Test II with Answer Key - Microelectronic Circuits | ECE 3040, Exams of Electrical and Electronics Engineering

Download Test II with Answer Key - Microelectronic Circuits | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040 Microelectronic Circuits Exam 2 March 28, 2008 Dr. W. Alan Doolittle Print your name clearly and largely: SD ly J; Oh s 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 a bonus problem 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: | observed an ethical violation during this exam: First 20% True /False and Multiple Choice - Select the most correct answer(s) 1.) (pote ase: The capacitance of a pn junction is larger in forward bias. 2.) (2-points) True( False; Diode leakage current increases as the semiconductor doping increases. - To oo se ‘AD 3.) (-poingf Ts) / False: The primary source of leakage current in pn diode is due to minority carrier driftTurrent, 4.) (2-point: y Tras) False: The steeper the energy band diagram slope is the greater (in magnitudé)the electric field. 5.) (2-points) Prue) False: A high voltage BJT must hold of a large base-collector voltage so the base quasi-neutral region width must be large in order to prevent “punch through”. 6.) (nth Tine ase The emitter current of a forward active BJT becomes minority carrier current in the base and majority carrier current in the collector. 7.) (2-points) If an engineer wanted to use a BJT to turn a motor (large current) on and off repetitively, which two combination of bias modes are best for this switching operation? a. Forward Active and Cutoff Forward Active and Saturation Saturation and Cutoff d. Inverse Active and Cutoff e. This is silly, a BJT cannot be used as a switch s B/( B+1) times the emitter current Ie . is o times the emitter current Ie 9.) (2-points) In a full wave rectifier, the current through a resistive load... = .. flows through only 2 diodes at any given time. b ..always flows in the same direction. c. --results in a stable DC output. d. -.consumes no power, e. None of the above 10.)(2-points) The law of the junction ... a. ...describes the balance between electrons and hole in equilibrium, np=n,’. >. )..seseribes the balance between electrons and holes at the depletion region edges as a function of the voltage across the junction. c. _...predicts that excess minority carriers will be present at the depletion region edges der reverse bias. @) predicts that excess minority carriers will be present at the depletion region edges under forward bias.. e. None of the above 13). 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: Viurn on=0.7 V and Ip=I,=25.9e-6A (a large power diode ). QU: Vian on=0.7 V, L=1.83e-14A, Boc=100, Va=15V . VinAC = ImV amplitude (i.e. 2mV peak to peak) at 1 kilohertz (period of 1 millisecond) Given the above input voltage, Vac, sketch and accurately label a plot of the output waveform Vout on the graph 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. 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 and amplitude of sin functions in your plot. Hint: Use the CVD/Beta analysis for the DC transistor solution. 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