In an n type semiconductor the Fermi level lies 0.4 eV below conduction band. If the concentration of donor atom is doubled, find the new position of the Fermi level w.r.t. the conduction band.

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written 8.7 years ago by

teamques10 ★ 69k

Given:-

Fermi level lies 0.4eV below conduction band

$E_{(F_1 )} = E_C - 0.4$

Concentration is doubled

$n_2 = 2n_1$

To find:-

New position of Fermi level $E_{(F_2 )}$

Solution:

We know that,

$n = N_C e^{-[(E_C-E_F)/kT]}$

$\frac{n}{N_C} = e^{[(E_F-E_C)/kT]}$

$ln(\frac{n}{N_C}) = \frac{E_F - E_C}{kT}$

$E_F - E_C = kT ln(\frac{n}{N_C})$

Initially,

$E_{(F_1 )} - E_C = kT ln(\frac{n_1}{N_C }) ------(i)$

After doubling the concentration,

$E_{(F_2 )} - E_C = kT ln(\frac{n_2}{N_C }) ------(ii)$

Subtracting (i) from (ii),

$E_{(F_2 )} - E_{(F_1 )} = kT [ln(\frac{n_2}{N_C }) - ln(\frac{n_1}{N_C })]$

$E_{(F_2 )} - E_{(F_1 )} = - kTln(\frac{n_1}{n_2})$

$E_{(F_2 )} - E_C – 0.4 = - kT ln(1/2) \ \ \ [ n_2 = 2n_1]$

Considering room temperature i.e T=300K

$E_{(F_2 )} = E_C - 0.4 - \frac{1.38 × 10^{-23}}{1.6 × 10^{-19}} × 300 ln(1/2)$

$E_{(F_2 )} = E_C – 0.38eV$

Fermi level is below conduction band by 0.38eV

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