Exam II with Answers - Microelectronic Circuits | ECE 3040, Exams of Electrical and Electronics Engineering

Download Exam II with Answers - Microelectronic Circuits | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040 Microelectronic Circuits Pala? am 2 March 16, 2004 Dr. W. Alan Doolittle Print your name clearly and largely: 5 ‘ GLUON 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. 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: I observed an ethical violation during this exam: First 40% True /False and Multiple Choice - Select the most correct answer(s) i.) (2-points) True Diffusion capacitance is due to majority carriers separated bya depletions region: 2) (@-poinsy(Fhie) False: LEDs, solar cells and photodiodes are all forms of diodes. 3.) (2-points)(Thue) False: Photodiodes are biased into reverse bias while solar cells operate in forward bias. 4) (eins tras) False: Base width modulation is when the base quasi-neutral region is partially consumed by the reverse biased base-collector depletion region resulting in asmaller base quasi-neutral width, 5) 2-point’ Try False: For a transistor operated in forward active mode, heavy doping i in the emi is needed to insure large current flow in the collector-emitter circuit. 6.) (5-points) If an engineer wanted to bias this transistor ar] p into saturation mode, which of the following is true? a. V3>V1_ and V1>V3 b. V2>V1 and V2<V3 c. V2>V3 and V1>V3 = V1>V2_ and V3>V2 : : | €. You cannot bias a transistor without resistors. 7.) (S-points) According to the “Law of the Junction” for a forward biased p-n diode, ... a,__...there are fewer minority carriers at the depletion region edges than in equilibrium . there are more minority carriers at the depletion region edges than in equilibrium €. ...the voltage on the diode should always be 0.7 volts d._... breakdown will occur at high electron concentrations near the depletion region edges e. ... forget this stuff, I will go work at Walmart™, 8.) (S-points) It is desired to have the small signal impedance (neglecting capacitance) of a diode with a saturation current, lo=1e-12 amps, equal 50 ohms to match the impedance of a RG-58 transmission line. Which of the following are true: a, The diode should be biased deep into reverse bias. The diode should be biased with a forward current of 518 uA c. The diode should be biased with a reverse current of 518 uA. The voltage across the diode should be ~0,52 volts The voltage across the diode should be ~0.605 volts g This question is completely unfair! @ The diode should be biased with a forward current of 14 mA Extra work can be done here, but clearly indicate with problem you are solving. Far Of Vt 224 r For 242 YL 4236 T- LT! +7" Te we Vi -a4 v, Tee Foy ~ V1 (0,001 + 0,001) — 24G,a) = vilaco)~o04 A for V4 >36 T- Ti+r" +" r 2 ~ Vi Vi-a4 yt-36 n at eT + RE Syy + ; ON FY my (0.003)-0. 04 - 0.9% T= 4 l.003)- 0,06 Anps Pulling all the concepts together for a useful purpose: \ A p= im Diode In=1 629411 10-15 A, “a V tumn.on =0.7 V fo Transistor: ue “ Beta=200 “ Viturn on = 0.7V Rt Saux RA z 15=4,3908586e-14 A 1k C2 ¢+———“k a 198 ct outa. tk } ea . oe save V tum on=0.7 V° | ae Sine qt a R2 =. a5 = c4 12k q oy a . 3} VinAc RG 13.) (40-points) What is the AC voltage gain, Vout/Vac? Assume: Bpc=200, Early voltage is infinite, and 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. 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. Hint: Use the CVD/Beta analysis for the DC solution. Then apply your results to convert to the small signal model for both the BJT and diode (i.e. do not ignore the small signal model of the diode). , Assume loreur active OC Sala tian . trode. Ip: [mA [forced by cuffent source ) aay TEER alla g L.A 1 fof Vik = 440 (oerae)= VV Ryh- Rifle, = $5 7 K lak Extra work can be done here, but clearly indicate with problem you are solving. 0 = Vi6 ~ Tp Rak - Vee — Ie Re -Lr Ka 13.7 = Te (4) + a7 +L (Rx+Ry ) 1D 2 Te (4,57k) + GMs (u751a5) 200 I _ ~B> “Gis7k +201 C500) = 114 uA Te Pla = 200 (Tg) = Gr Te ~ +e = RCO tg = 23.4 mA Vez Ven = Tq Rom = (3.7 ~ 11048 C9576) = ROTY Vi- ugv— R4Le = R4lV Vez Del sao) = Whar Rar R49 F, ward conve Vie ? Vg + Va? Ve — ver fad! Small sina l faramevers We Tez 2.008 = 5.9L m yous Te gm te = 238m - oA Uv Vr 0,0954 r Varvee _ wo because Vaz oO Te aia rps ga add - 217,62