Hi, Physically speaking, the concept of Fermi energy of a semiconductor does not make much sense, since you have a filled valence band and an empty conduction band. The Fermi level is defined by the condition that you have exactly n electrons occupying states below it. The smearing is then done according to this value. (Yes, the MP smearing is good in the sense that it gives quite accurate energies with relatively high values of electronic temperature, which is then not a physical quantity like in the Fermi smearing case, but simply a parameter)
While I'm not sure what this reported value is, I think it might be the chemical potential (which equals the Fermi energy in a metal, hence the confusion, I suppose). It does not affect any of your results unless it falls into some band, causing your system to be metallic. The total (electronic) energy is the sum over all one-electron energies, that is, the integral of your DOS from -inf. to Fermi energy (which is the energy of the highest occupied state, which is the valence band top), not the chemical potential. This is, by the way, the reason why DFT gives good geometries and energies despite the the fact that the DOS above the highest occupied band is, to a great degree, wrong. But at any rate, even for a metallic system, when you use a denser k-grid, you end up with a different set of one-electron states (the DOS) and a different Fermi energy, but you can still get the same total energy as in the previous case, which is what's important. Now, if your system has dangling bonds, that doesn't necessarily make it metallic. Furthermore, if your 'Fermi level' is in the gap, it isn't metallic. So you don't actually have to use lots of k-points (btw, a 10x10x10 grid is, to my opinion, quite small for a metal, although, if you have a large supercell, this might do). Finally, the 'Fermi level' position does not affect the total energy of an insulator at all (by definition of the total energy). Since forces are more or less simply a derivative of the energy, they aren't changed, too. So the main conclusion should be: you don't have to worry much about your Fermi level, as long as you get good energies or DOS or whatever you're calculating. Sorry for the cumbersome explanations and possible mistakes. 2007/1/11, Andrei Postnikov <[EMAIL PROTECTED]>:
On Wed, 10 Jan 2007, Oleksandr Voznyy wrote: | > The Fermi level is normally calculated by setting the cumulative occupation | > number of all bands to the number of valence electrons. | | As I understand this means that Ef in semiconductor would always be | at the VBM and not in the middle of the gap? Alexander - sorry, it seems that I was wrong. I just checked my old calculation for a wide-gap dielectric and see that Fermi energy = -6.866874 eV while the energies bordeing the gap -8.08 and -4.66. One should look into details of implementation... Yes in other band structure codes I know the Fermi energy is fixed by the last occupied band, i.e. it is set at the valence band top. Then if it technically lies higher, this must be due to the energy broadening introduced. But apparently in Siesta it is done differently. | By the way, how the bandstructure is calculated using only several k-points? The band structure as such (continuous bands) is not calculated, each k-point enters independently of others and contributes a (smeared) peak in the density of states. This summary DOS is (roughly speaking) integrated, and as the number of electrons is achieved, the Fermi level is set. (The details of implementation might be different). Best regards, Andrei +-- Dr. Andrei Postnikov ---- Tel. +33-387315873 ----- mobile +33-666784053 ---+ | Paul Verlaine University - Institute de Physique Electronique et Chimie, | | Laboratoire de Physique des Milieux Denses, 1 Bd Arago, F-57078 Metz, France | +-- [EMAIL PROTECTED] ------------ http://www.home.uni-osnabrueck.de/apostnik/ --+
