In the constant-current (pinch-off) region of an n-channel JFET, how does the drain current ID vary as the gate-to-source voltage VGS changes?

Introduction / Context:
In an n-channel JFET, the gate controls channel depletion. The constant-current (often called saturation) region describes operation where ID is largely independent of VDS but strongly dependent on VGS. Understanding how ID responds to VGS is essential for biasing and gain control.

Given Data / Assumptions:

• Device: n-channel JFET.
• Operating region: constant-current region (beyond pinch-off in VDS).
• Question: qualitative dependence of ID on VGS.

Concept / Approach:
Making VGS more positive (less negative) widens the channel and increases carrier density, causing ID to rise toward IDSS. Making VGS more negative shrinks the channel and reduces ID, reaching near-zero at VGS(off). Thus ID is a monotonic function of VGS in this region.

Step-by-Step Solution:
Start at some VGS < 0 with ID set accordingly.Increase VGS (toward 0): channel depletion reduces → ID increases.Decrease VGS (more negative): depletion increases → ID decreases.Therefore, the correct qualitative statement is: as VGS increases, ID increases.

Verification / Alternative check:
Transfer curves (ID vs. VGS) from datasheets always show rising ID as VGS increases from VGS(off) toward 0 V for n-channel JFETs, confirming the relationship.

Why Other Options Are Wrong:

• As VGS decreases, ID increases: Opposite of actual behavior.
• As VGS decreases, ID remains constant / As VGS increases, ID remains constant: ID is not constant with VGS; only weakly dependent on VDS in this region.

Common Pitfalls:

• Confusing VGS dependence with VDS dependence; “constant-current region” refers to the latter.
• Mixing p-channel and n-channel sign conventions.

Final Answer:
As VGS increases, ID increases.