Is this what I should be aiming for? I've currently designed a 7x6 (dual colour red/green) using 4 595's (which is working fine), but I just had a thought about joining the red and green cathodes, which should let me get by with 3, and far fewer traces...
Assuming it is, where should I put the resistors? On each led? (84 total) or can I get by with one per row or column. entire rows will be lit at the same time, but only one colour at a time would be on at any given point in the matrix.
I also know I will need transistors, but there's just way too many out there for my experience level to select the proper ones I saw a schematic that used a uln2803 (saves space, simplifies wiring). I am thinking I would need two, one per colour, but I am unsure to say the least...
On a related note, I have the datasheet (below) for the red leds, but I cannot for the life of me find the datasheet for the green, is there an average or 'safe' spec I can assume?
That is an odd way of drawing a matrix but I can just about see it now. The problem is that unless you multiplex the red and green one after the other then there is the possibility that you could have one or two LEDs on sharing the same cathode. This is a problem because you want to put the resistors in the cathode and then pull them to ground. As you only have one LED on per cathode you can connect it directly to a pin. The anodes on the other hand can supply up to three LEDs at the same time so you need some sort of current source like a PNP transistor or P channel FET. This applies more so as you extend the size of the array requiring more than 60mA.
The normal way to light a matrix to set a row high with all the other lows low and then ground the columns corresponding to the LEDs you want to turn on. Then go on to the next row. If you do that fast enough then they all look like they are on at the same time.
In that way you can put the current limiting resistors in the columns as only one LED will be have a current path through a column at any one time.
I was pointing out that the red and green have to be treated as septate rows and multiplexed in turn like all the other rows.