Determin e the Fermi energy level and the maximum doping concentration at which the Boltzmann approximation is still valid. Consider p-type silicon, at T = 300 K, doped with boron. We may assume that the limit of the Boltzmann approximation occurs when EF - Ea = 3kT. (See Section 4.1.2.)

Question

Determine the Fermi energy level and the maximum doping concentration at which the Boltzmann approximation is still valid.

Consider p-type silicon, at [latex]T=300 \mathrm{~K}[/latex], doped with boron. We may assume that the limit of the Boltzmann approximation occurs when [latex]E_{F}-E_{a}=3 k T[/latex]. (See Section 4.1.2.)

Accepted Answer

From Table [latex]4.3[/latex], we find the ionization energy is [latex]E_{a}-E_{v}=0.045 \mathrm{eV}[/latex] for boron in silicon. If we assume that [latex]E_{F i} \approx E_{ ext {midgap }}[/latex], 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

[latex]E_{F i}-E_{F}=k T \ln \left(\frac{p_{0}}{n_{i}} ight)[/latex] (4.68)

[latex]E_{F i}-E_{F}=\frac{E_{g}}{2}-\left(E_{a}-E_{v} ight)-\left(E_{F}-E_{a} ight)=k T \ln \left(\frac{N_{a}}{n_{i}} ight)[/latex]

or

[latex]0.56-0.045-3(0.0259)=0.437=(0.0259) \ln \left(\frac{N_{a}}{n_{i}} ight)[/latex]

We can then solve for the doping as

[latex]N_{a}=n_{i} \exp \left(\frac{0.437}{0.0259} ight)=3.2 imes 10^{17} \mathrm{~cm}^{-3}[/latex]

Comment

If the acceptor (or donor) concentration in silicon is greater than approximately [latex]3 imes 10^{17} \mathrm{~cm}^{-3}[/latex], 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.

Question 4.13: Determin e the Fermi energy level and the maximum doping con...

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