For those who are having trouble visualizing the circuit described
below, I have attached a schematic in PDF format.

Fr. Tom McGahee

************

Are you aware that you can use two optoisolators with their LEDs in series
to get an almost perfectly linear voltage transfer function?

You have to use an opamp to perform the linearization. The circuit looks like
this:

The output of an opamp is connected via a 47 ohm resistor to the Anode of
the first optoisolator LED. The cathode of the first optoisolator LED is
connected
in series to the Anode of the second optoisolator. The cathode of the second
optoisolator LED is connected to ground.

Each optosolator has its transistor's emitter connected to a 1.5k resistor
that connects to "ground". Note that each "ground" can be totally
isolated from the other. Each optoisolator has its transistor's
collector connected to a LOCAL V+. (Just as they each have a "local ground").

The output of the first optoisolator's emitter voltage is fed back to the
inverting electrode of the opamp. The output of the emitter of the second
optosolator's transistor will be a voltage that exactly matches the
input voltage applied to the non-inverting electrode of the opamp.

The circuit can track voltages from ground level up to the V+ of the
opamp, *minus* the diode drops of the two LEDs. So you will want the
V+ to be a few volts higher than the max desired output.

If using a rail to rail style opamp, you can get by with just a single
supply.

I also recommend using an opamp buffer to buffer the voltage out at
the emitter of the second optoisolator, unless the impedance of
circuit being driven is much higher than the 1.5k load resistor.

To get the absolute best matching characteristics, use a DUAL opto-
isolator instead of two separate devices. But test this out with
what you have... you may find it is more than adequate for
your needs even with two separate optoisolators.

How does it work? Well, the two LEDs are in series, so they have
absolutely identical currents. Since the light emitted is a function
of current, they emit identical amounts of light. Assuming that
they are fairly identical optoisolators, the emitter currents should
be the same. It helps if the V+ applied to both optoisolator
transistors is also the same.

BTW, all circuit resistor values may be varied quite a bit from the
values given, just remember to keep the emitter resistors identical.

The type of opamp used is not critical if you are interested in
voltages that are mostly DC. If you have a fast changing signal,
then you need an opamp with a decent slew rate. It must be capable
of furnishing the max current consumed by the LED(s).

Fr. Tom McGahee





Attachment converted: birth:Isolator.pdf (PDF /CARO) (00007613)