-----Original Message-----
From: Joe McCauley <pmcculey@TCD.IE>
To: PICLIST@MITVMA.MIT.EDU <PICLIST@MITVMA.MIT.EDU>
Date: Wednesday, May 27, 1998 9:05 AM
Subject: Re: Opto-isolator Question


>David,
>
>Any chance we could see a schematic?
>
>Joe


I hate ascii schematics...

Here's what you do:

Configure an opamp as a voltage to current converter, at 1V = 10mA
In the feedback loop, put the two opto diodes, in series. This means that
you have 10mA through
each diode.  Now connect the collector of one opto's transistor to your +
rail (+5V?).  Connect
the bottom of the diodes to ground, and the emitter replaces them to provide
feedback current.
(At this point, the current through the LEDs drops off to 10mA / the opto
transfer ratio)

If everything is working, you have 10mA through the transistor of the first
opto for 1V input.
The second transistor will also provide very close to 10mA regardless of
supply and load
(within broad limits)

This gives you the basics, if I explained it clearly :) and you can adapt as
you like.
I did find that dual optos were better matched than singles, but I didn't
have any trouble
with it even using different optos. The output current may be different if
the optos are
different types, but it is still proportional and linear.

Think of it as sort of an optically isolated current mirror :)