Determine the thermal-equilibrium electron and hole concentrations in silicon at T=300 \mathrm{~K} for given doping concentrations. (a) Let N_{d}=10^{16} \mathrm{~cm}^{-3} and N_{a}=0 . (b) Let N_{d}=5 \times 10^{15} \mathrm{~cm}^{-3} and N_{a}=2 \times 10^{15} \mathrm{~cm}^{-3} .
Recall that n_{i}=1.5 \times 10^{10} \mathrm{~cm}^{-3} in silicon at T=300 \mathrm{~K}.
(a) From Equation (4.60), the majority carrier electron concentration is
\begin{array}{c}n_{0}=\frac{\left(N_{d}-N_{a}\right)}{2}+\sqrt{\left(\frac{N_{d}-N_{a}}{2}\right)^{2}+n_{i}^{2}} \\ \end{array} (4.60)
\begin{array}{c}n_{0}=\frac{10^{16}}{2}+\sqrt{\left(\frac{10^{16}}{2}\right)^{2}+\left(1.5 \times 10^{10}\right)^{2}} \cong 10^{16} \mathrm{~cm}^{-3}\end{array}
The minority carrier hole concentration is found to be
p_{0}=\frac{n_{i}^{2}}{n_{0}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{10^{16}}=2.25 \times 10^{4} \mathrm{~cm}^{-3}
(b) Again, from Equation (4.60), the majority carrier electron concentration is
\begin{array}{c}n_{0}=\frac{\left(N_{d}-N_{a}\right)}{2}+\sqrt{\left(\frac{N_{d}-N_{a}}{2}\right)^{2}+n_{i}^{2}} \\ \end{array} (4.60)
n_{0}=\frac{5 \times 10^{15}-2 \times 10^{15}}{2}+\sqrt{\left(\frac{5 \times 10^{15}-2 \times 10^{15}}{2}\right)^{2}+\left(1.5 \times 10^{10}\right)^{2}} \cong 3 \times 10^{15} \mathrm{~cm}^{-3}
The minority carrier hole concentration is
p_{0}=\frac{n_{i}^{2}}{n_{0}}=\frac{\left(1.5 \times 10^{10}\right)^{2}}{3 \times 10^{15}}=7.5 \times 10^{4} \mathrm{~cm}^{-3}
Comment
In both parts of this example, \left(N_{d}-N_{a}\right) \gg n_{i}, so the thermal-equilibrium majority carrier electron concentration is essentially equal to the difference between the donor and acceptor concentrations. Also, in both cases, the majority carrier electron concentration is orders of magnitude larger than the minority carrier hole concentration.