Solutions for Exam III - Microelectronic Circuits - Fall 2007 | ECE 3040, Exams of Electrical and Electronics Engineering

Download Solutions for Exam III - Microelectronic Circuits - Fall 2007 | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040 Microelectronic Circuits Exam 3 April 23, 2007 Dr. W. Alan Doolittle Print your name clearly and largely: Solusions Instructions: DO NOT TAKE APART ANY PAGES OF THIS EXAM AND SHOW ALL WORK ON THE PROVIDED PAGES. Read ail 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 exams 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 cannot read it, it will be considered 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 30% Multiple Choice and True/False (Select the most correct answer) 1.) (6-points total) Identify the bias mode of the following MOS capacitors. (ay_tnversion (B) Acer lavion(C) Depletion 2.) (4-points) For the three capacitors in problem 1, which is true: a), They could be used to make a NMOS MOSFET ey could be used to make a PMOS MOSFET ‘c) The diagrams clearly show the Source-Gate bias d), The diagrams clearly show the Drain-Gate bias tre diagrams clearly show the Body-Gate bias 3.) (4-points) True enhancement mode NMOS MOSFET was invented after the depletion mode NMOS MOSFET and can conduct current in it’s drain-source circuit-even with Vcs=0 valts. 4.) (4-points) In the MOSFET transistor to the right, what is the voltage across the § D pinched off region? HF Cwy 7 a.) Vas-Vr 1 ! a l b) Vos Na aL \ c.) Vp, a { d \ i bs ; ; \ } e.) Not enough information given to Nae pe ne a a solve 5.) (3-pointsFrugy False: A well designed current amplifier should have a very low input resistance. 6.) (9-points) Name three improvements that feedback can do to a voltage amplifier a) Increase Bandwidyh ( freq. response) »)_Lnereage tapur resigrenee ) Vecreage uy put resistance od. Allows finite Galn — Crowes a virtual Ground qt0 Pulling all the concepts together for a useful purpose: 8.) (S0-points) Given the following circuit, (a) Identify the configuration of BOTH of the two stages (common ). (b) What is the AC voltage gain, Vou/Vin? You may assume all capacitors have infinite capacitance. You may assume all inductors have infinite inductance, Additionally consider the circuit to be operated at low frequencies where you can neglect all small signal capacitances. Grading will be based as such: part a~5 points, part b=18 points for DC solution (gate, source and drain voltages along with drain current), 9 points for the conversion to the small signal model and 18 points for small signal analysis. 10C 60uA WY Ra = 800K re R3 6" KH p tt 4 | vin 100k 4 a 3 100K J _~ cs Use the following parameters (note that V7 and ) vary with transistor type): h x For NMOS Depletion Transistors: Ky’=20 vA? Vaq=-4.0V A=0.0V7 Length (L)=10 um = Width (W)=10 um M 4 For NMOS Enhancement Transistors: Ky’=30 A/V? r= +0.75V vil Length (L)=10 um (W)=10 um For PMOS Depletion Transistors: K,’=40 uA/V?_ V7=+3.0V 4=0.0V Length (L)=10 um — Width (W)=10 um For PMOS Enhancement Transistors: K,'=50 vA/V? Vq=-1.75V 0 =0.1. V7 Length (L)=10 um = Width (W)=10 um fact a) S$ tage L Common Source Stage 2 Common Drain Extra work can be done here, but clearly indicate with problem you are solving. OC: av Staged’. oes 300 ue 100 ft 1) goo )= | -¢ « me 4 corFoor) +€ loots Jooty, ~ foo av Vrnye - TV Rey = Rt RNR, > 100K+ 198.8 i cy ' Gout “ vn. a ‘Leay => Ves, = vi rage Gouk ~ (2) 4 Crce4) (ese 873% +014 Gon = 15 e-6 Q 0157 14 Ol Vos) Assumption. (Nesange Wes Vr) = Q -a75)= Lapy < Vaca (56 Mage 3 V,2OV =~» Ves =(-1 + Tos Rs) = A-Ies Rs) Vp 249 Vos= Gy - (74 + Ins Rs) ve Vos = 19V -IpsRs Tos= (2) + (g0e-6) (Ves - £4). + 0 (ves)) (100-6 ) Wes * +8 Ves + 16) 2 uu te L Tyg = 100-6 (41-18 Lys Rs + Tos’ Rs $72 ~ Pos Re +16) Ips: 1,64¢3~ a6 Tos tio. 1000 O- 900 Tos*> — 3.6 Tos +16%e-% 2 Tos . >¢ « | BE) -4 i900 te} Q-NMOS ud a (1000) Tns.2 so5uA or 3M ne —~ - - —; Try $5508: Vege 3.45 V Vos = 12,45 v] Veg -t4) = 745V 2 Vos = Le OO Extra work can be done here, but clearly indicate with problem you are solving. Convers ve Srall sisne| Medel: Va = ~(9m: 35.) (Co, I]®y) Vaz grates, Re 4 ose = (V4 gris) Ysa Mout 2 + gm Vosa Re d Av V5s1 \ / Aa / 552 Peat AP in Ng51 a G5 > (') (-sntle) )G a) (or &) {abu-5) (uae Srl 00) wet ) (t. 469) | Av > 1662 vi | iF) u