JFET cutoff definition: Is VGS(OFF) the magnitude of gate-to-source voltage required to reduce the JFET drain current essentially to zero (turn the device off)?

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
Key JFET parameters include IDSS and VGS(OFF). Together they describe the transfer characteristic and are used in bias calculations. VGS(OFF) is the gate-to-source voltage at which the channel is fully pinched off and the drain current approaches zero (cutoff).


Given Data / Assumptions:

  • Normal operation keeps the gate-channel junction reverse-biased.
  • VGS(OFF) has opposite polarity to IDSS sign convention (e.g., negative for n-channel).
  • Shockley equation applies in the saturation region.
  • “Essentially zero” means only leakage remains.


Concept / Approach:
The depletion regions from the gate junction widen with increasing |VGS|, narrowing the channel. At VGS(OFF), the channel is pinched off across its entire length, stopping conduction apart from leakage. This is the end-point on the transfer curve opposite IDSS (at VGS = 0 V).


Step-by-Step Solution:

Start from Shockley: ID = IDSS * (1 − VGS/VP)^2; with VP ≈ VGS(OFF).Set ID → 0: occurs when VGS approaches VGS(OFF).Interpret polarity: n-channel has VGS(OFF) negative; p-channel positive.Conclude: definition aligns with the statement.


Verification / Alternative check:
Datasheet transfer curves mark cutoff at VGS(OFF) where ID is near the leakage floor, confirming the interpretation.


Why Other Options Are Wrong:

Incorrect: conflicts with standard definition.Only for MOSFETs: VTH is used for MOSFETs; VGS(OFF) is standard for JFETs.Synonym for IDSS: IDSS is maximum ID at VGS = 0 V, not cutoff.Drain–gate breakdown: a separate maximum rating (BVGD or similar), unrelated to cutoff.


Common Pitfalls:
Confusing VGS(OFF) with MOSFET threshold; missing sign conventions; using ID ≈ 0 without accounting for leakage in precision designs.


Final Answer:
Correct

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