chap017 - Elec Exercise Solutions E111 . -O.7— — I. 15...

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Unformatted text preview: Elec Exercise Solutions E111 . -O.7— — I. 15 = —R—s{—fl 5 -3.5 Rel => Rc; = 39; = 31:0- I'c; =1c2 =0.5|:I:LA= b' L m=1v _ (140-71—(«51 '5 _ 4.3 {:1 = i5 = um = 5 - (1.23}(3) =55: =1.31 V a: = 5 V P; = -l V in=lmA=llI°3 :53 = 2 V Pa] = 5 V E172 a. Q}: on . 1.5 -. 0.7 —— {—3.5} _ I; :3 RE — 2 =5 R: = 2.15 m icn:is=2mh=3ls_2 Ra: 9 RE: = 0.75 m 'c i uit Anal sis and Desi 2"" edition Solui :15 Man al Chapter 17 =1mA=RE=L3kQ =- iE = 1.23 1113. m b. vx=vy=2V=vQ1mngon 2*03- —3.5 4.8 . = = — = . I; RR 2.15 4 IE 2 23 mix 3.5 —'2 1.5 = ' _—_ _ = _ 7 Ba: I”? 233 a g, o a 3 kn E173 logic l= —0.7 V lemngonwlnnux=w=—0.TV -o.7 - 0.7 _- {-s.2) __ 2 R3 - =r RE = 1.52 kn is: .5 0-1— . .7! mnn=-1.5&Rcl= =r RE; = 320 {I 2.5 Vn=——-L5—0.7 Qflon=>ig=$iizl§£=2iim chszfiRgz=35TQ — .7— -— . 517.4 Pficxv + icx +1} + i1}(5-2) I. ux=w=lngicl :icxy = 3.22 IDA icn = 0 i3 = —---—_D'7 + 5'2 = 3 mA 1.5 — .4 . i. = g 2.53 mA P = (3.22 + u + 3 + 2.53)(s.2) d» E = fiélfi law b. w: = w = logico =>€cxv = D {ex = 2.92 M i; = 2.53 mA I} = 3 ILA P = [u + 2.92 + 2.53 + 3)(5.2) a» E = 55.2 mW NM}; = -0.70 - (-0.93) = NME = 0.23 V NM; = -1.17 - (—1.40) =? NM: = 0.23 V Emmi; Qircyi; Analxgig and Design. 2“d edition . Sglgg'ong Manual E116 P = IQ - Vcc :- 03 = Iq(1.T] :9 [5 =118 1AA anon =9'w=l.T‘Iqflc=1.T—O.‘l 0.‘ = —-—— = 3. => RC 0.115 =- 39 39 in VR=IJ§L3fiVE=L5V E17,? State A 3 C Q01 Q02 Qua Q1 Q: QR 1 0 0 0 of '03 03‘ ofl' on on 2 1 0 0 “on‘ off 05 off on on 3 0 l. 0 (IE on of ofi’ on "GE" 4 D 0 1 oi of! on on ufi' on 5 1 1 0 on on 03 nfi' on off 6 1 O 1 on ofi' on on off "03" 7 0 l 1 0E on on on off on 8 1 l l on on on on 0E 05 '(AANDC) DR (BAND?) h—F-v w—r mfor n-unfor mmfiunda statea31nd5 W Gummgoashighforfllese'4 sums 817.8 Hawk-gown) =(AGBHBC DPQHDOH F... r-I—‘Dr-ODOQ .— FOOGFHHO§ VCI . 5- . Ic= 101=4.9m:\ ._4.9fl _v;-—07 4.3 ‘B— 3 -0.163 mA- RB _-—; =-R3=26.4kfl E1710 (a) VI = v, =5V v! = V,,_.,(s.:u)+2Vr = o.s+2(o.7) = 2.2 V i,=§lf—1=o.70m4 in = Va: #2560!) ____ 5-401 2123"” P = (i1 HEW“ = (0.70 +1.23)(5) or P = 9.65mW (b) wx=vr =0 :>vl=0.70V Vac—v1 _5-0.70 E1111 i‘=—RI-———4-—=1.08mA P=i,-Vcc=(l.08)(5):> P=S.4 MW 8.. Input: =D.8+D.7+G.3=2.3V ;, = {5.1 = 5" 2‘3 = 0.675 um. 1411 = 0.5 + 0.7 + 0.7 =1.6 V 5- 1.6 = 1.7 mA. i: = 1.7 + 0.675 = 2.38 mA. 0.8 H=T€=fififlmh in: = 2.33 - 9.05 = 2.3 {BA 5 ‘ "'1 = 1.23 mA I'm;- = 5 ' “‘3 = 1.05 am £1 = Him = N5}. + inc (3mm) = N(1.us)+1.23 : a = g b. Ic.r.g.d = mA- 20 = Ni1+£3c = N(1.05} + 1.23 Y st {1' ha cr17: E c ‘ lti ilx = any = i1 - flu = = “1A 3. . lax a: lay = 0.1 a v; = 0.3 is: = liml = 3" + {Ex +i5y = 0.45 +2[O.l|45) u = 5 10.5 = 0'? “IA = lic1| = 0.54 MA um =O.B+ 0.1 =03 fl='\" _ - iz=ica=51:'9 3i; =|g3=2.13mA b. Sam: as pan (2.]. i3: = 2.73 + 0.54 = 3.27 mA c. vx=vy=5V=>y1=0.8+0.T+0.T='2_2V ii=§l§=a533mA il =I2 = 3—52.? 9 '1 =i->=0.467 mA im=3.27—D.533 . . . 5—0.1 . . ‘ in=21§=>ig=a053mA “a”: 2.2 :L—L—' 2' =2'23MA ig=i1—ifi=>ia=0.4l4ml\ 1:“in —D. ‘ - - ‘ {H.325 , laiflg=flflflmA fl=lfl=5m<flp=aamnmnon a 1m 0. 4 u9=fi.lV FOIQBI :¢_°=%=ual<flp=>qomummn 517.13 '9” ' 8. yx=lry=fl.lv 5 as E1115 £1 a: = 0.525 m t Gum m: P = i1[5 - 0.1.) = (0.525)(4.9) I: 5 _ Eo‘l + 0.8] 83 =>E=2§7mw IL: 5 90.5 icfimlx} = ,3: ' im = N11. 4- II: b‘ “x = "Y = 5 V' "1 = 2‘3 V mum-u = N(o.ssa) + 2.2: ._5—2.3 .1- a amass“. =AL_=_1fi. v¢1=1.6V=:-ia = 5;? =o.944 mA 1;. Outputhigh: in = 5 ‘6‘“ = 0.311 M q; jaws: naive ;; = 5‘“ =0.45mA P = (1': + £2 + incHVcc) = (0.333 + 0.944 + mums) W 5 " 2‘4 __ N -' £1: an .4; = {u.1)(u.45) a 0.045 m WUnin) = 2.‘ V : i;[mu) = 2.2 - IL 1.13 = N(n.o45) 517.” a. M It In: - BY = 0.1 V “B! = U.1+G.B = 0.9 V 5.17.15 ‘3‘ ' i1 = 5 ‘50-9 aw a». k: = w = 0.1 V : Mahmud = 1.5 V in z D Q; inverse Iain i3; =13, = o i; = 5 ‘6“ = 0.45 mA i... = in = o i}, = .8" - i; = (o.1)(o.45) = 9.045 b. ux-:w=3.fiv i23=2§1=m05m4fi "’1 " °" "' a" + “’7 = "3 V i; = (s + fiflim =(21)(o.05)=1.oa mA 5 - 2.3 it = in: = 5 am .4, = Ni}, =- 1.05 = moms} => 3321 ° ' ' ' in 2"" ed'tion So ' Manual b_ px = FY = 3.6 V, 0W1“ 10": 517.18 i. - 5' 2'3 = 3.45 mA 9‘ "' ‘2“??? . . ' = ‘ '3, =0.533 A m = (1 + 251:)" :51 6 =9: m im=0.5£m.A 3'01 =iaa=icn=0 {m =13“ = CI up: = 0.3 + 0.3 = 0.9 fizz-5-03:2.05ml'1 va.=0.1+0-7=0-8V ‘ 05 054 o'_8 ' - 2 06 A is = 0'1 =94.“ = 1.19 uA Im=2. + . —L5:-Im— . In Q (21”!) .1 = 5 ‘5” = 0.333 [ml ———-“" = “'5 “A 5,. a,” = 1v «'1, a. (20)(2.us) = M3333) W =' m €17.19 . 5 4.4 £17.17 a' “"3 ‘ 2.25 ‘2'“ m“? ., .04 .. I. rxxw=3-5V 1c=2-—+2—1-=>1£=3.67m.A ' 5 2 3 1 + YE in = ‘ ‘ =o.575 um = i3], = |ic1l=i32 4 -—————~— .. 3.67 13 = — = 0.36? mA pg. =0.a+o.1=o.9v o i235I23=156mA I m=2—_o.357=.ng=1.53mA b. ED=0aib=ia=2mA ' "'2 - M :i 1 61 A = -—-—— — = - r - - ‘3' {1 + firH-l} [311(4) ——--”‘ " 3:: = 3i; = (103(2) = 20 m = m + u MEL iL=20-2.0<L=>i1.§max}z18mA i3: = i2; = 0 =3» {£3 a i: = 2.56 mA 5,. = 2.45 + 0.675 — 93-5 a in! = 2.44 mA Unload: 5 ‘4” =3 emgmu) - 1.03 m b. vx=yr=0.1'V em a #:9- aim = 1.03 mA licxl =im = {ca :0 im=igg=u ii=£c°= Outputhishfln=0=ifil sic; =0 5 - £341.61 + 0.7 + (31);}..{4} #15, = 0.0342 mA 3;, = 1.03 mA Ice! is ' u'tAnal ‘s and!) i Chapter 17 Problem Solutions III .1 y; = —l.5 V; Q] 0E: Q: on —.'r——. . is: 0 5 35i=>.5=o.55mA £c1=0=avgm=lSV icz = is =9 m = 3.5 — in!“ = 3.5 — {0.56)(2) Ol' gm = 2.33 V b. V: =1.0 V; 01 on. Q: of (1 - 9.7)5- (-3-51 :3. MA ica=°=>fl3 l3: c. logic 0 It Inc; (low 1nd.) = 2.33 V Then [do] = 2.33 = 3.5 - "3.76)ch Ol' 52: = [.47 k9 17.2 (a) Q: on, v5. =—1.2-0.7 =-—-1.9V _]_ _ _ l‘l.r=a'cm =Jfl=132m 1": = '1 V = "E236: 1" '(sz-Rcz) Ru =0.758 kn (b) QI on, v, = —0.7-O.7 = 4.4017 -1.4 4-5.2) = 15 2 MA 25 v1 = "1 V =" CIRCI = "(Ijlxflca Re. = 0.658 K! 'z='c1= (c) For v“ =-D.7V, g on, Q1 off ='.wm =-0.‘10V v0, = 4-0.“! a v0, =—i.7V For v“ =-L7V, Q, off, (".12 on sum =-0.7V um = —1-0.‘1 .1) v0, = -l.TV 2"" edition Solutions Manual (:1) (i) For v... =—0.7V, 1‘, =152mA I... = ___—l‘7 '3(‘5'2) = m m is! =‘°-_7—3£:_5fl: L5,,” P = (i, Hm +ic,)(5.2) =(I.52 +1.17 +1.5)(52) or P = 213 mW (ii) For v”. = —1.7 V, ER = IJZmA . -o.7—(—S.2) [C4 = —--—3—= 15 MA - .7— —5.2 Q, = —i--3(—)= 1.17mi P = [1.32 +15 +1.17)(52) or P = 20.? MW 17.3 _ &T-07 ‘ I” — 05?+l.33 '15'“ V3 = mi. + v.r = [1.5)(1.33] + 0.7 or VB = 2.70 V b. logicllcvd=3.T-0.T=-m Po: 11x = My = logic 1. 3 - 0.7 0.5 93: = 3.7 - (2.37.5)(0.21) = 3.10 V =? = 2.4 V = 2.375 111A =t'nc1 Forux=w=logichaon 2-7-0.7 1'5: =2.5mA=ig¢,-3 My; = 3.7 — [2.5)(0.24) = 3.1 V =9- wzihfi' D! = 2.! V 17.4 Vn=hml+1° °=L+_°=o.5V 2 2 For in =1mA m=waflfl sum For 523 on. i _ V3- V3: —!—2.3! S —' R3 or 0.5—0.T+2.3 Rs: 1 afli=11kn Elggggmic gin-cullI agalgsis gag Design. 2“d edition Soiutigns Manual 17.5 V31 = Va. 1- Val = 0.5 +0.7 :- 1.2 V . 1.2‘-l.4 - 1—2.3! 3 I1 R2 or afiwaflfimm 3.: “:1'2=>n, =o.5 kn i3=m3fl=3=fifli=len i.=°;(fll‘3:§l=a=»g=o.rérkn FOIQRCII'I. unn=logico=flv=$ym=0JV i5=icn=lmA Sn Forvjslo‘inl:lv. i. = 1 ' 0‘7 ’ ‘2") = 1.233 mm 2.1 Formn=lofico=nm 93‘ = Then 1.7 — 0.7 a: = . [18 m Re: 1-235 =b Rm 0 8 Muimumiz forw=lngi=l=3.av Thalig=5um= “‘03 meahgic1=33V £n=3';:°=5ma 01' E: a 0.66 kn By lawman-y. R; = 0.65 kn 301' Q: 011. Kc: So RE; 5 0.12 m Ffl' as on. . - 0.? . I; I: 30.53 3 L423- link = 13:: m1 inc: = 4.423 = M Re: :3 RE; = 0.136 1:0 R8 :3- R5 = 0.52 kn [7.6 Neglecting base currents: (a) I,“ =(i.IIn =0 E“=5flc”::.~ I,,=l.72mA Y=0.7V' (b) r5l=§'lT°'7= Im=o.239mA 131:0 Lug—'03:» Ifl=l.72mA I’=D.7V - .1 (c) Im=fn=5—r:—:> 1m=183=0239m4 155:0,stv (:1) Same as (c). 1.7.7 (a) VR = -{1)(1)—o.7 :3 Va =_-l.7 V (b) QR off, then v01 = Logic] = —0.7 V QR on, than val = —-[1)(2) -0.7 => vm = Logic o = —2_7 1’ 91195.03; lhcnflfl Q; IQ; 011. than vm = —(])(2)_0'7:, Va: =L°SiCO=-2.7V W411i, Q! on, V: = 43-01;: V, = -2.4 V A=B=L°3fc1='°-7mon, V, =-o.7-o.7=> V, =—L4V “3 A=B=L°giémgug 0n. ,6 = = L57 "IA -o.7 — (-5.2) = — = 3 MA ‘3 15 P =(1.67+1+1+1+3){5.2)=> P =39.9mW A =B=Lagic0=-2.7V ic,=3mA.z'c,—_-L67m P=39.9nlW 17.8 I. ANDth b. 10E 0=0V s - (1.6 + 0.7) 1.2 V: = (2.25)(o.a) =- mg‘ L: 1.5 v Qami= 22.25mA Cha ter :P em utions c in: = 5:60.? =51“ :1 1.65 mA 1:. £3 = 0.367 _ 11.2— —3.1 . - . .7 . ._ _ I5.12:5 0172+0 a. =3mA [,3— 3‘3 _.1mA i:: = 0, 1;: =15 = 3 mA 1" = “0'2 $4” = 0.1179 mA -‘-—V =0 1" =(0.567+1+0.379)(0.9—(—3.1)1 5 1 a 11 7 or d. .5.=—Zi-éfi—+-‘—}=ifil=o.962mA E=Hgmw im=5““i+ml=nfl=2.25m\ 17.11 is: = a, {£3 =15; = 2.25 mA L i] = (‘0-9 '0'? ‘ ‘3) => 1', =1.4mA z 21.3 V 1 £3 2 __————(’°'2 " " (—3) :1: = 0.1-1 mm 17.9 “ ___ _.__L_.._.["°-2 ‘ 0'7 - (“3) =1- 1, = 0.14 mA V 3.5+“ 15 i. “=T—9'7*Z3i3.:5_v.. gq+gp=n+ia=1_4+o.1-1=1.54mA . F = = ' = 3.5 . b “010m” W 19ml v i,=%%:u’z=0.8mA is = 1.5-(01:+u.1) =D‘175 mA tn =01: mA 11115 04 ‘Lfl—u ="Mv Whaling-1: .2 =R.—#RE]=4.STKQ ‘3‘ b. i, = 1.; 111A - - Wil - - C. For Q: on. is = = 0.155 mA ‘3 = 15 a I} =1'n % '1: 0.153 IDA szim1=fl=fl=>nfl=smkn i=+in=ic=fim 2 RC: . IQ = D d. For By .- logic 1 = 3.5 v w = -(°-153)(0-5) =‘r = 4-9755 V 1 W - = im=¥=351nk 13:0.175mp. ‘- "l: 1 ‘m P =11.“ +is)(Vcc} = (0.35 + 0.175105} 5, ._. ‘93 ' ‘ '3 a i, = 11.14 mA aw i..=i;:i,=0.l4mA 17.10 ;,+.-,,=.', :4 =11.” mA I. Logic} a: 0.2 V 3—— i =01?! 10:11:11 = -o.2 v Ji— m, .-. —(u.14)(u.5} => 5 = -1107 v b. ig=w=ua ' - .T—D.T — —3 . R5 CL [email protected]'J$Il=l.5mA Sn _ — R§=3kn ia= 0‘01): 3):ii=fl.153mh I} =in=§i =0.153 IRA. . 0.3 0.4 _ I 4—*—'_ I _ = c- wm“m=T=R—1#§;_-__l_ltfl :1+:p=:.+u.-1.a+o.153 1.?53mA d. va=yy=hlinl=mzv i3=%=’i =OIBmA in: w=9357 “IA i2=0.953mA 0.4. 5:41.401! in: = T aim =D.4 am in; = 0.4167 111.1% Elfimnic Qimfl Angflgjg and Design. 2'“! edig'on Solutions Manual 1712 17-15 L L A-B‘=C-Dau=’Ql_Q'cumfi i. FOIPX=UY=0JVI5E=EI§E 5" 1‘1 = 5 ‘82‘2 :1. =0.525 mA VDD=213£1+V33+1332 i =i =0 Ind J--’— IE IS ii. Forux =vY=5V_ I :: __ a 1+” 51 u'=n.a+o.7+o.7:=>g'=z_gv . 5-—2.2 . Then 11= 3 :1. = 0.35 mA 0.8 . ' =' —— ,= . 7 2.522I5(2)+0,7+I_E.{15) I4 :1 15 :4- 029 mA 5‘15 i-5*"'1=>i;—204m11 3.5—0.1: 3— 2A , 5" [7.16 13:9.419 MA I.. Fat vx = 1134 = 5 V_ both ‘2‘ find Q: Mum into and saturation. 1" =15 -2{0.4]9}(2)=b Z = 9 gas 1 u; =fl.B +0.7 +0.8 =- y: = 2.3 V " A C 03 D 25v 1'I=‘5":"q=:~i=i =U.675mA 1" 3 = . = = . .1__£l_—_' ,. . 5—!O.B+D_7+o.1z _ '1: 2 :33=1.7 mA Now v m 25 07 13v i1=i31+i1af =2.375m\ HI: = . — . = _ 1 ad 15=%=i:=u.namA Y=93§+°J=$t=2§y 1 is: = i1 — £5 3‘ = 2.295 MA . 5 — 0.1 . u. Y=(AORB)AND(CORD) m= 4 #u=_1-22i91.é 17.13 fif—y—v t. ID V 1 = 0 V _ m b. .1 _—. 5 (0" “3"” =1.05 111A logic 0 = ~o.4 V 4 him“) =- 18in: = Nil}, +1.3 b. no. - A 5R 3 {20)(2.295} -_- N(1.os)+1.225 m2 - a 511 15 50 m = vs 5: ya u: a: ‘ a 17.17 D, and D, 01f, Ql forward active mode v, = us+o.7+o.7 =22V 5:11}; +51%, +v. and i.={1+fl)"z m=(AORB)AND(CORD} 17.14 So 5—2.2=' 1+ + I. FosCLOCK - high. In: flows through maleft side of 5“ mR‘ RI] mm.fl%ishigh.lpcflwsmshmcbflfl Assumefi=25 minor pulling low. If!) is low. Inc flows through . w 5-2.2 . _ the right R resistor pulling 0 low. ‘1 (“XL-,5) +2 3 '1 ‘ 00539 “A For CLOCK = law. Inc flow: though the tight side fl =(Hmi1 =(26x0‘0589): 1-| :15! MA nimemuumm;gmfimmmpmmus 5:51,: i,=1.47mA mm. 08 1,, :12 +1, -—'— = 0.0589+1.47-0.16 :3 b. .P = (In:+0-5Inc+0-lfnc+fl.lloc](3) 5 7 so 3 P 2 w i”=137m P= 1-7196‘13) -(11 H K }=' - 551! Q, in“ .on 5—0.1 1c, =—-6—=. 1c, =0.817m.4 __ Chapter 11; mulem Solutigng 17.13 17.20 p, L yxsw-O.IV.:nminumdon. I. Fervx=w=5V.Q.inmmflvemoda. -1 I . s-g0.a+u.a+0.'r) _ i] n 5 0.. g'. _ “IA |BL = s — IIIA #i I'm 3 is; + 2,531.3: = 0.45[l+2[0.11}= 0.54 mA - - ' - ELM-.5). _ 9x =w=5V.m Q1 mum-ruse Imemode. Icz= 2+ -2-°5mA Assume Q; and Q; in saturation. 0.0 ' i3,=(iag +£c=)——=0.54+2.05-0.533 _ 1.5 i1=uwsil=im=usm or ig=w:iz=2.05mh ‘ "'5 ', 0 533 m N" n a *9 I = - _ L5 .1 = = 0.533 mat in, = {em + .3) — .'. = 0.45 + 2.05— 0.533 “1‘” or icflma-X] = Brim = Ni}. in; = 3.97 u'LA or (20)(2.oe) = M0003) ia=%sii=2£3mh =Efifl . In. From above. for no high. I1: (0.1)(0.45) b' P“ Q" = 0.045 talk. Now i 2.23 , 5 — 4.9 (21)(0.1] afi'mfl-“d IL(m“)=(1+5’)(T)=‘T = 1.05 mA Ffl' Q: So i 2.05 :2—=BI—-=4.56<fl I;(mu)=NIi or 1.05 = M0045) Since (Ia/Ia] < ,3. mama-Insisth inuzunu'nn. =- E11 17.19 17.21 (I) v,=¥,=L03i¢1 (a) Vh=0.IV:QI,Sat:Q,,Qa,Cut0fi' '=0.s 2 0.7 = - . v s g; 1 22" .,.:_(2:Tt£fl.mm .=—-—= 0.35M , s P =:, (5-0.1) =(1.025){4.9)= 1. = a, -E =0.Js—0.0533=02961mA 202—“: I 15 (0)14, =5V, thnverse Active; 12an is =fl= 23.4 m Saturation 2.4 v” = 0.7 +0.s+ 0.7 = 22 V 5— (Hi-0.7 _ i,'_ = —-£3——)= 0525 am i, =—5 422 = 0.7m Assume 5= 25 in = fl, 4, =(0.l)[0.7) =o_07m.4 Then (25X02967): 2.04+N(0525) V“ =0.7+0.1=o.8V So N=10.2= N=10 _ 5—0.8 (b) New I: = -1—= MA 5 = 2.04 + M0525) P = (3‘, Ha, +i,}(5) = (0.7 +0.07 +42)(5) =0 SoN=5.64= N=S P=24.9mW Elgmnic gigggj; Analm; gm Qesign. 2"d editign Sgluligns Magma] 1722 b- In: = "Y = II: = 0.1 V I. l'x a yr Sill: = 0.1V ._ . 5 -(fl.1+o.sz , 1'91 = 5 “539+ 0'8 :4» I'm =1.05 mA '31 = 2 =5 :3; = 2.05 mA Then From pm (n). i1: fin - in; = (9.3)(025) = 0.015 mA Than h' px=yy=PZ=5v im= 5” =Mfl=£ =0.003TlmA 5 (08+08-H113 “'5'” 10‘ '" i” = I 3.9- =’ 11—i = “"92 “‘A 17.24 Then I.. vx=vy=vz=DJV is: = in = imfl + 353] = (0.592)“ +3[0.5]) 3" =ii =133m“ {81:5-10é1+0.8)+|.sl Bl . 5-!0.1+0.Bz . where .m_-_ m = 3 s“ a 2 05 um . g2 - 0.7) — (0.9} _ 0.4 I. = .. -I— 53: PM}: +I'c2- %- L73+2.05- 1.0 m R3; 1 . =3- iEa = 0.4 [ILA 3 IE. = 2.73 mA. 033:5;2‘1 :21“ mA mm - . 1.1 . - . . = —- 0.4 #131 = 1.5 mA 3"; = 5 (0319+ u 8 = [.05 mA '8' 1 + 1' =D=i = i9; = 1.29 mA I I _' Q: salmnnon 1c; = 51‘: For m high. “.23 . I. "x = W = "z = 1-5 V. Q: binned in the invm "'31 =0.8+0.7=1.5V=>Q§ of mem. rm=2—1.5=D.5mA —§ 1 " - " 0.2 0.5 =0.lmA is! = 2.5 + 0.7 at. = “LA I; .6313: =( ) in, = 621(1 + 353) = 0.250 -+- 391.31) a in = 0.475 InA Thu: arc; =D.5+0.1=0.9V {a =0.5 nut b. lax = w = If; =2 V 0.9 - !0.7 + 0.1! 0.1 in. = (1+ 5,131.5) ’ (mums) MP”, W = 0.00198 3111 (NegLigible) :31“ a: = 0.5 mA .-m.§_;_.;fl=4.ssm £E:=D=i£3 =5 £23 I 4.55 ILA. is: = £1.30 + 35..) = (0.5)(1 + 3[o.2]) in;=iag+im—E?=D.475+4.56—0.8 i _“m a- . =1L__i =4-235m icz=5i1.mdfmmpm(t). .1=1.5mA 80 mm g !11_E” a]. 17.25 I." a 1.1 -O.7 53—0.? (Mia-5) I 1 +1 = = U n n m 05"“ a. 'm = 0.00369 mA 5—(03—03) _ 5 — 1.1 [c_ID=__.l 1.33::W—5J3mh Now a {£3 a 5.13 mA JD=0.51~I,=0.51-5=—=o.51—1—° "“zimfic’ 5° Emmi-5531.41 Then lc-ID=IC- Ojl—I—c =1 Hi —051=46 50 C so ' ' 11.21 50 [c = $.01mA ._ Mm; mg outpm transistor Q; is a Sammy mm- ! — 1‘ 301:: 1 (1100an “m” m a_.———— = _ _ fi 50 ...‘______. w=o_4v‘;£=&.5__(§'4—M=u.5 ID = 051-01002: 1,, = 0.4093mA ‘" Va =0.4V “ I R 6 m _._._..._. m = 3. (b) I =0,V =V m! =o.w :8 OBOE CE( I) The“ I, = ' ' => 1, =G.5m.4 {m = 3-5 - (“v7 + 0-3) = .13 __. 0,273 mA 5 0| *— Rm- 3.6 I = _ ‘ = 0-7 c = [c 4‘9"”! i32=0j mA, 1'31 =0.5+a—7=l.59mA [31 = i3; + {(3 5 icy, =1.50 - 0.275 =1-222 111A L vx = yr = 0.4 V m is, = 2'5 - (2‘7 +0'” = 1.222 lnA C). 951 =04 +5011: Hm =].1V ‘=’ R = 139 i3| ! 2.8- 89-1}: =1.39 111A [L “x = yr = 0.4 v‘ va3=fl.4+0.4=tnm=0.flv ycg=2.5-O.T=-mj£y_ im=im=in=ico=im=ics Anmnmwmntsmm. = in = is: = 0 (N0 M) C. I’m = 1.5 V, 9:} = 0.8 V Current: admitted in put (I). 5 == imfiz +an + (1 + filimfis gag—ELM..— d. im=o.5m.ag=o.sm r “'7‘ + (“K”) (adult!) = mm = M; or (so)(u.5) _-. mus; => '3, = 0.0394 mA 30 in = firim H.351 =r in = 1.18 m)! 17.28 pg. - 5 — (0.0399015) I. For 9;: = w = 3.G v =- .fl____” -= 4-97 V vs, = 3(1),?) = 2.1 h‘ vx=w=uv =i31=5_2']=0.29mA 931 = u.7+o.7+u.a awn, = 1.1'V 1“ Va: = 0.7 +0.7 +0.4 = 1.8 V Ila =1.4 V . s — 1.5 W film= m =0.32mA. pg; = 1.1V . s -1.7 in = is; + in - M = 0.29 + 0.32 — 0.0933 :31 = 15 2.3 5° $1.21 3' 1.13 m in = in (I + 2311) = [.130 + 2[fl.li] in I: 1.42m Elmnig gm]: anglfijg gag Degigg. 2"6 edition Solutigng Mangal in 8 [L517 IIIA we: = 0.7+OA = 1.1 V 5 ' 1‘1 z: 0.951 mm. I'm- ... ac, _ = 0.517 + 0.951- 0.175 in: = is: or in. 21.293 um. Fora-o =04 V. 0‘3, =0.4+n.1=1.1v Then .. 1.1—0.1 -_ = 0.00000 A "‘ " (anus) '“ 51: 5 — 0.00000 0:61,: 0.39 um 50 icdmu} = His; = NI“; (30)(1.293) - M039) =b fl = b. P = [0.29 + 0.32 + 0.9mm + (99)(0.39)(0.4) P = 7.305 + 15.144 or W (Anna-mm; 99 load :imn'n which is 0mmny burn) 17.29 a. Aswan-.0019“. Fat vxslogieo-sOAV an a: M = 0.0915 mA 40 Madlynllofihhmzouwmdfmmvcc. P I in - Vac I (CLOTHES) s E I mfl b. in I a: 0.0725 M 40 in .- 5 " "'7 :3"! + a“ = 0.004 m 5 -!0.7+0.‘l! = 0.26 m i’“ a 15 P = (0.0725 + 0.004 + 0.20m) g. Ila-vault], vm=0.T+fl.(=1.lV . —l.1 . . ‘37: =5!” =78 MAS!“ 17.30 (a) v, = v0 = 2.5 V ; A transient situation 0mm”) =25-0.7 =1.8v VGAMN): 2.5—0.7 =1.8V =0 M“r in saturation vm(M,.) =5-(25+0.7) = usv vm(M,.)s-25 = 25v =:. M, in saturation in»: = Kn(”c.r.v ‘ VTN )1 =(0-1Xl3 "03]: :9 in” = 0.10144 im, = mum, w...)2 40095—03]: = ED, = 0.289 mA in = fiim = [50)(0.239} => fa = 14.45 mi! in = fiim = (SOXOJ) = if, =5mA Difference between is, and EM +1}, is a load current. (b) Assume im = 14.45 mA is aconstant l _ i -r V C VC=EI£CIJI=£ET=J=( CK) .‘CI 5 ISA-ID"I :=—(iE45 10"): {=5.19rls . x ' -——~-————— 5 lelO'” (c) r=———————(0)ggg 104): {=260m' . x ' — 17.31 Let R, = R2 =10kfl (a) in" =0.1mA JD}. =0.289 mA (Same as Problem 17.30) 1'“ zo—j-aom mA =03, =02s9-0m =0219mA 10 A in =(so)(0219) = in =10.95mA in = (:4; = 0.01 m = in = 0.14.07 = 0.03 MA in = (5010.03): in =15 mA (5X15x10"=) I=W= I=635M (c) t: 260": (Same as Problem l7.30) ...
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This note was uploaded on 04/20/2008 for the course ECE 332 & 333 taught by Professor Any during the Spring '08 term at Alabama.

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chap017 - Elec Exercise Solutions E111 . -O.7— — I. 15...

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