半導體物理與器件第四版課后習題標準答案

1、半導體物理與器件第四版課后 習題答案作者:日期:thSemiconductor Physics and Devices: Basic Principles, 4editionChapter 3By D. A. NeamenProblem Solutions73.1Chapter 3If ao were to increase, the bandgap energy would decrease and the material would beginto behave less like a semiconductor and morelike a metal. If a0 were to d

2、ecrease, the bandgap energy would increase and the material would begin to behave more like an insulator.Eu x exp j kx tThis equation may be written as2u xk u x 2jk3.2Schrodinger's wave equation is:22x,t2mV x x,tx,t t Assume the solution is of the form:x,t u x exp j kx - tRegion I: V x 0. Substi

3、tuting the assumed solution into the wave equation, we obtain: jku x exp 2m xE j kx t2u x 2mEu x .exp j x里 uxexp jkxEtwhich becomes22mjk2u x expE j kx - tu x2 jkexpxE kx t2u x .-exp j xkxSetting u equation becomes:2d u1 x2- dxwhere2jkUifor region I, thedu1 dx2mEk22-Ui x 0Q.E.D.In Region II, V x form

4、 of the solution:Vo . Assume the samex, t u x exp j kx tSubstituting into Schrodinger's wave equation, we find:2mjkexp j kx - tu x2 jkexpx2u x-exp j xVOu x exp jEu x exp jkxkxkxkxThis equation can be written as:2jk2u xTx2mVOux2mESetting u xu2 x for region II, thisequation becomes2d u2 x2jk dxdu2

5、 xdx,22 2mVOk2- u2 x 0where again2 2mEQ.E.D.j k B exp j k x22k A exp j k xB exp j k x 0Combining terms, we obtain22222 k k 2k k kAexp j k x22222 k k 2k k kBexp j k x 0We find that 0 0 Q.E.D.For the differential equation in u2 x and theproposed solution, the procedure is exactly the same as above.2k

6、Aexp j3.3We have,2.d Ui xdu1 x 22nz- 2jk k Ui x 0dx2dxAssume the solution is of the form: Ui xAexp j k xB exp j k xThe first derivative isdu1 xj k Aexp j dxk B exp j k xand the second derivative becomes2d Ui xdx22j k B exp j k xSubstituting these equations into the differential equation, we find2k A

7、exp j k x2k B exp j k x2jk j k Aexp j k x3.4 We have the solutionsu1 x A exp j k xB exp j k x for 0 x a andU2 x C exp j k xD exp j k x for b x 0.The first boundary condition is Ui 0U2 0which yieldsA B C D 0The second boundary condition is du1du2dx x o dx x o which yields k A k B k Ck D 0The third bo

8、undary condition is u1 a u2 b which yieldsAexp j k a Bexp j k a C exp j k bD exp j k band can be written asAexp j k a Bexp j k a C exp j k bDexp j kb 0The fourth boundary condition is du1du2dx x a dx x b which yields j k Aexp j k a j k B exp j k a j k C exp j k bj k D exp j k b and can be written as

9、 k Aexp j k a k Bexp j k a k C exp j k bk D exp j kb 03.5(b) (i) First point: aSecond point: By trial and error,a 1.729(ii) First point: a 2Second point: By trial and error,a 2.617c sin a,P cos a coskaaLet ka y, a xThensinxP cosx cosy xConsider of this function.dyd1 o. P x sin x cosx dysin yWe findP

10、-2dx1dx1 x sin x x cos x dydydx sinx sin ydyThendx1cos xP 2 sin xsin xsin ydyx2xFor y ka n ,n0,1,2,. sin y 0So that, in general,dxd ad0dyd kadkAnd；2mE2So1/ 2d12mE2mdEdk222dkThis implies thatd0dE,nfor kdkdka3.6(b) (i) First point: aSecond point: By trial and error,a 1.515(ii) First point: a 2Second p

11、oint: By trial and error,a 2.3753.8(a)1a3.72moE12athSemiconductor Physics and Devices: Basic Principles, 4editionChapter 3By D. A. NeamenProblem Solutions15Ei234 21.054 10 342m0E122moa23110 229.11 104.2 10193.4114 10 JFrom Problem 3.52a 1.729Ei2 一 一 34 21.054 10_31_ 10 22 9.11 10 31 4.2 10 102moEAt3

12、.4114 ka 0,1.72910 19 JBy trial and error, oa 0.859E21.729234 21.054 10Eo0.859234 21.054 10or E(b)3 a29.11 10 31 4.21.0198 10 18 JE2Ei181.0198 10196.7868 10196.7868 1010 10 22 9.11 102.51723110EiEo10 24.2 10 1019 J193.4114 103.4114 10192.5172191.6 10 1924.24 eVor E8.942 108.942 10 2010 1920J2mo E322

13、1.054E33129.11 104.234 210 3410 2101.6(b) At ka 2 ,E32m0E3219103a 221.0540.559 eV34 2103110 229.11 104.2 10181.3646 10 JFrom Problem 3.5,4a 2.617At ka181.3646 10 J.From Problem 3.5,2a1.7292mo E42.6172m°Eza 1.729E42.61721.05434 21029.11 1031 4.2 10 10 2E22 _1.7291.054 1034 22.3364E4 E310 18 Jor

14、E3.9(a) At ka2.336410 18 1.3646 10 18199.718 10 J199.718 1019 6.07 eV1.6 10 19or E1a3129.11 1010 24.2 101.0198E3 E21.36463.44741010103.4474 10191.6 1018 J18 1.0198 10 1819 J19一 2.15eV3.10(a)2m0E123.11(a) At ka ,1aE12 9.11 102 一 一 34 21.054 102m0E1314.210 10 23.4114From Problem 3.6,1019 J2a1.515E121.

15、05434 2103110 229.11 104.2 102m。E2a 1.515193.4114 10 JAt ka 0, By trial and error,E2_ _21.5151.05434 210 34312 9.11 107.830 10E2 Ei10 24.2 1019Joa 0.727；2m°Eo- 20.727or E(b)3 a7.830 10194.4186 10194.4186 10 19193.4114 1019JEo234 20.7271.054 10 342 9.11 10 314.2 1010 2191.6 1022.76eV1.803010 19

16、JE1Eo2m0E323.4114 10191.803010192 1.05434 2101.6084E3310"2 9.11 104.2 10or E1.6084 10101919 J1.3646 10 18J(b) At ka 2191.6 103a 21.005eVFrom Problem 3.6,4a 2.3752moE2mo E4E42a2.3752.375234 21.054 10 343110 22 9.11 104.2 10181.9242 10 JE4 E3E3At2a 1.5152 1.0541034 23110 229.11 104.2 10ka181.3646

17、 10 J,From Problem 3.6,1.9242 10or E18181.3646 10195.597 10 19J5.597 10 1919 3.50 eV1.6 10 192mo E2i -2- a1.515E21.515234 21.054 102 9.11 10 34 4.2 10 10 2197.830 10 19JE3 E21.36461819107.830 10一 _ 195.816 10 J一 _ 195.816 10or E 19- 3.635eV1.6 10Points A,B:directionPoints C,D:directionPoints A,D:dEd

18、kdEdkd2E dk2velocity in -xvelocity in +x3.12For T100K,Eg1.1704.73 10 4 100 2negative effective mass d2EPoints B,C: -d-Edk2636100Eg 1.164 eVpositive effective massT200K,Eg1.147 eVT300K,Eg1.125 eVT400K,Eg1.097 eVT500K,Eg1.066 eVT600K,Eg1.032 eV3.16_ 2For A:ECikAt k0.08Or_ _ _ 19E0.05 1.6 101010 m8 101

19、1, E 0.05eV21 J3.13The effective mass is given by1So 8 10 21C11.2510 2C1 0.08 10103810d2Edk2Now2CiWe have d2E dk2curve Ad2Ecurve B dk_ 34 21.054 103821.25 10314.44 10 kgor314.4437 10so that*m curve A m curve B9.11 1031mo0.488 m。For B: ECik23.14The effective mass for a hole is given by.1At k 0.08 10

20、10m 13.15*mpd2E dk2OrSo 8 10We have thatd2E dk2so thatCicurve AE 0.5 1.620C1 0.08371.25 10,E191010 2100.5eV208 10 Jd2E dk2curve BNow m34 2 1.054 102C121.25 10 37mn curve A mn curve B p por m.一 324.44 10 32 kg324.4437 10mo9.11 100.0488 mo3.17For A:2C2k0.025 1.61910_ _ 10 2C2 0.08 10or mFor B: E0.3 1.

21、6 10or mC22c23.186.25 10 3934 21.054 102 6.2539108.8873 10318.8873 10 31319.11 1019C231 .kgmo0.976moC2k2C2 0.0810 2107.51038 1.054 102c22 7.5 10327.406 10 kg327.406 10 L mo9.11 10m 0.0813 m0(a) (i) E or(ii)(b) (i)(ii)h_ 19E1.42 1.6 10h6.625 10 34143.429 10 Hzhc c 3 1010 一 I 14E 3.429 1058.75 10 5cm

22、875nm19E 1.12 1.6 10-34h 6.625 10_142.705 10 Hz10c 3 10_ _142.705 101441.109 10 cm 1109 nm3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positivearound k 0, and is negativearound k .23.20E Eo Ei cos k kOThendE Ei sin k ko dkE1 sin k kOandd2E 27Eicos k kodkThen11 d2E Ei 2-,22

23、mdk k koor2*m 2Ei3.21(a)(b)mdn2/ 34mdn21 /3mtmi2 /340.56momcnmtmi0.082m。0.082mo24.39momcn0.12m00.6098mo3.223/23/2 2/3(a) mdp mhh mih1/31.64mo1.64mothSemiconductor Physics and Devices: Basic Principles, 4editionChapter 3By D. A. NeamenProblem Solutions173/ 20.45m00.082mo3/ 2 2/3Also,X ax, y,z 00 . Th

24、enat xkx aa, so thatnxwhere0.301872/30.02348monx1,2,3,.mdp0.473 mo3/2/h、emhh(b) mcp -mhh3/2mlh172 mlhSimilarly, we havek2v 2 kyY yand2Z2zk23.23From the boundary conditions, we find3/23/20.450.0821721 / 2 mo0.450.082mcp0.34 m。kya wherenynyand kza nz1,2,3,. andnz 1,2,3,.From the wave equation, we can

25、write22kxky2mE2-For the 3-dimensional infinite potential well, V x 0 when 0 x a, 0 y a, and 0 z a. In this region, the wave equationis:The energy can be written2E Enxnynz 2m2nxas2ny22nz a2x, y, z-2 x2x, y,z-2y2x, y, z2 z3.24The total number of quantum states in the 3-dimensional potential well is gi

26、ven (in k-space) by2mEx, y,zgT2k dk 3 k dk 3 aUse separation of variables technique, so letx, y, z X xY y Z zSubstituting into the wave equation, we havewhere2mEYZ0xXZyXY JixT 2 zWe can then writek星Taking the differential, we obtain2mEXYZ 0Dividing by XYZ , we obtain12XXLet12YyXx2kx2Z 2mE2-X- kxX 0

27、xThe solution is of the form:X x Asin kxx B coskxxSince x, y, z 0 at x 0, thenX 0 0so that B 0 .111dk 2mdE2 . Em2EdESubstituting these expressions into the densityof states function, we havegT E dE3 a2mE1mdE32. 2ENoting thaththis density of states function can be simplified and written asthSemicondu

28、ctor Physics and Devices: Basic Principles, 4editionChapter 3By D. A. NeamenProblem Solutions193.25well,3.26gT EdE34 a 3/2 一2m E dEh33Dividing by a states so thatwill yield the density of3/24 2mh3For a one-dimensional infinite potential2mnE2-Distance between quantum stateskn 1knn 1 aNowNowThen(a) Si

29、licon, mn1.08mogc Egc3/22mnh34 2mn3/24 2mnh1gTdkgT3/2 EcEc2kTh3Ec.E EcdEEc3/23/ 22mnh3314 2 1.08 9.11 1034 36.625 103/ 2557.953 10 2kT3/ 2Ec 2kT2kT2kT3/23/22 dk k dk (i) At T 300 K,kT 0.0259eV1- -.2mn EE dE0.0259 1.6 104.144 10Then21J192mn dEgc 7.953105524.144 10 216.0 10193/22a 12mndEDivide by the

30、"volume" a, soSoor(ii) At T2mn0.067 9.11 10 31gc6.0 10400K, kT0.034533eV0.034533 1.6215.5253 10 JThengc 7.9535510or1910341.054 101.055 1018gE 、EE3J 1gc253m3cm0.0259400300_ _ 212 5.5253 109.239 10199.24 10 cm25 m33/ 2(b) GaAs, mn0.067 m。gc_314 2 0.067 9.11 106.625 10541.2288 103/234 33/22kT

31、23/ 2- 2kT3thSemiconductor Physics and Devices: Basic Principles, 4editionChapter 3By D. A. NeamenProblem Solutions_ _ _,_ 21(i) At T300 K, kT 4.144 10 J54213/ 2gc 1.2288 102 4.144 102339.272 10 m173or gc 9.27 10 cm21(ii) At T 400K, kT 5.5253 10 21J4 2 0.48 9.11313/210 312_ 3/2g3kT6.625 10552.3564 1

32、034 3_ 3/23kT3233.27gc 1.2288 1054 2 5.5253 101.427 1024 m 3183gc 1.43 10 cm3/2(i)At T300K, kT4.144 1021Jg552.3564 103 4.14410 21 3/23.26625310 mor19g 3.27 103 cm(ii)AtT 400 K, kT5.525310 21J(a)Silicon, mpp0.56mo3/ 22mph33/ 2,E E55213/2g 2.3564 103 5.5253 102535.029 1025 m193or g 5.03 10 cm4 2mp h33

33、/2,E3kTE dE2mp3/ 23.283/ 24 2mn (a) gc E 3, E Ech3kT4 2mp h33/ 24 2 1.08 9.11 10 313/23kT346.625 103/ 2 ；E Ec313/24 2 0.56 9.11 10 31233kT6.625 10 34355_ 3/ 22.969 10 3kT_ . _,_ 21(i)At T 300K, kT 4.144 10 Jg 2.969 1055 3 4.144 10 214.116 1025 m 3193or g 4.12 10 cm21(ii)At T 400K, kT 55253 10 J1.192

34、9 1056.E EcFor E Ec;gc 0E Ec 0.1eV;gc 1.509 1046 m 3J 1E Ec 0.2 eV;46312.134 10 m JE Ec 0.3 eV;46312.614 10 m JE Ec 0.4 eV;3.018 1046 m 3J 1gor _ 55_ _ 213/22.969 103 5.5253 1025 a6.337 10 m 3193g 6.34 10 cm(b) GaAs, mp 0.48m。(b) g3/ 24 2mph3313/24 2 0.56 9.11 10_ 34 36.625 104.4541 1055 E EFor E E

35、;g 0EE0.1 eV;4531g 5.634 10 m JE E0.2eV;7.968 1045m 3J 1E E 0.3eV;9.758 1045 m 3J 1E E 0.4eV;一 一 46311.127 10 m 3J 15kT ,f E11exp 5f E6.6910310kT,f E11exp 10f E4.54105(b) EEf(c) E Ef3.331 f E3.29(a)gcmn3/23/21.082.68gmp3/20.56(b)gcmn3/23/ 20.0670.0521gmp3/20.48expE EfkT3.30Plotor1f E11 expEfEkT(a)Ef

36、EkT, 1f E0.269(b)EfE5kT , 1f E6.6910 3(c)EfE10kT , 1f E4.5410 53.31(a) Wigi!10!Ni! giNi !7! 10 7 !10 9 8 7!10 9 81207! 3!3 2 1(b) (i) Wi12!10! 12 10!12 11 1010! 2 1Wi(ii)128! 12 8!6612 11 10 9 8!8! 4 3 2 1 4953.32f EexpE EfkT3.34(a)fFEEfexpkTEEc；0.306O qo d nt f exp9.32 100.0259EckTT f exp -0.30 0.0

37、259 22 ;0.02595.66 10 6EckT ;0.30 0.0259T f exp0.02593.43 10 6Ec3kT;Tf exp0.30 3 0.0259 220.02592.08 10 6Ec2kT;Tf exp0.30 2 0.02590.02591.26 10 6(a) EEfkT , f E 1 exp1f E 0.269(b)1fFkT2 ;kT ; 13kT2 ;f F exp2kT ;3.35Semiconductor Physics and Devices: Basic Principles, 4th editionChapter 3By D. A. Nea

38、menProblem SolutionsfF1 fFE Ef expkTkT0.255exp 6.43 100.02590.25 0.0259 2exp 0.025953.90 100.25 0.0259exp 0.025952.36 100.25 3 0.0259 20.025951.43 100.25 2 0.0259exp 0.02598.70 10 6fF exp exp Ec kT EF kTkTand1 fFexpEf EkTexpEf E kTkTEc kT EfSo exp kTEfE kTexp kTThen Ec kT Ef Ef E kTOr EFEcE2Emidgap3

39、.36En2 22nc 22maForn6, Filled stateE61.054 10 342 6 22312 9.11 1010 212 101.5044 1018JorE6181.5044 109.40 eV191.6 10Forn7, Empty stateE1.054 10 :34 2 7 227312 9.11 1012 10 10 22.048 1018JorE7182.048 10191.6 1012.8eVTherefore 9.40 EF12.8eV3.37(a) For a 3-D infinite potential well22mE222nxnynz-aFor 5

40、electrons, the 5 th electron occupiesthe quantum state nx2 2E5 一 nx2m2, ny 2, nz 1 ; so222ny nz 一 a1.054 102 9.11orE534 22210 31 123.7613.761 10101019191.6 102210 219 J122.35eVFor the next quantum state, which is empty, the quantum state is nx 1, ny 2, nz2 .This quantum state is at the same energy,

41、so Ef 2.35 eV(b) For 13 electrons, the 13 th electron occupies the quantum statethSemiconductor Physics and Devices: Basic Principles, 4editionChapter 3By D. A. NeamenProblem Solutions6nxE133,ny 2,nz 3; so1.054 10 34 22 3222 32or E132 9.11 10 31 12 10 10 29.194 10 19J9.194 10191.6 105.746 eVThe 14th

42、 electron would occupy the quantumstate nx 2, ny 3, nz 3 . This state is atthe same energy, soEf5.746 eV3.38The probability of a state at E1 Ef E being occupied isf1 EiexpE1EfkTexpexpE1 EfkTexpEi EfkTexpekTThe probability of a state at E2EfEbeing empty is1 f2 E21expE2 EfkT11 exp EkTexpekT1 exp 一 kTE1EfkTThis expression can be written asexpE1 EfkT0.01expE1 EfkT1 0.01orE1 Ef10.01 exp kTThenE1 Ef kTln 100orE1Ef 4.6kT(b)At Ef E1Ef 4.6kT