I've trying make a very simple 3 LED circuit with a red, blue and green LED.

I know my 3v battery will take the path of least resistance and only flow through the red, so I've added resisters to the green and red at 47 and 56ohms respectively, as instructed by this website:


I also added a 47 to the blue just to try and help it, because you see the blue is still barely lighting.

I don't know what I can do beyond trying to balance their resistances which I thought I had now done properly.

Thanks in advance for the help.enter image description here

  • \$\begingroup\$ Huh, I just assumed if their resistance was brought down they could all run fine. I've got a model being lit by 20 yellow SMD's in a 3v battery and it works great. Aside from the two greens not running at full power at the same time. But I assumed one of each colour could run as well as 10 yellows. \$\endgroup\$ Oct 27, 2018 at 0:10
  • \$\begingroup\$ Post a link to the data sheet for your LEDs, and add your schematic if you have one. \$\endgroup\$
    – Tyler
    Oct 27, 2018 at 0:13
  • 1
    \$\begingroup\$ White, blue, and green are usually 3V+. Red, orange, yellow, and some green-yelow are usually 2V+. \$\endgroup\$ Oct 27, 2018 at 0:15
  • 4
    \$\begingroup\$ Your statement "my 3v battery will take the path of least resistance" is incorrect. Current will take all possible paths, not just the path of least resistance. By Ohm's Law, the path having the least resistance will have the most current, but all other possible paths will carry current, in inverse proportion to their resistance (current in each path will be determined by the resistance of that path.) \$\endgroup\$ Oct 27, 2018 at 0:16
  • 3
    \$\begingroup\$ @MannyMcArthur: I just wanted to correct you on the "path of least resistance" comment - I often see that claim, but it is definitely incorrect and is misleading for beginners. \$\endgroup\$ Oct 27, 2018 at 0:47

3 Answers 3


A 3V battery does not have enough voltage for many blue or green LEDs.

You need more voltage. The battery voltage must be greater than the forward voltage of both the blue and green. Each LED should (must) have its own resistor

Keep in mind the actual voltage of the battery goes down with use.

This is the discharge curve of an alkaline battery.

enter image description here

A 3V CR2032 is not really 3V.

enter image description here

I never noticed the 8-15v info before. That's seems very high for tiny SMDs.

That's why you need a resistor, to drop the excess voltage.
With a 3V LED and a 8V supply, the resistor drops 5V.


The forward voltage of an LED depends on the composition of the LED, and on the colour of light that it emits. Red LEDs are about 1.8 volts, yellow about 2, and older green LED about 2.2 volts. Some newer green LEDs, and blue LEDs are up around 3 volts.

If you have LEDs of different colours, each colour (or really each LED) should have its own current-limiting resistor, selected to suit its forward voltage and desired current.

  • \$\begingroup\$ Not "older green" it's yellow-green. Lower wavelengths from violet to green are usually 3V gallium nitrides. From yellow to red are 2V gallium phosphides. That in between area (green-yellow) can go either way. \$\endgroup\$ Oct 27, 2018 at 0:29

Ok everyone, thanks for all the help!

I've definitely been wiring resisters in correctly, and balancing it between the different coloured LEDs.

The issue was my batteries not being a high enough voltage. I've just done a test with two 3v batteries and it's working great now. I'll need to find a way to accomodate the extra battery and adjust the resisters for 6v. (it's a very small build) .

I've got a lot to learn still but I think I'm doing alright for a beginner.


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