半导体物理课后习题(3)

由

n0?nD??ND?ED?EF1?2exp???kT0??????ND?EC?ED??EC?EF??1?2exp?exp?kT???kT00???

????得到ND=6.59?1019cm?3

20. 解：

① 由于Ec?EF?k0T，所以此半导体处于弱间并，此时

n0?Nc2?EF?EcF1/2??kT?0??221918?3????NF?1?2.8?10??0.3?9.48?10cmc1/2???????又n0?nD?ND?EF?ED1?2exp??kT0?????，所以

??EF?ED?ND?n0?1?2exp?kT?0???????E?ED???Ec?EF?????n0?1?2exp?c?????????k0T??????

??0.039?0.026??19?3?9.48?1018??1?2?exp??4.07?10cm????0.026??????② 由于杂质浓度不很高，半导体处于非间并状态，n型杂质全电离，

ni?1.5?1010cm?3，本征激发可以忽略，所以

n0?ND?4.6?1015cm?3

ni21.5?10104?3 p0???4.9?10cm15n04.6?10?Ei?EF又 n0?niexp???kT0????，所以 ???2?n0??4.6?1015?EF?Ei?k0Tln??1.5?1010???Ei?0.33eV ?n??＝Ei?0.026?ln????i??Ec?EF或n0?Ncexp???kT0????，所以 ???2.8?1019???4.6?1015???Ec?0.23eV ?＝Ec?0.026?ln?????NcEF?Ec?k0Tln??n?0③ 此时ND?4.6?1015cm?3、NA?5.2?1015cm?3，掺杂浓度不是很高，半导体处于非间并状态，杂质全电离，并且有杂质补偿作用，ni?1.5?1010cm?3，本征激发可以忽略，所以

p0?NA?ND?6?1014cm?3

ni21.5?1010n0??p06?1014??2?3.75?105cm?3

?p0??6?1014?EF?Ei?k0Tln??1.5?1010???Ei?0.28eV ?n??＝Ei?0.026?ln????i?或

?NvEF?Ev?k0Tln??p?0??1.1?1019??6?1014?＝Ev?0.026?ln???????Ev?0.26eV ?④ 当温度升高到500K时，查表得ni?4?1014cm?3，本征激发不可忽略，此时

N?ND?NA?ND??4np0?A?22?8?1014cm?32?2i?126?106?10??214??142??4?4?102?142??12

ni24?101414?3 n0???2?10cm14p08?10

??221. 解：

由电中性条件n0=nD+

即

n0?2?EC?EFNcF1/2???k0T???ND???n?D??ED?EF?1?2exp???kT0??????ND?EC?ED??EC?EF??1?2exp?exp?kT???kT00???????因为弱简并条件是0?EC?EF?2k0T 所以开始发生弱简并时有EC?EF?2k0T 所以：?? 所以：

EF?EC??2,F1/2????0.1 k0T??EC?ED??EC?EF??Nc?1?2exp?exp????kTk0T?0??????2?0.044??Nc?1?2exp??exp??2???0.1??0.026???ND?217???EC?EF??F??1/2???kT0???????

?1017??0.01?300?10??1?2exp??1.66?1017c??18??77?0.026?1.37?10??对于硅ND???2?0.044?18?3?2.8?1019?1?2exp??exp??2???0.1?7.81?10cm 3.14?0.026?????2?0.012?18?3?1.05?1019?1?2exp??exp??2???0.1?1.69?10cm 3.14?0.026???对于锗ND?

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