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240V indicator led at input of a power supply.

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_q12x_

_q12x_

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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.
[ElectriciansForums.net] 240V indicator led at input of a power supply.
[ElectriciansForums.net] 240V indicator led at input of a power supply.

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.
[ElectriciansForums.net] 240V indicator led at input of a power supply.
[ElectriciansForums.net] 240V indicator led at input of a power supply.

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.
[ElectriciansForums.net] 240V indicator led at input of a power supply.

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.
 
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Hello again.

First you need to calculate the power dissipated in the resistor - an amount in Watts. It is this power/electrical energy which is being turned into heat and warming up the resistor. The resistor needs to have a suitable power rating to handle this conversion of electrical energy into heat energy. The resistor must also be able to transfer this heat to its surroundings. Otherwise it will continue to rise in temperature until the materials from which it is made begin to decompose or even burn.

google how to calculate the power dissipated in a resistor.
 
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Cut to the chase - the power dissipated would be (230)² / 68000 / 2 = 0.4W. The division by two is due to the diode blocking the circuit for half the AC cycle. If the resistor is rated at 0.4W or more it should not burn out if it is allowed to cool in free air. Your idea about things coming back out of the LED driver causing damage is unlikely to be the cause. The voltage seen by the resistor is defined 99.9% by the 230V AC of the electrical supply, not by the LED driver. Although a PSU can send high frequency ripple back out of the input, the amplitude is very low (not really 'spikes') and ripple will not damage the resistor anyway.

The diode needs to be 600V or preferably 1000V rated, it has to block 325V minimum as this is the peak voltage of 230V AC. If the diode is breaking down under full voltage then current will flow through the resistor on both halves of the AC cycle and it will dissipate 0.8W instead of 0.4W. So one possibility is that the diode is not adequate for the purpose, it is breaking down, the resistor dissipation is doubled and the LED is also being damaged by the reverse current flow.

The other possibility is that the resistor is not 68kΩ - you picked a lower value up off the table by accident which is allowing much too high a current. We've all done it! By the time you realise there is a problem it is toasted and you never get to find out what resistance it actually was.
 
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Thank you both, mister @marconi and mister @Lucien Nunes for your answers. You are both right.
Hell, that image of the circuit with the 68K in it, I didnt even read it. Wow, what a mistake I did. I posted it only for the circuit itself, but I overlook reading it. Ha. Negligence from my part.
The problem is resolved now. Here is the final assemlement :
[ElectriciansForums.net] 240V indicator led at input of a power supply.
[ElectriciansForums.net] 240V indicator led at input of a power supply.

Now is working as intended with a 1/4W resistor in it but at a much higher value - of 220k.
In my original blowing up circuit, I used 10k and that was too small value. I built like 50 circuits like this one , but rarely and in time so I always forget the details. Im getting old.
[ElectriciansForums.net] 240V indicator led at input of a power supply.
[ElectriciansForums.net] 240V indicator led at input of a power supply.

Here is a test I did, putting my finger for reference, I used a 100k 1/8 resistor !!! and a 1N4148. and it run as the rest of these circuits that I build before. I know you dont believe me, because I am an artist, but try to.
So yes, problem solved and really, thank you for taking the time to respond to me.
 
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As a general rule you want the resistor to be generously rated for reliability. While many high-value resistors claim, say, 0.25W they often don't really deliver that long term. Also many smaller resistors are only rated to 250V or so max and mains power can have the odd 1-2 kV spike on it from lightning or just a big fault clearing, motor switching, etc.

For a small demo project like your that may not be an issue, but sometimes it is better to use two smaller ones in series, etc, to increase voltage handling (e.g. 2 * 33k instead of 68k, etc)
 
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pc1966 said:
The other approach is to have the small diode in inverse-parallel with the LED.
Click to expand...
The circuit I originally posted, I discarded very quickly and switched to this other one, that I forget to mention.

[ElectriciansForums.net] 240V indicator led at input of a power supply.


marconi said:
For q12 : You got off lightly this time
Click to expand...
No I didnt, I burned 5 leds, 5 resistors, I spent 1day and a half doing stupid and wrong experiments, all day long, until I realized the "simple solution" that I presented it here "like that". I even was desperate at some point because my other old circuits were working "miraculously" and mine was all the time blowing up. I couldn't remember this little detail, I forgotten it totalmente.
And in the end I did made a calculation of the working thing:
The voltage is 120V since its half of alternative wave that led is seeing. So:
V=120V, R=220000R (220k), results I=0.000545 and P=0.0654
I am used to rely more on the experimentation, I admit.
 
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Last edited:
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For q12 :

In a sinusoidal alternating current circuit the instantaneous power changes as the mains voltage goes up and down. The power is not bothered by the reversal of current flow every other half cycle. In a simple circuit of a resistor R connected across an alternating current supply of root mean square value(rms) V, the current flowing is V/I. Since power P = V x I, we can write P = V x V/R or Vexp2/R.

Since R is constant, P is proportional to the square of the applied voltage V - Vexp2.

If then P is proportional to V exp2 over all cycles of the mains, if we only use one half cycle the power dissipated in R will be halved.

We can write then Phalf = V* exp2/R where V* is the equivalent rms voltage over a full cycle.

Phalf = P/2 = Vexp2/2R = V*exp2/R

Therefore Vexp2/2 = V*exp2

V*, the rms voltage over a full cycle which equivalent to an rms voltage of V but just using every other half cycle (as caused by the diode) is

V* = square root of V*exp2 = square root of { Vexp2/2} = V/sqrt2 = V/1.414.

For a 220V supply V* is 220/1.414 = 155V.

Power dissipated in R is 155 x 155/220000 = 0.109W

Check 220 x 220/220000 = 0.22W for full cycle. 0.22/2 = 0.11W for half cycle.

Thus since P = V x I, and I = V/R, P = V x V/R. Halving the voltage also halves the current so the power is reduced by a quarter. Same logic applies if one considers current in P = Iexp2 x R - halve the current quarters the power.
 
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What can I say? Thank you in the first place. mister @marconi.
But you must understand that we on this Earth are 2 categories of people: Experimentalists and Theoreticians. I am definitely an experimentalist. I LISTEN to the theoretician, I can not do s*hit from what he is doing, (as you love to do your calculations) I love to do my experiments. I must admit that I am not good at math, is what I was told and I believe them, I am an artist and you probably know artists are 'humanism' inclined rather than 'realism', but I am the BLACK sheep in my branch that loves some aspects of 'realism' and that is the experimentation side. I am aware, that some calculations will help me speed up the things I do and experiment, but... I know what I know and I like what I like. What can we do? Right?
I also build a c# (programming) application that is using some of this formulas and make it simple and very quick to get an answer. I am using it sometimes, when I'm in absolute critical problem. I am not used to do every day calculations even with this simple and very efficient tool that I built myself.
I will try my best in the future to get into the calculation mindset, to do it more daily and routinely than I am doing it until now. I do not promise anything though, other than trying.
I do respect what you can do, but I can not do it like you. What is good, is showing me what's possible and what else is there. What I get from you, is there are other more detailed and specific variables that are complicating the simplistic "only half wave" rationament that I have. I did understand something, but not all and complete. You have a forever thumbs up from me. Haha.
 
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brianmoooore said:
If you require a LED indicator for 240V, I would strongly advise that you purchase a proprietary one with flying leads. No exposed joints at dangerous voltages.
Click to expand...
The O/P is wanting to build one, not go and buy it off the shelf.
He also seems perfectly capable of understanding the risks involved, as he's already experimented and seems to be still alive.
 
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mainline said:
The O/P is wanting to build one, not go and buy it off the shelf.
He also seems perfectly capable of understanding the risks involved, as he's already experimented and seems to be still alive.
Click to expand...
MCBs were quite expensive when I wired my property. I used 15 of them, and not one has ever tripped.
I'm still alive, and my house hasn't caught on fire, so what a waste of money they were.
OPs initial post states that he's an artist, not an electrician, and the fact that he hasn't electrocuted himself yet is no reason to expose himself to substantial risk.
 
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brianmoooore said:
MCBs were quite expensive when I wired my property. I used 15 of them, and not one has ever tripped.
I'm still alive, and my house hasn't caught on fire, so what a waste of money they were.
OPs initial post states that he's an artist, not an electrician, and the fact that he hasn't electrocuted himself yet is no reason to expose himself to substantial risk.
Click to expand...
MCBs are for cable protection, they wouldn't stop you getting killed from electrocution or from arcs causing a fire :)

Maybe you would have been better investing in an RCD 🤣


Look at the OPs previous posts, he obviously has an understanding of electrics.
 
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