= Charge neutrality with partial donor ionization
{title2=$n-n_i^2/n=N_D/[1+(2n/N_c)e^{E_b/(k_BT)}]$}
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 $N_D^+/N_D=[1+2e^{(E_F-E_D)/(k_BT)}]^{-1}$. The conduction-band relation $n=N_ce^{(E_F-E_c)/(k_BT)}$ gives $N_D^+=N_D/[1+(2n/N_c)e^{E_b/(k_BT)}]$. 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 $n\geq n_i$. In the freeze-out regime, holes are negligible and solving the quadratic gives $n=2N_D/[1+\sqrt{1+(8N_D/N_c)e^{E_b/(k_BT)}}]$. Strong <dopant freeze-out> reduces this to $n\simeq\sqrt{N_DN_c/2}\,e^{-E_b/(2k_BT)}$. An extrinsic plateau requires $(2N_D/N_c)e^{E_b/(k_BT)}\ll1$ and $n_i\ll N_D$, not simply $k_BT\gg E_b$.
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