My question pertains this circuit which is from this question, which in turn is a circuit straight from the data sheet.

Power supply with current limiter and stand alone current limiter

I don't understand the function of the FET in this circuit. Surely, with the gate tied to the source, it's an open circuit? It seems to me something clever is being done here and I would love to know what it is.


1 Answer 1


Yes that is a FET but what kind of FET ? It's a JFET and these have the property that some current flows when Vgs = 0

Look in the 2n5640 datasheet, you will see that IDSS, the Zero-gate-voltage Drain current is around 5 mA typical.

So this JFET simply behaves as a current source of roughly 5 mA. Ideal to power a LED at constant current :-) Or make a cheap-ass 1.25 V voltage dropper with 2 diodes like in this schematic. As long as Vout - Vss is large enough for the JFET to behave as a current source, this will work.

The JFET could be replaced by a resistor but then the circuit would become more sensitive to Vout - Vss.

  • \$\begingroup\$ light bulb goes on Of course! That makes complete sense. So all that bit does is create a stable adjustable 0-1.25V supply which is added on to the drop across R1 and allows you to pinch the current right down to zero. Awesome, answer accepted. \$\endgroup\$
    – izak
    Aug 19, 2015 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.