半导体物理与器件第四版课后习题答案5

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 Chapter 5 5.1 (a) 1519101300106.111dn Ne 808.4-cm (b) 208.08077.411 (  -cm)1 _______________________________________ 5.2 ap Ne  or 380106.180.119paeN 161096.2cm3 _______________________________________ 5.3 (a) dn Ne  dn N19106.110 From Figure 5.3, for 16106dNcm3 we find 1050ncm 2 /V-s which gives 16191061050106.1 08.10(  -cm)1 (b) ap Ne1 ap N19106.1120.0 From Figure 5.3, for 1710aNcm3 we find 320pcm 2 /V-s which gives 195.010320106.111719 -cm _______________________________________ 5.4 (a) ap Ne1 ap N19106.1135.0 From Figure 5.3, for 16108aNcm3 we find 220pcm 2 /V-s which gives 1619108220106.11 355.0 -cm (b) dn Ne  dn N19106.1120 From Figure 5.3, for 17102dNcm3 , then 3800ncm 2 /V-s which gives 17191023800106.1 6.121(  -cm)1 _______________________________________ 5.5 ANeLALALRdn or RAeNLdn    1.070102106.15.21519 1116cm 2 /V-s _______________________________________ 5.6 (a) 1610doNncm3 and 4162621024.310108.1oionnpcm3 (b) on neJ For GaAs doped at 1610dNcm3 , 7500ncm 2 /V-s Then  10107500106.11619J or 120JA/cm 2 (b) (i) 1610aoNpcm3 Semicondu ctor Phy sics and Dev ices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solu tions ______________________________________________...