Originally Posted by tcmtech

At least you admit to it. Thank you!

All you need to do is treat it like a very small relay. The input is a diode so it will only work in one direction. Assuming the Opto coupler diode has the common 1.2 volt drop across it and it had a minimum running current of 5 ma and you were powering it off of a 5 volt source
You would have 5 - 1.2 = 3.8 volts. You need at least 5 ma at the 3.8 volt drop so that means 3.8/.005 = 760 ohms.
being the IC I posted has a input current of 5 - 60 ma any resistor between 63 - 760 ohms would work.

The output side can be treated like a very small relay as well. just keep in mind that it has a 1 amp peak rating and a 100 MA continuous rating. and like any good relay designer you will need to put a snubber or voltage spike suppressor in parallel with the relay contacts.
For the very small current that a LED based light uses a .01 uf or probably even smaller will be sufficient. All it is doing is help knock down any voltage transients that may occur when the device switches off.

ok i think im starting to understand, how do i know which pins go where on that little chip? any chance i could have ya pop open paint and draw a cheap schematic and you would be my best buddy ever.

Thanks
Kevin