1. Mar 28, 2007

vgbraymond

In metal, fermi energy is the highest energy states that electron occupied at T=0. And so, fermi energy of a particular metal is an intrinsic value, and won't be changed.

Then for a semiconductor,
What is fermi energy, is there any concrete interpretation as of metal?
Why is it just in the middle of energy of VB and CB?
Besides adding impurities, would the fermi energy of a SC changed according to, say, temperature?

2. Mar 28, 2007

marlon

Check out : http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html

Indeed, the Fermi Level (FL) is a "metal concept". For a SC, the FL is actually defined as the chemical potential. In the case of an intrinsic SC, the FL is in the middle of the bandgap because if you apply the FL definition (based upon the Fermi Dirac distribution theory, just like in the case of metals), you wind up in the middle of the band gap. Doping (ie adding impurities) changes the position of the FL because you are adding or removing electrons. Adding electrons (ie adding electron donor atoms like P) makes the FL go up (ie go towards the conduction band edge), while adding holes (ie removing electrons) has the opposite effect on the FL.

Yes, the FL does NOT depend on T because it is defined based upon a condition that requires T to be 0 K ! The electrondistribution, obviously, changes with T as explained by the Fermi Dirac distribution.

marlon

3. Apr 2, 2007

xepma

The definition of the fermi energy is as follows:

At T=0 (ground state) it simply states that all states with an energy $$\epsilon$$ lower than the fermi energy $$\epsilon_F$$ are occupied, and all states with a higher energy are unoccupied:

At T=0:
$$\epsilon < \epsilon_F \rightarrow$$ occupied state
$$\epsilon > \epsilon_F \rightarrow$$ unoccupied state

For higher temperatures the distribution of particles will change according to the Fermi-Dirac distribution. But the Fermi energy does not change! It is, as you correctly stated, an intrinsic quantity. For example, an ideal Fermi gas has a Fermi energy which only depends on the particle density:
$$\epsilon_F = \frac{\hbar^2}{2m}\left (6\pi^2 n\right )^{2/3}$$