Connecting bi-color LEDs in series?

I’ve usually used single color LEDs and connected them in series for a 12V input. I’ve got some bi-color ones (red/yellow - three legs, common anode, both channels are 2.0V) is there a way to connect them in series??

I need to connect a few dozen of them so having a single LED to a 12V will draw too much power. Anything goes, helper IC’s, diodes, clever wiring etc.


The three wire (bi-color) LEDs are just like two separate LEDs that share a common ground. So, you can wire them the same as you do your single color LEDs except treat one wire as the color for one LED and one wire for the other color. There are also two-wire bi-color LEDs that use the diode behavior of the LED and expect you to reverse the current to change colors. That’s a little bit trickier to handle. But, you said yours have three legs so its simple.

I keep this little wizard in my pocket for calculating resistors values for LEDs. It will also show you the wiring for (single LED) series.

If that doesn’t make sense, let me know and I’ll throw together a sketch.

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@ ianlee74 - Very nice link to the LED wizard. Thanks for sharing.

I know about the LED wizard, use it all the time, although I think the bi-color is a bit more complicated. I need both colors to work (not at the same time), not just ignore one color if that is what you meant?

Could you provide a sketch?

It’s basically this. Insert whatever switching mechanism you want between the GND and the resistor. I’m assuming you want all the LEDs for a specific color on at once. Of course, you will need to make sure total current & V are within range. For 12V, you won’t be able to string more than 6 of the 2V LEDs together in series. Maybe even 5 depending on the brightness you see. Does this help?

That’s a parallel connection, essentially like this:

it draws 2A of current, which is way to much

my question was is there was a clever way to connect them in series???

this draws only 340mA

new LEDs? :clap:

what do you mean? yes they are new :think:

@ Darko - You have to choose one of the color components and loose the other or tie cathodes together on each LED.

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Sorry, I totally ignored that part… I’m not sure that’s possible with the 3-wire LEDs. You could certainly do it with the 2-wire bi-color LEDs but you’d have to fast switch them if you wanted to be able to light both colors at once.

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buy new LEDs. What you want to do isn’t really possible “simply”.

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ok, I was thinking of putting some mux or something on each shared wire but the power consumption gains that would potentially come out of it and also fewer led diode gains do not outweigh the classic way of simply doubling LED’s, the power consumptions is way higher in parallel when your talking about 400-600 LED’s.

I’m no EE…but my understanding of the math is that a 2V LED consumes roughly 20mA of current for a dissipated power of 40mW (P = VI). So, it doesn’t matter if they’re in series or parallel. Your total POWER consumed is based entirely on the number of LEDs. The only difference is do you want to supply more VOLTAGE or do you want to supply more CURRENT. It’s probably going to be a lot easier to plan this based on what your power supply can provide instead of worry about this too much. You’ll almost certainly end up with a combination of parallel & series.

Yes, but If I cant wire bi-color LEDs In series then I get a massive current draw. I have 52 yellow, 69 red, and 174 yellow/red, in parallel the 174 draw about 3A

If I just use 226 yellow (52+174) then it draws 920mA and 243 red (69+174) then it draws 980mA so using bi-color I gain only on space and fewer LED’s but not current, so this is scrapped for now.

@ Darko - Multiplexing is the usual answer to this problem. Have you considered it? If you multiplex into groups of LEDs that consume a smaller amount of current and then cycle through the groups very quickly then the eye can’t tell that you aren’t lighting all the LEDs at once. Also, consider the two-wire bi-color LEDs. LEDS are cheap. It would not be a big deal to just replace the ones you have.