Solution of Practice Exam 2 - Microelectronic Circuits | ECE 3040, Exams of Electrical and Electronics Engineering

Download Solution of Practice Exam 2 - Microelectronic Circuits | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040B Microelectronic Circuits Exam 2 July 3, 2001 Dr. W. Alan Doolittle Print your name clearly and largely: So i u + an $ 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. 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 can not read it, it will be considered to be a wrong answer. 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: J observed an ethical violation during this exam: First 25% Multiple Choice (Select the most correct answer) 1.) (S-points) When analyzing a circuit using the ideal diode model, which of the following is wae: The diode is replaced with either an open or a short circuit tJ The diode is replaced with a battery plus an open or short circuit c.) The diode has a small non-zero leakage current flowing in reverse bias d.) The diode has an offset voltage equal to the built in voltage of the diode e.) None of the above. 2.) (S-points) The emitter injection efficiency, y... a.) ...characterizes the ability of a diode to handle large currents. b.) ...characterizes the ability of a transistor to handle large currents. c.) ... characterizes the percentage of minority carriers in the base that make it to the collector. : (4) ... characterizes how effectively the emitter can inject carriers into the base =) ... characterizes how effectively the collector can inject carriers into the base 3.) (5-points) Which of the following bias diagrams is consistent with reverse bias (all three diagrams must be correct, i.e. the schematic symbols, energy band diagrams and the material drawing must have the correct polarity)? a) To) 2 —— [vt | pve =—_ iE X Lint b) -—— 21" yo! we oly: [ enme y x a. 7 | noe Xx X i c.) Yio = XY | “|” Bw a a retype a 7 D2 : ptyne Extra work can be done here, but clearly indicate with problem you are solving. b) Light’ T= T far te + Tear = -I, - GA (tn ++ Lp) Ge Wwe need Ww, W/ = 1 Dis bo / Ny + No . we) (Mei ~ Va Vi; = a ry (wae) Vy; = a val+s _ = falanGsste- 4 ef Lelt+le! =? Ww Vv: (60-19) (im £8 (on W O00038lem (864m) $o T= Ih - Ibe-l4 (.0016)" (Je-5em + 360, t Se-5 em) ley em Ye =- 193 e-4 Vout 2 ~I Re, {Vour = 183 mV (4° Section - 40%) Pulling all the concepts together for a useful purpose: 8.) (40-points) Given the following “video amplifier circuit” and BJT Parameters, what is the AC voltage gain, VoutAC/VinAC? Assume: Bpc—100, Early voltage is infinite, turn on voltages for all forward biased junctions are 0.6 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. Also, neglect all resistances that result from quasi-neutral regions. The diedes in his claus ale examples j = of over- valrase ingur a | Protection ag we di sau gsecd | Ine lass, { 0.6 Solarian ° hnane 10) ¢ DA are of RAzR URS Oz VK -I, Ran - Vee _ re R, ~ Tek, ao 7 q 5 Grio) = 10 (495) +006 ~ (Ori) Ig (a5 + 107 65V [2s = on 1ol(\oas) 15.4 “f | CS ee ee Extra work can be done here, but clearly indicate with problem you are solving. Vee 5-1 Rs = 5-(.54e-3\les) Ve= Bar v Ve = Van ~ Tg Ren > 5 (35m) - (154-6) 45 Ve= rarV Vs = Te (Rs +R,) = (éle-3) (105) > Ne 5a a Foruarcl Getive 's correct agsumphar 0, is of Ff since Vgc SV Da vs otf since Vg POV Small isha | (rameters Lee O0bin s gms Vr 00054 (2) ~- & . 100 rT = ge G.06m > 1,629 Ir Je ca) Va t Vee 00 Y = JS 'SE — 2 0° Xr mr OU Ip +35 _, +35 Dale: 94 = y= = = 0 |. Cs 577 =