Concentration of Holes in Valence band

Let dp is the number of holes in energy interval E and E+dE in valence band,

\[ dp = (1-F(E)) Z(E)dE \tag{1} \]

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. As f(E) is probability of electron occupancy, so the hole occupancy will be given by 1-f(E)

\[ f(E) = \frac{1}{1+\exp \left ( \frac{E-E_F}{kT} \right )} \tag{2} \] \[ 1-f(E) = \frac{1+\exp \left ( \frac{E-E_F}{kT} \right )-1}{1+\exp \left ( \frac{E-E_F}{kT} \right )} \tag{3} \]

In valance band E < E_f , so the exponential terms in denominator can be neglected as compared to unity

\[ 1-f(E) = \exp \left ( \frac{E-E_F}{kT} \right) \tag{4} \]

The expression of density of states is given by,

\[ Z(E) dE = \frac{4\pi}{h^3} (2m_h^*)^{3/2} E^{1/2} dE \tag{5}\] \[ Z(E) dE = \frac{4\pi}{h^3} (2m_h^*)^{3/2} (E_v-E)^{1/2} dE \tag{6}\]

Now, the number of holes in valence band,

\[ p = \int_{-\infty}^{E_v} Z(E) (1-f(E)) dE \tag{7} \]

Substituting Eqn 6 & 4 in Eqn 7,

\[ p = \int_{-\infty}^{E_v} \frac{4\pi}{h^3} (2m_h^*)^{3/2} (E_v-E)^{1/2} \exp \left (\frac{E-E_F}{kT} \right ) dE \tag{8} \] \[ p = \frac{4\pi}{h^3} (2m_h^*)^{3/2} \exp\left(\frac{-E_f}{kT}\right) \int_{-\infty}^{E_v} (E_v-E)^{1/2} \exp \left (\frac{E}{kT} \right ) dE \tag{9} \]

Let us assume,

E_v - E = x ⇒ dE = -dx

As E → E_v⇒ x→0

As E → - ∞ x→∞

Now the Eqn 9 becomes,

\[ p = \frac{4\pi}{h^3} (2m_h^*)^{3/2} \exp\left(\frac{-E_f}{kT}\right) \int_{\infty}^{0} (x)^{1/2} \exp \left (\frac{E_v-x}{kT} \right ) -dx \tag{10} \] \[ p = \frac{4\pi}{h^3} (2m_h^*)^{3/2} \exp\left(\frac{E_v-E_f}{kT}\right) \int_{0}^{\infty} (x)^{1/2} \exp \left (\frac{-x}{kT} \right ) dx \tag{11} \]

Using gamma function,

\[\int_{0}^{\infty} x^{1/2} \exp\left(\frac{-x}{kT}\right) dx = (kT)^{3/2} \frac{\sqrt{\pi}}{2}\] \[ p = \frac{4\pi}{h^3} (2m_h^*)^{3/2} \exp\left(\frac{E_v-E_f}{kT}\right) (kT)^{3/2} \frac{\sqrt{\pi}}{2} \tag{11} \] \[ p = 2 \times \left( \frac{ 2\pi m_h^*kT}{h^2} \right )^{3/2} \exp\left(\frac{E_v-E_f}{kT}\right) \tag{11} \]

The above equation can be written in simplified form as

\[ p = N_v \exp\left(\frac{E_v-E_f}{kT}\right) \tag{11} \] Where \[ N_v = 2 \times \left( \frac{ 2\pi m_h^*kT}{h^2} \right )^{3/2} \]

In intrinsic semiconductors, n = p = n_i. Using law of mass action, value of intrinsic carrier concentration can be calculated as,

\[ n_i^2 = n \times p \] \[ n_i = \sqrt{N_cN_v} \exp \left( \frac{-E_g}{2kT} \right)\]

Fermi level in intrinsic semiconductor

In intrinsic semiconductor, \( n_c = n_v \), \[ 2 \times \left( \frac{2m_e^* kT}{h^2} \right)^{3/2} \exp\left( \frac{E_f - E_c}{kT} \right) = 2 \times \left( \frac{2m_h^* kT}{h^2} \right)^{3/2} \exp\left( \frac{E_v - E_f}{kT} \right) \] \[ \exp\left( \frac{E_f - E_c}{kT} \right) = \left( \frac{m_h^*}{m_e^*} \right)^{3/2} \exp\left( \frac{E_v - E_f}{kT} \right) \] Applying logarithmic on both sides, \[ \frac{E_f - E_c}{kT} = \frac{3}{2} \ln \left( \frac{m_h^*}{m_e^*} \right) + \frac{E_v - E_f}{kT} \] Assume \( m_e^* = m_h^* \), \[ \frac{E_f - E_c}{kT} = \frac{E_v - E_f}{kT} \] \[ 2E_f = E_v + E_c \] \[ E_f = \frac{E_v + E_c}{2} \] Hence, the Fermi energy is in the middle of the valence band and valence band.