Semiconductors — Question 5

Question 5

An activated conductivity ratio reveals an energy gap

Over a limited temperature interval, use $\sigma(T)=\sigma_*e^{-E_g/(2k_BT)}$ with constant prefactor. Conductivity doubles from 300 K to 330 K. Use $k_B=8.617333\times10^{-5}$ eV/K.

1. Infer Eg in electron-volts.

2. Find the slope of a plot of ln(sigma) versus 1/T and explain how the intercept differs from that slope.

Question 5 - Solution

An activated conductivity ratio reveals an energy gap

(a) Method and calculation

$$\ln2=\frac{E_g}{2k_B}(1/300-1/330).$$$$E_g=\frac{2k_B\ln2}{1/300-1/330}=.394223\,\mathrm{eV}.$$

(b) Interpretation and further calculation

$$\ln\sigma=\ln\sigma_*-\frac{E_g}{2k_B}\frac1T.$$$$\text{slope}=-E_g/(2k_B)=-2287.39\,\mathrm K,\quad\text{intercept}=\ln\sigma_*.$$

Physical check

The gap inference assumes the prefactor stays constant; temperature-dependent mobility can alter the slope.