# Where does the factor of half come from, boltzmann distribution for bandgap energy

I have found that it is possible to calculate the conductivity of a semiconductor using the Boltzmann distribution:

The source is a slideshow presentation and doesn't list much information. The derivation given is as follows: For an intrinsic semidconductor: $\sigma=n|e|\mu_e+p|e|\mu_h$

$\sigma \approx C_1n=C_1p =$

$\sigma \approx Ce^{-\frac{E_g}{2k_bT}}$ where C is a constant and $E_g$ the bandgap energy.

My question is: Where does the half in the exponent of the boltmann distibution come from?

• "I have found that..." - Perhaps it would be a good idea to link the source. – Andrei Geanta Nov 5 '17 at 15:58
• From SRH theory assuming a mid-gap state for carrier generation/recombination. – Jon Custer Nov 5 '17 at 16:15

It's hard to say without seeing more of what you did, but if the trap state is halfway between the conduction and valence bands, then it would have energy $E_t = \frac{E_g}{2}$. The equilibrium fraction of the charge carriers in the trap state would then be given by $$n_t \propto e^{-\frac{E_t}{k_B T}} = e^{-\frac{E_g}{2k_B T}}$$
$$n_i \propto e^{-\frac{E_g - E_f}{k_B T}}$$ where $E_f$ is the Fermi energy. In many semiconductors, $E_f$ is precisely in the middle of the band gap, so $E_g-E_f=E_g/2$.