3
\$\begingroup\$

My brother and I were playing around with some new transistors we ordered (2n7000 n-channel MOSFETs), trying to make a simple switching circuit: connect a switch, LED turns on; open the switch, LED turns off. Here's a link to the circuit we were attempting to make.

enter image description here Here's a picture of what we implemented in a breadboard. There's one difference from the above circuit: the switch is not actually complete. We just have a single wire connected to the n-MOSFET's gate. So this is what we actually have in the breadboard.

As far as we can tell, from our limited electrical engineering experience, the LED should never be on, since there is no connection from the +5V power to the transistor's gate; the transistor should not let any current through. However, there were two interesting things: first, the LED turned on as soon as we powered on the circuit, even before we connected the transistor's gate to power. Also, whenever we touch/tap the wire connected to the gate (labeled by "??" in the circuit), the LED randomly switches between three different levels of brightness (on, medium, off), and will even "stick" at a brightness setting, even between power cycles of the entire circuit. The single transistor appears to act as a flip-flop!

So, the question is: why does touching the unconnected wire change the LED's state?

And, secondarily, how is the LED on, even if the transistor has no voltage source attached to the gate?

\$\endgroup\$
  • \$\begingroup\$ "The single transistor appears to act as a flip-flop!" - Actually, it acts more like a DRAM cell ;) \$\endgroup\$ – marcelm Oct 8 '18 at 0:06
5
\$\begingroup\$

A floating gate has a voltage. Any wire has a voltage. Until you put it in reference to something else, that voltage can be anything. Don't expect it to be 0V or any other known voltage.

That gate also has a capacitance. Your body is a capacitance. Every time you touch that gate, you're connecting two capacitors and a charge balance will occur. This will alter the voltage on the gate. Ergo, you've created a simple transistor capacitor memory scheme not terribly different than DRAM. Pretty nifty huh?

I'm surprised a little that the brightness remains the same between power cycles, but maybe that's attributable to having a high quality mosfet with a high gate impedance?

Lastly, although this is a great lesson, don't design a circuit like this with a floating node expecting it to be a specific voltage.

\$\endgroup\$
  • \$\begingroup\$ I am the brother that @feralin spoke of, so I feel qualified to explain: we actually did originally have a switch (as shown by the first circuit linked in the question) consisting of two wires that we could simply touch to each other. However, touching them did not seem to do anything (even though that completed a connection from +5V to the gate). At that point, we decided to experiment and got to the current situtation. So can you explain why, even with a switch, the transistor was not working as we expected? \$\endgroup\$ – Jashaszun Jul 21 '15 at 3:20
  • \$\begingroup\$ @Jashaszun You'll have to describe it a little clearer. You're saying when the switch was closed, the transistor did not light? \$\endgroup\$ – horta Jul 21 '15 at 3:22
  • \$\begingroup\$ I'm saying that whether or not the switch was closed, the transistor did light and exhibit the strange effects that we described in the question. \$\endgroup\$ – Jashaszun Jul 21 '15 at 3:22
  • \$\begingroup\$ @Jashaszun That sounds like your transistor is partially blown OR you have the transistor in backwards (source and drain flipped). If it was backwards, it's possible it is partially or fully blown now too. \$\endgroup\$ – horta Jul 21 '15 at 3:23
  • 2
    \$\begingroup\$ @Jashaszun If you look at the spec sheet: fairchildsemi.com/datasheets/2N/2N7000.pdf you'll see that the Rgs resistance is <=1Mohm which means that if you have 1M pulldown resistor, you actually create a voltage divider putting 1/2 of Vdd at your gate through the internal gate resistance. That means your transistor would always be on. Basically, your gate inherently has a resistor internally that connects to the source (and drain) that's less than 1Mohm. If your external resistors are anywhere near that, things are going to behave weird. \$\endgroup\$ – horta Jul 21 '15 at 3:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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