Calculate the thermal equilibrium concentrations of electrons and holes for a given Fermi energy.
Consider silicon at T=300 \mathrm{~K} so that N_{c}=2.8 \times 10^{19} \mathrm{~cm}^{-3} and N_{v}=1.04 \times 10^{19} \mathrm{~cm}^{-3}. Assume that the Fermi energy is 0.25 \mathrm{eV} below the conduction band. If we assume that the bandgap energy of silicon is 1.12 \mathrm{eV}, then the Fermi energy will be 0.87 \mathrm{eV} above the valence band.
Using Equation (4.11), we have
\begin{array}{c}n_{0}=N_{c} \exp \left[\frac{-\left(E_{c}-E_{F}\right)}{k T}\right] \\ \end{array} (4.11)
\begin{array}{c}n_{0}=\left(2.8 \times 10^{19}\right) \exp \left(\frac{-0.25}{0.0259}\right)=1.8 \times 10^{15} \mathrm{~cm}^{-3}\end{array}
From Equation (4.19), we can write
\begin{array}{c}p_{0}=N_{v} \exp \left[\frac{-\left(E_{F}-E_{v}\right)}{k T}\right] \\ \end{array} (4.19)
p_{0}=\left(1.04 \times 10^{19}\right) \exp \left(\frac{-0.87}{0.0259}\right)=2.7 \times 10^{4} \mathrm{~cm}^{-3}
Comment
The change in the Fermi level is actually a function of the donor or acceptor impurity concentrations that are added to the semiconductor. However, this example shows that electron and hole concentrations change by orders of magnitude from the intrinsic carrier concentration as the Fermi energy changes by a few tenths of an electron-volt.