> >OK, so the book suggests that I need to give each of the LEDs
> > 10mA to light it up properly and given that I am cycling
> > through each display in turn, each one is only going to get a
> > quarter of the time and therefore I should quadruple that
> > amount to 40mA for the short burst. Therefore the book
> > calculates an 82R resistor to be used on each of the LEDs. Now,
> > my question is that whilst none of the LEDs will get a
> > sustained 40mA current, surely if I'm cycling through all the
> > displays and they all have the same number on them, then some
> > of the pins on the PIC will be having 40mA drawn through them?
> > (cycling through the four displays, lighting up the same
> > segment, drawing 40mA each time and thus sustained pretty much)
> > And isn't that too much for the PIC?
>
> Yes, you're quite correct. Hobby books often contain circuits
> that should not see the light of day. Check out the limit per
> port as well. Either way, you're limited to 25mA.  Note that
> these

OK, thanks, good to know that I wasn't going completely crazy :-)
Also, I hadn't realised that there are port limits, I shall check
into those, thanks for pointing it out.

> The answer is either to use a lower average current (perhaps
> upgrading the display to something that give more mcd/ma) or add
> some drivers. In the case of the CC displays, you could add an
> octal source driver to drive the segments. Source drivers are
> generally more expensive and drop more voltage than sink drivers,
> however.

I've never come across these drivers before (like I said, I'm fairly
new to electronics really) so I'll look into them. I just did a
quick google but didn't find much - if anyone has any links to
simple example circuits using them I would be grateful.

> If you need super brightness, then a cheap ULN2803 will switch
> the segments and you can go up to much higher currents per
> segment (maybe 20-25mA average- daylight visible with the right
> filter).

So is the ULN2803 one of the things that you refer to as drivers
above? Or is that something different?

Thanks for your help, I appreciate it.

Matt

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Matt N. Marsh
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