Mike Harrison <mike <at> whitewing.co.uk> writes:

> As it's AC, a simple capacitive charge pump would give enough voltage to 
> drive a LED or optoisolator.  

Actually, no, because it would require Germanium diodes and would still provide
too little current for an opto. The easy way to sense it was as I said, using a
1:1 (can be other ratio) pulse transformer capacitively coupled. For ordinary
diodes something like this should work:

       +---C0---+
       |        |
A o----+-A-D1-K-+----o
       |        |
       +-K-D2-A-+
       |        |
       C1       R1
       |        |
       +-A-D3-K-+
       |        |
       +K-LED-A-+

With LED=ir (optocoupler, Vf=1V), D3-Schottky, C1=10uF and R1=5R.
C0=0.1uF, D1+D2 = 1N4001 etc

Peter P.



-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist