In an intrinsic semiconductor the energy gap *E*_{g}is 1.2 eV. 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_0 exp[E_g/(2k_BT)]`

where *n*_{0 }is a constant.

#### Solution

Energy gap of the given intrinsic semiconductor, *E*_{g} = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is written as:

`n_i = n_0 exp [- E/(2k_BT)]`

Where,

*k*_{B} = Boltzmann constant = 8.62 × 10^{−5} eV/K

T = Temperature

*n*_{0} = Constant

Initial temperature, *T*_{1} = 300 K

The intrinsic carrier-concentration at this temperature can be written as:

`n_(i1) = n_0 exp[- E_g/(2k_B xx 300)]` ....(1)

Final temperature, *T*_{2} = 600 K

The intrinsic carrier-concentration at this temperature can be written as:

`n_(i2) = n_0 exp[- E_g/(2k_B xx 600)]` ....(2)

The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.

`n_(i2)/n_(i1) = (n_0 exp [- E_g/(2k_B 600)])/[n_0 exp [E_g/(2k_B 300)]]`

`= exp E_g/(2k_B) [1/300 - 1/600] = exp [1.2/(2xx 8.62 xx 20^(-5)) xx (2-1)/600]`

`= exp[11.6] = 1.09 xx 10^5`

Therefore, the ratio between the conductivities is 1.09 × 10^{5}.