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I'm making a LED beatbox (seen on Kipkay.com). I'm using 8 LEDs (5 millimeters, I don't know if this helps) and my power source is 12 VDC 800 mA. Do I need to add a resistor? If yes, which one should I use?

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    \$\begingroup\$ Yes, you will need resistors. See this earlier thread. To determine the value of the resistor, we need to know 2 things: voltage drop across the LED (forward voltage drop), current through the LED. Post this info, please. This info usually comes from the LED's datasheet. Also, are you going to light up all the LEDs at once, or one at a time, or in groups? Are all of the 8x LEDs the same (color, model)? \$\endgroup\$
    – Nick Alexeev
    Aug 1, 2013 at 3:26
  • \$\begingroup\$ Related: electronics.stackexchange.com/questions/56910/… \$\endgroup\$
    – Anonymous Penguin
    Aug 2, 2013 at 20:44

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Yes you need a resister per LED. Assuming you want your LEDs to run at a current of about \$I_f\$ = 20mA current and your LEDs have a forward voltage of \$V_f\$ = 1.7V (this is a wild guess, check your datasheet), you'll want to size your resistors as follows:

\$ (12 - V_f) / (I_f) = R\$

So for the parameters I suggested that would be (12 - 1.7) / (0.020) = 515 ohms, and you would just pick a "common" (e.g. E12 series) resistor that is larger than that, so 560 ohms is your best bet. I gave you the equation form so you can plug in appropriate values from your datasheet, and obviously the 12 in the equation is your supply voltage. Keep in mind that the power your resistor will need to dissipate is:

\$P = I_f * I_f * R\$

So in this case 0.02 * 0.02 * 560 = 0.224 Watts, so definitely get 1/4W resistors, not 1/8W resistors.

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    \$\begingroup\$ Also why one led for each, when four 1.7v leds in series with a single resistor would work fine on 12v? \$\endgroup\$
    – Passerby
    Aug 1, 2013 at 8:41
  • \$\begingroup\$ @Passerby didn't realize they were in series, thought it was 4 individual LEDs. Wasn't obvious from original post, subsequent detail seems to have been added to clarify that point \$\endgroup\$
    – vicatcu
    Aug 1, 2013 at 14:05

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