Calculate the thermal-equilibrium electron and hole concentrations in a compensated p-type semiconductor.
Consider a silicon semiconductor at T=300 \mathrm{~K} in which N_{a}=10^{16} \mathrm{~cm}^{-3} and N_{d}=3 \times 10^{15} \mathrm{~cm}^{-3}. Assume n_{i}=1.5 \times 10^{10} \mathrm{~cm}^{-3}.
Since N_{a}>N_{d}, the compensated semiconductor is p-type and the thermal-equilibrium majority carrier hole concentration is given by Equation (4.62) as
\begin{array}{c} p_{0}=\frac{N_{a}-N_{d}}{2}+\sqrt{\left(\frac{N_{a}-N_{d}}{2}\right)^{2}+n_{i}^{2}} \\ \end{array} (4.62)
\begin{array}{c}p_{0}=\frac{10^{16}-3 \times 10^{15}}{2}+\sqrt{\left(\frac{10^{16}-3 \times 10^{15}}{2}\right)^{2}+\left(1.5 \times 10^{10}\right)^{2}} \end{array}
so that
p_{0} \approx 7 \times 10^{15} \mathrm{~cm}^{-3}
The minority carrier electron concentration is
n_{0}=\frac{n_{i}^{2}}{p_{0}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{7 \times 10^{15}}=3.21 \times 10^{4} \mathrm{~cm}^{-3}
Comment
If we assume complete ionization and if \left(N_{a}-N_{d}\right) \gg n_{i}, then the majority carrier hole concentration is, to a very good approximation, just the difference between the acceptor and donor concentrations.