selecting k points

[esi]

in zaz 6-6-20 structure with change in points current changes signifcenly.for instance

transmission_spectrum = TransmissionSpectrum(
    configuration=device_configuration,
    energies=numpy.linspace(-0.5,0.005,400)*eV,
    kpoints=MonkhorstPackGrid(1,1),
    energy_zero_parameter=AverageFermiLevel,
    infinitesimal=1e-09*eV,
    self_energy_calculator=KrylovSelfEnergy(),
    )
Current becomes -3.95849773196e-09 A

and for

transmission_spectrum = TransmissionSpectrum(
    configuration=device_configuration,
    energies=numpy.linspace(-0.5,0.005,401)*eV,
    kpoints=MonkhorstPackGrid(1,1),
    energy_zero_parameter=AverageFermiLevel,
    infinitesimal=1e-09*eV,
    self_energy_calculator=KrylovSelfEnergy(),
    )
Current becomes -2.91743529102e-09 A
please guide me to select correct k points.

[Nordland]

Can you perhaps send us a plot of the transmission spectrum?

[esi]

I send you plots of the transmission spectrum in 400 points and 401 points

[Anders Blom]

But what's the bias? If none of those big peaks are in your bias window, then your current is basically zero, and those two numbers are just various approximations of zero.

[esi]

I corrected range of transmission spectrum but currents in two k points of 400 and 401 were different.
voltage in right electrode is: -0.1V
voltage in left electrode is:   0.1V 
voltage in gate is:             -0.75V

when I considered points equal to 400 current became -3.83714017895e-09 A
and when I considered points equal to 401 current became -6.87924370431e-09 A

[Anders Blom]

From the pictures we see that the real contribution to the current would come from the peak at about -0.15 eV. But this doesn't come into play until around 0.3 V bias. So again, your current is very small and comes from the tiny peaks around the Fermi level (too small to see on the scale you are plotting, but you can zoom in to see them perhaps). If you really care about them (I wouldn't...), you will need to make a smaller energy interval (only the region of -V/2 to +V/2 plus minus a few kT contributes to the current) and perhaps add more points in that interval. Ideally an adaptive integration should be used, but again, once you hit about 0.3 V bias you get a real current (provided the peaks don't shift too much), and this 1e-9 A will just be invisible.