Question

Whenever a JFET operates above pinch-off voltage

• A. Drain current starts decreasing
• B. Drain current remains nearly constant
• C. Depletion regions become smaller
• D. Drain current increases steeply

Hint:

Once a field-effect device transitions past its pinch-off condition into saturation, the drain current flattens out, remaining virtually constant despite increases in the drain-source bias voltage.

Answer 1

The Correct Option is B

Concept: A Junction Field-Effect Transistor (JFET) controls conduction via a channel established between the source and drain terminals. This channel profile is modified by the reverse-bias voltage applied at the gate-source junction. When evaluating the output characteristic curve describing drain current (\(I_D\)) versus drain-source voltage (\(V_{DS}\)) for a constant gate-source voltage (\(V_{GS}\)), the operation is separated into distinct operational zones:
• Ohmic (Linear) Region: For small values of \(V_{DS}\), the channel acts as a relatively simple variable semiconductor resistor. As \(V_{DS}\) increases linearly, the current \(I_D\) increases in a linear, proportional manner.
• Pinch-Off Point: As \(V_{DS}\) continues to rise, the reverse bias near the drain side of the gate-channel junction becomes strong enough that the depletion regions meet, causing the channel cross-section to pinch off. The voltage at which this threshold occurs is defined as: \[ V_{DS} = V_{GS} - V_P \] where \(V_P\) is the intrinsic pinch-off voltage of the device.
• Saturation (Active) Region: When the operating voltage \(V_{DS}\) exceeds this critical pinch-off threshold level (\(V_{DS} \ge V_{GS} - V_P\)), the depletion layers do not fully choke off current. Instead, a narrow high-field region forms, and any further increases in \(V_{DS}\) simply increase the length of this localized pinch-off region. The voltage drop across the remainder of the active conduction channel remains fixed. Consequently, the electric field pushing charge carriers through the channel remains fundamentally locked. The current saturates and is modeled by Shockley's equation: \[ I_D = I_{DSS} \left( 1 - \frac{V_{GS}}{V_P} \right)^2 \] Because \(I_D\) is mathematically determined solely by \(V_{GS}\) in this region, it becomes highly independent of \(V_{DS}\). Therefore, the drain current remains nearly constant, allowing the device to behave as a voltage-controlled constant current source.