In a simplified model of an undoped semiconductor, the actual distribution of energy states may be replaced by one in which there are ${\mathbf{N}}_{\mathbf{V}}$ states in the valence band, all having the same energy ${\mathbf{E}}_{\mathbf{V}}$, and ${\mathbf{N}}_{\mathbf{C}}$ states in the conduction band all these states having the same energy ${\mathbf{E}}_{\mathbf{c}}$. The number of electrons in the conduction band equals the number of holes in the valence band.

1. Show that this last condition implies that $\frac{{\mathbf{N}}_{\mathbf{c}}}{\mathbf{exp}\left(∆E_c/\mathrm{kT}\right)\mathbf{+}\mathbf{1}}=\frac{{\mathbf{N}}_{\mathbf{v}\mathbf{}}}{\mathbf{exp}\mathbf{(}\mathbf{∆}{\mathbf{E}}_{\mathbf{v}}\mathbf{}\mathbf{/}\mathbf{}\mathbf{kT}\mathbf{)}\mathbf{+}\mathbf{1}}$ in which $\mathbf{∆}{\mathbf{E}}_{\mathbf{c}}\mathbf{=}\mathbf{∆}{\mathbf{E}}_{\mathbf{c}}\mathbf{-}\mathbf{∆}{\mathbf{E}}_{\mathbf{F}}$ and $\mathbf{∆}{\mathbf{E}}_{\mathbf{v}}\mathbf{=}\mathbf{-}\left(∆E_v-∆E_F\right)$.
2. If the Fermi level is in the gap between the two bands and its distance from each band is large relative to kT then the exponentials dominate in the denominators. Under these conditions, show that ${\mathbf{E}}_{\mathbf{F}}\mathbf{=}\frac{\left(E_c+E_v\right)}{\mathbf{2}}\mathbf{+}\frac{\mathrm{kTIn}(N_v+N_c)}{\mathbf{2}}$ and that if ${\mathbf{N}}_{\mathbf{v}}\mathbf{\approx }{\mathbf{N}}_{\mathbf{c}}$ , the Fermi level for the undoped semiconductor is close to the gap’s center.

1. In the undoped semiconductor, the number of states in the valence band is with energy $E_v$, while the number of states in the conduction band is $N_c$ with energy $E_c$.
2. The number of electrons in the conduction band equals the number of holes in the valence band.

We are given the condition that the number of electrons in the conduction band and the number of holes in the valence band are equal. Thus, using the probability equation in the density of occupied states equation, we can get the individual equations for both cases. This will imply the given condition. Similarly, for the second case, we can consider the condition that the Fermi level in the energy state is relatively large to the value kT.

Formulae:

The density of occupied states, $N_0\left(E\right)=N\left(E\right)P\left(E\right)$ (i)

The probability of the condition that a particle will have energy E according to Fermi-Dirac statistics, $P\left(E\right)=\frac{1}{e^{\left(E-E_F\right)}+1}$ (ii)

The number of electrons occupying the valance band is given by substituting equation (ii) in equation (i) as follows:

$$\begin{gathered} N_{ev}=N_{v}P\left(E_v\right) \\ =\frac{N_v}{e^{\left(E_v-E_F\right)kT}+1} \end{gathered}$$

Since there are a total of states in the valence band; the number of holes in the valence band is given by:

$N_{hv}=\frac{N_v}{e^{-\left(E_v-E_F\right)/kT}+1}.................................\left(a\right)$

Now, the number of electrons in the conduction band is given by substituting equation (ii) in equation (i) as follows:

$$\begin{gathered} N_{ec}=N_{c}P\left(E_C\right) \\ {N}_{hv}=\frac{N_c}{e^{-\left(E_C-E_F\right)/kT}+1}.................................\left(b\right) \end{gathered}$$

As we are given that $N_{hv}=N_{ec}$

Thus, using equations (a) and (b), we get that

$$\begin{gathered} \frac{N_v}{e^{-\left(E_v-E_F\right)/kT}+1}=\frac{N_c}{e^{-\left(E_C-E_F\right)/kT}+1}.....................................\left(c\right) \\ \frac{N_v}{e^{\left(∆E_v-E_F\right)/kT}+1}=\frac{N_c}{e^{\left(∆E_C-E_F\right)/kT}+1} \end{gathered}$$

where $∆E_c-E_c-E_F$ and $∆E_v=-\left(E_v-E_F\right)$.

In this case,$e^{\left(E_C-E_F\right)/kT}1$ and $e^{-\left(E_V-E_F\right)/kT}1$

Thus, using this condition in the above equation (c), we get that

$$\begin{gathered} \frac{N_V}{e^{-\left(E_v-E_F\right)/kT}}=\frac{N_V}{e^{\left(E_C-E_F\right)/kT}} \\ {e}^{\left(E_v-E_c+2E_F\right)/kT}=N_V/N_c \\ {E}_F=\frac{E_C+E_V}{2}=\frac{kTIn\left(N_V/N_c\right)}{2} \end{gathered}$$

Hence, the above condition is proved.