Question

In an intrinsic semiconductor the energy gap $E_g$ is $1.2eV$. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at $600K$ and that at $300K$? Assume that the temperature dependence of intrinsic carrier concentration $n_i$ is given by
$n_i=n_{o}exp(-\frac{E_g}{2k_{B}T})$
Where $n_0$ is a constant.

Answer 1

Given,

Energy gap in a semiconductor, $E_g=1.2eV$

For a temperature-dependent intrinsic semiconductor:

$n_i=n_{0}exp(-\frac{E_g}{2k_{B}T})$

Simplifying the equation:

$\frac{E_g}{2k}[\frac{1}{T_1}-\frac{1}{T_2}]$

$\Rightarrow \frac{1.2}{2\times 8.62\times {10}^{-5}}[\frac{1}{300}-\frac{1}{600}]=11.59$

Therefore,

$\frac{n_{i_1}}{n_{i_2}}=\frac{\frac{1.2eV}{e^{2\times k_B\times 600}}}{\frac{1.2eV}{e^{2\times k_B\times 300}}}$

$=e^{\frac{1.2eV}{2\times k_B}}(\frac{1}{300}-\frac{1}{600})$

$=e^{\frac{1.2\times 1.6\times {10}^{-19}}{2\times 1.381\times {10}^{-23}\times 600}}$

$\Rightarrow \frac{n_{i_1}}{n_{i_2}}=e^{11.59}=1.072\times {10}^5$