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Pinch off of N-MOST from the physics point of view

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dkace

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I am reviewing my basics on microelectronics and I have some questions that probably every book thinks are self explained; I will be more specific:

When the MOS is entering the saturation region, we say that the channel is closed in the drain side. And as the drain voltage is increasing, the current is remaining fairly stable - with a small slope.

Ok, I understand the mathematics. Can anyone explain it in plain physics? If the channel is closed to a pin or is closed -period, how can current go through and we modelized it as stable? Is the channel really closed or we asume that a small path is left that can't be closed any more? Meaning that although we had a small resistance increasing with the increase of the drain voltage, this is the maximum resistance we will have?
And if so, why all odels refer to this as " closing of the channel"?

Can anyone decribe it from a physisist point of view?

Thanks,

D.
 

Re: Pinch off of N-MOST

User "asdzxc" wrongly posted reply in report field. I quote it here:

asdzxc said:
Let Vgs be held constant at a value greater than Vt and Vds appears as avoltage drop accross the length of the channnel. That is as we travel along the channel from source the drain the voltage increases from 0 to Vds Thus the voltage between the gate and points along the channel decreases from Vgs at he source end to Vgs - Vds at the drain end. Since the channel depth depends on this voltage we find the channnel is no longer uniform depth; rather the chanell take the tapered form, being deepest at he source end and shallowest at he drain end. As Vds increasedthe channel becomes more tapered and it is resistance increases correspondingly. Thus the Id-Vds does not continue as a straigth line but bends.Eventualy when Vds increased to the value that reduces the voltage between gate and channel at the drain end to VT- that is Vgs-Vds=Vt the channel depth at the drain end decreases to almostr zeroand the channel is said to be pinched of. Increased Vds beyond this value has thoeorcaly no effect on the channel shape and the current through the channnel remains constant. Then we say the chanel is closed.The drain current thus saturates and mosfet is said to have entered the sturation region.
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    kunu

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Re: Pinch off of N-MOST

Let Vgs be held constant at a value greater than Vt and Vds appears as avoltage drop accross the length of the channnel. That is as we travel along the channel from source the drain the voltage increases from 0 to Vds Thus the voltage between the gate and points along the channel decreases from Vgs at he source end to Vgs - Vds at the drain end. Since the channel depth depends on this voltage we find the channnel is no longer uniform depth; rather the chanell take the tapered form, being deepest at he source end and shallowest at he drain end. As Vds increasedthe channel becomes more tapered and it is resistance increases correspondingly. Thus the Id-Vds does not continue as a straigth line but bends.Eventualy when Vds increased to the value that reduces the voltage between gate and channel at the drain end to VT- that is Vgs-Vds=Vt the channel depth at the drain end decreases to almostr zeroand the channel is said to be pinched of. Increased Vds beyond this value has theoricaly no effect on the channel shape and the current through the channnel remains constant. Then we says the channel is closed The drain current thus saturates and mosfet is said to have entered the saturation region.
 

Pinch off of N-MOST

>...chanel is closed.

The channel is not closed. It is pinched off. The channel shape becomes triangular. The point of pinch-off travels toward source when VDS increases beyond Vp. The channel triangle becomes a bit shorter, while the voltage drop over it remains the same, Vp.
The excess voltage VDS-Vp drops over the depletion region, between the Vp point and the drain. This region extracts the channel current (saturation). Shortening the triangular increases a bit its conductivity, and this increases the IDS (Early effect).
 

Re: Pinch off of N-MOST

Well, for sure the channel is not remaining the same after pinch off rather than creating a lower hight triangle zone which is far from reaching drain as VD is growing.
The best explanation is presented in WESTE: Although the channel of the carriers is closed the current we have as constant current after pinch off is from carriers travelling through the depletion layer. I can't understand why this movement is not represented with a fixed width carrier channel in the model. As far as I have understand, in representing models, there is always a reason we show or not show something.
Any explanation?

D.
 

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