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hw4 solutions - EEC 140A HW4 IDUE FEELS TYH 3.7 Si.9:1on...

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Unformatted text preview: EEC 140A HW4 IDUE FEELS * TYH 3.7 Si {.9 :1on a1“ Nd: \o‘ivfauol N630 « (a) fluff Hm efled‘mn concenfrai-t‘on Ho/NOL over Hm range. 450K 4 T 4. 600K ('9) Wha‘l- are. Hm *emwm'fura, ad“ which he = \IOENd and 1’01: LlNd 4 4-1 9% 4' 9 440:"ij hid-=0 } Lay-CLJNQ': O 9:. 4. \5‘ -X— 4- 3‘ (Admin Clu flack eXPyesstM Ler‘Hna fleci’m‘c 1;; fey—d 0HL§F 4.3L 4.41 a; 4.42 4. E70 % 140A_2008 TYU3.7 TYM3:7 %% y = n0/Nd; find T at n0 = 1.03*Nd; T = 526.2K clear, clf, format short 9 T = 300:1 :600; Nc = 2.8919*(T/300)."1.5; NV = 1.04e19*(Tl300)."1.5; Nd = 1e15; E9 = 1.12; k1" = 0.0259; y = 0.5 + sqn(0.25+Nc.*Nv.*exp(-(Eg/kT)*(300./‘|’))/Nd"2): % y = 0.5 + sqrt(0.25+10.79*T."‘3.*exp(-1.297e4fT)); plot(T,y,'r—'), grid on xlabel(’T[K]'), ylabel('n__0/N_d') axis([450 600 1 1.1]) TYU.%,‘] ‘ ______. S. ‘ $1 is. doped a‘r Md: {0 a? and hut—:0 PLoJf MM. @01an Omaha-rake». avg. HM range 4.50 4.14 (000%. (b) At— WLLQj- +£W1P—QAOXMM5I‘; 44,“ £91th umwmpk (a) nO/Nd ’TYM 3.7 1.1 1.09 1.08 , , 1.07 1.06 1.05 -- 1.04 1.03 1.02 1.01 450 TIK] 10—” =1.07; (LT SZLQM 4.1 A silicon semiconductor at T: 300 K 15 homogeneously doped with N4— _ 5 x 10'5 cm 3and Na— ._ 0. (0) Determine the thermal equilibrium concentrations of 3 I free electrons and free holes. (17) Calculate the drift current density for an applied 8- f; field of 30 Vlcm. (c) Repeat parts (a) and (b) for NJ— 0 and Na— — 5 x 1016 cm . 69) F0 Wc- gt, ) HL=| 5510' /0iM% <4 (“{d‘Nq)/Z 5 1"“ E Ed 574W E “14.1%“: Mo = (3‘10: hifinae r3:- “$0 t 1 q 6 Po: .VlL ____._ aoSXlOECAMB jalh‘f} '2‘- 3/4... “0 E} fir-OM 11:1"6:,4,% M4431 Nq‘l'N’d': S‘x ‘0 '____._—— Idh'fl c: (l,br~t5’q)(l7vo)LSX lot?) 30:. 40,5? A/Ma J (C) Nq" Nd = P0 '=- 57,1016”:- -P- ‘L-“fit g/C jaw: 7:::;) '~. /PNQE» 41m" Fig.4.?) (”4% 15' fl” =— l700 g5— 7.. Q‘s—i 3 5500 4.8 (a) The conductivity In a p— type silicon semiconductor at T— _ 300 K 13 cr— — 0 25 (El-cm) 1 .Determine the thermal- -equilibn'um values of electron and hole concentrations. (1’)) Repeat part (a) for an n— type GaAs semiconductor if the resistivity is p = 2 Q-cm. P—‘l‘jfl S1... 300 K 1 6’3 OIZS‘Q‘L‘QMY g = 4 32:94» :Ffi'td Y} Chad. E Net '9 A - Icsw'flsz 449ml win/id. $9946 INS—*3 ,. v_ 55* "‘3 . s 1:?“ ma New “at: 3-SXI0IF/CME’ 3 Izyo * 5108,93: D W WS- E’Qfiififig Tviocwfifs'ienif HAL Nifg'rm’r—l should be §= Biz-rem if“ (Now mink” ) w :7: ”.2: _.H....:T.a.:.~:;. 92.21.52; . I r _NC._ cam: an: EA: EA: gm: :0.— EO— EA: 2m: -'uy.f _ . . ==I?HII==-III_==III:=-III==-III__:-III_=m-III==-IIIan-III many-I.1PJI== I'll-unalllI-Z-i'II-nunnill2—u I'll-unu-I.II==IIII_=nlill annual... 'JJu-IIIII-IIIIII::IIII=-IIIII=IIIIIa-uaIIII-nunlllI...:lI II IIII IIIIII II III.III. 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When the conductivity is considered as a function of the hole concentration p0, (a) show that the minimum value of conductivity, am“, can be written as 251' (II-n “pol/2 (tum + up) 0min“ — where a; is the intrinsic conductivity, and (b) show that the corresponding hole concentration is p0 = 11,- (flu/pp)”2. 6 [-9 0°”st 9t (wt flat-'9“ a? haflGOMCEVtxr'fi-‘tbn IP¢. Tun-g [All ‘ V10 ‘-‘— pa (3“: €7an +e/Hoo-g effumo+flrPaE T0 Rho} 0W1 elé+flmumm Q- "'Gc—Uu'c-on/t 0; P 0/ gag/x WC.) .‘ch‘m danvifive. "gee FOKW Wfl‘r‘twgr _( Grandma); Inf} 2.- Fez" “a?” C...) _ 3° ‘5': ehptjflu’T/Uf) Of“ CV31,“ = M6; fir ;H\99+|'+a-E {n+0 gm {“3 I .’ 4.31 The total current in a semiconductor is constant and is composed of electron drift current and hole diffusion current. The electron concentration is constant and is equal to 1016 cm’3. The hole concentration is given by p(x) = 1015 exp (—Tx) cm’3 (x 2 0) where L = 12 um. The hole diffusion coefficient is DP = 12 cmzls and the electron mobility is p." = 1000 cmzN—s. The total current density is J = 4.8 A/cmz. Calculate (a) the hole diffusion current density versus x, (b) the electron current density versus .. x, and (c) the electric field versus x. q -- as ‘t‘eDf’E’j‘NDOO— i Oslo x1220?) “030:6 -"/[)=‘9"0”f6£1* 'j Izmo " m W69 W E; 1% _._ £15m) . ' -_ ~ + * “WW oat-21am _. ti 9M0 —( ems him/17.510 )on 10¢ng fl.) .- _{ I *6)» h8= | @740: (9.\x\ou x \0 3hr; ”$.qu [A - HomacswU-‘Al EVE EASE" - 5v SQUARES EVF-EASE‘ A 5 SQUARES 42-38l afiflaflanalmflmud 2223:; “ES: . EVE-EASE“ 5 SQUARES DU DUE-h I loI’IM 4'32 4.32 A constant electric field, 8: 12 Vlcm, exists in the +x direction of an n— type gallium arsenide semiconductor for 0 < x < 50 pm. The total current density' 15 a constant _ and 15 J- W 100 Alcmz. Al x: 0, the drift and diffusion currents are equal. Let 5:3 T x 300 K and u, : 8000 crnz/V-s. (0) Determine the expression for the electron " concentration n(x). (b) Calculate the electron concentration at x = 0 and at .z' x :- 50 pm. (6) Calculate the drift and difl'usion current densities at x. = 50 pm. {i E: \ZV/cm fiCGnsTCX) Wm, 6w- w r: an “exam fig y .4 Same" Cm & ‘ / de v; . I: Iook/m’l—kuflx) War 19:60“ a 78—5.». : 0 X170 I :3 I _ ‘ ... ) time?“ 499 fd—h egh__\l: =0 l=3~90 1-. —— D “glowed/v4 AX IQT Q n fln‘ an + Ah — g=0 it Mr 10: flxrloazFC-E d“ _(_ b X ___> ‘LbEXP()+QATA'\oD)<P()—B=o a}: __ L apt-”E (“U/t) WW = 0} X20 0| 0A~B= O {Vi/WE; =61)“ El: ad}? t I 22‘: “we "9 " «I “ m. 974mg (6+5)=eDm[—E) {Iii/4;: [lszio jgwoo) WHO E g C ill” (a: 015th W0 025‘? * ”“6 :er: EEK- ’12— =215§xl0c lien (.1?ng e—Ex [QT / 91-”; ":“io l5‘ [5 (x) -2 ’ '57— = 514?— 2‘3me — cw“ io—r d/;=3,2€€xlo M 1 in“) G I '9 ‘- 5’ —3' (5 Eta/L” 325)“? 60V! 7 _ {15‘ 3 ) V1 (5051M) :- o, — & 1&Pfi3wto H.)-1(9(l-— 432?,(105 #téJgjxlr‘?“ 2.1535500 [6) TV [time )[6'39XI05)[8000)£12)= ‘7’5A/cm7' ‘Jd'iW? Agni—oar: loo—TS”: 5 A/CMq-L , 72- ES 8| 50 SHEETS EYE EASE“ . SCUM-l EE‘ - S SQUARES - a iii-36? I SHEETSEYEAEA éufl'flflflal Brand 42-359 SHEETSEY AEASEWfiSOUAREG E 0:) 293 3h Arsenic is diffused into an intrinsic silicon sample and has the general profile shown the electric field. lit": A in Figure P4.4l. Sketch the equilibrium energy-band diagram. Show the direction of Arsenic doping A9 in (531$. Ga 1'9 & diner " ’_" 2_ Figure P4.4l I Figure for Problem 4.41. N461) AI'PFMsi‘am Pmfi'flL ) ‘fl/nuwg complain, {oninmu E {20 V [:00 is mph Genre/L bf) +£11- ..100MJ67¢{9 fflw’fifi’w) ”530) Owl Eva) as Show»: %L 00 Q2 ‘, ’2. 7 _ 4.42. Barman”; l‘wr-QaJ-Ml'n'i‘o amivfih’vgk gf sawrafll‘ The cgemewi FdeL is shat-09m ch Pg, 124,42. ék-ztcha HU— nui~2QA‘bVil/M enact-baud Afagraw, Baron M Q {s an qeafy‘for impqfiig; 7 \ B ‘ x —>— I . ‘ Figure P442 ! Figure for Problem 4.42. qu ' ' - - Boron doping =55500 ' Titoloum 4-50 4.50 Consider silicon at T = 300 'K. A Hall effect device is fabricated with the following geometry: d = 5 x 10’3 cm, W = 5 x 10*2 cm, and L = 0.50 cm. The electrical parameters measured are: IJ|E = 0.50 mA, VI = 1.25 V, and Bz = 650 gauss = 6.5 x 10‘2 tesla. The Hall field is E H = —16.5 mV/cm. Determine (a) the Hall voltage, (b) the conductivity type, (c) the majority-carrier concentration, and (d) the majority~carrier mobility. new E («my Fl [gut—‘72! 31:0 ‘ . "’7 jng-JUX) B=Bz 'Hxlm ’U’XB N36”) game—38 age-H - V =mfi‘52 w . “a” “*‘H Pf. Warhead»! : I 7" .21. 1‘. - 6’!) Y‘ —— V X dW K v, E . 46,9" ’55 (UK: ’1'." = Vii Gwmyfmmmngtheflaflefiect. dwem 32w .. +2 —‘ —§ den (‘0) V1_+‘1rl( éfilflrx VH <0 WA!" “’19?“ +°9gu39wv2nxam \vr th‘d‘re- IKBzw “"1" Ix Bz 03 = a ~f 60 wait-91‘0“? V1: “ ( iin Wt 0“} " 1 aleVu ataWHl I I ‘ ‘ “3+ 595;? its; 2): 82 ENG "-5- : 4:512‘“): eeafrms/wz’ 5 s as ‘3“ M g m: 4,421 l0 #1ng é _ 05$!“ ole-{lo '7‘} evaEx Wd CVI 5,; ml en VK S'No‘fc Sm vl- W0” H—lb‘wfl' 1: . 2W! .. 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