3
\$\begingroup\$

I am dealing with some three-way light switches that people are always forgetting to turn off.

Any suggestions/schematics for making a simple LED indicator at each switch to indicate if the circuit is on or off, which doesn't require an extra wire for the indicator? (The wiring is already installed and covered, so adding wires is not feasible.)

I would prefer mains-powered solutions (no batteries).

Update: Here's how it's currently wired. We use 14/3 with ground (4 conductors in the cable, not allowed to use the earth ground as a neutral). In the US we're required to have a neutral in all switch boxes, and are required to break the hot wire with all switches.

schematic

simulate this circuit – Schematic created using CircuitLab

These circuits are switching 120VAC 60HZ USA electrical mains.

Thank you!

\$\endgroup\$
  • 1
    \$\begingroup\$ Can you provide the circuit diagram of the wiring? \$\endgroup\$ – Whiskeyjack Dec 31 '15 at 6:32
  • 1
    \$\begingroup\$ For any Europeans, this is a two-way lighting circuit. Somewhere in American history these switches are named as 3-way - probably because they have three terminals and the wires go three ways? \$\endgroup\$ – Transistor Dec 31 '15 at 10:31
  • \$\begingroup\$ Wait, are you saying you've got access to a Neutral line at all of the junction boxes? \$\endgroup\$ – Daniel Dec 31 '15 at 17:36
  • \$\begingroup\$ @Daniel - yes you're correct. I updated the question with a schematic. \$\endgroup\$ – Ryan Griggs Dec 31 '15 at 17:43
  • \$\begingroup\$ In that case, this is a solved problem. You need to get a 3-way switch with a pilot light. See my answer edit for an example. \$\endgroup\$ – Daniel Dec 31 '15 at 20:39
3
\$\begingroup\$

The VERY old solution to this problem uses two resistors and a neon bulb.

I used to use a NE-2 bulb with 2- 100k resistors. One side of the neon bulb goes to Neutral. The other side of the neon bulb goes to one side of both 100k resistors. The other side of each resistor goes to the two switch legs.

The circuit makes use of the relatively high threshold voltage of the NE-2 bulb. This about 65 Volts AC.

When the switches are in a position that does NOT turn the lamp on, one resistor has 120 Vac applied, the other resistor is near 0 Vac. The neon bulb is NOT lit because the voltage across the bulb is less than the threshold voltage.

When the switches are in a position that causes the lamp to be lit, one of the 100k resistors is at 120 Vac, the other resistor is floating (not connected to anything). The neon bulb then lights.

You can place this circuit at each switch location. I used to drill a 1/4" hole in the cover plate and use silicone sealant to glue the neon bulb laying across the rear of the hole.

Note that this circuit also works with 4-way switches.

EDIT BY RYAN: added a schematic... Does this correctly show what you were discussing?

[Dwayne] Yes - exactly. In older houses without a neutral wire in the box, we would use appropriate voltage-rated resistors, then connect the Neutral side of the neon bulb to metal box so as to use Earth ground instead of Neutral. The Electrical Inspectors were okay with that because the current was so low (about 0.5 mA).

schematic

simulate this circuit – Schematic created using CircuitLab

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I was wondering if there was a neon solution... have seen these as used as indicators in switches/etc years ago. Thanks for great answer! \$\endgroup\$ – Ryan Griggs Jan 3 '16 at 0:51
  • \$\begingroup\$ I'm going to choose this one as the answer because it is the simplest and easiest to implement in my view. But all the answers were great, thank you to everyone! \$\endgroup\$ – Ryan Griggs Jan 5 '16 at 19:07
7
\$\begingroup\$

Edit: Since you have a neutral available, the solution to this is trivial. You need to use a 3-way switch with illuminated ON pilot light, like this Leviton model.


Original answer:

You could try to hook up a current transformer to an LED. It doesn't take much energy to make an LED glow if you can get the voltage above \$V_f\$:

schematic

simulate this circuit – Schematic created using CircuitLab

Note: I have never tried this. It should be taken as a circuit idea ONLY and should not be tried without knowledge of working with line voltages and adjusting the values of the components based on the load amperage and transformer ratio.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ +1 for plausible circuit .Make sure the CTs can handle the lighting circuit current .In NZ that is 6A .I hope the proposed CT will be small enough to fit in .You could use a bridge rectifier for steadier LED operation .Make sure the turns ratio is sensible like say 300:1 . \$\endgroup\$ – Autistic Dec 31 '15 at 9:00
  • \$\begingroup\$ Neat idea, especially if the 'transformer' primary could simply be the existing lighting circuit wire, clamp meter style - I wonder how many turns on what kind of core would be needed for the secondary? \$\endgroup\$ – nekomatic Dec 31 '15 at 11:08
  • \$\begingroup\$ One problem though, the transformer would need to be somewhat tailored to the lighting in use - I'm guessing the same specs won't work for both a five watt LED bulb and 500 watts of halogens. \$\endgroup\$ – nekomatic Dec 31 '15 at 11:11
  • 1
    \$\begingroup\$ Feasible circuit +1 (especially with small CTs and ferrite would work I reckon) \$\endgroup\$ – Andy aka Dec 31 '15 at 11:37
  • \$\begingroup\$ @Autistic It is implied that the primary should be the house wiring, going through the CT ferrite. There is not a symbol for that though. \$\endgroup\$ – Daniel Dec 31 '15 at 17:33
2
\$\begingroup\$

enter image description here

>

As your case you can try placing the 2 indicator LED circuits between A, B and C, D points in each switch. It will only show the off status of the bulb.

But applying the circuit can cause dimmer light, if it is a normal bulb and flickering, if it’s a CFL bulb even the lamp is in off status. Flickering can reduce the lifetime of the CFLs but it’s worth to try if you are using a normal bulb.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Good idea, we'll done. You should explain a bit better, perhaps use the circuit drawing tool to show two cases of on and off. \$\endgroup\$ – tomnexus Dec 31 '15 at 8:19
  • \$\begingroup\$ +1 for a fundamentaly good circuit .You could refine it by using small fullwave bridge instead of D1 and you could use 100nF X rated in series with 1K ohm @ watt to help the carbon footprint . \$\endgroup\$ – Autistic Dec 31 '15 at 8:47
  • 2
    \$\begingroup\$ You do realize LEDs are not rated for isolating mains voltages? \$\endgroup\$ – jippie Dec 31 '15 at 9:25
  • 1
    \$\begingroup\$ This only works if the wiring accessible at the light switch includes a neutral wire, though. I don't know about US practices but in standard UK wiring only the three connections to the switch itself, plus earth, would be brought to the wall box. \$\endgroup\$ – nekomatic Dec 31 '15 at 11:04
  • \$\begingroup\$ Yes in the US we are required to have a neutral wire at all switches. \$\endgroup\$ – Ryan Griggs Dec 31 '15 at 16:52
2
\$\begingroup\$

The off-the-shelf solution is a current indicator remote LED.

enter image description here

These are available in a range of sensitivities and, as with any current transformer, can be made more sensitive by increasing the number of turns through the core. Price about $12 each.

Put one on the common wire on each switch. Unlike some of the solutions proposed, this has full mains isolation and the LED is not connected to the mains.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ +1 for isolated, clean approach. Question: some of the switches control LED lights, which only draw about 5 Watts each. So current would only be about 50mA... not enough to turn on the indicator. Is there a fix for this? \$\endgroup\$ – Ryan Griggs Dec 31 '15 at 19:45
  • 1
    \$\begingroup\$ A "current transformer" is completely isolated. I had to draw it as a regular transformer due to a lack of symbols. This is noted in the comments. \$\endgroup\$ – Daniel Dec 31 '15 at 22:11
  • 1
    \$\begingroup\$ @Daniel, sorry, that was not my intention. I had been looking at them for another application but haven't purchased any yet. I thought they were useful because they avoided risks with open-circuit CTs. I was wondering if there if they are a deliberately 'poor' CT to allow them to handle over-current gracefully and not blow the LED. The data sheet doesn't elaborate. The isolation comment was to address the comment on binu's answer. You are correct that CTs are isolated - any that I've seen, anyway - but I've never seen a CT with a 20 mA secondary and I feared someone winding on a ferrite core. \$\endgroup\$ – Transistor Jan 1 '16 at 0:16
  • 1
    \$\begingroup\$ @RyanGriggs: no, I'm afraid I don't have a solution for the 50 mA LEDs at the moment. The danger with the CT approach is that you might set it up for 100 mA max and then someone plugs in a 100 W tungsten lamp. Anyone? \$\endgroup\$ – Transistor Jan 1 '16 at 0:19
  • 1
    \$\begingroup\$ @transistor If the core is only rated for 5A, it will probably saturate, and when that happens the secondary can't follow the waveform. I'm not 100% on this, but I believe you can take advantage of that to make the LED sensitive to low currents while not having to worry about overcurrents. OR you could just use a bidirectional zener to bypass the LED/resistor, since the secondary is a current, not a voltage. Nice find on the part, btw, a lot easier than making your own. \$\endgroup\$ – Daniel Jan 1 '16 at 0:42
1
\$\begingroup\$

Another option.

schematic

simulate this circuit – Schematic created using CircuitLab

Since your load is a 5 W LED bulb and load current is 0.05 A (so it must be a 120 V circuit) you could try the schematic above. The idea is that, when wired in series with the load, 2.1 V (3 x 0.7 V) is dropped across the diodes pretty much independent of current through the circuit.

For 10 mA through a red LED we would get a voltage drop of about 1.8 V on the LED. Setting R1 to 33 Ω should limit the current to a safe value while giving adequate light.

  • I have not tested this circuit.
  • The 1N4004 diodes are 1 A rated. If your lamp fails short circuit one or more could fail on over-current. Protect with a fuse.

Image from www.smallscalelights.co.uk

  • Taking @jippie's comment on-board (that the LED housing is not mains rated) you should use one of the LED clip / bezels for this rather than have the 'bare' LED protruding through the switch faceplate. Push the clip through from the front and the LED into the clip from the back. This expands the latches and prevents it falling out.
  • Using an anti-parallel LED (two regular LEDs mounted back to back in the one package) means that you get light on both half-cycles. No flicker and brighter.
  • You will drop mains voltage to the lamp by 2.1 V.
  • You can put one of these in each switch, in series with the common wire, and in your two-switch application you will drop 4.2 V to your lamp.
  • Bear in mind that if the LED lamp is replaced with a standard lamp the diodes will dissipate more heat.
| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ This is a great answer, but I have to say I'm not entirely happy placing anything in series because as you say, the current is limited. Who knows what bulbs may be screwed into the socket by anyone later. I don't want to risk lots of heat or a blown fuse/diode just because someone screwed in a 100W incandescent spotlight instead of a 5W LED bulb... \$\endgroup\$ – Ryan Griggs Jan 3 '16 at 0:57
1
\$\begingroup\$

The CR Magnetics Current Sensing Transformer with self powered LED can be brought up to the trigger current by multiple turns of the hot wire around the transformer and through the center hole. The device will not overpower if incandescents are used after setting the device up for LED's as it can take up to 20 amps (when normally your breaker would go). I checked this all out with CR Magnetics.

Use for 3-Way switch Lit when on Pilot Light for Master (hot feed) switch) without separate pilot light wire to Slave (Load) switch.

Note that the device is .32" thick. Other wires in a junction box should not interfere with the desired operation. Note that the LED saturates and does not get any brighter with higher currents. The LED is powered by the induced current in the secondary winding in the transformer.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

Just go buy a "Lighted handle" 3 way switch; they are made that way with an NE lamp under the toggle. Whenever the load (light fixture) is Off, the switch handle is glowing. Switch handle not glowing, the light is On.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ LOL that's the easy solution, but I wanted to know the "how" behind it. :) \$\endgroup\$ – Ryan Griggs Oct 17 '17 at 3:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.