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officialpracticefinal - A3 1 15 points Consider the CMOS...

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Unformatted text preview: A3 1 15 points Consider the CMOS logic gate shown at right. 5 pts (21) Write the Boolean expression 12D that describes the operation of the circuit (i.e., _0 Y in terms of the four inputs A, B, C and D). A Ms Mew-D- a Imam I __ I B—l[:JM7 Y Y awe Y: A.p.(B+Z) 5%KM4 Y =- 34. + A + D 10 pts (b) Determine the minimum size required for each transistor if this gate is to D M3 have propagation delays that are no longer 1' than a minimum-size symmetric CMOS inverter. Assam/<44 = 2.5)“? . A B M‘ M l Poll 0mm 2 boars‘L C458. is E IUMOS —T— T dau 7:425. l“ Sfii‘l‘é‘fiS J 3:: w w w A : a” "*7 w “:3, LJ. (bl <72?) “(1' z 3 27» r r-s t P UP , .Lp M; a? Mr, Lora (3/!) +l/Lare. Is. a, 5;,6l: pflos decide, POW?) Y (Mala. Thdcfigrc w w ) Lt); <73): 2’1; I‘ll M7§M§f6¢¢ an. M M5§Hé “a 6Q“ flag I I “j arc, 2, PMOS ale/urges lvx Sabres . “flaapsfe ‘ J at:- ; flavprxxfiéfi M (at: are (ebgztaa <95); =75 A4 2 10 points Briefly explain why the CMOS logic gate shown below will not work properly. A——H: Th5 $94.3“ will najr work bmse —‘P ~94; ?QH’UP % Pall<$own mc‘l’ubflfg Bot “it w “‘94 comply/Leafs, Tint. aU—u 4.5; 0—”: P P $ _._ Y 1 E76 ' (€42) A—‘K 3—”: We, paLl-"clawm (Ms‘l‘ot .— Y=~Alc+ @nD _ _ 3J4) F 5%15) r: A+~D+ at "51‘ A'C+‘ E99 A5 3 10 points Draw a schematic using only NMOS and PMOS transistors and inverters (if necessary) for a CMOS logic gate that implements the Boolean expression Y = Z + B(C +5) given the four inputs A, B, C and D. \! 1i A6 4 20 points Consider the amplifier shown below and assume that CIN and COUT are large coupling capacitors, CE is a large bypass capacitor, fl = 100, VT: 26mV, VCC = 10V, R1 = 6.7kQ, R2 = 3.3kQ, R5 = IIkQ, RE = 1kg, RC = 1kg and RL = 1kg). 2 pts (a) What is VBB? (I want a number, not an equation). V33“: R v : 33v (av) [2,»: l2; ‘5’ ,__-———::_.—-' 1 pt (b) Assuming you can ignore [3, What is VB? (I want a number, not an equation) A! _ VENUE?) _ 3.3V C av) 1 pt (c) What is VE? (I Want a number, not an equation) VExvg~a7 22,49! (Lav) 2 pts (d) What is 1C? (1 want a number, not an equation) m :VE. __ aev ._._ I“. IE E; - film ZAMA (llgmA) 1 pt ((1) What is VC? (I want a number, not an equation) I V¢:V%~:&R¢=10—2,ev =-' 7H\/ (3,7\/) 1 pt (e) Is Q1 in the forward active region of operation? (yes or no) \ EET Ts 13$ alt/Jacks 4294‘ ‘l’Lai‘s #0993966 a, 7;? (tiff/5 VO> V3 inf/(Tc: 5:0 35'3- is 93, J .r', Qfls FA) 3% A7 7 pts (f) Draw the equivalent circuit necessary to find the small-signal midband AC voltage gain (Va/vi) and then find it. Ignore r0. I want a number, but clearly indicate what you are doing or you will not receive full credit. {2104124 gs MB AC. eqoi‘xmfcmi c.5433? 7 ’UL' D0 + {\TF Eli-w @8447)” gang;- ’Up =1 -~5M<Q¢umwr -— 'awLCECHEJer v“ w r L (CXAM 13 3 Ac, _~ Z;émA :0 ‘1 9 3% V7, ZémV 3”“ a 0’5; Q‘JRL: Eben. 122 c’zg "U; "Do in" “L " 5'0 A8 3 pts (g) What is the small-signal midband AC input resistance, R,? (I want an equation, but provide all the numbers needed to plug into the equation — you just don’t need to do the arithmetic to find the final answer) Q°ZAQNQLHF F1A: 55—19:)ka a fir" L l Tr 8M , {XAMB PW: 34A R5: ZkHZngk 2: “75233.11” 2 pts (h) What is the small—signal midband AC output resistance, R0? Assume that r0 = 100 kg and give me an approximate answer. (I want a number, not just an equation) glazer as Q; \m gXAM g 15 4mg sou/He; A9 5 20 points Consider the amplifier shown below and assume that CIN and COUT are large coupling capacitors, CE is a large bypass capacitor, VCC = 10V, R1 = leQ, R2 = leQ, R5 = 6kg, RE = 10kg), rfl= 6kg), r1, = 1009, r0 = lOOkQ, RC = lOkQ, RL = 10kg, Cfl=1pf, and Cfl= SpF. 5 pts (a) Draw the high—frequency small-signal AC equivalent circuit. A10 15 pts (b) Find an approximate equation for (00H and provide all the numbers necessary to plug in and get a value — but you do NOT need to do the arithmetic. Clearly indicate what you are doing or you will not receive full credit. Show how to check any approximations you make. 694MB 7 a “12V 3 'V :5 \/ “~“%3v % I 2:50.93va V81; 10v Va: BE. ‘7 E g, V 2 :1: ” ’ng «fa .- a 3:25 {MA 8” “a?” if; . 23;: * EMUQHR) M23 : _l_: I am as 1355 t: —- 5’K3-“L/3(Zw) v0 1 0 2‘5 __.___ _ 0 UL < Z "Yé CM‘ 1 ZDYPF (/9 C ' : C :— 7 ClVIZZ—DéPF Mm % 11;) S, QM;:::KJ~L C3“; C’W'VCM. " 9249f: cu” Hans (Cm flail/5) PM) (away Fall pr mm ‘9‘: , 2m, 2w?» 914m (zwxoaléz 2-; EAL/44 law a 5‘45 b Q) l l , \ boo fiblélené 1’ H {1/ 2 v ‘— $ Milagrogfi/ C RMH'C‘M 2K6???) 146% n3 1 fl 5H3 WM 1300 [dc-Pr % 2.0 W. “€550 Wm M 3:9. 1—10 I a page) ‘M/tc Mllld‘afiPfigx A11 6 15 points Consider the six—transistor SRAM cell shown below and assume that VD = VDD = 3V. In order to read the value stored in the cell we first pre-charge the bit lines to VDD/2 = 1.5V. Then, at t = O, we set the word line high (VWL = 3V) to turn on M1 and M2. Ignore body effect and assume that Kn = 100uA/V2, Vt}, = 0.5V and C32 = 4pF. 2 pts (21) What region of operation is M2 in at t = 0+? Earle Tait“? a v”! 3 pts (b) Which end of M2’s channel (left or right) is acting as the source at t = 0+? ‘ l @3ch r 10 pts (c) Assuming that the current through M2 stays approximately constant at its value at t = 0+, use charge-control analysis to derive an equation for how long it will take the voltage on the right-hand bit line, vBL, to increase by 250mV (I have changed to using the total value, VBL, instead of the DC value, V32, because we are now dealing with a transient situation). Assume that vD = 3V during the entire read operation and that the inverter can source whatever current is necessary. Ignore all capacitances except for CBL. Make what you are doing clear or you will not receive full credit. More room is available on the next page. t 492 : Kwaiwh) i 1&5:wa Us: t’ W 5) C09“): 309% ACQCEL Z CgLA\/BL.: 69F>011g 1 lPC. f(AVeL15g‘9MV>/E AQC'BL : l—Eé :: lOnS A13 7 20 points Consider the common-emitter emitter-follower cascade shown below. 5 pts (a) Derive an equation for the small—signal midband AC resistance seen looking into the base of Q2, R12. Make what you are doing clear or you will not receive full credit. K . t i“. Dams impcciamca. We. iacy’nm : 5 pts (b) Derive an equation for the small-signal midband AC output resistance of the amplifier, R0. State any assumptions you make. Make what you are doing clear or you will not receive full credit. ! mp? { Assume, (\Di 5) 0M0? 05¢: tmpmafiwm» [\zi’. Esra-“newt :_ 2. 9 124+ pfi'Z \ Po QELLK fig}, / A14 5 pts (c) Derive an equation for the small-signal midband AC voltage gain of the emitter follower, vo/vcl. Make what you are doing clear or you will not receive full credit. 057% imoeAcMcc, l‘flFClQ‘;"l@/L agate/.24 [lu— : @30CEE2J12L) u” U o a, O EELl EL, 1" (\“Z (MW/gin) (fl Jr C l v Dl‘) Uélva +111. T'Maolél A15 5 pts ((1) Briefly explain the purpose of the emitter follower stage. Qél>12¢v ‘99 “H4; EAIH a"? 444:. Jar 5% F5 mo‘l’ Aowfl. Rt, 4 EL go «Hair (X0 Cam Ea closaécs 1., k It”; dose. ‘9: (fix/tea. $0 (ac: (Lo/CL Log; (UH f. m BF é+é~8fi boars CE 60 Lao Can» CLNUCL C» émH ‘65th V58f€§¥am LoHlegflf SacN‘Cfcifla +50 MUQL} aad n . A16 8 10 points Consider the common-source amplifier shown below. Assume that V] has a 2.5V DC component, VDD = 20V, R1 = 1001(8), RD = 10kg), V”, = 0.5V, K = 250uA/V2, Cgs = 4pF and ng = lpF. Ignoring channel-length modulation, derive the upper cutoff frequency of this circuit, may. I want both an equation and the number. Check any approximations that you make. Make what you are doing clear or you will not receive full credit. Vim/65: 2.5V :5 : KfivA5~UWf =~ MA VDD 56c}, RD VD — \JDD— IDQD :. [0V > V: ,‘, ’W‘msi‘sLG-i‘ if: in Wh‘l‘w‘bfl RI v0 05 02$$QVV€OQ 1/? “U l :. t: — %YHED V1 My — Jessavves / QDCMO : tons << lSOams / {a Milk/F “ppmx. is OK ...
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This note was uploaded on 04/16/2008 for the course EEC 110A taught by Professor Spencer during the Winter '08 term at UC Davis.

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officialpracticefinal - A3 1 15 points Consider the CMOS...

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