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What is the best way to light 10 POWER-LEDs with AC mains (220V)?
(I want to make an LED lamp.)
Here is more info:
I have 10 of these POWER-LEDs:
And I have soldered all of them in series (with heat sink).
If C1 was directly connected across L and N and takes a current of 80mA when the input voltage is 220V AC, then that is the limit of this circuit. Using a "transformer" is the only way you can increase this current.
By "transformer" I mean a conventional magnetic coupled transformer or a switching regulator that may or may not use a transformer.
If you are intent on pursuing this approach you'd need to have a capacitor value of at least 10uF - at 50Hz it's impedance is 318 ohms and with 220V across it there will be a current of about 0.7A.
You could use a directly coupled power supply if the unit is completely sealed. Any part of a directly coupled supply and the circuit using its power could be floating at line voltages, so this is not something you want to do for general purpose. As long as everything is sealed in the same unit that just has a line cord coming out, then these types of supplies can be appropriate.
A very simple circuit for driving two LEDs from 240 Vac 50 Hz line power is:
The capacitor will allow about 7.6 mA RMS to flow thru the two LEDs. Each LED protects the other from backwards voltage, and they light on opposite polarity half-cycles of the power line. Not only does the cap need to be rated for the indicated voltage, but it must also be rated for power line use.
This circuit is very quick and dirty in that it doesn't protect the LEDs from power line spikes. The LEDs will limit the voltage, so a power line spike will cause a burst of current thru the LEDs. If that happens too often, it will eventually degrade their lifetime. However, these LEDs are rated for 30 mA continuous and this circuit runs them at 8 mA continuous during normal operation. That will still be plenty bright at night. There is a lot of current headroom, and a occasional higher current spike of short duration really won't hurt them much. LEDs are also cheap and available, and the simplicity of this circuit makes it easy to just try it.
Again, everything needs to be sealed so that it is not possible to touch any conductive parts during normal operation.
The main advantages of such a direct coupled power supply is that it is simple and very efficient. A ideal capacitor doesn't dissipate any power. Just about all the power drawn from the line is used to run the LEDs.
One way to calculate the current in this circuit is by dividing the voltage accross the capacitor by its impedance magnitude. The impedance magnitude is:
R = 1 / (2 π F C)
When F is in units of Hz, C in Farads, then R is in Ohms. In this case the capacitor impedance magnitude, assuming 50 Hz, is 32 kΩ. Figure the LEDs drop about 2 V, so 238 V is put accross the capacitor. 238 V / 32 kΩ = 7.4 mA.
It should be obvious how to work this process backwards to find the capacitance that causes a particular current.
Best Answer
Ten in series will require 32V-34VDC at 300-350mA, or worst case just about 12W. You'll need a stepdown transformer from 240VAC to 36VAC at 350mA, and you'll need a half-wave rectifier (do NOT use a full-wave rectifier unless you use a 24V transformer!) good for 50V @ 350mA, and you'll need one stabilizing power resistor capable of dropping about 2V at 350mA (about 5.7 ohms), rated for 12W.
I mentioned a full-wave rectifier. You COULD use a full bridge rectifier with a 24VAC transformer secondary - that'd get you 24*1.414!=34V, but then you wouldn't have any room for the stabilizing power resistor so you'd be running on blind faith unless you removed ONE of those LEDs from the circuit and replaced it with an equivalent power resistor.
Do remember... this whole assembly will generate a significant amount of heat, and you have to get rid of that heat somehow or you'll overheat the LEDs and the resistor.