What is the Fermi energy of (undoped) graphene? All of the sources I have found for this online have been wildly unclear. Many use the phrase "Fermi energy" to refer to the "Fermi level" (which is emphatically not what I'm looking for; I want the Fermi energy as defined in this Wikipedia article: http://en.wikipedia.org/wiki/Fermi_energy ). Fermi energy is always greater than zero.
Has anyone ever measured the Fermi energy of graphene? Is there any way to calculate it from known quantities, such as the Fermi velocity, which is approx. $10^6$ m/s? i.e. is there a reason why the usual formula $E_F = \tfrac{1}{2}m_e v_F^2$ wouldn't work here? I read that the Fermi energy for undoped graphene is equal to the energy at the Dirac points, but I read elsewhere that that value is less than zero, which makes no sense because, again, Fermi energy is always greater than zero.
 A: You can find the band structure of graphene here: http://web.physics.ucsb.edu/~phys123B/w2015/leggett-lecture.pdf
Equation 9.
The minimum value for the energy is when both $\cos$ functions are 1 which gives an energy of $-3t$ which is the lowest occupied state.  The highest occupied state is the Fermi level which in this case is $E_f = 0$.  Thus the Fermi energy is $3t$.
The value of $t$ seems to be somewhere in the range of $2.5eV$ to $3eV$ (here with associated references) which would put the Fermi energy somewhere between $7.5eV$ to $9eV$.
I should point out that this value is not unique, but is by far the most commonly quoted result which stems from a tight-binding model that considers only nearest-neighbor interactions.  You can also consider second nearest-neighbor interactions as well as many others (e.g. spin-orbit) which will change the result slightly.
Finally, I would add we are just considering the $\pi$ band and aren't considering any of the more tightly bound orbitals that come from the interplane bonding (the sp2 orbitals).  I used to know the calculation for these energy levels as well but I seem to have misplaced my notes on the subject.  Finding this energy would give you the most correct value of the Fermi energy, but since these states are rarely accessed they are typically ignored.
A: TL;DR: graphene is a semimetal: if not doped, its Fermi level lies at the junction of the two conical bands, i.e., the Fermi surface degenerates into a point, and the Fermi energy is zero.
Indeed, Fermi energy and Fermi Level are not the same thing (see this answer for more details), although the two terms are frequently confused.
Fermi energy characterizes the highest occupied state in a metal, i.e., the position of the Fermi surface. In case of parabolic bands it is related to Fermi wave number as
$$
\epsilon_F=\frac{\hbar^2 k_F^2}{2m^*}=\frac{m^*v_F^2}{2},
$$
where $v_F=\hbar k_F/m^*$ is the Fermi velocity and $m^*$ is the effective mass.
Fermi energy can be evaluated from the electron concentration, by integrating the density of states with the Fermi-Diract distribution function at zero temperature (i.e., by integrating up to the Fermi energy):
$$
n=\int_0^{\epsilon_F}d\epsilon D(\epsilon),\\ D(\epsilon)=\int dk_xdk_ydk_z
\delta\left(\epsilon - \frac{\hbar^2(k_x^2+k_x^2+k_x^2)}{2m^*}\right)$$
(up to a possible factor, that I didn't verify - related to jormalization, spin, etc.)
The crucial differences between graphene and a metal with a parabolic band are:

*

*graphene is two-dimension, which will affect the integral for the density of states given above (no $k_z$)

*graphene does not have a parabolic band, i.e., one has to use different dispersion relation: $\epsilon(\mathbf{k})=v_0|\mathbf{k}|$ rather than $\epsilon(\mathbf{k})=h^2\mathbf{k}^2/{2m^*}$

*graphene is a semimetal: if not doped, its Fermi level lies at the junction of the two conical bands, i.e., the Fermi surface degenerates into a point, and the Fermi energy is zero. Note also that graphene is not doped as usual semiconductors, by adding impurities, but rather by applying a potential to an underlying metallic gate.

