 Solution

Assume uncompensated donor impurities, negligible intrinsic electron holes, and a nondegenerate conduction band. Then charge neutrality equates the conduction electron number density to the ionized-donor number density:The low-temperature condition relative to the band gap supports neglect of intrinsic carriers; one also needs $n_i\ll n$ and the Boltzmann approximation for the conduction electrons. Combining part (b)(i) with the effective density of states result givesEliminate the chemical potential by writing $e^{\beta (\mu -{ϵ}_D)}=(n/N_c)e^{\beta ({ϵ}_c-{ϵ}_D)}$. With the donor binding energy $E_i={ϵ}_c-{ϵ}_D>0$,Writing $A(T)=N_c{e}^{-\beta E_i}$, the physical root of the resulting quadratic equation isThis lies between zero and $N_D$. At semiconductor carrier freeze-out, $A\ll N_D$, it becomes $n\simeq \sqrt{N_D{N}_c}{e}^{-E_i/(2k_{B}T)}$; at essentially complete ionization, $A\gg N_D$, it tends to $N_D$. Including donor charge-state degeneracy would replace $A$ by $A{g}_{\mathrm{empty}}/g_{\mathrm{full}}$.