When drain voltage equals the pinch-off voltage, the drain current

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.