...and to add to this thread of information overload, I present my own version of a "datasheet":
I did what CrossRoads suggested a while ago. Also with the aid of some GIMP magic, I patched my photographed A6 page so it made a little more sense, hence the somewhat pathced-up look (Yes, my (not so) wonderful "all-in-one" printer-scanner will not scan any longer, for the lack of... tada... INK! $) - btw its heading for the parts bin ] Ok no more ranting about that printer).
Now I wont guarantee this is the same as the one you got. This one I got from Earthshine Design a while back, they are two-color Red Green with a common anode (so two cathodes pr. LED / dot). Also they are 32mm by 32mm.
Btw the "side with text" (in the drawing) is where I begun counting pin nr. 1, going around the device in the same fasion as is usual with IC's. As far as I know this was the only way to find pin nr. 1 (except testing it of course).
In addition I made a kicad parts library for those a while back as well. However, among other things due to the same ink shortage and my etching tank that got a hole in it (wrong ethcant, or wrong tank material probably), I haven't actually tested these yet (so no guarantee on the correctness etc, as usual).
so if i got this right, then the color of the LED (red or green) is controlled by the amount of Voltage that runs into the pins?
Not quite.. Maybe I'm nitpicking, but the brightness of a LED is controlled by the current that flows through it. Which again is controlled by a resistor in series with the LED. To calculate this resistors size, one must know how much voltage is over this resistor, and how much current will pass through it (and hence the LED, since they are in series connection).
R = U/I (or, resistance = voltage divided by current).
To get how much voltage is left over the resistor, one must know how much voltage is over the LED. If your current is not too far off from the "forward voltage" drop for the LED found in the datasheet, you can use that. Or look in the datasheet. Say Vf=2V. Then, if the total voltage to the LED-resistor series is 5V there will be 3V left over the resistor. if using 20mA, resistance should be 3/0.02 = 150 ohms.
The thing about PWM still holds, basically. That you can adjust the apparent brightness smoothly through the trick known as PWM:
However this gets a little more involved with a LED matrix.