Introduction: Simplest Capacitive Dropper Based LED Lamp

I made this really simple capacitive dropper circuit to make a LED light without much circuitry. One important thing to note is, that the output of capacitive droppers always can have the mains voltage potential to earth and therefore shouldn‘t be touched. I tested this circuit with a isolated power supply and if you also build it, you should only operate it on a isolated supply or in an enclosure. If one of the LEDs isn‘t properly connected or breaks, the other also will.

Build this circuit under your own risk.

Supplies

For 230V/50hz:

2 LEDs (or one LED and one diode, but the diode should be able to handle 0.5W at least)

A capacitor which isn‘t polarized and can handle 325VAC (should have anywhere between 0.1uF and 1uF which equals roughly to 5-50mA of current)

A resistor to discharge the capacitor (470k-1M, 0.25W are enough)

A resistor to limit the inrush current (470R to 2.2k, for <0.3uF a 0.25W resistor is enough, for anything above just go for a 1W resistor)

A fuse to prevent bad things from happening if the capacitor shorts out (250VAC, 0.2A or higher)


For 110V/60hz, the value of the capacitor can be anywhere between 0.1 and 2uF.

Step 1: Starting With the Soldering

Twist the legs of your capacitor and discharge resistor together, if you use SMD LEDs like I do, first check their polarity and solder the resistor legs onto the pads of the SMD LEDs. Do the same thing for the inrush resistor.


You are done. Test the circuit on a power inverter or isolated power supply first and add a fuse.