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Hello from cold Russia.

1)Don't use the grounding wire to connect N pin on your wall switch. If you don't want to win Darwin award. 

Look at the first picture.

2)Use capasitors to connect the switch and the bulb. 

Look at the second picture.

C1 - Capacity. Metal-film capacitor for 400 volts.

It works as a "quenching capacitor" or "resistance". In order for the "part of the current" to remain on the switch and feed it, in the switched on position of the key.

C2 - Capacity. Metal-film capacitor for 400 volts. 

Works as a "bypass" resistor. Serves to ensure that energy-saving gas-discharge or LED lamps do not blink (from the current passing through them supplying the switch) in the off position of the key.

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2 people have this question

Hi, How would @sshRage's solution work with 2-gang switches?


Hello. About Alex Dikaben... 

Sorry, no time for inadequate.

Here is a scheme in English.


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sshRage Thank you!

sshRage - do the value of the capacitors change for 60Hz AC mains (used in USA)?

ToMichael Ingraham. Clever question. Thank you. Yes. You are wright. I think that values of the capacitors will differ. Because resistive behavior of capacitors in case of "AC" depends on frequency.  I've got my results in case of 50 Hz. You've got to check it by yourself for 60 Hz. Please add my post with your results for different type of bulbs.

To maintain the same reactance in the circuit for the capacitor, the 60Hz equivalent should be 5/6 of the 50Hz capacitance; yes?


Michael, Yes, so in stead of 0,47uF you should use a 0,39uF. Since these components are dirt cheap, i’d order several values, just to able to do some trial & error. I’d very much appreciate to hear the results;) Ed

@SonOffJunky - I have several other home projects on my list before I plan on trying this out so it may be several months. Hopefully there are others (including US based makers)) who will give this method (and capacitance conversions) a try and report back.

@sshRage - in your English translation you added a few loads but also removed the 60W incandescent bulb entry. Is there a reason you did not include it? Thanks.

In order for this scheme to work there has to be a voltage drop to the bulb.  Just curious, have any of you trying to do this measured the actual AC voltage across the light bulb?

I suspect an incandescent (resistive load) light bulb would burn noticeably dim.  The LED and CFL bulbs probably work because they have a wider input voltage range.  I also wonder about the strain and heating load on the capacitor.  In HVAC motor circuits using series capacitors they use oil filled caps to dissipate the heat.  In radio circuits that use suppression caps across the AC line they require safety caps.

My gut says this whole scheme using capacitors is not a good or safe idea.  Run a neutral wire and do it right.


To Michael Ingraham and  SonOffJunky - your theory looks fine. I think it'll do. But please try it in practice. And add your results in this post.

To E W.

That scheme is for GAS or LED bulbs only. EW -  you are right. LED and CFL bulbs have a wider input voltage range. That's why they are working fine, and you will not sence the difference. Do not use that scheme for classic bulbs. Classic bulbs will burn darker.

And your guts have good "premonition". This is not a best solution. Best solution is to "run a neutral wire and do it right". BUT! If it is not possible - it is better to do that scheme, but not the one with the grounding wire. 

About heat - there are not heat at all on METAL-FILM capacitors. Try it by yourself. Use that type of condensors: (I don not know the right translation for this type). Please name it.


There are twomore things you should know about that scheme: 

1)If your bulb is dead - whole current will flow in C2. Like a shortcut. In that case your WiFi switch will power down in ON position. Then it power up again - and again OFF. In containious loop. 

2)If you will change the WiFi wall switch to the classic one, leave the C2 capacitor on the celling and you bulb will die - that will a problem. Whole current will flow to the C2. I guess it will burn.

So - dont leave a C2 capacitor on the celling in case you decided to change wall switch from WiFi type to the classic one.

Here is a solution scheme for multi-gang switches. 


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Very interesting Idea. 

The thing is I have 10 led spots of 5W each in the living-room, will the capacitor value still apply (or which one)?


I guess you are going to make them turn on/off simultaniosely. There will be one zone of ligh, not 10 zones.

And you will connect all LEDs in parallel connection. And you will use 1-gang wall switch to turn all of them on/off.
Then you will have to summarize power of your LEDs = 50W. You will need to use apox... 10 Micro Farades capacitor to make them work. And I thin that you will not need C2 capacitor. 

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Yes there is one zone with 10 LED spots, in parallel turned on/off by one switch.

So the C2 is actually the  "anti flicker" capacitor that counter the residual current from the C1?

I deduced (from yhe schematic and your answer) that the calculation is about 1uF/5W of bulb power for the C1 :) ?

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