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Prove that the Fermi level lies exactly at the centre of the forbidden energy gap in case of an intrinsic semiconductor.
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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.

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