Exam_S09_Final

Exam_S09_Final - ECE-255 FINAL EXAM May/9/2009 r”) . j...

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Unformatted text preview: ECE-255 FINAL EXAM May/9/2009 r”) . j ,z, . Name: “‘3?” (Please print clearly) ’ Student ID: INSTRUCTIONS This is a closed book, closed notes exam. Carefully 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. Perform all calculations to two decimal points HTS.“ 11:1.IISp‘Np21gn': I) Am the circuil shown below both diodes are identical and L: l()’H.~\. what is \"..',’ (\"'1:25111\") O 1 mA DI 132 V0 I‘ O (1) 0.7V > ) 0.61V mg; (3) ().65V fa :: 55$ [C /W WE (4) 0.82V " ' (5) (N I P (6) None ofthc above “ma 3 QM’DWM’ .ff N) Mao—“g .4. l {j x {of E V _ é Rpm? ' M H 'l 255. t IILII Sp 4N pilgc 3 2) For the circuit shown bele \vxx'hat is the power dissipath in each diode? Assume VBEMONFOVV and \"”t;t;2(0N)'0.6V 10V 201<Q Di D2 1. 5;; “9 ’° 1') e (1) PDIZO.30111W, Pm=0.26mW me) E; (2) Pm: Pm=0.30mW (3) PDIZPD2201T1W v ; ' ‘N‘fl (4) 'PDl: PD2=O.26mW (5) It depends on the reverse saturation current of diodes (6) Noneofthe above 9: j. w H 'i 755, I mul Sp H‘) mgr 4 3) For the circuit shown below. ifa sinusoidal \\-’L1\"CiSL1ppii€(i to the inpulwvhich one ofthe curves is the output Voltage? (VonZOVV, \/Z:=5V) 10k Vin N O V7oul Dz. VM 2 an?! L/ V1,:iov @ if m» 51/ Dz: ”% " } _— ,1, 5 "4“ 3 "f" “in ‘7‘ Von Vou ea x)” > w», I l f :9 a; ,4 -(1) 5.7V (2) 0.7V —> + t t Voul VOUAI (3) 5V \ (4/ v/ > > t 57V t —5V ' T ‘ Voul 0.7V > (5) -5V (6) None ofthese EClr-ZSS. Final Sp (N pngci 4) You are supposed to protect the digital circuit (D) from transient current surge such as shown below. Which one ofthe Circuits shown will do thejoh'? +5V— “5 V*—“ Vin . Voul ° V Vin . Vout V (4) ° 5V VOUl V Vin . . Voul (5) (6) None 0fthe above WW» 3m “fixg‘éflfiim ilzmm _ 5 D, «is» m '9 0&2» wfivfl Wm fi fi- év‘Uf ~ All!th EOE—25$ Finn} Sp (N page (1 5) What is V0? ifVi:O.725V and LEMMA (Assume \J’T:25m\/) 1 ()0!) MM [W ‘fl‘?3 ‘ I > v , r (1)0V L - w ’0 (2) -O.725V WV§9V W (4) -0.1V at“ (5) —2.5V (6) None ofthe above *7 5% V3 IKE-255. szii Sp W pugs "r 6) For the circuit shown below what is the value of 113‘? Assume for transistor [fig-=50 and VBE(on):0,7V. For ZenCi' diode VZ:6V, Rzzo. 15:10‘H‘A. $3ng 2M ' we; (1) 0 Vs iii. 9%» i (2) 1.92mA (3) lluA (4)_@«5fi”fiA @621 pA '(6) None ofthe above [112355. Finn? Sp 0‘! page N 7) For the bipolar Circuit shown below. [5 ofthe transistor is? VEB(OH):O. 7 VEE:5V RBBZSSOkQ VBBZI 42$ 0.48mA (1) 75 (2) 200 W (4) 50 (5) 250 (6) None ofthe above- él r a 4:: __ a” W % t§WLEW m kw ECILZSS. Final Sp (1‘) pug" ‘9 8) For the bipOlm‘ circuit shown below, If? [3:100: \"713I,5(011):O.7 gziyzxmm (6) None 0fthe above (1) 28.7}LA (5) I.435mA VC‘C:5V Rc:1.5kQ (3) 2.87mA S w 0.g‘»4.4 u r 9‘ % E» 3'74? $94; Mya- (4) 3.33mA EVE—255. Film! Sp “9 pugc H) 9) For the MOS circuit shown below: IDZ‘? v-IN: 1v, KN:().1 mA/v2 VDD=5v MI 5 t! “a. REZI (1 ) 0.1mA (2) 0.4mA (3) 0.5mA k {flngnA (5) lmA (6) None ofthe above 9 .V W hm» , fig $3 $2 a? A ("N N “~54 5% Q {A L”; is” b £1: {1’ A a , ‘ CA é, aimgvixfi £966 -’ 5“ egg 1‘ e "g a f «$ng 6% e" 15112-355.l~inniSprugu II 10) What is the configuration ofthe iiiiilti—smge amplifier shown below? 12 V/60 mA Voul (Lg) MN 335’?“ J x i (i flB—CC (2) CE-CC (3) CB-CB (4) CC—CC ) CB—CE (6) None ofthe above l;('[i-355. Finn! Sp W page 13 l 1) For the €11]itt€1'-fi)llO\-V€I' shown below, what is Rm? Assume [3:100 \"YBE(OH):O.7, \’_,\::/;, VT:25mV (1) ~1 2509 (5) ~500§2 {a U Lye-fl} _ Hg” VCC 8M2 20 HA _VEE Rm Ron! (2) ~ 00k Q \@ one Ofthe above Naif"? fig W + (39? 5 thl—ik (9%“ f fifiwg 1216? (3) ~8kQ m (4) ~4.8k§2 we; XI 9/ 155:“ [FIE-255. Final Sp H‘) pugc I} 12) What the output impedance (RU) for the source—Mllower ShOWH below? Assume VTN: l K362 , .7 A A mA/V", and /‘..=0.0l V I My (1) 4k§2 (2) w %009 (4) 50m (5) 09 (6) None ofthe above “1 % s u [KT-355. Final Sp (W page H 13) Figure below shows a cascade MOSFET. what is the mid-band gain (AstVn’Vl)? Assume 8m=12mAXV for both transistors (ii—20 (2) —400 (3) —2 (4) —100 My. @ L/(W (6) None ofthe above 5:: 1m. 3“ i3” If .23: ééfi? viz“ \ e; fax farm? ” WW flw'w’m‘ ECEQSS I ingll Sp W page ii 14)]:01‘ the differential amplifier shown below. what is Common Mode chcciinn Ratio (CMRR)? Assume 10:1mA and Rg-FRCZIC'RSL [32m VBE(ON):O.7V. \/’,\:1(,)()\-"_V and \"rZSmV (1) 46dB (5) oo £448 (3) 26dB (6) None of the above await fiwfifiwzg 9mg (4) 78dB Q”? " V H'LCSS. l’iiml Sp Il‘) page 1(1 15)Wha1 is the differential mode gain (IAdI) for the circuit shown below? 100 kg 100m (1) I? (2)100 (3)1 (4)25 _ MO IQ , " 50 (6) None of the above * mfiwwna 51 ie H EFL-E255. Finn! 5p IN pnuc I? 1(7) Which one ofthe current gain (Ai) frequency response curves belong to thc amplifier shown below? 5 , mum Ca :09 dytm Mfi‘mamf law“ G (:7 .4; a)" WW (5) Cll‘Clllt‘ can not provrde any a «tummy. QQQ current gam Q; (6) None ofthe above [VIE-255‘ l'inzll Sp H‘.’ pngc IN 17) For the amplifier Circuit shown below find fH‘.’ C1,:O.5pf (1) ~94MHZ 2) ~15MH2 (3) ~35MHZ (4) ~50MHZ (5) ~10MHZ (6) None ofthe above WW~ M a? a“! mitika : p. z“ *6 Co Hvfi/M C "' LA, (/¥‘¢0) " > 05(4I)D<(o/2+ l5, K10 VIQE ; 3 5&00 N I ’ g wfigq {26’2‘ # a / Offigwpvgg-y WES"??- “S‘ «.4 (‘3’ my“; l},("[:»355, qul Sp 0‘) page I‘) 18) For an amplifier, the gain transfer function is given by 141(5) = 6x10832 (siJ)(s+2)(y+1OOO)L9+2OOO) Mid—band gain (Amid) and lower cutoff frequency (03L) are? (1) Ainid:6X108, (OLZI rad/s (2) AmidZGXIOS, £052.24 rad/s (3) A id=600 (0L=2 rad/s Mifloa coL=2.24 rad/s (5) Amid=300, (DLZI rad/s (6) None ofthe above av(é): 2/ , ECITCFS. Final Sp 0‘) pugc 3H 19) For the amplifier circuit shown below cslimutcd value of‘I‘L is? For BJT assume [5:150 VA=75VT VBE(0n_):O.7. ZOHA 11512509 100le 1 Vin VEE (I) ~1kHz (2) ~10Hz [9/164112 (4) 01-12 (5) ~2.5Hz (6) None ofthe above a? l “.9 fl: ~ C. a f; EVE-255. liinul Sp ll‘) page: 20) A common—swine high frequency equivalent circuit is shown below. what is the upper cutoff 'l'i‘equeney ("19)? Assume gmr3mA/V ‘ it; $ MA A W-~ 17:1: ,.». "Mew ‘ . (1)~318kHZ ' I ~110kHZ (3)~1MHZ (4) ~3MHZ (5) ~50kHZ (6) None ofthe above gawk; 34 6%j' ia’gg‘ K '"’ gm : §6Ke€'2’"7w .5X/‘8/§’(//¢» fig) ~! . Mfiwm 2” H ...
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Exam_S09_Final - ECE-255 FINAL EXAM May/9/2009 r”) . j...

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