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jason_class

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using energy level and potential
Mar 04^th, 2006, 7:20pm

Hi Mrphy and All

I need some help in understanding the correct way to find the difference for energy level and also for potential difference.
Say for a diode, text books show that Vapplied =Vp -Vn where Vp=psi_p =quasi-Fermi potential for holes whileVn for electron.
This equation is used for both reverse and forward bias condition.
My question is why must Vp - Vn and not the other way around Vn -Vp?
What is the convention to use?

Then move to the example of p mos capacitor(or nmos),

Condition for inversion is Vs =2Vb.
It is said that Vs(or Vb) is postive for p mos capcitor(or nmos) while negative for n mos capacitor(pmos).
What is the correct way to get Vs from the below 4 choices which I could think of and teach me why it is so?
1) qVb = Ei - Ef Should qVb be writen as -qVb?
2) -qVb = Ei - Ef where -qVb is using equation E = -qV
3) qVb =Ef -Ei
4) -qVb = Ef - Ei

Kindly enlighten
Thank you Murphy and All

best regards
Jason

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Croaker

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Re: using energy level and potential
Reply #1 - Mar 5^th, 2006, 8:26pm

I'm not really sure what you are trying to ask in the first part.

The energy band diagrams are for electrons. Conventional potential is for positive charges. The electron has the highest energy where the potential for a positive charge (this is the convention for voltage) would be lowest. When going from electron energy to voltage, a negative sign must be applied.

E.g. if there is a band energy difference of 10 eV, an electron has to experience a 10 V drop in potential to move to the higher band

In other words, the electron energy band diagrams are inverted relative to voltage diagrams; moving up on the electron energy diagram is the same as moving down on the voltage diagram. *IF* voltage meant the potential of a negative charge, they would not have an inverted relationship.

-qV = E

For an NMOS device we have a p-substrate. We apply a positive voltage to the gate to bend the bands. Ef cannot change because the device is still in equilibrium (no current flows), so Ei bends.

For an inverted NMOS device, at the surface, Ei is below Ef by |Ei-Ef|; the total energy difference is 2*|Eib-Efb|, where Eib and Efb are the energy levels in the bulk.

With the way you've defined Vb:

Ei - Ef = -q*(Vi - Vf)
(Ei - Ef)/q = Vf - Vi = Vb

For what you presented, both 1) and 4) are equivalent.

The key is just understanding the inverted relationship between electron energy and voltage.

Cheers,
Marc

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« Last Edit: Mar 6^th, 2006, 5:20am by Croaker »

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jason_class Community Member Offline Posts: 39 singapore	Re: using energy level and potential Reply #2 - Mar 6^th, 2006, 2:44am Hi Murphy Your explanation helps a lot. In fact i figured out a few ways and each almost gives the same answer but with different sign. Therefore I hope someone in the forum can point out the exactly correct way to do it. Thank you Murphy! best regards jason
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Croaker

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Re: using energy level and potential
Reply #3 - Mar 6^th, 2006, 5:18am

jason_class wrote on Mar 6^th, 2006, 2:44am:

Hi Murphy

Your explanation helps a lot.
In fact i figured out a few ways and each almost gives the same answer but with different sign.
Therefore I hope someone in the forum can point out the exactly correct way to do it.
Thank you Murphy!

best regards
jason

I'm not sure what you mean by the 'correct' way to do it. There are only two ways, both are equivalent.

1) (Ei - Ef)/q = Vf - Vi
2) (Ef - Ei)/q = Vi - Vf

These are just opposite in sign; both are true. E.g. if Vf is 10 V above Vi, Vi is 10 V below Vf. It's the same as when you take the potential difference between two nodes in a circuit: Vab = -Vba.

Cheers,
Marc

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