In a Junction Field Effect Transistor (JFET), the gate is always reverse-biased. This reverse bias controls the width of the depletion region, which in turn narrows or widens the conductive channel. The narrower the channel, the less current can flow through from drain to source.
The reverse bias on the gate voltage (\(V_{GS}\)) directly controls the depletion width, and thus: \[ I_D = I_{DSS}\left(1 - \frac{V_{GS}}{V_P}\right)^2 \] This is the Shockley's equation for JFETs in saturation. Clearly, gate reverse bias is the controlling parameter.
Final Answer: (3) Gate reverse bias
In the circuit shown, assume that the BJT in the circuit has very high $\beta$ and $V_{BE} = 0.7$ V, and the Zener diode has $V_Z = 4.7$ V. The current $I$ through the LED is ________ mA (rounded off to two decimal places).


