半导体物理与器件第四版课后习题答案.pdf

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1、Chapter 3 If oa were to increase, the bandgap energy would decrease and the material would begin to behave less like a semiconductor and more like a metal. If oa were to decrease, the bandgap energy would increase and the material would begin to behave more like an insulator._ Schrodingers wave equa

2、tion is:txxVxtxm,2222ttxj, Assume the solution is of the form:tEkxjxutxexp, Region I: 0 xV. Substituting the assumed solution into the wave equation, we obtain:tEkxjxjkuxmexp22tEkxjxxuexptEkxjxujEjexp which becomestEkxjxujkmexp222tEkxjxxujkexp2tEkxjxxuexp22tEkxjxEuexp This equation may be written as

3、0222222xumExxuxxujkxuk Setting xuxu1 for region I, the equation becomes:021221212xukdxxdujkdxxud where222mEIn Region II, OVxV. Assume the same form of the solution:tEkxjxutxexp, Substituting into Schrodingers wave equation, we find:tEkxjxujkmexp222tEkxjxxujkexp2tEkxjxxuexp22tEkxjxuVOexptEkxjxEuexp T

4、his equation can be written as:2222xxuxxujkxuk02222xumExumVO Setting xuxu2 for region II, this equation becomesdxxdujkdxxud22222022222xumVkO where again222mEWe have021221212xukdxxdujkdxxud Assume the solution is of the form:xkjAxuexp1xkjBexp The first derivative isxkjAkjdxxduexp1xkjBkjexp and the se

5、cond derivative becomesxkjAkjdxxudexp2212xkjBkjexp2 Substituting these equations into the differential equation, we findxkjAkexp2xkjBkexp2xkjAkjjkexp2xkjBkjexpxkjAkexp220expxkjB Combining terms, we obtain222222kkkkkxkjAexp222222kkkkk0expxkjB We find that00For the differential equation in xu2and the

6、proposed solution, the procedure is exactly the same as above._ We have the solutionsxkjAxuexp1xkjBexp for ax0 andxkjCxuexp2xkjD exp for 0 xb. The first boundary condition is0021uu which yields0DCBA The second boundary condition is0201xxdxdudxdu which yieldsCkBkAk0Dk The third boundary condition isb

7、uau21 which yieldsakjBakjAexpexpbkjC expbkjD exp and can be written asakjBakjAexpexpbkjC exp0expbkjD The fourth boundary condition isbxaxdxdudxdu21 which yieldsakjAkjexpakjBkjexpbkjCkjexpbkjDkjexp and can be written asakjAkexpakjBkexpbkjCkexp0expbkjDk_ (b) (i) First point: a Second point: By trial a

8、nd error,729.1a (ii) First point: 2a Second point: By trial and error,617.2a_ (b) (i) First point: a Second point: By trial and error,515.1a (ii) First point: 2a Second point: By trial and error,375.2a_kaaaaPcoscossin Let yka, xa ThenyxxxPcoscossin Consider dyd of this function.yxxxPdydsincossin1 We

9、 finddydxxxdydxxxPcossin112ydydxxsinsin ThenyxxxxxPdydxsinsincossin12 For nkay, ., 2, 1, 0n0sin y So that, in general,dkdkadaddydx0 And22mE SodkdEmmEdkd22/122221 This implies thatdkdEdkd0 for ank_(a)a1aEmo2122103123422221102. 41011.9210054. 12amEo19104114. 3J From Problem 729.12a729. 1222aEmo2103123

10、422102.41011.9210054.1729.1E18100198.1J12EEE1918104114.3100198.119107868.6Jor 24.4106.1107868.61919EeV(b)23a2223aEmo2103123423102. 41011.9210054.12E18103646.1J From Problem ,617.24a617.2224aEmo2103123424102 .41011. 9210054. 1617.2E18103364.2J34EEE1818103646.1103364.21910718.9J or 07.6106. 110718.919

11、19EeV_(a)At ka, a1aEmo2122103123421102.41011.9210054.1E19104114.3J At 0ka, By trial and error,859.0ao210312342102. 41011.9210054.1859. 0oE19105172.2JoEEE11919105172.2104114.32010942.8J or 559.0106 .110942.81920EeV(b)At 2ka, 23a2223aEmo2103123423102.41011.9210054.12E18103646. 1J At ka. From Problem ,

12、729.12a729.1222aEmo2103123422102 .41011. 9210054.1729. 1E18100198. 1J23EEE1818100198.1103646.119104474. 3J or 15. 2106.1104474. 31919EeV_(a)a1aEmo2122103123421102.41011.9210054.1E19104114. 3J From Problem , 515.12a515.1222aEmo2103123422102.41011.9210054.1515.1E1910830.7J12EEE1919104114.310830. 71910

13、4186.4J or 76.2106. 1104186.41919EeV(b)23a2223aEmo2103123423102 .41011.9210054.12E18103646.1J From Problem , 375.24a375.2224aEmo2103123424102. 41011.9210054.1375. 2E18109242.1J34EEE1818103646.1109242.11910597.5J or 50. 3106.110597.51919EeV_(a)At ka, a1aEmo2122103123421102.41011.9210054.1E19104114. 3

14、J At 0ka, By trial and error,727.0ao727.022aEmoo210312342102.41011. 9210054. 1727.0oE19108030. 1JoEEE11919108030. 1104114. 319106084.1J or 005.1106.1106084.11919EeV(b) At 2ka, 23a2223aEmo2103123423102. 41011.9210054.12E18103646.1J At ka, From Problem ,515. 12a515.1222aEmo2103423422102. 41011.9210054

15、.1515.1E1910830. 7J23EEE191810830.7103646.11910816.5J or 635.3106.110816.51919EeV_ For 100TK, 1006361001073.4170.124gE164.1gEeV200TK, 147.1gEeV300TK, 125.1gEeV400TK, 097.1gEeV500TK, 066.1gEeV600TK, 032.1gEeV_ The effective mass is given by1222*1dkEdm We haveBcurvedkEdAcurvedkEd2222 so that BcurvemAc

16、urvem*_ The effective mass for a hole is given by1222*1dkEdmp We have thatBcurvedkEdAcurvedkEd2222 so that BcurvemAcurvempp*_ Points A,B: 0dkdEvelocity in -x direction Points C,D: 0dkdEvelocity in +x direction Points A,D: 022dkEdnegative effective mass Points B,C: 022dkEdpositive effective mass_ For

17、 A: 2kCEi At 101008.0km1, 05.0EeV Or 2119108106.105. 0EJ So 2101211008. 0108C3811025.1C Now 38234121025.1210054.12Cm311044. 4kg or omm31311011.9104437.4omm488.0 For B: 2kCEi At 101008.0km1, 5. 0EeV Or 2019108106.15.0EJ So 2101201008.0108C3711025.1C Now 37234121025.1210054. 12Cm321044.4kg or omm31321

18、011.9104437. 4omm0488.0_ For A: 22kCEE2102191008.0106 .1025.0C3921025.6C39234221025.6210054. 12Cm31108873.8kg or omm31311011.9108873. 8omm976.0 For B: 22kCEE2102191008.0106. 13. 0C382105 .7C3823422105.7210054. 12Cm3210406.7kg or omm31321011. 910406. 7omm0813.0_(a)(i) hE or 341910625.6106.142. 1hE141

19、0429.3Hz (ii) 141010429.3103cEhc51075. 8cm875nm(b)(i) 341910625.6106 .112.1hE1410705.2Hz (ii) 141010705.2103c410109. 1cm1109nm_(c)Curve A: Effective mass is a constantCurve B: Effective mass is positive around 0k, and is negative around 2k._OOkkEEEcos1 ThenOkkEdkdEsin1OkkEsin1 andOkkEdkEdcos2122 The

20、n221222*11EdkEdmokk or212*Em_(a)3/123/24ltdnmmm3/123/264. 1082.04oommodnmm56.0(b)ooltcnmmmmm64.11082.02123oomm6098.039.24ocnmm12.0_(a)3/22/32/3lhhhdpmmm3/22/32/3082.045. 0oommom3/202348.030187. 0odpmm473.0(b)2/12/12/32/3lhhhlhhhcpmmmmmom2/12/12/32/3082.045.0082.045.0ocpmm34.0_ For the 3-dimensional

21、infinite potential well, 0 xV when ax0, ay0, and az0. In this region, the wave equation is:222222,zzyxyzyxxzyx0,22zyxmE Use separation of variables technique, so letzZyYxXzyx, Substituting into the wave equation, we have222222zZXYyYXZxXYZ022XYZmE Dividing by XYZ , we obtain021112222222mEzZZyYYxXX Le

22、t01222222XkxXkxXXxx The solution is of the form:xkBxkAxXxxcossin Since 0,zyx at 0 x, then 00X so that 0B. Also, 0,zyx at ax, so that 0aX. Then xxnak where ., 3,2,1xn Similarly, we have2221ykyYY and 2221zkzZZ From the boundary conditions, we findyynak and zznak where., 3,2, 1yn and ., 3, 2, 1zn From

23、the wave equation, we can write022222mEkkkzyx The energy can be written as222222annnmEEzyxnnnzyx_ The total number of quantum states in the 3-dimensional potential well is given (in k-space) by332adkkdkkgT where222mEk We can then writemEk2 Taking the differential, we obtaindEEmdEEmdk2112121 Substitu

24、ting these expressions into the density of states function, we havedEEmmEadEEgT212233 Noting that2h this density of states function can be simplified and written asdEEmhadEEgT2/33324 Dividing by 3a will yield the density of states so thatEhmEg32/324_ For a one-dimensional infinite potential well,222

25、222kanEmn Distance between quantum statesaanankknn11 Now adkdkkgT2 NowEmkn21dEEmdkn2211 ThendEEmadEEgnT2212 Divide by the volume a, soEmEgn21 SoEEg31341011. 9067.0210054.11EEg1810055.1 m3J1_ (a) Silicon, onmm08.1cncEEhmEg32/324dEEEhmgkTEEcnccc232/324kTEEcnccEEhm22/332/332242/332/323224kThmn2/33342/3

26、3123210625.61011.908.124kT2/355210953.7kT(i) At 300TK, 0259.0kTeV19106.10259.02110144.4J Then 2/3215510144.4210953.7cg25100 .6m3 or 19100.6cgcm3 (ii) At 400TK, 3004000259.0kT034533.0eV19106.1034533.021105253.5J Then 2/32155105253.5210953.7cg2510239. 9m3 or 191024.9cgcm3(b) GaAs, onmm067.02/33342/331

27、23210625.61011.9067.024kTgc2/3542102288.1kT(i) At 300TK, 2110144.4kTJ2/3215410144.42102288.1cg2310272.9m3 or 171027.9cgcm3(ii) At 400TK, 21105253.5kTJ2/32154105253.52102288.1cg2410427.1m3181043.1cgcm3_(a)Silicon, opmm56.0EEhmEgp32/324dEEEhmgEkTEp332/324EkTEpEEhm32/332/332242/332/333224kThmp2/33342/3

28、3133210625.61011.956.024kT2/355310969. 2kT (i)At 300TK, 2110144.4kTJ2/3215510144.4310969.2g2510116.4m3 or 191012. 4gcm3 (ii)At 400TK, 21105253.5kTJ2/32155105253.5310969.2g2510337. 6m3 or 191034.6gcm3(b)GaAs, opmm48. 02/33342/33133210625.61011.948.024kTg2/3553103564. 2kT(i)At 300TK, 2110144.4kTJ2/321

29、5510144.43103564.2g2510266.3m3 or 191027. 3gcm3 (ii)At 400TK, 21105253. 5kTJ2/32155105253.53103564.2g2510029.5m3 or 191003.5gcm3_(a)cncEEhmEg32/324cEE3342/33110625.61011. 908.124cEE56101929.1For cEE; 0cg1.0cEEeV; 4610509.1cgm3J12 .0cEEeV; 4610134.2m3J13 .0cEEeV; 4610614.2m3J14 .0cEEeV; 4610018.3m3J1

30、(b)EEhmgp32/324EE3342/33110625.61011. 956.024EE55104541. 4For EE; 0g1. 0EEeV; 4510634. 5gm3J12.0EEeV; 4510968.7m3J13.0EEeV; 4510758. 9m3J14 .0EEeV; 4610127. 1m3J1_(a)68.256.008.12/32/32/3pncmmgg(b)0521.048.0067.02/32/32/3pncmmgg_ Plot_(a)!710! 7!10!iiiiiNgNgW1201238910! 3!7! 78910(b)(i) 12!10!101112

31、!1012!10!12iW66 (ii) 1234! 8! 89101112!812! 8!12iW495_kTEEEfFexp11(a)kTEEF, 1exp11Ef269. 0Ef(b)kTEEF5, 5exp11Ef31069.6Ef(c)kTEEF10, 10exp11Ef51054.4Ef_kTEEEfFexp1111 orkTEEEfFexp111(a)kTEEF, 269.01Ef(b)kTEEF5, 31069. 61Ef(c)kTEEF10, 51054.41Ef_(a)kTEEfFFexpcEE; 61032.90259. 030. 0expFf2kTEc; 0259.02

32、0259. 030.0expFf61066.5kTEc; 0259. 00259.030.0expFf61043.323kTEc; 0259.020259.0330. 0expFf61008.2kTEc2; 0259.00259.0230.0expFf61026. 1(b)kTEEfFFexp1111kTEEFexpEE; 0259.025.0exp1Ff51043.62kTE; 0259.020259.025.0exp1Ff51090. 3kTE; 0259.00259.025.0exp1Ff51036. 223kTE; 0259.020259.0325.0exp1Ff51043. 1kTE

33、2;0259.00259. 0225.0exp1Ff61070. 8_kTEkTEkTEEfFcFFexpexpandkTEEfFFexp1kTkTEEFexpSo kTEkTEFcexpkTkTEEFexp Then kTEEEkTEFFc Or midgapcFEEEE2_22222manEn For 6n, Filled state2103122234610121011.92610054.1E18105044.1J or 40.9106 .1105044. 119186EeV For 7n, Empty state2103122234710121011.92710054. 1E18100

34、48. 2J or 8.12106.110048.219187EeV Therefore 8.1240.9FEeV_(a)For a 3-D infinite potential well222222annnmEzyx For 5 electrons, the 5th electron occupies the quantum state 1,2,2zyxnnn; so2222252annnmEzyx21031222223410121011.9212210054. 11910761.3J or 35.2106.110761.319195EeV For the next quantum stat

35、e, which is empty, the quantum state is 2,2, 1zyxnnn. This quantum state is at the same energy, so35.2FEeV(b)For 13 electrons, the 13thelectron occupies the quantum state 3, 2,3zyxnnn; so2103122222341310121011. 9232310054.1E1910194.9J or 746.5106.110194. 9191913EeVThe 14th electron would occupy the

36、quantum state 3, 3,2zyxnnn. This state is at the same energy, so746. 5FEeV_ The probability of a state at EEEF1 being occupied iskTEkTEEEfFexp11exp11111 The probability of a state at EEEF2 being empty iskTEEEfF222exp1111kTEkTEkTEexp1expexp111 orkTEEfexp11122(a) so 22111EfEfenergy 1E, we want01. 0exp

37、11exp11exp1111kTEEkTEEkTEEFFFThis expression can be written as01.01expexp111kTEEkTEEFForkTEEF1exp01.01Then100ln1kTEEForkTEEF6.41(b) At kTEEF6.4,6.4exp11exp1111kTEEEfFwhich yields01. 000990. 01Ef_ (a) 0259.050.580.5expexpkTEEfFF61032.9(b) 060433.03007000259. 0kTeV31098. 6060433.030. 0expFf(c)kTEEfFFe

38、xp1kT25. 0exp02.0 or 5002.0125. 0expkT50ln25. 0kT or 3000259.0063906.050ln25.0TkT which yields 740TK_ (a)00304.00259.00.715.7exp11Ef or %(b)At 1000TK, 08633.0kTeVThen1496. 008633.00.715.7exp11Efor %(c)997.00259.00.785.6exp11Efor %(d) At FEE, 21Ef for all temperatures_(a)For 1EEkTEEkTEEEfFF11expexp11

39、Then611032.90259.030.0expEf For 2EE, 82. 030.012.12EEFeV Then0259.082.0exp1111Ef or0259. 082.0exp111Ef141078.10259.082.0exp(b)For 4. 02EEFeV, 72. 01FEEeVAt 1EE,0259.072.0expexp1kTEEEfFor131045.8EfAt 2EE,kTEEEfF2exp10259. 04.0expor71096.11Ef_(a)At 1EE0259. 030. 0expexp1kTEEEfFor61032.9EfAt 2EE, 12. 1

40、3. 042.12EEFeVSokTEEEfF2exp10259.012.1expor191066.11Ef(b)For 4. 02EEF,02.11FEEeVAt 1EE,0259.002.1expexp1kTEEEfFor181088.7EfAt 2EE,kTEEEfF2exp10259.04 .0expor 71096.11Ef_1exp1kTEEEfF so2exp11kTEEdEEdfFkTEEkTFexp1 or2exp1exp1kTEEkTEEkTdEEdfFF(a)At 0TK, For00expdEdfEEF0expdEdfEEF At dEdfEEF(b)At 300TK,

41、 0259.0kTeV For FEE, 0dEdf For FEE, 0dEdf At FEE,65. 91110259.012dEdf(eV)1(c)At 500TK, 04317.0kTeV For FEE, 0dEdf For FEE, 0dEdf At FEE,79.511104317.012dEdf(eV)1_(a)At midgapEE,kTEkTEEEfgF2exp11exp11Si: 12. 1gEeV,0259.0212.1exp11Efor101007. 4EfGe: 66.0gEeV0259.0266.0exp11Efor61093.2EfGaAs: 42.1gEeV0

42、259.0242.1exp11Efor121024.1Ef(b) Using the results of Problem , the answers to part (b) are exactly the same as those given in part (a)._(a)kTEEfFFexpkT60.0exp108 or 810ln60. 0kT032572.010ln60.08kTeV3000259.0032572.0T so 377TK(b)kT60.0exp106610ln60. 0kT043429.010ln60.06kT3000259.0043429. 0T or 503TK_(a)At 200TK, 017267.03002000259.0kTeVkTEEfFFexp1105.019105. 01expkTEEF19ln017267.019lnkTEEF05084.0eV By symmetry, for 95.0Ff,05084.0FEEeV Then 1017.005084.02EeV(b)400TK, 034533.0kTeV For 05.0Ff, from part (a),19ln034533.019lnkTEEF10168.0eV Then 2034.010168.02EeV_