I have a bedside LED light embedded into the wall. When turned off, the LED are still at around 5%, not so bright, but similar to the fire exit sign you would see at the cinema. Immediately, I guessed power leakage via a capacitance effect, but I was yet unable to get rid of the 5% glowing. Any suggestions?
The symptoms are similar to LEDs stay on (very dim) when switch is off, except that the OP had night lights on the switches, I do not.
Here is a photo of the overall system: mains supply, wall switch, appliance, and shunt towards the switch + another of these appliances.
It seems to be made of a series of LEDS (12 n.o.) which are connected to an AC/DC transformer. This transformer sits at the back of the structure of the lamp, where it is embedded into the wall. It indicates "LED Power Supply" and is also TUV certified, so I take it this is a high-quality device. The LEDs panel is bolted into the structure of the lamp so the two most likely came together. It also looks like the transformer came already wired to the LEDS panel, but I could not confirm as I moved in whilst they were already installed (new).
The transformer is rated "AC 220-240V 0.1A" Neutral + Live, and "DC 10-21V 300mA (CONST)". It also says "No-Load: 32VDC Max Prated 6.3W" (not sure what this is for?).
The power supply
We are in Spain. Grid supply for the flat is AC 130V x2. That is, no neutral, 2 live cables of 130V each when measured against earthing; happens often in Spain and other parts of Europe I read.
This is not a fluorescent effect: when I disconnect the house from the grid, it stops glowing immediately.
The fact that the lamp is a bedside light is mostly irrelevant: the lamp is activated by a wall-mounted switch, like any typical ceiling lamp would be. This switch is unfortunately acting on only one of the two phases. Changing this switch would be ideal, but unfortunately very difficult: (i) not easy to find a dual-wire wall switch, (ii) I would need to rewire an extra cable into a duct which is already crowded, and (iii) the extra shunting would have to be done behind the embedded light and there isn't a lot of space there either.
Note I have this same 5% power problem with all LED appliances of the house: 2 bedside tables and 3 ceilings mounted appliances of another brand. For obvious reasons, the bedside appliances are the most annoying ones and my priority right now.
I understand the problem most probably comes from the fact that my house configuration is L+L, hence rendering it impossible to isolate the appliance from a live cable when the switch is on the OFF position. Also, as mentioned the appliance expects L+N; however I cannot see how the appliance would be able to tell the difference between 130+130V and 0+230V (the transformer has no earthing terminal), so I guessed the problem is elsewhere, would you agree or am I missing something? I suspect the problem is with the wire connecting the appliance to the switch: it must leak, probably through a capacitance leakage effect, where the switch plus its 2 cables act as a small inline capacitor.
What I tried
To counter-act this leakage effect, I tried adding a capacitor in parallel, whose terminals bypass the terminals of the appliance (the AC side of the transformer). This would let the AC flow through the capacitor instead of the transformer/LEDs. I tried the following configurations:
- 1 capacitor of 330 nF
- 2 capacitors of 330 nF (appliance + 2 capacitors all in parallel)
- 1 capacitor of 470 nF
- 2 capacitors of 470 nF (appliance + 2 capacitors all in parallel)
- 3 capacitors of 470 nF (appliance + 3 capacitors all in parallel)
The 470nF capacitors were rated non-polarized, 275VAC, and X2. The 330 nF capacitors were rated non-polarized, and 400V.
Unfortunately, it had absolutely no effect at all. The LEDs did not even dim by a fraction.
What I could try
I am considering adding a classic resistor, but would this be wise? I guess it would be safe for the resistor to shunt the DC side of the transformer including because the transformer indicates there is a "short circuit and overload protection". Then how could I calculate the optimum value of this resistor?
I could also rewire so that the switch acts on the DC side of the circuit, but the transformer would end up being constantly live, which in addition to sucking up idle power through heating the coil, might also reduce its lifetime.
Are there additional testings I could do to further diagnose the problem? For example, I was thinking I could make sure the wall switch is not leaking any power, but dismounting it is complicated; can I verify this directly from the shunt you can see on the photo of the overall system? Here is a better zoom.
Disconnecting one of the AC cables
I ran a series of tests by disconnecting one of the cables plugged at the driver's AC terminal under various configurations:
- Disconnect the live cable that goes to the wall-mounted switch -> the LEDs glow 5%
- Disconnect the regular live cable and turn the wall-mounted switch ON -> the LEDs glow 5%
- Disconnect the regular live cable and turn the wall-mounted switch OFF -> the LEDs are at 0%.
Whether the connected/disconnected cable was on the N or L terminals of the driver had no impact.
I concluded the problem was not with the electrical installation of my home (beside having to half Lives instead of L+N), neither caused by a bad quality switch or induction effects.
Connecting both AC terminals to the same cable
I connected both ends of the driver's terminal (N and L) to the same live wire. That is, I short-circuited the driver. To my greatest surprise, the LEDs were at 5%!
It becomes clear to me that adding capacitors, resistors, or anything on the AC side is hopeless. I believe what is happening is a capacitance effect directly within the LED driver; most likely permitted through the coil of the voltage transformer.
I am now looking for a solution on the DC side, and see 3 options:
A. Add a resistor in parallel to the LEDs. Advantage - Very easy to do since it can be added by just inserting the resistor between the existing DC terminals. Disadvantage - When under low-power mode (5%), some current will still go through the LEDs, so the resistor needs to be small enough, but then this becomes a problem when working under normal conditions (100%) because that resistor will drain most of the power normally dedicated to the LEDs, which additionaly could also overload the driver.
B. Add a resistor in series to the LEDs. Advantage - A small resistance should suffice to absorb the low-power mode, whilst it also fulfils the purpose of not draining too much power when working under normal conditions. Disadvantage - Need to create a terminal between the resistor and the cable going to the array of LEDs (is it safe to use electrical tape?).
C. Add a Zener diode in series to the LEDs. Advantage - Resistance extremely small, even better than option B. Disadvantage - Same as B, but also maybe hard to find Zener diodes which sustain 6 Watts? I guess I could add several of them in parallel to spread the load, but then I would most likely also need to add a resistor to "buffer" the transition phase (5% to 100%) in order to avoid overloading and burning the Zeners one by one, and thus this would defeat the added advantage of option C when compared to option B. Is this correct?