2
\$\begingroup\$

I have 220V AC LED lamps (bought from eBay) connected to a 1000W pure sine wave inverter, which itself is connected to a car battery. There is a ususal on/off switch between the lamp and inverter on the "hot" wire (called L or phase). Strangely and unexpectedly, the lamp glows dimly (at about 10%) even when off.

The LED lamp has a ceramic capacitor in it, along with another unidentified component (that looks like an IC or transistor.) I assume the capacitor (and other component) is to smoothen "flutter" due to mains frequency, but my assumption is that no non-negligible current can flow through it as long as one of the wires (be it phase or neutral,) is disconnected.

I know parasitic capacitances can theoretically suffice to enable energy transfer even over a single wire, but I thought very high frequencies (atleast in the kHz range) and/or voltages are necessary for this. What am I overlooking here?

I know that the mains wires have stray capacitance due to their length, but this too should only be able to "charge up" when the phase line is connected, but it is definately disconnected here. Moreover, the issue affects all LEDs in all sockets, not just a specific one, so I doubt it's a defect/failure in the wiring. The ground wire is not connected to any of the LEDs (they only have 2 terminals.)

A few updates after reading some of the comments/answers:

  1. Definately not afterglow. It disappears when the inverter is off, and remains indefinitely when on.
  2. Affecting all switches/LEDs in the house. No way there is still a connection, since they light up normally once the switch is turned on. I'm using the usual power lines of the house (disconnected from the utilities though,) with the inverter plugged into one of the house sockets with a specially confected "male-male" cable.
  3. Definitely not parasitic capacitance (in any way or form), since the effect isn't present when using utilities/generator power instead of the inverter. Likewise, it's not due to the presence/lack of any other components, since it only appears when powered by the inverter as opposed to utilities/generator. It definitely seems to be a peculiarity of the pure sine wave inverter.
  4. I tried connecting the ground of the inverter to the ground of the socket (thus house,) but it doesn't affect/resolve the issue. I'll post a picture of the glow later with dis-/connected inverter, so everybody can see it's not a visual misinterpretation.

At this point I want to make aware of useful videos: https://www.youtube.com/watch?v=1uEmX5XClPY https://www.youtube.com/watch?v=_bgUy6zA0ts

The large amount of cabling in Steve Mould's video acts as a capacitor, but not a usual capacitor. If he had used a discrete component instead, it would not have worked. This is kind of like a parasitic "series" capacitance. This effect is relevant, but insufficient on its own. The ElectroBoom video also provides useful hints, showing that even microamperes of current can make the LEDs light. This fact, combined with the MHz harmonics due to the pure sinewave inverter, would allow sufficient AC current to pass to the LED over a single wire, causing it to glow.

I answered my own question, please see it below.

P.S: Excuse the random capitalization, it's a bad habit I keep forgetting, given that in German, all nouns are capitalized.

enter image description here

\$\endgroup\$
8
  • \$\begingroup\$ Isn't it just the the afterglow of the phosphorus surrounding the actual LED you are seeing? \$\endgroup\$
    – ocrdu
    Jan 3, 2022 at 16:15
  • 2
    \$\begingroup\$ Add another series on-off switch to prove it'd definitely disconnected. Maybe the glow fades after a few tens of seconds. Do some more testing. \$\endgroup\$
    – Andy aka
    Jan 3, 2022 at 16:26
  • \$\begingroup\$ @ocrdu Phosphors used in LEDs are designed to have very short lifetimes, since in the excited state the phosphor cannot absorb photons. Thus a long relaxation time is inefficient. I've measured values in the upper nanoseconds to low microseconds on random white LEDs. \$\endgroup\$ Jan 3, 2022 at 17:34
  • \$\begingroup\$ @user1850479: I own a few (mains) LED lamps that show this effect for up to 20 seconds, and a "LED filament" variety that has an ever longer afterglow. All are have a "very warm" colour temperature, but that may be a coincidence. The decay in light intensity is clearly visible, though. \$\endgroup\$
    – ocrdu
    Jan 3, 2022 at 17:42
  • 1
    \$\begingroup\$ 1) definately not afterglow. It disappears when the inverter is off. And remains indefinately when on. \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 17:53

5 Answers 5

4
\$\begingroup\$

Definately not parasitic capacitance (in any way or form), since the effect isn't present when using Utilities/Generator power instead of the Inverter.

I don't think you can make this conclusion. For two reasons.

Firstly the output waveform of the inverter may have substantial high-frequency components in addition to the fundamental sinewave (I know it was advertised as "pure sinewave", but without numbers as to exactly how pure I'd take that claim with a huge grain of salt). High frequencies couple more strongly than low ones (though even 50Hz can couple strongly enough to light some LED lights if long cables are involved).

Secondly the relationship of the supply to ground may be different for your inverter than for your utility and generator supplies. This may lead to ground wires in your cables providing effective screens against capacitive coupling for the utility/generator supplies, but not for the inverter supply.

I also suspect your "10%" number is way off the actual difference in light level, human visual perception is highly non-linear.

\$\endgroup\$
3
  • \$\begingroup\$ Moreover, utilities GND and N in some systems are actually tied together at some point. \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 18:40
  • \$\begingroup\$ "Moreover, utilities GND and N in some systems are actually tied together at some point" exactly, therefore in a system with a tradtional utility grounding setup a ground wire placed between a permanent live wire and a switched live wire will provide effective screening against capacitive coupling. OTOH if the "neutral" is not actually at earth potential then said ground wire may actually be a source of capacitive coupling. \$\endgroup\$ Jan 3, 2022 at 19:15
  • \$\begingroup\$ Yes but GND isn't relevant to this setup, since when on the inverter, all of these lines are disconnected from the household. And 60Hz is too small for capacitive coupling. More likely this is due to the inverter using a MHz PWM signal to achieve the sine wave. \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 19:25
3
\$\begingroup\$

It is because of the parasitic capacitance. LEDs need very little current to glow and such low current flows even when one power line is disconnected because of the parasitic capacitance. Electroboom made a Video about this.

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Please see updated Question \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 18:00
  • \$\begingroup\$ thx for the reference to ElectroBoom. I checked out the video: youtube.com/watch?v=_bgUy6zA0ts.But honestly, it doesn't seem satisfying, since he also comes to the conclusion of "capacitance somewhere in the wires" even though the only thing changing in my case is the power source, moreover, any filter capacitor of any other device would already supposed to provide enough capacitance to make this effect always happen, but it doesn't. Moreover, the light intensity I'm seeing is a bit stronger than in the video. \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 18:59
  • \$\begingroup\$ Moreover, although I can't be sure, I feel like I already tested connecting the plug the other way around. IE I tried with both the switch being on L as opposed to N. Neither made a difference, whereas ElectroBoom seems to only handle the case of L (live Phase). I'll double check this \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 19:04
2
\$\begingroup\$

Definately not parasitic capacitance (in any way or form), since the effect isn't present when using Utilities/Generator power instead of the Inverter. Likewise, it's not due to the presence/lack of any other components, since it only appears when powered by the Inverter as opposed to Utilities/Generator. It definately seems to be a peculiarity of the pure sine wave inverter.

My guess is parasitic capacitance across the switch or wires (rather than within the light itself). When you're running off of clean mains power there is too little capacitance to couple the 50/60 Hz through, but when you turn on the inverter, the KHz or MHz frequency generated by the power fets easily crosses small parasitic capacitance.

\$\endgroup\$
3
  • \$\begingroup\$ All though mains power is anything but clean, there is some good information in your post, as it hints the inverter might be creating high frequency harmonics. As I recognized in the question, such high frequencies can explain the energy transfer. What isn't clear is if such harmonics generated by the inverter are strong enough, and whether this is unavoidable in its design, or a sign of low quality \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 18:43
  • \$\begingroup\$ What's also strange though, I've powered LEDs using 10kHz and 100 kHz many times on a breadboard, even on those designed as DC Converters. Never did any harmonics on any wires cause the LEDs to turn on/glow like in this case. "Glow" here doesn't mean like Glow-In-the-dark paint, it means more like "a fraction of when fully on, not too much, but also clearly not off". \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 18:48
  • \$\begingroup\$ Mains power will be clean compared to the output of an inverter. It has to be since very high frequency harmonics will not propagate long distances on power lines. I think what you are seeing is probably fairly normal, although a higher end inverter might be less noisy. You'd have to perform real measurements to know. \$\endgroup\$ Jan 3, 2022 at 18:49
2
\$\begingroup\$

Ok, After some research, I'd answer myself as follows:

A pure sine wave inverter creates the sine wave using a high-frequency PWM signal. If you connect a good-quality oscilloscope to the inverter, you would detect these high frequency harmonics on the output. Moreover, the LEDs in question even light up with a few microamperes, as shown in the youtube videos. Moreover, the N and L channels are both floating when running off the inverter. Thus the high frequency harmonics still reach the LEDs over the N wires. Additionally, as it is a real physical system, there is always a small amount parasitic "innate" capacitance of both N & L wire segments after the switch (this is not the same effect as attaching a discrete capacitor component in parallel with the LED, nor the same as attaching said capacitor in serial irrespective if 1 or both terminals would be connected. Rather, this is due to the innate capacitance any conductive object, even a single piece of wire, has. For more info, see the videos).

These 4 facts COMBINED (not any one by itself), explain why the LEDs light up on the inverter, but not on utilities/generator, since the wiring in the house creates sufficient capacitance at high-frequencies, but not mains frequency. "Sufficient" here only applies to the LEDs, not to other appliances, since these are so sensitive, and only applies since the N line is NOT grounded (not because it is Grounded as Peter assumed).

So the solution becomes to Tie one of the Inverter Output lines to GND, while keeping that GND connected to the house GND or mains N/GND. Either case would suffice to keep the N line to which the LED is attached to, stably at 0V, ie it absorbs the high frequency harmonics, such that even the LEDs wouldn't be sensitive to it, and wouldn't glow. Connecting the Inverter's GND to the House GND doesn't suffice to prevent harmonics from appearing on the N line. The N line also needs to be connected to the GND. Although this defeats the safety purposes of having GND, it isn't an issue if done only when connected to the inverter, rather than utilities. The GND line of the Inverter is probably only present to detect faults, it doesn't bind N to 0V.

Also, this phenomenon only appears with the very simple/cheap (not necessarily a bad thing, I love cheap) LEDs, since the ones with more complicated circuitry (and often include galvanic isolation through an embedded transformer) insulate the LED from the harmonics or absorb them instead.

\$\endgroup\$
1
\$\begingroup\$

I have seen same behaviour in domestic LED, with long wires parallel to other powered wires. You can check connecting a incandescent small lamp (used microovens or refrigerators) with your LED lamps, that load will discharge this residual voltage. To solve this issue, sometimes I have wired in parallel with LED´s a snubber (resistor + capacitor) used in universal motors or contactors coils, you need to make some trials to find proper values.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Please see updated Question \$\endgroup\$
    – mo FEAR
    Jan 3, 2022 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.