'+',}b" 't;5 0. ECE 3040 Microelectronic Circuits. Exam 2. October 26, Dr. W. Alan Doolittle. t70 fa +1(; h.. "5
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1 '+',}b" 't;5 0 ECE 3040 Microelectronic Circuits Exam 2 October 26, 2009 Dr. W. Alan Doolittle Print your name clearly and largely: t70 fa +1(; h.. "5 Instructions: Read all the problems carefully and thoroughly before you begin working. DO NOT REMOVE ANY PAGES FROM THIS EXAM. You are allowed to use 1 new sheet ofnotes {l 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 grading). 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! 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. 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 ofthe two following cases: I DID NOT observe any ethical violations during this exam: I observed an ethical violation during this exam:
2 First 20% True!False and Multiple Choice - Select the most correct answer(s) 1.) (2-points@ False: The diffusion capacitance ofajunction is due to minority carriers separated across the junction. 2.) (2-points False: Diode leakage current, 10, has two terms summed together. Each of the 0 terms come from minority carriers drifting (leaking) across the junction, so one term is from holes and one term is from electrons. 3.) (2-points) True Avalanche breakdown of a diode is always a destructive process that will' the diode if left uncorrected. 4.) (2-pointsFalse: Ifone wants to use a transistor to "switch a motor" coil on and off, the tr s or would primarily be switched from saturation to cutoff. 5.) (2-points) True BIT with a thin base quasi neutral width is best for high voltage (large V CE pplications. 6.) (2-points(!!,;l False: Zener diodes are used as voltage references but their "noisy" avalanche process can introduce undesirable noise. 7.) (2-points) Ifan engineer wanted to bias this transistor into forward active mode, which ofthe 4.) lowing is true? +1 V2 + 1 V3 a. Vl>V2 and V2>V3.. V2>Vl and V2>V3 c. Vl>V2 and V3>V2 1 ± J d. VI <V2 and V3> V2 e. None ofthe above.
3 ... 8.) (3-points) For the 6 circuits shown below, indicate what the function/name ofthe circuit is. Place letter of answer in the blanks to the right of each circuit. Choices are: a. Current mirror circuit b. Voltage Reference circuit ( t') c. High current gain (super beta) circuit etr ti...< t, '" ) d. Non-functonal (Auburn engineered) circuit. R9 Q2N2222 VIA \.J R10. Vou! Cln 12V _ 'VV'v Rload 1k 01. I b Cin 12V _ 1 R13 02N2222 Vou! M \.J 10 Q1 Rload f 02 D1N750 I 0 J
4 ts) The common base current gain, (loc......depends on the number ratio of majority carriers to total carriers emitted from the emitter into the base.... depends on the collector-base voltage in a real transistor. (t'5e """; Jt. "" fr ".,«ItA. C( fl... depends on the fraction of minori carriers in the base that can "survive" long enough to be collected in by the bcollector reverse biased junction d....depends on the fraction of minori carriers in the base that can "survive" long enough to be collected in by the bollector forward biased junction e. None of the above /J t t'i\
5 . 10.) (20 points total in 3 parts) A silicon diode with the following parameters is biased into reverse bias with a 5 volt bias nj= lew cm- N D =le19 cm- N A =le15 cm (a - 4 points) Which side ofthe depletion region, Xn or xp, is larger? Note: you do not have to calculate Xn and xp to answer this (although you could) but you do need to justify your answer. )J0. 'If:: IIJX n.!!if... If r.. I e. I-,vel - 7f :; l"et't Xf >"> x" (b - 6 points) Draw the energy band diagram ofthe diode under this reverse bias. Be sure to label the quasi-fermi energy separations (Efn-Efp), and the conduction and the valencc,ban"d. e..."".,.,,..11(... Ev fft' """-'>'. /,\ V I,V )... A::-(" / '..'...--",'-...,.."' ,,---".-- Ert\
6 (c 10 points) ) For each side of the junction, determine the equilibrium minority carrier concentration and the minority carrier concentrations under thelpecified reverse bias (4 total numeric answers)g... j.e tlepl 'I}dh. <rejf.(!)1!'"\ -pd,ef.. Ll Y\, Jhf hi. Ie h;a.. /.Vi )!)I'h = -No r.4,k,) lvl+ 1'- ( -I no -8 g )0 ao ( -5VfoIO;'1_ I) t - }Oa r -:::z.. bplt io i5/&;o2j1. \ 10'6 e - I S- c-:] o= l' fo,0 1'1 '- - 'J /:\ -{--tj-"'"'1.i=--()t.3--!j ""f :... - c> B jf 2 - (0 [r:z 0]
7 11)(20 points) The following circuit has diodes that have a tum on voltage of 0.6 volts and a breakdown voltage of 32 volts. You may assume the diodes have no resistance in forward bias nor in breakdown. Draw the output voltage waveform (vout) clearly labeling the minimum and maximum voltages and the time axis. v=120sin(21t6ot) volts i R=10,OOO Ohms e,..l. 2ee(' $ tes " E:er t7'1 -u, If Of'" 4=,l V out -=- 0.6 T " off' 1 G r rofq. ej. IIiIN'.l... "Y=9 :r., -. v \}.D /' / I / I :l.' - )J.;& "" / () h r", f(.te e,, t if' t/ \.ts.'!if"., eer S (J f.f' x -IO 7 J, V B('''e J.v,, I",re.ị ti '"'\
8 ... C.1 12). 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: V turnon =O.7 V and I o =I g=1.83e-14a Ql: V tumon =O.7 V, Ig=1.83e-14A, floc=200, V A=200V VinAC ImV amplitude (i.e. 2mV peak to peak) at 1 kilohertz (period of 1 millisecond) 01' R1 100k > -' R3.. I 9V 1k. + on ;:()Tl ",):9 C2 I 0 R2 100k C4 R4 1k + 9V Given the above input voltage, Yin, sketch and accurately label a plot of the output waveform Vout including the DC average output and the magnitude and phase of the signal. To do this you must solve the DC and AC solutions ofthe 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 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 of transistors and diodes. Also, neglect all resistances that result from quasineutral regions. For full Credit, be sure to check your assumptions on the mode of operation ofthe transistor and to clearly label the axes ofyour plot. For full credit, do not ignore the diodes. OL '. 0;".ks'. b"", are re vu'5e./. '" s...1 {I:... '1e - L...f.,! eel 'I- r/jdlv'er ''''I'ply) It..(..(." ect..c (_,\ I l'0, -=.I'it- -:. fj - Ls \...'1.- OJ
9 Extra work can be done here, but clearly indicate with problem you are solving. Oc C(J}""'J '. R.d,:7 IOOf\/( ItJotr \" e 1'\'ll-',,a ([,ti. I " YJo, -,, 1 -;:: CJ t\ 1: (eo f\ V-r-k by s,/",,-e+l'y VTk"' 0 --'-7 bv<+ ;f ytj )i)" tj1 9"" ;'-' t ItJ() (( & -1V -+ IOOt\ 1: -f- (otrr - 9V: 0 I = 10.J.{A Vi J... ':. - q V + IC?O {( I Vt"" -= ()V A-c.t,.., e f A..I!3 50 t\ +- &.1 ' +,l'e (11<;") -<[ =0 19 (5()" + 'Jot{ I k)) '".3 V t ({?+1.) Ie: "3"3 M. A T G-= ('18:' h. bl,...,ft C t,,,,,k "Sur>P j- '. J: E =(j:l ;- I )II3 (;.6 "'b "" 4 - r6-: 0 + f6 Sorr"".:! L6SV V -:: qv -IEUK) = :J,-lf v VG Z - 4V -I- -Ie- (IIrJ -.'t V V F ) V(3 ) Vc. V -tfl... rp)
10 Extra work can be done here, but clearly indicate with problem you are solving. {C9 t\. V f'+ to.-t ({ $.tsj... a (.. I _ " 'i _ IA 51) 9 - Vr - C?,,,,'t - fy'l f :I.- t9tj l;' 6 )L... (11 -:;z.- - 5"-' (!). ;). "" - Vit + V.! _?o(j... {;;z,'t I.f,,!):.. 30Q17 rt? =' J' c - t;, 6f"",4 :J l II R 1<; 1 4:", 1'<5 f; t9 I t:d'. be? t.:: Al': ""' RI "R:t II rit" RI {{((ij.iiv-1f f- f<, A. t::: - :J ;1/-t;e ((' II R,) v= /\F;;utflF,' j :) (h) Z:[-;l so)( ",. - 0, " ( Ol 77'II r0 -;).?V!V
11 Extra work can be done here, but clearly indicate with problem you are solving. \ \-7\V _ \ MV -:L 16<0 -.q.v
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