What is the purpose of an inversion layer in a MOS device?

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The discussion centers on the role of the inversion layer in MOS devices, specifically its formation and significance in charge movement. An inversion layer forms when a sufficient voltage causes the Fermi energy to exceed the intrinsic energy level, attracting electrons to the oxide/semiconductor boundary. This layer is crucial as it establishes a conducting channel that allows for the movement of charge carriers, essential for device operation. The conversation also touches on the derivation of a formula for calculating dark current, with references to Fowler-Nordheim theory, indicating a broader interest in understanding the underlying principles of MOS technology. Overall, the inversion layer is vital for enabling current flow in MOS devices.
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[SOLVED] inversion layer in a MOS

Hello,

I have some questions about MOS devices, and CCD

In a MOS or MIS after applying a voltage on the metal, the valenceband and the conduction band bend downwards (in the usual band diagram) when a positive voltage is applied. (p-type semiconductor).
When the voltage is large enough, the fermi energy (in equilibrium) is above the intrinsic energy level, so that an inversion charge will build up.

The first thing I don't understand is why the electrons make an inversion layer in the first place. I mean, why would the electrons come so close to the oxide/semiconductor boundary in the first place (I understand the electrons are drawn to this boundary by the applied voltage, but why is it such a big deal that the Fermi energy level is larger than the intrinsic level?)

The second thing I don't understand, is what is the use of this inversion layer? Why is the inversion layer necessary in a CCD?

Any help would be greatly appreciated,

thanks in advance,

Oh, and a bonus 'question': I've seen a useful formula for calculating the dark current, but I can't find a reference. It's about this formula:

amount of dark current generated = CT^{1.5}e^{-E_g/2kT}

I've seen it in some books, but none of them make a reference to where it is derived, and I don't have a clue how to derive it myself.
(Maybe this question isn't in the right place here, then I'll try a different subforum).
 
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Theraven1982 said:
Oh, and a bonus 'question': I've seen a useful formula for calculating the dark current, but I can't find a reference. It's about this formula:

amount of dark current generated = CT^{1.5}e^{-E_g/2kT}

I've seen it in some books, but none of them make a reference to where it is derived, and I don't have a clue how to derive it myself.
(Maybe this question isn't in the right place here, then I'll try a different subforum).

Look into the Fowler-Nordheim theory for field emission current. The derivation to get that particular equation isn't trivial and makes many simplifications and assumptions.

Zz.
 
Thanks a lot for your answer.
It is not necessary that I know exactly how it is derived, but just wanted to know where it came from. I've seen the name 'Fowler-Nordheim' before in a paper, and now I at least understand where it comes from.

Any takers for the other questions? ;)
 
The inversion layer is necessary to establish a conducting channel through which majority carriers can easily move.

- Warren
 
Thanks.
But why is this conducting channel needed? This inversion layer is above the depletion region which fills up with charge; when the device is clocked (for measuring how much charge is in the well), the charges move from well to well. I don't see the need for this conducting channel.
 
Theraven1982 said:
Thanks.
But why is this conducting channel needed? This inversion layer is above the depletion region which fills up with charge; when the device is clocked (for measuring how much charge is in the well), the charges move from well to well. I don't see the need for this conducting channel.

The inversion layer provides the majority carriers (electrons or protons in nMOS and pMOS case) which make up the current. The inversion layer IS the conducting layer

marlon
 
The inversion layer provides the MINORITY carriers not the majority carriers.

My mistake.

marlon
 
I'm sorry for the (very) late reply, but thanks a lot! I now fully understand. Sometimes I need to ripen new information ;).
Thanks again all!
 

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