What Causes Semiconductors to Lack Fermi Surfaces?

[Niles]

Homework Statement

Hi all

I can't seem to figure out, why semicodunctors do not have Fermi surfaces. At T=0K, there are no electrons in the conduction band, and thus there is no Fermi surface - all OK here.

But at T > 0K, there are electrons in the conduction band. Why is it then that it is said that "semiconductors do not have Fermi surfaces"?

 

[jdwood983]

If the Fermi energy level falls in the bandgap (where no electron states can exist) of a semiconductor, there is no Fermi surface.

 

[Niles]

But the Fermi energy is the energy of the highest occupied state - so if we have T>0 K, there will be thermal excitation up to the conduction band. This implies that the Fermi energy lies in the conduction band?

 

[jdwood983]

I'm sorry, I wrote Fermi energy level but it should be just Fermi level which is not the same thing.

Also,the Fermi energy is the energy of the highest occupied state at T=0K.

 

[Niles]

Oh, you mean the chemical potential. But does it make a difference that it lies in the energy gap? I cannot see the connection between the chemical potential and the Fermi surface. And thank you for helping me; I really appreciate it.

 

[jdwood983]

Oh, you mean the chemical potential. But does it make a difference that it lies in the energy gap? I cannot see the connection between the chemical potential and the Fermi surface. And thank you for helping me; I really appreciate it.

Not quite, the chemical potential is just \mu. The Fermi level comes from the Fermi-Dirac distribution:

<br /> f(E)=\frac{1}{1+\exp\left[\frac{E-\mu}{k_BT}\right]}<br />

In relation to the conduction band, this is expressed as

<br /> f(E)=\frac{1}{1+\exp\left[\frac{K-\zeta}{k_BT}\right]}<br />

where K is the difference between the Fermi and conduction energies and \zeta is the Fermi level:

<br /> \zeta=\zeta_0\left[1-\frac{\pi^2}{12}\left(\frac{k_BT}{\zeta_0}\right)^2-\frac{\pi^4}{80}\left(\frac{k_BT}{\zeta_0}\right)^4+\cdots\right]<br />

with \zeta_0 being the Fermi energy at 0K. So, depending on the temperature, the Fermi level can/will be in the bandgap, hence saying that semiconductors do not have Fermi surfaces.

 

[Niles]

According to http://en.wikipedia.org/wiki/Fermi_level, then \zeta is the difference between the Fermi energy and the conduction band energy. Is this wrong?

I haven't heard of the "Fermi level" before, and it is not intuitive for me, unfortunately. Is there a physical explanation of why semiconductors do not have Fermi surfaces (apart from the fact that their Fermi level is in the band gap), just like there are explanations of why metals have Fermi surfaces (i.e. they have free electrons)?

 

[jdwood983]

According to http://en.wikipedia.org/wiki/Fermi_level, then \zeta is the difference between the Fermi energy and the conduction band energy. Is this wrong?[\QUOTE]

Yes. K was defined as the difference between the conduction band energy and the Fermi energy while \zeta_0 is defined as the Fermi energy with the relation between the Fermi energy and the Fermi level, \zeta, as above .

I haven't heard of the "Fermi level" before, and it is not intuitive for me, unfortunately. Is there a physical explanation of why semiconductors do not have Fermi surfaces (apart from the fact that their Fermi level is in the band gap), just like there are explanations of why metals have Fermi surfaces (i.e. they have free electrons)?

As far as I know, the reasoning is because the Fermi level is in the bandgap. There could be something more to it that I am unaware of, but I am not sure one way or the other.