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LED for AC

bogdanfirst

New Member
Here is wahat i use for leds when i connect them to high voltage ac, line directly to the mains.
here is how the circuit works.
on one semiperiod the zenner conducts in one way, thus having 2.7v(or the zenner value) and lights the led.
on the other semiperiod, you get 0.6-0.7V(like to a normal diode, because the zenner conducts like a normal diode in one of the way), so the reverse led voltage is not too high to damidge it.
 

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bogdanfirst

New Member
ok, ill answer your questions...
from what i claculated this circuit is good in using for ac voltages over 20v, this is when the effficency is better than using a single resistor.
you might not want to put the resistor, R1, it is for current limiting, but for a green led, it works fine without it, but since for ir leds, and small voltage leds, it is a little bit hard to find a zenner diode for the right voltage for led.
and, finally, the value of C1 depends on what voltage you use the circuit for(sory i firgot to mention it), but make sure that you need the right voltage capacitor, depending the ac voltage you use the circuit in.
i have something made like this, but with a highter voltage zenner diode, and 4 12mm yellow leds.
and another thing, use a 1W at least power for the zenner. it is not that much power disipated, but i found out than 0.5W burn, quite fast, i think it is because i used quit old ones, but i am not sure.
 

kinjalgp

Active Member
I have a doubt with the circuit suggested by bogdanfirst.
The capacitor has a property of passing AC and blocking DC. Here you are applying AC to the circuit so the capacitor won't come into the picture as it offers very low impedance to AC. Also during the negative half cycle LED is reversed biased but zener diode is forward biased thereby allowing full AC current to pass through it which may cause zener to go into avalanche breakdown and burn out after some time. So I think zener current limiter is also required.
 

bogdanfirst

New Member
well, it works fine with a 1.3W zenner diode and with voltage less than 300V. i have something to indicate the power of a 220V fan and i have not problem in using it.
 

pebe

Member
You don’t need the zener or R1. Let’s look at a bit of basic theory.

1. An LED requires unidirectional current. The unwanted ½ cycle can be lost in a reversed-biased parallel diode when fed from AC.
2. It needs to be current – not voltage - driven. A typical mean current is 10mA.
3. So with a 50% duty cycle the supply current from AC mains needs to be approx 20mA.
4. When a capacitor is used in series with the LED and fed from 200v, the capacitor constitutes almost the entire impedance to the supply.

From that we can calculate that a 330nF capacitor will give an impedance of 9646ohms. That will give a current of 22.8mA – a mean of 11.4mA – which is near enough.

To recap: LED in parallel with reversed biased diode like a 1n4001. Put that combination in series with a 330nF cap. Choose a polycarb or similar with an AC rating for mains voltage.
 

bogdanfirst

New Member
no, i dont think that your calculatios are quite corect, or maybe they are..
and the way you need the zenner is for limiting the current trough the led.
maybe it works with what you say, but it is safer like this. (for the led)
who knows, you do it your way, i do it mine.
now, i dont say that it doesnt wirk your way, but the voltage on the led is not so stable.
 

Gervey

New Member
I agree with your analisys of the circuit pebe. I'm doing something similar but with more current (280mA) and more frequency (400Hz) and I have to do the math in order to use the correct capacitor. :shock:
 

bogdanfirst

New Member
look, i am not saying that i don do the calculations, but it is what i used since now. when i posted the last messaakes it was about 3-4a.m. in the morning so i really wasnt in the mood of making calculations.
i totaly agree that you can use only a normal diode, but it works with the zenner too. and the component number remais the same. so...
 

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