written 3.5 years ago by |
It can be shown for intrinsic semiconductors that fermi energy level lies midway between conduction band and valence band. The proof is as follows:
At any temperature (T>0K)
ne = number of electrons in conduction band
nv = number of electrons in valence band
We have, by definition of intrinsic semiconductors:
The number of electrons in conduction band is given by
$n_e=N_c e^{-\frac{(E_C-E_F )}{KT}}$
The number of electrons in valence band is given by
$n_v=N_v e^{-\frac{(E_F-E_V )}{KT}}$
Where,
Nc = effective density of states in conduction band
NV = effective density of states in valence band
For best approximation, we consider Nc = NV
For intrinsic semiconductors, nc = nv
$\therefore N_c e^{-\frac{(E_C-E_F )}{KT}}=N_v e^{-\frac{(E_F-E_V )}{KT}}$
$\dfrac{N_V}{N_C}=\dfrac{e^{-\frac{(E_C-E_F )}{KT}}}{e^{-\frac{(E_F-E_V )}{KT}}} $
$\dfrac{N_V}{N_C}=e^{-\frac{(E_C-E_F-E_F+E_V )}{KT}}$
$\dfrac{N_V}{N_C}=e^{-\frac{(E_C-2E_F+E_V )}{KT}}$
As NC = NV = 1;
we get,
$e^{-\frac{(E_C-2E_F+E_V )}{KT}}=1$
Taking log on both sides, we get,
$-\dfrac{(E_C-2E_F+E_V )}{KT}=0$
$\therefore E_C+E_V=2E_F$
$\therefore \dfrac{E_C+E_V}{2}=E_F$
Thus, the Fermi level lies exactly at the centre of the forbidden energy gap in case of an intrinsic semiconductor.