1、Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_1Chapter 33.1If were to increase, the bandgap energy oawould decrease and the material would begin to behave less like a semiconductor and more like a metal. If were to decrease, the obandgap
2、energy would increase and the material would begin to behave more like an insulator._3.2Schrodingers wave equation is:txVxtm,22tj,Assume the solution is of the form:tEkxjutxep,Region I: . Substituting the 0Vassumed solution into the wave equation, we obtain:tEkxjjkuxmep2tjtEkxjuEj epwhich becomestjx
3、jkm22tEkjujeptxjx2tEkjEuepThis equation may be written as022xumxjkxuSetting for region I, the equation xu1becomes:021212 xukdjdxwhereQ.E.D.2mEIn Region II, . Assume the same OVxform of the solution:tEkjutep,Substituting into Schrodingers wave equation, we find:tkxjujkmep22tEjjtkxjxuep2tEjVOtkxjEuepT
4、his equation can be written as:22xjkxu02umEVOSetting for region II, this xequation becomesdxujkdu2202xumVOwhere againQ.E.D.2E_Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_23.3We have0212112 xukdxujdxAssume the solution is of the form:jAu
5、ep1xkjBThe first derivative isjkjdxe1xkpand the second derivative becomesjAkjdxuex221xkBpSubstituting these equations into the differential equation, we findxkjke2pjAjBxkjke20xpCombining terms, we obtain222 kjAe222k0xpxjBWe find thatQ.E.D.0For the differential equation in and the u2proposed solution
6、, the procedure is exactly the same as above._3.4We have the solutionsxkjAxuep1Bfor anda0jC2xkjDepfor .xbThe first boundary condition is021uwhich yields0DCBAThe second boundary condition is0201xxduwhich yieldsCkBAk0DThe third boundary condition isbua21which yieldsakjBkjAexpexpbjCjDand can be written
7、 asakakjexpexpbj0The fourth boundary condition isbxaxdu21which yieldsakjAkjepBbjCjxkDand can be written asakjAkepBxbjC0_3.5(b) (i) First point: aSecond point: By trial and error,729.1(ii) First point: Second point: By trial and error,6.a_Semiconductor Physics and Devices: Basic Principles, 4th editi
8、on Chapter 3By D. A. Neamen Problem Solutions_33.6(b) (i) First point: aSecond point: By trial and error,51.(ii) First point: 2Second point: By trial and error,37.a_3.7kaaPcossinLet , ykxThenyxcossinConsider of this function.dyyxxPsincosin1We find dyxdyssi12iniThenyxxPdyx siicosin12 For , ka.,200nSo
9、 that, in general,dkdyx0And2mESodkEdk2/12This implies thatfor E0an_3.8(a) a1Emo221031421 .0.95aoJ14.3From Problem 3.5729.1aEmo210313422.0.5J8912E194. J9076or eV2.1.8(b) 23aEmo210313423.0.95J864From Problem 3.5,7.24a1.Emo210313424.0.956J834E18164.6.2J907or eV7.81_Semiconductor Physics and Devices: Ba
10、sic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_43.9(a) At , ka1Emo22103141.0.95J94At , By trial and error,ka85.o21031342.0.9oEJ957o119157.4. J208or eV.6.919E(b) At , ka322mo21031343.0.95EJ864At . From Problem 3.5,ka79.2a12Emo21031342.0.95J823E1819.64. J07or eV5.19_3.10(a) a1E
11、mo22103141.0.95J94From Problem 3.6, .2a51.2Emo210313422.0.9J98712E194. J9046or eV76.21.8(b) 23aEmo210313423.0.95J864From Problem 3.6, 7.4a35.224Emo21031344.0.97J8234E18164. J9057or eV5.6.1_Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_53.
12、11(a) At , ka1Emo22103141.0.95J94At , By trial and error,ka7.ao202Emo2103134.95oJ908oE11918.4. J96or eV05.01(b) At , 2ka3Emo210313423.0.95J864At , From Problem 3.6,ka5.21.2Emo210343422.0.95J198723E198.76. J1905or eV635._3.12For K, 10T10637.4. 24gEeV.gK, eV20K, eV3T25.K, eV40971gEK, eV56.K, eV603_3.1
13、3The effective mass is given by12*dkEmWe haveBcurveAcurve22so that m*_3.14The effective mass for a hole is given by12*dkEmpWe have thatBcurveAcurve22so that mpp*_3.15Points A,B: velocity in -x direction0dkEPoints C,D: velocity in +x directionPoints A,D: 02dknegative effective massPoints B,C: 2Eposit
14、ive effective mass_Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_63.16For A: 2kCEiAt m , eV108.05.EOr J211986.5So 2123810.Now 3824125.Ckg34.or omm10.97o8For B: 2kCEiAt m , eV10.5.EOr J201986.5So 01208371.Now 37241205.Ckg324.or omm1.97o08_
15、3.17For A: 2kCE210198.06.5. 32392421.mkg087.or om31.96For B: 2kCE210198.0.33825738242105.7Cmkg2406.or om31.9o8_3.18(a) (i) hEor 34190625.1Hz493(ii) 14.cEhcm nm5107.887(b) (i) 34962.hHz1(ii) 4075.ccm nm99_3.19(c) Curve A: Effective mass is a constantCurve B: Effective mass is positive around , and is
16、 negative 0karound .2_3.20OOkEcos1Thendkin1OksandEdkco212Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_7Then212*1Edkmoor21*E_3.21(a) 3/123/4ltdnm3/164.08.oodn56(b) ooltc .2.123om098.4cn1._3.22(a) 3/22/3lhhdp3/2/08.45.0oom3/2417odpm3.(b) 2
17、/12/13lhhcom2/12/13308.45.ocpm0_3.23For the 3-dimensional infinite potential well, when , , and 0xVaxy0. In this region, the wave equation azis:222 , zxyxy0,mEUse separation of variables technique, so letzZyYxXzyx,Substituting into the wave equation, we have222zZXYyZxY0mEDividing by , we obtain21122
18、 zZyYxXLet022 XkxkThe solution is of the form:BAxXcossinSince at , then 0,zyso that .Also, at , so that ,ax. Then where ank,321xnSimilarly, we haveand 2ykY 221zkZFrom the boundary conditions, we findand ynakzznawhereand .,321.,321zFrom the wave equation, we can write02mEkzyxThe energy can be written
19、 as22anEzyxnzyx _3.24The total number of quantum states in the 3-dimensional potential well is given (in k-space) by32adkkgTwhere2mEWe can then writekTaking the differential, we obtainSemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_8dEmdEmd
20、k 2121Substituting these expressions into the density of states function, we havedadgT 2123Noting that2hthis density of states function can be simplified and written asdEmhadEgT 2/334Dividing by will yield the density of states so thathg32/4_3.25For a one-dimensional infinite potential well,22knEmDi
21、stance between quantum statesanakn 11Now adkgT2NowEmkn21ddThenEaEgnT21Divide by the “volume“ a, somnSoEEg 3134 0.967.02105. m JE1831_3.26(a) Silicon, on08.1cc EhmEg32/4dkTEcnccc/kTEcnch2/332/42/332/kmn2/334/10625.9814T2/37kT(i) At K, eVT.1906.J214.Then 2/315093.7cgm263or cm1.c(ii) At K, 40T0459.keV3
22、196.J25Then 2/310.1093.7cgm253or cm4.c(b) GaAs, on62/3342/1025.97kTgc 2/18kTSemiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_9(i) At K, J30T2104.k/35128.cgm379or cm.c(ii) At K, J40T2105.k/34128.cgm73cm.c_3.27(a) Silicon, op56.0EhmEg32/4dkTEp
23、332/EkTph32/32/42/32/mp2/334/10625.94kT2/1kT(i)At K, J321./35049.gm20643or cm1.(ii)At K, JT215.k/3039.2gm21076or cm4.(b) GaAs, op8.2/3342/10625.9kTg2/3kT(i)At K, J30T2104.k2/35316.2gm2or cm97.(ii)At K, J40T2105.k/3316.2gm295or cm._3.28(a) cnc EhEg32/4c342/10625.981cFor ; cEcgeV; m J1.4619.31eV; m J2
24、0c 0eV; m J3.46.31eV; m J4cE8(b) Ehmgp32/342/10625.941For ; EgeV; m J.4513.31eV; m J2009687eV; m J3. 45.31eV; m J4E2_3.29(a) 68.25.01/32/3pncmg(b) 051.4.7/32/3pnc_Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3By D. A. Neamen Problem Solutions_103.30Plot_3.31(a) !710!iiiNg
25、W238978910(b) (i) 1!02!i6(ii) 234!8!91!8iW495_3.32kTEEfFexp1(a) , F1expf269.0f(b) , kTEF55f31.f(c) , F010expf54.Ef_3.33kTEf Fexp11orkEfFexp1(a) , TF269.0Ef(b) , 531(c) , kE1054.f_3.34(a) kTEfFFexp; cE61032.905.f; 2kc.expFf61.; kTEc 0259.3Ff64.; 23c.expFf6108.; kTEc 259.3Ff6.(b) kTEfFFexp1; E0259.exp1Ff 51043.6; 2kT.2f51.3; kE 029.exp1Ff56.; 23kT0259.3exp1Ff14.;kTE20259.exp1Ff617.8_
