Chapter 3If a o were to increase, thebandgap energywould decrease and the materialwould beginto behave less like asemiconductor and morelike a metal. If a o were todecrease, thebandgap energy would increase andthematerial would begin to behavemore like aninsulator._______________________________________Schrodinger's wave equation is:2 2 x, tV x x, t2m x2j x, t tAssume the solution is of the form:x, t u x exp j kx E tRegion I: V x 0 . Substituting theassumed solution into the wave equation, weobtain:2E2m xjku x exp j kx tu xexp j kx E txj jE u x exp j kx E t which becomes22 u E2mjk x exp j kx t2 jk u xexp j kx E tx2 u xexp jEtx2kxEu x exp j kx E tThis equation may be written ask 2 u xu x 2 u x 2mE2 jkx2 2 u x 0xSetting u x u1 x for region I,the equationbecomes:d 2 u1 x 2 jk du1 x k 2 2 u1 x 0dx2 dxwhere2 2mE2In Region II, V x V O. Assume thesameform of the solution:x, t u x exp j kx E tSubstituting into Schrodinger'swaveequation, we find:22 u Ejk x exp j kx t2m2 jku xexp j kx E tx22xexp ju kx E txV O u x exp j kx E tEu x exp j kxEtThis equation can be written as:k 2u x 2 jk u x 2u xx x22mV Ou x2mEu x 02 2Setting u x u 2 x for region II, thisequation becomesd 2u 2 x du 2 x dx 2 2 jkdxk 22 2mV Ou 2 x 02where again22mE2We have22 jk du 1 xd u 1 x k 22u 1 x 0dx 2dxAssume the solution is of the form:u 1 xA exp jk xB exp jk xThe first derivative isdu 1 x jk A exp jk xdxjk B expjk xand the second derivative becomesd 2 u 1 x jk2k xdx 2A exp j2 B expj k jk xSubstituting these equations into thedifferential equation, we findk 2 A exp jk x2B exp jk xk2 jk j k A exp jk x jk B exp jk xk 22A exp jk xB exp jk xCombining terms, we obtain22 k k22kkk22A exp jk x22 k k 22kkk 22B exp jk xWe find that0 0For the differential equation in u 2 xand theproposed solution, the procedure is exactlythe same as above._______________________________________We have the solutionsu 1 xA exp j k xB exp jk xfor0 x a andu 2 x C exp jk xD exp jk xfor b x 0 .The first boundary condition isu 1 0 u 2 0which yieldsA B C DThe second boundary condition isdu 1 du 2dxx 0dxx 0which yieldsk Ak Bk Ck D 0The third boundary condition isu 1 a u 2 bwhich yieldsA exp jk aB exp jk a C exp jkbD exp jkband can be written asA exp jk aB exp jk aC exp jk bD exp jk bThe fourth boundary condition is du 1 du 2dxx adxxbwhich yieldsjk A exp jk ajk B exp j k ajk C exp jkbjk D exp j kband can be written ask A exp jk ak B exp jk ak C exp jk bk D exp j k b 0_______________________________________(b) (i) First point:aSecond point: By trialand error,a 1.729(ii) First point:a2Second point: By trialand error,a 2.617_______________________________________(b) (i) First point:aSecond point: By trialand error,a 1.515(ii) First point:a 2Second point: By trialand error,a 2.375_______________________________________Psin a cos a coskaaLet kay ,a xThenP sin xcos x cos yxConsiderd of this function.dyd P x1cos xsin ydy sin xWe findP1 x2sin x dx x 1 cos x dxdydy sin xdxsin ydyThendx P 21sin x cos xsin xsin ydy xxFory ka n ,n 0, 1, 2, ... sin y0 So that, in general,dx 0d a ddyd kadkAnd2mE2Sod 11/ 2dE2mE2mdk2 22dkThis implies thatd 0 dE forkn dk dka_______________________________________(a)1a2m o E 12a2 221.054 10 34 2E 12 9.11 10 31 4.2 101022m o a 23.4114 10 19JFrom Problem2 a 1.7292m o E 2a1.72921.72921.054 10 34 2E 22 9.11 10 314.2 10 10 21.0198 1018JE E 2E 11.0198 10 18 3.4114 10 196.7868 10 19JorE6.7868 10 194.24 eV1.610 193 a 2(b)2m o E 3 22a2 2 1.05410 34 2221.054 10 34 2E 310 31 4.2 10 10 2E 310314. 2 10 10 22 9.112 9.111.3646 10 18Jka1.3646 10 18JFrom Problem ,At. From Problem ,4a 2.6172a 1.7292m o E 42.617 2m o E 2a1.7292a22.6172 1.054103421.72921.054 10 34 2 E 4E 210 314.2 10 10 22 9.11 10 314.2 10 10 22 9.112.3364 10 18J1.0198 10 18JEE 4 E 3E E 3E 22.3364 10 18 1.3646 10 18 1.3646 10 18 1.0198 10 189.718 10 19J3.4474 10 19Jor9.718 10 196.07 eVorE 3.4474 10 19 2.15 eVE 1.6 10 191.6 1019_______________________________________ _______________________________________(a) At ka, 1a2m o E 1 a221.054 10 34 2E 12 9.11 10314.2 10 1023.4114 10 19JAt ka 0 , By trial anderror,oa 0.8590.85921.054 10 34 2E o2 9.11 10 31 4.210 1022.5172 1019JE E 1 E o3.4114 10 19 2.5172 10 198.942 10 20J8.942 10 20orE0.559 eV1.6 10 19(b) At ka2 ,3 a22m o E 3a22(a)1a2m o E 1a221.054 10 34 2E 110 31 4.2 10 10 22 9.113.4114 10 19JFrom Problem , 2 a 1.5152m o E 2a 1.51521.515 21.054 10342E 22 9.1110314.2 1010 27.830 10 19JE E 2E 17.830 10 19 3.4114 10 194.4186 10 19J or E4.4186 10 192.76 eV1.610 19(b)3a 22m o E3a 222 2 1.054 10 34 2E32 9.11 10 31 4.2 10 1021.3646 10 18JFrom Problem , 4 a 2.3752m o E 4a 2.37522.375 2 1.054 10 34 2E42 9.11 10 31 4.2 10 1021.9242 1018E E4 E3J1.9242 10 18 1.3646 10 185.597 10 19Jor E 5.597 10 193.50 eV 1.6 10 19_____________________________________(a) At ka , 1 a2m o E1a22 1.054 10 34 2E131 10 22 9.11 10 4.2 10ka 0 3.4114 10 19JAt , By trial and error,or1.6084 10 191.005 eVE10 191.6(b) At ka 2 , 3 a 22m o E 3a 222210 342E 31.05410 31 4.2 10 1022 9.111.3646 10 18JAt ka , From Problem ,2 a 1.5152m o E 22 a 1.5151.515 2 1.054 10 34 2E 210 34 4.2 10 1022 9.117.830 10 19JE E3 E21.3646187.8301910 105.816 10 19Jor5.816 10 193.635 eVE1.6 10 19_______________________________________For T 100K,4.73 10 4 2E g1001.170100636o a 0.7272m o E oa 0.72720.727 2 1.054 10 34 2E o31 10 22 9.11 10 4.2 101.8030 10 19J E g 1.164 eVT 200 K,T 300 K,T 400 K,T 500 K,T 600 K,E g 1.147 eVE g 1.125 eVE g 1.097 eVE g 1.066 eVE g 1.032 eVE E1E o3.4114 10 19 1.8030 10 191.6084 10 19J _______________________________________The effective mass is given by11m *d 2 E2dk 2We haved 2 E curve A d 2 E curveBdk2 dk 2so that m* curve A m* curve B_______________________________________The effective mass for a hole isgiven by1 d2 E 1m*p2dk 2 We have thatd 2 Ecurve A d 2 Edk 2 dk 2curve Bso that m*p curve A m*p curve B_______________________________________Points A,B: dE0 velocity in -x dkdirectionPoints C,D: dE0 velocity in +x dkdirectionPoints A,D: d 2 E0 dk 2negative effective massPoints B,C: d 2 E0 dk 2positive effective mass_______________________________________For A:E C i k 2At k 0.08 10 10m1, E 0.05eV OrE 0.05 1.6 10 19 8 10 21 JSo 8 10 21 C1 0.08 10102C1 1.25 10 382 1.054 10 342Now m2C1 2 1.25 10 384.44 10 31kgor m4.4437 10 31m o9.11 10 31m 0.488 m oFor B: E C i k 2At k 0.08 10 10m1, E 0.5 eV OrE 0.5 1.6 10 19 8 10 20 JSo 8 10 20 C1 0.08 10102C1 1.25 10 3721.054 10 342 Now m2C1 2 1.25 10 374.44 10 32kgor m4.4437 10 32m o9.11 10 31m 0.0488 m o_______________________________________For A: E E C 2 k 20.025 1.6 10 19 C 2 0.08 1010 2C 2 6.25 10 3921.054 10 342m2C 2 2 6.25 10 398.8873 10 31kg8.887331or m10m o9.11 10 31m 0.976 m oFor B: E E C 2 k 20.3 1. 610 19 C 2 0.08 10 102C 2 7.5 10 3821.054 10 34 2m2 7.5 10 382C 27.406 10 32kgor7.406 10 32m om10 319.11m 0.0813 m o_______________________________________(a) (i) Eh10 19 orE 1.42 1.6h6.625 10 343.429 1014Hz(ii)hcc3 1010E3.429 10148.75 10 5cm 875nm1019(b) (i)E 1.12 1.6h6.625 10342.705 1014 Hz(ii)c3 10102.705 14101.109 10 41109 nm cm_______________________________________m *22E 1_______________________________________(a) m dn42/ 3m t21 / 3m l42/321/ 30.082m o 1.64m om dn 0.56m o (b)3 2 1 21m cnm tm l0.082 m o 1.64m o24.39 0.6098 m om om cn 0.12m o_______________________________________3 / 23/22/3(a) m dp m hhm lh(c) Curve A: Effective mass is a constantCurve B: Effective mass ispositivearoundk0 ,and is negativearoundk.2 _______________________________________mdp(b) m cpmcp0.45m o 3 / 23 / 2 2 / 30.082 m o0.30187 2 / 3m o0.023480.473 m o3 / 23 / 2m hh m lh1/ 2 1/ 2m hhm lh3 / 23 / 20.450.082 m o1 / 21 / 20.45 0.0820.34m oE E OE 1 cos k k OThendE E 1 sin k k OdkE 1 sink k Oandd 2E E 1 2cos k k O dk 2Then1 1 d 2EE 12 m * 2dk 2 k k o2or_______________________________________For the 3-dimensional infinite potential well,V x 0 when 0 x a , 0 y a ,and0 z a . In this region, thewave equationis:22 2x, y, z x, y, z x, y, z x 2y 2z 22mEx, y, z 02Use separation of variables technique, so letx, y, z X x Y y Z zSubstituting into the waveequation, we have2X2Y2ZYZ2 XZ2 XY2x y zwhere22mEk 2We can then write2mE2XYZ 0k2mEDividing byXYZ , we obtain12X1 2Y12Z2mEX x2Yy2Z z22LetTaking the differential, we obtain1 2Xk x22Xk x 2X 0X x 2x2The solution is of the form:X xAsin k x x B cos k x xSincex, y, z0 atx 0 , thenX 0 0so that B 0 .Also, x, y, z 0 at x a , so11 1dk2mdE2 ESubstituting these expressionsinto the densityof states function, we haveg T E dEa 32mE 132Noting that1mdE2Em dE2EthatX a 0 . Thenk x a n x wheren x 1, 2, 3, ...Similarly, we have221 Y k y 2and1 Zk z2Y y 2Z z 2From the boundary conditions, we findk y a n yandk z a n zwheren y 1, 2, 3, ... and n z 1, 2, 3, ...From the wave equation, we can write2222mE0 k x k yk z2The energy can be written as22E E n x n y n zn x 2 n y 2 n z 2a2m_______________________________________The total number of quantum states in the3-dimensional potential well is given(in k-space) byk 2dka 3g T k dk3h2this density of states functioncan besimplified and written asg T E dE4 a 3 3 / 2E dE32mhDividing by a 3 will yield thedensity of states so that4 3 / 2g E 2m Eh 3_______________________________________For a one-dimensional infinitepotential well,2m n En 2 2k 22a2Distance between quantum statesk n 1 k nn 1naaaNow2 dk g T k dkaNow1 k2m n E1 1 2m ndEdkE 2Then(ii) AtT 400 K,400 kT 0.02593002a 1 2m n g T E dEdE2EDivide by the "volume"a , so0.034533 eV0.034533 1.610191 2m n g EESog E12 0.067 9.11 10 311.054 1034Eg E1.055 1018m 3 J 1E_______________________________________(a) Silicon,m n 1.08m o4 2m n 3 / 2g c EE E ch 34 2m n3 / 2 E c 2kTg c E E c dEh3E c4 2m n3 / 22E c 2kTE E c 3 / 2h33E c4 2m n3 / 223 / 2h 32kT34 2 1.08 9.11 10 31 3 / 223 / 26.625 10 34 33 2kT7.953 10552kT3 / 2(i) AtT 300 K, kT 0.0259 eV0.0259 1.6 10194.144 10 21JTheng c7.953 10552 4.144 10213 / 26.0 10 25m 3org c6.0 10 19 cm 35.5253 10 21JTheng c7.953 10 552 5.5253 10 213 / 29.239 1025 m 3org c 9.24 1019 cm 3(b) GaAs, m n 0.067 m o4 2 0.067 9.11 10313 / 222kT 3 / 2g c6.625 1034 331.2288 10 542kT 3 / 2(i) AtT 300 K, kT4.144 10 21Jg c1.2288 10542 4.144 10 213 / 29.272 10 233mor g c9.27 1017 cm3(ii) AtT 400 K, kT 5.5253 10 21 Jg c1.2288 10542 5.5253 10 213 / 21.427 1024 m 3g c1.43 1018 cm 3_______________________________________(a) Silicon,m p 0.56m o4 2m p 3 / 2g EEEh343 / 2Eg 2m pEEdEh 3E3 kT4 2m p3 / 22EE E3 / 2h 33E3 kT4 2m p3 / 223 / 2h333kT31 3 / 24 2 0.56 9.11 1023 / 2 33kT6.625 10 3432.969 10553 / 23kT(i)At T 300 K, kT 4.144 10 21Jg 2.969 10 55 3 4.144 10 21 3 / 24.116 1025m3or g 4.12 1019 cm 3(ii)At T 400 K, kT 5.5253 10 21Jg 2.969 10 55 3 5.5253 10 21 3 / 26.337 1025m3or g 6.34 1019 cm 3(b) GaAs, m p 0.48 m o4 2 0.48 9.11 10 31 3 / 22 3kT3 / 2g 36.625 10 34 32.3564 10 55 3kT 3 / 2(i)At T 300 K, kT 4.144 10 21Jg 2.3564 10 55 3 4.144 10 21 3 / 23.266 1025m3or g 3.27 1019 cm 3(ii)At T 400 K, kT 5.5253 10 21Jg 2.3564 10 55 3 5.5253 10 21 3 / 25.029 10 25m3or g 5.03 1019 cm 3_______________________________________4 3 / 2(a) g c E 2m nE E c h34 2 1.08 9.11 10 31 3 / 2E E c6.625 10 34 31.1929 10 56 E E cFor E E c; g c 0E E c 0.1eV;g c 1.509 10 46 m 3 J 1E E c 0.2 eV;2.134 1046m3J 1E E c 0.3 eV;2.614 1046m3J 1E E c 0.4 eV;3.018 1046m3J 14 2m p3 / 2(b) g E Eh 34 2 0.56 9.11 10 313 / 2E E6.625 10 3434.4541 10 55 E EFor E E ; g 0E E 0.1 e V;g 5.634 10 45 m 3 J 1E E 0.2 eV;7.968 10 45m3J 1E E 0.3 eV;9.758 10 45 m 3 J 1E E 0.4 eV;1.127 1046m3J 1_______________________________________g c 3 / 2 3 / 2m n 1.08(a) 3 / 2 2.68g m p 0.56g c 3 / 2 3 / 2m n 0.067(b) 3 / 2 0.0521g m p 0.48_______________________________________Plot_______________________________________(a) W i g i ! 10!N i ! 7! 10 7!N i ! g i10 9 8 7! 10 9 81207! 3! 3 2 1(b) (i) W i12! 12 11 10!10! 12 10 ! 10!2166(ii)12!121110 98!W i8!12 8!8!4321495_______________________________________f E1E E F1 exp kT(a) E E F kT ,1f E1 exp 1f E0.2691 (b)E E F5kT , f E1exp 5f E 6.69 10 31 (c)E E F10kT , f E1exp 102.08 10 6E c 2kT ; fF exp0.30 2 0.02590.02591.26 10 6(b) 1 f F 11E E F1 expkTexpE F EkTE E ; 1 fF exp 0.25 6.43 10 50.0259f E 4.54 10 5 E kT ; _______________________________________1 f E 11E E F1 expkTor210.25 0.0259 2f F exp0.02593.90 10 5E kT ; 1 fF exp0.25 0.02590.02591 f E1E F E1 expkT(a) E E kT , 1 f E 0.2692.36 10 E3kT ;25F(b) E F E3 5kT , 1 f E 6.69 10(c)E F E 10kT , 1 f E 4.54 105_______________________________________(a) f F exp E E F kTE E c; fF exp 0.30 9.32 10 60.0259E c kT ; fF exp 0.30 0.0259 20.025925.66 10 6E c kT ; fF exp 0.30 0.02590.02593.43 10 6E c 3kT; 2f F exp 0.30 3 0.0259 20.02590.25 3 0.0259 21 f F exp0.02591.43 10 5E 2kT ;1 f F exp0.25 2 0.02590.02598.70 10 6_______________________________________f FE EF E c kT E Fexp expkTkTand1 f F expE F EkTE F E kTexpkTSoE c kT E FexpkTE FE kTexpkTThen E c kT E F E F EkTOrE c EE FEmidgap2_______________________________________2 22E nn2ma 2 Forn 6 , Filled state1.05410 34222 E6610 31 1210 1022 9.111.5044 10 18J1.504418orE 6109.40 eV1.6 10 19Forn 7 , Empty state1.054 10 34 2722E 710 311210 1022 9.112.048 10 18JorE2.048 10 1812.8 eV 71.6 10 19Therefore9.40 E F 12.8 eV_______________________________________(a) For a 3-D infinite potential well2mE22222n x n y n zaFor 5 electrons, the 5thelectronoccupies the quantum state n x 2, n y2, n z 1 ; so22n x 2n y 2n z2E 52ma1.054 10 34 222 2 22 122 9.11 10 31 1210 10 23.761 10 19Jor3.761 10 19 2.35 eVE 510 191.6n x1, n y 2, n z 2 . This quantumstate is at the same energy, soE F 2.35 eV(b) For 13 electrons, the 13thelectron occupies the quantum staten x3, n y 2, n z 3 ; so1.054 10 34 223 2 2 2 32E132 9.11 10 31 12 1010 29.194 10 19JorE 13 9.194 10195.746 eV1.6 10 19The 14 th electron would occupy thequantum staten x 2, n y 3, n z3 .This state is at the same energy, soE F 5.746 eV_______________________________________The probability of a state atE 1 E FEbeing occupied isf 1 11E 1EE 1 E F1 exp1 expkTkTThe probability of a state atE 2 EF Ebeing empty is11 f2 E 2 11 exp E2 E FkTE1exp1kTE E1exp1 expkTkTor1 f2 E 21E1expkT(a)so f 1E 1 1 f 2 E 2energyE 1 , we wantFor the next quantum state, whichis empty, the quantum state is1 1E1E F E1E Fexp 1 expkT kT0.0111exp E1 E FkTThis expression can be written as1E1 E Fexp kT1 0.01E1 E FexpkT or1E1 E F 0.01 expkTThenE1 E F kT ln 100orE1 E F 4.6kT(b)At E E F 4.6kT ,f E11 1expE1 EF 1 exp 4.6 1kTwhich yieldsf E1 0.00990 0.01_______________________________________(a)f F exp E E Fexp5.80 5.50 kT 0.02599.32 10 6(b) kT 0.0259 7000.060433 eV 300f F exp 0.30 6.98 10 30.060433(c) 1 f F exp E F E kT0.02 exp0.25kTor0.25 150exp0.02kT0.25ln 50kTor0.250.063906TkT 0.0259ln 50 300which yields T 740 K_______________________________________(a)10.00304f E7.15 7.01 exp0.0259or %(b)At T 1000 K, kT 0.08633 eV Thenf E 1 0.1496exp 7.151 7.00.08633or %(c) f E 1 0.9976.8517.0exp0.0259or %(d)At E E F, f E1for all2temperatures_______________________________________(a) For E E1f E1expE1 E FE1 E F kT1 expkTThenf E1 exp 0.30 9.32 10 60.0259For E E2,E F E2 1.12 0.30 0.82 eVThen1 f E 110.821 exp0.0259or1 f E 11 exp0.82 0.0259exp0.82 1.78 10 140.0259(b) For E F E 2 0.4 eV,E 1 E F0.72 eV At E E 1 ,f Eexp E 1 E F 0.72kTexp0.0259orf E8.45 10 13At E E 2 ,1 f EE F E 2 expkT0.4exp0.0259or1 f E 1.96 107_______________________________________(a) At EE 1f EE 1 E F0.30 expexp0.0259kTorf E9.32 10 6At E E 2 , E F E 21.42 0.3 1.12 eVSo1 f EE FE 2expf EexpE 1 E F1.02kTexp0.0259orf E187.88 10 At E E 2 ,1 f EexpE F E 2kTexp0.40.0259or 1 f E1.96 10 7_______________________________________EE F 1f E1 expkTsodf EE EF 21 1 expdEkT1 exp E E FkTkT or1 exp E E Fdf EkTkTdE21 expE E FkT(a) At T 0 K, ForE E FexpdfdEE E FexpdfdEAtEE FdfdE(b) At T300 K, kT 0.0259 eVkTexp1.12For EE F ,dfdE0.0259or1 f E1.66 1910(b) For E F E 2 0.4 ,E 1 E F1.02 eV At EE 1 ,ForE EF ,dfAtE EF ,dE11df 0.025919.65 (eV)dE21 1(c) At T 500 K, kT 0.04317 eVFor E E F, df0 dEFor E E F, df0 dEAt E E F,1df 0.04317 115.79 (eV)dE2 1 1_______________________________________(b)Using the results of Problem ,the answers to part (b) are exactlythe same as those given in part (a)._______________________________________(a) f F expE E FkT10 8 exp 0.60kTor 0.60 ln 10 8kTkT0.600.032572 eVln 1080.032572 0.0259T(a) At E E midgap ,f E1 1E EF E g1 exp 1 expkT 2kTSi: E g 1.12 eV,300 so T 377K(b)10 6 exp0.60kT0.60 ln 10 6kTf E 11.12 kT 0.606 0.0434291exp2 0.0259 ln 10Torf E 4.07 10 100.043429 0.0259300or T 503 K_______________________________________Ge: E g0.66 eV1f E0.661 exp2 0.0259 (a) At T 200 K,kT2000.0259 0.017267 eVorf E 2.93 10 6300f F 0.05 1GaAs: E g 1.42 eV1f E1.421exp2 0.02591 exp E E FkTE EF 11 19 expkT 0.05orf E 1.24 10 12E EF kT ln 19 0.017267 ln 190.05084eV By symmetry, for f F 0.95,E EF 0.05084eVThen E 2 0.05084 0.1017 eV(b)T 400K, kT 0.034533 eVFor f F 0.05 , from part (a),E EF kT ln 19 0.034533 ln 190.10168eVThen E 2 0.10168 0.2034 eV_______________________________________。