LED pilot light powered from 230V

[ericgibbs]

I'm thinking wich one is more reliable, this circuit or other one witch include a small encapsulated transformer rated at 6VAC and 80mA plus a voltage regulator, bridge and a filter cap.

I'm talking reliable in terms of continuous use. and power consuption.



What to you think?

Thank you,

hi,
Work out 230V * 0.02 * 24 * 365..... then tell me what you think.

 

[adrianbodor]

I don't understand what your saying, i'm talking reliable in terms of circuit failure after long period of continuous use.
And BTW the current through the LEDs is 10mA max. not 20. and the circuit power consuption is about 0.25W using the wattmeter at the mains measured with multisim, but according to my calculation P=V*I -> P=230*0.010 so it is 2.3W an hour !!!!!!!

Am I missing something? Measured power consuption after rectification is 5.6*0.010=56mW an hour

I'm a little confuse because 2.3W is huge to drive a LED from mains

Thank you

 

[ericgibbs]

I don't understand what your saying, i'm talking reliable in terms of circuit failure after long period of continuous use.
And BTW the current through the LEDs is 10mA max. not 20. and the circuit power consuption is about 0.25W using the wattmeter at the mains measured with multisim, but according to my calculation P=V*I -> P=230*0.010 so it is 2.3W an hour !!!!!!!

Am I missing something? Measured power consuption after rectification is 5.6*0.010=56mW an hour

I'm a little confuse because 2.3W is huge to drive a LED from mains

Thank you

hi,
The high power usage surprises most people, say its 2.3W/hr multiply that by 24hrs and then by 365 days and see how much per year.

 

[adrianbodor]

Then why multisim shows me 212mW? BTW the circuit is converted into DC and limited by zener diode. It's impossible to drive two LEDs from the mains at 2.3W

https://i186.photobucket.com/albums/x300/darks0ul1981/Electronics/PowerConsuptionLED230V.jpg

Why a GU10 18LED 230V downlight consumes 1W? Is the same principle

I'm in fog right now

 

[Diver300]

The current taken from the mains by these circuits is nearly 90° out of phase with the voltage, so the power in Watts is very different from the Volt Amps.

2.3 VA and 0.212 W sounds about right.

(and can we please stop writing Watts / Hour. That is a meaningless unit. It is 0.212 W and the time makes no difference)

 

[adrianbodor]

Thank you, now i understood. I was scared about that.

 

[carbonzit]

carbonzit is a very bad design. Firstly there is nothing discharging the capacitor except the reverse voltage through the LED. This will damage the LED.

BS.

I say this because as I type, my little power-line powered LED with no bleeder or other diode is happily emitting light as it has been for the last week. So are you telling me it actually doesn't work, and I'm just hallucinating?

I'm sure your suggested circuits work fine too, but so does mine.

As another member's sig here says, prove me wrong.

 

[tcmtech]

I have been running a circuit similar to what Collin has posted in #17 but I have four banks of 10 green LEDs and the capacitors are resized to supply the correct current for that number of strings.

I have no bleeder resistors or over voltage protection on the main filter capacitor and my circuit has been on since around the early 2000's now, so long I have forgot what year I built it out of old junk copier parts.

I just used common poly or mylar type non polar capacitors on everything to get the long long working life.

Relating to Carbonzits circuit yes it will work with many of the more rugged red, orange, yellow, or green LED's which are not overly sensitive to being driven in reverse with a limited current.

 

[carbonzit]

Relating to Carbonzits circuit yes it will work with many of the more rugged red, orange, yellow, or green LED's which are not overly sensitive to being driven in reverse with a limited current.

Thank you for confirming that I'm not completely crazy. (At least I hope so ...)

 

[carbonzit]

hi CZ,
If you are asking me.??

Yes.

The discharge resistor is a UL requirement, the same as the 15R.

1. Do you know for a fact that these are UL requirements?

2. Even if they are, who gives a ****? Sure, if I were designing something to be mass marketed, I sure as hell would follow their rules, and then some: safety first. But if I'm designing this for my own use, I would simply just ignore them. As I'm sure most other folks here would as well.

 

[Diver300]

I've just found a data sheet that shows an LED as containing a bi-directional zener in parallel with the LED. The same data sheet said that the maximum reverse voltage is 3 V.

If a capacitor limited circuit is used without a reverse parallel diode, then the reverse current in the LED will be the same as the forward current. That will certainly exceed the maximum reverse voltage, so that will be running the LED outside its specification, and I have no idea how much damage that will do. If the bi-directional zener is there, it won't do any damage, but I can't work out why the datasheet doesn't reflect the presence of the zener.

I have seen that sort of thing in an opto relay. It turned out to behave as though a bi-directional zener was included. I don't know if there really is one, but inductive spikes were limited by the relay.

 

[tcmtech]

1. Do you know for a fact that these are UL requirements?

2. Even if they are, who gives a ****? Sure, if I were designing something to be mass marketed, I sure as hell would follow their rules, and then some: safety first. But if I'm designing this for my own use, I would simply just ignore them. As I'm sure most other folks here would as well.

From my long time understanding of private hobby level electronics and electrical practices UL has about as much jurisdiction over that as the French embassy has over what I do in my back yard here in the USA.

 

[carbonzit]

If a capacitor limited circuit is used without a reverse parallel diode, then the reverse current in the LED will be the same as the forward current.

That's utterly ridiculous. Did you even think before you typed that?

If you did, you'd remember that it's a diode. The only time the reverse current will be the same as the forward current is if it breaks down, and of course by that time it will be completely toasted.

The reverse current in my line-operated LED, which is currently happily operating as I type (in case you missed that before), is far below the forward current.[/QUOTE]

 

[tcmtech]

The reverse current in my line-operated LED, which is currently happily operating as I type (in case you missed that before), is far below the forward current.

[/QUOTE]

I would be curious to see what your LED's reverse breakdown voltage is. Do you have an O-scope? And if so have you ever looked at your resulting wave form to see what all is going on with your LED?

So who did you piss off to get the red square?

 

[Diver300]

I am talking about a circuit with a capacitor in series with an LED. I think that is what you are using, as in post no. 10 in this thread you said:-

"It's just a 0.56µF cap, a 1KΩ resistor and a red LED in series"

Now capacitors can only pass as much charge in one direction as the other. If the LED is in series with a capacitor, there has to be the same current in the LED and the capacitor.

I realised that the LED shouldn't be able to pass reverse current, but the capacitor shouldn't be able to pass DC.

So we have various possibilities.

1) If the LED does not conduct in reverse at all at the peak-peak voltage of the supply, the capacitor will charge up to the peak voltage, and then no more current will flow. That will not let the LED light, as no current will flow after the first few cycles while the capacitor is charging.

2) The capacitor is leaking some DC current. Then some current will flow and the LED will light, but the circuit is limited by the leakage current of the capacitor, so a big resistor would have the same effect.

3) The LED passes some reverse current.

4) Some combination of the 2) and 3)

Without taking measurements, I don't know exactly what is happening in your circuit, but you have to be getting reverse current in the LED or significant leakage in the capacitor.

 

[Diver300]

I have just tested an LED. My multimeter has a resolution of 0.1µA and I couldn't measure any reverse current at 40 V, which is as high as my power supply goes to.

The datasheet for the LED says that 5 V is the maximum reverse voltage, at less than 100 µA. They could obviously put a much larger voltage on the data sheet.

 

[carbonzit]

I am talking about a circuit with a capacitor in series with an LED. I think that is what you are using, as in post no. 10 in this thread you said:-

"It's just a 0.56µF cap, a 1KΩ resistor and a red LED in series"

Yes, that's correct.

Now capacitors can only pass as much charge in one direction as the other.

Hmm, not sure about that: how about the case of a capacitor in series with a diode? (Of course, something you'd normally never do.) In this case, it would be receiving pulsating DC. What would happen then?

If the LED is in series with a capacitor, there has to be the same current in the LED and the capacitor.

Yes, indeed.

I realised that the LED shouldn't be able to pass reverse current, but the capacitor shouldn't be able to pass DC.

So we have various possibilities.

1) If the LED does not conduct in reverse at all at the peak-peak voltage of the supply, the capacitor will charge up to the peak voltage, and then no more current will flow. That will not let the LED light, as no current will flow after the first few cycles while the capacitor is charging.

2) The capacitor is leaking some DC current. Then some current will flow and the LED will light, but the circuit is limited by the leakage current of the capacitor, so a big resistor would have the same effect.

3) The LED passes some reverse current.

4) Some combination of the 2) and 3)

Without taking measurements, I don't know exactly what is happening in your circuit, but you have to be getting reverse current in the LED or significant leakage in the capacitor.

Interesting, and now I can't say I really know just what's happening in that circuit. I have no scope, unfortunately.

The circuit does work; the LED does light, with what seems like must be a lot more than leakage current through the cap. So what is happening here?

So who did you piss off to get the red square?

Heh; didn't even notice that until now. Guess I have to work my way up here.

 

[Diver300]

You could put a diode in series with the LED, in the same direction. Use a diode which has a reverse voltage larger than the peak-peak voltage of the supply, so at least 400 V for a 120 V AC supply. If the LED goes out, then it was the leakage current in the LED was what was discharging the capacitor each cycle.

If you put the diode in reverse parallel with the LED, and the LED lights brighter, then it was capacitor leakage that was discharging the capacitor.

 

[carbonzit]

But what if nothing is discharging the capacitor?

This is what I suspect is the case. When I unplug the light and put a voltmeter (DMM) across the capacitor, I read a (rapidly declining) 70-80 volts. So it's retaining a charge in one direction. Which says that the LED is doing what one would expect it to do: rectifying the AC.

Any other ideas? As I said, this does work. Now I think it would be interesting to determine just how it works. It doesn't seem to be immediately obvious, either.

I'll try a diode in reverse-parallel as you suggested and see what happens.

 

[killivolt]

So who did you piss off to get the red square?

hi CZ,
If you are asking me.??
The discharge resistor is a UL requirement, the same as the 15R.

Yes.
1. Do you know for a fact that these are UL requirements?

2. Even if they are, who gives a ****? Sure, if I were designing something to be mass marketed, I sure as hell would follow their rules, and then some: safety first. But if I'm designing this for my own use, I would simply just ignore them. As I'm sure most other folks here would as well.

A few of these, might do it?