I work in a bar and we want to install UV lighting with UV LED strips. Currently, we have a simple circuit that consists of a 12V transformer and wires which go all around the bar, with regular white LED strips placed at key locations.

Now, buying another transformer and running another circuit would cost some money and also would be a lot of work, so I was wondering if we could reuse the circuit already in place, since the regular lighting and the UV lighting should never be used together at the same time.

The idea is to add next to each of the existing LED strips a new UV strip, then wire it in parallel with the existing wiring, but in reverse. Then, add a switch next to the transformer that reverses the polarity of the wires coming out of it.

This way, only one set of strips would be active at a time, and they can be easily switched, and the installation would be quite easy and cheap.

Would this work?

Also, if both sets of LED strips are about 60W each, and the transformer is 72W, it would not be overloaded, since the current that flows through the inactive LEDs is next to none, is that correct?

  • \$\begingroup\$ So, your looking to see what the reverse breakdown voltage of a typical 3 led + resistor circuit is. \$\endgroup\$
    – Passerby
    Commented Nov 16, 2015 at 3:23
  • \$\begingroup\$ @Passerby yeah, it seems so. But I also want to know if there is something wrong with my reasoning. \$\endgroup\$
    – Godkiller
    Commented Nov 16, 2015 at 3:27
  • \$\begingroup\$ The overload issue is not an issue. If they do reverse breakdown, it will be a quick spike and dead led strips, so the current overload will only be for a few seconds at best. If the strips will survive a constant 12V reverse bias, thats debatable. You can get a small strip and test, or add a diode for reverse protection (one on each end, sized for 14V and how ever many amps as a precaution). A quick google search didn't provide any examples of people using strips this way, so you may be the first. It is a good thought though. \$\endgroup\$
    – Passerby
    Commented Nov 16, 2015 at 3:46
  • \$\begingroup\$ you imply that you have several (many ? ) strips of LED's. Adding to what Passerby has commented on, are you sure the LED strips are all in series connection ? \$\endgroup\$
    – Marla
    Commented Nov 16, 2015 at 3:49
  • \$\begingroup\$ @Passerby yeah, well, I have some background in low-level programming and you do stupid stuff like this all the time, reusing and refitting what you've already got... anyway, testing seems like a good idea, I suppose 24h should be enough. But I would like to know: if they survive, would the reverse current wear them down more quickly than if they were properly connected and left on? \$\endgroup\$
    – Godkiller
    Commented Nov 16, 2015 at 4:00

1 Answer 1


better option is to work with 2 positive rails and shared negative, using a switch to choose which positive rail to turn on.
negative from leds goes directly to power supply, positive goes to switch.

  • a two position switch (on-on)
  • a three position switch (on-off-on)
    connect the power supply in the middle terminal, one positive going to each led rail on side terminals.

but you can do what you are thinking and reverse the voltage, just use a two or three position switch with double terminal and connect like the image (the motor there will be the leds).

about the supply, it's correct what you said, leds consumes no current in reverse.
enter image description here

  • \$\begingroup\$ OP is looking for a solution that doesn't require rewiring. They only have two conductor wire already in place. \$\endgroup\$
    – Passerby
    Commented Nov 18, 2015 at 22:11
  • \$\begingroup\$ did you read all the answer? "but you can do what you are thinking and reverse the voltage, just use a two or three position switch with double terminal and connect like the image (the motor there will be the leds)." \$\endgroup\$
    – Daniel Y.
    Commented Nov 21, 2015 at 3:28
  • \$\begingroup\$ Did you take reverse bias breakdown failure into account? \$\endgroup\$
    – Passerby
    Commented Nov 21, 2015 at 3:36

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