Author Topic: Simple GNR transmission question  (Read 6671 times)

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Offline simCity

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Simple GNR transmission question
« on: June 13, 2013, 11:21 »
Hi all,

I followed the tutorial on GNR transmission spectrum and simulated the following code with 0V and 0.1V biases. Obtained transmission spectra are also attached. When Vbias=0V, the transmission spectrum is OK as given in the tutorial but when Vbias=0.1V, the spectrum changes a lot, being zero around Ef-lowering the conductance a lot as you can see from the second figure. Why does the transmission spectrum around Ef lowers this much when GNR is biased?

Thanks in advance,
Jim

Offline nori

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Re: Simple GNR transmission question
« Reply #1 on: June 13, 2013, 15:03 »
when Vbias=0.1V, the transmission at Ef is from the valence band of left electrode to the conduction band of right electrode.
Maybe this transmission is prohibited due to symmetry of wave function.

Offline Anders Blom

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Re: Simple GNR transmission question
« Reply #2 on: June 13, 2013, 16:36 »
If you look carefully at the band structure of a zigzag graphene nanoribbon, it's quite complex in the area close to Z=pi/a - it's actually a big wiggly (this is shown in the tutorial). This is why T(E) at zero bias is 4 and not 2. What will happen when a bias is applied is that rather than make the conduction and valence band tangle up, as in the zero bias case, they will split and create a small gap. Actually the conduction and valence bands will attach themselves to the Fermi levels in the two electrodes, so as the bias is increased, the bands will be pulled apart by about the same amount as the bias (this is precisely what Nori wrote, just in other words).

It should be noted that applying a bias to a perfect, infinitely long ribbon is a bit artificial, and actually rather hard to describe with the kind of models implemented in ATK. A more appropriate (and interesting) case is when some defect or distortion is introduced in the central region, or when the two electrodes are different, e.g. p/n doped.

Offline simCity

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Re: Simple GNR transmission question
« Reply #3 on: June 14, 2013, 08:23 »
Hi,

Thanks for your answers. So according to ATK, GNR displays a high resistance due to the splitting up of the valance and conductance bands when a bias is applied. Is this a computational fact? I couldn't find any experimental result or paper in the literature discussing this.

Regards,
Jim

Offline Anders Blom

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Re: Simple GNR transmission question
« Reply #4 on: June 17, 2013, 16:30 »
No, that's not what it means. It just means the hole and electron levels separate, but that's normal. It should again be stressed that it is artificial to apply a voltage drop in a perfect conductor under purely ballistic conditions, at least with a large bias, and also keep in mind that we are not computing the resistance here, only the conductance (i.e. there is no inelastic scattering).

Offline simCity

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Re: Simple GNR transmission question
« Reply #5 on: June 18, 2013, 06:50 »
Dear Anders Blom,

Thanks again for your reply. However I also did the simulation of graphene sheet (not GNR) as attached in the py code, its transmission spectrum at 0.1V bias is also attached. As you can see, the transmission is not zero between E_right and E_left as it was in GNR simulation. Why is there such a difference between GNR and graphene sheet?

Thanks in advance,
Jim

Offline Anders Blom

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Re: Simple GNR transmission question
« Reply #6 on: June 18, 2013, 21:51 »
Your self-consistent simulation is only performed at the Gamma point, so it doesn't mean much. But even if you were to increase the k-points for the SCF loop (say, to 9 or so) as well as for the transmission calculation (100-200 would be a good number) you can't expect to predict much from this calculation. As explained many times, the NEGF model is designed primarily for the case which is not perfectly periodic in the Z direction. Trying to apply a finite bias across a metallic-ish part of the center, arbitrarily cut out from the perfectly periodic crystal, and looking at ballistic, elastic transport, is just not a particularly well-defined (or interesting) problem.