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hw2and3sol - E LF E EV Egob:31 c Plat Farmifimctiau/ar Er...

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Unformatted text preview: E. LF E.- ,. EV Egob. :31 (+c Plat Farmifimctiau/ar Er - 1 2V. {(5) =[1+ ,(s-Emr 1.. (twin-54%” We will choose E in eV and therefore use kI' = 0.0259 E 3 5-5; kT ‘ 0.75 0.90 0.95 0.98 1.02 1.05 '1. 10 1.25 4.6525 4.3510 -1 .9305 .0.7722 +0.7722 +4 .9305 ' +3.35 10 +9.5525 0.99994 0.97939 0.87330 0.68399 0.31 600 0. 12669 0.02061 0.00006 b7 an alth £16174 , 5% as u eke/6°“ 9f 11.04,, wk“ caudadcd. Hang/x 4 gap-{mild 9P- ! Y’H'. 7%.; Ef'blfldx. 25 W4 q (15414414.!- V; Ma JoCKfiL-m,‘ I 40 E:— /¢V = AL 170-19 A J 7;: Mn he; 5‘ inn”- kirdvz.‘ 04.577. — 1—.— / wig-c M, .-_ rad—ma: ’91:. ‘LW = ‘ftlLuo—glg‘ / . 7—Y(L .J“ _ =- /1. .— \/>[ .’ “0 win] —_ Signal £1! up” #2 5a, hob. 2.6 Find L/ar ”0an 12 k'zV elzétrcru. Comment on :— microscape: compared to visible ‘ light. .4..- h h 6.63x10'3‘ m v=./25'/ , x=-—-= a 5- '" 725m [2 mV x9.u><u>‘“]“z For wow: 4.9x10'” 1/: ‘ -9 u: -Io - hmnwr =123x10 [1001' =113x10 m=l.23A ForleeV: A. = 1.23x10’°[12x10‘]'”7’ a 1.12x10‘”m =0.112A Visflale light is about 0.5m = 5000A, so the molution of electron microscopy is much better. 5‘) Ce) (9) ,2; 8.»: [ring— 6 =- LGDLUD‘H 400‘ 1.27:: AZaluo'zLJC-x aopk 6 K (~LDLXID'H4oal ® 37 44%;» +0. Man—L1, ,1: 430403 a. alt-rm d—wh Fermi (cm! efigwrzfl :4 i Z- . ___. ': 0.025“? W“; f . w: :— ‘E‘ci Eur], Em] tic 55146.35". @) 5-5 15mm W07 .5 Lfig-EZ-Ib— = 1c. 3 2 JV .5 9’21“: .5 mag—E ”aoksangqv 5 16, 2E? x3: /. 381u10'” 02¢!“ k— fiv I I _ -p(Er-r2£r) ’-~~_-Z§Mw__&) w: m”:— w_._. w. (1. 6 [ +5. ’1: OJH m§ , (a) Explain why holes appear at the top of the 'valence band. Electron energy is plotted “up”‘in band diagrams such as Fig. 3-5. Thus conduction band electrons relax to the bottom of the conduction band. Holes, having positive charge, have energies which increase oppositely to that of negatively charged electrons. That is, hole energy would be plotted “doWn” on an electron energy diagram such as Fig.3-5. Holes therefore relax to the lowest hole energy available to them; i.e. the “top” of the valence band. - ~ (b) Explain why Si doped with 1014 cm'3 donors is n-type at 400K, but Ge is not. According Fig. 3—17, the intrinsic concentration n,- at 400K is , n. (400K) ~ 10” cm'3 for Ge ~ 1013 cm'3 for Si Thus at this temperature, Nd » m for Si, Nd « ni for Ge. Prob. 3.10. . For Si with N4- Na = 4 x 10” cm'3, find E; and Kg. "0 = Nd — Na = 4X10” = nie(EF-El)/kT . 15F — E. = kT1n(no/n,-) = 0.02591n(4x1015/15 X10”) = 0.324eV RH = -(qn0)" = -(1.6 x10"19 x4 x 10‘5)-l = —1562.5 cm3/c “W #7, 56L I OMS ma) '4; L5¥L013 50“-: ['flam. 7M) Shae. N975 m, UR, -_ . III ._ I; 7‘: A}: A +[L :v‘yfnlb = 4207 “DH-CM; .1‘ 5 [9; LL!” : 2.07] X10! Guy—3 71 . (5,) Ax.— 910'b (W'ijL) 71; >7 A4 I :p Y1 =71; “ {Klolbzm'x ‘umpL¢L $9M; " 2447949 I a, [Kc-5 A 54;);on m: 44' Silfbon. a+ fbaM Mt), M: 4:54! 1‘17 aflbv er 2:“ I." {V'I’fiW-C. 6" J V133 . Vin/j qua—lam [3—24) for n 1 l9) -(zc~ "-J 4 44:44 M¢ C kg! = /‘/V c 497' 44.3 :‘n #0 Frvlalam. 6(55 ‘5‘ —EV Hal/i7— EMA): :: - /:‘.¢ ‘f‘EV 1—,, i 2 E: " Er. ’E/ éiT/Zx/zfléi z/mj Cyan-111‘"; (3-14,) and (3—22), (ZWT/a (if? A 1(277'Mfi7a— A4 a»: 4A ewva msgpe 4* =0.5/M. ML [Mam/m, I (h. w) out be 2m ‘9m 63% 54¢” 41w. M531 19 54:02“ fiat—+534 41‘“: +L¢ Ferm‘ ("Backm in’cld #4 Cm-n‘ar Concac‘raim 4,5 4 fid-wn 0';— amy. 77m} FrnJuul— raw/+5 in “(59475" Mar Cat—L band «543:. J 07) F/‘oM I‘Qwrc, ivfl I A[ fiv’ 6'; @ 4001:? {1: Ar- 5; @ watt/0'2“ ZkIDIrC-u‘a. 5», 4+ T= 400K N!) >7 YL.‘ fir S,‘ vhfr‘i. Mk“ 7% 5: 4-4%. :24}— Ar (1 ”4 4001!: , [JD 44 71,; I 5: 4r #113 Cam, +144. 51: is [31521. Ca) ,4+ 3001: , m—ano’kw’ 4;, S“ sincc M: “MD J 13 and “5“‘4, ,4 g MD = 0'2...” I’m U7) '1; " lino" emf" . 1/1“an “ML 14/07/44., 7° 4- A - we,» ’1‘“ 4’5 = Aza’xzo" cm" F (9) ”a 4 "’0 arc. W >7 VII“. M: '{A’ld- am &- /:s; 4%“ 0M. 74-. Alb-Mo. = m’iw’ aan n4 majm‘lxdc p‘F ow. mafia? 5o fll‘ 5' _5 [Din -=2.&5_X10 cu Ch {2: mm— I ma- azrxmm- I So . 55C [‘3 Pvfirgg ' V (5-55!) ’ p: A c ”r I 53—15: JrAfiJ R 5-1541=90Lfie/Afl%x(ov)=D-547¢‘/ ‘ 6b. )\ [Ill all ——.—~_‘ fig . a, €54”ch M’— " 50 it. {S K >z10 II) .-. 2,LS_\<IO;5~," —6= ineffi— =9. mafia/A ' = 0&an ,._‘.._ ‘wgmflt. ‘H [I {2/0/ I (“1.45" 5“? 0-1 «cm 19'!!! 4841.99 11m: «in Ct9§€,§.€£ti0nal, areajs. doped with 10’3 cm‘? antimony;w F ind ”the current at 300K with [OI/applied. From Fig.3-23, 11,. = 700 cmzN-s o .= qpnno =1.5x10“9 x700x10‘7.=112(Q-cm)" = p‘1 p =0.0893Q-cm R = pL/A = 0.0893 x0.1/10‘6 = 8.93 x103 12 I = V/R =10/(8.93 x103) = 1.12 mA Repeat for a length of 1 pm. Now 8 = 10V/ 10'4 cm = 105 V/cm, which is in the velocity saturation regime. From Fig.3-24, v5 = 107 cm/s I = qAnvS =(1.6x10'19)(10‘6)(1017)(1o7)= 0.16A (b) How long does it take an average electron to drz'fi 1 pm in pure Si at an electric field of I 00 V/cm? Repeat for 1 OSV/cm. From Appendix III, 11,. = 1350 cm2 -s I low field: vd= p08 =1350 x 100 = 1.35 x 105 cm/s t=L/vd =1o4/(1.35 x 105) = 7.4 x 104° 5 = 0.74 ns high field: scattering-limited velocity V; =107 cm/s (Fig. 3-24) t=10'4/107 = 10'” s =10 ps ‘ ¢5Sun'-"#¢: SC. [9 A‘J‘lffi- 7—1/4. Fofihlfib Vep Cowswi deo‘i‘rbns +0 filo.” 'p/‘mx {T #9"'C1, (2:35:15 "GUif'M-‘i- 4o filo-u prom C +0 D 71¢ &$¢+WZ’9¢5’7 p‘F clajnns 1"; D’t 4L4, ~>< dinolwns.‘ 71" 'q’k‘MMI “FF/”A ”War“ £ch 7; ,\ +5. +a—.;:i-"So“fir\- me‘ik Lora/4% hirer, ofue-i-wnf ;= b (513-) . ij E resuH-s D\ a veal-or ijflfi/ .7V,=!~‘re42§1m.; [101‘ . 3 )9 Alva—'70; £9!“ elech’w, 50 [53,3 run-K4, .-\_1 Ail—aim. 711i Antonia: 7,?ng ,Wlfiere‘gre. Cat/5e; 656650” +0 file VP 4W; nae .19.va _g.'4<7a 14¢ sh. Cmfirj a magnify, V4.5 ‘57,,“ ane Wk; WWJG ~7¢9fii£4éélns~ms 5:). Ac . x. ,, . , . _- qr ,_ ;.- ‘ Wu; #3 7» mum/S .U’) fin“ fig (0 {as-Ia WW +%/W$ (9) A? >(asn_'&¢-“'Seem‘ 14am +L< ,a may; Tjra/l: grovt'JeAiz-Irnoéfihly skmsuihuff-é/s in 4&4ij 'zfemloeva‘ére. n W a 5 sap/77w; "A“ 1;; ”44+ meal, A L ”Anni“, V- 546 07) W5 CM x: 1cm 'Mééil'ii‘y jalali fromllui aéem WM 74% bum/l:- #14}- M‘p‘éfi/h‘y dean/am as 49?"? increases. 77155 6"?“ 4/” 6" “‘9‘ '7" 97"" 5—25 in -H-< 42rd: ‘72:.‘5hl'k'th/fztdd iii-inns}; 5; mi» Carrier Maw; m no.4— ‘ . 143 51530163304 M Mm 5.+.3 57A 7155+ M ”f?” batsiLI-fi‘pcgfluo'F'tfica'i'Ir-flj Wnt'Sut: W 554%»? ml ; 4;.” away. 147594 554%, is Jvc, VCém m" ”7544 4.164% 24:1- 315+; 7'1 M'Mnsfl I dopeJ- 5L; [@7949 Sakai} ,'> cfvc 7‘9 ‘Mc MWn O'F'Fé‘v’riffsmwifls I‘Drtieul imp/rig at‘l’W‘) anal only fiwfsfilh “Z .... $.C. 50, eloFl’ij cream #( 4M Shiloh}; Meals-M. WW Jeorures m6)l.¥7. &' (this is clearly an open—ended problem with several alternative solution approaches and many possible solutions. What follows is only one possible approach and solution.) (a) The choice of sensor dimmsions is totally Irhil'rary. A relatively small cross section would be advisable to permit amore rapid response no temper-am changes. (One might consider sawing up a wafer to fabricate the sensor.) To be spedfie. let us assume the sensor is bar-like, with length I: 1 an and cross—sectional area A = lmrnxlmm = 1041:1111. Considering the resistance of die sensor, we know R ‘ pI/A or p = (AIM where the resistance is restricted to m s R $10000 Butwehaveohosen All = 10"- cm We therefore require iO'ZQ-em S p S ion—em _ ‘59 at all measurement unwary-s The doping must chosen such the p falls within the cimd range over the operating nempenmre mge (—30°C 5 T5 40°C). At room lempennu'e and taking the Si to be n-type. reference to Fig. 3.8(a) indicates that we must have 4.5 x 101‘lcm3 5ND 545 ><1013/¢:m3 Anticipating changes in the allowede range at higher andlawa temperatures. let us . ohoose N]; in the middle of the above notedrange, say N0 = 101‘lcm3. and check to see tf the prequirement is met at the nemperann-e exnemes. Referring to Fig. 3.7(a), we conclude . at T =—30°C = 243 K, ”Q ~ 1800 cmz/V-see andp = 0.347 Q-crn atT= 40°C = 3131114. - 1100 cmzN-sec and p-0.568 Q-cm The computed extrema lempenmre vain: fall comfortably within the allowable p-range. Previous y boldfaced items are then the acceptable design solution. Prob. 3.20 Calculate the number of electrons, holes, and n; in the unknown semiconductor with E; 0.25eVbe10w EC; Incomplete ionization : 1 . E = ‘ =0.1267 f( d) l+e$5u5 n = (1- f)Nd = 8.733x10‘4cm‘3 5,—5, Also, n = Nee *7 E: -EF NC = ne [‘7' = 8-733x1014 Xe0.25/0.0259 =1.359x10‘9cm-3 = N» EF-EV p = Nye [CT = 1359 x1019 Xe-(l.l-0.25)/0.0259 n,. = J; = 8.142 x.109cm"3 = 7.591 x10“cm‘3 Prob. 3.21 Referring to Fig. 3. 25, find the type, concentration and mobility of the majority carrier. Given, 3, = 10‘4 Wb/cmz From the sign of VAB‘, we can see the majority can-lets are electrons. -411, (Io-Mo“) n .. _—-—\=3.125x1017cm-3 ° qt(-VAB) 1.6x10"9(10‘3)(2x10‘3) —3 Flaw: 0'1/10 _3=O.OOZQ-cm L/wt L/wt 0.5/0.01x10 1 '1 ' m.“ — =x=1mooo cm2 ~s)" pqno (0.002)(1.6x10 19)(3.125><1o”) (V (f Wm C93 "PUP” va) Tb mm 5” ~~ 5%? 01> EU 1:2, 5/ \ E; [‘7 afim 51/065 \ EWTUPBWM (owl) rrms __- _- _\_‘\> Exts'r‘. M Rania»: 3mm 6w.) 56A 1"? (-9 05 a“ Dvyw “€235;st .' . 5‘ .fl— M,» r. J3 AW 7 4.3—3 )c Jimc‘mB ' , [:1/ '— olvx be) \ .‘n \_,v «v by " m*vgeak(= Epeak — EC Setting m* = mg, we obtain — kT _ (0.0259)(1.6X10-19) ”peak ‘ W —[K 1/2 31 J =6.75x104 m/sec = 6.7sx106 cm/sec 9.11x10' (a) for all cases. The semiconductor is concluded to be in equilibrium because the Fermi level has the same energy Value (it is constant) as a function of position. EF-Ei vs. x. Under equilibrium conditions, n = niexp[(EF—Ei)/kT] and Diagram (a) Diagram (b) 3-16 Diagram (c) , A. a. “kuww m P i H1 Diagram (0 Diagram (e) Diagram (d) . 2, \II . kc .3}; band, 3-17 Prob. 4.5 If no = Gx, find 8(x) for no >> n,-. We also assume Ep remains below EC. At equilibrium: dn J» =qunn8 mung-=0 Prob. 4.9) . 6 separation of the quasi-Fermi levels and the change of conductivity upon shining light on a Si sample. The light induced elecu'on-hole pair concentration is determined by: 5n = 8;: = gap: = (1 019x104) = 1014m-3 << dopant concentration of 1015 cm“3 => low level n =10ls +10l4 =1.1x10‘5cm‘3 2 10 2 _"- _(1-5><10 ) 14~ 14 -3 p°+6p-:+6p-_———1015 +10 ~10 em up 1.1x1029 Fn—Fp=kT1n —2 =0.0259ln ——-fi =0.518eV a ni 2.25x10 0,, = 1300 cm2/(V - s) from Fig. 3.23 D up=—”-———12—-=463cm2/(V-s) kT/q _ 0.0259 A6 = q(l.l,,5n + upsp) = 1.6 x10“9(1300 +463)(10‘4) = 0.0282 (9 - cm)'l vuwvaa‘m I»: but. .i a. an 35 (Caleul e the separation in the quasi-Fermi levels and draw a band diagram for an n- type Si being steadily illuminate . The induced electron concentration is 6n = gap: = (1021x104) = io‘Scm“3 which is comparable with Nd = lOlscm‘3. => this is not low level and 5n2 cannot be neglected. gap = W105" +ar5n2 (If: 1 =——_31—15=10’9cm3s_ Tuna (10 )(10 ) => 1021 = (10'9)(1015)6n +10'96n2 SolveforSn => 8n=6.18x10‘4cm‘3=5p 15 14 Fn—Ei=kT1n "0+5" =0.0259 1° ”'18:,” =0.300eV n- ' 1.5x10 1 l l4 5,. — Fp = ”111(k): 0.02591n[w10—]= 0.275eV ni 1.5x10‘° n 1015 EF—Ei=kT —° =0.02591n ——w =o.2ssev n,. 1.5x10 36 i WWWfimamrmmli In: t. Mu .v v 5-. Prob. 4.12 51,05) Find Fp(a:) for an exponential excess hole distribution. For 6p > p0, ‘ Ap 22(2) 2 .st = Ap e4”? Apes/L, = n; 6(Ei-FPVI‘T 6p I Ef—Fp 2 [CT 1n — "i E: = kT In 92 can” /-7 E . F Tll ‘ +/ E Ap :1: — - / - —_-_ " = k _. _ .— T [In 7" LP] / ’FP EV We assume the excess minority hole concentration is small compared to no throughout, so no band bending is observable on this scale. 41' 7 (3) gram Eb 5&5;ij , L41 5.): 0,32% @5947”, £5 __ - ‘ ' 2’ 2, . 3.. no — M‘ a ‘ ': /0 Ofi :9 67‘ /¢075X201 w‘g [L/{L / 20 ’ % F” ‘ 7r " p n,— =-- 7,52er arm-5 " / Mew/p“ CF)! "ébZZr—w: 0 31;” n =~m C, / /0/0€ "”929 l4?k/¢7f 3 (fi'éXflf pf; F , ,oztz _ {7: VI. 6 = mmg p a : /,p73xzolpm 3 .—. , 2.. N016,» A? t n; on wwflkb vNDfioL IELPvlL/Bzwm (puurnazs " - /,073 X/0 MGM SF [5 1331‘ 4¢ Yb , $0 Low-«4945; wawzo/v’ alumna/vs 170 Mar“ 5X57"? a. mm a, MC!) wywur 27m: [aw/JD, 5,4. :5 Pvfifi. 5o, vowel/7° a) é‘éjflr“ Av "72"? '- 7WL . = / 9‘ ” —— A 7 10’2“” Am a ,. I. ’6 ’ s it;- a. JD K. :- 7.3xm‘ m" Agn ___ “ yin “El 117‘! ' [/JYHD ) 26 'ax" £4 ’2) {Z - 4' fig ”53% - /pp 9:51%23T‘2',;¥ID’:“_ 7kg flog” 577”“ ubSUK/IJ- fl‘ 5M1 smau‘mmv‘ ‘ a 71415 News SA :5 m- ,4- pupa-w off/$9 if: 50. 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