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.
EF=EV+KTln(NvNA)EF−EV=0.3eVT=300°KEF−EV=KTln(NvNA)0.3=300Kln(NvNA)Kln(NcNA)=0.3300Kln(NcNA)=0.001(a)T=350°KEF−EV=KTln(NvNA)EF−EV=350Kln(NvNA)EF−EV=350×0.001EF−EV=0.35eV(b)T=400°KEF−EV=KTln(NvNA)EF−EV=400Kln(NvNA)EF−EV=0.40eVE_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=0.40eVEF=EV+KTln(NANv)EF−EV=0.3eVT=300°KEF−EV=KTln(NANv)0.3=300Kln(NANv)Kln(NANc)=3000.3Kln(NANc)=0.001(a)T=350°KEF−EV=KTln(NANv)EF−EV=350Kln(NANv)EF−EV=350×0.001EF−EV=0.35eV(b)T=400°KEF−EV=KTln(NANv)EF−EV=400Kln(NANv)EF−EV=0.40eV
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