[DFTB-Plus-User] ohmic resistance

[Gabriele Penazzi]

On 12/09/2014 02:52 PM, Sascha Thinius wrote:
> Thanks Gabriele,
>
>> as you can apply a bias and calculate the current, you can calculate
>> trivially the conductance.
> That is clear as daylight.
>
>> The conductivity (or resistivity) on the other hand can be calculated
>> indirectly, as you can extract the mean free path from the transmission
>> probability, similar to linear response theory.
> Not clear at all how this should work. How I can extract the mean free
> path from a transmission probability of the form T(E)? But note, I am
> a completely newbie in this field.

Ok, first be aware that not always you are able to extract a
resistivity, in fact you can derive a mean free path in the limit of
weak scattering. If you have strong disorder you may build up resonances
and localization, and loose the ohmic behaviour.

In a nutshell, if interference effects are negligible you will have
linear relation between the linear density of scatterers and both
resistance and transmission, and the whole point is about neglecting the
effect of the contact resistance. You'll have l_0 = T/(d*(1-T)) where
l_0 is a characteristic length (mean free path in a single band system),
d the linear density of scatteres, T the transmission . Or,
equivalently, you can write R(L)=R0+R0L/l_0 where R is the resistance
and R0 the quantum contact resistance. As the conductivity depends
linearly on the mean free path, you can extract it from the slope of
R(L), as you write below.

I am not sure what L and L0 are in your formula.

Anyway, I hope this helps a bit to give you a direction but I strongly
suggest you to go through the literature as 'nutshell' answers can
generate confusion. If you want to give a look to some nice papers where
conductivity and mfp are calculated from transmission and linear
response, here's a couple of examples (1D systems, though, as I'm more
familiar with these systems):

10.1103/PhysRevB.74.245313
10.1088/1367-2630/11/9/095004

There is also a bunch of free online lectures on nanohub.org

Cheers,
Gabriele
 


>
>
>> You may want to give a look to S. Datta or M. Lundstrom books, among
>> the others.
>> There is also quite a lot of literature where Landauer is used to
>> calculate mean free paths.
> I found a formula of the form
>
> G-1 = L / sigma W + L0/sigma W
>
> Where 'W' represents the width of the conductor.
> Doing at minium two calculation with different length I should be able
> to calculate the conductivity 'sigma' from the slope and the mean free
> path from the axis intercept.
> Is that OK, or bad practice?
>
>
> Cheers,
>
> Sascha.
>
> On Mon, 08 Dec 2014 16:24:03 +0100
>  Gabriele Penazzi <gabriele.penazzi at bccms.uni-bremen.de> wrote:
>> On 12/08/2014 03:34 PM, Sascha Thinius wrote:
>>> Dear all,
>>>
>>> Is it possible to extract an ohmic resistance or conductivity from a
>>> dftb+negf calculation?
>>> Does somebody know how?
>>
>> Dear Sascha,
>>
>> as you can apply a bias and calculate the current, you can calculate
>> trivially the conductance.
>>
>> The conductivity (or resistivity) on the other hand can be calculated
>> indirectly, as you can extract the mean free path from the transmission
>> probability, similar to linear response theory. You may want to give a
>> look to S. Datta or M. Lundstrom books, among the others. There is also
>> quite a lot of literature where Landauer is used to calculate mean free
>> paths.
>>
>> Gabriele
>>
>>
>>>
>>> Thanks,
>>>
>>> Sascha.
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