Exam2Spring2007 - ECE 3040 Microelectronic Circuits Exam 2...

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Unformatted text preview: ECE 3040 Microelectronic Circuits Exam 2 March 31, 2007 Dr. W. Alan Doolittle Print your name clearly and largely: 50 {u ‘I' I’dflf f Instructions: Read all the problems carefully and thoroughly before you begin working. You are allowed to use 1 new sheet of notes (1 page front and back), your note sheet from the previous exam as well as a calculator. There are 100 total points in this exam plus two bonus problems at the end of the exam. Observe the point value of each problem and allocate your time accordingly. SHOW ALL WORK AND CIRCLE YOUR FINAL ANSWER WITH THE PROPER UNITS INDICATED. Write legibly. If I cannot read it, it will be considered to be a wrong answer. Numeric answers without supporting work will be counted as wrong. Do all work on the paper provided. Turn in all scratch paper, even if it did not lead to an answer. Report any and all ethics violations to the instructor. Good luck! Sign your name on ONE of the two following cases: I DID NOT observe any ethical violations during this exam: lohsuw'ed an ethical \iolalion durng [Ins exam: First 20% True {False and Multiple Choice - Select the most correct answer(s) 1.) (2-points) True I‘ The depletion capacitance of a junction is due to minority carriers separated across the junction. 5 e a I M I 2.) (2-pomts) True if Diode leakage current is proportional to the area of the device and the voltage across the diode. 3.) (2-pointSKfiB/ False: Zencr diodes can operate based on tunneling when an electron jumps through a very thin energy barrier even though it does not have enough energy to go over the barrier. 4.) (2-points)}1=ue A transistor biased into saturation is turned on as hard as the circuit allows and us makes the best amplifiers. 5.) (2-points)@z’ False: A BIT with a thin base quasi neutral width results in high common emitter current gain. 6.) (2-points @ The base current of the B] T is dominated by the majorityr carriers int c 'n- ot the minority can-iem injected from the emitter. O1 7'.) (2-points) If an engineer wanted to bias this transistor into saturation mode, which of the following is true? —- a. V1>V2 and V2>V3 . . . b. V2>V1 and V2>V3 ( )V1>V2 and V3>V2 V1<V2 and V3>V2 e. None of the above. 8.) (2 points) The law of the junction relates a— the leakage current of one side of the junction to the leakage current on the opposite side of the junction voltage produced across a junction to the excess carrier concentrations on both sides of the junction c. capacitance of the depletion region to the diffusion current in the junction. (3- the doping concentrations on either side of the junction 9.] (2-points) When a diode is forward biased to near the built in voltage... as“ ...the dominant current is drift current. - bi the dominant current is diffusion current. C- the dominant capacitance is diffusion capacitance.- d- the dominant capacitance is depletion capacitance- ‘f- the energy bands are strongly sloped © the energy bands are almost flat. 3- Nnne of tho ahnvn 10.) (2-points) The current flowing in the collector-base junction of a forward active biased transistor ...is mostly minority carriers in the base that were stripped out by the large , electric field of the base-collector junction. ...is mostly current originating in the emitter. c. ...is mostly drift current originating in the base d. ...is small due to it originating as base current. e. None of the above 12.) (20 points total in 2 parts) Anode A thyristor is a family of power devices that consists of 4 alternating p-n-p—n type regions creating 3 metallurgical junctions as pictured here. If the anode (see figure) is positively biased relative to the cathode, junction J2 (see figure) is reversed biased while junctions J1 and J3 are forward biased. Under these normal conditions no current flows from the anode to the cathode. However, if a voltage or current - pulse is applied to the gate, the reverse biased junction J2 can go into i $3M avalanche breakdown and a huge current can flow between the anode and cathode. These modem thyristors can switch large amounts of power (up to megawatts). Gm .. (a — 6 points) Draw the energy band diagram of the thyristor in equilibrium. Be sure to label the anode, cathode and gate and the fermi energy, and conduction and valence band. .._._.—-— u..._______' (b - 14 points) ) Draw the energy band diagram of the thyristor biased as described above in the conducting state. Be sure to label the anode. cathode and gate and the and to:iductiohijLalemebtfllflntjidgttijy the region oi‘tlie device that has the that—is in avalanche llI'CElkdBWILHJ-r — —- : ._"—' am _ __ l/ 6‘“? c E //‘:— \_ ._ “Iv—Eff; E? x aha/e _—-— 13) (20 — points) For the following circuit, the output voltage is desired to peak at a frequency the FCC (Federal Communications Commission) reserves for use by industrial “noisy” equipment, 13.56 MHz as shown below. The voltage source, has a DC plus AC combined voltage equal to Vin=VDc+sin[t (2n) (l3.56MI-Iz)] volts. If the zero bias junction capacitance, Cjo = 4 pF (i.e. 4e—12 F) and the built in voltage of the diode is 0.9V, what is the value of VDC required to obtain the output voltage spike (resonance) at 13.56 MHz? You may ignore all resistances of the diode in reverse bias but not in forward bias. (Hint: The magnitude of the voltage at 13.56 MH7 is irrelevant only the frequency is meaningful.) .L1 We? ’VTA L '-‘- '“ “mm .. 1% ng-Ec find-w; lvaC .27 ffiéonlfl‘f '-" {a}; W, L-C. we: 2W€fi.y‘hflz) L-FIQ-H' H -12. w 1 v: has“ C rot/9’5? 5,15) 14). Pulling all the concepts together for a useful purpose: {40-points total: DC solution = 12 points, conversion to small signal model = 12 points, AC solution = 12 points and 4 points for accuracy of the graph) For the circuit below: Diodes: me 011:0} V and Io=Is=l .83e-14A Q1: me “=03 V, I,=1.83e-14A, BDC=200, VA=200V VinAC = lmV amplitude (i.e. 2mV peak to peak) at 1 kilohertz (period of 1 millisecond) 10c l ' . . . , R1 R2 "r 173'” ’ 181K _ R: ——_ 9V Iso 500 V, J m _ VoutDC cs _ | L . _‘End RES C4 I :2 Cl l 01‘ , VoutAC Rout ‘ I! . "5 'r— ‘ — o — o -- r _— V'InAC ' | h | .- _.._v3 H _ _ ' v2 - I ,' 02 | Re ' Grid ' cs —— '33..s3c cw — — —' I' K .. _ _. _ _ _._ _ .4 _ ._ ._ _.._ _ Given the above input voltage, VinAC, sketch and accurately label a plot the TWO output waveforms VoutAC and VoutDC on the ggaph paper provided on the next page. To do this you must solve the DC and AC solutions of the circuit. Assume the turn on voltages for all forward biased junctions are 0.7 V. You may assume all capacitors are very large values and are thus, AC shorts and any inductors are very large values, and thus AC opens. Additionally consider the circuit to be operated at low frequencies where you can neglect all small signal capacitances of transistors and diodes. Also, neglect all resistances that result from quasi- neutral regions. For full Credit, be sure to check your assumptions on the mode of operation of the transistor and to clearly label the axes of your plot. Hint: Use the CVDI’Beta analysis for the DC transistor solution. Then apply your results to convert to the small signal model for both the BIT and diodes (i.e. do not ignore the small signal model of the diode). ’ 0C CII((H:+ ‘ _.___.—--""——‘-— 60ml}I IT —‘1v + I. R‘ +0.1 +152: --‘W=0 I73 : 1230?. + 13“)RE) 129: 461.7419 I¢=915=9fi+q8 In- m A Extra work can be done here, but clearly indicate with problem you are solving. VGra‘ EA. (“aye : V 2 JW 4. 3:32? gage -Em:"\‘€f F; Farawqu v 6.5069}. c Safe- *0 v Ice/(Crier av + an; RC 2: -- Lrv j "5 “9"‘1’5’9‘59/ 4m” ‘ I 5"de (gal/éffilrafi W Answer Page ...
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This note was uploaded on 08/23/2011 for the course ECE 3040 taught by Professor Doolittle during the Spring '11 term at University of Florida.

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Exam2Spring2007 - ECE 3040 Microelectronic Circuits Exam 2...

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