I am confused about the MOSFET parameters. Consider a depletion N-channel MOSFET. Suppose the gate-source (Vgs) threshold is Vth = -2V, in its off position. Suppose we have not reached the threshold and we are at Vgs = -1 V. This means that the MOSFET will conduct when a voltage Vds is applied (let Vds = +3 V to drain). There also exists the pinch-off voltage Vp beyond which saturation occurs.

Actual datasheets seem to suggest that Vth and Vp are independent of each other, yet I have seen statements such as Vds > Vgs - Vth. I am not sure if Vp = Vgs - Vth, as some discussions seem to suggest.


1 Answer 1


The threshold voltage is the gate-source voltage \$V_{GS}=V_{Th}\$ at which drain-source charge flow begins.

The pinch-off voltage is a drain-source voltage \$V_{DS}=V_{PO}\$ at which saturation occurs. Definitely independent from \$V_{Th}\$.

(updated) However, the pinch-off voltage is dependent on \$V_{GS}>V_{Th}\$ as shown in the image clipped from Dr. Bakr's lecture notes below. Clip from Dr. Bakr lecture notes

  • \$\begingroup\$ It is stated that the pinch-off locus constitutes Vgs-Vth, but I am not convinced it is actually true. \$\endgroup\$
    – Roger
    Commented Nov 23, 2022 at 2:57
  • \$\begingroup\$ Let me give you an actual example. Suppose I pass Vgs=-1V to the gate (let Vth=-3.5V). Would that mean that if I use Vds=4V (positive on the drain), I would in effect be supplying -5V to the gate? \$\endgroup\$
    – Roger
    Commented Nov 23, 2022 at 3:10
  • \$\begingroup\$ @Roger: Not at all. The voltage \$V_{DS}\$ has no effect on \$V_{GS}\$. All \$V_{GS}\$ does is modulate the channel resistance. There is no \$V_{DS}\$ unless there is drain current or a supply voltage \$\endgroup\$
    – RussellH
    Commented Nov 23, 2022 at 3:16
  • \$\begingroup\$ If I keep Vgs to 0V and apply a positive voltage to the drain, will saturation occur at when Vds reaches -Vth ? If Vth is -3.5V, then would applying Vds at 3.5 give saturation? \$\endgroup\$
    – Roger
    Commented Nov 23, 2022 at 4:53
  • \$\begingroup\$ @Roger: To both your comment questions, you are correct. See my updates and view the link. \$\endgroup\$
    – RussellH
    Commented Nov 23, 2022 at 13:21

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