On Thu, 6 Jan 2000, Jeff King wrote:

> OK, I got some food  in my stomach so I can think a little
> more clearly now.
>
> You'll only need 4 pins to give you 12 LED's with this
> method. You can light any LED at any time OR if you
> need all on, you'll need to multiplex them.
>
> The pins that are not actively driving the LED in question
> (only two pins are active at any one time) need to be set
> to tri-state (inputs).
>
> Let me try and draw a picture:
>
> First,    each LED is back to back with the other, such as:
>
> ->|-     and
> -|<-   meaning each anode is hooked to the cathode of the other
> in each two LED group. I will represent this as a -**- in my drawing,
> where each "**" represents two LED's back to back.
>
> O is a PIC PIN
>
> So here we go:
>
>
> |------------**------------|
> |                          |
> |        |-------**--------|
> |        |                 |
> |-------**--------|        |
> |        |        |        |
> |---**---|---**---|---**---|
> |        |        |        |
> O        O        O        O
>
> 12 Leds from 4 pins.
>
> I left out the currently limiting resistors, but you'll
> need to add them directly at the PIC pins... don't
> forget the current will be going through two resistors,
> so make sure each is half of the final value.

You know, I've been doing a lot of programming lately - too much in fact.
But the last time I played with leds you couldn't get current to flow
through a 'back-to-back' configuration. Maybe I should just stick to
programming. :).

Scott