Here is what I have understood. Please correct me if I'm wrong.

The depth of induced channel depends on the voltage V. Since the voltage across the gate and the path along the channel decreases from Vgs at the source to drain end to drain end, the channel is not uniform an is tapered. If the applied voltage Vds is increased to some value which causes the voltage between gate and drain end to be lesser than the threshold voltage, the channel is said to be pinched off.

If the channel is pinched off, how does the drain current flow? It's given in the book that the MOSFET enters saturation region and will have a constant value of drain current. But I don't understand how does the current flow when the channel no longer exists at the drain end.


4 Answers 4


The current can still flow through the "substrate" even though the channel is pinched. The reason why it saturates is that there will be a region of higher resistance of size proportional to the Drain-Source voltage, and therefore the resistance of this region will be proportional to the same voltage.

But as current is voltage/resistance, the dependence will cancel out and you'll get "constant" current.

From Wiki (emphasis mine):

Even though the conductive channel formed by gate-to-source voltage no longer connects source to drain during saturation mode, carriers are not blocked from flowing. Considering again an n-channel enhancement-mode device, a depletion region exists in the p-type body, surrounding the conductive channel and drain and source regions. The electrons which comprise the channel are free to move out of the channel through the depletion region if attracted to the drain by drain-to-source voltage. The depletion region is free of carriers and has a resistance similar to silicon. Any increase of the drain-to-source voltage will increase the distance from drain to the pinch-off point, increasing the resistance of the depletion region in proportion to the drain-to-source voltage applied. This proportional change causes the drain-to-source current to remain relatively fixed, independent of changes to the drain-to-source voltage, quite unlike its ohmic behavior in the linear mode of operation. Thus, in saturation mode, the FET behaves as a constant-current source rather than as a resistor, and can effectively be used as a voltage amplifier. In this case, the gate-to-source voltage determines the level of constant current through the channel.

Also, from the MOSFET operation description, under saturation:

Since the drain voltage is higher than the source voltage, the electrons spread out, and conduction is not through a narrow channel but through a broader, two- or three-dimensional current distribution extending away from the interface and deeper in the substrate. The onset of this region is also known as pinch-off to indicate the lack of channel region near the drain. Although the channel does not extend the full length of the device, the electric field between the drain and the channel is very high, and conduction continues.


@clabacchio explained right, but still we can justify it with electrical field. Basically in mosfet, two electric fields are generated by gate voltage and drain voltage in channel and perpendicular to each other. When drain voltage reached Vp( piunch off voltage), channel tapered and disconnected with drain (N+) terminal. Now electrical field comes in picture:


here by tapering of channel, near source terminal channel remains same but near drain terminal it totally disappears, so electrical field near drain is:- E(gate )= V/0=infinte or very high. and electrical field is directly proportional to conductivity :- Sigma= nquE
if E is very high then even with out channel, a higher constant current flow from drain to source. Remains theory u can add from upper answer like I= V/R, v increasing and r is also increasing so we get constant voltage. plz let me know if i am wrong.


The problem is one of semantics.

"Pinched off" does not mean pinched closed. Instead it's a technical term; the label for an operating mode. "Pinchoff" means "device operates in constant-current mode." We could have chosen a different term, such as BJT's saturation/cut-off, but now we're stuck with it.

First, with the gate voltage reduced and the channel wide open, the channel has minimum resistance, with near zero voltage between drain and source; acting like a short circuit. Next, increase the gate voltage, so the depletion region invades the channel from the side. The channel behaves like a physically narrow material. It's resistance increases.

With a drain resistor present and gate-voltage increasing, voltage along the channel increases with increasing channel resistance. Finally it grows large enough to rival the gate voltage which produced the depletion region. At that point the linear "resistor behavior" becomes nonlinear, and the channel begins to behave like a constant current source, rather than a simple resistor. The constant-current mode is called "Pinch-off." But the channel is only narrow, not pinched closed or "off."

To remove confusion, perhaps personally label it "nonlinear mode?" "FlatPartOfTheGraph?" Other better labels?

During constant-current mode, the narrowed channel has gone into breakdown, and it's being shaped by negative feedback processes taking place between different regions along the channel. (If it was positive feedback, the channel would exhibit instability, "turbulence" and become an oscillator.)

With the channel narrowed and breaking down, if Vds is then increased, the pattern of channel-voltage and the borders of the depletion zone dynamically change, causing the narrow channel to become longer. Longer channel has higher resistance, with a self-regulating effect that produces a constant current independent of Vds changes.

Heh, increase the gate voltage sufficiently, the channel narrows to nothing, and the drain current falls to zero. The device enters "pinch-closed" mode!

Is there a hydraulic analogy? How about two balloons being pushed together. If we blow compressed air between a pair of balloons, that's a pair of depletion zones with a conductive channel between. Push the balloons together and, when they seem to be nearly touching, they'll distort, and make a loud raspberry sound. That's an unstable channel. A real FET channel wouldn't oscillate, instead it would remain narrow, with flattened balloons forming its parallel walls. Increase the compressed air feed, and the balloons change shape, making the channel longer, with length increasing just enough to keep the flow constant. Now push the balloons together harder (higher Vgs,) and the channel becomes uniformly narrower. Pushing the balloons 'programs' the constant-current effect for lower current; decreasing the flow.

For balloons, a higher "Vds" pressure-feed can never force a larger "Id" flow through the channel. Instead the balloons distort, widening the flat region, which increases the channel length proportional to increased Vds pressure. At some voltage the balloons would rupture. A Vds punch-through.


I wonder if real balloons would give a constant-Id characteristic? Even if they were making raspberry-sound oscillations, increasing the compressed air pressure would change their shape and their oscillation, and might end up lengthening the channel, preventing increased flow. Go and measure the loud, wobbulating balloons' Vds pressure, Id flow, and Vgs pressure. Ignore the unstable AC part, and you might be able to sweep the pressures and experimentally obtain a set of characteristic curves for the very first Field Effect Balloonistor (FEB.)


The channel gets pinched off near the drain region, this happens because as the VDS increase the drain body depletion region increases(due to reverse bias), now the electrons flowing through the channel are injected into the depletion region at the pinch-off region and are collected in the drain. So, the channel still conducts and current will flow.

You can also check, punch-through effect which occurs in short channel Mosfets, here the drain body depletion extends until the source body depletion thereby eating up the channel. Now, the electrons travel to the drain from the depletion regions.


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