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

Equation (4.46) can be written as

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

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

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

n_{0}=\frac{2}{\sqrt{\pi}}\left(2.8 \times 10^{19}\right)(2.7)=8.53 \times 10^{19} \mathrm{~cm}^{-3}

Comment

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

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