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In an intrinsic semiconductor the energy gapE_(g )is1.2 eV. Its hole mobility is much smaller than electron and independent of temperature. What is the ratio between conductivity at 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration n_(i)is givenby n_(i) = n_(o) exp (- ( E_(g))/( 2K_(B)T )T) where n_(0) is a constant . |
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Answer» Solution :Given , INTRINSIC carrier CONCENTRATION `n_(i)=n_(0)e^(-Eg//2k_(B)T)` and Energy gap`E_(g) = 1.2 eV`. `K_(B) = 8.62 xx 10^(-5)eV //K ` For` T=600 k ` `n_(600) =n_(0) e^(-Eg//2k_(B)XX600)` ...(1) For `T=300K` `n_(300) =n_(0) e^(-Eg//2kB xx300)` ...(2) Dividing EQN (1) by Eq(2) we get `(n_(600))/( n_(300))= e^([(E_(g))/(2K.B)((1)/(600) -(1)/( 300))])= e^((E_(g))/(2K.B)((1)/(300)- (1)/(600)))` `= e^((1.2)/( 2 xx8.62 xx10^(-5))((1)/(600)))``( :'e=2.718) =e^(11.6)=(2.718)^(11.6) =1.1 xx 10^(5)` Let the condcutivity are `sigma_(600)` and `SIGMA _(300)` `( :' sigma=enmuc)` `( sigma_(600))/(sigma_(300))= ( n_(600))/( n _(300))=1.1xx 10^(5)` |
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