The gate voltage in a JFET at which drain current becomes zero is called 

Explanation:

Shockley's equation is given as:

\({I_D} = {I_{DSS}}{\left( {1-\frac{{{V_{GS}}}}{{{V_P}}}} \right)^2}\)

​Where,

VGS = Gate to source voltage

IDSS = Drain to source saturation current

VP = Pinch-off voltage

According to Shockley's equation and characteristics curves for a P-channel JFET, the Drain current ID decreases with an increasing positive Gate-Source voltage (VGS).

The Drain current is zero when VGS = VP. For normal operation, VGS is biased to be somewhere between VP and 0. 

The characteristics curves for a P-channel junction field-effect transistor are shown below

Case I: If VDS = 0 and VGS = 0, the device will be idle with no current i.e. IDS = 0.

Case II:

• Now consider VDS to be negative while VGS is 0.
• At this state, the current flows from the source to the drain (as per conventional direction) as the holes within the p-substrate move towards the drain while being repelled from the source.
• The value of this current is restricted only by the channel-resistance and is seen to increase with a decrease in VDS (Ohmic region).
• However once the pinch-off occurs (VDS = VP), the current IDS saturates at a particular level IDSS, during which the device acts as a constant current source.

Case III:

• Next, let VGS = positive while VDS is negative.
• Here the effect exhibited is similar to that in Case II with the fact that the saturation occurs at a faster rate as the VGS becomes more and more positive.
• Here the current stop or ceases to flow as the value of VDS becomes equal to VP, turning the device into OFF state.

In JFET after pinch off the drain current becomes almost constant.