ECE340_L15_S14_Distribution

# Fermi level fp will generally depend upon me t unless

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Unformatted text preview: = no = 1014 = ni e 2 i ( E f − Ei )/ kT 20 = (1.5 × 1010 )e ( E f − Ei )/ kT E f − Ei = 0.0259 ln(6.7 × 10 3 ) = 0.228 eV 4 ) τ n = τ p = 2 µ sec Therefore δ n=δ p=g optτ n = 2 × 1013 5 ) n=n o + δ n = 1.2 × 1014 cm −3 = (1.5 × 1010 )e( Fn − Ei )/kT , kT=0.0259 eV Therefore Fn − E i = 0.0259 ln(8 × 10 3 ) = 0.233eV 6 ) p=p o + δ p = 2 × 1013 cm −3 = (1.5 × 1010 )e ( Ei − Fp )/ kT Therefore E i − Fp = 0.0259 ln(1.3 × 10 3 ) = 0.186 eV Note: np≠ni2 when excess carriers are present! 14 Fermi Level Comments •  The separa\on of the electron and hole quasi Fermi levels (Fn- Fp) provides a metric for the devia\on from equilibrium •  The individual devia\on Fn and Fp from EF is a measure of how far the electron and hole popula\ons are from equilibrium •  Depic\ng the quasi Fermi Levels on a band diagram gives a good representa\on of how a sample is locally devia\ng from equilibrium )( ( E − F / kT np = ni e( Fn − Ei )/kT ni e( i p ) E − F / kT = ni2 e( Fn − Ei )/kT e( i p ) 2 ( Fn − Ei + Ei − Fp )/ kT =n e i F − F / kT = ni2 e( n p ) Valid for carrier densities such that quasi-Fermi level is 3kT away from band edge 15 ) Posi\on and Time Dependence •  The quasi Fermi level can depend upon Fn posi\on (x) •  The quasi Fermi level Fp will generally...
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