Determine the Fermi energy level and the maximum doping concentration at which the Boltzmann approximation is still valid.
Consider p-type silicon, at T=300 \mathrm{~K}, doped with boron. We may assume that the limit of the Boltzmann approximation occurs when E_{F}-E_{a}=3 k T. (See Section 4.1.2.)
From Table 4.3, we find the ionization energy is E_{a}-E_{v}=0.045 \mathrm{eV} for boron in silicon. If we assume that E_{F i} \approx E_{\text {midgap }}, then from Equation (4.68), the position of the Fermi level at the maximum doping is given by
| Table 4.3 | Impurity ionization energies in silicon and germanium |
||
| Ionization energy (eV) | ||
| Impurity | Si | Ge |
| Donors | ||
| Phosphorus | 0.045 | 0.012 |
| Arsenic | 0.05 | 0.0127 |
| Acceptors | ||
| Boron | 0.045 | 0.0104 |
| Aluminum | 0.06 | 0.0102 |
E_{F i}-E_{F}=k T \ln \left(\frac{p_{0}}{n_{i}}\right) (4.68)
E_{F i}-E_{F}=\frac{E_{g}}{2}-\left(E_{a}-E_{v}\right)-\left(E_{F}-E_{a}\right)=k T \ln \left(\frac{N_{a}}{n_{i}}\right)
or
0.56-0.045-3(0.0259)=0.437=(0.0259) \ln \left(\frac{N_{a}}{n_{i}}\right)
We can then solve for the doping as
N_{a}=n_{i} \exp \left(\frac{0.437}{0.0259}\right)=3.2 \times 10^{17} \mathrm{~cm}^{-3}
Comment
If the acceptor (or donor) concentration in silicon is greater than approximately 3 \times 10^{17} \mathrm{~cm}^{-3}, then the Boltzmann approximation of the distribution function becomes less valid and the equations for the Fermi-level position are no longer quite as accurate.