Concentration of Electrons in Conduction Band
Let dn is the number of electrons available between E and E+dE in conduction band,
\[ dn = Z(E) f(E) dE \tag{1} \]
Figure 1. Energy band diagram of intrinsic semiconductor
Where Z(E) is the density of states in E and E+dE interval, f(E) is the Fermi distribution function. The number of electrons in conduction band,
\[ n = \int_{E_c}^{\infty} Z(E) f(E) dE \tag{2} \]Where Ec is the bottom of the conduction band. The expression of density of states is given by,
\[ Z(E) dE = \frac{4\pi}{h^3} (2m_e^*)^{3/2} E^{1/2} dE \] \[ Z(E) dE = \frac{4\pi}{h^3} (2m_e^*)^{3/2} (E-E_c)^{1/2} dE \tag{3} \]Probability of an electron occupying an energy state Ef is given by Fermi distribution function,
\[ f(E) = \frac{1}{1+\exp \left ( \frac{E-E_F}{kT} \right )} \tag{4} \]Substituting Eqn 3 & 4 in Eqn 2,
\[ n = \int_{E_c}^{\infty} \frac{4\pi}{h^3} (2m_e^*)^{3/2} (E-E_c)^{1/2} \frac{1}{1+\exp \left ( \frac{E-E_F}{kT} \right )} \tag{5} \]Assume E-EF ≫ kT,
\[1 + \exp\left(\frac{E - E_{f}}{kT}\right) \simeq \exp\left(\frac{E - E_{f}}{kT}\right)\] \[n = \frac{4\pi}{h^{3}}(2m_{e}^{*})^{3/2} \int_{E_{c}}^{\infty} (E - E_{c})^{1/2} \exp\left(\frac{E_{f} - E}{kT}\right) dE\] \[n = \frac{4\pi}{h^{3}}(2m_{e}^{*})^{3/2} \exp\left(\frac{E_{f}}{kT}\right) \int_{E_{c}}^{\infty} (E - E_{c})^{1/2} \exp\left(\frac{-E}{kT}\right) dE \tag{6} \]Let us assume,
E-Ec = x ⇒ dE = dx
As E → Ec⇒ x→0
As E → E ⇒ x→∞
Now the Eqn 5 becomes,
\[n = \frac{4\pi}{h^3}(2m_e^*)^{3/2}\exp\left(\frac{E_f}{kT}\right)\int_0^\infty x^{1/2}\exp\left(-\frac{(E_c + x)}{kT}\right)dE\] \[n = \frac{4\pi}{h^3}(2m_e^*)^{3/2}\exp\left(\frac{E_f - E_c}{kT}\right)\int_0^\infty x^{1/2}\exp\left(\frac{-x}{kT}\right)dx\]Using gamma function,
\[\int_{0}^{\infty} x^{1/2} \exp\left(\frac{-x}{kT}\right) dx = (kT)^{3/2} \frac{\sqrt{\pi}}{2}\]The final equation becomes,
\[n = \frac{4\pi}{h^3} (2m_e^*)^{3/2} \exp\left(\frac{E_f - E_c}{kT}\right) (kT)^{3/2} \frac{\sqrt{\pi}}{2}\] \[n = 2 \times \left( \frac{2m_e^* kT}{h^2} \right)^{3/2} \exp\left(\frac{E_f - E_c}{kT}\right)\] \[n = N_c \exp\left(\frac{E_f - E_c}{kT}\right)\] Where \[N_c = 2 \times \left( \frac{2\pi m_e^* kT}{h^2} \right)^{3/2}\]