N-channel MOSFET switch behavior: When VGS = 0 V on an N-channel MOSFET used as a switch, there is no ________ formed between source and drain.

Introduction / Context:
An N-channel enhancement MOSFET requires a positive gate-to-source voltage (VGS) above its threshold (Vth) to create an inversion layer (channel) that enables conduction between drain and source. Understanding this gate-controlled channel formation is critical for using MOSFETs as digital switches.

Given Data / Assumptions:

• The device is an enhancement-mode N-MOSFET.
• VGS = 0 V (gate tied to source potential).
• No body diode conduction is assumed in the forward-blocking direction for this question’s intent.

Concept / Approach:
At VGS = 0 V, the MOSFET is “off” because the gate electric field is insufficient to invert the channel region. Without inversion, there is no conductive channel linking source and drain, so the idealized switch is open. Only leakage currents flow. To turn the device “on,” VGS must exceed Vth (and often be driven higher for low RDS(on)).

Step-by-Step Solution:

Verification / Alternative check:
Examine MOSFET transfer characteristics (ID vs. VGS): at VGS = 0 V, the drain current is essentially leakage-level only, confirming no conduction channel.

Why Other Options Are Wrong:

• Voltage drop: Even when off, small leakage may exist; “no voltage drop” is not the defining concept.
• Capacitance/Inductance: Parasitic capacitances/inductances still exist; they do not vanish at VGS = 0.
• Body diode: Many MOSFETs include an intrinsic body diode; the question focuses on the channel, not diode conduction.

Common Pitfalls:
Assuming off-state equals an ideal open circuit with zero leakage; overlooking the body diode in high-side/low-side arrangements; forgetting that logic-level MOSFETs require sufficient VGS for low RDS(on).

Final Answer:
Conductive channel