Question

In a P-Type semiconductor, the Fermi level in 0.3 eV above the valance band at a room temperature of 300 oK. Determine the new position of the Fermi level for temperature of a (350 oK) and (b) 400 oK.

Answer 1

$E_F=E_V+KT\ln(\frac{N_v}{N_A})\\ E_F-E_V=0.3eV\\ T=300\degree{K}\\ E_F-E_V=KT\ln(\frac{N_v}{N_A})\\ 0.3=300K\ln(\frac{N_v}{N_A})\\ K\ln(\frac{N_c}{N_A})=\frac{0.3}{300}\\ K\ln(\frac{N_c}{N_A})=0.001\\ (a)\\ T=350\degree{K}\\ E_F-E_V=KT\ln(\frac{N_v}{N_A})\\ E_F-E_V=350K\ln(\frac{N_v}{N_A})\\ E_F-E_V=350\times 0.001\\ E_F-E_V=0.35eV\\ (b)\\ T=400\degree{K}\\ E_F-E_V=KT\ln(\frac{N_v}{N_A})\\ E_F-E_V=400K\ln(\frac{N_v}{N_A})\\ E_F-E_V=0.40eV$