1. May 23, 2012

### John1397

I built an array of 14 LED's the 20,000 mcd's ones maybe drawing 25 ma at 3.3 volts with each two LED's powered from 125 volts AC then to a .22 mfd capacitor in series feeding a full wave bridge rectifier so there are 7 capacitors and seven full wave bridges feeding all 14 LED's and they draw 6 watts power on AC and if you figure just the power consumption of 14 LED's at 25 ma on 3.3 volts they should only use a little over one watt is this an acceptable amount of current loss?

John

2. May 26, 2012

### Staff: Mentor

So this is a cheap and nasty method to power some LEDs for lighting, is it? Did you get the circuit from a publication, or is it your own plan?

Everything needs to be treated as though it is live at mains voltage, as indeed it only too easily can be. All wiring insulation and the LED mountings must be suitable for providing safety as though at 125VAC.

https://www.physicsforums.com/images/icons/icon2.gif [Broken] How did you measure the 6 watts?

Last edited by a moderator: May 6, 2017
3. May 26, 2012

### John1397

I seen this searching for ways to power LED's from AC it seemed easy has no transformer which wastes current to, but I see LED lamps that have 42 LED's and only draw 3 watts. I plan to switch them all to run off 12 volts DC so this will eliminate all the capacitors and diodes then I will be able to see how much current they draw.

4. May 26, 2012

### Staff: Mentor

Using DC will make it simpler to measure the power, but won't much change the current the LEDs need for a particular intensity of output. You didn't answer:
It may well be that your measurement technique is wrong.

5. May 26, 2012

### John1397

Used a digital power meter even plugged in a 7 watt Christmas bulb and showed 7 watts so I know it is right seems most of the leakage is is on all these diode bridges.

John