hi neyhmor, thank you very much for your quick response and for the links you gave me, I already read some papers, but what i want to know exactly as follows: for the calculation of the energy of an transition level (0 / q), we must find the energy of the system in the neutral charge state (0) and the enrgy at charge state (q), it is the latter that intrigues me, is this energy that of the defect level or it must change few things in the input file of siesta and recalculate the energy?
slimane; --- En date de : Dim 6.2.11, N H <[email protected]> a écrit : De: N H <[email protected]> Objet: Re: [SIESTA-L] semi metalic caracter À: [email protected] Date: Dimanche 6 février 2011, 21h36 The answer for the first question is : No For the second you can take a look on: Van der Walle CG, Neubauer J J. App. Phys 2004 95, 3851. Janotti, A.; Van de Walle, C.G. Phys. Rev. B. 2007, 76, 165202. Lany, S.; Zunger, A. Phys. Rev B 2008, 78, 235104. Makov, G.; Payne, M.C. Phys. Rev. B 1995, 51, 4014-4022. Lany, S.; Zunger, A. Phys. Rev. B 2005, 72, 035215. Lany, S.; Zunger, Phys. Rev. B, 2010, 81, 113201. Gerstmann, U.; Deak, P.; Aradi, B.; Frauenheim, Th.; Overhof, H. Physica B 2003, 340, 190-194. Moreira, N. H.; Aradi, B.; Rosa A. L.; Frauenheim, Th. J. Phys. Chem. C 2010, 114, 18860-18865. >From the beginning be aware that calculating charged defects is a very tricky >subject! On Sun, Feb 6, 2011 at 5:02 PM, Slimane Haffad <[email protected]> wrote: Dear siesta users: I calculated the band structure of nitrogen doped ZnO nanowire and showing a acceptor defect level just above the highest occupied state, knowing that the Fermi level is below the highest occupied level. I want to know if this looks like a semi-metallic character and how to calculate the energy of the system at charge state q (-1) in siesta. I'll be very grateful if someone can help slimane,
