Compensated semiconductor: electron and hole concentrations

When both donors and acceptors are present, they compensate each other and the larger one decides the type.

Given: $N_d = 10^{16}$ cm$^{-3}$ (P), $N_a = 10^{17}$ cm$^{-3}$ (B). Assume Si at 300 K, $n_i = 1.45\times10^{10}$ cm$^{-3}$, full ionisation.

Since $N_a > N_d$, the sample becomes p-type. The donors' electrons fill $10^{16}$ acceptors; the rest of the acceptors create holes.

$$\begin{aligned} p &= N_a - N_d = 10^{17} - 10^{16} = 9\times10^{16}\ \text{cm}^{-3} \\ n &= \frac{n_i^2}{p} = \frac{(1.45\times10^{10})^2}{9\times10^{16}} = 2.34\times10^{3}\ \text{cm}^{-3} \end{aligned}$$

Answer: $p = 9\times10^{16}$ cm$^{-3}$, $n \approx 2.3\times10^{3}$ cm$^{-3}$ (the sample is now p-type).