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After reading What is negative voltage? and the answer given by kellenjb I wondered if I understood the topic sufficiently, and started theorizing. If the + or - sign is just a convention and the voltage is the potential difference to ground, and my PSU has a +12V, -12V and GND connector, would it make no difference wether I connected a circuit to +12V and GND or to GND and -12V? In other words, say I had a 12V LED, could I connect the anode to GND and the cathode to -12V and still have it work same as when I connect the anode to +12V and the cathode to GND?

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  • \$\begingroup\$ +12V, GND and -12V are all relative terms. Here you assumed GND to be 0V. Someone else might consider your -12V to be GND. At the end it all comes down to the potential difference across the two points. \$\endgroup\$ – abhi Jul 11 '17 at 12:20
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Yes, the LED would work.

Think of it like this: Voltage is basically the difference between the potential of two nodes.

So if you have the anode to +12V and the cathode to GND (which is basically our 0V reference point) then the LED will see a voltage of +12V-0V=+12V

But then, if you have the anode to GND and the cathode to -12V the LED will see a voltage of 0V-(-12V)=+12V. As you see, there is no difference in practice.

Also you say:

the voltage is the potential difference to ground

That is true for an absolute value of voltage, like the +12 or -12V of your PSU; they are referenced to its GND terminal. But as I said before, voltage is actually the potential difference between two points. In this sense, the absolute voltage of the anode and cathode of the LED is not important.

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the voltage is the potential difference to ground

Voltage is the potential difference between two nodes and has nothing specifically to do with ground unless one of those nodes happens to be ground.

You can connect a circuit that requires 12 volts between ground and -12 volts for sure but make sure the polarity is right i.e. positive terminal of the target circuit connects to ground and the negative terminal of the target circuit connects to -12 volts.

In other words, say I had a 12V LED, could I connect the anode to GND and the cathode to -12V and still have it work same as when I connect the anode to +12V and the cathode to GND?

Yes.

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    \$\begingroup\$ The only catch is that ground shuld be able to source current, which probably is the case since he has a negative power supply. \$\endgroup\$ – Vladimir Cravero Jul 11 '17 at 8:43
  • \$\begingroup\$ @VladimirCravero Yes it's a good point; if in normal operation the positive rail is designed for a heavier load the net current will flow into the GND pin on the power supply whereas, in the scenario proposed, the "ground" current would flow away from the power supply and this might make a difference. Unlikely but cannot be ruled out. \$\endgroup\$ – Andy aka Jul 11 '17 at 8:58
  • \$\begingroup\$ @VladimirCravero Could you expand on that? What do you mean with "ground should be able to source current" and how does a negative power supply fit into this? \$\endgroup\$ – Bas Jul 11 '17 at 10:13
  • \$\begingroup\$ With positive supplies, current goes from the supply into ground, while with your setup current goes into the negative supply, flowing out of ground. \$\endgroup\$ – Vladimir Cravero Jul 11 '17 at 10:25
  • \$\begingroup\$ @Bas provide a link to your proposed power supply and I'll take a look but it would surprise me a lot if it was a problem. It's a theoretical problem but quite probably not a real problem. \$\endgroup\$ – Andy aka Jul 11 '17 at 10:35
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Short answer:yes.

Longer: think, would the device know the difference. If it only has two wires on input, only voltage between them matters. But in practice sometimes it is not that simple.

Sometimes this GND is somehow still in system as GND. For example as enclosure potential. Then you may have limitations.

Bottom line- convention may be almost as strong as physics, especially if you have no tools or knowledge to check, like it happens in complex systems.

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As long as you have a potential difference there will be a current flow. In your case the potential difference is

           (0 - (-12)) volt  = 12 volt

so your 12 volt load will work.

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