Calculate the electron concentration using the Fermi-Dirac integral.

Let $\eta_{F}=2$ so that the Fermi energy is above the conduction band by approximately $52 \mathrm{meV}$ at $T=300 \mathrm{~K}$ .

Question

Calculate the electron concentration using the Fermi-Dirac integral.

Let [latex]\eta_{F}=2[/latex] so that the Fermi energy is above the conduction band by approximately [latex]52 \mathrm{meV}[/latex] at [latex]T=300 \mathrm{~K}[/latex].

Accepted Answer

Equation (4.46) can be written as

[latex]\begin{array}{c}n_{0}=4 \pi\left(\frac{2 m_{n}^{*} k T}{h^{2}} ight)^{3 / 2} \int_{0}^{\infty} \frac{\eta^{1 / 2} d \eta}{1+\exp \left(\eta-\eta_{F} ight)} \ \end{array}[/latex] (4.46)

[latex]\begin{array}{c}n_{0}=\frac{2}{\sqrt{\pi}} N_{c} F_{1 / 2}\left(\eta_{F} ight) \end{array}[/latex]

For silicon at [latex]T=300 \mathrm{~K}, N_{c}=2.8 imes 10^{19} \mathrm{~cm}^{-3}[/latex] and, from Figure 4.10, the Fermi-Dirac integral has a value or [latex]F_{1 / 2}(2)=2.7[/latex]. Then

[latex]n_{0}=\frac{2}{\sqrt{\pi}}\left(2.8 imes 10^{19} ight)(2.7)=8.53 imes 10^{19} \mathrm{~cm}^{-3}[/latex]

Comment

Note that if we had used Equation (4.11), the thermal equilibrium value of [latex]n_{0}[/latex] would be [latex]n_{0}=[/latex] [latex]2.08 imes 10^{20} \mathrm{~cm}^{-3}[/latex], which is incorrect since the Boltzmann approximation is not valid for this case.

[latex]\begin{array}{c}n_{0}=N_{c} \exp \left[\frac{-\left(E_{c}-E_{F} ight)}{k T} ight] \ \end{array}[/latex] (4.11)

Question 4.6: Calculate the electron concentration using the Fermi–Dirac i...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions