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The board I created uses the WS2811 which outputs a 12V RGB signal, with the high signifying an inactive LED and the low being full brightness. The components connected to this receive a 12V power signal in addition to these three 12-0V signals.

The RGB LEDs I bought have the exact opposite. Their high signal makes the LED active and have a common ground instead of a 12V power. Therefore the current essentially goes the opposite direction.

I.e. for a white RGB LED with 25% brightness:

Spec    | Common (V) | R (V) | G (V) | B (V)
--------------------------------------------
WS2811  | 12         | 9     | 9     | 9
RGB LED | 0          | 3     | 3     | 3

So essentially I want to subtract every signal from 12V. I feel there should be a simple soltuion for this problem, but my knowledge doesn't seem to reach as far.

Thanks for any help!

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    \$\begingroup\$ You are basically out of luck. The WS2811 is a low side driver requiring either common anode LEDs or discrete ones that can be wired that way. Putting in external transistors or FETs would defeat the compactness of this - you need to change either your LEDs or driver chips to match the other. \$\endgroup\$ – Chris Stratton Jul 15 '18 at 17:28
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Unfortunately your out of luck. As Chris said you either need to change your RGB LEDs to compatible Common Anode or change your driver. Alternatively you can potentially save yourself some money and use transistors, FETs or some other type of signal converter.

This instructable shows a simple common anode to common cathode converter that would likely solve your problem. There are also discrete transistor ICs like this that might be less obtrusive.

If you go down this road you would have to drive your LEDs like you said:

Spec              | Common (V) | R (V) | G (V) | B (V
---------------------------------------------------------
Normal            | 12         | 9     | 9     | 9
Using Transistors | 0          | 3     | 3     | 3

However, The WS chip might have trouble sinking high currents, as it is only designed to source current.

Ultimately the most logical and practical solution is to change LEDs or change driver.

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