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Jun 15 '21 at 9:15 comment added Ama The driver is acting even more weird: if I have the appliance all plugged "normally", then disconnect one of the DC cables from the driver, and turn it ON, well... the LEDs turn ON as well (5% glow). I have four of these drivers and all seem to show the same symptoms.
Jun 15 '21 at 9:12 comment added Ama @dandavis yes I understand 1k is too low, I run the test only with the driver turned "off" (one of the AC terminals disconnected). I did this test to see if an extreme scenario would at least have some effect. Thanks for clarifying the lower resistance due to extremely low current.
Jun 14 '21 at 22:04 answer Mavaddat Javid timeline score: 1
Jun 14 '21 at 17:47 comment added dandavis if a 1k across the LEDs doesn't suck up stray off currents, it sounds like the driver's not really off. Could be defective... The stray voltages we usually see "around here" are induced from live wires running next to lighting wire run and have virtually no current, in which case a 100k resistor provides a lower resistance path then a faintly glowing LED. A 1k is too low, that sucker will get finger-scorching hot. It sounds like an isolation failure in the driver, which might indicate it dangerous, but does indicate it doesn't operate how you want; it should be replaced.
Jun 14 '21 at 15:33 comment added Ama @dandavis I did what you suggested but it did not work; I went as low as 1.0Kohm and the LEDs kept glow at the same brightness (5%). I suppose my feeling about most of the current travelling through the LEDs was right.
Jun 8 '21 at 19:09 comment added Ama If the resistor is so large, doesn't it mean that most of the current (albeit pretty insignificant) will go through the LEDs instead of the large resistor? Or do LEDs have a very large resistance when given small currents?
Jun 7 '21 at 18:37 comment added dandavis Label says 21v max nominal under operating conditions. (21v/4700Ω)*21v = 0.09W; under 100 milliwatts worst-case. You won't notice a thing when it's on as 0.1W out of 6.3W is negligible. You don't need to calculate anything having to do with forward voltages unless the resistor is in series with the load, which you don't need to do as your driver is constant current. The point of the LED-parallel resistor is to consume the tiny amount of off-state leakage that excites the LEDs as (a tiny amount of) heat instead of annoying light.
Jun 7 '21 at 18:27 comment added Ama @jay613 indeed! Hence my separate question here: electronics.stackexchange.com/q/569590/286916
Jun 7 '21 at 14:10 comment added jay613 Fascinating experiments. Shattering assumptions about how a real-world system behaves like a theoretical paper circuit diagram. I guess you need more info about the internal design of the driver before you can say your are "short circuiting" the A/C side. I mean, you are obviously short-circuiting something, but we don't know what, and you've shown in a fascinating way that that is where things get interesting.
Jun 7 '21 at 11:18 comment added Ama @dandavis thanks for the suggestion. How did you calculate the 4.7k? I understand the driver should not be asked to deliver more than 21V (6.3W). I also understand LEDs do not have a linear resistance, so should I turn the appliance on, measure the voltage provided by the driver and this way calculate how much voltage is left for the resistor? For example, if I measure 15Volts, then this is 21-15=6 Volts left, and at 300mA this is a maximum resistance of 6/0.3 = 20 ohms?
Jun 7 '21 at 11:13 comment added Ama @jay613 see my updated question for the full details. It is weirder than I expected.
Jun 7 '21 at 11:11 history edited Ama CC BY-SA 4.0
Added further investigation and thoughts
Jun 1 '21 at 14:51 comment added dandavis i use small bleeder resistors on the dc side to suck up those tiny stray currents. Lost of newer bulbs parallel the LEDs with a tiny resistor just for this reason. try 100k, then less as needed, but stay below 100mw, whatever that is on your output; ~4.7k if the label is to be believed...
May 31 '21 at 21:19 comment added Ama @jay613 I have now tested to disconnect cables. The conclusion is: whichever wire (switch or mains) and whichever terminal of the LEDs transformer/driver (neutral or live), the appliance glows even when one terminal is connected to one of either live cables. Of course, if connected to the switch wire and the switch is OFF, then it does not glow, and when it's ON then it glows 5%. Tomorrow I will update my question to reflect these new findings.
May 31 '21 at 16:25 comment added Ama I added a capacitance on the AC side, and am considering adding a resistance on the DC side (much safer than playing with the AC side, IMO). Tonight when it's dark I check what happens if I disconnect the blue cable that connects the appliance.
May 31 '21 at 16:04 comment added Ed Beal Many people don’t understand that with AC circuits and you have power on a line that is parallel that is inductive coupling. It actually creates a voltage we call phantom voltage because it normally has almost no current with high efficiency LED’s there may be enough to create a glow. Shielding the wires may be a possible solution but adding capacitance on the dc side will do nothing. Have you tried opening both conductors , not just killing power to the house. Is your wiring in conduit? It sounds like it and another circuit may be the cause of the induced voltage causing the glow.
May 31 '21 at 15:00 history tweeted twitter.com/StackDIY/status/1399380144356265985
May 31 '21 at 14:48 comment added Ama Tonight I will try to disconnect one of the live cables and see if it keeps glowing.
May 31 '21 at 14:47 comment added Ama No I don't, and the switch is probably less than 1m worth of cable away from the appliance, which puzzles me. As commented in my answer to your question, maybe this is because most of the capacitance comes from the coil of the LED transformer itself, but then it would mean that the system would always glow, even under the normal scenario anticipated by the designers, and that would not make much sense; not for an appliance with a TUV certification.
May 31 '21 at 14:10 comment added jay613 This is getting interesting. You answered the question that I asked on EE to help answer this question. :). Do you have two-way switches controlling this lamp?
May 31 '21 at 13:44 comment added Ama @user263983 as indicated there are no neutrals, both wires are live at the "half" of the tension of a "regular" 230V live. This is the way power is supplied by the grid in my street. However, I did try switching the wires, just in case and it did not work.
May 31 '21 at 13:41 comment added Ama Thanks @jay613. I investigated this matter for weeks and read dozens of articles on the matter, so I thought I would save people the redundant work. I am looking for suggestions on how to get rid of the remaining 5% glow of the LEDs, as indicated in the introductory paragraph. :) In order to do so, I guess the other question you mentioned would need to be addressed, as well as my sub-questions.
May 31 '21 at 13:11 comment added jay613 This question has been asked several times but never with this much detail and it has never been answered very well. Even though you've provided LOTS of amazing background, you haven't asked a very precise question. I think, "How can I avoid this" would be a good question. A double pole switch would not be a good answer, as you note. Another question is "how and why does this happen?". I have taken the liberty of asking that on the EE Stack Exchange: electronics.stackexchange.com/questions/567760/…
May 31 '21 at 11:59 comment added user263983 Where is the switch? Try to reverse connection. Your switch may be on neutral wire, but phase should be disconnected.
May 31 '21 at 11:47 review First posts
May 31 '21 at 14:00
May 31 '21 at 11:40 history asked Ama CC BY-SA 4.0