30 LED in 240V

I am making blocks of 30 LED to be conected to 240V using a Full-wave rectification, and a resistor to drop voltage.
This resistor produce much heat.
How can calculate and use a capacitor insted of the resistor?

Regards

Elba

Dear Elba
Pls Share the Schematic .
1:-what is the forward curent rating of LED's.
2:-What is the emiting color of LED.

Elba,

you can use a capacitor having a capacitive reactance equal to the dropping resistor value. The capacitor is placed on the AC side of the bridge rectifier.

One issue is when the AC is switched ON at the peak of the AC waveform. With the resistor the current is limited but with the capacitor the peak current will be quite high and may damage the LEDs. To limit the current during this event a series resistor can be use with the capacitor. A resistor of perhaps 1/4 the reactance of the capacitor would work.

To make this circuit reliable with commercial power it must be designed for the expected line-to-line spike voltage. I don't remember what this is for 240 VAC but for 120 VAC it is 1 kV. So I would design the circuit to withstand 2 kV spikes. The diode voltage is limited by the LEDs but the capacitor voltage is not. It must be rated for 2 kV. And the series resistor must be rated for a peak of 2 kV. The series capacitor should have a bleeder resistor to discharge it when power is removed. And there is the question of can the LEDs survive the spike current? With the resistor-only circuit the current is limited to some value but with the capacitor and 1/4 value resistor circuit the current is four times higher during the spike.

Using a capacitor introduces several new issues but does reduce the power dissipated. It's all an engineering tradeoff.

Here's a capacitive reactance calculator (and formula):

http://www.electronics2000.co.uk/calc/reactance-calculator.php

Does that mean you have 30 LEDs in series with 240V (which is more like 240 * 1.41 = 338V peak (less the forward drops of the rectifiers), If so, for LEDs with a forward drop of 3.4V (assuming standard 20ma white LEDs [you didn't say], that's 338V - 1.4V - 30 * 3.4 = 234.6V that must be dropped by that series resistor. No wonder it's getting hot!

I.e. a series string of standard white 20-30ma LEDs is going to drop, typically, only 96V to 105V, depending on the forward drop of the LEDs (which can vary a bit from part to part, and over time).

The 30 LED are connected in serie, they are White, 20 mA.
But if you have a better idea, I will appreciate.
Thanks in advance

Thanks so much

Thanks Steve.
Regards

Thanks Steve, It is very helpful
Regards

Elba

With my experiments i have found that a capecitor of value 1mfd can provide a forward current of 60 mA. Select the capecitor value accordingly.
The capacitor should be rated to atleast 630 V AC and it will be good to provide a bleeder resistance accross that.
this combination should be put infront i.e. before the bridge rectifier.

You mentioned that there will be more than one string of 30 LEDs. There is enough [Voltage] "head room" to drive two of those strings in series (and almost three, but there needs to be enough voltage left over to provide a "current regulation" effect and you should follow the manufacturers spec which usually sets the MAX LED forward voltage at 4V (though that is usually when driven at 30ma so you may be able to get away with 3.6V max but check the spec [which doesn’t always give such details].

Another possibility is a little two pin IC called the "CL2" :
http://www.mouser.com/Search/Refine.aspx?Keyword=cl2n3 [Digi-key doesn't have it]
http://www.supertex.com/pdf/datasheets/CL2.pdf

The CL2 is a constant current source/sink [doesn't matter where it is put in the series string]. This is the BEST way to drive LEDs [with a constant current]. But you will need to make sure that the voltage across the CL2 never exceeds 90V -- so there needs to be enough LEDs in series with it to drop the voltage down to that max (i.e. not even 60 will be enough, so add one or more zeners to drop the rest (use more than one to distribute the power)).

AND MAKE SURE THESE LEDS ARE ISOLATED/INSULATED SO THERE IS NO SHOCK HAZARD!!!

Above I provided a link to the Mouser catalog page for CL2N3. A better choice is the CL2K4. It has a TO-252 (D-PAK) package and can take 2W (the CL2N3 is a TO92 and can handle only 600mW -- not enough for your application, unless you take up the slack with zeners).

http://www.mouser.com/ProductDetail/Supertex/...=Wgksyu2ekN8hKHaZfvdGQw4JQgWWFy1KikRrMe7IubE=

But, they only have two left (with 3Week lead time) so:

http://www.alibaba.com/trade/search?SearchText=cl2k4-g&IndexArea=product_en&fsb=y

Happy hunting

Also, try calling SuperTex (the manufacturer). See if they can locate a distributor for you. Their contact info is on this document:

http://www.supertex.com/pdf/misc/CL2PSS.pdf

Hi Steve, first, thanks a lot.
I never heard about that CL2, and really looks great.
But I mention, I will connect them to 240V AC, Do you have any suggestion how drop the remain voltage?

And again thanks

Regards

Actually, just use a resistor (I suggested one or more Zeners, but because of the constant current, a resistor will behave like a constant voltage device, once the CL2 begins to regulate).

Use the formula: R = (V-5)/20ma where V is the excess voltage you want to drop. Its V-5 because the CL2 needs a minimum of 5 volts to operate properly (i.e. it current regulates over a voltage range of 5 to 90 volts).

If that will result in a higher wattage resistor than you desire, then also include a series capacitor on the AC side (as described in other replies). Divide the resistance R between the resistor and the reactance of the capacitor.

Thanks. It sound great
I will let you know the result
Regards

Elba

Thanks David, it will help me
Regards

I´m taking note

Thanls