Transmission spectrum and DOS

**carbon9** · Nov4-09, 11:18 AM

Hi all,

I am in doubt on a subject. The density of states function of a metallic CNT, is given as

in the book "Introduction to Nanoelectronics" of Hanson. On the other hand, in the same book, it is said that metallic CNTs have 2 conducting channels. As I understand, it means that DOS must have the value of 2 at the above plot. However, it is less that 1. Could you please give an idea on this confusing point?

Regards,
Carbon9

**sokrates** · Nov4-09, 03:25 PM

No, no conducting channels probably means number of subbands, not the number of states.

The signature of a subband in a density of states plot is like a new step, so in your plot I see two such steps between 0 eV and 1 eV ( similarly 0 eV and -1 eV).
Bottom line:
Number of modes (or conducting channels) is not the same as density of states.

**carbon9** · Nov4-09, 03:34 PM

Thank you sokrates.

The signature of a subband in a density of states plot is like a new step, so in your plot I see two such steps between 0 eV and 1 eV ( similarly 0 eV and -1 eV).

So, for example, if we inject electrons having 0.5eV, then does it mean that there will be only one conduction mode for these electrons?

**sokrates** · Nov4-09, 11:38 PM

That is correct. Since:

$LaTeX Code: G (E) = \\frac{ q^2}{h} M(E) T(E)$

q = electron charge
G (E) = conductance at energy E
M(E) = number of conducting channels ("modes") at energy E
T(E) = transmission at energy E

So if your Fermi Level is referenced at 0 eV (which is in your diagram) and you inject 0.5 eV electrons, there will only be 1 conducting channels per spin (based on the DOS you gave). Furthermore, if you know the transmission at that energy, you can calculate the current. Remember this is the Landauer model of transport.

Also notice that even for 0 Kelvin, where all your electrons have energies equal to zero ( Ef ) there's still an open channel for conduction. What does this tell you? --- That you have a metallic material. A semiconductor would not have an open channel at 0K because there would typically be a band-gap that would prevent carriers from moving to the "conduction" band. At the DOS this gap would correspond to a lack of states between the gap energies.

**carbon9** · Nov5-09, 03:39 AM

Thank you again, Sokrates.

Yes, according the plot given, there seems only 1 conduction channel. Hence, the conductance of the CNT will then be

$LaTeX Code: G= \\\\{2q^2}/{h}$

totally for spin up and spin down electronic current. However, in the literature (and in the book I mentioned) it is written that

For a SWNT, when ballistic transport occurs, the resistance of the tube is length independent, and is, theoretically approximately 6.5kOhm. This resistance value results from having two propagation bands (called pi-bands) forming parallel propagation channels where each channel has resistance equal to the resistance quantum R=12.9kOhm Thus

$LaTeX Code: G= \\\\{2Nq^2}/{h}$

$LaTeX Code: G= \\\\{4q^2}/{h}$

Hence, I still understand that according to the plot, the conductance of the CNT is half of the value given in the literature?

Very thanks,

Regards

**carbon9** · Nov5-09, 03:42 AM

Thanks.

**sokrates** · Nov5-09, 03:28 PM

Yes, because "the steps" we just discussed come in pairs in Carbon Nanotubes.

This curious fact is directly related to the boundary conditions. Since Periodic Boundary Conditions (PBC) is naturally imposed in carbon nanotubes, the subband spacing in k-space is 2*pi/ c (c : circumference).

In PBC, in contrast to Box Boundary Conditions (BBC) where the wave function has to be zero at the edges, the states are more widely spaced. BBC is naturally imposed in graphene, because the edges are suddenly cut. And if you impose BBC - the state spacing in k-space is pi/c -- half the length as compared to PBC! So it seems like for the same width, a rolled up graphene sheet has less states than an unrolled one. But physically for the same graphene sheet, rolled up, or laid out, you would expect the same number of modes because after all, you are just changing the boundary conditions. Therefore, you can conclude that if the states are more widely spaced in PBC, then they must come in pairs to make the total number of states equal...! So when you increase the energy and hit a subband in CNT's, in fact you hit 2 (excluding spin) because they are two-fold degenerate.

Bottom line: Graphene conductance is usually quoted in 2q^2/h in the literature (including spin) and CNT conductance is reported in units 4q^2/h ... (including spin)

You can look into PBC and BBC more in detail, but this is the basic reason.

**carbon9** · Nov7-09, 11:07 AM

Thank you very much Sokrates.

**carbon9** · Nov12-09, 10:24 PM

Hi again,

In the same book of Hanson, the equation for the DOS of 1-D wires are given as

How can I determine the Enx,ny values for this equation for the given DOS plot at the first post? DOS function seems to start from 0 however it is greatet than 0 for metallic wires as shown in the plot.

Regards,
carbon9