E_field, LVTOT: Why voltage drop is only one half ??
Posted: Thu Sep 02, 2010 8:14 am
Hi
I have made VASP calculation of local potentials using LVTOT= .TRUE. when applied electric field along Z direction is 0.0 and 0.05V/A, where the periodic direction is Y. Let V(z, E) means the local potential in LOCPOT averaged over X and Y axes at a given value of Z under Efield = E. If I make a plot of V(z, E = 0.05) – V(z, E = 0), it will give voltage drop across Z direction. For this, I have two questions.
(1) Does LOCVTOT file give the potential energy at a point for unit negative charge or positive charge?
(2) I found that the total voltage drop is only one half (= 0.7V) of the applied voltage (= 1.4V) calculated from E_field(= 0.05)*Lz, where Lz is the supercell size along the Z direction. Can you comment what’s going on here? What am I not understanding correctly?
I have made VASP calculation of local potentials using LVTOT= .TRUE. when applied electric field along Z direction is 0.0 and 0.05V/A, where the periodic direction is Y. Let V(z, E) means the local potential in LOCPOT averaged over X and Y axes at a given value of Z under Efield = E. If I make a plot of V(z, E = 0.05) – V(z, E = 0), it will give voltage drop across Z direction. For this, I have two questions.
(1) Does LOCVTOT file give the potential energy at a point for unit negative charge or positive charge?
(2) I found that the total voltage drop is only one half (= 0.7V) of the applied voltage (= 1.4V) calculated from E_field(= 0.05)*Lz, where Lz is the supercell size along the Z direction. Can you comment what’s going on here? What am I not understanding correctly?