Question

The energy gap of pure Si is 1.1 eV. The mobilities of electrons and holes are respectively $0.135m^2{V}^{-1}{s}^{-1}$ and $0.048m^2{V}^{-1}{s}^{-1}$ and can be taken as independent of temperature. The intrinsic carrier concentration is given by $n_i=n_0{e}^{-\frac{E_0}{2kT}}$ where $n_0$ is a constant, $E_g$ the gap width and k the Boltzmann's constant whose value is $1.36\times {10}^{-23}J{K}^{-1}$. Find the ratio of the electrical conductivities of Si at 600 K and 300 K in the form of $n\times {10}^4$. And report the value of n, where n is an integer -

Answer 1

The total electrical conductivity of a semiconductor is given by
$\sigma =e(n_e{\mu }_e+n_h{\mu }_h)$
For an intrinsic semiconductor,
$n_e=n_h=n_i$
We can thus write for the conductivity
$\sigma =e({\mu }_e+{\mu }_h)n_i$
or $\sigma =e({\mu }_e+{\mu }_h)n_0{e}^{-\frac{E_0}{2kT}}$
As the mobilities ${\mu }_e$ and ${\mu }_h$ are independent of temperature, they can be regarded as constant. The ratio of the conductivities at 600 K and 300 K is then,
$\frac{{\sigma }_{600}}{{\sigma }_{300}}=\frac{e({\mu }_e+{\mu }_h)n_0{e}^{-\frac{E_0}{2k\times 600}}}{e({\mu }_e+{\mu }_h)n_0{e}^{-\frac{E_0}{2k\times 300}}}$
$=e^{\frac{E_0}{2k}(\frac{1}{300}-\frac{1}{600})}=e^{\frac{E_0}{1200k}}$
As per given data, $E_0=1.1eV$
and $k=1.38\times {10}^{-23}J{K}^{-1}$
or $\left(\frac{1.38\times {10}^{-23}}{1.6\times {10}^{-19}}\right)eV{K}^{-1}$
$k=8.625\times {10}^{-5}eV{K}^{-1}$
We thus have
$\frac{{\sigma }_{600}}{{\sigma }_{300}}=e^{\frac{1.1}{1200\times 8.625\times {10}^{-5}}}$
$\approx e^{10.63}\approx 4\times {10}^4$