Show that for an intrinsic semiconductor, fermi level lies midway of conduction band and valence band.

In an intrinsic semiconductor, the electron concentration in the conduction band is

\[ n=N_c\,e^{-(E_c-E_F)/kT} \]

and the hole concentration in the valence band is

\[ p=N_v\,e^{-(E_F-E_v)/kT} \]

where \(N_c\) and \(N_v\) are the effective densities of states.

(Simple probability form: number of electrons in the CB \(n_c=N\,f(E_c)\), and number of holes in the VB \(n_v=N[1-f(E_v)]\), with \(f(E)=1/(1+e^{(E-E_F)/kT})\).)

Intrinsic condition: every electron that moves into the CB leaves a hole in the VB, so \(n=p\):

\[ N_c e^{-(E_c-E_F)/kT}=N_v e^{-(E_F-E_v)/kT} \]

Taking logarithms: \(-\dfrac{E_c-E_F}{kT}=\ln\dfrac{N_v}{N_c}-\dfrac{E_F-E_v}{kT}\)

\[ 2E_F=E_c+E_v+kT\ln\frac{N_v}{N_c} \Rightarrow E_F=\frac{E_c+E_v}{2}+\frac{kT}{2}\ln\frac{N_v}{N_c} \]

If the effective masses of electrons and holes are taken as equal, \(N_c=N_v\) and \(\ln 1=0\). (Even when they are not equal, \(kT\) at room temperature is only 0.026 eV, so the correction is negligible.)

\[ \boxed{E_F=\frac{E_c+E_v}{2}} \]

Measuring energies from the top of the valence band (\(E_v=0\), \(E_c=E_g\)): \(E_F=E_g/2\). Hence the Fermi level of an intrinsic semiconductor lies exactly midway between the conduction band and the valence band, at the middle of the forbidden gap.

 ───────────── Ec  (conduction band)
 - - - - - - - EF = (Ec+Ev)/2
 ───────────── Ev  (valence band)