Detect 240V with a PIC

[dhouston]

How do i know what value resistors to use?

Look at the datasheet for your optoisolator - there's an Absolute Maximum Ratings section with an entry for Continuous Forward Current. Ignoring the forward voltage drop of the LED, you need about 4000 ohms total for 60mA @ 240V. The lower values in the example probably factor in the forward voltage drop. In this case, you want the current to be near the max to improve the coupling. If using the PIC pin directly you can limit the current to much lower levels, hence the larger resistor I recommended.

I take it the diode is just to stop reverse voltage going into the optoisolator. Would it work the same if it was in series with it?

I would stay with the example or use an optoisolator designed for AC.

I think im missing some of the document. skimask mentioned page 77 but all i see is a single page with a short writeup and a link to a PDF file. I dont see any kind of navigation for multiple pages either

I don't see it either.

[The Master]

About 4,000 ohms is what i got. 3,975 taking into account the voltage drop of the opto. The confusion is because the example shows 130K resistors. Together thats 260,000 ohms. I work that out to be a voltage drop of 15,600 (i think)

[dhouston]

They may also be taking into account that it's AC so the average current can be a bit higher as it's not steady state (i.e. continuous).