LED lights don't turn off completely [closed]

Question

I recently installed some LED lights in my house. They don't shut off completely and glow all night long. Will a Zener diode solve this issue and how do I choose the right one?

Answer 1

Modern LEDs can be glowing with a very low current, in the range of a few microAmps, even when they normally are operating at full specs with hundreds of milliAmps.

The producers of some driver ICs (I remember one note from TI) recommend using a relatively large resistor in parallel, to achieve a complete turnoff discharging quickly the capacitors. It all depends on the number of LEDs in series in your lights (i.e. the operating voltage of the series); roughly, you could employ something that goes from 100 kOhm up to many hundreds (sized for a few hundred microAmps leakage current), in parallel to the LEDs.

Answer 2

I don't see how a Zener diode would be used (let alone be code-compliant).

You could change the switch to a 3-way and do a bit of rewiring if you have neutral available in the box. Not sure that would be code-compliant but it would work and would infinitesimally reduce energy consumption.

The usual suggestion is to keep at least one of the bulbs on each circuit as incandescent, but of course that increases energy consumption.

Answer 3

The problem you're encountering is typically caused by capacitive coupling in elongated cable runs (for a detailed explanation, refer to this video: https://youtu.be/1uEmX5XClPY).

I believe using a Zener diode might not prove successful in an AC context. This is because even with Zener diodes rated for higher voltages, a short circuit could potentially occur during half of the AC cycle. An alternative approach could be linking two similar-rated high-voltage Zener diodes in an inverse-parallel arrangement, or perhaps using a Transient Voltage Suppression (TVS) diode. However, these options may not rectify your issue. These methods are more suited for creating a surge protectors or spike guard, rather than reducing the glow of your light bulb.

A possible solution is to experiment with high-value resistors, ranging from 100k to 1M ohm or even 2.2M ohm, with the exact value depending on the intensity. Personally, I would opt for resistors in the range of 0.5W to 1W. The reason for this is that many 1/4W resistors are not rated for 250VAC (or 350VAC peak to peak). However, if your local voltage is lower, such as 110VAC as it is in many countries, a 1/4W resistor might work fine. You'll need to consider power dissipation using the formula P = (V * V) / R. For instance, with 250V and a 220K resistor, the calculation would be (250 * 250) / 220 = 0.28 Watts, necessitating a resistor > 0.28W. In a 110V situation, you would be converting 0.055 watts into heat.

This resistor will drain the excess voltage, which should prevent the bulb from glowing when the switch is off. One point to consider is that while the bulb glowing when switched off incurs a minor cost, this power is at least visible as light. With a resistor, the power is converted into heat instead. This heat dissipation will be minimal when the bulb is switched on as we calculated before. After you switch off the bulb, this already minuscule power wastage will decrease even further, to the microwatt range or less. However, it's worth mentioning that even though it is a small amount, this will still be an extra cost on your electricity bill.

An alternative solution is installing a switch close to the bulb, this way you effectively eliminate the chances of capacitve coupling via lengthy wires.

Yet another better but expensive solution will be a complete rewiring of your home's electrical system!