chap005 - Electronic Qircuit Analysis and Design, 2'“1...

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Unformatted text preview: Electronic Qircuit Analysis and Design, 2'“1 edition Solutions Manual Chapter 5 Exercise Solutions E5.1 (a) Vm = [.2V, VG, =2V Vn,(sar) =VG, —v,,, =2— 1.2 = 0.3 V ('1) VM 2 0.4 :5 Nonsaturation (ii) VD, = 1:: Saturation (iii) VD; =5 =3 Saturation (0)11", 2-1.2v. v6, =2v Vmbat) = VG, —v,.,, = 2 —[—1.2)= 3.2 v (i) VD, = 0.4 = Nonsaturation (ii) V,” = I =) Nonsaturation (iii) VD, = 5:: Saturation 155.2 C Kn = Wflfl I'll 2!. 3.9 8.85x10'“ c,, =fig =fl——~4———1=7.67210" F/cm 1“ 450x10 100 500 7.67 10" K, =(—l(--)(—x—)=:. It, =0.274mA/V2 2(7) — (b) v,,, =12v. VG, =2v (i) VD, = 0.4 V :5 Nonsaturation 7,, = (0.274)[2(2 —12)(0.4)—(0.4}1] => 1,, = 0.132 mat (ii) Va, :11! => Saturation 1,, = (027409-12)2 =9 1,, = 0.175 m (an VD, =5V =5 Samm— 1,, =(0.274)(2-12)1 =5 1,, = 0175 MA V,,, =—12v. VG, =2v (i) V,” = 0.4 V = Nonsaturation 1,, = (0.274)[2(2 + 12)(0.4) — (0.4)‘] = 7,, = 0.658 M ' (ii) V,” = IV :5 Nonsaturation 7,, =(0.274)[2(2+1.2)(1)—(1)1]=7 7,, =1.43 M (iii) V“ = S V = Saturation I, = (0.274)(2+ L2)1 = 7,, = 2.31M E53 v”, = IVA/,3, =3v .v,,, :45: V,” =45>v,,,(:ar)=vc, —v,,, = 3—1=2v Transistor biased in the saturation region 1,, = K,(v,,,— I,,)‘ = 0s = K,(3—t)‘ = K_ = 0.2 m N” (a) V5, = 2V, VD, =45V Saturation region: 1,, = (0.2)(2 ~ [)1 =- 1,, = 0.2 mA (0) V6I =3V. VM=IV Nonsaturation region: 1,, ={0.2)[2(3a i)(l)-(1]’] => 1,, =0.6mA BSA (3.) V”, = —2 V, V56 = 3V v,,,(sa:) = v“, +v,, = 3—2 =1v (i) 'Vm = 05 V =9 Nonsaturation (ii) V5,, = 2V :7 Saturation (iii) V5,, = S V :9 Saturation (b) VTP =05V, V“, = 3V V5,,(sat) = V56. +Vn, = 3 +05 = 3.5 V (i) V3,, = 05 V => Nonsaturation (ii) Vm = 2 V = Nonsaturatiou (iii) Vm = 5 V :2 Saturation E55 (a) A = 0. V0, (rat) = 25 --0.8 = 1.7 V For VD: = 2V. VD, = iOV =9 Saturation Region 1,, = (0.1)(25— 0.3)2 => 1,, = 0.289 m (b) J. = 0.02 v“ In = Ka{VGS _ VW)2(1+ 1V0!) For Va: = IV I, = (0.1)(25— 03)1[1+(0.02)(2)] =9 7,, = 0.300 M v,,=10v 1,, =(0.1)[(25—o.3)‘(1+(0.02}(10))] =9 1,, = 0.347 m (c) Forpart (a), 1:0: .- =N For part (b), 2. = 0.02 V-l . r, = (Va, ~ 14,)2 14 = [(0.02)(0.1)(25— 0.3)1" or r, = 173 kfl Electronic Qircuit Analysis and Design, 2"d edition Solutions Manual 85.6 V”, =Vm+y{ urn/flI —Jfi:] 2¢f=o.7ov. vmo=1v (a) v“ =o=.vm=1V (b) 1/“ =11! . vm =1+(0.3s)[40.7+1— 0.7 :1 vm=uev (C) V” =4v. Vm =1+(035)[J0.7+4 -Jfi]= V”, = 1471/ 135.7 _ 1'11 _ = 40 ~ VG—[R!+Rq](l0) 5 [4mm 10) s VG=-1V V: = 10115—5 Then V0, 2 V5 —v5 = —I—(IDR, —5) = 4— IDR, Assume transistor is biased in saturation region r. = w..— 0124;“: 5 4-116, = (0.5)(0.1)[sz m 1]2 = 0.514; —3.s=0= VG, = 2.65V I, =(05)(2.1si5—1)2 = 1,, = 1.3611111 VD, = 10-1002, +30): 10-(136)(l+2)=1- VD, =592 v V.” >le:sar) , Yes 55.8 . R: V = H) -5 a (3&3?) 1 309 = - 5 — 0.714 V vk=3—1DRE=5-(L2)In 50 V51: = V: - Va = 5 —- (1.2119 — 0.714 = {.236 —(1.J]ID 4.236 - V55 1.3 1., = 11,0!” +19.)2 19: £286 — V35 = [l.2){0.‘35}>< mic - 21351—11 +1—11’) 1.235 — via = (0.311336 — 0.51}; + 0.3 0.313% + 0.413.; —3.905 = 0 -o.4 :1; 1110.412 + 4(0.3)(3.936) 2(03] Must use + sign :1» V55 = 3.04 V V31: = I; =10.25){3.04 - 1]:2 :19 =1.o4 mA V“, = 10 — IDOL: + R9} = 10 —(1.o-1111.1 +4) =1 V52 = 4.59 V V” > 1150(5111), Yes 155.9 1 ID = K.(Vcs 'Vm) 0.4 = 0.2511335 — 11.1112 =1 1x03 = 2.06 v . R2 . ._.= — 11 Pa- (Erna?) on R: -- _ - 2.05 = =? R2 — 53-1 kn EM V35 :4: VDD —IDRD 7.5 -4 RD ' 0.4 V55 > VM(mr] . ch => R5 = 5.75 kg €5.10 ID = “CS—Eglfl and v5 = -b'ss s 5 -'Vas 5° R5 ‘ 0.1 I. = am... — m)“ 0.1 = {0.0301(1/55 - 1.2)1 => V65 = 2.32 V SoRs=5flfl=~£§=—26£fl VD; = — Vs =9- VD =V35+Vg= 4.5—2.3? VD =2.13 51—h; 5—2.13 ‘ = = R2 =35.Z kn RD In 0.}. '= V“, > Vm(mt) . Yes E5.” I1: = i? and ID = K4135 +14.)2 0.12 = {0.05010}; — 11.812 V5.5- : 2.35 V R: = M :9 R.- = 53.75 m 0.12 V55 = 3 =20- [MIKE-PHD) 5 =20 2- (0.12)(sa.75) 40.121115 Rn = 20 - (O.12!(63.75) - B 0.11 :2? RD 2 36.25 kn Test ur ndc ta din 55.12 V V [o=_2%£§W 10=K.(V5:'Vm)1 : 10 — VG, = (10)(0.2)(vgs - wax!“ + v3") 10 .- Va: = 2v}; — 5V5: + B ZVéS —7Vc,-s —2 =0 7 a: Mm? + «(232 2(2) Us: 4- Sign: VEE -_.- V2: = 3.77 V 10 -3.7' ID = m-fi—J— 9 I2 =n.a23 mA Power = In V95 = [D.623)(3.TT} = w =9..' V" V55 = 15.5.13 For VD, =22V _ 5-2.2 5 {a : KI(VGJ -VTN)2 056 = K" (22- 1)2 ID = [0:056mA K. =0-389mA/v=%._#ucu 2 W _ (339)(2) w L" (40) g 15—195 155.14 (a) The transition point is Van ‘Vm+vm[l+4Kw/K¢j 1+,me/KIL a — 1 +1(1 + \/0.05,’D.01) _ 1 + 30.115.10.01 1236 . _ ‘—‘P VI: — 2.24. \r v" = v0, =v,,-vm =224—1=> V0,:L24V (b) We may write 1,, = K_,,(Vm - mm)“ = (o.os}(224 w 1)’ :9 I, = 76.9 M 55.15 V = V” —vm +vm(1+./Km/K,L) " 1+:in/K4 25_5-1+i(l+m - Hump/Ki Cha ter 5: Exercise Solutions 25+2 Km/K.L=S+ Kan/Ki: WK. 4:53.167. Km mi = 2.73 b. For V; = 5. drivcr Ln nansanuatcd mgzon. foo = In; KID[2(VI _ Vnw )Va 4%1 1 = Kchm. "er. )1 K fawn -i-r'].,,,,)1/lg 41;] = [ch - v0 AIM]: at. 2.7312(5 - m1, — v3} = [5 — Va — 111 22.24% — 2.731;: = (4 _ Va)“ =15 - 8% + K? 3.75%? - 30.24% +16 2 0 30.24 *1/[30243‘ — 4(3.73]{16} v° = um] 135.16 If the n'ansistor is biased in the. saturation region 2 2 ID = K.(Vos _v7?l) = Ka(-VTN) ID = [0.251(15)’ =. ID =1.56 1113‘. VD: = Van — IoRs =10 - [1.56}[4} =5! Vpc = 3.76 VDS_>VGS -Vm, =-—Vm 3.76 > -[-2.5} ‘15.: - biased in the saturation region pa“; = Iprs = (1.56}[3.76) am 55.17 We have Va, 2 1.2V <VG, «V1w = —V1w = L8 V Transistor is biased in the. nonsaturation region. ID = KI[2(VGS ‘Vm We: '44:] and [a =MA=§£= R, 3 ID = 0.475 m 0.475 = K.[2(0-(-1.8))(12) —{12}‘] 0.475 = K_(238) = K, = 0.: 55 m I v1 K = E .ffiz. " _ L 2 W_(l65)(2) W_ L— 35 = —9.43 icotronic Circuit Anal sis and Desi 2nd edition Solutions Manual 55.18 (3.} Transition point for the load transistor - Driver is in the. saturation region. loo = In; 2 z Kchso ‘Vmo) = Kchsn _VTNL) V0511”! ) = Vast. " Vrm. = "Vrm. avm=vbrva =2v Than v0, =5-2=3v. v0, =3v fifm — 1)=(-vm..) 0.08 _V ——1=2=: V 2139‘! 0.01( " ) -"———— (b) For the driver: Va- =Vn ‘Vmo V” = IBQV, Va, =039V £5.19 (a) For V, = 5V , Load in saturation and drivcr in nous aturation . Ion = for. Km [2{Vi _ an )Va '16]: KIL(_VTNL )2 K.» [1(5_1)(0.25)—(0.2s)1]= 4 = E = 2.06 K..:. 1 (b) [a], = u(_me)2 : 0‘2 = Kill- [_{—2)] Km: =50yAIV‘ and Kw =103pA/V' E5 .20 (a) first = Kuawan ‘Vmiz = KuWan “er V6.“ :21? =VG“ =3V K a K 1 2—i'=—-‘—3—1 _-:.=_. ( ) Kali )= K“ 4 CD} ’9 = K.1(Vcsz 'Vm )2 But van =vw =2v 0.1: K_1(2— 1)‘ = K,2 =o.imA1v’ (c) 01: 1<,,,(2-1)1 = K_, zozmrvz 02 = rifle—1)! ==> K“ = 0.05m1v1 E521 For RD =iom.v,,,, =5v. and V, = w ID=S——1-=0.4mA 10 1,, = Kn[2[VG, -Vm)VD, 43:5] 1,, =0.4 = K_[2{5-—1}(l)-(1)2]: K“ =o_057mA/v‘ P = 1,, 4105 = [0.4)[1) = P =o.4mw E522 ID = K_[2(Vm — mm, 413,] = {ousonzm — o.r](0.35) - {035?} ID=U.319mA R _"'D_D-_V°_1°-0-35 9’ ID “ 0.319 => RQ = 303 kn E5 .23 (a) Transistor biased in the nonsaturation region ID = 5—1.5—V'2E =12 R In = K.[3(Va: "Vm )Vns 'V55] 12 = 4[2(5- 03m, Jugs] 4v; —33.6VD, +12 = o = v” = 0.374 v Then R=§Zflia R=251§1 12 -——~-- 55.24 3- Vl=5vu W=01M16m0ff=$f22= s-v ID = Ku[2(V, -V,.~)Vo—V;]=—-R—Q- . D (o.os}(3u)[2[5 —1)V3 — v.3} '= s - v; 1.514,” -13'va + s = u 1- . oiL___. War/amino v“ " 2(1.S) Ig= Im = 3—3343 =- IB =12. =0.l$3 mA [3. V; = V; = 3 V S-V R ° =2{Ic.[2(v, «mm-v31} n a _ vcl = 2(0.05](30)[2{5 —1)'VE: - va’] 3v: -2514. +5 = o Te§t Your [Anderstanding ghaptcr 5: Exercise SQIutions I Vo=w=w 2(3) IR = Listing: 15 = 0.160 mA In] ._ IQ: = 0.680 mA E525 s—v (a) n, = ° = K.[2(V2“Vm)Vo—V02] D s-(mo) 25 = Kfl[2(5—1)(0.10)—{0.l0)1]=> K, =0.248mA Iv“ a-Vn 25 5 — v9 = l2.4[3["u - W} 12.40;.” - 100.201, + 3 = u v 100.2 d: ./(l00.‘2)2 — 4412.4}(51 " em.“ = v9 = 0.0502 v b. = 2{0.24a){2(s n 1M - V92} E526 V056“) = V65 - V}: = —l.'.’ - [—‘i.5] => Vpshat) = 3.3 V Va.- ‘ (-1.0 3 I =I 5 1- = — D “( VP) “(1 (—40) 2. IQ = 6.45 111."; IU Ln E527 2 In =Iass(l - Va.— 3 _ 1.2 = 2 1- = Pr.- = -—0.564 V. Jun—— E528 \ ‘I ft; = 1053(1 - VP . 0.0 ’ 3 -I95$(1' Vsptsau = H: — V55 = 3.8 - 0.8 Vgafsar) = 3.0 V‘ E529 [D = KM: _V:w)1 a. VG;- = 0.35 = ID 25(035 — 035]2 :5 ID = 0.25 {LA 1:. Vas = 0.30 :19 = 25(050 — 03.15.].2 ==» In =1.56 pA E530 Assume. th: transistor is biased in the saturation region. V65 : ID—Ipss(l- VP) 3—1150 V“ )231/ - 117v ' {—3.5} ———G“— ' = v; = 4’55 21.17 VD =15 — (0x03) = 3.5 V5; = 8.6 — (1.17} = 7.40 v v93 = 7.43 > v55 — Vp = —1.17 - (—3.5) = 2.33 Yas. the Hammer is biased in the saturation mgion. 55.31 V5 = [53.5 — 5- : (2.5)(D.25} - 5 V5 3 -4.375 V95 = 5 =:. VD = 5 — 4.375 =1.625 - _ ,_ RD = a ___1_5-° =, 3;, =1.35 m 2.; - — (20)“ _ _ 814-82 —'2=>R:+Ra-—-200kfl H; = V55 + V; = —1.42 — 4.375 = 4.795 . * R, _ pa .. (R1+Rz)(20}—10 -— 5.795 = (20] — 10 => Electronic ircuit naI sis and Desi n 2“‘1 edition E532 O—Vg V _:- V5=—VG; In: H- =% Va: : In—Inss(1—T};—) Vac V65 2 'I 4: i’és“ -—-—6(1— 4)_5k1 +1) 0.375ng - was + a = o 4 :l: 15 - 4(u.375)(51 2(0375} V55- : 3.85 or VQE =1.51V w..._.~...—r VG: = impossiblc V :3 V0 = I035 - 5 = (l.81](0.4} —- 5 = —4.275 Vsp = V; — Va 2 —l.8: - (—4.275) =1.81mA =- VEQ = 2.47 V Vsnuat) = Vp - Va: = 4 — 1.31 = 2.19 SD V5.0 > Vso[sat} 55.33 R: R; R". = R] = m: =100 kn 19.3 = 5 mA. V: = -Ians == -(5)(1-2} = “‘5 V VSDQ =12 V = V:- = Vs - Vsoq =—5—12=-—18V Rp=w=>flg=ndlkfl V 7‘ v I5q=lnss(1-—Gs-) =95= (1—E V5;- = 0.333 V 'G = Yes + Vs = 0.838 — s = "5.162 _ L - — 5.162 = RLuocu—za} a R. = m m 1 R13; 31+ R: [337 -100)R; —.= (100}(357) = 100 => [357)R; = 10:1(337) + wait: = R; = i135 kn Solutions Ma 11:11 55.34 In = ((0/6, —v,,,)’ =9 5=so(v6, 43.15)1 =~ Vfi: = 0.466 V V,- = (0.0051(10) = 0.050 v =- VGG = V65 + V: = 0.466 + 0.050 =:. v9; = 0.516 \' V]; = '5 — (UJDSHIUU) a V; = 4:5 V VD; = VD — V; = 4.5 - 0.050 => Vii = 4.45 ‘5’ 135.35 I Jrr; = KIEWG: ‘Vrn )Vns 'Vnzrs] _—_ 100[2(o.7 - o.21(o.n— (01):] IngpA 2.5-0.1 =— = 1 RD 0-009 =>Rp 26 kn Electronic Circuit Analysis and chign, 2'” edigign Soluti0__ns Manual Chapter 5 Problem Solutions 5.1 (a) VD, =6V >VG, ‘vm =5—15=35‘v Biased in thc saturation region 1,, = {(491333sz1 =(o.25)(5—15)1 = I5 = 3.05 m (b) VD. = 25V <Vm(sar) = 35v Biased in nonsaluration region [a = K.[2(Vas —V1N)VDG “viii-1 1,, = (o.2s)[z(s—15)(25)—(25)‘]=a ID = 231 m ’.3§+03=VGS =9 VG, =2.38V V” (sat) = VG; —Vm = 233—03 =9 Vm(sar) = 1.53 V (b) 15%? sxva, 43): ’15 E+03=V6§ =9 Vm=354V vp,(saz) = 354-03 = mum) = 2.74 v 5.3 I. Vcs = U Vm[mr) = Va, 4’”, = 0—(—25) = 25 V L V9; = 0.5 V =7 Biased in amnion ID = (1.1mm - (-2.5})(u.5} — (05):} = IQ = 2.43 M ii- Vn: = 3.5 V =- B'uscd in summon ID = (1.1)(0 - (—2.50)3 : I2 = 6.33 [113- Vps=5v Smu(ii)=>12=6.8BmA II. Vas=2V Vadm) = 2 - (—2.5) = 4.5 v i. VD; = 0.5 V a Nonsnruration In = (1.1mm —. (—2.5})(o.5] - [0.5%] =5 12 = 1.63 mA V95 = 2.5 V =:» Nonsmdon ID = (1.1)[2(2 — (-2.5))(2.5] - (2.5)”) => ID =17.9 mA V55 = 5 V :Satuxation ID = (1.1)(2 - (—2.5);2 2 I0 = 22.3 111A 5.4 VD: >VG, —Vm =O—(—2) = 2 V Biascd in the saturation region k; w I” =3‘E(VGS_VTN )1 0.080 w 2 w 15: —' “- 0- -1 => —=9.375 [ 2 L1 ( L 5.5 C _ to. _ (3.9}(3.35 x 10-“) 5' °' - I; _ 4.50 x10-—' 2 = 7.57 x10" [=7ch to: K" =-———#"C" 2 L l 64 ==— . 10" —— 2(550)(7 67x )[4 J K, =o.399mArv= b. V55 = V55 = 3 V =5 SRMEDIDII ID = [cums — V,” )2 = (0399)(3 —03)‘ a’ In =1.93mA 5.6 _ (a, _ (3.9)(a.as x 10"“) "' C” ’ to, ‘ son at 10-! =- :43- = 5.75 3-: it!"l Finn2 I}: K; gala—9E 2 L =%(500)(5.7le0* 1%”): K,=o2ssmA/v* Electgggic Circuit Anglggis and Design, 2"” cdition Sglutions Manual b' ‘L Vas=01VD$=5V I.. Vp=0=§V55=5V Mas-(sat) = D — (-2) = 2 V =1> Biased in snturlfion Biased in saturation ID = 2(5 — 0.5.)3 => 12 = 40.5 m 1,, a [memo --[—'2))2 =- 19 = 1.15 um I). Vp=-2V=$V5D=JV ii. Vas=2V. VDs=1V =3 Nomamminn 14mm): 2 - (—2) = a V ID = 2[2(5 — 0.5)(3) — (333] Nomarundon =:- IQ = 35 mA 1;, ={O.233)l1(2 — (—2)}(1)- (1)2] $115M c. Vn=4V=>VSD=1V 5.7 I =5 Nam“ ID = 2[2(5 - 0.5)“) - (1)2] (a, (3.9)(a.ss x 10-“) Cu=-;;=W :1g=16mA = 3.63 x10" cmm’ d. VD=5V#VSD=O=»IE=0 K “ 2 L 5.10 = i(600}(3.63x104)[%) Vm(sar) = V55 + VT; 2 (a) V (mt)=--I+2=9Vm(mr)=lv K —(1 04x10“)W 5” " ‘ ' (b) Vm(5ar)=0+2=>Vm[Jar]-=2V (c) V5.0(sa1‘) = 1+ 1 =1: Vmfiar] = 3 V 10:32.0, "' )2 k' W 2 k' W 2 :s m s I” fit“: “’11) =E'I'F’wimfl 1.2 x10“ = (1.04 x 10‘ )W(5 .. 1)” 0940 2 a—W = 7-21 m (a) I» {—2 1510) = 0.040 5.3 (b) 1,, = [T 5}(2)‘ = ID = 0.43 m Biased in the saturation region in both cas . k’ cs (1:) 1 = 99—19 exa)‘ => 1, =1.08 mA I =4-EW 1-19,)1 " 2 —--—— D 2 L 36 (1) 0225=(L'2‘“’11E](3+v,,)' 5.11 Vm(sat)=vm +14, =0+2=2V (2) 1.40 = [1”?ng + 11',.,.)2 V = V . ' Take ratio of (2) to (I): (a) m 1 I‘io1'1530.1r111:mn2 L40 (4 W"): 1,, = K,[2(V,G +v,,)vm '11155] ———= 6.222 = -— 0225 (3+v,,)‘ In =(05)[2(0+2)(1)—(1)‘] = 1., =15 m {——— _ _ 4 +14, _ 6222 9 149 - 3+Vn. = v" — _2'33 v (b) V“, = 2 V . Saturation Then 1,, z K412” +11”)1 = [05)(0+ 2)1 => . 0.040 w 1 w - 0225= __ _ _ __= I *ZmA [ 2 IL}3 2.33) =: L 25.1 D _ 5.9 (1:) V” = 3 V . Saturation 51111151501). ID=2mA Vs=5V, Va=fl:V5a=5V m v". =—05v =Vm(sar)=V,G +V,, =5—0.5=45V _ Chaptcr 5: Problem figlutions “ e“ _ (1950.85 :1: 10"“;- 10, ‘ 500 11-110“ = 6.90 x 10" F/cm’ 1:, = (11.60,) = (5751(090 x104) => 45.5 uA/v‘ 2;, = (11,0...) = (375305.90 3410-8) :- 25.9 FAIV" _ 0.0259 " ’ 2 Want K,l = Kll r k' =_L[.“i) =41.3 2 L ,, 2 L , 40.3 W I W _ N — 41.3 5 — 1.71 :- L=4um WN=T.051.1m 5.13 VG; = 2 v. I5 = (0.2)(2 - 1.2)“ = 0.120 mA 1- 1 1 : r '8] kn — 2 ———-—- = [ ° .11; (0.013(0123} J—-—-—-—- vcs :4 v. I; = (0.2)(1 -1.2)‘ =15? ma 1 ‘ (0.013(1571 1 1 ' =—= V = 0,. A (m) =1» d 100v n: :0: Ta = 63.7 [(0 5.14 10 = [£280- 4)(3--0.0)2 =(0.11s)(3—013)1 =2 I, = 0.774 m ,. .. L =, = L = .___l ' 111,, r,1,, (200)(0.774) Mmax} = 0.00646 V " V‘ - 1 = I = V‘(mm) = )(319): KP =41_3;1A/v1 = Kn 5.1 s V... =vm +4.12%. +0., -,(2¢,| AV“, = 2 =(02)[ 251, +125, —J2(0.35)] 25+ 0037 = ./2(o.35)+v,, = V" =710.4 v 5.16 Mm = y[ 201, +03, “[E] 1.2 = y[.f2(037) + 10 - ./2(0.37)] = 3142.42) Then 7/ = 0.496 v "1 5.]? a. Va 2 c.3111, = (5 x106)(2?5 x 10-“) Va = 16.5 V 5.18 Wan: VG =1311241= can. =(6 x 10‘)!» 10, = 1.2 x 10" cm = 1200 Angstroms 5.19 . R: 13 . = = = . v V“: (R.+R,)VW (18+32)(1°’ 36 Assume transistor 01:01:11 in saturation fission V V —V 1 I =_..s =-—G 05:11:01 - ) D R: R: 65 IN 3.6 — v.35 =10.5)(2)(V55 - 0.11)2 = v3.5 - Lava: + 0.54 V55 — 0.01/65 — 2.95 = 0 0.5 0.53 42.90 V“: :1“ }+( ) 2 :9 V5: = 2.05 V VG — V55 _ 3.5 - 3.05 Rs - 2 Va: = V1212 — 10(30 + Rs) =10 — (0.775114 + 2) In: =>fp=0.775mA = VD = 5.35 V V9,- > V55[Sll) Electronic Circuit Analysis and Design,l 2'“1 edition Solutions Manual 5.20 R ' 6 V = 20 -10: _ _ ” [R,+R1 ) [n+6 20) 102° Vc=-4V Assume transistor is biascd in saturation region ’0 =T=T=KJVGVVWV K, = [93299.](60) =9 [.3 mA/Vz —4 - Va, + [0 = (1.3)(05)(v55 - 2)2 = 0.91%, — 3.6V“ + 3.6 Than 0.9%; - 2.5%, — 2.4 = o = 2.6: (2.5? +4(0.9)(2.4) VG, W: v6: = 3.62V Va-V§E+10 = —4—3.GZ+10 Rs 05 In =4.76mA VD, = 20— [DMD + R3) = 20—(4.76)(l.2+05) => VD, =11.9V VDs =11.QV>VG, 4/," =3.62—2=1.62V ID: :9 5.21 IO—VS Rs Msume. transistor biased in sanitation region R; ) = - 10 V" (R: + R: (20) 22 = (a + 22) [20) — m =3 Va =43? V ID = = KP(VSG +1)??? V; = Va- + V56 10 - (4.57 + Vsa1= (1)(0-5)(Vsc - 2): 5.3a — V5.5 = 0.5(V5’G - Wsa + 4) 0.5ng _ V55 — 3.33 z u 1: Ju}: + 440530.33) 2(0.5} => V39 = 3.77 V Vsa = Ip=m_“.%i3_'711=12=3.12m V39 = 20 — 19(35 + Rn} a 20 - (3.12)(n.5 + 2} => Vgg =11: V '50 > V506“) 5.22 vs = 0,. VSG = Vs Assum: sum-man mgion In =04: {CAVE +V,.,,)1 0.4 = (0.2)(V3 — 0.3)” 0.4 V5=HE+03=>V5=121V Va = Infla — 5 = (mus; — 5 = —3 v V59 = V5 -— VD = 2.211 — (*3) =5 Vsn = 5.21V VSD > V55 (55.!) 3.23 VDD = I‘DQRHHJ",m +1,,QR_T (l) 10=1m[5)+5+v65 and k; W Ina = [$IE}VGS " VTN )2 or (2) 1m =[9:9:_°K%)(vfl —1.2)2 Let va,=25v Then from(l). i0=IDg(5)+S+25=> IU=05mA Then from (2), 05 =[$(;£91%}25— 12}1 = W —~=9.36 L {MRS=VGS=RS=Y-‘i=3-§=a R5=5kfl I” 05 —— = (05)(o.os) = 0.025 M Then RI+RI=FIO~:—5=400kfl [RI—:32:va = 2v“ :1. [rig-Jan) = 2(25) => RI=R1=2OOI¢Q 5.24 R: VG = (R! +R2){10)— s 5.5 ._ (Imam) - s _ —2.25 V V -V -—5 z ="—“'—"'6 Z: ( )=K.(Vas" m) Ass:me mists: biased in ma.an region In - 2.25 — Va: + 5 = (0.5)(n.5)(vos — (-1])' 2.75 - V65 = [0.3)(Vég + was +1} Cha ter 5: Problems lutions 0.3%; + 1.6Vas — 2.45 .—. o —1.5 :h “(1.5)” + «0.3)(245) V55 = 2(1):.) :3- VEE =1.24 V -2.25- 1.21 5 ID = —:—=:-JD =2.52mA 0.5 V95 = 10 — Ia(Rs + Rn] = 10 — (2.523015 + 0.3) => VQE 1‘: 5.47 V V95 3* Vpsfilt) 5.25 20: {MRS +V5M HDQRD (I) 20:11” +10+IWRD EDI-example. let {be =08mA and V5,; =4V —.——_._—'—-' Then 0.3 = {Eli}: -31 = .“L = m 2 L L IMR5=me(o3)R5=4=» Rg'zsm From(1}20=4+10+[0.3)Rn.-—: Ro=75kfl _ 20 —m+m I, =(os)(o.1)=> R, + R, = 250 m [ETR1}(20)=2V,G =(2){4) .‘L = ~ - 250(20) s=> 3,-100m. R2—1som 5 .26 (a) (i) IQ=SO=SDO(VG,—L2]1=> v63=1516v VD, =5—(—l516)== VD, =6516V (iv) In 2 1 =(05)(VG, — 12)’ = v“ = 2.51V v0, =5—{—2.61) :3 VM = 1611’ (b) (i) Same as (a) VG, =1!” =1516 V (iv) V,” = VD, = 2.6! V 5.27 In = _VTN)2 n.25 = (0.2)(Va5 — M)a V55: %%+0.6=-V55=1.72V =9- V: = —1.72 V VD = 9 “ [0.35)(24] =5- VE = 3 V S .18 5—1/5 5—1 . In= =M= 5M Rn RD VG=0 ID = {go/Gs 4),”) => 1/; = 4.11 V __ —3.11—(—-5) In —0.8—- R5 1 => 0.3 = (0.4)(VG, - 1.?)1 5.29 V0,, =Vm +1.30}? 9 = 25+(0.1)R = R =65 m k’ w 1 [De = +Vfl ) (0.1) =[92—151%](25- 15]1 = 1E— = 3 Thenfor L=flm. W=32fim 5.30 5 = 1903, Wm = :W(2]+25 1m =I.zsm l0 RI+R2 100 = Igor” +Vn.)1 = (125)[0.1) =5 R, + R2 = 30 m In: 1.25 = 0.50!” +15)‘ :9 1:255 — 15 = v“; vm=o.osuv ' Va =11, -v,c =25—o.03u= 2.42v _ Ra , s 2.42 =[fi](1o)— s = so R, = 59.4 m. R. = 20.6 m 5.31 19",: o. [—g’fl 20):: 0.30m1v' In: Kp(VEG+VTP)2 05=ojo(v,a - 1.2)2 :1. v,,S = 2.491; v, =Vm =2.49V 5~v - . = R , 3R: = 5 249 S v - _ _ = D ( =«5-—3=I Eur—4m I, 05 —— 1,, = R, = 5.02 m Ru Electronic Circuit Analysis and Design, 2"" edition Solutigns Manu___a1 5.32 _—Vsp—!—-10} __—fi+10 10- RD -"-'*D-- RD =5 RE=Ofikn 1,, = K,(V,G urn)“ = s = 30;” 433‘ V5.5 = «if-+1.75 = 3.0-1 V 3 V; = «3.134 R: = 10 - 5 = —3.04 VG (R: + R:)( ) R“. = mm: = so Inn _1_- (sumo) = s —3.u4 = R. = 403 m R: “’33” = so a 1:; = 99.5 1:11 405 + R: 5.33 (a) K... =[%9)(4)=120er1 If“ =[%)(1)= 30m1v’ For v, = IV . M I Sat. region. M 2 Non-sat region. 1m = 101 343('Vm X5 " Va} — (5— var] = 120(1‘ 0'3): We find v; -6.4v0 +7.15 = 0: v0 = 4.955V (b) For u, =31! . Mt Non-sat rcgion, M2 Sat. region. I“ = InI 30[—(-13)]’ = 120[2(3-os)va 4;] We find Jeni-17.6110 +324 =0: v0 = 0.19311 (c) For v, = 5V . biasing samc as (b) 3 4—43)]1 = 120[2(5-os)va 43] We find 4v}, - 33.6% + 324 = o = v0 2 0.0976 V 5.34 For v, =5V . M. Nomsat region. M1 Sat. region. ’01:“): [%1[2(5 - US$115) - (0.1 5?] = (1)[<J - (-33]! which yields = 3.23 1 5.35 21. M1 and MI in saturation KuWasn "mef = Knnwcr: ‘Vm1) K.“ = Kuzv Vm = Vm: =3- Vgfl == Vgsa = 2.5 V, W; = 1.5 V ID = {15)(4a)(2.5 - 0.3? .—.> IQ = 1.73 mm. W W _ _ V <V b. (L)1>(L)z=fi (:51 as: warm — o.a)’ = (151mm - 0.2.)“ V051 = 5 - V651 Law/gs. - 0.5] = (a — V551 — 0.3) 2.63VGSi = 5.50 =- Vga = 2.09 V 1 V25: = 2.91 V W: = V531 = 2.09 V ID = {15)(15)(2.91 — 0.3)“ =- 19 = 1.0 mA 5.36 Each transistor biased in sum-anion. Ma: V1=Vc=sa=2V W ID = 0.5 = 0.01%?) (2 — 1)1 J W —- = 7. =o-(L) EB M2: V652=W-V;=5-2=1V ID = 0,5 =0.013(E) {a - 112 L 2 W =5 2 - 6.94 M1,: Vasx=ID-V;=ID-5=5V 1;, =0.5 -_- o.n13(£) (5 — 1)2 L 1 =3. = 1.7! 5 .37 M L in saturation M n in nonsaturatic-n [Elwm—vm)‘ L = [Iglpwm — 14,4»)an - Vgsoi (1}(5 — 0.1— 0.3)1 = D [2(5 - 0.3m.” — [0.11:] 15.31 = D[o.sa] (wan-a Qhaptcr 5: Problem Solugions 5.38 M L in saturation M D in nonsaluration W [EJLWML Jim): 4% (nu—sf w = — 25—0.s o. —— _ ‘1 (L)D[( )( 05) (005]: 32¢ = I331.411'5] 5.39 [up = “(Van ‘er = Kach “Vm )1 V651 = 5 - Vasa 400 fifs - V653 - l) = (V55; — 1) 2.4“!ng = 6.56 => V953 = '2_76 V Vgfl = 2.24 V V522 = VQEJ = 2.76 V IRE: = “(Von ' Vm )2 = (0.2)(2.76—- 1)2 [up = 0.620 m If; = 1?qu -v,,,)‘ =(o.1)(2.76—1)’ IQ = 0.310 ml I9 = Kama-t 4’11”: => 0310=(o.03)(va,l — 1)’ 0310 Tim Van = 11% + 1:: V6,. = 2.97 v 5.40 JD=Vgp—Va 3—0.1 =--—-—-=0. 9 RD 10 4 mA Transistor biased in nonsnmrarion 5.41 5: JORD+VY+VM s=[12)R, + l.6+0.2=> R0 = 267 Q h; w in = (Exam, — vm)‘ 0.040 W 2 W 12: —.-.—..— _ 5_, _= [ 2 IL 03) = L 34 5.42 5=V_u,+l'p!§tb+V:r 5=0.lS+(lS)RD+l.6=: RD=217Q k’ W 2 ID=[TPII VSG+VTP) W 5:993?- Eja—osfa —=35 2 L L 5.43 Vpsfilt) =VGS —Vp 50 V05 2- Vpsfisat) = —Vp, In =-" 1255 5.44 V3583!) = V53 -— VP = Va: + 3 = Vaduz) ll. V63 —1 2 b. In —Inss(1—V—P) —-6(1-_—3) =>In =16? [11A -2 2 c I3 = 6(1‘ —3) :2- IE =D.66’rmA d In :0 5.45 2 In = Ipss(1— F 1 2 8=I 1——- 1 Dss( VP) ;/ = 2.055 :5 V: = 3.97 V P l 1 2.8 = 1555(1 — = Ipssflljfic) a lei-525.0 mA 5.48 5.46 Elcc ronic ircuit Anal sis and Desi n 2‘“I edition 5.50 1/5 = -V135. V513 = Vs - Vno WI!“ V50 2 V5.0(Slt) = VP - V55 V5 - V131: 2 1"? - V05 — V55 — V513 _>_ VP "" v6.5 => V213 <_ ‘VP 50 Van < -'2.5 V V 2 1'9 = 2 = 113550 - 1,23) V95 2 . 1 . ._. ._ —— V :: .D \I 2 (1 2.5 ) ‘4 as 1 5 =7 V: = -I.OG V ID = K.(Vos ‘Vm)2 185: K,(035—v,,,)2 862: ic_.(c1_'5—1-',,,)z Then 5.5! 2 1 5 OBS—V i-=112145=£-—--~——Ifi—)T:a 1:,M zcmw 3‘52 (DSO—Vm) --—-—— 135: 1r(,,(-:1.35—11.221)2 :5 K, = Luau/v1 ID = [(0,155 - m): 250= («oars—0:14)2 =5 K2096] mAle 5.49 V55 2 _ V _ Vcs ’9 ‘I"“(“T¢) - R.- ' Rs Va: 2 _ VG: l0(1- :? — - .2 2Y0: V315) _ = -V 2(1+ 5 + 25 as l 25 V35+§VGS+2=0 5.52 2V3, + 45%; + so = o —45 :1: 1/115)“ - 4mm) Q V9: = —l.].7 V Vas 1.17 g _— = ——-- I = 5.85 I“ R5 0.2 5-9-———-m‘ VD = 20 — (5.55)(2) = 3.3 v VD5=VD — V5 = 3.3 — 1.17 => V2: = 7.13 V Solutions Manual Vas=VDD-Vs 8=lD—V5-=>V5='2V=IDR5=(5}R5 ). 2 ) Let [1:155 =1C' mA =5- 3: =04 kfl _ V In =1055(I - as V;- 2 ) 3V}=—3.41V 03+Vs=d1+3=lv VG=V . Hz \ . 1 1/ = = _ G (R,+R:)LDD R1 (500}{10) => R] :5 Mn ‘Rm ~ Von V5 = 20 -(5)(0.5)=17.5 V VG = V5 + V55 =17.S +0338 213.3 V R: . 1 V = V = —- m - V G (R1+R:) on El R DD 111.: = RL{100)(20)=> 111 =109 m I 1093.: 109 + R: mum V65 7: VP 2 ) =1. Vgc = 0.465 V ID = 1.955(1— VGS 5:7(1 3 Vsn =VnD-ID(Rs+R-DJ e =12—[5)[o.3+ R9) =- Ru = 11.9 1111 V5- =12 - (5)105) = 10.5 v VG = v; + Vcs = 111.5 + 0.455 =1o.965 V R: ) = v VG (R1 + R: DD 10 965 — (%){12} =1- R; = 91.4 m ____.__.__ Qhagtcr 5: Emblem Solutions 5.53 R: ) 60 V = = 5 (Hall: "D (140+sn)(2fl) :9 Vi = 6 V V65 V35 16(l+ 2 +76— —5—-Vas vés+9vas+m=n —9i m9)2 -4(10} 2 => V95: —l.30 -1.30 W 19:3(1-({_“) =IQ=3.65mA Vns = VDD - [0(35 + Ra) = 2a — (3.55}(2 + 2.7) V1): = 2.35 V V95 3- V9563!) _-. V65 — V; = —1.30 - (—4) = 2.7 v (Yes) 5.54 Vns = Wan - 15(R5+Rp} 5 =12 —- ID(U.5+1) =5 IQ = 4.67 mA V5 = I535 = (4.57)(0_5) :5, V; = 2.33 V . R: . 20 P = ----—— 1; = — a (314-32) DD (650+ZU){12) =. VG = 0.311 V was = VG -— V5 =1 -' 2.33 :- Vg: = -132 V 2 ID = 1055(1- 2 4.57 = ID(] -( 1-32)) => V}: = —5.75 V VP 5.55 V I ID = fuss-(l - VG: =0 P IQ = 12:: =4 mm 5.56 Vsn =VDD—InRs lO=20—[I}Rs=> RE: 10 kn V05 '20 R = ___.. = __.. ._. 1 + R: I 0'] 200 kn V65 2 ID = j (1- _) ass VF V05 2 1:21—T =>Vgs=D.SEISV Va = V5 + Va: = 10 + 0.535 =10.53'5 R V; = ( 1 )Voo R: + R: - _ R: __ 10.336 _ (200)00) => R: - 106 m R: = 94 kn 5.57 VDs = Von - IDIRS + RD] 2 = 3 - {o.o4n)(1o «1- RD] => RE =15 m [D = ICU/GS 'Vm }2 4n = 2500/55 — [1.20)2 =. V9; 2 0.50 v VG = V55 + V; = 0.50 + [0.040)(10) = 1.0 V R: . V = P 6 (R1 + R?) DD R: 1= TEE (3)=‘Rz=50kfl R1 = 100 5.58 For Va = 0.70 V = VD, = 0.70 > Vm (sat) = V55 — Vm 0.75 — 0.15 = 0.6 Biased in the samration mgion _ Von—VD, = 3-0.7 I _ :9 I =46 " RD 50 9—.fl 1,, = qufl 4 m)‘ =: 46 = Maris-0.15)1 = K=123pArv1 ...
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chap005 - Electronic Qircuit Analysis and Design, 2'“1...

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