Hello, I was messing around with a circuit when I accidentally discovered this circuit. The LED stays off until you touch the yellow wire. Can someone explain how this works? I don't understand how just one connection is making the LED light. I know that you can use a transistor to amplify current to pass through skin, but that takes two connections. Can anyone explain how this circuit is working?
The resistor is 470 ohms, and the transitor is 2N2222 A338, the input voltage is 9 volts.
The short answer is, you are an antenna. You have built a simple radio receiver.
The long answer, you're also one "plate" of a giant capacitor, the other "plate" are the electrostatic charges on the objects in your environment. Due to the ubiquitous presence of electrical service, there's a lot of 50/60 Hz electrostatic fluctuation in those.
You're picking up a mixture of RF from the ether, and electrostatic charges from the environment.
You'll be picking up mainly capacitive coupling to the mains wiring in the building, a much
stronger signal than most RF sources.
If the circuit is battery powered you might see much less or no effect as it will also pick up
capacitive coupling with you, if a similar amount the voltage difference won't be enough to
turn the transistor on.
MarkT:
You'll be picking up mainly capacitive coupling to the mains wiring in the building, a much
stronger signal than most RF sources.
I was trying to communicate a concept in a very simple way. I knew the explanation would get drawn out like this.
You can do a simple experiment to see the difference. Attach a very small valued capacitor to the transistor base (maybe 100pf or so). now touch the other lead of the cap. It will block the 50/60Hz and pass RF. See if the LED comes on (or how brightly). Now connect the cap from the transistor base to the emitter. Touch the base. Now you should see mostly the 50/60 Hz component.
And as mentioned above, it makes a big difference whether circuit ground is connected to other grounds.
Thank you, that helps a lot!