The electron concentration in an $n$-type semiconductor is the same as hole concentration in a $p$-type semico

Question

The electron concentration in an $n$-type semiconductor is the same as hole concentration in a $p$-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.

Accepted Answer

The current through a semiconductor is $I = neAv_d = neA\mu E$. The ratio of currents is $\frac{I_n}{I_p} = \frac{n_e e A \mu_e E}{n_h e A \mu_h E} = \frac{\mu_e}{\mu_h}$. Since $\mu_e > \mu_h$, therefore $I_n > I_p$.

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