Homemade LED Bulb

WARNING! PRESENTED CIRCUIT IS CONNECTED TO DANGEROUS MAINS VOLTAGE WITHOUT GALVANIC ISOLATION. DO NOT TOUCH THE CIRCUIT WHILE IN OPERATION. DO EVERYTHING AT YOUR OWN RISK, AUTHOR DOES NOT TAKE ANY RESPONSIBILITY.

This is a simple LED bulb which uses well known capacitive dropper to power LED's directly from mains voltage. I've used orange 3mm LED's in E14 base lamp to make nice decorative lamp which consumes less than 2W and it gives nice orange hue.

1x X2 or suitable rated unpolarized capacitor (I've used 470nF X2)

1x 1MΩ resistor

1x fusible resistor (fuse and inrush resistor at same time I've used 10Ω from another LED bulb)

1x bridge rectifer at least 500V rated (or 4 discrete diodes)

1x 2.2 - 4.7u electrolytic capacitor 400V

1x 100kΩ resistor (to discharge electrolytic capacitor)

1x 100 - 680Ω resistor (to limit LED current)

3 or 5mm LED's

Firstly, I have opened a cheap E14 supermarket LED lamp and take its guts. I didn't took fusible resistor out and simply reused it. Then I soldered and glued 53 3mm orange LED's to the LED cap to prevent electric shock when touch the bulb and to spread the light from the LED's.

The power supply is well known capacitive dropper which uses capacitive reactance to limit LED current. Unlike a resistor this method does not dissipate the power in form of heat. But it has a poor power factor which stress wiring if excessive currrents are drawn. But this circuit draws 15-16mA so there is nothing to worry about. Using capacitive reactance formula and some simple math one can calculate current draw of this circuit easily or can use online calculators. You can alter the values of my design using your countrys mains voltage and frequency. If done properly and LED's are not driven at their maximum or beyond this circuit can power LED's for many years.