hw5 solutions

hw5 solutions - J __ .. . W-ag=wf ‘ - '---

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Unformatted text preview: J __ .. . W-ag=wf ‘ - '--- 2xlT7a(a)DdufimficflmLeqnflflfimlwmonmeof v-v . (b)mmwmhmrmfimmdmifmm . maau=1ou"3m. ' ' _ gamrmbm; __ _ Q4. . . _ . _ loaf-09"“ "mo_ ""—'..‘-'-‘----—--"—-:\0 <<P° _ . . j ; 421m 5USHEETSEVEEWE‘«ES€2UAFES . n - 42-55:- 1mSHEETS EYE-EASEN5 SQUARES * National RM!” 424% Manama mamas". 5 mamas _ isZ x10" m‘3-s-',mmbemegmhmepufichmdmity ofholcsinfiisdifiauniilvohmeisg, = 10"" can—’4" Musically—Whitman? 49-381 [0 BHEE‘IE 42-392 101) SHEETS ‘EJ-MS- 7n“ RHFE‘IS - -%Natfaaal“flran'd 94‘ _ . _ . 1 . -. Lfi: M '- ‘w "mm _ _ ' ./—\' - . 3xleum-‘uraamx.mwmwumAm.mcw _j " ' . - - Wmfieufldhmg'mhmmlfismmxmfiw -1 - '__WW—Itié0i¢3p{0}_'=firfo) ' . 7:..1 .7 m , 1 a h ' I r -' Irma-JSX'IIO'H' up=mufm,- iio=1x1°"'s, (ammnuhrmnmmflhohmuafmcfimm ;. 3 _ rjM-MEBWGJWflEWWWMHym w - - _- 313mm“ ofx. . 7 - - - " '- rt: 355m _. ____.____.L.____m__ ._-___.Afi,_.--__.____7M._.__...i_fi._______.__ ifknloflem €1,2J _ ’ *8.21 I}Consider the semiconductor described in Problem 8.16. Assume a constant electric I field 80 is applied in the +x direction. (a) Derive the expression for the steady—state excess—electron concentration. (Assume the solution is of the form 6“”f .) ([9) Plot 5:: versus x for (i) 80 = 0 and (ii) 50 = 12 V/cm. (6) Explain the general characteristics of the two curves plotted in part (b). 5W) > E- c (9 “i7 {Tamovort 6.5L. 0? 238%., wildom'i‘Y . 9,; is: __ n 555 3W 5:: w» P “t l iP‘HF‘Q W2 " k ii/lOWtOO-QHEOU «Idolaefil 7‘79 i 0 is. 8N0) :[0 Egg ' Q3607?) wufmrthxl garners Owe, éba-fvw6 ‘ l i )4“: (200—??Abe Corrferg' 228 a? 3 5:2 2 9w fworgr‘m S I :Dm 512:7“ 80 8x H+ u m: at l 5/“ 400w? :45 TM: Margin; a (—25:53 + - Efl, a 50:03” l2 V/em 9x7" hT 3" L: ggtw'iagtfi7§gazo ~ gm = Age 133’; B?“ we A=o§ Sg’txm) menses Lemma 0W1 Be glflfflr BE»??? 23,0”): 0 x Gubs'i-"t'iv'i—e $5M“: State“;in [Lt-{0%} e '2 x: {4H mt.) “INN/1U 0 .5 _ 01“ ifinglz‘r 1L;- “1% ‘l—iv{ fior Swiod“"70* gm” b>Q m'g‘s cl must be F03; ‘ iahpil CO” 5‘92"?” [Vesseflg 4 CUM»: §m(o)acr)[_(q+to)x] EX “CH”; 5 t P24 gmm = Ewe) err; [_(EE°+\J(E§G)L+ L >X‘( WT m L2: # ;(0} C01’ fo’O = grpwo) WV (fl X/Lh\) 5%) E“ 50?”) 01 : @+b) > I/LH ) 90 H decays, «C3519 its” Cargo 1: 0' m 5'9"“? #59 01"; M 491th (IS OVFaa‘i-f Chad 'H/LL efifg¢fvt dfflusim‘ Sf M warm” {.6 0"':mm a, gm Fm: 7x i £19H¢Wmhwmhk w4789W&flmWW”WW°“““smv . l/‘QLQIS O’Md aquarom [S V gF?:W09TEfl¥3 @gm‘aQu-a '53 r 94 _m_?§“_____u*mfifim_m__;9; : ! z—Lni—J_%_E2\1+LL=_,Q:)LJ fi RT ‘ k-T} Lip C : L _\D To >0: £«q+b F F F a) 53 *3, fig gflfl WW - b><L 1/ Th! efifpal’s'v4 ci/("6(us.1‘am 1’9 Embnmcedl (De (1 ): (a) 2/ *8.23 onsider the n—type semiconductor shown in Figure P823. Illumination produces a i constant excess carrier generation rate, GB, in the region —L _< x < +L. Assume that the minority—carrier lifetime is infinite and assume that the excess minority—carrier hole concentration is zero at x = —3L and at x 2 +312. Find the steady-state excess minority-carrier concentration versus x, for the case of low injection and for zero ; applied electric field. i, . a 4 (L‘ 3‘“ 1' ‘h—prut, WOO/n win orih wwfus are hams 4.. 2 fl = o «COr steed, 94:01 at r Qvem Qumcii‘ov of K 31/6) {g git/1H: {—or- .—L 4 x 4 L I 300:0 Lor- LélxlégL Qt . ‘D "Lisa. gs '_ gr __ 2 sp inr 05 st', @182. ,’{o)__: J hfll Q ; 1: ii rapxle-l- %l—o- Us %fi_ %Px+C: t or Lélflég i t n 3x11 0 v gar. (SIX—5C?“ éaiwfi bolas/idan Bough-£1va 2 : f, g : .Md I = (g X L ‘7‘ (6‘92 0 clip dx P I m I 1 {1 t in) _ f6 L7H“ CO = Ct LII—ca. FZDthszihg—ri. ‘3L P] Co: _€_ (BLl-LLi) " it. - Ci / DP, 22 C = 2t it 5(3) 0 _ acct—w; co Wyn ‘ ‘l=_fl%la sz+f£ ELE- _. . .7 ' '- - -' 3.25 Consider-the'fimcfioh fee, 1) 3=' Wyn—m éxp(_—x2 [41»). (a) Show that this _ _ _ ' ' ' 1 .I -' function is a'solution to the difi‘erential'equatidn D(82f/3x2) z: {if/3:. (b) Show that 2 _. ‘f‘ . _ _ ' the integral of the function f(x, 1) overx fmm ~00 to +00 is unity for all values of ' ' ' ' ' time. (c) Show that this function approaches a 6 function as t approaches Zero. ; | I semiconductor at T = 300 K doped at Na = 5 x 1015 cm"? fthe Fermi level with respect to the intrinsic Fermi level. hat the excess carrier concentration is 10 per- ier concentration, Determine the quasi- (c) Plot the Fermi level and 3.28\: Consider a p—type silicon (0) Determine the position 0 i (1;) Excess carriers are generated such t ‘ cent of the thermal—equilibrium majority-can Fermi levels with respect to the intrinsic Fermi level. quasi-Fermi levels with respect to the intrinsic level. Eer,(9,b‘f} QXCfi’H inofll rah", 0161 afihllrJ-R-A QUJA W 9"? 300K) Y’"1\{P32 Na: 0“” “(WPCE Fri—TE} 3 Mo: _ IS -3 to Z ' {99 Slotsm’ W 5 no: “Effie”: = 4'5K'0LFW—3 i K — o ’ ‘5‘ (a); by: EF = hT%(—%): 0.025% a€£¥369= (9.31% av .3: 0L0; GM WQJOr‘r‘i—H, GQTTFQVHTHP~+HW Waterfaeé anaeé— F18er £5“,ng from 9.70; 4 ILL; -,: Err,” EFL == hTQMCV‘jf”); V10 = 4am << gm: @393: Sam E — I L. tarts,“ POW"! , w. F? L MC )) Pod; _ hips iw __ m [:‘Fh‘ l1-1:; I: {QTL/t ( 2' 0.02‘3‘iximfw \= O. ZO?7eV T: V“ amt” _____—— '. E _ 2: . ’- S‘ Fp Ln -=— leTflmggsrffltD), _ 0.25 7:18,, 0mg} J: __ os—glgd/ i L Lyme" (C) Ec 1 t 0.2m EFVI WIMWH‘“ EFn t Er- w , W tr 2 W Ah Ev # i . wmflfwbm 5’. Bo _._- _l_ m mew . a; i i 8.30 / A p-type gallium arsenide semiconductor at T = 300 K is doped at Na = 1016 cm*3. The excess carrier concentration varies linearly from 1014 cm‘3 to zero over a distance of 50 ,um. Plot the position of the quasi—Fermi levels with respect to the intrinsic Fermi level versus distance. w m; e l l LQE-Q mature} s (3.9:: Name 6w 5 hc=l.g¥l(9 mam vyn‘hoh‘i“! cawfnr'; are eQacI'vouc, -; ghbd 01¢ch Wifhorr‘i‘g Cam'ers 4191:24wa Hm fiuasf- FeriMF 130an EF M35130? K i l a ; n I ' \ g9! 8M6): gm)— E‘fflx = wow-- it) x r K g d l a 0L "to 1 ' It? _ 34(0): ‘0 W 3 + A : 50115497.”. :1" “I???” 0 81m)ch fig “3 EF; - EFF: KT E: E “lewfiw(mo+ SSWQU-jj); : V" Fr" ' ' r " 3“ Era-<9”) EFn—EF, czar-M éérucc cQonw-lnob Pairs Creali)%P——=gh qwd EV ER r all E in: C almev Sfuq Pp>> Sn(0)[l* £3) .. “ ._ __ __ F3 fl EFF fi-kTMtgi-sd-fl- “F o.5s°\o.v :—0,02§9* 22.4w 1 Ev “EFL Z __£____fl_fl______ F 7x _ _ W ‘X a“ tin—eff = aozsmgi GU- ix} J EE E , 10.49H + 0.029%?»0' EDQV n‘ Lfi Prflb‘QJZL/m 96 2/1 0.48 0.46 ' 0.44 ml: 0.42 ~ 0.4 ' EFn - 0.33 0.36 0.34 ...
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hw5 solutions - J __ .. . W-ag=wf ‘ - '---

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