At cut-off, the JFET channel is _______

Question

Examoogle · Accepted Answer

Completely closed by the depletion region

Explanation: At the cut-off region, the gate-to-source voltage $V_{GS}$ of a JFET is equal to or more negative than the pinch-off voltage $V_P$. This causes the depletion regions from the two gates to meet and merge, effectively sealing off the conducting channel between the drain and the source.

Working Principle: The JFET operates by controlling the width of a depletion region via a reverse-biased p-n junction. As $V_{GS}$ increases negatively, the depletion region expands into the n-channel. When $V_{GS} = V_P$ (the pinch-off voltage), the depletion regions from opposite sides touch, blocking the path for charge carriers and causing the drain current $I_D$ to drop to zero.

Key Formulas: $V_{GS(off)} = V_P$ — The gate-source voltage at which the channel is completely depleted; $I_D = I_{DSS} \left(1 - \frac{V_{GS}}{V_P}\right)^2$ — The Shockley equation defining drain current

Key Points: At cut-off, the drain current $I_D$ is approximately zero.; The pinch-off voltage $V_P$ is a constant parameter for a specific JFET.; JFETs are depletion-mode devices that are normally 'ON' at $V_{GS} = 0$.

Advantages: Extremely high input impedance; Low noise performance

Disadvantages: Small gain-bandwidth product; Temperature sensitive characteristics

Applications: Low-noise amplifiers; High-impedance buffers

Additional Info: In N-channel JFETs, the pinch-off voltage is always negative.. Option D is technically true that the gate is reverse biased, but Option B describes the specific physical state of the channel at cut-off, making it the most accurate answer.

Answer: B is correct — At cut-off, the gate-to-source voltage has expanded the depletion region enough to completely close the conducting path between the drain and source.

Exam Tip: Always remember that JFETs are voltage-controlled devices; $V_{GS}$ dictates the channel geometry, which in turn determines the current flow.

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