91影视

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

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

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

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

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

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:

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

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

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

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

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

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

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

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

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

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

Times more than the minority carrier.

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