半导体物理与器件第四版课后习题答案4

1、 . . 17/17Chapter 44.1where and are the values at 300 K. (a) Silicon(K)(eV)(cm) (b) Germanium (c) GaAs(K)(cm)(cm)_4.2 Plot_4.3By trial and error, K(b)By trial and error, K_4.4 At K, eV At K, eV or or eV Now so cm_4.5 For K, eV For K, eV For K, eVFor K,For K,For K,_4.6 Let Then To find the maximum va

2、lue: which yields The maximum value occurs at (b) Let Then To find the maximum value Same as part (a). Maximum occurs at or_4.7 where and Then or_4.8 Plot_4.9 Plot_4.10 Silicon: , eV Germanium: , eV Gallium Arsenide: , eV_4.11(K)(eV)()(eV)_4.12meVmeV_4.13 Let constant Then Let so that We can write s

3、o that The integral can then be written as which becomes_4.14 Let for Then Let so that We can write Then or We find that So_4.15 We have For germanium, , Then or The ionization energy can be written aseVeV_4.16 We have For gallium arsenide, Then The ionization energy is oreV_4.17eVeVcmHoleseV_4.18eV

4、eVcmeV_4.19eVeVcmp-type_4.20eVcmeVcmeVeVcm_4.21eV cmeVcmeVeVcm_4.22p-typeeVcmeVcm_4.23cmcmcmcm_4.24eVeVcmHoleseV_4.25eVcmcmcmeVeVcmHoleseV_4.26cmeVcmeVcmcmeVeVcm_4.27cmeVcmeVcmcmeVeVcm_4.28 (a) For , Thencm (b) cm_4.29 So We find eV_4.30 (a) Then cm (b) cm_4.31 For the electron concentration The Bol

5、tzmann approximation applies, so or Define Then To find maximum , set or which yields For the hole concentration Using the Boltzmann approximation or Define Then To find maximum value of , set Using the results from above, we find the maximum at_4.32Silicon: We have We can write For eV and eV we can

6、 write orcm We also have Again, we can write For and eV Then orcmGaAs: assume eVThen orcm Assume eV Then orcm_4.33 Plot_4.34cmcmcmcmcmcmcmcmcmcmcm_4.35cmcmcmcmcmcmcmcmcmcmcm_4.36Ge: cm (i)orcm cm (ii)cmcmGaAs: cm (i) cmcm (ii)cmcm(c) The result implies that there is only one minority carrier in a vo

7、lume of cm._4.37 (a) For the donor level or (b) We have Now or Then or_4.38p-typeSilicon: orcm Thencm Germanium: orcm Then cm Gallium Arsenide:cm andcm_4.39n-typecmcmcmcm_4.40cmn-type_4.41cm So cm, Then cm so that cm_4.42 Plot_4.43 Plot_4.44 Plot_4.45 so cmcm_4.46p-type Majority carriers are holescm

8、 Minority carriers are electronscmBoron atoms must be added So cmcm_4.47n-typecmelectrons are majority carrierscmholes are minority carriers so cm_4.48 For Germanium(K) (eV)(cm) and cm(K) (cm)(eV)_4.49For cm, eVcm, eVcm, eVcm, eVFor cm, eVcm, eVcm, eVcm, eV_4.50cm so Now By trial and error, KAt K,eV

9、 At K,eVcmeV then eVCloser to the intrinsic energy level._4.51 At K, eVK, eVK, eV At K,cm At K,cm At K,cm At K and K, cm At K,cm Then, K, eVK, eVK, eV_4.52 (a) For cm, eVcm, eVcm, eVcm, eV (b) For cm, eVcm, eVcm, eVcm, eV_4.53 oreVImpurity atoms to be added soeV (i) p-type, so add acceptor atoms (ii) eV Then orcm_4.54 so orcm_4.55Silicon(i)eV(ii)eVcmcm Additional donor atomsGaAs (i)eV (ii)eVcmcm Additional donor atoms_4.56eVeVFor part (a);cmcm For part (b):cmcm_4.57cm Add additional acceptor impuritiescm_4.58eVeVeVeV_4.59eVeVeVeVeV_4.60 n-typeeV_4.61eVcmcm Now SoeV_