# Relation between band structure, dispersion, density of states, and the Fermi energy and Fermi level

Despite the long title, this question is mostly qualitative (although I am interested in quantitative results if possible).

Say you have an electronic band structure (energy as a function of "k") for a given system, obtained from the tight binding method, DFT, or whatever method.

1) How does this band structure relate to the Fermi energy (or Fermi level)?

2) I assume that density of states comes into play to answer this... Could someone shed some light on how exactly?

Thanks so much for your time.

The Fermi energy is the energy of the highest occupied state at absolute zero of a system of non-interacting (or mean-field interacting) fermions. So for a band structure, the Fermi energy is the energy of the highest-occupied level after you have filled al your bands with electrons. Note that each (spin) band can hold $N_{\textrm{cells}}$ electrons, where $N_{\textrm{cells}}$ is the number of unit cells.
The Fermi level is the energy (not necessarily corresponding to an actual energy level of the system) at which the occupation probability given by the Fermi-Dirac distribution is $1/2$. The Fermi level of a metal at zero temperature is the same as the Fermi energy of that metal (see also this question). This is not true for insulators, where the Fermi level lies inside the energy gap.
The density of states $\rho(E)$ at energy $E$ is defined such that the number of states between energies $E$ and $E+dE$ is given by $\rho(E) dE$. For a discrete spectrum, the density of states consists of a number of delta peaks at the energy levels of the system with weight given by the degeneracy of that level. In a band structure, the states are labelled by $\mathbf k$ and the band index $n$. For example, near energies where a band is very flat at finite $\mathbf k$, the density of states is going to be large, because in this case there are a lot of different $\mathbf k$ values (states) for almost the same energy.