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How does conductivity of a semi conducto...

How does conductivity of a semi conductor change with the rise in temperature ?

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In n-type (extrinsic) semiconductor, the number density of electrons in conduction band is approximately equal to that of donor atoms and very large as compared to number density of holes in valence band. Thus,
`n_(e) ~~ N_(d) gt gt n_(h)` …(i)
where `N_(d)` represents the number density of donor atoms. If `n_(e)` is the number density of conduction band electrons in the semiconductor and `v_(e)` the drift velocity of electrons, then `I_(e)= e n_(e) A v_(e)` ...(ii)
Similarly, the hole current `I_(h)= e n_(h) A v_(h)` ...(iii)
Using equations (ii) and (iii), the equation (i) becomes `I= e n_(e) A v_(e) + e n_(h) A v_(h)`
or `I= e A (n_(e) v_(e) + n_(h) v_(h))` ...(iv)
If `rho` is the resistivity of the material of the semiconductor, then the resistance offered by the semiconductor to the flow of current is given by
`R= rho (I)/(A)`
Since V= RI, from equations (iv) and (v), we have `V= RI = rho (I)/(A) xx e A(n_(e) v_(e) + n_(h) v_(h))` ...(vi)
or `V= rho I e (n_(e) (v_(e) + n_(h) (v_(h))))`
If E is the electric field set up across the semiconductor, then `E= (V)/(I)`
From equations (vi) and (vii), we have `E= rho e (n_(e) (v_(e) + n_(h) (v_(h))))`
or `(1)/(rho) = e (n_(e) (v_(e))/(E ) + n_(h) (v_(h))/(E ))`
On applying electric field, the dirft velocity acquired by the electrons (or holes) per unit strength of electric field is called mobility of electrons (or holes). Therefore, mobility of electrons and holes is given by
`mu_(e )= (v_(e ))/(E ) and mu_(h)= (v_(h))/(E)`
`therefore (1)/(rho) = e(n_(e) mu_(e ) + n_(h) mu_(h))`
Also `sigma = (1)/(rho)` is called conductivity of the matrial of semiconductor.
`therefore sigma = e (n_(e) mu_(e) + n_(h) mu_(h))`
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