Difficulty: Easy
Correct Answer: the value of VDS at which further increases in VDS will cause no further increase in ID
Explanation:
Introduction / Context:
An n-channel Junction Field-Effect Transistor (JFET) is a depletion-mode device whose channel conductivity is controlled by the reverse-bias on its gate-to-channel PN junction. A key operating point is the pinch-off condition, where the channel is sufficiently depleted so that increasing the drain-to-source voltage (VDS) does not significantly increase the drain current (ID). This question checks your grasp of the JFET output characteristics and the definition of pinch-off in the saturation (constant-current) region.
Given Data / Assumptions:
Concept / Approach:
For a given VGS, as VDS rises, the reverse bias along the channel increases and the channel narrows. At the pinch-off voltage Vp (for that VGS), the channel is just pinched near the drain end and the device enters the so-called saturation or constant-current region: ID remains approximately constant even if VDS increases further.
Step-by-Step Solution:
Verification / Alternative check:
Inspect a JFET family of output curves (ID vs. VDS for several VGS). Each curve shows a knee where it flattens: that knee corresponds to the onset of pinch-off (saturation region). Measuring ID beyond the knee shows minimal change with VDS.
Why Other Options Are Wrong:
Common Pitfalls:
Final Answer:
the value of VDS at which further increases in VDS will cause no further increase in ID
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