How to fix LED fixture not turning off completely when the switch is off

[drakejest]

I have LED ceiling lights that are not completely turning off even when the switch is off.



Sometimes the brightness of the glow is just to much for me and finally decided to do something about it.

tried to explain what is going on but did not offer a solution on how to fix the issue. The pcb on 8:11 looked really similar to what mine uses. The solution to the problem though is on another video by Electroboom saying to place a 470K ohm resistor on the output of the rectifier.



The circuit that electroboom is using is not the exact same circuit as i have. So will it still work? and how do i calculate the value of the resistor for my circuit

[Vovk_Z]

I think an additional load value has to be a size or  bit higher than  parasitic leakage. I don't know but I would start with 0.5-1 mA possible value of a leakage current. 1 mA is 220-240 kOhm for 220-240 VAC.
I don't think there is a strong need to disassemble a lamp, you just may put resistor in parallel with a whole lamp.

Some people use film caps, for example X2 type, about 10-100 nF. They waste much less active power than a resistor but still make a load.

[drakejest]

I think an additional load value has to be a size or  bit higher than  parasitic leakage. I don't know but I would start with 0.5-1 mA possible value of a leakage current. 1 mA is 220-240 kOhm for 220-240 VAC.

So there is no specific value then as the wires on a room could have a very higher parasitic leakage than on another room.

I don't think there is a strong need to disassemble a lamp, you just may put resistor in parallel with a whole lamp.

The power supply is on an external enclosure, so it is very easy access, and i think this is the best place to put it also so there is no splicing on wires.

Some people use film caps, for example X2 type, about 10-100 nF. They waste much less active power than a resistor but still make a load.

Hmm X2 is not in the search filters where i am buying from (lcsc.com). There are 600V+ rated  Polypropylene Film Capacitors (CBB). Will that do? Also what value would be the most 'catch-all' value?

I will need to order different values else shipping cost will bankrupt me. The components are very cheap anyway.

Is the difference of the amount of wasted power between the cap and and resistors great? A 1.2ma leeak would waste about 1/4 Watt of power, that seems a lot. I do not know how to model the film caps into my simulation though so i cant tell how a capacitor would do. How is it connected? parallel between L and N ?

[Zero999]

A capacitor theoretically wastes no power at all, since it has no resistance. In reality, it will use some power determined by the resistance of the capacitor and cable.

The power can be calculated in a resistor using Ohm's law.

P = V²/R

I'd use a 10nF capacitor, based on what I know about the capacitance of cables. 600V should give plenty of margin for 230VAC operation.

[themadhippy]

Or a  neon indicator  across the lamp terminals.

[drakejest]

A capacitor theoretically wastes no power at all, since it has no resistance. In reality, it will use some power determined by the resistance of the capacitor and cable.

The power can be calculated in a resistor using Ohm's law.

P = V²/R

I'd use a 10nF capacitor, based on what I know about the capacitance of cables. 600V should give plenty of margin for 230VAC operation.

So is a capacitor better? wont the resistor be guaranteed way of controlling of how much the leak will be? or its a thing that  a 10nF capacitor will do the job the same but waste less power?

[Zero999]

A capacitor theoretically wastes no power at all, since it has no resistance. In reality, it will use some power determined by the resistance of the capacitor and cable.

The power can be calculated in a resistor using Ohm's law.

P = V²/R

I'd use a 10nF capacitor, based on what I know about the capacitance of cables. 600V should give plenty of margin for 230VAC operation.

So is a capacitor better? wont the resistor be guaranteed way of controlling of how much the leak will be? or its a thing that  a 10nF capacitor will do the job the same but waste less power?

Either a resistor, or capacitor will work. A capacitor will not waste as much power, as a resistor.

[drakejest]

A capacitor theoretically wastes no power at all, since it has no resistance. In reality, it will use some power determined by the resistance of the capacitor and cable.

The power can be calculated in a resistor using Ohm's law.

P = V²/R

I'd use a 10nF capacitor, based on what I know about the capacitance of cables. 600V should give plenty of margin for 230VAC operation.

So is a capacitor better? wont the resistor be guaranteed way of controlling of how much the leak will be? or its a thing that  a 10nF capacitor will do the job the same but waste less power?

Either a resistor, or capacitor will work. A capacitor will not waste as much power, as a resistor.

I do not understand how the capacitor is able to solve the problem could you please explain it to me? Are we relying on the internal resistance of the capacitor for it to work?

I forgot to ask how is the capacitor connected>? is it in inline with the live terminal or parallel between live and neutral?

[BeBuLamar]

You switch your light on/off using a solid state device? If you use an ordinary switch I can't see how the LED stays on.

[Zero999]

A capacitor theoretically wastes no power at all, since it has no resistance. In reality, it will use some power determined by the resistance of the capacitor and cable.

The power can be calculated in a resistor using Ohm's law.

P = V²/R

I'd use a 10nF capacitor, based on what I know about the capacitance of cables. 600V should give plenty of margin for 230VAC operation.

So is a capacitor better? wont the resistor be guaranteed way of controlling of how much the leak will be? or its a thing that  a 10nF capacitor will do the job the same but waste less power?

Either a resistor, or capacitor will work. A capacitor will not waste as much power, as a resistor.

I do not understand how the capacitor is able to solve the problem could you please explain it to me? Are we relying on the internal resistance of the capacitor for it to work?

A capacitor works on the same principle as a resistor. It passes a small current determined by its impedance.

Z = 1/(2πFC)

Where:
Z is the impedance in Ohms
F = frequency Hz
C = capacitance Farads.

The capacitor doesn't dissipate any power because the current leads the voltage by 90º, so when the current is at the peak, the voltage across it is zero and power = v*i. The capacitor charges and discharges, causing power to be transferred between the capacitor and back to the mains again. The net power dissipation is ideally zero, yet in reality it's a tiny amount due to the resistance in the wiring.

Here's a simulation showing the power dissipation in a 10nF capacitor connected across the 240V mains (a peak voltage of 340V). Note the phase shift between the voltage and current and that the power dissipation cycles between positive and negative, as power is transferred to and fro between the capacitor and the mains. The average power dissipation will be zero. I've included a resistor for comparison. Note how the power dissipation in the resistor is always positive and the current is in phase with the voltage.

Look up power factor.

I forgot to ask how is the capacitor connected>? is it in inline with the live terminal or parallel between live and neutral?

In parallel with the LED lamp.

[themadhippy]

If you use an ordinary switch I can't see how the LED stays on

capacitive coupling in the switch cable,if you wire your lighting circuit so the feed loops in and out to the switches problem solved.If however you loop your feed in  and out at the light fitting with a switch line from celling  to the switch on the wall,as was  standard in uk domestic wiring until recently, the switch cable acts as a capacitor and lets enough current through to tickle  the leds

[radiolistener]

I have LED ceiling lights that are not completely turning off even when the switch is off.

just remove neon lamp from the switch and issue will be solved.

[Vovk_Z]

X2 caps are placed in the "Suppression Capacitors" group at lcsc.

[drakejest]

A capacitor works on the same principle as a resistor. It passes a small current determined by its impedance.

Z = 1/(2πFC)

Where:
Z is the impedance in Ohms
F = frequency Hz
C = capacitance Farads.

The capacitor doesn't dissipate any power because the current leads the voltage by 90º, so when the current is at the peak, the voltage across it is zero and power = v*i. The capacitor charges and discharges, causing power to be transferred between the capacitor and back to the mains again. The net power dissipation is ideally zero, yet in reality it's a tiny amount due to the resistance in the wiring.

Here's a simulation showing the power dissipation in a 10nF capacitor connected across the 240V mains (a peak voltage of 340V). Note the phase shift between the voltage and current and that the power dissipation cycles between positive and negative, as power is transferred to and fro between the capacitor and the mains. The average power dissipation will be zero. I've included a resistor for comparison. Note how the power dissipation in the resistor is always positive and the current is in phase with the voltage.

Look up power factor.

Ahhh i get it now ! so basically the capacitor will draw power by charging itself and the charged power will be pushed backed to mains so ideally i am not billed for anything since the power consumed is returned back to grid. Unlike a resister where the consumed power is turned to heat.

If you use an ordinary switch I can't see how the LED stays on

capacitive coupling in the switch cable,if you wire your lighting circuit so the feed loops in and out to the switches problem solved.If however you loop your feed in  and out at the light fitting with a switch line from celling  to the switch on the wall,as was  standard in uk domestic wiring until recently, the switch cable acts as a capacitor and lets enough current through to tickle  the leds

Can you please rephrase is it so that i can understand it a little better, so that i will be aware where to properly place the switch in the circuit? As far as i know how its done in my place is its

Live  from the breaker -> long wire to switch -> long wire to LED lights -> long wire Neutral to back to breaker

just remove neon lamp from the switch and issue will be solved.

No neon light on the switch. it does have glow in the dark plastic which is not electrically connected in any way.

X2 caps are placed in the "Suppression Capacitors" group at lcsc.

Ohh thank you ~! lcsc has some very cheap stuff but their search filters are so terrible you have to really know what you are looking for. When i typed x2 a few days ago it said it didnt exist. which i have quite a few similar experience in the past.

 What are X2 by the way? there are also X1 caps. I would assume X2 are better? There is even a Y2.

[themadhippy]

Can you please rephrase is it so that i can understand it a little better,

maybe.

The above is the almost standard way uk lighting circuits were wired for years.Wire A (loop in )goes to the fuseboard,wire B (loop out ) goes to the next light and wire C is the switch wire.The problem is wire c  always has one conductor live, and the way the cable is made it acts like a capacitor so a tiny bit of current bleeds across,often enough to cause the leds to glow slightly.
One solution is to wire the lighting circuit so the loop in (wire A ) and loop out (wire b) go to the switch,not the light  fitting ,getting rid of the long switch wire (c) and instead running a cable from the switch to the light

An added advantage is it also brings the neutral to the switch which can be usefull in the future

[Vovk_Z]

What are X2 by the way? there are also X1 caps. I would assume X2 are better? There is even a Y2.

You don't need neither X1 nor Y1/Y2. Use an X2 cap, in parallel with a lamp.

[drakejest]

Can you please rephrase is it so that i can understand it a little better,

maybe.

The above is the almost standard way uk lighting circuits were wired for years.Wire A (loop in )goes to the fuseboard,wire B (loop out ) goes to the next light and wire C is the switch wire.The problem is wire c  always has one conductor live, and the way the cable is made it acts like a capacitor so a tiny bit of current bleeds across,often enough to cause the leds to glow slightly.
One solution is to wire the lighting circuit so the loop in (wire A ) and loop out (wire b) go to the switch,not the light  fitting ,getting rid of the long switch wire (c) and instead running a cable from the switch to the light

An added advantage is it also brings the neutral to the switch which can be usefull in the future

the second image really made it all understandable for me, So what im getting that is to minimize the length of wires that doesnt have a neutral along with them. It would be best that all Live wire should a acompanying neutral traveling along with them.

[drakejest]

What are X2 by the way? there are also X1 caps. I would assume X2 are better? There is even a Y2.

You don't need neither X1 nor Y1/Y2. Use an X2 cap, in parallel with a lamp.

Can you tell me what those rating means ? like for example if you were to rank them from best to worst where would X2 be and what would be at the top?

[drakejest]

A capacitor theoretically wastes no power at all, since it has no resistance. In reality, it will use some power determined by the resistance of the capacitor and cable.

The power can be calculated in a resistor using Ohm's law.

P = V²/R

I'd use a 10nF capacitor, based on what I know about the capacitance of cables. 600V should give plenty of margin for 230VAC operation.

If i could not get a 10nF capacitor, is it better get the next value that is higher?

[Vovk_Z]

Can you tell me what those rating means ? like for example if you were to rank them from best to worst where would X2 be and what would be at the top?

link #1
link #2
link #3
link #4
I don't rank them. I just use X2 caps in between the line and neutral. X1 caps are usually rare or unobtainable so they are not cheap. And similar thing with Y1/Y2 caps - easily obtainable ones usually have small rated capacitance, so they are mainly used when capacitance is lower than 10 nF in places where there is really a need for Y-class type.
If you want to rank them then you may rank them by Peak Impulse Voltage - the highest value is 8 kV for Y1 caps.