Question: - (a) Compare equation (10-11) evaluated at room temperature for a silicon band gap of 1.1 eV and for a typical donor-state/conduction band gap of 0.05 eV.

(b) Assuming only one impurity atom for every 10³ silicon atoms, do your results suggest that majority carriers, bumped up from donor levels. should outnumber minority carriers created by thermal excitation across the whole 1.1 eV gap? (The calculation ignores the difference in density of states between donor levels and bands, which actually strengthens the argument.)

The ratio of electrons excited at T > 0 o the total number in the valence band at T = 0 is

$\frac{N_{excited}}{N_{\text{V}}}=\frac{k_{B}T}{\Delta E_{\text{V}}}{e}^{\frac{E_{gap}}{2k_{B}T}}$

Here T is the temperature, K_B is Boltzmann constant, is energy bandgap and$\Delta E_{\text{V}}$ is the width of the valence band.

The value of$k_{B}T$ is:

$$\begin{gathered} k_{B}T=\left(1.38\times {10}^{-23}\text{J/K}\right)\left(300K\right) \\ =\left(4.14\times {10}^{-23}\right)\left(\frac{1\text{eV}}{1.6\times {10}^{-19}}\right) \\ =0.026\text{eV} \end{gathered}$$

The width of the valence band is

The ratio of electrons excited at to the total number in the valence band at T=0 is:

$\frac{N_{excited}}{N_{\text{V}}}=\frac{k_{B}T}{\Delta E_{\text{V}}}{e}^{\frac{E_{gap}}{2k_{B}T}}$

Substitute 0.026 for K_BT, 10 for Ev, and 1.1 for E_gap in the above equation:

$$\begin{gathered} \frac{N_{excited}}{N_{\text{V}}}=\frac{\left(0.026\right)}{\left(10\right)}{e}^{\frac{\left(1.1\right)}{2\left(0.026\right)}} \\ =1.69\times {10}^{-12} \end{gathered}$$

The ratio of electrons excited at T > 0 to the total number in the valence band at T = 0 is:

$\frac{N_{excited}}{N_{\text{V}}}=\frac{k_{B}T}{\Delta E_{\text{V}}}{e}^{\frac{E_{gap}}{2k_{B}T}}$

Substitute 0.026 for K_BT, 10 for Ev, and 0.05 for E_gap in the above equation:

$$\begin{gathered} \frac{N_{excited}}{N_{\text{V}}}=\frac{\left(0.026\right)}{\left(10\right)}{e}^{\frac{\left(0.05\right)}{2\left(0.026\right)}} \\ =9.94\times {10}^{-4} \end{gathered}$$

Hence, the result of silicon is$1.69\times {10}^{-12}$ and for doped semiconductor is $9.94\times {10}^{-4}$.

If the doping is 1 in ${10}^5$, the majority carrier will decrease by ${10}^5$, becomes $9.94\times {10}^{-9}$ which is

$\frac{9.94\times {10}^{-9}}{1.69\times {10}^{-12}}=5880$

Times more than the minority carrier.

Hence, the majority carrier bumped up from donor level is 5880 times more than the minority level.