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第三章习题和答案1.计算能量在E=Ec到2C*2n100EE8mLh之间单位体积中的量子态数。解:2.试证明实际硅、锗中导带底附近状态密度公式为式(3-6)。2222CC3231*22301001003881*2203EE23*22332()4()ZZV21Z()4()100224()8L310003ccnnnChhEEmlmlnCcnCncmdZgEVEEdEhdmgEdEEEdEVhhEmEEmhEL()单位体积内的量子态数()()22222222221/221/22()212(())[]2(())[]xyzCtltClCkkkhEkEmmxyzabcmEkEabhmEkEch导带底附近()对于椭球方程:则:1/22221/23/234V=abc32(())2(())4[]34(8)(())3tclctlCmEkEmEkEvhhmmEkEh椭球体积公式:21/21/2321/21/2343(8)(())322(8)(())tlCtlCdvmmEkEdEhmmEkEdEh3.当E-EF为1.5k0T,4k0T,10k0T时,分别用费米分布函数和玻耳兹曼分布函数计算电子占据各该能级的概率。费米能级费米函数玻尔兹曼分布函数1.5k0T0.1820.2234k0T0.0180.018310k0T4.画出-78oC、室温(27oC)、500oC三个温度下的费米分布函数曲线,并进行比较。y=(1.38065e-23)*(273.15-78)*log(1./x-1);(图中红色)y=(1.38065e-23)*(273.15+27)*log(1./x-1);(图中粗蓝色)21/21/2321/21/2321/21/233/221/22*23~dZZ2VdV2=2V(8)(())1s22V(8)(())Z()=S4(8)(())(2)(8)tlCtlCtlCdntlndntlEEdEdmmEkEdEhmmEkEdEdhgEdEdESVmmEkEhmSmmmmsmm在空间内的量子态数又因为,导带极值不是个,而是个,则:若令:则有:13FEETkEEeEfF011)(TkEEFeEf0)(51054.451054.4y=(1.38065e-23)*(273.15+300)*log(1./x-1);(图中细蓝色)5.利用表3-2中的m*n,m*p数值,计算硅、锗、砷化镓在室温下的NC,NV以及本征载流子的浓度。Ge:Nc=1.05×1019cm-3Nv=5.7×1018cm-3Ni=2.0×1013cm-3Si:Nc=2.80×1019cm-3Nv=1.1×1019cm-3Ni=7.8×109cm-3GaAs:Nc=4.5×1017cm-3Nv=8.1×1018cm-3Ni=2.3×106cm-3evEmmmmGaAsevEmmmmSievEmmmmGeeNNnhTmkNhTmkNgpngpngpnkoTEvcipVnCg428.1;47.0;068.0:12.1;59.0;08.1:67.0;37.0;56.0:)()2(2)2(2000000221232023206.计算硅在-78oC,27oC,300oC时的本征费米能级,假定它在禁带中间合理吗?所以假设本征费米能级在禁带中间合理,特别是温度不太高的情况下。7.①在室温下,锗的有效状态密度Nc=1.051019cm-3,NV=5.71018cm-3,试求锗的载流子有效质量m*nm*p。计算77K时的NC和NV。已知300K时,Eg=0.67eV。77k时Eg=0.76eV。求这两个温度时锗的本征载流子浓度。②77K时,锗的电子浓度为1017cm-3,假定受主浓度为零,而Ec-ED=0.01eV,求锗中施主浓度ED为多少?3022302222331022331030223/211223/21122212()22()5.110223.39102222()()()()()()()ncpvcnvpncCCVVkTmNhkTmNhNhmkgkTNhmkgkTkTmNhNTTNTTNTTNTT解:()根据得(2)根据:有同理:'3193183'318317377771.05101.3710/30030077775.7107.4110/300300CCVVNNcmNNcm()()()()0000101010022022033-2:1.08,0.593ln(3-30)2430.591950.016,ln0.007241.0830.593000.026,ln0.0116641.085730.0nppCVFinSiSimmmmmEEkTEEmkTmTKkTeVeVmkTTKkTeVeVTKkT的本征费米能级,表当时,当时,当时,0330.59497,ln0.02241.08kTeVeV00001220.6712300191813320.76127718177320000(3)()(1.05105.710)1.9610/77(1.37107.4110)1.4210/exp()12expCFDFEgkoTicvkikiEEkTCFcCDDDEEkTnNNenecmKnecmEEnnNekTNNNnn室温:时,则有同时有0001717173001801212100.01(12exp())10(12exp())1.6510/1.37100.0067DcCFDDEEEEEkTkTCDDCNneeNnENncmNkT8.利用题7所给的Nc和NV数值及Eg=0.67eV,求温度为300K和500K时,含施主浓度ND=51015cm-3,受主浓度NA=2109cm-3的锗中电子及空穴浓度为多少?0E'20421213321''1632002002004.77410(0)00.745000.60235300()2.010/500()1.610/+()(ggkTggggEkTicViCVDADiTEEKEeVEKeVTKnNNecmKnNNecmnpNNnnNNnpn时:时:根据电中性条件:导带电子由有效施主和本征激发共同提供2122202001530103015301530)0()22510/3007.6810/9.8410/5004.8410/AiDADAiinNNNNnnnpnncmTKpcmncmtKpcm时:时:其中:300K时,DADNpNNn00=5×1015cm-3310152130201068.7105)1096.1(cmnnpi9.计算施主杂质浓度分别为1016cm3,,1018cm-3,1019cm-3的硅在室温下的费米能级,并假定杂质是全部电离,再用算出的的费米能级核对一下,上述假定是否在每一种情况下都成立。计算时,取施主能级在导带底下的面的0.05eV。1930103016163191818319191932.810/ln,300,1.510/ln,1010/;0.026ln0.212.8101010/;0.026ln0.0872.8101010/;0.026lnCDFcCiDFiiDFccDFccDFcNcmNEEkTTKNncmNEEkTNNcmEEEeVNcmEEEeVNcmEE时或00190.0272.810(2)0.0590%,10%110%112190%12DFDFcCDDEEDkTDEEDkTEeVEEeVnNenNe施主杂质全部电离标准为占据施主是否或没有全部电离全部电离小于质数的百分比)未电离施主占总电离杂全部电离的上限求出硅中施主在室温下)(不成立不成立成立317191831716317026.005.00'026.0023.019026.0037.018026.016.0026.021.0161005.210,101005.210/1005.2exp21.0,026.005.0exp2%10(exp)2(2%10%9.822111:10%5.322111:10%42.021112111:10cmNcmNcmNNNNTkENNDeNnNeNnNeeNnNDDCDCDDCDDDDDDDEEDDDCD10.以施主杂质电离90%作为强电离的标准,求掺砷的n型锗在300K时,以杂质电离为主的饱和区掺杂质的浓度范围。11.若锗中施主杂质电离能ED=0.01eV,施主杂质浓度分别为ND=1014cm-3及1017cm-3。计算①99%电离;②90%电离;③50%电离时温度各为多少?[解]未电离杂质占的百分比为:DDDDNNcDTkETkENcND2_lnexp2_00=;求得:TTTkED116106.11038.101.019230;323153230*/)(102)2(2cmThTkmNcn∴)_10ln()2102_ln(2_ln11623152315TDNNTDNNcDTDDD(1)ND=1014cm-3,99%电离,即D_=1-99%=0.013.2ln23)10ln(116231TTT之上,大部分没有电离在,之下,但没有全电离在成立,全电离全电离,与也可比较)(0DFFDDDFFDDFDDFDFDEEEEcmNEEEEcmNEEcmNTkEEEE026.0023.0;/10026.0~037.0;/10026.016.021.005.0;/102319318316''19300.01270.0127191730.0260.0260.0127,1.0510/3002exp()20.012710%exp0.0260.10.11.05103.2210/22DCDDCDCCDDiAsEeVNcmKAsNEDNkTNNNNeecmAsNGen上限上限的电离能室温以下,杂质全部电离的掺杂上限掺杂浓度超过的部分,在室温下不能电离的本征浓度133141732.410/5~2.410~3.2210/iDcmAsnNcm上限的掺杂浓度范围,即有效掺杂浓度为即:3.2ln23116TT,T=37.2K将ND=1017cm-3,D_=0.01代入得:10ln4ln2310ln116234TTT即:2.9ln23116TT,T=536K(2)90%时,D_=0.131410cmNDDDNNcTkE21.0ln02314231510ln21021.0ln116TNTNTDD即:TTln23116,T=24.3KND=1017cm-3得:10ln3ln23116TT即:9.6ln23116TT;T=160.5K(3)50%电离不能再用上式∵2DDDNnn即:)exp(21)exp(21100TkEENTkEENFDDFDD∴)exp(4)exp(00TkEETkEEFDFDTkEETkEEFDFD004ln即:2ln0TkEEDF2)exp(00DFcNTkEENcn取对数后得:NcNTkTkEEDDC2ln2ln00整理得下式:NcNTkEDD2l
本文标题:第三章习题和答案
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