I want to make an AC powered 5mm LED string. I searched a lot but I never found a working one.

I just tried one experiment from the Instructables site.

But when I connected it to AC power, one resister got blown. Anyone please tell me how can I do this easily?


closed as unclear what you're asking by PeterJ, Daniel Grillo, Matt Young, Michael Karas, placeholder Jun 2 '14 at 15:04

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  • \$\begingroup\$ Sounds like you didn't do your math. \$\endgroup\$ – Ignacio Vazquez-Abrams May 24 '14 at 11:26
  • \$\begingroup\$ Which resistor and precisely what part number did you fit? \$\endgroup\$ – Andy aka May 24 '14 at 11:29
  • 1
    \$\begingroup\$ Please don't vote to close this question. A good answer should provide a warning to inexperienced people who want to play with mains electricity. \$\endgroup\$ – Rocketmagnet May 24 '14 at 16:06
  • \$\begingroup\$ Those are the components which I added to the bcb bord... \$\endgroup\$ – RAFEEK CJ May 25 '14 at 0:14
  • \$\begingroup\$ Those are the components which I added to the bcb bord. 1N4007 diode ,1mega ohm 1/4 walt resister,220 ohm 1 walt resister, capacitor 0.47 mf 450v, capacitor 100 mf 50v \$\endgroup\$ – RAFEEK CJ May 25 '14 at 0:38

All parts of this circuit should be regarded as being at mains voltage at all times. It can kill anyone who touches it.

Note that in the comments someone says

  • Capacitor in parallel to R1= 0.47 Microfarad 450V

Note that it should be a 450V cap for 230 VAC mains.
And it should be an X rated cap (across mains use).
The capacitance will affect the current provided.

Current will VERY APPROXIMATELY be V x 2 x Pi x C
At 110VAC, 60 Hz
Ima ~~~= 40 x CuF So for 1 uF I ~= 40 mA.
I have ignored R2 as impedance of the capacitor dominates.

Diode polarity matters.


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