When drain voltage equals the pinch‑off voltage in a JFET/FET, the drain current remains constant with further increase in drain voltage (i.e., it saturates and does not significantly increase or decrease).

Direct fill‑in answer

When drain voltage equals the pinch‑off‑voltage, then drain current remains constant with the increase in drain voltage.

This is because the device has entered the saturation (or constant‑current) region, where changes in drain voltage no longer cause appreciable change in drain current.

Quick Scoop: Why does it stay constant?

  • At pinch‑off, the channel is just constricted enough that any further increase in drain voltage mainly shifts the pinch‑off point slightly along the channel instead of pulling more current.
  • The FET behaves like a current source in this region, so IDI_DID​ is set primarily by VGSV_{GS}VGS​ (via Shockley’s equation) rather than by VDSV_{DS}VDS​.
  • This “flat” part of the output characteristics is what we call the saturation or constant‑current region in JFETs and MOSFETs.

Mini numeric/visual story

Imagine plotting drain current vs. drain voltage for a fixed gate‑source voltage:

  1. For small VDSV_{DS}VDS​, the curve rises almost linearly (ohmic region).
  1. At VDS=Vpinch‑offV_{DS}=V_{\text{pinch‑off}}VDS​=Vpinch‑off​, the curve bends and starts to flatten.
  1. Beyond this point, the line is almost horizontal: IDI_DID​ stays approximately the same even as VDSV_{DS}VDS​ grows.

That flat horizontal part corresponds to the statement “drain current remains constant with the increase in drain voltage.”

Key exam facts (bullet style)

  • Correct option: remains constant.
  • Region name: saturation / constant‑current region.
  • Typical device: JFET or depletion‑mode MOSFET.

Information gathered from public forums or data available on the internet and portrayed here.