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I am a self-taught hobbyist, and I have a question about an experiment that I did. I got two ATX power supplies, a and b, and hooked them up as follows:

schematic

simulate this circuit – Schematic created using CircuitLab

I was not expecting current to flow, but to my surprise, the LED lit up. This is my first question: Why does current flow from PSU a to PSU b? Is this because the PSUs are connected to the same power strip, and are not actually insulated?

So after seeing the results of that, I had another idea. I wondered if the PSUs could sink current from a 9v battery. So I connected (-) of the 9v battery, and 12v from the PSU according to the following schematic:

schematic

simulate this circuit

Again, the LED lit up, but just barely. I could only just see a tiny bit of light from the led. Why does this happen?

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    \$\begingroup\$ rather than posting such information as a comment, edit it in to the original question. \$\endgroup\$ Sep 17, 2017 at 15:47

2 Answers 2

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Why does current flow from PSU a to PSU b?

If they're ATX supplies then the output GND is connected to protection earth, so both GNDs are connected together to the power strip's earth (or the house wiring). So, no surprises here.

So after seeing the results of that, I had another idea. I wondered if the PSUs could sink current from a 9v battery. So I connected (-) of the 9v battery, and 12v from the PSU according to the following schematic

This sounds like your switching PSU is leaking a bit of high frequency common mode noise. This is normal and expected for a switching PSU, although the actual current should be very low to meet with EMC/EMI regulations.

To have current flow, you need a closed circuit. In your case, what closes the circuit would be the parasitic capacitance between the 9V battery and the ATX power supply metal case. If I'm right, the LED illumination will vary if you bring the 9V battery closer to the ATX supply. It would also work if you replace the 9V battery with any metallic object (like a kitchen spoon, whatever) since what matters isn't the fact it is a battery, but rather a conductive object large enough to form a capacitor with the PSU case.

Try it ;)

I find this question interesting, because the first part is really basic, while the second is rather high-level ;)

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  • \$\begingroup\$ OK, if you hold the battery, you'll increase its capacitance. If your naked feet touch the floor, then you're part of the circuit. This iis like a test screwdriver electronics.stackexchange.com/questions/96101/… \$\endgroup\$
    – bobflux
    Sep 17, 2017 at 16:02
  • \$\begingroup\$ Also, never attempt to connect a battery to two wires of a computer supply unless you are sure the circuit is safe - computer power supplies can deliver plenty enough current to catastrophically destroy a battery... \$\endgroup\$ Sep 17, 2017 at 21:44
  • \$\begingroup\$ hmmm.... I may have to try that.... MWUHAHHAHAHA (safely, of course.) \$\endgroup\$
    – user163229
    Sep 26, 2017 at 12:24
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The ground on non-isolated power supplies is connected to earth ground, so for the first circuit the current was flowing out through the power supply B ground and completing the circuit through the power supply A ground.

For the second circuit, if there were no other connections to the 9V battery there should not be any current flowing and the LED should not light up at all. If it does light up, then there is some connection between the 9V battery and the power supply or earth ground (possibly leakage through the case of the battery and something conductive on the bench?). You might put a push-button switch in series with the circuit and make sure the LED turns on and off as you push the switch (LEDs can appear to be on because of the lensing of ambient light when they are really not on). You might also try eliminating the resistor in the second circuit and see if it gets brighter.

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