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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?

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • \$\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
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    \$\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
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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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