Enhancement-only channel formation: Which device type has no physically formed channel at zero gate bias and becomes conductive only when properly biased at the gate?
Correct Answer: n-channel E-MOSFET (enhancement)
Introduction / Context:Choosing the correct FET requires knowing whether the channel exists at zero bias. Enhancement-mode MOSFETs are normally off, while depletion-mode devices and JFETs typically conduct at zero gate bias (subject to pinch-off with reverse gate bias). This question targets that fundamental distinction.
Given Data / Assumptions:
- Comparing JFETs, depletion-mode MOSFETs (D-MOSFETs), and enhancement-mode MOSFETs (E-MOSFETs).
- Zero gate-source bias as reference condition.
- Silicon devices at typical operating conditions.
Concept / Approach:In an E-MOSFET, no channel exists at VGS = 0. Applying a gate voltage of suitable polarity induces an inversion layer, creating a conductive channel. For an n-channel E-MOSFET, VGS must exceed a threshold VTH > 0 to conduct. By contrast, JFETs have a doped channel that conducts at VGS = 0 and is pinched off by reverse bias; D-MOSFETs also have a channel at VGS = 0 that can be depleted or enhanced by gate voltage.
Step-by-Step Solution:
Check channel presence at VGS = 0: E-MOSFETs — none; JFETs and D-MOSFETs — present.Assess polarity: For n-channel E-MOSFET, apply positive VGS ≥ VTH to induce channel.Conclude the correct device: n-channel E-MOSFET.Verification / Alternative check:Datasheets list threshold voltage VTH for E-MOSFETs and show negligible ID at VGS = 0. JFET and D-MOSFET characteristic curves show significant ID at VGS = 0 that falls with reverse bias.
Why Other Options Are Wrong:
- p-channel JFET (option a) has a physical channel and conducts at VGS = 0.
- n-channel and p-channel D-MOSFETs (options b, c) are depletion-mode, so a channel exists at VGS = 0.
Common Pitfalls:Assuming all MOSFETs behave the same at zero bias; enhancement devices are normally off.
Final Answer:n-channel E-MOSFET (enhancement).