Problems in applying electric field onto nanoribbon structure.

[donmbringer]

Hi,

I have two questions when I built the model of GNR with applied field.

1. In the attachment, figure 1 is the device I built in builder, figure 2 the the model of junction that has already been included in customer builder (the nanoribbon is not the same, but this question is concerning the applied field). clearly, the way of applying electric field seems to be different, but can they get the same results if the nanoribbons are the same for both? in addition, how to add the dielectric in my model?

2. If I want to calculate the conductance vs. gate voltage in my model, can I just use the same program shown in tutorial "graphene junction device, chapter 4" after I get the transmission spectrum using the scripts generator?

Or, the code is only suiltabl for the model insider customer builder? If so, what should I do?

Thanks ahead!

[Anders Blom]

1. To add the dielectric you do the same thing as when you inserted the metal gates. In the "Spatial Regions" widget you can choose if the region should be metallic or dielectric. You model is a bit different since you have a top and a bottom gate. If that is the situation you want to model, then you should set it up like this - there is no right or wrong, it's a choice you make. The results will be somewhat different.

Do note that applying an electric field to a single layer of graphene has no effect - you need a bilayer structure. In your case there will anyway be an effect since the field is only applied in a certain region in space (not in the electrodes), but it will probably more resemble a doping effect than anything else (a more or less constant shift of the onsite potential for all atoms in a region in space). But our example junction pretty much works that way too.

2. The general ideas apply to any system, but I can't promise all the scripts will work right away, you may need to adapt them (but only very little, if your system is similar).

[donmbringer]

Thanks for such a detail reply.

May I ask, If I build the top gat (only) and the dielectric by myself using the spatial region and I, for example set the voltage to be 2V. Will the electric field goes perfectly downwards like figure 1F, or, some non-uniform side field will also be included (like 2F)?

In addition, I feel confused about this point. If we do not include the botton electrode, we then do not have the reference (the surface with V = 0). How can we get the strength of the electric field with V/nm?

Thanks ahead

[Nordland]

Really nice graph.

If the gate extends the entire cell, it will be like your first graph. If the gate do no extend entire cell, then the second picture is more correct.
I have a post on this somewhere on the forum, let me dig it out.
(found it http://quantumwise.com/forum/index.php?topic=1597.msg8232#msg8232 )


If you dont have a gate at the bottom, you have to set the two options - you can set Dirichlet Boundary Condition which is this case will have the same effect as
setting the bottom gate. You will then be able to calculate the field simply by taking the gate voltage and divide it with the length from the edge of your gate to other side.

You can also go with Neumann Boundary Condition, then you will have to calculate the electrostatic potential on the bottom and then you can substract this from the gate voltage and divide with the length to get the field.

[donmbringer]

I see. Thanks for such a clear explanation! 

[donmbringer]

Sorry. Another problem arise. When I test the I-V curve of the configuation in Figure 1, after I get the transmission spectrum, I run the program in the attachment, which was copied from tutorial. However, I get the following warning, and the simulation has taken more than one day, and haven't been finished. Could you please help me to have a look on this problem?

################################################################################
# WARNING                                                                      #
#                                                                              #
# The computed multigrid residual is greater than the required accuracy.       #
#                                                                              #
# Computed residual :   7.91855e-012                                            #
# Required accuracy :   1.00000e-012                                            #
#                                                                              #
################################################################################

[Anders Blom]

As long as the error is so small, it's nothing to worry about. If it gets larger, however, like 1e-5 or 1e-6, you may have a real problem.