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From Linear Optical Isolation for Safe Sensor Operation:

isolation circuit

  • Are the three grounds indicated all connected?
  • Are the four Vcc indicated all connected?
  • If the above is all true, how could this be an isolated circuit? The grounds aren't isolated.
  • If the above is false, wouldn't U1 fry at the first hint of significant common-mode potential to ground?
  • Are U1 and R1 considered pre- or post-isolation?
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The grounds on Vin and at the bottom of R1 must be connected, or U1 won't work.

If the optocoupler is to be useful, the ground at the bottom of R2 must be isolated from the ground on R1, and the Vcc on U1 and U2 must be isolated.

Pin 3 of the opto should be connected to U1's Vcc, and pin 6 of the opto should be connected to U2's Vcc.

The labelling of that diagram is very misleading if it is intended to show optical isolation.

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  • \$\begingroup\$ Are the connections for pins 3 and 4 backwards? As it is, the left diode will be reverse-biased. \$\endgroup\$ – Reinderien Aug 30 '17 at 2:03
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    \$\begingroup\$ I think the diode between pins 3 and 4 is being used as a light sensor, rather than as a LED, so it may be OK. \$\endgroup\$ – Peter Bennett Aug 30 '17 at 2:07
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    \$\begingroup\$ Former employee of a company that makes a similar product. Was worried that was our drawing. Luckily not, but we also had a lot of crap datasheets. \$\endgroup\$ – The Photon Aug 30 '17 at 2:56
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That is not a very good illustrative diagram, and it's copied directly off the IL300 datasheet:

Here is how it should be connected:

enter image description here

The LED has an output that varies with temperature and decreases with time, whereas the photodiodes are quite stable with time and will track each other well. So one photodiode is used in a feedback loop to control the LED brightness, and the other responds similarly on the other side of the isolation barrier.

As the LED decreases in efficiency, the feedback loop will increase the LED current to maintain similar emitted light, which is seen by both photodiodes. The photodiodes are in reverse bias.

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The grounds between isolated sections may be different. You have also to make it different on PCB with a groove or enough distance between traces.

What's the point? With such, you break the ground loop. For example: the sensor can be grounded at distant point, while your DAQ is grounded in the cabinet. Also all the EMI or overvoltage spikes on the sensor side won't go further to the DAQ side.

VCC and GND are different on each side.

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