Trying to split the current output of an LED driver

Question

I am currently using an 8 channel LED driver wher I have set the outputs to produce 20mA of constant current. Problem is I have 11 LEDs to drive. So I have decided to connect two LEDs in paralelled to each output, and then finally on the 6th output connect a single LED. I have 100Ohm resistors in series with each of those parallel resistors, and my LEDs can handle up to 30mA of current. However I want all 11 LEDs to have the same brightness, more or less. How do I get that 11th LED that is alone to split its current in half? Can I put a resistor in parallel with that LED? if so how do I calculate the resistor value to get it as close to 50% as possible, or should I use dummy LEDs to put in parallel?

Answer 1

The LED driver regulates current, so it adjusts the voltage until the current is 20mA.

If you put two LEDs in series, the LED driver will still regulate to 20mA, but the voltage required for that will be the sum of the LED forward voltages.

The downside is that you need a high enough supply voltage, and any channel that has only a single LED will need to drop the voltage in the driver. If it's a buck converter, fine, but a linear regulator will heat up quite a bit.

Answer 2

If you have enough voltage headroom, you should put the LEDs in series, rather than parallel. This way each LED is getting 20mA.

Parallel LEDs, and a resistor in parallel is not a good solution. LEDs are not resistive devices, they behave like diodes where they have a certain voltage drop that doesn't change very much with current.

When you have your LEDs in parallel, tiny differences in their forward voltage drop will cause them to share the current unevenly. This will make their brightness not very uniform. The 100Ω resistors will help with this, though.

Remember, when devices are in series, they have the same current through them. When devices are parallel, the current is different through each branch.