Calculate the thermal equilibrium concentrations of electrons and holes for a given Fermi energy.

Consider silicon at $T=300 \mathrm{~K}$ so that $N_{c}=2.8 \times 10^{19} \mathrm{~cm}^{-3}$ and $N_{v}=1.04 \times 10^{19} \mathrm{~cm}^{-3}$ . Assume that the Fermi energy is $0.25 \mathrm{eV}$ below the conduction band. If we assume that the bandgap energy of silicon is $1.12 \mathrm{eV}$ , then the Fermi energy will be $0.87 \mathrm{eV}$ above the valence band.

Question

Calculate the thermal equilibrium concentrations of electrons and holes for a given Fermi energy.

Consider silicon at [latex]T=300 \mathrm{~K}[/latex] so that [latex]N_{c}=2.8 imes 10^{19} \mathrm{~cm}^{-3}[/latex] and [latex]N_{v}=1.04 imes 10^{19} \mathrm{~cm}^{-3}[/latex]. Assume that the Fermi energy is [latex]0.25 \mathrm{eV}[/latex] below the conduction band. If we assume that the bandgap energy of silicon is [latex]1.12 \mathrm{eV}[/latex], then the Fermi energy will be [latex]0.87 \mathrm{eV}[/latex] above the valence band.

Accepted Answer

Using Equation (4.11), we have

[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)

[latex]\begin{array}{c}n_{0}=\left(2.8 imes 10^{19} ight) \exp \left(\frac{-0.25}{0.0259} ight)=1.8 imes 10^{15} \mathrm{~cm}^{-3}\end{array}[/latex]

From Equation (4.19), we can write

[latex]\begin{array}{c}p_{0}=N_{v} \exp \left[\frac{-\left(E_{F}-E_{v} ight)}{k T} ight] \ \end{array}[/latex] (4.19)

[latex]p_{0}=\left(1.04 imes 10^{19} ight) \exp \left(\frac{-0.87}{0.0259} ight)=2.7 imes 10^{4} \mathrm{~cm}^{-3}[/latex]

Comment

The change in the Fermi level is actually a function of the donor or acceptor impurity concentrations that are added to the semiconductor. However, this example shows that electron and hole concentrations change by orders of magnitude from the intrinsic carrier concentration as the Fermi energy changes by a few tenths of an electron-volt.

Question 4.5: Calculate the thermal equilibrium concentrations of electron...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions