Question

Two semiconductors are identical except that one has a band gap of 1.2 eV, while the other has a band gap of 1.1 eV. The room temperature (T = 300 K) intrinsic carrier density of the 1.1 eV material is ni1(300K) = 1.0×1019 m−3 and that of the 1.2 eV material is ni2(300K).1)Calculate the ratio of intrinsic carrier densities for the two materials at room temperature. ni2/ni1=7.56×10−21

Answer 1

To solve this problem it is necessary to apply the relationship given by the intrinsic carrier concentration, in each of the phases.

The intrinsic carrier concentration is the number of electrons in the conduction band or the number of holes in the valence band in intrinsic material. This number of carriers depends on the band gap of the material and on the temperature of the material.

In general, this can be written mathematically as

$\eta_i = \sqrt{N_cN_v}e^{-\frac{E_g}{2KT}}$

Both are identical semiconductor but the difference is band gap which is:

$E_{g1} = 1.1eV$

$n_{i1} = 1*10^{19}m^{-3}$

$E_{g2} = 1.2eV$

$T=300K$

The ratio between the two phases are given as:

$\frac{\eta_{i1}}{\eta_{i2}} = \frac{e^{-\frac{E_{g1}}{2KT}}}{e^{-\frac{E_{g2}}{2KT}}}$

$\frac{\eta_{i1}}{\eta_{i2}} = e^{\frac{E_{g2}-E_{g1}}{2KT}}$

$\frac{\eta_{i1}}{\eta_{i2}} =e^{\frac{(1.2-1.1)(1.6*10^{-19})}{2(1.38*10^{-23})(300)}}$

$\frac{\eta_{i1}}{\eta_{i2}} =e^{-1.932367}$

$\frac{\eta_{i1}}{\eta_{i2}} =0.145$

Therefore the ratio of intrinsic carrier densities for the two materials at room temperature is 0.145