Question

In an n-channel JFET, if the gate voltage is made more negative than the pinch-off voltage, what happens to the drain current?

• A. Drain current increases linearly
• B. Drain current saturates
• C. Drain current drops to near zero
• D. Drain current reverses direction

Hint:

When \(V_{GS}\) is more negative than the pinch-off voltage \(V_P\) in an n-channel JFET, the channel is completely closed, and the drain current drops to zero.

Answer 1

The Correct Option is C

Understanding the Concept: A Junction Field-Effect Transistor (JFET) is a voltage-controlled semiconductor device where current conduction through a conducting channel is modulated by adjusting the depletion region width at the reverse-biased gate-channel p-n junctions. For an n-channel JFET, the conducting channel consists of n-type material surrounded by p-type gate regions. Applying a negative voltage to the gate terminal (\(V_{GS} < 0\)) increases the reverse bias across these junctions, expanding the depletion layers inward into the channel.

Step 1: Understanding the pinch-off threshold condition.
The pinch-off voltage (\(V_P\)) is defined as the specific negative gate-to-source voltage value at which the depletion regions expand completely across the channel and meet in the center. When \(V_{GS} = V_P\), the continuous conducting path for electrons between the source and drain terminals is completely obstructed.

Step 2: Determining current behavior beyond pinch-off.
When the gate voltage is made even more negative than this pinch-off threshold value (\(|V_{GS}| > |V_P|\)), the channel remains thoroughly blocked and choked off. With no open conducting channel remaining for carrier transport, the total drain current (\(I_D\)) drops down to near zero (except for a negligible, minute leakage current). This condition represents the cut-off state of the JFET, which matches Option (C).