Voltage-divider biased JFET: Is the drain current ID at its maximum value when VGS = 0 V?

Introduction / Context:
Bias curves for JFETs show how gate-to-source voltage (VGS) controls the drain current (ID). A common memory point is that the maximum ID under normal operation occurs at VGS = 0 V and is denoted by IDSS. Recognizing this helps designers quickly estimate operating points in voltage-divider bias networks.

Given Data / Assumptions:

• JFET is operating in its normal region with the gate-channel junction reverse-biased.
• VGS controls channel pinch through depletion width.
• IDSS is defined as ID at VGS = 0 V.
• Shockley equation models behavior: ID = IDSS * (1 − VGS/VP)^2 for VGS between 0 and VP (negative for n-channel).

Concept / Approach:
As VGS becomes more negative (for n-channel) or more positive (for p-channel), the channel is pinched more strongly, reducing ID. Therefore, when VGS is at 0 V, the gate exerts the least depletion effect, giving the maximum drain current IDSS for a given device and temperature.

Step-by-Step Solution:

Verification / Alternative check:
Examine ID–VGS transfer curve in any JFET datasheet: the peak of the curve at the left (for n-channel) corresponds to VGS = 0 V with value IDSS, confirming the claim.

Why Other Options Are Wrong:

Common Pitfalls:
Mixing up conventional current with carrier motion; forgetting that IDSS assumes operation in the saturation (active) region; assuming IDSS at any VDS without checking datasheet test conditions.

Final Answer:
Correct