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A semiconductor has an electron concen- tration of 0.45xx10^(12)m^(-3) and a hole concentration of 5.0xx10^(20)m^(-3). Calculate its conductivity. Given electron mobility =0.135M^(2)v^(-1)S^(1), hole mobility = 0.048m^(2)v^(-1)S^(-1). |
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Answer» Solution :The conductivity of a semiconductor is the sum of the conductivities due to electrons and holes and is given by `SIGMA= sigma_(e)+ sigma_(h)=n_(e)emu_(e)+n_(h)emu_(h)=e(n_(e)mu_(e)+n_(h)mu_(h))` As per given DATE, `n_(e)` is negligible as compared to `n_(h),` so that we can write ` sigma=en_(h)mu_(h)` `=(1.6xx10^(-19)C)(5.0xx10^(20)m^(-3))(0.048m^(2)V^(-1)s^(-1))` `=3.84Omega^(-1)m^(-1)=3.84Sm^(-1)` where S (siemen) stands for `Omega^(-1)` |
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