240V indicator led at input of a power supply.

Thread Starter

q12x

Joined Sep 25, 2015
908
Hello,
I am the artist, (I'm not an electronist or electrician).
I have a weird problem that I never encountered.
I have a small power supply led driver for a power-led wall light.
1645319677862.png20220219_163800.jpg
I want to attach an indicator led running at 240V to its input ( directly on the 240V wires). You can see the 240V input in the picture, in the left side, as brown and blue wires. I marked with red the location where I soldered this circuit.
20220219_163606 Copy.jpg220v led indicator circuit for mains.jpg
Here is my circuit. The resistor burned to black totally the first time I mount it and it took the led with it, but it didn't had the diode. Now with the rectifying diode included, is still fumigating and get extremely hot, but I don't let it to burn totally like it's predecessor.
Here is the weird part: You see this extension cord? I keep it on the table, and IN IT I plug in the cord for this power supply with it's burning indicator light at 240V. Do you see that green led already working inside the extension cord? It is the same circuit I used as well. And it never burned out. But this one I am using here, NEAR the power supply, is always burning. VERY strange, right? If this power supply is kicking back and burning my led and its resistor, immediately at it's input, then it should burn the led inside the extension cord, correct? This is the strange part I am very intrigued. Or maybe I am using the wrong diode? I didnt used a 1N4007 (for 1000V) like in the circuit, but a "smaller" one.
20220219_164533.jpg
I believe there are some sort of spikes coming back from the power supply, especially from that big violet filter capacitor.
- Do you think I should put another circuit?
I want it to be on the 240V side because if something will broke, at least I know for sure is getting 240 on its input.
I was thinking to put it in the output of this power supply, on the DC side with 72V@120mA. But I am very intrigued of this failure that I encounter for the first time, or at least I am aware of it and I can properly formulate it. Very weird.
This is not my first indicator at 240VAC. I put like tens of these indicators along the time. Mostly inside extension cords like you see the one on my table. And in a few power supplies but in their output usually. I dont recall putting in the input as I am doing right now.
I think is a fascinating problem. I really hope this is a common problem that you guys are aware of it. For me is the first time.
Thank you very much, and I really am curious what you will tell me.
 

dl324

Joined Mar 30, 2015
14,457
LED to mains used to be a disallowed topic, but it appears to be okay now...
Or maybe I am using the wrong diode? I didnt used a 1N4007 (for 1000V) like in the circuit, but a "smaller" one.
It looks like you're using a signal diode that's probably breaking down at 100V or so. Instead of relying on the reverse breakdown voltage of the diode, you could put that type of diode anti-parallel to the LED.

1645286632996.png
The problem with the resistor is that you're probably exceeding it's maximum working voltage of around 250V. Use 2 or 3 in series to reduce the voltage that each resistor sees. Or use a 1W or larger resistor.
1645286901425.png
 

Tonyr1084

Joined Sep 24, 2015
6,884
220VAC ÷ 68000Ω = 0.003A (3mW)
0.003A x 220VAC = 0.718W (718 milli Watts).

What wattage resistor did you use? To handle 718mW I'd go with a 2 watt resistor. A 1 watt resistor would work but it would be too close to the peak wattage handling ability. I'd opt for a margin of safety.

When you ran your circuit did the LED glow at all? Even briefly?
 

AnalogKid

Joined Aug 1, 2013
10,052
Others have addressed your circuit and power dissipation problems. Still, two things:

maybe I am using the wrong diode? I didnt used a 1N4007 (for 1000V) like in the circuit, but a "smaller" one.
Why? What do you mean by ""smaller""? What diode did you use? Complete part number - ? Photo - ?

You see this extension cord? . . . It is the same circuit I used as well.
How do you know that? More importantly, how do we know that? Photo of the inside - ?

ak
 

Audioguru again

Joined Oct 21, 2019
4,666
Most LED circuits operating from a high voltage AC use a capacitor in series to limit the current without producing any heat.
A low voltage little diode must be reverse-connected parallel to the LED.
Of course the capacitor must have a high voltage rating and you must calculate a suitable reactance for the current that is needed.
Usually a high voltage resistor is connected parallel to the capacitor to discharge it when the circuit is not powered.
 

Tonyr1084

Joined Sep 24, 2015
6,884
The numbers I gave are down and dirty numbers. 220VAC can be anything from 218 to 240V
We don't know the exact forward voltage of the LED, nor do we know what diode you used.

ASSUMING the numbers I gave, the Vf of both diodes is practically a nil subject since they are so low. If we assume an absolute 220VAC and a resistance tolerance of 1%, the numbers still come out virtually the same.
(220 - 2Vf - 0.6Vf) ÷ 68000 = 0.003197A
220 ÷ 68000 = 0.003235A. A difference of 38.3nA (Nano Amps). Nothing worth being concerned about at this time. However, assume the voltage is 240VAC.
240 ÷ 68000 = 0.003529A. Even then there's not much to be worried about. But at that amperage (multiplied by 240VAC) = 847mW. Thats 129mW closer to 1 watt. If the resistor is a 10% (to the low side - assume 61.2KΩ)
240 ÷ 61200 = 0.00392A (call it 4mA)
0.00392A x 240V = 914mW. Much too close for comfort for the resistor sake.

While someone may say 'you forgot to account for tolerances in voltages and resistances and forgot to subtract for forward voltages' the numbers are still reporting the same thing: 1 watt resistor is too small. Yes, you're below 1 watt. But you leave yourself virtually no margin for error or tolerances.
 

Thread Starter

q12x

Joined Sep 25, 2015
908
I just tested 2 things:
1- I changed the diode into a 1N4007 and 2- I changed the circuit like mister @dl324 mentioned :
1645328389501.png
None did any better. Not even the new circuit which left me contemplate at it, in atmost "dafak?" stupidity. The resistor still fumigated, (I unplug as I see a brief smoke from the resistor). I also believe that smoked resistor is much lower wattage now, after being boiled a couple of times. I measure it and is still 10k as it was when new, but he clearly and visible smoked a couple of times. Is slight brown-ish. Not brown yet. He has some life in it though.

I also got very intrigued and stupefied and I opened up my long lasting indicator led from inside the extension cord.
20220219_192817.jpg20220219_192822 Copy.jpg
It is made by me as I told you before,as you may probably know me, building every circuit on cardboard. And you can see a BLUE 1/4 Watts resistor inside there, a 1N4007 diode and you must use your imagination for the led that is pointing the other side of the case.
The circuit I used in there is exactly as mister @dl323 circuit:
1645329143375.png
I believe you when you say I must put a higher Wattage resistor in it, but ALL the indicator LED's I build so far, ALL of them worked perfectly with 1/4W resistors, and even with some experimental diodes, like 1n4148 or even a transistor that was blown but behaving as a diode, and they worked perfectly, no resistors blowing up or even warming up.
The source of the problem I am intrigued to discover here. The cause.
My initial idea was to insert a fixed capacitor (somewhere) to really lower the current. But I should probably listen to mister @dl324 and insert first 2(20k) or 3(30k) 1/4W in parallel to help the power dissipation.
A) To answer to mister @ericgibbs - the 240V direct Neon indicator lamp, is a very nice idea. But I dont have them and a led is brighter and smaller, and I like them better.
Like I mentioned before, I build ---tens--- of these indicator led circuits and no problem with any of them - ever.
B) To answer to mister @Tonyr1084
What wattage resistor did you use? - I used 1/4 W resistor @ 10k (because these are super bright LEDs and they work fine with 10k). I use them for years like this and they are awesome. I know the old GaAr leds were optimal with 1k to 500R. But not these new ones.
When you ran your circuit did the LED glow at all? Even briefly? - Very good question! Yes the led lit UNTIL the resistor burned OFF totally. So...like 5seconds or more. In the second test here, with the Hot resistor that is fumigating, the led is fine.
C) And for mister @AnalogKid - I already answered you in the entire context of this response.
D) to mister @Audioguru again : Yes, you are refering to a capacitive dropper circuit, that I also built a good bunch of them. I did thought on including a capacitor if the more power resistors will not work. I have to try them first and the capacitor dropper or even less complicated capacitor circuit, will be plan B. Because its a lot more space for a capacitor dropper circuit than it is for a simple indicator led circuit.
- I am still baffled by the fact the circuit inside the extension cord didnt burn EVER, to anything I plug in, and trust me, I plugged in some very power tools in that extension cord and that little circuit with its led never failed me. But this one, is just retarded. Hahahaha..but whyyyyy?
 

Audioguru again

Joined Oct 21, 2019
4,666
Your 240VAC has a peak voltage of 240 x 1.414= 339V. BUT your resistor breaks down with a voltage of 250V and higher.
To reduce the voltage in some resistors you connect them in series, not in parallel.

The circuit in the extension cord used a high voltage resistor with a resistance that is high enough that it does not get burning hot, just warm but allows enough current for the LED.
Your 68k resistor with a peak voltage of 339V across it limits the peak current to 339/68k= 5mA but its average current is 3.5mA. Then the resistor heats with 3.5mA squared x 68k= 0.83W. A large 1W high voltage resistor will get very hot.

A capacitor needs a high voltage resistor in series with it (try 15k) to limit its peak charging current.
I calculated 47nF (0.047uF) at about 500V should work.
 

Tonyr1084

Joined Sep 24, 2015
6,884
B) To answer to mister @Tonyr1084
What wattage resistor did you use? - I used 1/4 W resistor @ 10k (because these are super bright LEDs and they work fine with 10k). I use them for years like this and they are awesome. I know the old GaAr leds were optimal with 1k to 500R. But not these new ones.
When you ran your circuit did the LED glow at all? Even briefly? - Very good question! Yes the led lit UNTIL the resistor burned OFF totally. So...like 5seconds or more. In the second test here, with the Hot resistor that is fumigating, the led is fine.
10K 1/4W on 240VAC (ignoring peak voltage (1.414)) is 24mA. @ 24mA 240V is 5 3/4 watts. Numbers don't lie. If you've used 10K resistors at 1/4 watt it must not have been 240VAC. Correct me if I'm wrong.
Your 240VAC has a peak voltage of 240 x 1.414= 339V. BUT your resistor breaks down with a voltage of 250V and higher.
I wouldn't have thought to use 1.414 without a cap. Does that still apply? We're rectifying the current but not capturing the peak voltages with a cap. I'm not convinced on 339V. Not saying you're wrong, just saying I'm not convinced.

The point about a resistor being rated for a high voltage is something definitely worth considering. I hadn't considered that when I looked at this problem, so - - - .
 

dl324

Joined Mar 30, 2015
14,457
I believe you when you say I must put a higher Wattage resistor in it, but ALL the indicator LED's I build so far, ALL of them worked perfectly with 1/4W resistors
It sounds like you've never put them on mains. If you operate the LED at 5mA, you'd use about a 47k resistor. Power dissipated in the resistor will be over 1W:

\( \large P = \frac{V^2}{R} = \frac{240^2}{47k} = 1.23W \)

A 2W resistor would give you the power dissipation and voltage rating you need. If you put 1/4W resistors in parallel, you would need to use a series+parallel combination to reduce the voltage across each parallel combination to a tolerable level.

For the power calculations, you can use RMS values, but for working voltage, you should use peak voltage.

EDIT: Regarding the reverse breakdown voltage required for the diode. It's a don't care. The signal diode will see a reverse voltage equal to the forward voltage of the LED. The LED will see a reverse voltage equal to the forward voltage of the signal diode. Your only concern is limiting the currents to something the diodes will tolerate and keeping voltage and power dissipation of the resistors within their capabilities.
 
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Thread Starter

q12x

Joined Sep 25, 2015
908
I bloody find the problem, haha.
I told you, I build tens of these little indicator leds in my life. Probably 40 or 50. I dont even recall. But I had large time gaps between making them, so I always forget their circuit, I have to get back to my notes. Damn I get old and forget things.
Here is the working indicator led, using as I originally design it, with a 1n4148 diode in it, and an even smaller power resistor, this time at 1/8W !!!! Hahahaha. And no smoke this time, all is fine.
I will leave you to take a guess what I did. A little hint, mister @dl324 was close to the mystery in it's thinking. Not on the point precisely but very close. I think you can guess from the image what I used there. I will tell you after you will take 1 or 2 guesses.
Hehe, its very easy. It's that easy that I forget it completely.
(By the way, it is working for 10minutes on the table as it's older brothers that are working for years).
Also, I put my finger in the image to have it as reference for the size of the components I used. And also folow the black cord to see it is going straight into the mains. That white switch is just a switch. I also added a white led (also at 240Vac) on it and you can see it. Is no trick.
Im very curious if you can figure it out. I will tell after 2 guesses, no worries.
I am upset on myself that I forget it. Blah.
20220220_000314.jpg
 

AnalogKid

Joined Aug 1, 2013
10,052
Based on this schematic in #10, the LED is on during every other half-cycle. Since it's power already is half-wave rectified, putting another 4007 in series with the string will cut the resistor power dissipation by 50% but not affect the LED brightness.

ak

1645309801289.png
 
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