In one line
In an n-channel JFET, a negative gate-source voltage widens the depletion regions and narrows the channel, reducing drain current; at pinch-off the current stops increasing with drain voltage and is controlled only by the gate voltage.
Explained simply
Back to the garden hose: with no squeeze (\( V_{GS} = 0 \)), opening the tap more (raising \( V_{DS} \)) first increases flow, but beyond a point the hose near the nozzle collapses and the flow levels off. Squeezing harder with your hands (more negative \( V_{GS} \)) lowers this levelled flow; squeeze enough and the flow stops completely.
Technically, the gate-channel junction is reverse biased. Its depletion layers extend into the lightly doped channel, reducing its width and increasing its resistance. The drain end is more reverse biased than the source end, so the channel is narrowest near the drain.
How it works
- \( V_{GS} = 0 \), small \( V_{DS} \): channel behaves like a resistor; \( I_D \) rises linearly (ohmic region).
- Increasing \( V_{DS} \): depletion near the drain grows; at \( V_{DS} = |V_P| \) the two layers almost meet. This is pinch-off. \( I_D \) now stays nearly constant at \( I_{DSS} \) (saturation region).
- Negative \( V_{GS} \): channel already narrower, so pinch-off occurs at a lower \( V_{DS} \) and the saturation current is less than \( I_{DSS} \).
- \( V_{GS} = V_{GS(off)} = V_P \): channel fully closed, \( I_D \approx 0 \) (cut-off).
- Very large \( V_{DS} \): the gate-drain junction breaks down; avoided in use.
- Transfer characteristic: \( I_D \) versus \( V_{GS} \) in saturation, a parabola-like curve from \( I_{DSS} \) at \( V_{GS} = 0 \) down to zero at \( V_P \).
Key formula
Shockley's equation: \( I_D = I_{DSS}\left(1 - \dfrac{V_{GS}}{V_P}\right)^2 \)
\( I_D \) = drain current (A), \( I_{DSS} \) = drain current with gate shorted to source in saturation (A), \( V_{GS} \) = gate-source voltage (V), \( V_P \) = pinch-off voltage, equal to \( V_{GS(off)} \) (negative for n-channel, V).
Quick example
For a JFET with \( I_{DSS} = 10 \) mA and \( V_P = -4 \) V, at \( V_{GS} = -2 \) V: \( I_D = 10(1 - 0.5)^2 = 2.5 \) mA. So a 2 V change at the gate, with practically zero gate current, controls the drain current. (Illustration only.)
Common mistakes
- Saying pinch-off means the drain current becomes zero; at pinch-off \( I_D \) saturates, it is zero only at \( V_{GS(off)} \).
- Applying positive \( V_{GS} \) to an n-channel JFET (it forward biases the gate junction).
- Calling the flat region 'saturation' and thinking it is the same as BJT saturation; in a FET it is the region used for amplification.
How MU asks it
Numericals are not expected in this module (MU syllabus), so questions are theory with diagrams. Typical: "Explain the working of an n-channel JFET with its drain and transfer characteristics" (5–8 marks). For full marks draw the structure showing depletion regions for \( V_{GS} = 0 \) and negative \( V_{GS} \), draw the drain characteristic family marking ohmic, saturation (pinch-off) and breakdown regions, draw the transfer curve, and write Shockley's equation.