chap016 - ectr ' iu't si d at 2"“cdiin Chapter...

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Unformatted text preview: ectr ' iu't si d at 2"“cdiin Chapter 16 Exercise Solutions E161 3. Drive: in maturation; _ V -- v k' W 'n = A”)? = [Ell-£1203 ‘ Vm )Va " Efiflgfl = §ii(5}[2(5 _ u.s)(u.15: - (Er-151’] 4.35 .. _ Ti;- = 8r.5[1.2373] =3- Rg = 44.8 kn b. From Equation (16-10): 2 3.920(V}. - 0.3)“ + (V1. — 0.31— 5 = n V 03 —1:t.,/1+4(3.92}(5) n -— - = ———-—-—- 2(3.92) V" - 0.8 =1.D =4?“ = 1.8 V V“ = 1.0 V 516.2 a. i. w=0=vo=4V w=4V.driverh1nmsmfiou (% amu— Mr = 190W“, —Vm)vm, —v:m] 2(5 - m — 13' = (1mm — 1)». —v§1~ IG-Bvo-i-Ug = Mew—ug) 9»: —56m +16 =0 56 4.: was)” — 4(9)(1s) 5'“ = 2(9) w a: 0.30 V b. P = in - Vpp . 35 2 In a 2—(2)[5 — 0.30 — 1) = 479 “A P = (0479)“) => E = 2.; mW (w) [5}(44.5)(V;. - 0.3)2 + (W. - 0.3) - 5 = U olu ' ual E163 750 P=in-Vpp:ip=—5—=150}u‘ 35 W 2 = —- -- - 0.2 -D.B 2 ( L )5. . 150=2BG(-H-l) =D.536 L L L I . k’ W ’° =[?" ELF“: -Vm)vo ~23] W .5. = %(T)a[2(4.2-Mum-(0.212] 3. Load in saturation; drive: in mammalian; k’ W 2 '- VM} k' W = EJDPUQID " Vmo )Vnso ' V350] 2[—[-1..=.])2 = (5)[2(5 - on)” - 1.3] 4.5 = 661.5». - .3) Eu: — 51.69.: + 4.5 = o 51.6 :i: ./(51.5}' - 4(6){4.5) ”° ‘ Ms} W = 0.0881 V b. Load. Va = Va” 4- Vm Equation (16.2600) =5-1.5=>vo.=3.5V From Equator: (16 — 2303)) : Em.- m)=—Vm gm, -01} = —(-1.5) = 1.5 Load: 9“ = 1.57 V m = 3.5 V Driver: v“ = 1.57 V Em = 0.37 V 3.5.. 2 P = In - V99 :3 (73.75)“) =5 P= 39‘ Ew c. in = (2}(1.5)’ = 73.75 ,uA in = D[2(5 — o.s)(o.os) — (mosfl W W 70 a: 7.31- D = — 9.58 516.6 From Equation (15 - 35) 2 5— . V13 = 0.35+ 0 85 Lil-gum I} ‘16 :71 +3{16] '- —"'« Via = 1.81 V Then from Equation (16-34): (5 — u.as)+15(1.31- 0.35} V = “w (1+2(15)) VOLU .-. 0.591V V“, = 0.35 Vnflu = 4.15 50 NM; =‘V1L — VOL” =0.35-0.591 =5 NM; = 0.259 V NM}; = V633 -V13 =4.15-1.81 #NME =2.34V 1316.7 Prom Equation {16-35}: VIL=I+ 1'7 3V“ =1.3]V 3(5)“) Than from Equation (16-37): Vina = (5 -1.T) +(s)(1.31 $1) = 6.55 V From eqmion (16-41): Vm = 1+ 4—213) a (3K5) VIE =1.38 V Than fl'om Bqualion (16-40): 1.88 — 1 2 NM; = Vu. — Vow =1.31 — 0.44 VOLU = : Vow = 0.44 V => NM; = 0.37 V NMH = Vast! - V13 = 4.85 — 1.83 =$ NME=197V E163 a. i. A=lazic1=10V, B=logiso - “A” drive: in nonsuumion. "a" driver o5 [%I%).“”m)’ = [39420. 4mm 4%] 2(3)? = (ln}{2(1o —- 1.5m. — 9132;] 9 = 5(17Vu1. - V021.) 5V9“. — 35%;. + 9 = o 35 i\/[85)1 ~ 4mm var. = ‘""—““—"_— 2(a) => %L_=M_[LT_‘L A=B=logicl = [%I%)D[2(vj — may“ -V;L] 2(a)” =‘(2)(101[2(m - 1.5m. - v1.3.1 9 = 10(17; — W...) 101/02,, — 170V“. +9 = o 170 i Mum): u 44mm) 200) => V“ = 0.0531 V ‘51.: h. Both cases. 35 in = 74mm“ =315 mt.» P=in-Vpp =5- 2=;,1§mw E169 P=iD'VDD =>ip= %= 160.151“; - _ _§. 5'. =— 1 :0—150— 2 (L)L(l.4} —34.3(L)L “4.66 L - . ‘ W m = 160 = (T)D[2{5 - 0.3)(o.12) — (n.12fl W =9 --9.20 BIGJO EDD P=io-Vno=>in=-g—=1603A -. -E E 1 213—160.. 2 (LL-(1.4) =5 = 4.55 .....—..L__ in = [5015A _£.l. E _ =1 _ 2 3 (Llama mania—{212)} =5 = 27.6 E1611. 3. From the load mister: Jr: W Int. =(‘f I]L(VM‘VM)J = 55540.5)(5 — 0.15 — 0.731 or In; =15G.T “A Maximum Va occur: when either A or B is high and C is high. For the: two MOS in suits. the. effecfive kN iscutiuhalf.so I... =§[[%I{-]D}2(vw- WWW-I231 150.7 = 9] [2(5 —o.7)(o.15)— (0.15)”) which yields =13.s h. P = in - Vpp = (150.7](5] =5 P = 753 Ew £16.12 I. uo[mu)aucur:whanA=B=1de=D=Dor A=B=OlndC=D=l [21mm ==a£242<vwr2 (0.5)(L2)2 = ( 0.72 = (0.634%?) D 2.» (I) = 1.14 b. .-,, {flag-my =[? tun—(4211‘ in =12.6 14A P = in - Vpp = (12.6)(5) => P = 63 EW 316.13 a. KJK, =1 v": W3“. =5v 1+1 Vm=fl+lVrP|=5+2=Vm=7V V“, =P’n-Vm =5——2=>Vm =3V 3;) [2(5 — n.7)(o.15) —(u.15)'] D tel: xemic l ‘ons b. K_/K’=O.5 __,, v“ = W W=>V1‘=5.52V 1+ 0.5 V6”: 752" Wyn =3.52V c. K‘IK,=2 10 -2+\/52 =>Vn= 1+fiU=>Vn=L49V yup: V Van-[=149V 516.14 a. KH=K’=50MIV1 1”“: 2.5V infirm) = m. 42)‘ = smzs—o-s)‘ =a-1'p(ma.x')= 145 FA 2. K_=K, doom/I” v“ = 2.5 v {Dunn} = (200)(2.5 — 0.2)2 :9 {fit-nu) = 57% ,uA E1615 P = f . c; - V3], (2.1:: x 10“) = f(0.5 x 10'“)(:*.)I f = 2.22 x 10‘ Hz => {= 22.2 kEz 516.16 I. KJK' = 200/80 = 2.5 3v": m‘u' "5'5 2}=>1&.=4.32v 1+ 2.5 ‘41P: =6,32 v v“... = 2.22 v 10—2—2 2.5 = - -1 b“ V” 2+ 2.5-1 [2 2.5+2 ] :9- V“ = 3.39 V Vin-:0 = %{(1 + 2.5)(2.29) + w — (2.5)(2) +2} Vaflv = 9.43 V 10—2—2‘[ 2(2.5) I] V = 2 — ”‘ + 2.5 — 1 :73(2.s}+ 1 »W V _ (4.as)(1 + 2.51— m — 9.5}(2) + 2 “L” ‘ 2(2.5) 23w = 0.502 v ‘ ' ‘ ' chsi 2“cl editio ti n M c. NM; = v” - V“... = 3.39 — 0.302 : NM: =2.59V NM}: = Wm" -VJ'H = 9.43—4.86 =- NMH :45? V 516.17 _ 5 — 2 + (l)(6.B) “ V“ ‘ 1+1 V“ = 1.9 v WP! = V VON: = L] V b. V“, =o.s+%-[5—2flo.a] :9 V” = 1.63 V Vnm: a: %{2(1.63)+ 5 —- 0.8 + 2} W)"; = 4.73 V V13 = 0.8 + %[5 - 2 — 0.5} =5 VIE = 3.18 V vs“, =%{2(2.1s)- 5 — 0.3 + 2} Vogu = 0.275 V C- NHL = V1; - Wm: = 1.53 -—- 0.275 : NMI = 1.35 V NM}; = Vogu - V”; a 4.73 — 2.18 :9 NMH 32.55V 1516.18 a. AaGaMgu (amaze-0mm) .45 z) 2 L N g = 5. BE) =1.[£]..{Z) 1' 2 L, 2 2 LN K 1 2 K,=2KN=E"—= F From Equation (16—55). 5-1+v’0.5 l) V =——A—QV =2.76V " 1+J63 II Vnpg = 3.76 V “IN: = 1.76 V b. From Bqunt'mn (16-58011) Edi-'3'“) = Eng/1': ' Vm )1 -35 Ev... 3— so_ 2(L Mucus. 1} _54.. L Iv W -— =0.2 =>(L)~ 93 W .. (T) n = [3)(0323) = .33 . [E] Afilfl) L ."2 2 3 L , Or (Wm! £16.21 Or =. m = 3 (Wing 9 £16.22 mos: Mm,Mm in series :5 [%)=2 Mm.Mm in parallel = g}: MM, in series with MM,an :9 = 2 . . W . Efi'ectwe oomposste =1 for each side. .n ' Exe ie 'ons PMOS: £16.25 Wm! the effective composite of each side to be 2‘ The NMOS part of the circuit is: M" ,Mm in series :9 [%)=4 MM,MP, in parallel =- :4 L s MFD,M,., in series :- (-pE—)=8 £16.23 a. w=¢=5Vafl=4V h. v;=3V.da=5V=>gg=3V c. v;=4.2V,é=5V=>£9=IlV d. w=SV,¢=3V=:>ug=2V E1627 (3.) v, =8V,¢=l0V=>vm=8V M,J innonsatursfion Kaboom " mall'o ' V3] Kzlyw “'o ‘lez K - 1 1 13.16.24 film;— 2)(05)—(05) ]= [10—05— 1] MNMOSpmofthecircuitis: 3 fig)” KL ‘ (b) v,=¢5=8V::-vm=6V %[2(5— 2x05) -(05)‘] = [10 — 05 - 2? K 2') -—”= 15 KL £15.23 Exclusive-OR A a f D 0 0 l. O 1 0 1 l a 1 1 o J_ '4 a J— 3; bl ' d Desi 2"d ed'tion s lutions a! £16.29 N El 6.32 _L_ From Equetion (16.93) , 5 (we. 2 worm-3r; (WILL. {Von " nu )1 12 q, = 2(25}(0.4)- again)2 = 0526 (25 - 2[O.4)) : C‘- From Equation (16.95) —I - (WILL _ k; 2(Vme) - We; a v —'———-'— NMOS conducting for 0 s w 5 4.2 V QNMOS Conducting: 0 g t g 8.4 5 mos Cutoff: 3.4 515103 PMOSctttnffforflgwglzv =PMOS Cutofl': o g r 5 2.4 5 mos Corneal-in; 2.4 g r g 10 s +(fl: e‘) ’0 E1630 (a) 1K = 32:32 array Each row and column requires a. 5-bit word :5 6 transistors per row and column. => 3216+ 32x6 = 384 transistors plus 1311fo transistors. (b) 4 K 3 64x64 array Each row and column requires a 6-bit word: 7 transistors per row and column => 641:? +64x‘7 = 396 transistors plus buffer transistors. (c) 16K=1283128 array Each row and column requires a 1-bit word: 8 transistors per row and column => 128:8 + 12818 = 20-48 transistors plus buffer transistors. E1631 16 K => 115384 cells Total Power = 125 MW = (2.5))“, =>Ir==SOJnA 50w! Mmfioreeehoell,!-m=>l—105M J’m lazii Now,I= R orR— I 3‘05: R=032W (W113)... — 1‘; (Von +17»)! _ W = 425i (2.5-0.4)1 l m So of transmission gate devise must be < 0526 times the of the NMOS transistors in the inverter cell. The of the PMOS transistors . W . . must be <131 times the of the transmissmn gate devices. Then the of the PMOS devices must be < 0.589 times ot‘NMOS devices in cell. E1633 Initial voltage across the storage capacitor = rim—V” =3—05= 25?. Now W I -—I=C— V=-—-r K (it Of C + 25 where K:25V_ t=l.5m.r, r=?=usr, and C = 0.05 pF . Then 1(15x10'3) —-—— z: (0.05xw'“) I = 4.17:10‘“ A = I = 41.? p4 1.25:2.5- Elcc nic ircuit nal si andD i 2“Cl cditio “Mons Manual Chapter 16 Problem Solutions 5° '11" Enifl‘é—‘i P = Ip[mu} - V59 (3) AVm z —C“‘"!‘['Jz¢p+yn “42¢fi] or E = QIQIZS yaw, 3.9 8.35 10-“ CI=E_=LL:T_)_;167I104 [6.3 1.. 450x10 a. pm Equation (16.10). the minor puim is fauna] JR 5, N. from _ _ Ir: = [2(1.6x10 ")(11.7)(s.35x10 ")(sxio“)] £an Jury +0," _ Vm )4“, = 0 =5.15x10" K. gsomwfi RD=20m,Vm =03V Then 2 0.05)(20)(V —V ) +01, -Vm)—-S =0 5.1 51:10" ( I. IN AV”, _ W- [J2(0.343) + VS, — ,}2(0.343)] For V 21V: —1i,71+4{0.05}(20)(5) B V" 4"” = 2(005)(20) arm = 0.57 ![Jl.686 u JOfiSG] :9 Mn. = 0.316 V - For V$=2V: =1.79V50V1.=2.59V AV”, = 0.671[Jz.686 - Jesse] 2) AV”, = 0.544 V I0. = 1.79 v r = 5 v ‘ dumminad fr - (b) For Vcs =15 V. V55 =5 V, transistor biasediathe or w '3 0mm marina mama. _ 2 yo = 5 — (0.05)(20)[2[5 — 0.3)”. — pg] ‘ AID—KARE- m) u§~9Avo+S=0 For V53 = 0. f2—__ ID = 0.2(2.5 — 0.0)1 = 0.575 um 9" * (9‘4) ' “mm x 0.500 v So v0 = 2 1 Fa V53 = 1' ( ) ID = 0.2[2.5 -[0.s 4.0.3101)” = 0.303 M b_ For RD g m m. For Vsa = 2r In .-.. 0.2(25 - [0.0 + 0.0401)”I = 0.207 M (V V J _ -1i Jl+4(0-°5)(200}(5) 1‘ ’7" ‘ 2(0.05)(200) 16.2 V —V = 0.559 V 50 V“- = 1.45 v (a)! = 9" "=10, V -V v -v' n D [2(GI m)a 01 Vu=nfisgv s-(m) =K. 2(5-0.s)(0.1]|—(0.1)1 ‘fOxlO’ [ 1 m, = 5 — (0.05)(200)[2(5 - 0.5)»0 - "31 or K, =1.476x10" AIV' = 8x10 [9:] or 1003 - 05:0 +5 = 0 W 2 L f 2 So that (T) = 3-09 _ 35 5‘ (.__.—35) ' mans] = 0 0592 v _....._.._..._ ”° ‘ 2(10) ' b. Pram Equation (16.10). mam. ~Vm1‘ + V. -Vm]—Vm =0 (0.1476)(4D][Vn - 0.0]3 + m. — 0.3] — 5 = 0 —-1 sin/(1)1 +4(0.1470)(40)(5) or [Vrr - 0-5] = 2(0.1‘t‘75}(4°) or [10. - 0.3] = 0.039 ° ' ' al ' and!) i 2“ edition Soluti 115 Manual 16.4 16.5 P=in‘Vm l=iD(5)=>fn =02mA Now in = K-[2(V: ‘Vm)"a ""31 0.2 = K_[2(s-' 0.0)(02) -(0.2)’] Also or K" =0.122m.4w= {9.99 E'. 2 L): 11‘, = Van -K_RD[2(1I, -Vn. )vlg — 11;] 02 = s — (0.122)R,[2(s - 0.3](02) — (02)1 ] => RD=24kfl 1310:113qu (16.21): 300 a- d...’ 1 2+2(1+ 50 V“: “W Va :15, 4m =4.67—2= Va, =2.67V 200 1+— Pm Bquuion (16.23): 200[2(a — 2M - "3 4121;. ~v§j - [a — Va]: = [64 — 1st + ya] 'I_ J .. 5v} -64w+64'=0 v0: «5:139" 2(5) 50 500.0 - 1m — 2]2 04 :1: £54)“ - «(5)0540 16.6 (a) From Equation (16.23) %—[2{3 - 0.5)(025) - (0.2s)‘] = (3 - 0.25 — 0.5)2 :9 fi =4.26 K1. (b) %[2(25 — 0.5)(025) - (0.231] = (3 — 0.25 — 05)“ :3 52— = 5.06 KL 2 (c) in = KL(VGSL ‘Vm1.)2 = KJVDJJ " “a "VI-ML) = [g 13(3 — 0.25- 0.5)2 => 1'.D = 0203 m P =1‘D-Vm = (0.203)(3} = P = 0.608 mW for both parts (a) and (b). 16.7 (a) P=O.4mW=iD-Vm=in(3):5 I'D =0.I33mA 1-,, = KAI/m —v,, *Vm): 0.133 = [M Z] (3- 0.1-0.5}2 = (0.2304)[E) 2 L L L L W S -- = 0.577 ° iii—[205 -05)(0.1) - (0.1)’] = (3 — 0.1 — 05)’ K0 W —=14.a that — =ss4 L D V 3—05+05{1+J14.3) fl [+4143 or V” =10: V, Va, =0.52V 03) W 1. = K11. "Vow NMH = Vaflu 'KH From Equation (16.35) I, _05+(3-05) (1+2(14-31')_l : m" 14.8 m Vm =l.10V VOW =3.0-0.5=2.5V NMfl =VOW 4!,” = 2.51—1.10: W” = 1401/ Va =Vm =0.5V chapter 16: Problem §glufions (Von ' my?“ 'an) mm: 1... El {Ki 1'. = (3 - 0.5) +14.s(1.1— 0.5) 1) 1 + 2(143) Vow = 0.372 V Then NML = V,,_ — Vow = 0.5 — 0.3 72 2:. AWL = 0.123 V 16.8 We have K “KARO: “an)"o ‘Vclvl = (Von ' “a “V1140: 1. [26,013 J?" - Vm)(0.08Vm) —(o.03Vm)’] = (V0,, — 0.0mm —Vm )2 (WILL, (WILL [2(Vm, — 2(0.2)Vw)(o.oww) — 0.110154%] = [(092 - 1):)11'90]2 = 0.511341%, (WILL, _ (WILL, _ (WILL [0.096] - 05134 :1» _ 54 (WILL ' 16.9 VM=V,-Vw=Logicl So (a)V,=4V= Vofl=3V (b)V,=SV::> Vm=4V (c) V,=6V=> Vm=SV (d) V,=7V:> Vm=6V For 1:]' = Va" Kb 20’: 'Vr)"a “V1131: KL[VJ ' “o ‘Vrr Then (a) (1)[2(3 — 1)Va - V01] = (0.4)[4 — VOL — 1]1 = V“ = 0.557 V (b) (1)[2(4 — 0V“ — 14;} = (0.4)[5 - VOL - 11' => VOL = 0.791 V (c) (1}[2(s - 1)Vm -V&] = (0.4)[5 — VOL a 1]2 = V“ = 0.935 V (d) (1)[2(6- 1):!“ —Vg,,] = (0.4)[1 — V0L - 1]1 = VOL = 1.03 V 16.10 I. Furload Va=Vm+Vm=sa2=3V ’K E? "(Va “ yawn} = 'Vm 500 —--V -u.3 =— —2 1001!: 1 ( 1 =} V], = 1.69V Load VOI=3V Driver: V",‘ = V" ‘Vm = 1.69 - 0.3 = 0.39 V VI. =1.69 V Driver VB. = 0.89 V b. From Equmion (1629(1m: %- [21s - 0.3m — as] = 1—1-2)? Syg—i2ua+4=0 _ 42 :t 1/112)“ — 4(5)(4) __ _________..=a,. 920.0963‘.’ Ia'u 2(5) U c- [a = KL(—Vm)2 = mo[‘{—2}]2 =’ in = 400 114 16.11 [$02123 - o.s)(o.1) - (0.1)‘] = (—F’m)2 So (.Vm)‘ = 49 :) Vm = —2.21V 16.12 (a) P =fD'Vw wows-(3)21, =50” in = KL("VM)2 m1? V11+1= %[2(3 —o.s)(o.1}-(o.1)’] = [-(-1)]1 KB _ (WILL, = 104 z, [130 = 255 E ‘ (Wth For the Load: V0, =Vm+Vm =3-1=> Va, =2V J2_04(V,, - 05) = [-(-1)] 2 VI, = 121) V For the Driver: V0, = V" —Vm =1.20-05= Va, = OJOV Vn =l.20V ' ' d i 2'"cl edi 'on olu '0 Manual 3(204) Then 1/0,,“ = (3 — 1) + (2.04}(0.902 — 0.5) = 2.112 V Vow = EELS] = 0.405 V NM: = 0.902- 0.405: NML = 0.497 1’ NM" = 232—1312: W” = 1.511! 16.13 a. Plum Eqmfinn (1629M): (1?) 9120.5 — 0-00-05) - til-051’} = (%)LH-1)]‘ 00:1 W 111E = 5.06 b. in = 1111-1-11? fl' is = ‘10 EA P as in - Vpp = {40)(2.5) =31 P = 190 EW 16.14 I. L 01=0.5V=9-ip=0=:ofi=g iL y, = 5 v, Plum mam (15.12), m, - 5 .. (u.1)(20)[2{5 - 1.51m: — v31 21': — 15w + 5 = 0 15 : 1:05)” — 4123(5) In; = 2(2) #5 =0.35V 5-0.35 = . 25M 2“ 0’23 P =i1: - Von = (03325113) => W ip= b' i. »,=0.25v=9ip=0=>g=9 u: = 4.3- V, From Equnlion (16.23]. 100{2{4.3 — 0.1M — .0} = 10[5 — u. — 0.1]2 10[7.2»o - 1-3} = 10.49 — 8.6110 + 03 Then 11»: — 80.6110 + 13.49 = 0 _ 30.5 :1: “(30.6): —— 1(11)(1a.49) y" ‘ 2(11) :3- fl ..-.. 0.237 V 111m in =1o[5 - 0.237 41.7]2 :05 .011 P: in ‘VDD 2- (155)(5) => P: 325 pW C. L vr=0.03v=>in?9=>2=0 w=5V 1.. = 01-10.? 4101—1-01 =40 m P: in - Vpp = (40H5) =9- P =‘ZDU 11W 16.15 From Equation [16.35) VIE =0.5+5_"{°'_B). { 1+3119) 1} 10 ' 311-300)— 21 V = CI. 0.4 - — “1' B + 2 {5.57 1} O! V”! = 1.96 V M5; in non-samfinn‘region. Then Kn[2("an ' V17: )Vm —v;fl] = K1[Vm ‘voz -Vm]: "as: = "on Sfld V652 = V”; = 1.96 10[2(1.90 — 0.3m, — 113,! = [5 - you — 0.1112 23.2w: - 1M, = 17.64 — 3.4m: + 0:, GI 111-3, — 31.00“ + 17.04 = 0 _ 31.6 *1/(31033 - 4(11){17.641 "2 ‘ 2(11) 9‘ a: = 0.753 V ha ter 16: 1e luti s Nome inman 11m 2 Kn[v, - m] =KL[VDD -'1q.,,—1r',,,l,,]I J10 . (w — 0.3) = 5 — 1.9-6 — 9.3 = 2.24 Then VI = 1.51 V 16.16 a. From Eqult'mn (16.41.). _ 214-2)] V”; -I].8+ m => Mm in nan-summit»: and ML: in 5mm Kn[2("m ‘Vm)vo: " “31] = KLc-VM ): 4[2(l.95 — 0.3)902 ‘ "31] = 1111—1—31: 41,3, — 9.2%.: +4 = G 1/”! =1.95 V = Hg: 9.2:: 1/92}: 400(4) an: = 2(4) =3» yo: = 0.582 V Bod: Mm and. ML: in stflnfinn region From Equa- dun (16.28033). fi- (u: - 0.5} = —(-2) or HI = [.3 V (+2) 7W M9: in mutton. M3: in am-samtiou KD[v01-Vm]: = x.[2(-vm)(s w...) —(s—»-...:'] {(1.25 — 0.3)” = mus — U021-(5 - m)2 (a - an)“ — 4(5 - 90211-1131 = o 2 - . 5 - "a: = 1d: (4) “mo 81) = 0.214 V 2(1) 1:. Va. a 0-5-1- = 1.25 V = Mn Io m = 4.735 V To find w: 413-»: - 0-3}1 = 1111-1—21): Va] —°.B = l um I: 1.3- V :_ V13- : 1.95 V, V”, = [.25 V 16.17 Li. Neglecfin; this body street. W=Vnn~Vn MWVDD=5V. thenao=L2V Taking the body cfi'ac: into account: From Problem 16.1. Vm = Vmo + 0.671[Jo.636 + V” - J0.686] 1nd V53 = “n Then 1-. = 5 - [0.5 +0.571(m- m] V. = 4.756 — mum 037140.536 + 1'0 = 4.756 - vn 0.450(1636 + an)’ = 22.52 - 9.51% + .13 u: — 9.95». + 22.3 = u 9.96 :1: ‘1’ (91913.)2 — «22.3) 2 33:31:01! Ila: b. PSpioc results similarto Figure 16.180). 16.18 Results similar to Figure 16. 18(1)). 16.19 a. Mx on. My cumfi. me Equation (16.291111): %[2(5-o.s)(oz)-(02)']=[-(-2)]' or %=2.44 h. Favx=vy=5v 2(2.44)[2(5 — ma)». — v3] = Héz)!’ 4.53:1: - 41.0». + 4 = n _ tum-11)“ - «4.331(4) W ' 2(4.as) m = 0.095? V c. in = (gym—(4)]: = 161.1 uA P= (1so)(5) =- P=_soofl for both mm (1) mm. 16.20 (I) Maximum value of v9 in low state- when only one input is high, then. §f[2(a—osxo.n)-(on’]=[+0]“ K0 — = 2.04 K1. (b) PHD-V“, 111:1“.,(3)=:u'D = 33311,: 333 = [8?" fir-1H4}? = = 0.3325 Then [i] =1.?0 L .o (c) 3(2.04)[2(3 — 0.5)v9 — vg] = [—(~1)]’ 3 v0 = 0.0329 V 16.21 I. P = in - Van 2511 = 19(5) => in = so 1111 in =111111-vmr so = "MM-2)!” So that = 0.417 _L_¥£L_ %[2("1 ' who " v3] = [‘Vm]: L %a[2(s — 03)(o.15) -(o.15)’]= [-(-2)]‘ I. Kn: E.) =135 °' Tr" 32” [L m. I. h. Fuvx=uy=0=bvog=5muo3=4.2 Then Kmlzb'm ' mayo: ""3:] +Kna[2("o: ' Vm)”oa - "31] = KL1[_VM2]: Km $8,113, x8_Ku ucl a[2(5 - on)”: - #32] + s[2(4.2 - 0.8m: — #33} = [Ill-(4)}a 67.2w; - 31:3, + 54.1”: - avg, = 4 Then 15»: — 121.601: +4 2 a 121.5 i fluzm)’ — «153(4) “‘1 5 2(16) So a: = 0.0330 V I. W: con write xx[2("x ’ Vm )“mr " vim] = Ky[2("r “’9: 'Vm)"nsr "01:19] = KL[VDD “v0 "VTNT Where P1: = Upsx + vnsv Wehavo v.Ir =15. =9.2V,VDD=10V,VTN =0.8V As a good first Approzdmnlion. neglect the v3“- and vi,” mum. Lot m z zygsx.1‘hcn from tho firs: and third term: in the. above equation. 9{2(9.2 - 0.3%“) 5 (1)110 - 21:55,: — 0.3)“ (151.2)m5x ; 84.64 — 36.81455): So that 1'95; = 0.450 V From the. first and second norms of the ohm: nqundon. 9{2(9.2 — 0.5)vnsx] E 9[2[9.2 — Irng — 0.3}.Vysy] 01' (15.3)(o.45} = 2(9.2 - 0.45 — 0.3)UQSY which yields 1; = 0.475 V Than yo = 3115;: + unsy = 0.450 + 0.475 or = 0.925 V We have #55: = 9.2 V l-lld- !!ch = 9-2 - 9135:: = 9.2 v 0.45 or Hair = 5.75- V 0. Sins: m is closem ground potential. the 1:11:11}.I efi'nct will have minimal afloat on the mulls. From 11 P550113 mllysis: For ran-1(a) : “on: = 0.462 V. rosy = 0.491V, Mo = 0.9536 V, UGSX = 9.2 V. and PGSY = 3.738 V For part (b) : vnsx = 0.441.", rpgy = 0.475 V, 90 = 9.9154 V. Pasx = 9.3 V, and was? = 8.759 V 16.23 I. ‘ We can write. K41": ‘ m)"m ’szm'] = K,[2(Vr “Von "' who“ " v3.17] = [id—FM? From 111: first and third harms. (11631th vEsx). 4[2(s —- 0.31115”) = (HM—1.5)]I 0" W From the second and third tun-ms. (neglect Firs-v). 42(5 — 0.057 — 0.0)upsy] = (1)[—-[-—1.5)]I 0' 2w Now #65:: = 5. vasv = 5 — 0.067 s Hasy = 4.933 V and Va = 9:352: + Unsr =- P9 = 0.135 V Sinmunisclusewpoundpmmmebody-efiem halitzlnefiectonuzcreaulta. 16.24 ComplanentafCBANDCIORA :(B-C)+A 16.25 Considering 0 and: able. we find A B . Y 0 0 0 O 1 1 1 0 l. 1 1 0 whicth flutthedmuitpaforma theme fun:- than. 16.26 (a) Carly-out = Ao{B+C)+B.C 140’ 7" Vain“ a ter 6: 1: tie (5)170" Val = W=02V %[2{5- 0.8)(02) _(02)‘]= {+15}? = W W F — = .h -— =13? «(Ll [100(12)D So, for M6: =13? 6 To achieve the required composite conduction parameter, For M. -M,: =2J'4 l-S 16.28 I. From Equafim (16.55). 5-0.3+0.3 1+1 =VEE=2.5V p-chnnnal. 1’”. =2.5 —(—-0.51 =- Vge, :33 V' nwchmmLVum =2.5-0.8=> V“! = 1.7V ¥I=Wt= 0 Form =2 V.NMOSinutImfionmdPMOSin nunatumiou. Fran Eqmfim (16.49). (2 - 0.3)1 = [2(5 — 2 — 0.0)(5 - Va) - (5 — pm} 1.44 = 4.4(5- vo} - (5 ~ 00)2 So (5 — we): — 4.4(5 - p.)+1.44 = 0 2 - I I5 _ W) I 4.»; $1114.41 4mm 44) 2 O! 5-m=0.356=>g=4.64v BYM.MIQ=3V, 5:0.356V 16.29 (a) x_ = 2) =00 mm K, = [%}4) = 80 MN” K 1’ +1' + —-"--V * "° "' 4K, "’ 3.3—0.4+[1)(0_4) (i) V” - H K. - 1+ 1 KP V" = L65 V PMOS: IQ, = P}, - V". = Les—(4.4) = Va, = 2.051! mos; _ Va =1}, 4/,” = l.65—(0.4}=> Va, =11er (iii) For 1-,, = 0.4 V: NMOS: Non-sat: PMOS:Sat Ku[z(yam — WV.“ _ V1719] = xiv” 4'le: 2(v, — o.4)(o.4) — (0.4)“ = (3.3 * v, - 0.4)2 :a v, = 1.89 V For v0 = 2.9 V , By symmetry 1:, 21.55 - (1,39 «1.65):- v, = L41 V (b) K,=[? 2)=so,u4w= K, = [329)” = 40 mm? 33—‘141' 59 -(o4) a} V_= 4” =3 Vn=144V 80 1+ -—-— 4o PMOS: Va = L44-(—o.4) = Va, = 1.34 V NMOS: V0, =1.44—o.4= 14,, = [.04V (iii) For v0 = 0.4 V (80)[2(v, -o.4)(o.4) -(o.4)' ] = (4o)[3.3 — v, - 0.4]2 :5 9, =15: V For WWOSIS-at, PMDS:Non-sat (so)[v, — cm]1 = (4o)[2(33 — u, - o.4)(o.4) - (o.4)'] :9: v, = 1.16? 16.30 a. For 'vm =0.6<Vm :> um =5V N. in nonmminn and P, in saturation. From Equation (16.57), [am - o.a)(o.a) - (as?) = [5 - u. — 0.31“ 1.2" —-1.32 = 17.64 - 3.4»; + u} 01' u} - 9.3»; + 15.95 - o 9.6 :I: “(9.6): — 4(1)(1a.95) 1': = 2 v: = 2.73 V 5- Vom S "on 5 Von Fm symmetry. VI. = 2.5 V WP: = 2.5+ 0.5 = 3.3 V I131 WIN. = 2.5 — 0.3 = 1.7 V 501.7(53 (3.3V ition Solut't ns Manual 16.31. a. van: 5 I'm 5, Von By aymmctry. V" =15 V Var: = 2.5 + 0.8- = 3-3 V andVDN‘ = 2.5 —-0.8 =1.7 V So 1.1' < am < 3.3 V b_ For v02 =0.6<Vm 2 v0] =5V N2 in nonsaturation and P1 in saturation. From Equation (16.57), [2m2 . o.s)(o.s) - (0.5)? = [5 — um - 0.3]2 1.2m — 1.32 =11.“ - 3.4m + p}, or v.3, - 9.6m +1335 2 0 50 W2 =zgz._=_.3fll Put ya; =2.78. both N1 and P1 in SEMBfiDn. Then vi = 2.5 V [6.32 '- Jipfl = JEN 'Vm) Jim: = M- (2.5 —o.a) = 0.533 (MW 1:. Jr... = Jul [7.5 — 0.3) = 2.12 (mum 16.33 (a) K,=(§29 2)=so;:.4nrz1 K =[% 4):.5011411/3 r In.” = K-("I "Farr = sqzj'os]: or [A]... =1445pA a ter : Probiem lutio s (b) K_=SOMIV'.K,=25,MJV’ 3" _ It, From Equation (16.55), i = fiLg .2(Vw + pm} 5n '3' 9 4v 5-0.8+ —(0.8] or V" = 25 22.2w 1 [SO r“ = i+ — W Them 25 k; I'd—JV”: W") P I...M = K_(V,, 42m)1 = 50(2111—013)2 or JD_M=99.4;;A (b)For[E) =2'[E) :4 L _ L , 16.34 I m - :> 22.38 a. P = fCr. V59 r“ (so)(2){5 —03) r“ For Von = s v, P = (10 x 10') (0.2 x 1|:"’)(5)2 l = -—--—— ~ 2.33 m or P = 50 ,uW r“ (25)(4)(s—o.a) 2"“ For Van =15 V. P = (m x1o'){o.2 )u('lU—n)(15}:I F w ‘ OrP=450pW Dr I!” '- 1 b. For Von = 5 v, P = (10 x IG‘HM x 10"?){51z C: = W3C; =4-75m or P z 50 ,uW Now, for NMOS: . . v 05 . 1&35 1:41:33“ orr‘=—r-_f-=-f3—§:a 5:021)“ (a) P = {Qt/3,, ={150xm‘)(o.4x1o"‘)(s)’ For PMOS: = 15x10" Wlinverter For I." = V Total power: P, = (2x10‘)[15x10") a r', = r—“= m :3 i, = 0.21 mA :1 ' — P, =3000 W!!!! For r“ = 4.76m. (h) For f=3oomz :,="4=4l%= i‘=0.105m.4 r . — 15:10” =(300x10‘)(o.4x10"1}I/;, z: " V”, = 354 V 16.37 Flora Hawaiian (16.73) 16.36 (a) For v, 5V“, . NMOS in nonsamration V”, = 1.5 + 3-110 - 1.5 -1.5)=a- M in =Kn[2(v.r ‘ m)"m 'W’n] “d "as 50 mfiqmfim (16.72) so i=§LEK-[2(Vw_ym)] 14”“; = [3(4.125}+10 - 1.5 + 1.5] or r. n: or film =9.125 v 1 ct = kl W From .13) W»... or " Vm=1.5+:--(10—1.5—1.5)=>W , _ 1 1:41:13un {15.73) d _——F'—'_——'-_ l k;[%) .(Vm_ym) vow = i-[2(s.515)-— 10 — 1.5 + 1.5] " or V9“: = 0.375 V For 1:, 50, PMOS in nonsaturation in = Kp[2(VDD 'v: +Vzr)vm ‘vinl and VIDEO for 1:, ED. :11 Si csi 2“d dition New NM]. = V". — Va“; I 4.125 -0.375 :- NMI = 3.25 V NM" = $4,” _ v“, = 9.125 .- 5.375 =>NME=3.25V 16.38 me Eqmtion (16.71) zoo _. _5- . — V1L=L5+t10 l 15) 2 L—l 100 193+3 (5‘1 so = 1.5 + 7[2(0.632) - 1} V“ = 3.343 V P1011! (16.70) 1 100 Van” — {(1 + + 10 100 -(fi)(1.5} -l- L5} or V = 9.372 V From Equation (16.77) (10—1.5—1.5) 4%) (% ‘ 1) 3(152) +1 a 1.5 + 111.51 — 13 V13 =1.5 + VIE = 5.07 V Fm Eqmdon (16.76) (5.07)(1+ 1-5003) -1n — (gym) + 1.5 50 100 2(a) Now N1"; = V”. - Vega = 3.348 - 0.9275 Vow = or V9“: = 0.9275 V or NM = 2.42 V NMH = Vagu- - V”: = 9,272 - 5.07 OINME :42!) V oluti ns M 16.39 I. VA=II3=5V N1 andN; 011.50 ugslzupnzflv P1anquofi' So we have a P; - N: CMOS inverter. By symmetry. u: = 2.5 V (Intuition Point). b. Forug=vazucEw c. We have xn=£a§1=r§13m K.- {2419, Then fi'om Equation (16.55) s+(—o.s) + “K” .(03) K VI, = ’ K. 1+ . .1. Now i: = (1)8) Then V“ =W: V” =1.65V [6.40 BydnfiflfiomNMOSisonifgmvoltagcissvmtsofi fimvmbov. Stan: N1 N: N: Na. M _W_ $TVTT o of of on on of 0 0E 05 on 5 on on of on on 0 1 2 3 an on 4 (ax on w) @[vx AND vz) Exclusive 01?. of (In: 011 w)witl1{vx AND v5} ghapter 16: Emblem Solutions 16.41 16.42 (a) A classic design is shown: (a) A classic design is shown: _ _ (b)[ A .B .C signals supplied through inverters. [ _ (b) For Inverters, =1 and = 2 " W " 16.43 For PMOS in logic filnction, let = 1, then L r —"‘—”'{A on B) AND c for mos in Logic function, = 225 L . 16.44 Let = 1 for each PMOS: Composite PMOS I [%]= 5. Want composite = 2.5 for NMOS, So that = 5(25) = 125 for each NMOS. edition 16.45 W , . W (a)Let = l for each NMOS. Composute ofNMOS = 6. Want composite ot'PMOS = l2. Then L P =| and :2 for each transistor in L l L ’ = 6(12) = 72 for cash PMOS. Let invetter. (a) For 3~input NOR: Let =1 for each NMOS. Composite of NMOS = 3. Want composite ofPMOS = 6. Then = 3(6) = £8 for each PMOS. ’ For 2-input NAND: Let :1 for cock PMOS. Composite of P PMOS =- 2. Want composite of NMOS = l. W Then = 2 for eachNMOS. Sizes ofPMOS transistors in (b) are substantially less than those in (a). 16.46 By NMOSoEtfptevolug==o NMOSoniflltevfllllfle=5V PMOSoflifutevolnge=5V PMOSonifptevoltnge=o Slit: N1 P1 NA Na Na: V01 N: P: 1‘02 1 off on off of ofl' 5 on of! G 2 on off on of! off 5 on 03 0 3 off on 03' of! OH 5 on ofi 0 4 on DE DE ofl' on 5 on. ofi' 0 5 off on off of ofl’ 5 on ofi 0 6 on ofi' off on on 0 (:5 on 5 Logic human: m =(u4 OR 93) AND In: W1 5 0 5 5 5 0 W2 5 0 5 0 5 5 "03 0 S 0 5 0 0 v03 :00: OR Hz) AND VY 16.4? State 1 2 3 4 5 6 Logic flirtation: 16.48 CLK 16.49 (LR _____..__.. chapter 16: Problem Solutions 16.50 V2 1.7 I C: .2557? __fli 2!—— C d! 30 } AV =——.— .f C ) For AVc=—0.5V 2(2xlu-") - : —0.5= -—-——_l—;—- :3 t: 3.125 ms 25.!!!) —— 16.5! (a) (i) v0 = 0 (ii) v0 = 42 V (iii) v, = 2.5 V (b) Va = (a) v9 = 32 V (in ya =25V 16.52 Neglect the body efi'ect. I. vufloyie I) = 4.2 V. l) = 5 V b. vg=5V=>u091=4LzV M1 in nonsmdon and M; in setunfion. From Equa- tion [16.23) [i§)n[2(vm Java. - v3.1 {53me - 4m? [201.2 - n.s)(o.1) - (0.132} D = (1:[5 — 0.1 —0.a]2 Dr W W (7)9(0-67) = 16.31 =9 (73.) = 251 New PM =4-2V =9 965': =4.2V M; in nonsmtion and M4 in smuatinn. From Equg. tion (162903)). W W [205353 -Vl'h'D)v02 'an]=[f] [—VIM]:I D L [2(42 — u.3)(o.1) .—[o.1)’j = (2)[_(-1.5n= o «In: — 1 '15 (L )Dmm) _ _... Or = 3.36 16.53 For¢=l'$=uummY=B. Andfnrm=0,3=1_ then Y = A. A multiplexer. 16.54 A :5 Y 0 0 1 D L 0 1 0 0 1 1 0. 1 :Mduerminate Without the top transistor. the circuit perfomu the exclusive- NOR (Intermix. 16.55 A B Y TTT 1 0 1 O 1 I 1 1 O Exclusive-OR function. 16.56 This eheuit is mien-ed m u I memes: circuit. Idenfieel minimum-sized mister: em be used mrwghour. When é: 'ulow,C: ischugedtono. Then when m is highmdmislow.Memm50n. IfA: B=D.lhen MamiM. mufisoC;remainsehuzedmd.-m =hig|1 Whenéazoeshignthenvm happliedmthegateeofMg ande. Theeileuilpetformsfl'LeOR my: funaian. 16.57 Thiseircuitisrefenedmlsutwo-phasemdoedc‘arcuit. The same mam-mung]: ratios between the driver and load transistors must he maintained I: discussed previmsly with the enhancement lead inverter. Whendu ishigh. m beenmeudzeemnplememofuz. When 95; goes high. then Ira become: the complement of um or is mesa-nan: w. Theekeuitlaashifileginm. 16.53 Let Q=O-md Q's-Ins S increases, 3 decreases. When @- reedtes the trm'ttion point of the M, - MG inverter. the flip-flop with change state. From Equation (16.28119), fK V11: EL‘(—Vm)+Van D where KL = K; and KB = K,. Then V" = g-H—znu 2 V}, =5: 2.095V This is the region where both MI and M, are biased in the saturation region. Then S=E.(—-VM)+VM= %.[_(_2)]+1 or S 21.77]! This analysis neglects the effect of MI starting to turn on at the same time. 16.59 Law-all. vx=$. Vaz=Q, Induu1=6. Assume VT...” =0.5V andV-mp =—fl.5 V. Fat-S =0. II: have the following: Vanggv ‘69 = 51/ M;— Hr. .. ' a G G. G M, l— E. Ifwemttteswitctfingtooecurfior.fl=2.5 V.then beuuseofthenmaymmenjrbetweenmetwodrwitx.we marmama'uomuuumzjv. SetR=Q=15demumeagoeabm Fu-flIeM1-M5 invu'ter.M1inmuurefionandM,-,in When K,[2(25 «05)Q - G 1] = K,[2.s —o.51’ Or For the other circuit, M2 —- M ‘ in saturation and M"i in nonsaturation. Then K,(2_5 — 05.)2 + K45 * 0.5)1 = K,[2(5 — Q' — 0.5)(25) - (2.5I )] Combining these equations and neglecting the (j 3 term, we find "Q" 14V d K' 09 2. an —-'= . — K V39 1 16.60 #31 if? a. Poaitiveedgetriggeredflip-flop whm_CLK = 1. output utmtmmnfimmeuq=fi=a b. For example. put a CMOS unamiflinn game between unautputandmepteefm drivenbyaCEKpulse. 16.62 FwJ=1.K=O.lndCLK=1;miamakesQ=lmd §=0. FUJ=0. K=LIMCLK=LmdlIQ=LEh=nm circuilisddvensoflut0=llmd§=L Huddully. Q =0, mentheeircuil is driven so that thereis noclmngeandQ=Dmd§=L 1:1.h’=1.andCLK=1.andifQ=1.thnnd1e 'CircuirisdxiveantQ=0. Uinitially.Q=0. d1enfl1eeirc1fltisdrivensethalQ=L So if I = K = I. the output changes state. 16.63 Forl=vx=1,K=uy=0,a.ndCLK=uz=L mauve—10. For]=vx=0,K=vy=11deLK=vz=L menuo=1 Nowoouaidch=K=CLK=LMthvx=vz=Llhe nutputiulmysm =0.Someoutpu1doesnurchangestate wheuJ=K=CLK=L 'l'hisisnuucmsllyIJ—K flip-flop. 16.64 64 K :> 65,536 transistors arranged in a 256x256 array. (3.) Each column and row decoder required 8 inputs. 0’) ('1) Address==01011110 so input = w‘ 5545351117“. (ii) Address = 11101111 so input = ETESESGIEIEIEIED (E) (i) Address= 00100111 so input = aflrfisaca 35151 Ea (ii) Addressa-Ollllml so input = “TESESE#EI3GE‘?IEO [6.65 Put 123 words in a 8x16 array, which means 8 row (or column) address lines and 16 column (or row) address lines. 16.66 Assume the address line is initially uncharged, then JV 1 I [=C—i V=— =——. d! or c CIIdr Ct Va .c _ (2.7)(5.8x10'”) 1' 2501110“ 1': 626x113" 53 62.6»: Then I: Cha ter 16: ble olutions 16.67 (a) 5 '10'1 = [3251%12(5 — 0.7)(01) — (0.1)‘] W —- =0.329 arm (b) 16K:>16,384 cells % #4 Power pereell =(2m)(21r) =4 1111' Total Power = P, = (4 pWX16384) :> P, = 65.5 mW Standby current = (2 MX16_334)=> I, = 323 m 16.68 16 K 216,384 cells P, = 200 mW :3 Power per cell '1: : 200 512.2111? 16,334 P 12.2 V 2.5 '=——=——-=4.ss E‘fl:——:> '9 V”, 2.5 “A R R R=0.512MQ Ifwe want v0 = 0.1 V for a logic 0, then [12' W 1In = T)[2(Vnn - VIN )vo ' 1’3] 4.88 = [%I%12[2_5 - 0.7)(01) — (o.1)’] W - -— 20.797 So [L] , 1 So T5=Logic1=5V 10 11 D 11 Avery short time aftertheruwhasbeen addressed, D remainschargedat Vw=SV. Then Mn, MA, and Mm begin to conductand D decreases. In _ steady-state. all three transistors are biased in the nonsatursfiun region. Then Kps[7-(Vms +Vmays ‘Vésjl = u[2(Vau ' Wynn 'Vdm] = K.I[2(Vosl " V1111)le - Vél] Or Kps[2(Vm + VmXVm ' D) "(Van "091] = “[2le-Q-Vm)(D-Q)-(D*Q)2] =K..[2(Vm—Vm)Q—Q‘] m Equating the first and third terms: [? 1)[2[s — 03x5 — D) —(s— of] = [%}2)[2(s—0.a)g —Q‘] (2) As a first approximation, neglect the (5— JD)2 and Q’ terms. We find Q = 1.25 —0.250 (3) Then, equating the first and second terms of Equation (1): [% 1)[2(s—os)(s-D)-(s-D)‘] 40 =[7 11215-0—o.s)(D—Q)—(D-Qr] Substituting Equation (3). we find as a first approximation: .0 = 2.!4 V Substituting this value of D into equation (2), we find 3.4(5— 2.14) -(s- 2.14}2 = 4[s.4Q- Q’] We find Q: 050V Using this value of Q, we can find a second approximation for D by equating the second and third terms of equation (I). We have 20[2{¢2-Q)(D~0)-(D—Q)‘] = 4o[2(420) — Q=1 Using Q= 050V, we find D= 1.791! [6.70 Initially MN. and M,l turn on. MNHNonsat; Mpsat. KM[Voo '9'“er = K,.[2(Vm - Vm')Q— Q1] [430 1)[5 - Q — 0.8]1 = [5&9)2)[2(5 — 0.8}Q — Q1] which yields Q = 0.771 V Initially 7M].2 and M3 turn on Both biased in nonsaturation reagion szizipnn + VrPIHVm _ m _ (V00 - m2] = Kn,[2(Vm —Vm)§-‘§ 1] [$14120 — 0.2}(5- g) —(5— 91’] = 1)[2(5— 0.s)'Q‘ -52] which yields 9' = 3.73 V Note: [W] L) ratios do not satisfy Equation (16.95) 16.71 For Logic 1, v]: (5)(0.05) +(4)(1) = (1+ 0.05)vt =. v, = 4.0475 V VI: (s)(0.025) +(4)(1) = (1+ 1.025):2 :2: v, = 4.0244 V For Logic 0, v1 : (D)(0.DS) +(4)(1) =_(1+ 0.05)v, :> v, = 3.8095 V {55(0025) +(4}(1) = [1 +0025)»2 = v, = 3.9024 V ...
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chap016 - ectr ' iu't si d at 2&amp;quot;“cdiin Chapter...

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