Charge neutrality with partial donor ionization ($n-n_i^2/n=N_D/[1+(2n/N_c)e^{E_b/(k_{B}T)}]$)

For a nondegenerate semiconductor with donor density $N_D$, no acceptors, and donor energy $E_D=E_c-E_b$, an occupied donor has two electron spin states while an empty donor has one. Its ionized fraction is therefore $N_D^{+}/N_D=[1+2e^{(E_F-E_D)/(k_{B}T)}{]}^{-1}$. The conduction-band relation $n=N_c{e}^{(E_F-E_c)/(k_{B}T)}$ gives $N_D^{+}=N_D/[1+(2n/N_c)e^{E_b/(k_{B}T)}]$. Combining charge neutrality, $n-p=N_D^{+}$, with the semiconductor mass-action law, $np=n_i^2$, proves the displayed equation.

Its left side increases strictly for $n>0$ and its right side decreases, so there is one physical solution, with $n\ge n_i$. In the freeze-out regime, holes are negligible and solving the quadratic gives $n=2N_D/[1+\sqrt{1+(8N_D/N_c)e^{E_b/(k_{B}T)}}]$. Strong dopant freeze-out reduces this to $n\simeq \sqrt{N_D{N}_c/2}{e}^{-E_b/(2k_{B}T)}$. An extrinsic plateau requires $(2N_D/N_c)e^{E_b/(k_{B}T)}\ll 1$ and $n_i\ll N_D$, not simply $k_{B}T\gg E_b$.