1

I bought a string of lights made of white LEDs (SÄRDAL by that famous Swedish company) that doesn't have an on-off switch. The AC/DC converter directly plugs into the mains (220V) and I wanted to add a switch to turn it off without having to unplug it.

I bought a very simple switch from the hardware store and installed it on the leads that make up the string of lights, after the adapter since this was my only option. There are two leads on the string, and they run on 24VDC.

The switch is made so that one of the leads is interrupted mechanically when in "off" state, but the other lead is directly connected from one end to the other end of the switch by a strip of copper.

I thought this should be enough to interrupt the circuit and turn the LEDs off, which is mostly does, except some of the LEDs still light up very faintly in the "off" state.

When in the off state, I measure a very slight negative DC voltage between the two leads with my multimeter (as opposed to positive 24V when on).

Why is that? I thought it might be due to the two leads being twisted in a double helix way along the length of the string of lights, acting as inductors? Or maybe the electrical insulation between the two leads is not thick enough?

I tried connecting the negative lead to the terminals within the switch that are directly connected together (and the V+ wire on the interrupted side), as well as connecting the V+ lead to those and interrupting the negative wire, but both options still produce a faint light.

Thanks.

EDIT: this might not be important, but the LEDs are connected in parallel (I think), since there are 3 leads between any two LEDs, even though only two leads come out of the adapter.

beeb
  • 111
  • 5
  • 1
    Do your plugs allow you to reverse the connection? (I'm curious if the problem goes away when you mechanically interrupt only the _opposite_ lead.) If so, there is an earth ground problem of some kind, I imagine. – jonk Jan 31 '17 at 08:10
  • Hey @jonk. The switch allows me to interrupt either lead, and I did as stated in the question, but to no avail. – beeb Jan 31 '17 at 08:15
  • there will be a large CM E field with a floating DC SMPS from the conducted noise filter to line and ground to reduce conducted noise. However the ungrounded DC has crosstalk in transformer creating this CM high impedance voltage at the switching rate such that a few pF is enough parasitic capacitance to conduct up to a few hundred uA enough to create dim light – Tony Stewart EE75 Jan 31 '17 at 08:31
  • 2
    You MAY be getting capacitive coupling. Try adding a load resistor across the LEDs on the "cold" side of the switch. Probably a 1k resistor should be low enough - or a 24V light bulb or 2 x 12V bulbs. (These would light when powered but serve to demonstrate if a load helps. – Russell McMahon Jan 31 '17 at 08:31
  • Can you add a switch which cuts both lines, this might help to really switch of the LEDs? – auoa Jan 31 '17 at 08:36
  • Thanks for your comments. @RussellMcMahon what do you mean "across the LEDs" ? I have a bunch of 1k resistors laying around so I wanted to try that. – beeb Jan 31 '17 at 08:42
  • @auoa this would probably solve the problem but not help understanding how it works. Also, this might be hard to find in a format that can be put on a cable. – beeb Jan 31 '17 at 08:44
  • 1
    the very slight negative DC voltage between the two leads with my multimeter indicates ac hum or CM AC field is present and the +V side is conducting thus lower peaks leaving a slight negative DC false reading. (not true RMS), grounding DC is one way to fix it another is a 0.1uF cap or R across LED strip. – Tony Stewart EE75 Jan 31 '17 at 08:53
  • Thanks @TonyStewart.EEsince'75 . I just noticed the LEDs are mounted in parallel (I think) so I would have to put a R or cap across every LED? This won't be easy since they are encapsulated in injected plastic. I would have to strip the leads around each LED what a pain, let alone the easthetics aspect. How would I go about grounding DC? – beeb Jan 31 '17 at 09:05
  • no just after the switch. Cap across 24V rail. which means at least 7 LED is in series if white. forget about gnd. – Tony Stewart EE75 Jan 31 '17 at 09:17
  • By "across the LEDs" I mean, between the unswitched lead and the switched lead at a point where 24VDC is applied only when the LEDs are powered. It is electrically connected from LED -ve bus to LED +24V bus. The aim is to load the 24V unpowered rail when no DC is formally applied. Finding what stops the "problem" is a step in understanding it. IF this works it indicates that energy is reaching the LE power rails and that the load of the resistor is enough to reduce the resultant voltage to an acceptable level. Tony is suggesting the same thing as I did but also suggests a capaitor may work. – Russell McMahon Jan 31 '17 at 09:55
  • 1
    A schematic would really help here. – Dmitry Grigoryev Jan 31 '17 at 10:13
  • Thanks for all your comments. I tried putting a 1k R across the two wires on the LED side of the switch, didn't help (even made the problem slightly worse I think). Same with the 0.1uF capacitor across both wires on the LED side: didn't make it much worse but didn't help either... This thing is weird – beeb Feb 10 '17 at 09:50
  • Related: https://electronics.stackexchange.com/questions/237111/led-lamps-keep-glowing-when-dimmer-is-turned-off/237116 – winny May 16 '17 at 20:20
  • LEDs produce a voltage when light shines on them. Do they also light up when your room is dark? – Christian Jun 16 '17 at 08:16
  • @Christian yes they do light up in the dark, that's actually the reason I want to fix this. – beeb Jun 16 '17 at 08:53
  • Other options include a power strip with a switch, a single outlet switch, a rf or wifi controlled outlet (it's Christmas time so super cheap ones are in stores right now). A dpdt switch so you switch both sides of the cable. – Passerby Dec 02 '17 at 22:26

2 Answers2

1

While this doesn't answer why this is happening, you could wire your switch like this to stop the LEDs illuminating when switched off.

schematic

simulate this circuit – Schematic created using CircuitLab

HandyHowie
  • 4,030
  • 1
  • 15
  • 22
  • 1
    If the OP has bough a typical inline switch (torpedo switch) such as might be installed on a mains lamp, it wont have the contacts to do this. I've never seen an SPDT switch in such a package (though there's probably a reaons for them to exist) – Chris H Jan 31 '17 at 10:40
  • Exactly what he said, I can't do it with the switch like it is currently. I guess I could modify it, I might try this at some point when I have time. – beeb Feb 10 '17 at 09:34
  • Is there some reason with your set up that you can't install the switch in the mains power cord? I am assuming the LEDs don't glow when you unplug them? – LsD Jul 20 '17 at 13:53
0

schematic

simulate this circuit – Schematic created using CircuitLab

Another solution is add ferrite donut or clam shell core ferrite like that used on Laptop charger DC cables, Video cables etc.

Tony Stewart EE75
  • 1
  • 3
  • 54
  • 182
  • Thanks for your answer. I finally got time to check this out, and adding a ferrite donut didn't help unfortunately. I'm gonna try other solutions now. – beeb Feb 10 '17 at 09:31
  • It should have beeen large enough to wind many turns around donut, then small ceramic cap to shunt noise for series paired wires in donut – Tony Stewart EE75 Feb 11 '17 at 01:46
  • Okay I will try with the cap, but the donut I have can only hold the wires twice around. I'll see if I can get my hands on a larger one. – beeb Feb 12 '17 at 19:33