11 Questions Solved for Microelectronic Circuits - Exam 2, Fall 2003 | ECE 3040, Exams of Electrical and Electronics Engineering

Download 11 Questions Solved for Microelectronic Circuits - Exam 2, Fall 2003 | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040B Microelectronic Circuits Exam 2 March 20, 2003 Dr. W. Alan Doolittle C Print your name clearly and largely: Sa [ AyGns Instructions: Read all the problems carefully and thoroughly before you begin working. You are ailowed 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 10 points bonus (all or nothing, no partial credit), 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. Do all work on the paper provided. Tum 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: l observed an ethical violation during this exam: 0 1 2 3 4 5 6 7 8 0 20 40 60 80 100 120 Test Score N um be r o f E xa m s hifi 9.) (20-points) For the following questions concise answers are required. Points will be deducted for wordy or unclear statements. In 3 short sentences or less, using simple diagrams if necessary to explain, answer the following: a.) (9 points) In a BJT, what is the “base width modulation effect”? Be sure your answer includes a description of which yoltage change generates this Dele A hein magniv de 2 rhe ve verse b 'as@ Luse. ~ llector val tage / Veg| , pr ereages the |ncreagse gf lexior (eG Fj? Con GUE S ‘CoG of ih hase ~ tf re 104, This uces ¥4e base pes neu a wilh, ame [entnerl% pabSeun V, collecvar FEg ON Aled ane Or) b.) (points) What consequence does this “base wit tion effect” have on the common emitter current gain, B? . phe crouse A dH, res ltlns 118 te, A oon | Veg) resale Jn Q leper G ot ) Thus, ? be come s a PGS at lhel 6 a > | Veg) or [Veal ¢.) (5 points) What consequence does this “base width modulation effect” have on the slope of the Ic vs ole ‘utrent voltage cl teristics? The base wi el, moc a latlg 4 isl ose / oftect yvesulrs (nr a One wa Slye +0 the Le us Ke Carve , JN L Thas, Va is Eintye. {4° Section - 40%) Pulling all the concepts together for a useful purpose: 10.) (40-points) Given the following video amplifier circuit (takes a video signal from a 75 ohm transmission line represented by R6 and amplifies it into another 75 ohm transmission line represented by R5), what is the AC voltage gain, Vout/Vin? Assume: Boc=100, Early voltage, Va is 10V, turn on voltages for all forward biased junctions are 0.7 V, the saturation current, Is= 1,=Se-15A for D1 and D2. 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. Hints: Use the CVD model for DC solutions. 2 } Beta= 100 Vee BV f ng tang (stor is*5e-15 7 Ve=10V fi 75 78 R6 Vin Since. OL isin 4 AB and connects 40 “3Y oa gral | 04 magt be ott, — ( ‘sin Gories wont 4 Since DA 15 1K ” BIN ev, Da must be off. LRA ancl COANE EHS KO Raz Ralls = $0 Sr Vek = 5V- TR, =- 2 5y-G77ed lao 1.6 V 41 mA Extra work can be done here, but clearly indicate with problem you are solving. Ve >, - wd Vad Ras gore x te a sy ¢ *( | Lév + Tg Rk + Yee tte = SV : Te Rr + Ie (br) Ra = 5V-1 6 ~O7 ~ 2AM Ts = 90 + 101L 1500) j > 16V + ors Ve: —svaeTeR- Vex S¥-Tefe te ee ‘ 1.18.e-3) 200 Woe —— Yon bY Vex wis} Check: Ve> Ve o 3 Foru-ero” Aerie ap Vc 2 VO mall Siena l facemerees "* V 0 a gns Se le Pree, oz V4 Vee gm = 0.0857 arp et, TL, 2 + Asst Gn 508 v [rns 14552] = 10 £a3sel lo =] 127F0]