Pinch-Off and Saturation Of MOSFET

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Hi,

I don't understand pinch-off in MOSFETS.
I have two big issues:

1.
VDS is pointing from drain to source, so the voltage works in the horizontal plane.VGS on the other hand,is in the vertical plane.
So how they can cooperate? How does increasing VDS till VGS - VTh get you in pinch-off.
Why does increasing VDS above VGS - VTh result in the moving of the pinch-off point to the left?
In what sense are VGS and VDS interacting with each other, I don't see it.

2.
I don't get how electrons are swept from the shortened channel bye the electric field of the depletion regio towards the drain.
With all the voltages from Gate and Drain and built in voltages of depletion regio, I don't know where the electric field is pointing at.

Here's a sketch, a messed up sketch from me in Paint:

7tyPfhx.jpg


Hope some people can explain it to me.
Thanks in advance.
 
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Hi.

Change the point of view of the voltages. Instead VDS, analyze VGD. So, you have VGS, VGD (so VDS = VD - VS = VGS - VGD).
Think about this: A MOS transistor is, in principle, purely simmetryc. (look into the schematic, you can switch S with D).

If VGS > vt, and VGD > vt , (then VDS = VGS - VGD > 0) the MOS works in triode mode (the channel is entirely formed across L).
Now, if you want to analyze another mode, for example saturation:

VGS > vt, VGD < vt (channel isn't formed in the drain-side) then:

VGD = VG - VD < vt ---> VS + VG - VD < vt + VS --> VSD < vt + VSG --> VDS > VGS - vt
(It's easier connecting S to 0 V)
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Why does increasing VDS above VGS - VTh result in the moving of the pinch-off point to the left?

This happen because you're forcing no channel forming en the D-Side.
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I don't get how electrons are swept from the shortened channel bye the electric field of the depletion regio towards the drain.

The Electric Field betweed the Drain and the channel is very large, and the density of electrons it's huge in the channel, so, it's a electrostatic force driving the electrons (some of them, see the ID vs VDS curve).