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The datasheet is here http://www.ssousa.com/pdf/SLC800.pdf . I need help on the following:

  1. Where is the maximum operating DC voltage is specified? It appears to be 500VDC but I am not sure.

  2. I need to wire it to a battery bank of maximum 1050VDC, across a shunt of 1miliohm with a full scale deflection of 200mV.

Can it handle common mode voltage or do I need some margin? Can someone more experienced with this guide me?

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  • \$\begingroup\$ You say you intend to wire the input across a 1 milliohm resistor where the voltage drop will get up to only 200mV. From what it sounds like this device will not be too useful to you in that manner. It takes a voltage drop of nominally 1.2V across the input LED to get appreciable current through it. This device is not a voltage sensing device but a current transfer device. \$\endgroup\$ – Michael Karas May 24 '14 at 17:53
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The first line of the features section on the first page of the data sheet says:

enter image description here

The section were you saw 500V is just a test condition where they were measuring the input to output isolation resistance.

enter image description here

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  • \$\begingroup\$ Thanks for your reply, but it cannot keep working on 1500Vrms ? , It is only 1min rating. \$\endgroup\$ – Adi May 24 '14 at 17:47
  • \$\begingroup\$ Indeed not. It is rather common to find parts rated this way. (One.) It looks good on a data sheet as a high number. (Two.) You can use this as a guide for designing in applications where there are momentary high voltage surges possible. If you need a guaranteed continuous rating then you need to search out a manufacturer that would specify a coupler with a suitable continuous rating. Short that you would have to find one with a high enough surge rating that you would then test and validate to be suitable at a lower continuous operating level. \$\endgroup\$ – Michael Karas May 24 '14 at 18:10
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The part is (allegedly) tested for 1500V RMS AC voltage (one minute), which is 2.1kV peak but there's no indication of the rating for continuous operation. That's not really a very impressive test voltage for 1050V DC continuous operation. You should probably take precautions to limit the damage from a failure. More likely they actually test it for a much shorter time at a higher voltage so the probability of it passing a 1 minute test at 1500VAC RMS is very high.

The similar Vishay (nee Infineon, nee Siemens) IL300 datasheet is here. It is tested for 1s at 5300 VRMS (about 7500V peak).

You can see some typical application circuits in the above-linked datasheet that will help answer your question about working from the shunt. You need op-amps on both sides, and you need power supplies on both sides.

enter image description here

The (transfer) accuracy of the part you chose is much better than the IL300, but even so you might want consider a more expensive isolation amplifier that has some of that stuff built-in and perhaps uses a different isolation method that is explicitly rated for the service you have in mind.

Edit: If you really want to guarantee reliability and cannot find an isolator with an appropriate rating and guarantee from the manufacturer, you can use a fiber optic cable to isolate the signal (digitize it first). Provided the materials used are appropriate, almost any reasonable isolation voltage should be possible.

Or, lose the idea of the shunt entirely and use a DC closed-loop Hall effect pickup with appropriately insulated wire. That would be my best suggestion!

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  • \$\begingroup\$ Thanks for the reply, but common mode voltage is still an issue :-( . \$\endgroup\$ – Adi May 25 '14 at 17:26
  • \$\begingroup\$ Is ISO122 applicable \$\endgroup\$ – Adi May 25 '14 at 17:36

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