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This is probably a really silly question, but I can't seem to find the answer anywhere.

Say I have two totally separate DC power supplies running from AC power, and another piece of equipment that requires 12v DC. Why is it not possible to take 12v from one supply, to the device that needs powering, and then take the return to the ground on the second supply? What is it that stops the current flowing?

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  • \$\begingroup\$ Would your question be the same if you used two 12 Volt batteries or is your question specifically about the DC side of two AC to DC power supplies operating from common AC supply? \$\endgroup\$ – RedGrittyBrick Nov 11 '13 at 17:12
  • \$\begingroup\$ It would not be the same. I understand why it wouldn't work with batteries due to the chemistry involved. What I didn't understand was why the two supplies did not share a common ground at the AC outlet. I've updated the question to clarify. \$\endgroup\$ – Cameron Ball Nov 11 '13 at 17:33
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Sometimes you can, if they're both grounded power supplies which have a low-impedance connection to a common earth point.

If they aren't, then "ground" on one is not at all connected to "ground" on the other, and there is no loop for current to flow.

Sometimes they're both supposed to be ground, but are actually at slightly different potentials, which causes all sorts of problems. Especially to audio systems ("ground loop"), resulting in everything from annoying hums to arcing and damage.

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  • \$\begingroup\$ What if they are running of the same AC wall socket via a powerboard? Does that count as a "common earth point" since ultimately all current must be coming from that same original outlet? \$\endgroup\$ – Cameron Ball Nov 11 '13 at 16:04
  • \$\begingroup\$ That depends entirely on the design of the power supply, which will probably be galvanically isolated. In that case it's not the "same" current but a current which has been induced on the other side of a transformer. \$\endgroup\$ – pjc50 Nov 11 '13 at 16:08
  • \$\begingroup\$ (That is, you can easily design 2 power supplies that will let you do this, but fail modern safety standards.) \$\endgroup\$ – pjc50 Nov 11 '13 at 16:11
  • \$\begingroup\$ Ahh OK so it is entirely possible for this to happen, it's just that for the most part it's a bad idea and supplies are generally designed to not allow this? \$\endgroup\$ – Cameron Ball Nov 11 '13 at 16:12
  • \$\begingroup\$ Yes, because having a direct link from mains 'neutral' to low-voltage ground exposes the user to risks of electrocution in certain failure conditions. \$\endgroup\$ – pjc50 Nov 11 '13 at 16:16
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You mean ...

schematic

simulate this circuit – Schematic created using CircuitLab

There is no potential difference established between the +ve of V2 and the -ve of V1. voltages are not absolute values, they are differences between two points.

If you measure the voltage with respect to the unconnected end you would see that there is no voltage across R2 - hence no current will flow through it.

If there is a "hidden" connection between the negative sides of the DC supplies, then a potential difference exists across R2.

schematic

simulate this circuit

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  • \$\begingroup\$ I'm not sure I entirely understand. Say I have two computer PSUs, and an LED that runs at 12v, if I take a yellow wire from the PSU, put it on the + leg of the LED, then attach a black wire from the other PSU to the - leg, why won't it work? In my head I see the AC coming in, then going through the DC supply, through the LED, then back through the other PSU. Obviously this is not the correct way to picture it, but I don't understand why not. \$\endgroup\$ – Cameron Ball Nov 11 '13 at 16:02
  • \$\begingroup\$ @Cameron: There is no force to push electrons through R2. \$\endgroup\$ – RedGrittyBrick Nov 11 '13 at 16:16
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What is it that stops the current flowing?

Air, probably. Most DC power supplies are galvanically isolated. That means they pump charge between their two terminals and aren't attached to anything else.

Consider these two circuits:

schematic

simulate this circuit – Schematic created using CircuitLab

V3 is able to develop a current, because it can pump charge around the circuit and through the load.

V1 and V2 are not able to develop any current, because they are pumping charge into the end of a wire. There's no place for current to go. The current doesn't flow for the same reason that it doesn't come shooting out of the power supply terminals with nothing connected: air is a poor conductor.

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  • \$\begingroup\$ Galvanic isolation was what I needed to know about. In my head, I assumed both supplies would end up with a common ground back at the wall socket - which I now know is not the case. \$\endgroup\$ – Cameron Ball Nov 11 '13 at 17:31
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If your DC adapter has only two plugs, then it's going to be isolated if it meets safety standards. Many don't, and they're quite dangerous.

Enough home outlets have the neutral and hot wires incorrectly reversed that it's a massive safety problem to assume neutral is ground.

DC adapters with three prong plugs, however, usually tie the DC side ground to the ground pin on the outlet. Usually.

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In circuits theory we know that the current only flows in a CLOSED circuit. If you do not close, you cannot make the power supply to work. A simple way to understand is to think in a simple battery (chemical reaction as we see in school). If just one side supplies current, you cannot maintain the chemical reaction so your voltage still working.

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