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Exam1Spring2008 - ECE 3040 Microelectronic Circuits Exam 1...

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Unformatted text preview: ECE 3040 Microelectronic Circuits Exam 1 February 20, 2008 Dr. W. Alan Doolittle Print your name clearly and largely: /b( - ‘3 P"\ g’ 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) as well as a calculator. There are 100 total points. 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: H First 33% Multiple Choice and True/False (Circle the letter of the most correct answer or answers ) an i; l .) (3- -points) True arious energy bandgaps can be eproduced 111 research laboratories T A i S {rd/,1 W WBE tools .hut these tools are not suitedfoflarge scale manufacturing due to the' ‘ rowth rates resulting from the atom by atom layering they implement. True r False: The bandgap results from the sp 1tting of energy evels, typICally the s als s. 3.) (3-points) True c6? The term (1 -f(E)), where f(E) is the fermi distribution function, descrlbes $1321 1 1ty that a state at energy E IS occupied. {‘5 Fa ‘56 4.) (3- -points Tru lor False: Since both energy and momentum must be conserved when electron- hole recom mation happens indirect semiconductors require a phonon (lattice vibration) to allow the transitio - ; occur. 5.) (3— p—ointsgr False. Strong atomic bonds lead to large energy bandgaps. 6.) (3- -points True r False: The density of states predicts that at energies equal to the bandgap energy (E: e there is no available state for electrons to occupy. 7.) (3-points True r False: Impact ionization occurs under very high electric fields when electrons slam into atoms and knock free another electron. Select the ME answer or answers for 6-10: 8.) (4-points) The minority carrier diffusion equation a. can predict drift currents . b. ). .cannot be used when drift current is present can determine the majority carrier concentrations if one also uses the law of the junction to :)relate n to p. is a simplification of the current continuity equation .is something I really do not understand and thus I might not pass this exam! 9.) (4-points) The valence electrons ...? a5} Eisappear when the crystal is formed. " Lx articipate in bonding the crystal’s atoms together. flan be captured by an acceptor thus creating a free hole. Cannot be promoted into the conduction band unless they are in a direct bandgap material. lO.)(4-points) The following energy band diagram indicates the material is: a.) Degenerate and n-type EC b.) In equilibrium / egenerate and p-type ------------------------------ N 11 low level injection .) Non-degenerate p-type E.- """""""""""""""""""" g.) In steady state Second I 7% Short Answer (“Plug and Chug”): For the following problems (11—12) use the following material parameters and assuming total ionization: ni=le-14 cm‘3 (not a mistake) NA=1el6 cm'3 acceptors m¥,*=0.8mo m,,*=0.1m0 EG=3.4 eV Electron mobility, um = 1200 cmZ/Vsec Hole mobility , up = 50 cmZ/Vsec Temperature: 27 degrees C Nate '1 I A mm by “ll/\lfi Wc/‘cclloi I'M/0] +46 "T‘i’le/‘E’i/ l 3 p/ngfl 67an m «(17%; as 145/”01”) §O)(Vef/vé/EMWI{4£\ A: Fy} (4.) (1" 12.) (7-points) What is the resistivity of the semiconductor? ;_ I / i 6/201th + AW?) ‘ ‘ i 94—1 = le-mefl ‘ m l.6xr0"°‘(lwm e~l+Lt + 9‘0 ((616)) JM / "on W_n__ —-~" '* i 7AM 13.) (20-points total) A 0.2mm x 6 um x 1cm semiconductor resistor is made from the semiconductor from problems 11 and 12. It is biased on two opposing sides (longest dimension) with 120 volts. a) (3 points) What type (diffusion, drift, electron dominated, hole dominated, etc...) of current results? , 7 , fl, Hd/e a/apyqfnmsecl: {WW 7” t i" tiff/‘64 1- b) (7 points) Draw the 1 dimensional energy band diagram in the direction of the electric field indicating the actual slope of the bands (numeric answer). E; // E/ / / ‘J E ~ L val EV __’ ,/ Cl” 01X" 3 ((9 I” 611/126; 3/;> c) (10 — points) Determine both the electron and hole current density and currents flowing in the device. «*- [:1qu : A 3(001 X61404) /7f’ 7 1.-” L: fi_ .g x 55% 1 h ., cA w ., (7; A“ A0,, .3 )1; ((0 /F/cm '1, (/i jet-Def ‘10! 40(uf'0" ‘ ’5. 7h. ‘1 6 Pulling all the concepts together for a useful purpose: l4.) (30-points) An infinite length of GaN semiconductor is grown in a Clemson University epitaxy reactor. In the middle of the growth, the Clemson student drops his gold wedding ring in ‘ the growth system creating midgap impurities, enhancing Shockley- Read x='1 "m A /7 2 0 F0 - Hall recombination, resulting 1n a region of very high (recombination. In this defective region, the minority carrier lifetime 1s 0 0 seconds indicating zero excess minority carriers in this region. Next to this defective region is a region of ver_y high minority carrier lifetime,_l_0 ”C h 3 [1’74 g nanoseconds The sun has been shining uniformly on the entire semiconductor (including I X: infinity ) for more than 10 seconds and is uniformly absorbed, generating 1el7 extra C 17 electron— hole pairs per cm3. per second 1n the defect free region. If the semiconductor is held "‘- : W ‘5 at room temperature (27 degrees C), determine the minority carrier diffusion current density 0‘14 at all positions in the semiconductor Assume a minority carrier mobility of 4 0 cm 2_/Vsec. +oo 311nm“ New; 3 we (051:4 >2> 10115 J 20W , LU“ /V5C< M a?2 An], An I’M 0" 2 A" é":‘1105éém%kc Given: 0 = D" p———” General Solution is: Anp(x= A’ev/l’ + Be #W .7 W dx T" \ . I 2 " ’ * - 5* , , V\ 21/. d An An x +1; 1 z ‘7’ Given. 0: D 2 ’7 ——”+GL General Solution iszAnp(x)=Ae /I‘" +Be /' +GLT” [/‘6‘1?/ 8/” dx Tn _ “ii?"A'np ‘ Given: 0 = D” 7 General Solution is: An (x) = A + Bx dx' '7 191;] “73.4 . . dZAnp . . 2 :‘7' Gwen. O = D" 2 + G1, General Solution is: An], (x) = Ax + Bx + C C 1’62 5RJJW‘ dzAn Given: 0 = D" d 2 ’1 + GLOf(x) General Solution is: An], (x) = lr— gm ”f(x)dx] + Bx + C x r) r, N 3 1 ' _ dAn An J/ Ijééyo P ’7 G 131" A()A 0/ . 1ven: = — enera o ution Is: 11 t = n t: e T” £100 X dt T" p ”( ) —~‘ “A” “MG”.-. NMVWWU. .1 General Solution is: Anp = GL1” ”/0 [v r‘ a/éff’ol’ we 095, cm ,41 ,K':C7 49:0 /“ \ 5 - l g. x - / / » / I '1 (A .1 "r C/‘LL X -1 ,4: IVMAO/ 17L), Lat/Pier 0/ {lure/e” 51 6714.52,)‘w‘c147 \L [/11 (4.6%» LI ’3 96/1/6 0\/ [,1 2 41‘ 1‘4 7 g 1 \\ ‘1 m 6‘ "71 L/ .-1 _ ‘1‘ 1:“ L. 7m:“// [Ail/5,7 '_ L. CLIP] C" {.x a (1:17 -1, a (5 1c3'C’X‘IC' ) ;, . ,‘l [(2 - 22 Use this page for additional work. \ M,” + X/Lh ~ Amy(x):fle + 562 +éL3». M79 M56, “we 6? X26) flm/(X): O Ala/1(X): 0: l4 '1' B + 6L 145 $13456 é/‘XZpO flnfly(X)t 64‘ Z‘k/ 3:767 7(61‘ Am/(x) w 46 IC‘th. -\ , ‘ /\ ’oyej’lxer :7 (4 2 ‘6LCK gun/(x3 65A (‘ €ng -5 Amflx): “0410— 9 )cm (sh/(AK no a I‘m: 0 15W X50 nfimflsn 7,0,391 Far X767 5,: : iii" =37 mg): 0 ‘waéc, M‘W’ffi‘fw” M» A O (Ca/’Xéo ¥ 7" (X): 5(5155/‘4/(“498‘X/0'Ewm firm >0 ><‘»=0 i w ...
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