“OFF” versus “ON” devices: Among the listed components, which one is considered a normally-OFF device (no conduction at zero gate bias under ideal conditions)?

Introduction / Context:
Switching applications and bias planning require knowing whether a device conducts by default. Depletion-mode devices (JFETs, D-MOSFETs) are normally ON and must be biased to reduce conduction, whereas enhancement-mode MOSFETs are normally OFF and need a gate drive to conduct. BJTs require base bias to conduct but are not categorized as depletion/enhancement devices.

Given Data / Assumptions:

• Ideal device behavior at VGS = 0 for FETs.
• No leakage or parasitics considered for the conceptual answer.
• Room-temperature, low-frequency operation.

Concept / Approach:

An E-MOSFET lacks a physical channel at zero gate bias; it needs a gate voltage above threshold to induce an inversion layer and allow current. JFETs and D-MOSFETs have pre-existing channels and conduct at VGS = 0, requiring reverse gate bias to reduce current. BJTs are current-controlled devices; without base current they do not conduct significantly, but the “normally-OFF” classification is conventionally applied in the FET context to enhancement-mode MOSFETs.

Step-by-Step Solution:

Verification / Alternative check:

Datasheets list ID(off) for E-MOSFETs at VGS = 0 and specify threshold voltages where conduction starts. Application notes for power switching emphasize the fail-safe OFF state with no gate drive.

Why Other Options Are Wrong:

• JFET and D-MOSFET: normally ON (conduction at VGS = 0).
• BJT: needs base bias; classification context here is FET “OFF” vs. “ON.”
• Vacuum diode: conducts with correct polarity and sufficient voltage; not a gate-controlled OFF device.

Common Pitfalls:

• Assuming “no gate drive” equals “no conduction” for all FETs; depletion devices contradict this.
• Confusing leakage currents with purposeful conduction.

Final Answer:

E-MOSFET (enhancement mode)