Difficulty: Easy
Correct Answer: Correct — with the drain at a higher potential than the source (VDS > 0) and the gate reverse biased, the n-type JFET channel conducts.
Explanation:
Introduction / Context:
JFETs are voltage-controlled devices where the gate-to-channel junction is normally reverse biased. Understanding which terminal polarities allow conduction is essential for biasing and small-signal design. This item focuses on the polarity requirements for current to flow in an n-type JFET channel.
Given Data / Assumptions:
Concept / Approach:
In an n-type JFET, the channel is n-type semiconductor. Electrons are the majority carriers. With the gate reverse biased, depletion regions pinch the channel to a degree set by VGS. Drain current ID flows when VDS > 0, pulling electrons from source to drain (conventional current from drain to source). The current magnitude follows ID = f(VGS, VDS), saturating when VDS exceeds a threshold often called VDS(sat) in the JFET context, with control primarily through VGS.
Step-by-Step Solution:
Verification / Alternative check:
Datasheets show characteristic curves ID versus VDS for different VGS ≤ 0; conduction increases with more positive VDS until saturation. This confirms that a positive VDS enables current for an n-channel device.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing JFET polarity with MOSFET conventions, or assuming VGS must be positive for all field-effect devices. JFET gates are reverse biased; MOSFET gates are insulated and can take positive VGS for n-channel operation.
Final Answer:
Correct — with VDS > 0 and the gate reverse biased, an n-type JFET conducts.
Discussion & Comments