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In my application I need to measure a voltage that is in the area of 50V, but my microcontroller has to be isolated from my power circuit. I came across the ACPL-C87A-000E IC. I followed the typical application circuit found in page 12 of the datasheet, but at first, I am not using an op amp. The circuit I am now using, converts the 50V to around 1.9V via a voltage divider, which is sent into the IC. This is my circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

By looking at figure 13 of the datasheet, page 9 (Vin vs Vout+,Vout-), I thought that I could send the 1.9V in the IC and by measuring just the Vout-, I could get a value that is proportional to the voltage measured (to be exact, in figure 13, Vout- and Vin seem to be inversely proportional). I left the Vout+ pin unconnected. Also, let me point out that all the voltage measurements are taken from digital multimeters.

However, there are some problems with the above-mentioned idea.

  1. Firstly, immediately after connecting the voltage divider's output voltage to the Vin of the IC, there seems to be a drop in that voltage (goes from 1.9V down to ~1.88V). The IC's Vout- value however seems to be correct for the given Vin, around the one expected from figure 13.
  2. Furthermore, the voltage divider's output voltage continues to change slowly, going from 1.84 to 1.89, as long as it is connected to the Vin of the IC. Obviously, the Vout- changes accordingly.
  3. After leaving the circuit on for some time, the voltage divider's output suddenly dropped significantly, to around 0.3 Volts, which I think may be the result of the IC going bad for some reason.

Another option I am considering for voltage measurement is the LTC6992-1, as suggested here. However, I don't know how to implement the last part, of measuring the PWM's duty cycle and converting it back to an analog voltage that can be read from my microcontroller. I am using a Beaglebone Black.

I would be really grateful for any other suggestions, perhaps simpler solutions, for measuring an isolated voltage with my microcontroller.

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    \$\begingroup\$ Can you provide a schematic? \$\endgroup\$ – Long Pham May 4 at 14:18
  • \$\begingroup\$ And a direct link to the datasheet, please. \$\endgroup\$ – carloc May 4 at 17:25
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    \$\begingroup\$ A relaxation oscillator is a very low power/current method and converts a voltage to a pulse frequency. This can easily drive an opto-isolator for isolation. Another method is to use a sawtooth (easily generated) and a comparator with a divided voltage to get PWM, instead. But I see no range of voltages over which this must work nor a specification of the needed precision, limits on drift over time, accuracy, etc. So details matter. Could you add more? \$\endgroup\$ – jonk May 4 at 18:39
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    \$\begingroup\$ One option is to use linear optoisolators. They are an LED and a pair of photodiodes in the same package where one photodiode is used as a feedback element to stabilize and linearize the other. Examples of so-called "high linearity optocouplers" are the IL300 and HCNR200/201. Check out the application circuits in the datasheets since they are all transferrable between components, especially the HCNR200 datasheet. Also read IXYS AN-107. \$\endgroup\$ – DKNguyen May 4 at 19:45
  • \$\begingroup\$ I personally think using linear optocouplers is simpler than converting to a digital square wave and transmitting across an isolation barrier for reconstruction and nor are you t tied specifically to a rare optical (or capacitive) isolation amplifier IC that might suddenly stop being produced. \$\endgroup\$ – DKNguyen May 4 at 19:51

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