Why would there still be current present when the circuit is off?

[Gavthesparky]

A valued customer has replaced all his Halogen GU10's with LED replacement lamps. For some reason three lamps on one switched circuit remain on but very dim when the switch is off. This room is switched from 3 points and depending on which switch turns off the lamps this has an effect on the amount of light produced in the off position.

I hope this makes sense.

Things I have tried already.
Replaced the switches.
Checked to see if there is a borrowed neutral.
Tried other lamps. It is fine with halogens but I imagine the current is only slight so you cant see any difference.
Turned off other circuits to see if there proximity had any effect.

 

[polo1]

Induced current, use a snubber on the switch.

Regards.

 

[trev]

Comment removed

 

[Inteificio]

Ok I should save this answer as a script.

LEDs get more efficient the less electricity going through them.
This scales the whole way down to 1 electron dropping through a photon will be given off.

So to say they need a small amount of electricity to run is an understatment.

So where is the the energy getting in to that circuit.

My guess for this one is parallel circuits and induced current. As you switch different amount of wires in, the 'ariel' changes length, hence the change in brightness.

Even if the whole house is isolated you will still get some light. Normally a cap charging will build up enough charge to pulse a circuit.

Normally is better when I write this, but so many times written and so warm!

The fix (not that it needs one) is simple. A resistor to even up the potential difference.

My simple line is to tell the customers that the lights are so efficient that stray magnetic fields light them. It is free light.

 

[telectrix]

Induced current, use a snubber on the switch.

Regards.

snubber needs to go across L ans N. bit of a bugger if there's no N at the switch.

 

[Knobhead]

Put it in the fitting

 

[Moses]

A snubber across the L-N will not cure this fault,
as the snubber network is designed to limit the rate of rise of voltage across semiconductor devices,
to stop them failing catastrophically due to rate of rise of voltage across them.

You need just a resistor, to bleed the electromagnetic pickup from the switched live, down to the N-wire,
possibly at the ceiling rose.

 

[Inteificio]

A snubber across the L-N will not cure this fault,
as the snubber network is designed to limit the rate of rise of voltage across semiconductor devices,
to stop them failing catastrophically due to rate of rise of voltage across them.

You need just a resistor, to bleed the electromagnetic pickup from the switched live, down to the N-wire,
possibly at the ceiling rose.

Am glad someone agrees with me!

 

[Farmelectrics]

i gather a snubber is a resistor what size would you get for the above problem and do you need to put it at end of line

 

[polo1]

Sorry, I posted in a hurry last night - meant to say switch wire and neutral (at fitting/ceiling rose). Likewise, loose/improper use of the term snubber - resistor (Must try harder, must try harder).

Regards.

 

[telectrix]

my understanding is that a snubber can be a resistor, capacitor, or a combination of both. in this situation, trial and error to fit the correct type is in order.

 

[Knobhead]

The problem with a resistor is heat dissipation when the circuit is in use. The more I think of this the less I like the idea.

This is a bit unconventional but it would work.

View attachment 12820

 

[telectrix]

my thoughts as well, tony. heat. a 330k ohm resistor would have to dissipate around 1.6 watts.

 

[telectrix]

tony, might be a brain fart on my part, but , referring to your excellent diagram, could a "drain" resistor be incorporated between the N and L1 of the switch?

 

[Knobhead]

This is even more unconventional but again it would work.

View attachment 12821