Yes, exactly! It's very old-school, but if you can work within all the
limitations it isn't bad at all.

Cheerful regards,

Bob

Forrest W Christian wrote:
> Bob Blick wrote:
>  > Not if the voltage is about a volt or more(an IR LED drop). Opto #1
>  > LED sees the measured side. Basically the output of the two optos
>  > fight it out for the opamp's input - the opamp drives opto #2 LED.
> 
> Ok, my brain hurts... but I think I've got my brain twisted around what 
> you two are trying to explain to me.
> 
> All of the linear opto schematics I had seen had a single LED driving 
> two internal photo transistors.   On the input side of these schematics, 
> there was typically an opamp with a separate power supply which 
> basically completed the feedback loop...   That is, the opamp's feedback 
> loop went through the LED and one of the transistors so that the LED was 
> turned on enough that the phototransistor was turned on enough that the 
> voltage out of the "input side transistor" matched the input voltage. 
> This of course needs a power supply for the opamp on the input side. 
> On the output side transistor there was typically a unity gain opamp set 
> up just as a buffer.  The theory being that both transistors will be 
> turned on the same amount so once you "balanced" the input side, the 
> output side would also be producing the same voltage.
> 
> What I'm hearing you say (or at least I think I am hearing you say) is 
> to take a more-traditional dual optocoupler (perhaps matched) and feed 
> the signal of interest into one of the LEDs.  Then effectively wire both 
> phototransistor outs into another opamp which then drives the second LED 
> in the optocoupler.  The theory being that both sides are going to react 
> the same at the same current level through the LED, so that when the 
> opamp is stable, the voltage out of the opamp (and into the second LED) 
> on the output side is going to pretty much match the input signal level.
> 
> Is this more or less correct?
> 
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