Calculate the Fermi energy...

Quote:

drk wrote:
From college textbooks, to calculate the Fermi energy, you need to
know the electron density and the (effective) mass.
But what if I only have a bandstructure - the Fermi level is always
set at zero eV in these images.

I think you may be misinterpreting some other feature as the Fermi
level. The thing to remember about band structures is that no theory
gives you an absolute energy scale. (Getting such information requires
a theory of the surface, and surface physics demonstrates very clearly
that this value can be changed by surface treatments.)

How do I extract the Fermi energy from a bandstructure with for
example one parabolic band. (Do I have to set the energy of the band
minimum at zero, making the Fermi energy >0? - Or is this too simple?)

The most general expression for the "Fermi energy" (the term is usually
used by the metal physics community to refer to the solution at T=0,
and usually with an implicit assumption that the energy of the bottom
of the band structure is known) is that E_F which satisfies:
\int_{-\infty}^{E_F} D(E) dE = n,
where n is the electron density and D(E) is the density of states. The
band structure determines the density of states. A somewhat useful
expression for the density of states is
D(E) = 1/(4\pi^3) dV/dE,
where V is the volume of k-space enclosed by a constant energy surface
at the energy E.

The reason for the -\infty lower limit to the integral is that I am
very carefully hiding the question of the energy scale in the D(E),
which will, of course, be zero below the band minimum. For a simple
parabolic band D(E) = \sqrt{2m^3(E-Emin)}/(\pi^2\hbar^3) .

The term "Fermi level" is more often used in semiconductor physics and
is (a) understood to be very sensitive to details of the sample, and
(b) taken at finite temperature. In this context, the general implicit
expression will be:
\int_{-\infty}^{+\infty} D(E) f((E-E_F)/kT) dE = n,
where now f is the Fermi-Dirac distribution function. The reason why
the Fermi level in semiconductors is so sensitive is that D(E) becomes
negligible over a finite energy range (the band gap), leading to an
ill-conditioned implicit problem.

But, from another point of view, the Fermi level in a semiconductor
is very well defined if there is an ohmic contact to that sample which
is connected to external circuit node. Then E_F = -q V where q is the
elementary charge and V is the node voltage. The thing that needs to
be calculated is how the band structure has arranged itself with
respect to this known energy! (The failure to appreciate this point
causes no end of confusion for students and instructors in semiconductor
device courses.)

And what about having a bandstructure with two parabolic bands with
different band minima.
If the above procedure is correct, then the electrons in the two bands
have different Fermi energies. Is that correct? (Because the band
minima are not at the same energy.)

No, there is only one Fermi energy, but two bands make two contributions
to the density of states. The key thing to understand is that one
always has to "solve" for the Fermi energy (or level). Any
"evaluation" of a closed-form expression is an approximation that is
limited to some particular case.

- Bill Frensley