Looking for some help, I am trying to figure out the run time of my LED circuit from my two 18650 3500 mAh batteries wired in parallel. I have them running 5 wide angle 5 mm green LEDs wired in parallel with 10 ohm resistors.

When running the forward voltage is 3.5 V and the current according to my multimeter is .159 A. My working is 3.5 x .159 = .556 mAh. I used the website LED Calc and it said it should be around .370 mAh. Does this sound correct?

I am trying to get at least 12 hours of runtime from them without dimming much.

If I add a photocell which says it operates between 3 - 5 V will this consume much power and reduce runtime much?

Apologies for the rookie questions but I keep going over it and trying to work out the best method. I had tried using 12 V LED running a boost converter from the batteries with the LED wired in series. It seemed to consume even more.

  • 4
    \$\begingroup\$ If you are asking about a circuit, you should post it's schematic. \$\endgroup\$
    – Eugene Sh.
    Commented Feb 10, 2022 at 15:54
  • 6
    \$\begingroup\$ ”My working is 3.5 x .159 = .556 mAh” That’s not correct. You are multiplying current and voltage. The result is power, not capacity. \$\endgroup\$
    – winny
    Commented Feb 10, 2022 at 16:10

1 Answer 1


Back of the envelope: you are drawing about 160mA. A single 3500mAh battery will power this for \$T = \frac{3500\,\text{mAh}}{160\,\text{mA}} = 21.875\,\text{h}\$ . two batteries will run for twice as long if you manage to keep the current drain symmetrical between the two batteries.

Subtract 10%-15% fudge factor and you have an ok estimate.


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