midterm_solution_2011

midterm_solution_2011 - V4095 5'igjflx: +~ ‘ ‘9 {ii-1r:

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Unformatted text preview: V4095 5'igjflx: +~ ‘ ‘9 {ii-1r: /~"‘+‘~:’/‘ no}, u ,4"<f,2“~’x D j: \4 N‘ Mg I759 1/ ,9 3 r124; WW Li H 1 ,\ ~- ;/;.r.:;w w :73 cm nggs g3 / MW. M ’ij E ‘ " m i In, ’ r-fi/ , , r r '5‘ V, rm f,” w ’4‘ I < . .7 v v If” r’; “ml-we; ‘M—wfi/Ci ' "M9; ‘ 7 ~13 ; ‘ /' : F1 ‘ vu »v«-’O- 04‘ "‘"M‘ \ """ " “.3: ' 7 “IRMA ' fi i __)"~§ U V a w my 3’ KM) (0“? ".1 ——~N—-— am“ u w r V, ‘,,,A.._.-~W..o{ ' ‘—«— a ' I ~ ~‘ .L \x..:‘\“‘ fl“ 71,”; f a 4 r x , g .' :2 C3539 (5i J5) {SJ/‘3 (7 no Vfirj K i e ' V 24‘) Q 1' '6 71,, K, / M ~~ , ~ .2 l . U ““ if _‘ .k '3' Mg; . \i. V; J; 1 r 7 r“ e “mug: «qr; ‘6 C { ‘ ‘L J w 51‘ v w ; (‘1 1mm ”5 Q. “’tcfi‘: “ ' ”; 'x" -r c — D “‘2 V, , 1/0 /‘ ” '{28 s; 1101:1131 de 9113, sseamd Inc/C u; 091V 'sdog JO Kemp? seq 17 J10 filo—me; ma qqgm 1911mm; ma g1qu 119MB sdggfi S sq qsnm Joqereduloo 9m J0 £91913 [12101 eqo, 31 eweq noK LIED $93213 Xmaux AAOH (squgod ()1) ('8) QSIAAISIIQO 0 pm; [0:{g]g=[0:{g]v ueqm I we sandqno _ .IOTBJ’BCIHIOO [12113113 V 'Joqvemdmoo {@8113 ggq—gg 19 ufigsep 01 peqs'e am noA 11311318913 [99,1ng saueuuoyed 118m 12 SB sn (EL sdqu 111mg (1qu sq: paugof Knueoal amaq noA squgodgg :1 uogqsenb IIOZ ‘8 ISQWQAON WBXH “1193mm 885/281 {HOE ECE 123/223 ' ‘ Midterm Exam (Continued) November ’8, 2011 (b) (5 points) Sketch a CMOS implementation of your comparator, M35: gm 1w (3‘ .«rww “:11! «xx—74¢ wen—p—mm-vw—v'w“ .» a 3 g g a Page 2 of 10 o J ‘ I — I v ‘ 9,,»3“ A 1‘9 _ 1 ff“? WA; mug; bur 4w? olmvae- ($4.9. (AM * Quito, 3d” Juana/1+ a! G2»: 2 :~ (y’wvu OHM lug/(p? Pam r '1 a? v “if i n :8“ i JCIV "f E‘QJI Can‘l 9 "2 r’ 4" fl i" h} rm fJZ/CM/ it EgE342‘;an + “2’5 9442,? Ceirw—fi /gé°[i+27cf’w 1 \ ‘1 m, A.» 4: “a; 47‘ .J +310‘:[§1+2f}cm + Stir/99115” "'7 H J an" i (9% H925 + Sflfin-slawv- : W900 7, ’ Ljaé'fw Wind :HGOLy Cm»! ECE 123/223 Midterm Exam (Continued) November 8, 2011 (C) (10 points) N OW you realize that the load that is presented to you is 100 times that of a unit sized inverter with equal rise and fall. How much power can you save given this constraint? You may assume an activity factor of 0.1. kgpz 4:, 1'” M'Wme‘itfifiJ 9:27 his?) fir-21 Jjggd ngwwrfl‘LLpW—F‘QW y "jg: 15:9 C m1 r *‘ "4...; r ‘7‘ 1 ’\ r’ 3 c: w» ,r / K, VFW/2% 2 _ P ‘ ’ " ‘ V x] ‘77,,»‘3‘; 1‘? - “Avm I K“ «saw/'2‘, ; Chan) C a ,,;. A (S " w (n i : (7," VS“ hing; U firm; $1” 4: ’crvfirr 1’ [‘1‘ IV if Wk L.v’?n’r‘2}«‘6 * "' ’--”‘ ’” n ,L- \_. r t 9‘ .a Cl“)? ‘ li'x Chanda! Jr C“ arm, Jr Cflmd ” a i” ‘“ s r) r" V n / ~ 2%)Qihv 4—2 ‘Jflv m ‘ -—r ,_ M / <‘ {33, ("/1 r 4 r" / u fl "L .r ’F M f OW ’L .4. gm] #3. W ‘Jr/ _, ,_ J, / n , mm] L ' ; f / I.» (n K, M, V, / «r W W m I - r“ r "‘ - A” V v ~ tag» » K0 «3. A}? i f 24 “Jr , J‘ {)3 dm‘, ,3 u r 4’ by .927": 1’ * ‘ " M #4- y M: ‘ m a. a, . 1"" k A .Ign m . w H V ‘ a c ( ,1 f. “ was; Cw '2' 5 ; 1 3 1 Q; I k / \ ’5 0 ‘3‘ 6 /g, l I WM I ‘ ’ ' (WM. L w ._ y. I ~ r «I; an) ‘th ‘ J y 4 (3 g I i“ ' AW" g M N (‘er £1,141,400 F ,i l” yl H Irich A] :1” a r 3, W L! r N '73 ' la“ V ~ * \ H ' e L" f Filo-*1 /” ‘ “:3, 2);; 9} L730 “« 0 WM 3323 " n» “ y . FM“ g5 WWI—M ‘ ” "W‘Tr—"l :8 V vvvvv " “PM ,,,,,, , V‘fl‘ ‘am 1 ') ("JL/i “ "M’ "’j (m L" “1/ M_./ “m” i {y} ; guy i“ -.:—\ 3 :rmy—J - K“ aim—«.mum .«mfirzcflfid, a v 7” fie: /7 __. ’_ \ '” “: Q _ an,“ '5 fl n A >_ h “L44, : ‘ijLI'Q’Q: W A" "45‘ “A Z“ :3 A: 7:: / r z » r o F” (:3 J / . #v _‘ '5 M‘ __\ W J “kw 83L“ gag] ” w an / 7 ‘3 I" .4 /’ E "‘ ’3‘ M a . y "7 A » m , K r & r x v 5N L39”) 9 l, a» n ,Jr'irj K v w "L ,G 35: u a“ It .1 ' ’4 a W m.” r ‘ a h . 1 . Gm, ,* r v , {F ! »., l: a: my we ~ m A» , O (Ly/y: A A‘ Inf“)>f’\ » [9:43; a: v 1 K POI/JJJ n , ‘ Page 3 of 10 ECE 123/223 Midterm Exam (Continued) November 8, 2011 (d) (10 points) Graduate students only: What is the smallest delay that you can have in you comparator given that the load is a 100 times that of a unit inverter with equal rise and fall. x ,. v" . a z y I I F .,x‘ ‘ “youth—1 Q ) ‘4‘ . no ' r. A. < < f l» H}, L a '5 3% ox; . —“‘«~. ROM r"\»9 _. :7" Ni,“ N, < v ,2 ~ \ fl _W_~_ be" , ,_ ._/ J; /r’§" / ° /23 ---- —~“ 4’ ‘ 5* 'Mw‘“ l ‘ t i w ,. 2/ , a M , ; fl / f v- “a 5 ‘ .f A [3' ¥ r \w m a v mew_ , - “N i x 1 L WW2 a 55L A &, A ’ D n" 9,U Ur? 69119 1 4 231% i H i "z x. r y [QC 2 .4 z s , ‘i‘ ,4 + J ” / {41‘ 5’3 5 s ’r { r. \4 O fir » ‘2 5‘: . ,M H n .2 ,. / a f j, a, P '” [0 die ’1 /0 "’L ’ a , “1 : c 60 94 I' 1.. 1p; 2:: ‘5 a." Ser "I 141/" I x”. n!» “f ‘ )’ W m D441 H: g Q“ ‘0 M \‘1 Ta”. <3“ a 1 r . r 3 "w «a 3 y ‘i””‘“” “g «I w ; f 1 1 : Elna/w} 2 j: I 2— )” 0;»; f I -: ' l, _ m 1 : I : 5" "WW" 4‘ } p, ‘r F N Q -: g 5 j 'l d“ flop: :4 ; A? 1 I T" J I; N M Lmr.» :3” 2/ C i ; a. 49,. q, :7 1‘9"! {24 if“ \ 6 "F I a K i“«/ k l.‘ “.012! C .4 >4 .24”. ’ Page 4 of 10 ‘ ECE 123/223 Midterm Exam (Continued) 2. that she can do a better design than a CMOS figure 1 we! Vddh u Vin H Figure 1: Question 2 November 8,. 2011 Question 2: 20points One of your friends Anna Logue goes through your digital design class notes and decides gate and she shows you the design in (a) (5 points) What is the value of Vddh so that the output swings to Vdd When the input in low. 90 fighy jJ-u. ./ Bail/J 3C f dal- ’ A J -, M!“ g) > r“ , [3’31 e r « a Jig a z {7 - 7 a A‘A) 9w “I PDQ} 12:31.9 my Ln. . 72' N; f g» 5;} i ‘ t‘/O [4’11" QJNQJ h}, /) l, _,:, Mug—L!» r/ rff‘ 5 i (9., ,,,, a L 1 (Ci \ Q If! I ‘ [(3/5 / ‘ . a" A lur {Jr [5. :3": : C} i. {Q “T W13 lfiix‘n‘zfi. 2' ‘j’a ‘“—a€;m I f V - f .F r N 5, ‘ ‘ I ‘ "i l‘ v ' r = «* ' 0 V‘olfl/‘llx, ” Vtwi / l» W J m1 4 a M} I‘ t 9 f . / - i ‘x m’ u V5WF‘: “L ‘5 1‘ . l ,‘g ‘\ ,? L,}‘~\‘ «W‘ V. -f"}') r 4 i one \ c i/U’f} ) ‘/d:;’m ~—-- /Ay(€.} jinn l\ I’D. —. i “a ,7 I” , A. .i it %",M:i:?,_i:f* fig-i , Waldo. , S/ov‘ml“ ~ 5/44,“ :: \LDLQ-:~.L’{E1{ { ; r/fliclir ‘" V,” Aw” ‘ "Hf—“fl ,1?“ Vow : VDD Made 5» be; 27 3/1?er . y". i (FUK 2 o V0141 I ” o a 5&3”? 5H: V {lair/m "m Law?) :éu’ l bird?" " '5 "ID/D 31%?” C ECE 123/223 Midterm Exam (Continued) November 8,2011 (b) (5 points) Did Anna come up with a better design? If yes Why, if not Why not. AM 7:” rm a did mp 1' (fir! ust 3 bflfiflu 64147 n, 5414,12" mg ‘ 1 K‘ I? , w (7 :3, x . ' » M 0m 3 (1291:2412; «Lia WK 5" 94‘ WJ" 4”” ’r’ “" "‘5"? {"3" «L x, .a w r : I a V a f! y y, ‘ :5 Ir :4; .- I 4- ‘ ,J’y‘l 4 ‘d'lgf I ‘fM/l' (frff‘pfl {‘6 1rd): h f *4 2.5 _ , A _ f’mmflg 3’3 {fwd I _ I r; _ A! . . x r : , t Mix v.1 (fee Mares/3r? a ’ _ 91‘, g. f 5 Jr» A; Ct Skit"? 57" 5‘ ‘! "3V4" ’ k" 1»! f r ‘ QAQ a}: ax; “A: I r Q, » y ’ at; ‘ fl k7”: I’m l :16? Wu. 47 s N '3‘, @WJ: @Gafia Page 6 of 10 ECE 123/223 Midterm Exam (Continued) November 8, 2011 (c) (10 points) Graduate Students only'Given the above value of Vddhl'what should the ratio of the NMOS devices be so that the output swings at least to Vm when the input is at Vdd. WM $53 writ/m? Vim : Wm; : Wm igmfi 1'“ —' i d‘ A’V‘l T I” A -‘—-—»««-1 W M wk fl Wm WM "MW i‘irb L t 5* A «~ “’ '.., *1 l2]: C ‘ \ , [a ,9, "1/ l ’ w U r" — w: are if»: r 1:435 "F * "" u Mr, “J .1” » s I» ‘7' 9 -' . i< 3,5! V78» ' Vi» : gm” o a QWLV: f, ‘ ‘ \ 1 LA) H a I Villa wife; 522,311 $1,421,513 Page 7 of 10 ECE 123/223 Midterm Exam (Continued) November 8,2011 3. (25 points) Question 3: 25points Consider the logic path shown in figure 2 Figure 2: Circuit for Question 3 Size the path so that it has minimum delay using an optimal P:N ratio. Given that up : um = 3 : 1 and the input gate is the smallest size possible. n x 7 " fl 1' t £V,«!:q‘}f_{x Ux ' : ‘1“ I ,r *3 >2 4 . axon/drab ’1’ a ‘ -"z . i7 f *v all p h f’ z "’ O calm: , my ELMM [pm " 4’ W’Y‘ a; U ) V gravel "* w M_ a, . yr p E- . r’ r x m= ~ L, a!” , v 52,. 1" F1, ‘44; l’ ( ‘ z‘r" Clu ’ V 5 /3J 1‘), 3 0/“) l H Ill: IL quT 0’9 M -~ — ‘ fl 4 fl 1... W I G r Q F'c ./ r,“ (—1 r 001 / 3'5”) 00‘ w 6‘“? 'w $13 #1,! C :7 T7 r 0 ’ w (€011: 031i “m” 9‘1? r .i " 1:“ W ha y a is”: Cir! .... ” ‘rbn M f {:jr i v~ / ' z 3" r j QM L ‘1” / r—o' (5’ 0&4; {’5 e/r A; l C b W" s V" .6»~ r I’ 2: l l ‘ r s v ,4 J {7’ I m’ w« ‘ Q r” i p 0: 457 ‘ «p " "" «’ a k ’ it.“ : , r "x :.r _. (r w ’x C ' K i (m C y.“ ,1 | .- K g w ~ *”‘ x.) J“ 05.7,}; ’ l 5 r 5 1 w 90: y l 5/ l I» For x . A e 531 m» C, T j; ML) ‘ ~r r :1: a” v 55. \W. , 'A ‘K., 5 ~ w ° :1 “'n gm; ,vevéwx , in n 2L Kim/:23! d z ‘4 ” "w :7 ,r/ ~ z, (,5- , C I 14‘ 1 ’ #1 r1 w ~ C, rv v’é 9:1, {I 3 (’9 I w, :r "" U6 I; 3'? i l0 a"? 1395*? w id; 3:512. W mm Page 8 of 10 U ECE 123/223 Midterm Exam (Continued) Novémber 8, 2011 contd... a ‘6} 0,96 42:ir{ gHstE‘ 35:35" r I'l’ 5“ I 9.9 1’ ,5 Q C? 1" (I 3 Q .. U ("’1 , a"? rpm] ’5”: 4’ »"‘" 6,!aa,2 ha“ a , n ,; {J} by , Q: c 243. 2?!“ K; C“ ’“ ‘ . ch ~\ 5€2fi___,2~l : r31, m, M (via?! a Injé : w a \ ’ 3E3 ((2 ~. rim-’3" ‘ r/w ' ,n ' (“JM :2, ‘ Jr: WEE !; _{ A w 5 k - » r~ n. 7+3‘ p » {SQ-sf“; \ r LAnN/ , do *1rx/ Z) (6 v (GD/'1,» '” 00W 222222222 m. ,. \ on ‘ ‘ I’l'wf; , V " .2 f <7 ‘ r l g ' J'f / ’/ ..)\'7q~ :3 5‘ 9 wow 5H , 3 k vi? ré -“' (z , Q .1 . f {'3 m 3 Cw x '5 , 3 0r / :3 {356:} 2 (k4‘n ,fl ; H" “’ \ 2. \V 17 8 ’ 4 "1 f7, . , (7v, -~ 4 :r 1 r /2 In, (:1 ‘r'v/ /.: (“-16.an 'v i '«H‘””““"‘ ' r "‘ (2 r ,1 Page 9 of 10 0-0111 ...
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midterm_solution_2011 - V4095 5'igjflx: +~ ‘ ‘9 {ii-1r:

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