Determine the required donor impurity concentration to obtain a specified Fermi energy.
Silicon at T=300 \mathrm{~K} contains an acceptor impurity concentration of N_{a}=10^{16} \mathrm{~cm}^{-3}. Determine the concentration of donor impurity atoms that must be added so that the silicon is \mathrm{n} type and the Fermi energy is 0.20 \mathrm{eV} below the conduction-band edge.
From Equation (4.64), we have
E_{c}-E_{F}=k T \ln \left(\frac{N_{c}}{N_{d}}\right) (4.64)
E_{c}-E_{F}=k T \ln \left(\frac{N_{c}}{N_{d}-N_{a}}\right)
which can be rewritten as
N_{d}-N_{a}=N_{c} \exp \left[\frac{-\left(E_{c}-E_{F}\right)}{k T}\right]
Then
N_{d}-N_{a}=2.8 \times 10^{19} \exp \left[\frac{-0.20}{0.0259}\right]=1.24 \times 10^{16} \mathrm{~cm}^{-3}
or
N_{d}=1.24 \times 10^{16}+N_{a}=2.24 \times 10^{16} \mathrm{~cm}^{-3}
Comment
A compensated semiconductor can be fabricated to provide a specific Fermi energy level.