Do I have magic fingers?

[Hextejas]

I am playing around with a silly little breadboard project consisting of a jfet, a potentiometer, and a led.
It was not working so I started pushing wires and the led increased and decreased in brightness as my finger got closer or farther. Not touching but just getting close.
Kinda freaky and got me to wondering why.

[kalel]

Being a novice and not having used JFETs before, I'm not sure how they work.

I do know that even a BJT NPN (especially if you use something like a darlington pair or use 3+ of them) can be affected by presumably the electrostatic field when you come close to the gate. Also if you unroll some tape or move some piece of plastic or other similar material nearby.

Perhaps your electrostatic field is affecting the FET as well, but I'm not sure about that.

[TomS_]

[Hextejas]

Thank you Toms, that is a perfect explanation

[Zero999]

Very good video. I think mains hum heavily contributed to this effect. It would have been rectified by the gate's PN junction, causing a negative charge to build up. He said another power source is required to provide a negative voltage to turn the FET off, which is not strictly true. If he has exchanged the positions of the FET and LED, then it could be made to turn off, by connecting the gate to 0V.


It's not just FET circuits which can behave like that. A BJT circuit can do that to, if it has enough current gain. I experienced it today with a circuit composed of an NPN-Darlington, connected to a PNP BJT, forming a super-Sziklai pair. The current gain was so high (2 to 10 million) things normally regarded as insulators, such as plastic, turned it on.

[Yansi]

Very good video. I think mains hum heavily contributed to this effect. It would have been rectified by the gate's PN junction, causing a negative charge to build up. He said another power source is required to provide a negative voltage to turn the FET off, which is not strictly true. If he has exchanged the positions of the FET and LED, then it could be made to turn off, by connecting the gate to 0V.

How is this supposed to turn off? I think it won't. The diode drop is too small for that to happen.

[Zero999]

Very good video. I think mains hum heavily contributed to this effect. It would have been rectified by the gate's PN junction, causing a negative charge to build up. He said another power source is required to provide a negative voltage to turn the FET off, which is not strictly true. If he has exchanged the positions of the FET and LED, then it could be made to turn off, by connecting the gate to 0V.

How is this supposed to turn off? I think it won't. The diode drop is too small for that to happen.

You're right, it won't turn off completely, but 1V across a red LED won't be enough to make it light enough to be seen, at normal room lighting levels.

[Yansi]

Why one volt?  I think it will be pretty much still turned on at that voltage.

[james_s]

Why do you think that?

The forward voltage depends largely on the LED color, many types of green and blue LED chips won't light at all below 2-3V.

[Yansi]

Because the transistor has no reason to settle at Vgs -1V  in such circuit. Please proove otherwise? 

[Zero999]

Because the transistor has no reason to settle at Vgs -1V  in such circuit. Please proove otherwise? 

1V was just an estimate. The exact voltage will depend on the J-FET. Some will be higher, others lower. Unfortunately, I don't any have components handy or a decent red LED model for LTSpice, so have substituted it for three silicon diodes, which should be close to a red LED. Different J-FETs, give different forward currents. The 2N3819 gives a forward current of 1.8mA, with a gate voltage of -1.8V, so the LED will be on and the 2N4117, a forward current of 2.5µA, with a gate voltage of 938mV, so the LED may only be visible, in a really dark room..

[Yansi]

Well, never had a hand on a FET with such low cutoff voltage as 0.9V.  But those 1.8mA I'd consider pretty much ON state.

Conclusion: Depends on exact specimen.

[Zero999]

Well, never had a hand on a FET with such low cutoff voltage as 0.9V.

My point was, because most of the FETs, shown in the video, cut-off at 1.2V, the LED would be off, if it were connected as a source follower, with the gate at 0V.

But those 1.8mA I'd consider pretty much ON state.

Conclusion: Depends on exact specimen.

I agree.

[David Hess]

How is this supposed to turn off? I think it won't. The diode drop is too small for that to happen.

The forward voltage drop of an LED is large enough to provide enough voltage between the gate and source to shut off some JFETs.  If it did not turn off, then the JFET would provide a constant current through the LED which could be useful in itself.