ ' 1 =:» Mag ‘ {0.015)(0.501} 1'0: r0 = 133 RQ c. A, = —gm[rulifln) = “(0-335)(133HS} =-.~ A, = —6.30 6.7 Kati; = K_[v,, 5mm]1 _ K 1/1 sin m: l ‘1 sin1 (or = 1[1 —cos?.wt] 1 So Knv:,= K—‘f” [—1— cosZcut] Ratio of signal at 200 to that at 01: K v1 up ~coslax ZKHVm—wm), sinax The coefficient of this expression is than: _"e__._ 4(vGm - v," ) 5.8 0.01 = —Y£---- 4(VGSQ ' Vm) 50 V9: = (0.013(4}(3 — 1) => VI' = 0.03 V so .- - .-= -V.-= - R,||R,+R,, (son) (0962)“ I; = —g_(0.962)(r;[Rn) = ~(1)(0.952)(50l10) =9 :1, = -B.02 :RQ=BkQ Electronic ircuit Anal is and Desi n 2"cl edition 6.10 A» = -9m(rollfinl — 1a a -gm(100||5) =3! 9... = 2.1 mAfV 6.1] R 2 = V a. VG (121+ R2) DD 200 “G = (mill?) * 4-3 V V Va: 2 In: -£-—-—= K..(Vas" 1'») RS 4.8 - Vas ={1)(2)(V§3 — ”’65 + 4) ZVés — 7V6; +3.2 = o 7 i ‘Im' — ¢{2)(3.2] _ 2 95 V 2(2) In = (mus - 2)’ =3» I; = 0.920 mA VD: = Von - IDERD + 35') —.— 12 .- {o.92)(3 + 2) =2- V2: = 7.4 V (b) V. _ -Igul 51(RDIRL) 1+gnR5 whm V113 |R . zoolzon R,|]R,+R5, =3oo|l200+2 ' 120 — V . 4 120+2 (093 ) Then 1+ g_R, g_ = 2(1)(2.96-2) = 1.92 M IV 30 A. ___ -(1.92}(o.934)(3||3) [1.033(2) = fl Va: {-ng 1.5.1 \ 51.4525: 11—. = ‘ —1 &' = - 'D 3.5 m Ay“ Aip = 0.92 mA => {Aral = (o.92)(3.5) = 3.22 a 6.44 V E-flm 1143+? Ex. Solutions Manual 6.I2 a" [DO a 3 mA :- V5 =11;qu = (3)035] = 1.5 V [m = 1mg"55 41m): 3 = (ZNVGS - 2)“ =- Vas = 3.22 V Va = Va: +-Vs = 3.22 +1.5 = 4.72 v R: l = v = .._. m VG. (RI + R2) DD R: R. 'VDD 4.72 = Riemann) =:. 3.: 535 m 1 636R: 636 + R: 'leRDIiRL) 1+gm-RS 9". = 2(2H332 - 2) = (.55 nah/V - "(4'88)” '5) =-.» A. = —2.03 ” ‘ 1+ (4.353015) b. Av: 6.13 From Problem 6-1.1: ID = 0.920 mA '55- = 7.4 V gm =13? mAIV Av = —SM(RD||RL{R_IHJ;I—1|T—Rs:] =4L1:2)(3|3)[2:0"———°(:§l030032 ]= 422310.934] .4, =-2.83 c232. -l 1.5 k9 :- IAmsl = (D.92)(1.5) = 1.38 =:> 2.76 V pubic-peak output voltage swing big = - (3.91:5. Alp = 0.92 mA 6.1-4 Chapter 6: Pmblcm Solutions Vnsq = W - IDGRD — (-Vasql Output Value: Swing- — Vnsq - V5593] =[v* ~1 m11,,+1»'m]—( 0"ch —v m) =V’-[DQR,,+Vm 5° Ian=laa =‘_“' AVosl l kk‘-l=—r[v*—1m2 R +vm] 5kg. for: = 5 THIS ‘- J'IJ<.'1(1‘:")~1- 1] = 1.2-2IDQ 2190 =9 [no =04 mA=IQ 3,. = 2.111.159 = 2.!(05}(0.4) = 0.894 mAI‘ v 1 1 '° 11m”{n.o1)(o.1)"25°m Av = —Qm(-RD "35“,“) = _{n.891}{1ol|1o|12511} = A. = -1.38 6.15 3- 1r1) = 51'an: ""21111)2 2 = “v.” - (—1))3 V55 .1 -0.293 V :- Vs = 0.293 V = InRs = (2)35 =5 R; = 0.146 H2 VD = VD: + V:- = 6 + 0.293 = 6.293 R1: =W#RE =1.85kfl _ A” = -9m(finllflL] b l+ymRs 1-321? (V6.1' 111:) g," = 2(4)(— —o.293 + 1) = 5.66 mA/V A. = —(5.ss)(1.35| 2} =1» A. = —2.93 1+ (5.561(0146) 2.2-1.4 A... = (151911111552) = (211125112) =1.92 v 1.92 .'=—-=D.645=>V.=U.645V A” 2.95 _ 6.1.6 1- V1251: = V121) - [130(R1: + Rs) 2.5 = 5 -— 190‘? + Rs} I _ 2.5 no _ 2+3; —V Im=K.[Vcs‘Vm)2=_E§i I _ 4.535 = Va: — --IoaRs — 2 + Rs 2 K“ 4.512, “Vm = 25 2+R5 2+R, 2 . 4 1 _ 2.5R5 = 2.3 2+Rs 2+3: 4 2+R3—2.SR_.7 ’_ 2.5 2+3; _ 2+ Rs (2—1.51133 4— =15 2+Rs 4(4 —- 685 + 2.25113.) = 2.5(2 + R3} 9121. — 26.53.; +11 = 0 25.5 :1: 1/065)" - 1{9y(11) R5 --——-——-——— = 2(9) R5 = 0.5 R“ or 2.44 1:9 But 12; = 2.44 119 => VG; = -1.37 Cu: 05'. =9- Rg = 0.51:9, [pg =1.D wit. 1.. A. = -9«-{RDIIRLJ l +9m-RS 3,. =2 I11!m =1.I(4)(1) =4mAiV .. = ~(H21I2) a A. = —1.33 1+ (1)(11.5) 5.17 I I. 5 = Ians +Vsna +IDQRD “5 5 = Iqus-I-E +Ipq(10} -5 1 1. Inq=R +10 ——._._._E.._.... Vs = Vsoq + Iano - 5 -"- Vsaq 2. 1+ Inc-(10) = Vsco 3. _ I Inc — K,(Vm ’2) Electronic ggircuit Analysig and Dcsigg, 2"I1 egjtion Sgllmggs Mgual 4 Choose. 35:10 knafna = 20 =0.‘2Dmt\ Vscq = 1 +{o.2)[1'(1) = a v 0.20: K,(3—2)‘ = K, =o.20nw152 b. log = (0.20m — 2)2 = 0.20 mA g_ = 2 KP!” = 21/(o.2)(0.2) = 0.4 mA I v A, = —gm(.flp"RL) = -(D.4)(IDHIO) 3 A. = -2.0 c. Choose. R5 = 20 kn =3 Ipq = :7] = 0.133 mA "5.5:; =1+{fi.133}[10)= 2.33 V 0.133: 1:42.33-2)‘ = K, =122mmv1 g... = 2 (1.22){o.133) = W A, = -(o.aoe)(m||1o) = A“ = —4.03 A Large: gain can be achieved. 6.18 L :m =1: K,(vm 1114,.)2 1 = 50/59.; — 1.5)3 => V564 = 1.95 V R; = 5 —11-95 => RE '2 3.05 kn Vgnq = 10 —- IpQ(Rs + Ru) 5 =10 — (1)1335 + Rn} 9 RE 21.95 kn b. g_=2 K,1m=L/(5)(1)=4.47m1v A, = —y..[Rg[[R;,} == —(4.4T)(1.95}|2) =3» A" = -—{.41 c. 1 0.937 m => Impsl _-. (1)0351} = 0.937 v Swing in output voltage = 1.91 V pubic-peak Ain = - -Al¢gs 5.19 Ina = Kn(VGIa ”VTN) g_ = 2 If"!m 2.2 = 2% = K. = 0.202 "1.11le 6 = o.202(z.a Wm)” =1 vm = £2.65 v 2 Vpsq =13 -IDQ[Rg+ RD] RS+RD=13;'° =13: 20:35:1113—39 A = _9m(RDIIRL] ” Hymns Rp-l '2‘2(Rn +1) = 1+ (2.2)(1.33 - RD] 2.2RD 1+Ro (3.93 -- 2.235)“ + R5) = 2.2511) 3.93 + 1.7311; — 2.21::2 = 2.2129 2.21:}, + 0.4735 — 3.9: = o —o.47 +1/(0.4T)’ + 4(2.2)(a.93) RD _ 212.2} =a~ R9 =1.23 kit a; = 0.10 m vc=vfi+vs= 2.3+{s)(o.1)=3.4v 1 1 V = __ . "I . — __ — _ 5 R1 R. Van RI. (100)08) 3 4 => R1 = 529 kn 5293: 529 + R: 1 + 3.93 - 2.25:; = 2200: R2 =123 kfl ’" / /;1 i 4 1435:") lg,“ V. V1 = 9+Vsa~ VIA-7m} = Vsa *‘Vrp V -V Vina =J—:EIM+VSD(5‘") =(9+Vm)”(vsc+vrr} 2 = 9 +21.5 + Vsc _ 1.5 Vqu = 3.75 + V55 = 9 + V35 - Inqfln 2 I” = KAVSG +Vn.) + (V50 + V11») Cha ter 6: Problem Solution Set RD : 0.1.3.1, = 2 H? 3.75 = 9 — Ipq[2) => 1m; = 2.625 11141 b; g_ = 2 m1” = 2.!(2)(2.625) = 453 MIA IV 1 1 ° = um; = (0.0l)(2.625) Open circuit. A» = *9m(Rnll"01 A. = —4.55(2:1as.1)==» A, = 4.70 l' = 38.1kfl a. With R L A. = -4.58(2|120||38.I)=» A. z -195 = Cums: = 5.62% 6.21 ‘- ’09 = Kr-{Vsc +Vrr)2 2: (0-5NV5'G + 2‘]: =2: V53 = G V 10 —0 R = 5 2 Vn=u—vm=o—s=>fip=:5_—(fl q .- =>R5=5kfl #RE=21(Q Av = -§mtRD u. g. =2J.r(,,.'De =2,a‘(05}(2) =2mAW A. = 42m; =- A. = -4.0 6.22 A“: —Fm(RDHRJ—} '95.: == Van - Inq{Rs+ RD] 10 =20-[1}(R5+ RD] =5 Rs +Rp =10 kn LGIRQ=8HL Rc=2kfl A... = -10 = —g.,.[8|[20) g, = 1.75m1v =2 Kg” =2.{Kl(1} => K_ =o.755mrv= V: = IDQRS = (HR) = 3 V :m = Riva. - V". )' =5 l: 0.766(VG, — 2)2 : V55 = 3.31 V VG = vas+v5 =3.31+2 =s.31 1 VG=fiT'RIn‘VDD 1 - _ =3! R—1(200)(2{Jj = 3.31 =5 RI = :53 kn T53R1 m- —200=fi R3 =27? kn 6.13 gmru (“(50) A” = _--- = —-—-—- v = _ 1+ gmrn 1+ [4)(50) ” LE5 Ra=illro=i 50:30:0249m 9m 4 “—_“— A = m 2 «5012.5; ” 1+ammnRs} 1+4(50||2.5) 4(233) l+4(2.35} R0 = L 9m =~ Ra = 0.225 m z A, = 0.905 I. fa R5 = I 50 2.5 6 .24- R2 _ 396 "' VG ‘ (R. +R2)VD° _ (396+124o)(10] => VG = 2.42 V 10~(vm+2.42) “=0 “—3— S 7.58 — V55 = (2)(4...
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