tien_midtermsol1

tien_midtermsol1 - EEC 110A N C Tien Winter 2002 MIDTERM...

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Unformatted text preview: EEC 110A N. C. Tien Winter 2002 MIDTERM EXAM Friday, February 22, 2002 NAME: SoLo‘TRO N This exam has 4 problems (weighted differently). SHOW ALL WORK to receive consideration for partial credit. You may use the back side of each page, if needed. V» Diode, ’- j) 2 I; (3 [Ar-— l} Mew; VT ‘= 5%: Vs _ _ . EXT Iv Tc : T3 6 E/VT' IS: [)6 lOlS'A Q ) 71 300K ll ‘ “Cox W 2' _ NOSPET ./ 19 1/4 T: 6%?wa Meg K“ 2, L. “L z. wit“ EMH r 2,, = was—m 6 W1») w) W0 “4‘ UV‘ Problem 1: (30) Problem 2: (25) Problem 3: (20) Problem 4: (25) TOTAL: (100 pts) Problem 1. For this amplifier (given that Vent = 2.5 V), Kc " WLQ" to W) a) Calculate lo, 13, and VB- 3 C’VOKT’ _ _ 5 -Veut LC .. T :. 2 m A E fl ‘ :3 \/A 1% V89. ' I61: 22.5x\65A :\/T ,6“ =— 07+sz Draw the small signal equivalent circuit of this amplifier (hybrid 1: model) and find the values for each element \0‘ {7V3 1’) in c) What is the small signal voltage gain of this amplifier? \0 e are = ~jm “MM/Ra) ’U'o ’g7H’5 VTZZQWV 2r” 2 ‘3'" (Wm _ {-4091 '4 \fT=25m\/ VP» 2 51/ V.”l Problem 2. Given the folloWing current mirror, I‘ \L K‘ JEOKT W) ‘3‘” a) Calculate R1, if Iom=2 mA H‘ M2 and that transistor M2 has a gate width, W2 that is twice that of M1. In other words, (W/L)2 = 2X (W/L)1 ,- (w ‘ FOK M I 1 -' K = 100/” " 2 LL : = 1 2 V2 .Lour Il 2WA > I V+w=0.gv L_ I z 1‘ 7‘ z o -L\ : Kl(\/Cs$|~ Mn) :9 VGS‘ h V+L‘+ T‘T zg'é‘év v —\/ (‘2‘ : Ml : It 5% new W? b) Now assume thatx=0.5 V'1 and that Vout=5V. Whatisthelyalue oflom? L same as {ounA m (3) (40¢ ML) VDSZ: ou'r : 5 V R; = 2- RI V65, =\/esz =aéev (90M Pasta») 1001': Ibz _ K2. V 1, {L1 H-AVDS; _ ( 1‘ Ibl Kl _ .HA '— 2 louT : ‘71:: : qu 31, root : Kz (V65; Why-(l + A V55) = = 2 K: (VestHy—(H—A Vow “MA Problem 3. For the resistive biasing circuit, Rt ED v» z ‘W in a) Whatisthevalue ofVG j ’2‘ .; [OMIL 6? RL RS R1 ‘~ 3 MO— ? (ED -. 44:9— VG = \/bb R :R: vs €614,3- = 5\/ Viz-fife K E ‘ MA L b) Let’s say In decreases by 10% because of external environmental factors. \0 What is the new value of Vs with respect to the old Vs. [Looking for (V Snew/VSold)] lni-lfaugt Wl‘llvt. 113M: 0.? Ibvgd CiVB'p across ”R5 olea‘easeS : Vs = Jib/RS V5 : O'Cl V9014 mm /b) c) Given the new V5, does ID increase or decrease and why (qualitatively)? a . 9 v5 4, um 1“. fig Woes mfia‘bve ¥eeclbada. Vpp = W Problem 4. For the common source amplifier, E\ Rb : a) Calculate ID and V1). 421 ' Va v6 = v95 m; = 2-6 v at we. - ’A = O VGS = V6 : Q'bv ; _. 2 ~11) : K (V55, V—tlay 7‘ ZMA V1) : VJ)vath = SV ’55) b) Find Rim Rout, and small signal gain, av. \69 1Qeye: RJ/Rz = 4. 1? Mn. ...
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tien_midtermsol1 - EEC 110A N C Tien Winter 2002 MIDTERM...

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