I know this is somewhat of an odd question, but I'm curious about whether or not a small (or maybe matching*) voltage on a wire can represent a closed switch.

** more about this in a moment*.

First of all, take a look at this terribly simple (and poorly drawn illustration) of my garage door opener set up.

garage door opener and switch

Okay, so when I press the wall switch button the circuit is closed and the garage door opener is activated.

Here's what I'm curious about:

If I took the ground (neutral) wire and plugged it into a bread board and applied 5v directly to that wire, would the garaage door activate? Would that represent a closed circuit to the system? Is that dangerous to do? I would think not, since it is only 5v.

Not Mains Power

Also note, those wires represent just the activating switch, not the mains power. That's why even though this is connected into a US home (wired as 110v) the line shown is only about 5v.


I know I could attach a relay to a bread board, then connect both the hot and ground to the bread board through the relay (via NO) and then apply voltage to the relay thus closing the relaying switch and completing the circuit.

However, I'm wondering if in this case (low voltage) if you could just apply 5v to the ground to simulate the circuit closing.

I'm not sure if what I am saying is dangerous or there is another way to do this or if the answer is simply use a relay and forget about other methods.

EDIT - Added another diagram and more notes

So imagine that the bread board in the diagram is used and the ground is now connected to another power source of 5v. Now I can apply the 5v from the other location or I can close the original switch. Can you explain why this is a problem? A short when both are on maybe? But if they were guaranteed to only be one on at a time would it work?

separate power source

Edit 2 - added schematic

Okay, take a look at the schematic. We're talking about the momentary switch section. It's 12v. That's fine, change my diagrams above (in your head) to 12v. :)
There are comments that say "don't do it!!!" That's fine, I believe you, but can you explain why or why not? And don't just say, "you'll die!" :)

schematic with momentary switch

Edit 3

So I guess it was this that I was thinking of -- either switch being able to turn it on -- but I guess that does use the same source voltage so maybe that's the thing: two way switch

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  • \$\begingroup\$ I don't see the actual question, but judging by your circuit the 5V are supplied by the opener, and it is just expecting to get shorted. So, please, don't provide it 5V from elsewhere. And don't mess with this with the current level of understanding. For your safety. \$\endgroup\$ – Eugene Sh. Jun 27 '16 at 17:33
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    \$\begingroup\$ Voltage is not measured "on a wire" or at a single point. Voltage is a measurement between 2 points. Without knowing more about the internals of your garage door opener it is impossible to say how it would behave if you applied an external voltage source to it. There's a reasonable chance that you'd break it. \$\endgroup\$ – brhans Jun 27 '16 at 17:43
  • \$\begingroup\$ Do you have a datasheet for your garage door opener? Schematics? User manual? Make and model? Anything? \$\endgroup\$ – Dampmaskin Jun 27 '16 at 17:45
  • \$\begingroup\$ I kind of think I figured it out now, because if you replace the garage door in my diagram with a light bulb it might get you all to think a bit more abstractly about whether another switch could switch the bulb on. That's kind of what I was talking about although I didn't know it. Thanks. \$\endgroup\$ – raddevus Jun 27 '16 at 17:59
  • \$\begingroup\$ @daylight Garage door openers work nothing like a simple light bulb... Its all computerized. \$\endgroup\$ – Passerby Jun 27 '16 at 18:26

This is how the garage door opener probably works assuming a simple 2 wire switch with no extra features (I am not liable if you blow something/yourself up):


simulate this circuit – Schematic created using CircuitLab

Just a simple input.

Could you add a random 5V? No. One, without a common ground it won't really work. Two, Directly connecting 5V to ground would be a dead short and Bad News™.

Simplest Solution, wire another switch (or relay powered by a different source) in parallel. Done.


simulate this circuit

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  • \$\begingroup\$ That's a good well-written and explained answered. Thanks very much for the good discussion. agreed: use the relay. :) FYI - just so you know, that last diagram is exactly what I have begun to implement but was curious on this other question. \$\endgroup\$ – raddevus Jun 27 '16 at 19:18

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