Outpur pin maximum current

The spec says the output pin current is 40 mA (max).

I need to drive an automotive relay that is also spec'd at 40 mA.

Can the output pins be connected together to increase the output to 80 mA (max)?

No. Use one pin and a transistor.

This is a good question--can you parallel pins to get more current capacity? For a low-cost implementation, I can see the utility. If you aren't using the other pins, why not use them etc.

I know that if you run a pin configured as an output into a pin configured as an input, and you set the output pin high, I think you can toast something, but I have never tried it.

, I think you can toast something,

I think you're confused.

AWOL:
No. Use one pin and a transistor.

DUH!

You can use a IC!
You can use a IC relay!
You can use all sorts of things, but that was not the question!

Does anyone have an real answer and a rational for not using two pins connected to gether like you can with an IC such as a CD4049 Inverter.

It is unwise to do this. For a start 40mA is the point at which damage starts to happen not a safe limit.
So you will need to use three pins.
Second using digital write for each pin will result in the pins being at diffrent leves thus frying the chip. You get round this by direct port access, but this limits you as to what chips you are using.
There is no mention of current sharing on output pins in the data sheet.
You are one step closer to the processor with back EMF issues.

All in all I would not advise you to do it this way.

I

know that if you run a pin configured as an output into a pin configured as an input, and you set the output pin high, I think you can toast something, but I have never tried it.

Then what you know is a falsehood. A output pin wired to an input pin would result in almost zero current being drawn from the output pin. Perhaps you just meant wiring two output pins together and setting them to opposite states would result in a short circuit Vcc to ground through both pins?

Lefty

yes that's what I meant.

That's what I wanted to know. Thanks for your assistance.