I have some (white) LED pucks that each run on (3) AAA batteries in series, or 4.5v. These are the kind of pucks that you would put underneath a kitchen cabinet, push on/off. Each puck has (3) white surface mount LEDs wired in parallel. I don't have data sheets available, and so I don't know the forward voltage drop for the LEDs, and also don't know how the equation changes when there are (3) LEDs in parallel. So the best I can do is measure the battery current thru the puck when using the (3) AAA batteries. The measured current is 62ma. I have a 12vdc power supply in the project (which powers other 12v things), which I am guessing should first be run through an LM regulator to get the voltage closer. I did do a test using 5v directly from a USB cable (no current limiting resistor), and I fried one of the LEDs. So, I know that I need minimally a current limiting resistor. Can someone help me calculate the value of that resistor?

  • \$\begingroup\$ A battery is a DC power source. \$\endgroup\$
    – Hearth
    Apr 9 at 3:59
  • 1
    \$\begingroup\$ Chances are battery voltage with almost fresh 3S AAA is about 4.2 V@62 mA (measure!), and \$V_f\$ is around 3.2…3.7 V (again). I'd start with an additional 22 Ω for 5 V or 150 Ω for 12 V and proceed as suggested by sai. With three pucks lighted in lockstep, try putting them in series for 12 V. (Consider explicating in your question: What is an LM regulator?) \$\endgroup\$
    – greybeard
    Apr 9 at 7:15
  • \$\begingroup\$ Thanks, I will try the solution starting with 22 Ω. The (3) LEDs in parallel are in a single puck (surface mounted). Then there are 4 pucks, also in parallel, for which I would need 18v if in series. \$\endgroup\$
    – svenyonson
    Apr 9 at 13:56
  • \$\begingroup\$ @greybeard - Sorry, don't know where the LM came from, somehow I thought there was an M in the prefix - but I meant a 5V regulator something like the L7805 \$\endgroup\$
    – svenyonson
    Apr 9 at 14:14

1 Answer 1


Resistor value = (Vsource-Vf)/ILED. Since you do not have Vf, you can start out with resistor value = Vsource/ILED and then reduce the resistor value in small steps till ILED is equal to what you want.

  • \$\begingroup\$ Thanks guys, works perfect. 22 Ω is a little high on the current side (80ma for 3 LEDs in parallel), but 27 Ω was too much, just a glowing coal on the LEDs. I'm guessing white LEDs are more tolerant, usually higher forward voltage? This application will only light the LEDs for about 3 seconds, then a delay which should allow a little cool down. And only occasional use at that (this is for backlighting for a DIY large format scanner) \$\endgroup\$
    – svenyonson
    Apr 10 at 3:53

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