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I would like to know if the following is possible, and if it is possible, roughly how the wiring should be done ...

I have 2 light bulbs (BULB1 and BULB2) connected to SEPERATE switches. I want to install BULB3 so that BULB3 turns on when ...

  1. BULB1 is on, or
  2. BULB2 is on, or
  3. BULB1 and BULB2 are both on

and turn off when ...

  1. BULB1 and BULB2 are both off

I have access to the wiring at BULB1 and BULB2, and no access to the wiring at the switches. BULB3 will have to draw power from either BULB1 or BULB2.

The idea is to add a light to a dark passage where there are bulbs on both sides, but none in the middle.

Thanks for any suggestions.

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    \$\begingroup\$ Is there any particular reason not just to put a motion-sensing light in the middle of the passage? \$\endgroup\$ – Nick Gammon Mar 25 '18 at 5:16
  • \$\begingroup\$ Just use double pole switches. \$\endgroup\$ – Hot Licks Mar 25 '18 at 19:51
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another relay option.

schematic

simulate this circuit – Schematic created using CircuitLab

with appropriate choice of relay this layout could be used in the case where both switches are on different phases, but not all relays that can switch 240V are suided to that task.

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  • \$\begingroup\$ Nice simple solution and easy to wire. \$\endgroup\$ – RoyC Mar 25 '18 at 13:28
  • \$\begingroup\$ Thanks for the guidance ... this got the most votes but for some reason my installer decided to go with the guidance from @Transistor. I will do a little reading up on this to see how this works. \$\endgroup\$ – A-S Mar 26 '18 at 1:18
  • \$\begingroup\$ @A-S: This is more efficient than mine. It needs only one relay. \$\endgroup\$ – Transistor Mar 26 '18 at 6:00
  • \$\begingroup\$ @A-S Transistor's circuit has the advantage of not blinking the power when switch2 is operated. also the installer makes more profit on parts mark-up. \$\endgroup\$ – Jasen Mar 28 '18 at 7:26
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@Transistor has some good ideas. Perhaps a couple of big 4-diode bridges would allow LAMP3 to be driven, if you only have access to LAMP1 & LAMP2. Haven't shown the power sources to LAMP1 and LAMP2. They are assumed to be wired up to AC source via their respective switches. Makes my head hurt to imagine if phasing is important.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ This should work just fine. Phasing is NOT important, assuming that Neutral is common to both bulbs (as required by the Electrical Code). \$\endgroup\$ – Dwayne Reid Mar 24 '18 at 22:52
  • \$\begingroup\$ This assumes standard center-tapped Distribution Transformer. \$\endgroup\$ – Dwayne Reid Mar 24 '18 at 22:58
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    \$\begingroup\$ Transistor had considered this solution too but decided against it as being too risky for the level the OP appears to be at. :^) \$\endgroup\$ – Transistor Mar 24 '18 at 23:05
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    \$\begingroup\$ I think phasing is important here -- in the extreme, if the two lamp circuits share a neutral but have 180° opposite live phases, then lamp 3 will get twice the voltage (minus two diode drops) anywhere in the cycle, \$\endgroup\$ – Henning Makholm Mar 24 '18 at 23:25
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    \$\begingroup\$ @Transistor I agree that there are risks. OP asks if it is possible. Is it OK to say, "yes, possible, if done with care". Your switch and relay solutions are safer. \$\endgroup\$ – glen_geek Mar 24 '18 at 23:33
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The simplest solution is to use 2-pole switches. Middle lamp will turn on if either SW1 or SW2 turns on.

Unfortunately this doesn't suit you.

schematic

simulate this circuit

Figure 2. The next best thing is to put two lamps in the middle.

schematic

simulate this circuit

Figure 3. The relay option.

If either lamp turns on its associated relay turns on too. Contacts of the relay light the middle lamp.


Truth tables

By the way, engineers use truth tables to succinctly specify the required logic for applications such as this. Yours would look like this:

SW1    SW2    Bulb 3 (middle)
off    off    off
on     off    on
off    on     on
on     on     on

This is a classic "OR" logic arrangement.

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  • \$\begingroup\$ Problem states no access to wiring at the switches... \$\endgroup\$ – Dean Franks Mar 24 '18 at 22:28
  • \$\begingroup\$ Using phototransistors, he might even be able to avoid rewiring the existing bulbs (although the commodity on-when-dark circuits would be the inverse of what's requested) \$\endgroup\$ – Ben Voigt Mar 25 '18 at 1:11
  • \$\begingroup\$ Thanks a ton for this! I sent this link to the electrician and he will try to implement Figure 3, and the fallback is Figure 2. \$\endgroup\$ – A-S Mar 26 '18 at 1:10
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Half-in-earnest-answer, ideas that are more appropriate for equipment design than house wiring, just to explore theoretical possibilities... and highlight the traps and safety hazards in some ideas that one might come up with.

There is a simpler version of the circuit with the bridge rectifiers, BUT I can make no claim to whether it would actually be legal under your code. Would probably need additional measures taken to be safe, and might put extra stress on your switches due to capacitor charge current. Might create undue harmonics in your AC. Will likely not work with CCFLs or some LED bulbs. Some of these constraints/safety concerns are also valid for the bridge rectifier circuit. So, what I am suggesting here is a way to do it that is probably an acceptable way if you are building an appliance of some sort, but not a good idea to put into your house wiring.

Simply feed a correctly sized filter capacitor via diodes (half wave rectifiers), with the lamp parallel to the capacitor. The capacitor would need to be dimensioned for a given wattage of lightbulb, since the effective voltage the lightbulb sees depends on it. A much too large filter capacitor would feed the lightbulb 1.4 times mains voltage, a much too small one 0.5 times. Someone reaching into that socket if it is ever unpopulated might have issues with 400V DC, though. Also, a diode failing short could blow an electrolytic filter capacitor to pieces.

Instead of using any capacitor, using a lightbulb rated for half your mains voltage would also work (easy if it is a 240V system - get a 120V lightbulb. Disclaimer: A 120V bulb on a 240V system might unconditionally fail code since it is plainly a component not rated for the nominal mains voltage of your house wiring). There is, of course, the risk of someone "borrowing" the lightbulb from that lamp and putting it into another socket).

Even simpler, there are LED bulbs that are specified for 110V-240V input - some of them might very well accept half-wave rectified 240V AC without any complaint (YMMV. Don't try if you aren't CERTAIN the bulb can handle it fine! Some PSMPS circuits might accept DC input but degrade or overheat. You would need a schematic of the bulb to judge that, and make sure nothing else is inserted in that socket.)

...

House wiring in many countries is made out of clunky things in grey plastic cases, with huge screw terminals and more hieroglyphic approval marks than tattoos on a gangster, that do the same things as much smaller and cheaper equivalents found in appliances and on circuit boards. That is because reasons, mostly because there is an entirely different level of fool-proofing, clueless-next-buyer-of-that-house-proofing, and assuming-the-other-guy-will-do-it-wrong required compared to an in-device environment....

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There is no simple way to do this (without relays, etc). I would recommend putting two bulbs in the middle and wiring one in parallel with bulb 1 and one in parallel with bulb 2.

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  • \$\begingroup\$ +1: this could be the most simple and effective way to achieve an equivalent solution, if the OP is OK with having two bulbs instead of one in the center. \$\endgroup\$ – Lorenzo Donati supports Monica Mar 25 '18 at 12:03
  • \$\begingroup\$ Thank you for this suggestion ... this is the fallback plan now. \$\endgroup\$ – A-S Mar 26 '18 at 1:08
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This might be a good use case for some home-automation tech. It’s a bit expensive (maybe $150?) but doesn’t require opening walls to run extra wires, etc.

There are several different kinds, but the approach using Insteon would be

  • replace the two switches with Insteon switches. They directly control their respective lights.

  • replace bulb3 with an Insteon bulb

  • teach each switch to control bulb3

If you want, you can even have a setup like that control all three from each end.

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It's not clear why you'd ever want to only have two of the three bulbs turned on. All of the answers, so far, seem workable, but they all show that the two-bulb condition requires some special circuitry, that has to meet electrical codes and mount somewhere. You can connect all three bulbs in parallel and let either switch turn on all three, if both existing bulbs are on the same phase.

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  • \$\begingroup\$ ... and on the same circuit breaker, ... and if you don't mind the confusion of how to switch them all off. \$\endgroup\$ – Transistor Mar 25 '18 at 13:12
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    \$\begingroup\$ I would assume interconnecting two things directly (not via relays etc.) that are fed off two different breakers would make most electrical codes violate YOU! \$\endgroup\$ – rackandboneman Mar 25 '18 at 15:34
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Hate to be overly simplistic - have a bulb 4 next to bulb 3! Then you can simply wire directly from bulb 1 to 3 (so when 1 is on so is 3) and when bulb 2 is on so is 4. Basically a direct power connection, no electronics, just an extra bulb.

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    \$\begingroup\$ This is unintelligible, Lee. There is no bulb 4 in the original question. You need a schematic. There's a button on the editor toolbar. Be aware that answers float up and down by votes and user sorting preferences so you can't use this site in the same way you might use a forum. Welcome to EE.SE. \$\endgroup\$ – Transistor Mar 27 '18 at 19:35

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