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 600 K and that at 300 K? Assume that the temperature dependence of intrinsic carrier concentration $n_i$ is given by $n_i=n_0\text{exp}(-\frac{E_g}{2k_{B}T})$ Where $n_0$ is a constant.

Answer 1

Step 1: Find intrinsic carrier concentration at temperature 300 K.
Given, energy gap, $E_g=1.2eV$
Temperature $T_1=300K$
Intrisic carrier concentration is given by,
$n_i=n_0\text{exp}(-\frac{E_g}{2k_{B}T})$
Find intrinsic carrier concentration at temperature 300 K.
Here, $n_0=\text{constant}$
$k_B$ = Boltzmann constant $=8.62\times {10}^{-5}eV/K$
So, $n_{i1}=n_0\text{exp}(-\frac{1.2}{2\times 8.62\times {10}^{-5}\times 300})$
$n_{i1}=n_0\text{exp}(-23.20)$
$n_{i1}=n_0(8.40\times {10}^{-11})$

Step 2 : Find intrinsic carrier concetnration at temperature 600 K.
Given,
Energy gap, $E_g=1.2eV$
Temperature, $T_2=600K$
Intrinsic carrier concentration is given by,
$n_i=n_0\text{exp}(-\frac{E_g}{2k_{B}T})$
Here, $n_0=\text{constant}$
$k_B=$ Boltzmann constant $=8.62\times {10}^{-5}eV/K$
So, $n_{i2}=n_0\text{exp}(-\frac{1.2}{2\times 8.62\times {10}^{-5}\times 600})$
$n_{i2}=n_0\text{exp}(-11.6)$
$n_{i2}=n_0(9.16\times {10}^{-6})$

Step 3: Find the ratio between conductivity.
Hint: Ratio between conductivity is equal to the ratio between the respective carrier concentration at that temperature.
Ratio between conductivity at 300 K and 600 K is equal to the ratio between the respective carrier concentration at that temperature
$\frac{n_{i1}}{n_{i2}}=\frac{n_0(8.40\times {10}^{-11})}{n_0(9.16\times {10}^{-6})}$
$\frac{n_{i2}}{n_{i1}}=1.09\times {10}^5$
Final answer : $1.09\times {10}^5$.