Maximum Common Cathode RGB LEDs Connected in Series to Arduino I/O Pins?

Hey all!

I have been thinking about making a homemade strand of RGB LED lights, but I’m kind of a noob when it comes to LEDs.
Can someone please explain what the maximum number of common cathode LED’s I can connect to the Arduino in series is? Also why is the number limited to begin with?

I’ve noticed that other people on this forum trying to create long strands of LEDs always use an external power supply and most seem to agree that connecting too may LEDs in parallel is a bad idea. I would understand this if LEDs were like light-bulbs and had a resistance. With each additional bulb the equivalent resistance of the series decreases and thus the current drawn increases. LEDs are diodes though, and have no resistance. It would seem I could just put a resistor between the cathode line of the LEDs and the ground pin of the Arduino, then the current through the series would be a function of only that one resistor regardless of how many LEDs were added. I tried this configuration with 1 through 6 RGB LEDs (200 Ohm resistor). The drawn current gradually increased with each LED, but the increase was very small.

I’m not saying this explanation is correct. I know this is wrong somehow (and it’s probably really obvious). I just want to have a better understanding.

Thanks,

Data Sheet for the LEDs: http://www.scribd.com/doc/153463372/YSL-R596CR3G4B5C-C10-pdf

Forward Voltage: Green,Blue 3.2V
Red: 2V
Forward Current: Red,Green,Blue: 20mA

Screen Shot 2014-01-01 at 6.46.09 AM.png

US3R31:
LEDs are diodes though, and have no resistance.

Yes they do.

US3R31:
Forward Voltage: Green,Blue 3.2V
Red: 2V
Forward Current: Red,Green,Blue: 20mA

If they have voltage and current then you can calculate the resistance using Ohm’s law.

Warning: Diodes don’t have a constant resistance, it varies with the voltage/current applied to them.

Thanks,

Just to be clear: Let's say I have one RGB LED. The 3.2V, 2V forward voltages will cause a 3.2V,2V drop across the diode over a large value of currents. The forward current (20mA) is the minimum current required to output visible light. I don't want to supply more current than what is necessary so by Ohms law V=IR => 3.2V=(20/1000)R => R=160 Ohms.

Are you saying that the LED will supply this 160 Ohms by itself? I always though of diodes as virtual short circuits and would have added the 160 Ohm resistor in series to get the required current.

Also, I said the LEDs were in series when I actually had them in parallel. The diagram in my original post is correct and also shows them in parallel. Sorry for the confusion.

Thanks again for the help,

US3R31:
Just to be clear: Let's say I have one RGB LED. The 3.2V, 2V forward voltages will cause a 3.2V,2V drop across the diode over a large value of currents. The forward current (20mA) is the minimum current required to output visible light. I don't want to supply more current than what is necessary so by Ohms law V=IR => 3.2V=(20/1000)R => R=160 Ohms.

Correct.

US3R31:
Are you saying that the LED will supply this 160 Ohms by itself?

Yes.

It's the LAW.

US3R31:
I always though of diodes as virtual short circuits and would have added the 160 Ohm resistor in series to get the required current.

As I mentioned earlier, they vary their resistance with applied voltage. Their resistance can drop away to almost nothing but it's always there.

A LED close to its forward voltage has quite a lot of resistance, as you've calculated.