半導(dǎo)體物理與器件第四版課后習(xí)題答案

1、ifao were to increase, thebandgap energywould decrease and the materialwould beginto behave less like a semiconductor and morelike a metal. Ifao were todecrease, thebandgap energy would increase and thematerial would begin to behave more like aninsulator.Schrodingers wave equation is:22.x,t2mx2V x x

2、,tx,t tAssume the solution is of the form:x, tu x exp j kx - tRegion I:V x 0. Substitutingtheassumed solution into the wave equation, weobtain:jku x exp j kx 2m xChapter 32u x-exp j xEu x expj kx -This equation may be written2jk-u-xas2mE uu x IEexp j kx 一xkxwhich becomes2m2jk u x exp j kxu x . . E .

3、2 jkexp j kx txSetting the equation becomes:.2d u1 x dx2 whereIn Region II,u1 x2jk 也。 dx2mEV xVo .sameform of the solution:x, t u x exp j kxfor region I,k22u1 xAssume theSubstituting into Schrodingers waveequation, we find:22m2jk u xexpkx ju x _2 jkexpx2u x2 xexp jVOu x expkx - tkx tj kx tEEu x exp

4、j kx 一This equation can be written as:2u xk u x 2jkx2mVo2-2u x2-x2mEx u xSetting ux U2 x for region II, this,2d u2 xdx22jk 虹x dxk22mVOU2 x,2d u1 xdx2equation becomeswhere again2 2mEWe have2d Ui x du1 x . 225 2jk kUi x 0dx2dxAssume the solution is of the form:u1 x A exp j k xB exp j k xThe first deri

5、vative isdu1 x j k Aexp j k x dxj k B exp j k xand the second derivative becomes2j k Aexp j k x2j k Bexp j k xSubstituting these equations into thedifferential equation, we find2k Aexp j k x2k B exp j k x2jk jk Aexp j k xj k B exp j k x22k Aexp j k xB exp j k x 0Combining terms, we obtain22222 k k2k

6、kkAexp jk x22222 k k2kkkBexp j k x 0We find that0 0For the differential equation inu2 xand theproposed solution, the procedure is exactlythe same as above.We have the solutionsUi x Aexp j k xBexp j k x for 0 x a andU2 x C exp j k x D exp j k x for b x 0.The first boundary condition isUi 0 U2 0which

7、yieldsA B C D 0The second boundary condition isdu1 du2dx x 0 dx x 0which yieldsk A k B kCk D 0The third boundary condition is u1 au2bwhich yieldsAexp j k a Bexp j k aC exp j k b D exp j k b and can be written asAexp j k a Bexp j k a C exp j k bD exp j kb 0The fourth boundary condition isdu1 du2 dx x

8、 a dx x bwhich yieldsj k Aexp j k aj k B exp j k aj k Cexp j k bj k D exp j k band can be written ask Aexp j k ak B exp j k a k C exp j k bk D exp j kb 0Fory ka n(b) (i) First point:Second point: By trial and error,n 0,1,2,. sin ySo that, in general,dxa 1.729(ii) First point:Second point: By trial a

9、nd error,dyAnddk2mEa 2.617So2mE1/2dk2m-2dEdkThis implies thatdkc dE 0dkfor(b) (i) First point:Second point: By trial and error,a 1.515(ii) First point:Second point: By trial and error,(a)1aa 2.3752m0E1一2a234 21.054 1031102 9.11 104.2 10cosa coskaLeta kay,ThencosxcosyConsiderof this function.dy3.4114

10、 10From Problem2 a 1.7292m0E2E2a 1.72921.7291.05419 J34 210 343110 229.11 104.2 10g p dyWe find1sin xcosxsin yE2181.0198 10 JEi181.0198 102sin xdxdycosxdxdyor6.78686.7868 10Thendxdy1 人sin x xcosxsindx x -dysin xsin ysin y(b)1.63a 21910一一 193.4114 1010 19 J19一 4.24 eV2mo E32 a_2_ 34 221.054 10E329.11

11、 10 31 4.2_2 _ 34 221.054 1010 10 2E32 9.11 10 31 4.2 10 101.3646 10From Problem ,4a 2.61718 JAt ka181.3646 10 J.From Problem ,2 a 1.7292m。E42.6172moE21.729E42.61721.05434 2103110 22 9.11 104.2 10E21.729234 21.054 102 9.11 10 314.2 10 10 2or182.3364 10 JE4 E32.3364 10 18 1.3646 10 18199.718 10 J1.01

12、98 E3 E2 1.3646 10 183.4474199.718 10191.6 10 196.07eVor(a) At ka1a10 18 J3.4474 101.0198 10 1810 19 J19191.6 10 192.15eV2m0E12 a(a)1aE1234 21.054 10 343110 229.11 104.2 102m0E12At193.4114 10 Jka 0 , By trial andE1234 21.054 102 9.11 10 31 4.2 10 10 2error,oa 0.859Eo234 20.8591.054 102 9.11 10 31 4.

13、2 10 10 23.4114From Problem ,10 19 J2a 1.515or2mE1.5152.5172E1Eo3.4114 10 198.9428.942 10(b) At ka 22mo E3210 19 J1.6 10 193a 22.5172 10 1910 20 J20一 0.559 eVE2or(b)3 a1.5152 9.11E2E17.83021.054 1031107.830191034 24.2 101910 J3.411410 21910194.4186 10 19 J194.4186 10 19191.6 1022.76eV2mo E32 a 2or_

14、191.6084 10E3234 221.054 101.6 10 191.005 eV2 9.1110 31 4.2 10 10 2(b) At ka 2 ,3aFrom Problem ,1.3646 10 18J4a 2.3752m0E32mE2.375E321.05434 2103110 22 9.11 104.2 10E42.375234 21.054 103110 22 9.11 104.2 10At181.3646 10 Jka , From Problem ,2a 1.515or1.9242E4 E3181.9242 101810 J2m。E21.5155.597181.364

15、6 105.597 10E191.6 10101919JE21.515234 21.054 103429.11 104.2 1010 2(a) At ka1a2mo EiEiAt3.50 eV7.830E3 E2181.3646 1010 19 J7.8301910or195.816 10 19 J195.816 10 19191.6 103.635 eV234 21.054 10 343110 229.11 104.2 10193.4114 10 Jka 0 , By trial and error,oa 0.7272moEo0.727Eo0.727234 21.054 10 34ForEg

16、Eg 1.164 eV100K,424.73 101001.170 6361002 9.11103110 24.2 101.8030 1019 JT200K,Eg1.147 eVT300 K,Eg1.125 eVT400K,Eg1.097 eVT500 K,Eg1.066 eVT600 K,Eg1.032 eVE1Eo3.411410191.80301910191.6084 10 19The effective mass is given by1 d 2E 122dkWe haved2Eso thatA d2Ecurve AT curveBdk2*m curve A mcurve Bor4.4

17、437 1031m 3 m。9.11 10The effective mass for a hole ism 0.488 mo2For B:e Cik101_ _ .At k 0.08 10 m , E 0.5 eVOr1920E 0.5 1.6 10 198 10 20 Jgiven by* mpWe have thatSo8 10 20C10.08 1010 2C1371.25 10Now2m 2G34 21.054 103721.25 10d 2E2 curve A dkd2E2 curve B dkso that*mn curve A mn curve B ppor4.44 10 32

18、 kg4.4437 10 32m 3 mo9.11 10m0.0488 modEPoints A,B:-0 velocity in -xdkdirectiondE 一 .Points C,D:0 velocity in +xdkdirectiond2EPoints A,D:d 0dk2For A:E E C2k0.025 1.6 10 1939C2 6.25 102c21.054C2 0.08 10102 6.25 1034 21039negative effective massPoints B,C:2 0dkor318.8873 10 kg318.8873 10319.11 100.976

19、 moFor B:C2k2positive effective mass0.3191.6 10C210 20.08 10C27.5 10 382For A:e Ci kAt k 0.08 10 10m1, E 0.05eVOr_ 19_ 21 .E0.05 1.6 108 10 J2m 2c21.054103834 2SoC1Now_ 21_ _ 108 10 C1 0.08 10or m1.2510382C134 21.054 1021.25 10 38._ 314.44 10 kg2 7.5 10327.406 10 kg327.406 10mo9.11 10 31m0.0813 mo(a

20、) (i) E hE 1.42 1.6 10or _ h 6.625 10 34143.429 10 Hz/.、hc c 3 1010(ii) 而E 3.429 10(a) mdnE1 22/ 34mt21 /3mi2/ 340.082m。21/31.64m058.75 10 cm(b) (i) Eh(ii)875nm_ 191.12 1.6 10346.625 102.705 1014 Hz c 3 10102.705 1014mdn0.56m0(b) mcnmtmi0.082mo24.39 0.6098momomcn 0.12mo1.64mo41.109 10 cm 1109nm(c) C

21、urve A: Effective mass is a constant Curve B: Effective mass is positivearoundk 0,and is negativearoundk 2(a)mdp3/ 23/2 2/3mhhmlh0.45mo3/ 23/2 2/30.082mo0.301872/ 30.02348momdp0.473mo(b)mcp3/ 2 mhhm3/21/2 mhhm1/ 23/ 20.453/20.082 m1 / 20.451/2 mo0.082mcp0.34moE Eo Ei cosk kOThendE E1 sin k kO dkE1 s

22、in k kOandd2E2 i, i,2E1cos k kOdkFor the 3-dimensionai infinitepotential well,V x 0 when 0 x a, 0 y a, and0 z a . In this region, the wave equationis:222x, y, z x, y,z x,y,z222xyz2mE2-x, y,z 0Then2_211 d2E Ei*222mdk k ko orUse separation of variables technique, so letx, y, z X xY y Z zSubstituting i

23、nto the wave equation, we have2XYZ xXZ2Y-2 y2Z XY2 z2mEXYZ 0Dividing byXLet2X-2xXYZ2Y2y,we obtain2 . 一1 Z 2mE2X-2xkx2X-2 x2kxX 0The solution is of the form:X xSinceAsin kxxB cos kx xx, y, zat0, thenwhere,2 2mEk We can then writeTaking the differential, we obtain1111 mdk 一 2mdEdE2 . E2ESubstituting t

24、hese expressionsinto the densityof states function, we haveX 0 0so that Also, thatgT E dE3a 2mE 1 mx, y,0.02EdEata , soNoting thath0.Thenkx anx wherenx1,2,3,.Similarly, we have21 YY 7,2.ky and2Z2z2this density of states function can besimplified and written asgT EdEFrom the boundary conditions, we f

25、indky a ny andkzawherenzDividing by density of states so that/34 a 3/2 一3 2m . E dE h3a will yield theny 1, 2,3,. and nzFrom the wave equation, we can write1,2, 3,.3/24 2mg E3- Eh22kxk22mEThe energy can be written2asEEnxnynz2m2nx2ny2nzFor a one-dimensional infinite potential well,2mnE2-Distance betw

26、een quantum stateskn1 knThe total number of quantum states in the3-dimensional potential well is given(in k-space) byNowgTk dk2 dkgT k dkk2dk 3-3- aNow-；2mnEdk2mndEE(ii) AtT 400 K, kT 0.0259400300ThengT2a 1 dE 2mndEE0.034533eV0.034533 1.61019Divide by the volumea, so2mn5.525310 21 JThengc7.9531055 2

27、 5.5253 10213/ 2orgc9.239 109.24 1019 cm 325 mSo1341.054 10.2 0.067 9.11 10 31(b) GaAs,mn4 2 0.067gc 0.067m。313/29.11 10 316.62534 3102kT3/21.05518103J 1(a) Silicon,mn 1.08mo3/2gc E42mnh3gc4 2mnh33/2 Ec 2kT-EEc dEEc541.2288 103/ 22kT(i) At T 300K, kT 4.144gc 1.2288 1054 2 4.1449.272 1023 m3or(ii) At

28、gc_ 17gc 9.27 10T 400 K,1.2288 1054gc3 cm211010 21J3/2kT 5.5253 102 5.5253 10 212431.427 10 m1.43 1018 cm 321J3/23/24 2mnh33/24 2mnh32 1.08 9.11 10 3134 36.625 10 343/2EcEc 2kTEc(a) Silicon,m 0.56mo p2kT3/23/2_ 3/22kT7.953553/210 2kT4 2mp h33/22mph33/2-E EE.E3kTE dE(i) At T300 K, kT 0.0259eV _ 190.0

29、259 1.6 103/ 24 2mphE 3/ 23kT4.144 10Then21J3 /24 2mph3gc 7.9535510213/22 4.144 102 0.56 9.1131103/23kT3/ 2orgc2536.0 10 m6.0 1019cm 36.625 10-3kT 33/ 255, _ 3 / 22.969 10 3kT(i)AtT 300 K, kT 4.144 10or(ii)Ator(b)(i)At55212.969 103 4.144 102534.116 10 m4.12 1019 cm 321J3/ 2T 400 K, kT 5.5253 10GaAs,

30、or2.969(ii)Ator6.34mp1055 3 5.52536.337 1019310 cm0.48mo310.48 9.11 106.625 10 34552.3564 10T 300K,102121J3/23kTkT25m3/23/24.1442.3564 1055 3 4.144253.266 10g 3.27 1019 cm 31010T 400 K, kT 5.52533 /23kT21J213/22110 J55213/22.3564 103 5.5253 10 212535.029 10 m1935.03 10 cm3/24 2mn(a) gc E3E Ech -313/

31、 24 2 1.08 9.11 103E Ec6.625 10 341.1929 1056,E EcForgcE Ec;Ec 0.1eV;461.509 10 mgc 0Ec 0.2 eV;2.134 1046m3JE Ec 0.3 eV;2.614 1046m3J3J 1E Ec 0.4 eV;3.018 1046 m3J 13/ 24 2mpg 3h4 2 0.56 9.11 10 3134 36.625 104.4541 1055 1For5.6343/2, E E0.1 eV;4510 m0.2 eV;457.968 10E E45：9.758 10 mE E3J 10.3eV;3J

32、10.4 eV;1.127 1046m 3 J 1(a) 7gc(b) 一g3/2 mn37?mp3/2 mn37? mp3/ 21.080.562.68Plot(a)(b)WiWi(i)(ii)0.0670.48gi!Ni! gi Ni !10 9 8 7!7! 3!3/20.052110!7! 10 7 !10 9 8 120 2 1Wi12!12!12 11 10!10! 12 10!6612 11 1010! 2 19 8!8! 12 8!4958! 4 3 2 12.0810 6(a)(b)(c)or1 exp EE kTEc2kT ;exp0.30 2 0.02590.0259Ef

33、EfEfkT , f E0.2695kT,6.6910kT ,4.5410101.2610 61 expl(b)fF1 expE EfkT1 exp 5expEf EkT1 exp 10exp 0256.43 10 50.0259fFkTexp0.25 0.0259 20.02591 exp EE kT3.9010 5kTfFexp0.25 0.02590.02592.3610 5expEfEkT3kT(a)EfEkT,1 f E0.269(b)EfE5kT,1 f E6.69(c)EfE10kT,1 f E4.5431010 5fFexp0.25 3 0.0259 20.0259(a)fFe

34、xpE EfkT2kT1 fFEc;fF0.306exp 9.32 100.0259EckT0.30 0.0259 2exp 0.0259EckT ;fFEc3kTfFexp5.66 10 6andexp3.430.30 30.30 0.02590.025910 6So0.0259 20.025951.43 10 50.25 2 0.0259exp0.02598.70 10 6exp正kTEc kT Ef exp kTfFexpexpkTEf E kT exp EckT EfkTkTexpEf E kTnx 1, ny2, nz 2 . This quantumkTThenEckT EFEf

35、E kTOr EFEcEE midgapstate is at the same energy, so Ef 2.35eV(b) For 13 electrons, the 13 electron occupies the quantum statethFororFororEnE6E7nx 3, ny222n2ma2n 6, Filled state34 2221.054 106E13or1.054 102, nz34 23 ; so2 c 223231102 9.11 1012 10199.194 10 19J3223110 22 9.11 1012 10181.5044 10 J18The

36、 14thE61.5044 10191.6 10 199.40 eVn 7, Empty state 一 34 2 _ 221.054 1073110 22 9.11 1012 102.048 10 18JThereforeE13199.194 10 19191.6 10electron would occupy the5.746eVquantum state nx 2, ny3,53This state is at the same energy, soEf 5.746 eVE7182.048 101.6 10 199.40 Ef12.8eV12.8 eVThe probability of

37、 a state atEiEfEbeing occupied isf1 Ei 1. EiEf1 exp kT1E1 exp 一kT(a) For a 3-D infinite potential well22mE 2222-n x n y nzThe probability of a state atE2 Ef Ebeing empty isf2 E2For 5 electrons, the 5occupies the quantum stateth electron1 彳E2 Ef1 expkTnx 2,ny 2,nz1; soorE522-nx2m1.0542ny2nzor1E1 exp

38、一 kTE expkTE1 expkT34102222 9.11 10 31 12 10 10122f2 E23.761 10 19J3.761 10 19 x/ E 519- 2.35 eV1.6 10 191 exp E kT(a)soenergyf1 Ei 1 f2 E2Ei , we wantFor the next quantum state, which is empty, the quantum state is11Ei Ef exp kT1 expkTor0.25 exp - kT，50 0.020.010.25kTIn 501 expkTorThis expression c

39、an be written askT0.25In 50T0.0639060.0259 -300or1 exp EEfkTwhich yieldsT 740 KEi exp 0.01ThenEiEforEi(b)AtEff EiEfkT1 0.01Ei Ef exp kTkTln 1004.6kTEf 4.6kT ,(a)or(b)AtThenor1 expEiE f1 exp 4.6kT(c)7.15 7.0 exp 0.02590.003041000K, kT 0.08633eV1 exp7.15 7.00.14960.08633which yieldsf Ei 0.00990 0.011 exp6.85 7.00.0259(a)fFexp上身kT5.80 5.50 exp 0.0259(b)(c)0.9979.32 10 6kTfF7000.0259 3000.060433 eV0.30 exp 36.98 100.020.060433expkT0.25 exp kTor(d)AtEf ,1 for all2temperatures(a) For e EiThenEiForEfE2ThenexpexpE1 EfexpEi EfkTkT0.309.320.0259610E2,1.12 0.30 0.82 eV1 exp0.820.02591.0