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exam1sol - ECE—ZSS Exam I February/l 1/2010 Name(Please...

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Unformatted text preview: ECE—ZSS Exam I February/l 1/2010 Name: (Please print clearly) Student ID: INSTRUCTIONS This is a closed book, closed notes exam. Carefiilly mark your multiple choice answers on the scantron form. Work on multiple choice problems and marked answers in the test booklet will not be graded. Nothing is to be on the seat beside you. When the exam ends, all writing is to stop. This is not negotiable. No writing while turning in the exam/scantron or risk an F in the exam. All students are expected to abide by the customary ethical standards of the university, i.e., your answers must reflect only your own knowledge and reasoning ability. As a reminder, at the very minimum, cheating will result in a zero on the exam and possibly an F in the course. Communicating with any of your classmates, in any language, by any means, for any reason, at any time between the official start of the exam and the official end of the exam is grounds for immediate ejection from the exam site and loss of all credit for this exercise. ECE—ZSS, Ex 1 Sp 2010 page 2 1. What is the input impedance of the inverting amplifier shown below? Rin (1) R1+R3 (3) R31|R2 (5) oo (6None of these W‘Q/ bath 3! (1&37161755’7 WI“ ELM \\ (4) R2+R3 ECE-ZSS, Ex 1 Sp 2010 page 3 2. For the circuit below, What is the OpAmp output current if Vm=1V? 101(5) 1k!) Vin mm? (1) 401.1110 (2) 1mA (3) —99mA (5) IOOmA (6) None of these fi ,‘Nfrf yd mad” V0 ( 5:; W W U 0 ~v / W A /o I: w [/0 : IO V66}? , IO WM ’ WWW” IO”? 09 I)? ‘L_ mw e ; WWW (4) 101mA ML“ ‘3 ) Nut? ‘ H1- .» i652} mA 1303-255, Ex 1 Sp 2010 page 4 3. Which one of the frequency spectrums shown below describes the response of the following amplifier, AVO=1+R2/Rl is the closed loop DC gain (347 Gain //Gain Ave ' Avo (1) (2) Gain 1; {I Gain AVG Avo (3) (4) f l "z f l Gain AVO 2% \9/ LE7 (6) None of these 1 Ra'x. f i {/Lm I}, UL‘M V “MM—i WWW 7M + M Q“ 3100C Am a??? gé% l—K @ \m 0‘: 1‘1 ECE—255, Ex 1 Sp 2010 page 5 4. Figure below shows a cross section of n—type silicon wafer (reigion-l, ND=IO15 cm'3) into which we diffuse p—type dopant (region 2, NA=5xlO16 cm'3). Then we diffuse n—type dopant (region 3, ND=1019 cm‘3) into the area that we had introduced p-type dopanw hat is the electrical conductivity (o) in region 2 if hole mobility in that region is 500 oni2/vn (nfifé‘ffififiw m.__.‘.wmmmmw“wwwmwwm «w w- _,‘_w__‘_,_uwmwwwn swimsuime w Region 1, n—type N9? 5’ if)”; WWWafi (1) ioooni (2) 0.089.crn (3) sooooin 4) 3.92ooni / (5) 129-0111 (6) None of these W" i Q l ‘" 0 "° we) karzdmw 2% N D ; p 0 '5' Mam NA 9 ijm: [Sf i z" I N 3 I50 K! C) w 10 WW W” ti 6 a, :5 lg“ Mm N9 :' 4 Cf >< mm «WW 5’ C! X i of f5; ( Mo. 3 l N .1; j W” ‘6’ g. i) win : 967 990?) 9“ IO g Cm lofty ECE—ZSS, Ex 1 Sp 2010 page 6 5. If the hole concentration in a silicon sample (hole/cm3) is described by: “‘ m a? p(x)=105 +1019 exp ~i i: DP LP Where Lp is the diffusion length. What is Jp(X) 2 D x D x (i) p(x)=1019q~fflexp —— (2) Jp(x)=105—1019qL—pexp —— f, - v 12 LP x p Lp 19 DP x (3) 0 (4) Jp(x)=10 qTexp —L— >4 x L" P D 2 (5) Jp(x)=1019q—L£exp[—f€-] (6) None ofthese i P p I' 6') ECE-ZSS, Ex 1 Sp 2010 page 7 6. What is ID is the circuit shown below, assume V0n=0.7V. Q (Kw‘gd‘w a?) valdefi) \ f} ‘ v‘ fi 455'“ ./ z 8 \mx (1) 0.033mA (2) 0.023A 0’ (4) 3.92 (5) 0.015mA (6) None of these ECEQSS, Ex 1 Sp 2010 page 8 7. What is V0 in the circuit shown below, assume Von=0.7V $§9€wlwyflm~iw { (I) 4 '33} (2) -5V (3) 0V \‘"Wm,,,w,,,ww,w (5) ‘4-3V (6) None of these % 1 Qt;- QLT‘C”) fig”? 0‘? : 5v WT ., 5%)? [L4 (4) 0.7V @512de aging/12:23 a?) é! , ECE-ZSS, Ex 1 Sp 2010 page 9 8. In the circuit shown below, D2 is ??? and its current is ??? Use the constant voltage drop model V0n=0.7 10V r g“ - 0 ‘ Xfirflg mwmwymmwwrvwewy‘ M 7 T w 4’5 + 017W” C 41 > 6 my gag» 5 \ i: 9W” ’ (1) on, ID2=142.5;1A Q2) on, ID2=125pA (3) off, ID2=0 A O wrofigfi (D I (4) 011: ID2210'7-5MA 01’1, ID2=357.5uA (6) None of these More” 0 e x g ,g " , / W ‘ 2% 345% row? 05 Gaming“ WM 30? we a” is C; .: WLPWVMW M ....... .. Q 01:; O W “ 2 “W” / (a? : Ox? $.5 TL? 40 /(,_ ECE-ZSS, Ex 1 Sp 2010 page 10 9. A full wave bride rectifier is shown below 4 ,fl 5 k; a R f “r 2 «W- US=Vpsinwz :— T i kt" : )W‘rfl D1, D2, D3, and D4 have turn on voltages of 0.7 volt. If Vp=10V, What value capacitance you need to choose for a ripple voltage of 0.1V if the input frequency is 1 KHZ and R=lOkQ? 1; i W 2U. (1) 9.3”}: C W am 0 7"“? (2) 4.65uF WMW fl (3) 4.75m: UV 02 fl: (4 2.15m: 1 (5) ( wl‘qESiZifle‘iofthe above C: to a» ledge p / X ) MA a 3 pg x laid” 3 4 13» M“ 44. 2 ll! 2’3 EOE-255,13): 1 Sp 2010 page 11 10 For the circuit shown below, if a sinusoidal wave is applied to the input, which one of the curves is the output voltage? (Von=0.7V, VZ=SV for both diodes) 10k Vin Vout D1 ” Vp=10v $9 % We .24»! 0‘36 D2 00 Mic-gm, __ 147i} 0 t C/ej; %§J 53m r“ Vout ’ 5" K/’ Vout ’ (1) 5'7V (2) 0.7V I Vom Vout (3) 5V t t ~5V VOUI 0.7V i (5) .,5V F : Z (6) None of these ECE—255, EX 1 Sp 2010 page 12 11. For the voltage regulator shown below at 1ft). lmA, the Zener resistance RzleQ, what is the output voltage under no-loa‘dwoondition (RLZOO) l, 510 2;; 61% M j X Id? R :05 m X '0'} (1) 5.6V V, b )9/ (2) 4.4V f K I, (3) 5 52 g + / I X '0 ("(4) 5.6%?) (6) None of the above ;“ 5 ‘4 0 l «a g (C) I 0" L: I C} K C0550 “ 2: 10/ 5. 6 z ECEQSS, Ex 1 Sp 2010 page 13 12. What is the value of V0, V0n==0.6V and 1510444. (VT225mV, and supply voltages for the OpAmp are :1 0V) ,,,,,,,,,,, V, _ _,. ¥ ‘. ,, v (1) 0.6V (2) -0.6V (3) 0 w”"€4y)"flfifls?m<}*§§>t0 the supply rail (iIOV) L K (5) T67, None of the above Vic)» ‘2» {W «:L'NMM.~,V._.w.,,,,w.:w I X (*0 Q \‘\ 372 5;; .‘éx i ‘70 m § 13%; "‘30 a4 (h <3 \\ ’22 ECE—255, Ex 1 Sp 2010 page 14 ¢j : VT 1n[NA]2VDj Junction voltage “i _ 26s _1_ _1_ ' - ' wdo _ q ( NA + ND )qfij Depletion layer Wldth at zero blas wd = wdo 1+ 2—1? Depletion layer Width at reverse bias (VR) 1' C jo = 8580 Junction capacitance at zero bias W0 C 0 . . . C - = " J unctron capac1tance at reverse bias of (VR) ni=1010 cm’3 at room temperature 85:11.9 so 80:8.85X10"14 F/cm q=1.6x10'19 C ...
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This note was uploaded on 01/12/2012 for the course ECE 255 taught by Professor Staff during the Fall '08 term at Purdue.

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exam1sol - ECE—ZSS Exam I February/l 1/2010 Name(Please...

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