QuantumATK Forum
QuantumATK => General Questions and Answers => Topic started by: Shan on July 6, 2015, 16:41
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Hello QW team,
what is the default metal it takes when we insert the metallic region using 'spatial region' option in atk builder. There is an option to change the dielectric constant when we insert the dielectric layer, but there is no option available to tune the work function or conductivity of the metal.
Then what is the default metal used here... and how to change the metal type.
thank you...
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The metal work function at zero gate bias is defined by your potential, sort to speak.
Please, have a look at this tutorial which explain how to tune the gate bias to get the desire metal gate work function:
http://quantumwise.com/documents/tutorials/latest/InAsDevice/index.html/
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Dr. Umberto Martinez,
I had almost a similar query, regarding the spatial region.......
1) If we take ....a spatial region...say oxide...on top of that a metal....and then apply a voltage (like a back gate)...what actually happen to the transmission...?? {because in such a case, there is no concept of lattice mismatch, surface states...etc. etc. / only that I include is the dielectric constant value...!!! }
2) The voltage that we have applied....; how does that get counted..??!! Is it including a potential term in the Hamiltonian??
3) Is there at all any diff. ...... if I take a point...or large slab of metal.... while applying a bias..??!!
Thanks & Regards_
Dipankar
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1. It becomes a transistor (3-terminal device), but gate current is not taken into account
2. Yes, in the electrostatics
3. The thickness of the metal doesn't matter, but don't make it too thin or you risk that there is no grid point inside the metal
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.....but gate current is not taken into account
May be I didn't quite get..what you tried to mean....!!!
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Let, me give you an example_
Firstly, I didn't included any gate...so it was merely a two port struct. / For a particular bias at one of the electrodes.... I get the Tr. spectrum...as show in Fig.1_a . / A self consistent I_V of that two port device...is shown in Fig. 1_b...../ Then, keeping the same right electrode bias.... I applied a gate voltage (say 1V)...through the spatial region..... My Tr. spectrum (Fig.1_c) got completely changed..... so the self consistent I_V (Fig. 1_d)/ But, there after.....even if I took diff. gate voltage values, .....I didn't find any change in the I-V (as shown in Fig. 1_d) !!! :o
What might be the reason??!!
Thanks & Regards_
Dipankar
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Even if the gate current is not taken into account and the dielectric is just a change in epsilon, with application of different gate bias there should be change in the transmission spectra, as the bands would be shifted up/down accordingly. ???
But even with different "shape" of transmission spectra if the area under the drain bias window is the same then it would give the same current. :o
Maybe that is happening?? We should check that. Possibly an increase in the drain bias would enlarge the window to give a better understanding.
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It's not really the matter of the bias window, I guess...!! Because....when you are finding the I-V...self consistently...you are varying your right electrode voltage for a sufficient range.... (see it's 0 to 1 V).... :)
Say,
Vg = 0.5 V : Now, varying right electrode voltage from 0 to 1 ...will of course change the bias window in 5 diff. cases. So, your current varies....and you get the I_V plot.....
Next,
Vg = 1 V : ---do the same----
Vg = 3 V : ---do the same----
If in all the cases....you have the same (exactly identical) I-V curves.....;
that means ...... there is hardly any change with the change in Vg !! ..... Is it not??
Regards_
Dipankar
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Not quite sure I understood the real problem. Anyway, please add all your input files so we can heck what your guys are actually doing.
thanks.
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Dr. Umberto Martinez,
Yes...I will send those to you.... :)
Thanks_
Dipankar
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Your metal region is outside the cell, hence it is not taken into account in the calculation.
Please, increase your vacuum region both above and below your system and add your regions into the cell.
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Okayyy...there was problem...!!! Let me redo the whole...... :)
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But, this in turn....leads to another interesting point....
If the applied Vg was not at all coupled to the channel....then the oxide slab would have been merely a support....
How come that isolated region is altering Tr. spectrm. (See Fig1_c) hugely??
Anyways, thanks a lot Umberto..... :)
Regards_
Dipankar