According to my book and teacher :
It is the Gate voltage at which the depletion region closes the channel.
According to some people on the internet and Youtube videos:
It is the Drain to source voltage, keeping Gate voltage zero, at which the drain current becomes constant.
So which one is the pinch-off voltage? Vgs or Vds ?
Confusing, because the "pinchoff state" is the second one. This state appears when the Vgs voltage is large enough that the channel becomes a constant-current source (because it's now being length-modulated by Vds voltage. The channel-resistance is changing so as to keep Id constant.)
So, call it the FET "pinchoff-state voltage."
On the other hand, that first one; the Vgs voltage required for FET cutoff state, is entirely different. It has little to do with the FET pinchoff-region. Perhaps call it the "pinch-closed voltage."
(Does your textbook author really have this noob misconception? Could be. There are plenty of bad books out there, where authors are teaching their own misconceptions to thousands.)
As a kid I was very confused about these, and couldn't understand how FETs could even work, if the linear region only happens when the channel is entirely "pinched," meaning closed off. I don't know where I picked up this wrong idea. Maybe a textbook wasn't clear enough. Or maybe an author really thought that Pinchoff means Cutoff or Closed Off.
DOH, "Pinchoff" is not pinch-closed! Now I understand evvvvrything!!!
In the set of FET characteristic curves, the flat, constant-current VI curves are the "Pinchoff region," while the sloped curves going through the orgin are the "resistive region." We can draw a parablolic curve to separate the two regions (upside down parabola.) This curve is often called "Pinchoff voltage," and it's a Vds drain-source voltage. And of course it's a different voltage for each value of Vgs gate voltage.
Analogy: if you have a flow of water, then try to stop it by pushing two balloons in from both sides. It acts like a resistor. Push the balloons close together, and suddenly they flatten! The water isn't stopped. Instead, suddenly the width of the gap between the balloons becomes constant. That's "pinchoff region." It's a weird fluid-dynamics mode. The water-flow in the gap between balloons starts acting like a slab of constant thickness. The gap stops behaving as a resistor. If you increase the water pressure coming from above, the gap region becomes longer, and the net flow stays the same. Very weird, no? That's Pinchoff operating mode.
But if you push the balloons together much harder, you can pinch-closed the water flow, reducing it to zero, and putting the FET into cutoff.
In actual FETs, during Pinchoff mode, the channel behaves oddly because it goes into avalanche breakdown, starts emitting a visible glow, and dissipates significant heat. Applying higher voltage to the ends will just make the channel grow longer. It starts acting like a wire of varying length (where the length of the channel doubles if you double the Vds voltage, but the thin width stays the same.) And if viewed under a NIR microscope camera, all the mosfets on the chip will have a dim infrared glow, if they're operating in linear analog mode (Pinchoff region.)
For JFETs IDSS occurs at Vgs=0 and then pinchoff is alway stated in specs as simple OFF threshold for Vgs. It is rated for a fixed Vds, Id leakage current (e.g. 10nA) at pinchoff = \$V_{GS}=V_{GS(off)}\$
e.g.
They are actually both correct.
The more natural definition is the voltage that cuts off the changel and reduces the drain current to an extremely low level.
However the pinch off voltage also sets the voltage at which the device transitions from the linear region to the saturation region. As the drain voltage increases it tends to close off the channel and cause the current to remain constant rather than increasing linearly as in the triode region.
