半导体物理与器件第四版课后习题答案9

Semicondu ctor Phy sics and Dev ices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solu tions ______________________________________________________________________________________ 1 Chapter 9 9.1 (a) We hav e dctnNNeVeln 206.010108.2ln0259.01619eV (c) 01.428.4mBO or 27.0BOV and 206.027.0nBObiV or 064.0biVV Also 2/12 dbisdeNVx  2/116191410106.1064.01085.87.112 or 6101.9dxcm Then sdd xeNmax  14616191085.87.1110`1.910106.1 or 4max1041.1V/cm (d) Using the figu re, 55.0BnV So 206.055.0nBnbiV or 344.0biVV We then find 51011.2nxcm and 4max1026.3V/cm _______________________________________ 9.2 (a) nBbiV0 dctnNNV ln 1519105108.2ln0259.0 =0.2235 V 4265.02235.065.0biVV (b) 161910108.2ln0259.0n 2056.0V 4444.02056.065.0biVV biV increases, 0Bremains constant (c) 151910108.2ln0259.0n 2652.0V 3848.02652.065.0biVV biV decreases, 0Bremains constant _______________________________________ 9.3 (a) 09.101.41.50mBV (b) nBbiV0 dctnNNV ln 2056.010108.2ln0259.01619V 8844.02056.009.1biVV (c) 2/12dRbisneNVVx (i) 2/116191410106.118844.01085.87.112nx 510939.4cm Semicondu ctor Phy sics and Dev ices: Basic Principles, 4th edition Chapter 9 By D. A. Neamen Problem Solu tions ______________________________________________________________________________________ 2 or 4939.0nxm snd xeNmax  14516191085.87.1110939.410106.1...