Question

Please clearly answer the following question showing all work, each step, and the reasoning thereof.

16.1-7 Fermi Level of an Intrinsic Semiconductor. Given the expressions (16.1-12) and (161 13) for the thermal equilibrium carrier concentrations in the conduction and valence bands: (a) Determine an expression for the Fermi level E of an intrinsic semiconductor and show that it falls exactly in the middle of the bandgap only when the effective mass of the electrons me is precisely equal to the effective mass of the holes me. (b) Determine an expression for the Fermi level of a doped semiconductor as a function of the doping level and the Fermi level determined in (a).

Answer 1

Fermi level in intrinsic Semiconductor we know that for an intrinsic semiconductors No. of elections - No of holes in - P Hole concentration & Electron concentration n = -CEC Nee KT P no-CEF-EV) = Nye KT – Ep + Ey the EF e KP ve - Ect Ex - 2 Ef KT Ectv – 28 logne = KT in Ne - Eater-268 KT EF 2 KT No Ecto - 21 - No ECTE, -26 - KT LINE FF - Feteo - Kenmore - Eetru _ k l m n o No and Ny are material constcents

we know that Ne = material constant and is a function, of temp Neo 482x16 (mg) 3/2 3/2 me effectère mass of election Ny = 4.82 x 101 me ] 312 me effective mass of electron it me -me, then we=Nr . put @n0 = ECHEV – KIRA mer – fetEV – KT en un en - ECTEN - Kenci) - ECTEU – ITX (0) EF - Ecteu no E = ECTEU Ferma levelet falls exactly in the middle of the bandgap only

b) Perioni level in N-type Semiconductor (dopred Semiconductor In N-type N=N No Doping concentration Mee (EET) - No No = e RT In Ne - ECE KT Mb KT EC-EF - en re E-EF = ket lanter T EC-KINONE Here fermiterel is a femetion of doping level. se doping concentration and temperature