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A semiconductor has an electron concentration of 0.45 xx 10^(12)m^(-3)and a hole concentration of 5.0 xx 10^(20) m^(-3) Calculate its conductivity. Given electron mobility =0.135 m^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_ee mu _e + n_he mu_h= e ( n_e mu_e+n_hmu_h)` As per given date, N is negligible as compared to n, so that we can write ` sigma= e n_hmu_h ` `= ( 1.6 xx 10^(-19)C )/ ( 5.0 xx 10^(20) m^(-3) ) ( 0.048 m^2 V^(-1)S^(-1))` `= 3.84 Omega^(-1)m^(-1)= 3.84S m^(-1)` where S (siemen) stands for `Omega^(-1)` |
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