Jinx wrote:
> You can eliminate half of the resistors. Each pair of LEDs are
> commoned at their cathodes, so you put the resistor there (as
> Olin mentioned the other day), rather than one per anode

I'll have 5 times as many resistors if I put them on the cathodes, 
reason being that I'll need to put them on the cathode for every column.

Think about it:

Quantity LED's if they go on anodes =  quantity LED's per row = 14 resistors

Quantity LED's if the go on cathode = (quantity LED's per row / 2) * 
quantity columns = 14 / 2 * 9 = 7 * 9 = 63

Unless I'm missing something.. ?


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