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Show that for intrinsic semiconductors the Fermi level lies midway between the conduction band and the valence band. Draw the energy level diagram as a function of temperature for n-type of semi-c

Mumbai university > FE > SEM 1 > Applied Physics 1

Marks: 8M

Year: May 2015

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Let,

$n_e$ be the number of electrons in the semiconductor band.

$n_v$ be the number of holes in the valence band.

At any temperature, T>0K

$n_e = N_c.e^{-(E_c - E_F)/kT}$ and

$n_v = N_v.e^{-(E_F - E_c)/kT}$

Where, $N_c$ is the effective density of states in the conduction band.

$N_v$ is the effective density of states in the valence band.

For best approximation, $N_c = N_v$

For an intrinsic semiconductor, $n_c = n_v$

$N_c . e^{-(E_c - E_F)/kT} = N_v . e^{-(E_F-E_c)/kT}$

$\frac{e^{-(E_c - E_F)/kT}}{e^{-(E_F - E_c)/kT}} = \frac{N_v}{N_c}$

$e^{-(E_c + E_V + 2E_F)/kT} = 1 \ \ \ ( N_c = N_v)$

Taking ln on both sides,

$\frac{-(E_C + E_V - 2E_F)}{kT} = 0$

$E_C = \frac{E_C + E_V}{2}$

Thus, Fermi level in an intrinsic semiconductor lies at the centre of the forbidden gap.

Energy diagram as a function of temperature for intrinsic semiconductor-

enter image description here

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