At 10:47 AM 6/29/2008, you wrote:

>Let's say I have an LED with a Vf of 1.8 V and I want to put about 40 mA
>through it. The cathode goes straight to ground and the anode goes to a
>PIC pin.
>
>I can go from 5 V on the PIC pin, through a 80 ohm resistor into the
>LED, but of course the problem is that I'll be drawing more than the
>maximum current for a PIC pin.

Of course there are other problems.

>But I'm wondering if I could use two PIC pins together. Let's say I
>connect RD0 and RD1 together and then do the following in C code:
>
>PORTD |= 3u;
>TRISD &= ~3u;
>
>The idea would be that 20 mA would come from each pin.

PIC outputs are not designed to reliably source even 20mA with low voltage
drop. *You* still must prevent them from seeing more than 25mA, but that's no
indication that they'll work in your application.

>This a go-er?

Nope. See Figure 18-29 of DS41291E. Max/min is not guaranteed,
only 3-sigma distribution, so it's actually even worse than that. At least
this data sheet is complete and *has* the curves (not all do), so it's left
up to you to read the datasheet and understand it.

I have to disagree somewhat with Olin here- the outputs will typically share
current  quite well (they are well matched -- Idss and Vth-- because 
they're on the
same  chip), and 20% derating for matching is probably enough for 
sharing (especially
with such a huge voltage drop), but there are other problems (such as the huge
voltage drop).

Was it not the Bard himself who said:

If current be the food of light, source on
give me excess (capability) of it, that buffering..
that the LED brightness may not sicken and so die

?

>Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com



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