Difficulty: Easy
Correct Answer: Incorrect
Explanation:
Introduction / Context:Metal-oxide-semiconductor field-effect transistors (MOSFETs) come in depletion and enhancement varieties. Enhancement-type MOSFETs are normally off at VGS = 0 V and require a gate voltage exceeding a threshold to create (enhance) a conductive inversion channel. Misstating this behavior leads to biasing errors and device misuse.
Given Data / Assumptions:
Concept / Approach:For an E-MOSFET, conduction begins only after the gate-to-source voltage induces an inversion layer. Saying it turns on “when the channel is depleted” confuses the depletion device mechanism with enhancement behavior. An enhancement device cannot rely on a pre-existing channel; it creates one by field effect once |VGS| reaches VTH and grows with further |VGS| increases.
Step-by-Step Solution:
Classify device: enhancement vs depletion.Recall condition: E-MOSFET needs |VGS| ≥ |VTH| to form a channel.Interpret the claim: “channel depleted” describes turning a depletion device off, not turning an enhancement device on.Conclude: statement is incorrect for enhancement MOSFETs.Verification / Alternative check:Look at ID–VGS transfer curves: for enhancement devices, ID ~ k*(VGS − VTH)^2 (simple square-law model) for VGS above VTH, and ~0 below threshold—reinforcing that an induced channel is required.
Why Other Options Are Wrong:
Correct: contradicts E-MOSFET physics.Depends only on ID rating: ratings do not change the turn-on mechanism.Only true for depletion MOSFETs: this clarifies the confusion but is not the given statement.Works with gate shorted to source: with VGS = 0 V an E-MOSFET remains off (neglecting leakage).Common Pitfalls:Confusing depletion and enhancement types; assuming any MOSFET conducts at VGS = 0 V; ignoring threshold spread and temperature effects.
Final Answer:Incorrect
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