The partition of two-probe system into left+scat+right rely on good screening of electrodes. If silicon is used as electrode, you should include huge number of layer into the scattering region. Actually, I think transiesta is not the best choice for your problem.

best,
Yun-Peng

On 11/27/2011 09:10 PM, Roberto Guerra wrote:
Hi Peng, thanks for the answer.

Actually I'm using silicon for the electrodes, that may not that efficient
in screening like metals.
I've verified that I get different AVTRANS plots when I change the cell
side along z, or when I passivate the electrodes with hydrogens, or when I
use a periodic structure (both left/right electrodes are bulk-Si).

Actually, the "starting charge density for transiesta" seems to influence
the final AVTRANS.
For this reason I cannot figure out what's the proper way to build the
left+scat+right system.

RG


On mer 23 nov 2011, Yun-Peng Wang wrote:
In my opinion, the purpose of left+scat+right structure is to build a
starting charge density for transiesta run, especially charge density in
the scattering region. Due to good screening property of metals or other
conducting electrodes, the dangling bonds at ends of left+scat+right
structure influence little on charge density in the scattering region.
In transiesta run, the left+scat+right structure extend to
left(infty)+scat+right(infty), again the dangling bonds matter nothing.
Overall, I think inserting vacuum is an easy way to deal with different
left/right electrode problems.

Yun-Peng

On 11/23/2011 05:48 PM, Roberto Guerra wrote:
Hello,

I cannot figure out what's the correct way of setting up the
scattering region in the case of electrodes made by different
materials. After I do the calculation for the left and right
electrodes, I build the system with left+scat+right regions. The
problem is that in defining the cell size along z I have to introduce
some vacuum between the replica, otherwise the electrodes will
interact and change the results. But then I have to passivate the
electrodes due to the dangling bonds! What's the proper way to treat
different electrodes at the z boundary of the cell?

Responder a