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in an n-type semiconductor ,the fermi level lies 0.3eV below conduction band at 300K.if the temperature is increased to 330K,where does the new position of the fermi level lie?

and what is fermi level?

Dear Student,

Please find below the solution to the asked query:

For an extrinsic semiconductor the situation is slightly more complicated. At absolute zero in an n-type semiconductor, the chemical potential must lie in the centre of the gap between the donor level and the bottom of the conduction band. At low temperatures in such a semiconductor there are more conduction electrons than there are holes. If the donor level is more than half full, the chemical potential must lie somewhere between the donor levels and the conduction band. At higher temperatures, when the donor level is completely depleted of electrons, and the contribution from intrinsic electrons to the overall electrical conductivity becomes more substantial, the chemical potential tends towards that for an intrinsic semiconductor, i.e., halfway between the conduction and valence bands, and therefore in the middle of the band gap.

The Fermi level for n-type semiconductor is given as

${E}_{F}={E}_{C}-{K}_{B}T\mathrm{log}\left(\frac{{N}_{C}}{{N}_{D}}\right)$

Where,

E_{F} is the fermi level.

E_{C} is the conduction band.

K_{B} is the Boltzmann constant.

T is the absolute temperature.

N_{C} is the effective density of states in the conduction band.

N_{D} is the concentration of donar atoms.

So, given that at T= *300K*, the difference between E_{C} and E_{F} is *0.3 eV*.

Therefore,

${E}_{C}-{E}_{F}={K}_{B}T\mathrm{log}\left(\frac{{N}_{C}}{{N}_{D}}\right)=0.3\phantom{\rule{0ex}{0ex}}\Rightarrow {K}_{B}log\left(\frac{{N}_{C}}{{N}_{D}}\right)=\frac{0.3}{300}={10}^{-3}$

So, at *T = 330 K*, the new position of the fermi level is,

${N}_{C}-{N}_{C,new}={K}_{B}T\text{'}log\left(\frac{{N}_{C}}{{N}_{D}}\right)={10}^{-3}\times 330eV=0.33eV$

Hope this information will clear your doubts about the topic.

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