(As a follow-up to an earlier question answered by the community - thanks!) So I've got the test circuit setup using the HCPL 7520 (datasheet) now and planning to use an Arduino Uno to measure the voltage at the VOUT.

I've done some tests now though and the VOUT seems to be too off. Below, I've got vRef voltage-divided to 4.41V and the VIN+ to 154mV. The rough VOUT I was expecting was VIN * GAIN = .154 * (4.41/.512) = 1.32V. But I read 3.05V at VOUT and, once adjusted (by -vRef/2 by looking at Fig.5 in the datasheet), I get 0.845 which is completely off.

(Pls ignore pin numbers in the diagram)

enter image description here

Another attempt was VIN+ = 104mV, expected = 0.89V. But I read 2.73V at VOUT, which adjusted becomes 0.525V. Again, not right.

I'm clearly doing something wrong - could anyone correct me or point me to the right direction?


  • \$\begingroup\$ Do you get 2 volts out if you short out R2 or open circuit R1? Have you used the pin numbers in the diagram for the chip as per above? Where did the diagram come from? \$\endgroup\$ – Andy aka Aug 14 '20 at 10:36
  • \$\begingroup\$ I got 2.26v for shorting out R2 and 2.16V for open circuit R1. Apologies, those pin numbers in the diagram are wrong so pls ignore pin numbers (I created the diagram with a generic IC part and didn't change pin numbers) \$\endgroup\$ – nchaud Aug 14 '20 at 12:04
  • \$\begingroup\$ Maybe you need to show your actual circuit. 2 volts is about right with the resistor checks I suggested. Have you confirmed the input voltage with a multimeter and the reference voltage also (just in case of resistor value error). \$\endgroup\$ – Andy aka Aug 14 '20 at 12:13
  • \$\begingroup\$ Your C1, C2 and C3 are still in backwards. If you draw your schematic conventionally with positive on top this becomes instantly apparent. Also if R1 and R2 were drawn vertically and in line with each other it becomes instantly apparent that they form a voltage divider.. \$\endgroup\$ – Transistor Aug 14 '20 at 12:20

Look at figure 5 in the data sheet: -

enter image description here

If your input is about 0.15 volts, for a 4 volt reference you will get about 3.1 or 3.2 volts out. I'm not saying the device is perfect but I think you are misinterpreting the results somewhat.

  • \$\begingroup\$ Totally agree that, even though my vRef is different to 4V, going by this chart is more accurate than my calculations but can you see what's wrong with my calcs at all? I thought V_IN * GAIN gives V_OUT. (and my Gain I've calculated at 4.41/.512) \$\endgroup\$ – nchaud Aug 14 '20 at 15:12
  • \$\begingroup\$ @nchaud all I can say is that 2 volts is the neutral position as per my comment below the question. Double check with meters that you are getting 4.416 volts and double check your input is 154 mA. \$\endgroup\$ – Andy aka Aug 14 '20 at 15:18
  • \$\begingroup\$ @nchaud The your reference voltage is 4,09V instead of 4,41V (Vdd2=5,39V). However, the measured values ​​are approx. 30% smaller. Try to measure the value of the output voltage at zero input, there you should see the half of the reference voltage. If that's okay, there's a good chance that the optocoupler deteriorated slightly during the first experiments.(?) \$\endgroup\$ – csabahu Aug 14 '20 at 16:11
  • \$\begingroup\$ @Andyaka, I double-checked as all my voltages were from a multimeter (and all resistors used are high-precision) - vRef was slightly different at 4.383 but that still makes VOUT far away from what's expected. The y-axis in your chart goes up to 4V but they say their vRef is 4.0V whereas mine is 4.41V so I have to know how they drew this chart if the gradient isn't the gain...? \$\endgroup\$ – nchaud Aug 14 '20 at 17:10
  • \$\begingroup\$ @nchaud it looks to me like they've used a 4.096 volt reference i.e. the y-axis range is 0.4 volts to 3.6 volts for an input range of -0.2 volts to +0.2 volts or, 3.2 volts / 0.4 volts = gain of 8 = 4.096 volts *0.512. \$\endgroup\$ – Andy aka Aug 14 '20 at 17:22

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