How to use isolated amplifier to measure voltage

Question

I want to read the cell's voltage with a microcontroller, but the cell does not share the same GND with the microcontroller. I want to keep them separate. I read some suggestions for this problem, and I try to use an isolated amplifier (I.A). I am considering using this isolated amplifier AMC1200 to measure the voltage of a solar cell. This is the first time I work with I.A., so I want to ask if my design looks reasonable.

Problem Description

In the schematic, Vcells+ is the positive terminal of the cell, and CGND is the negative terminal of the cell. The maximum value of the cell's voltage is 7.2V. Since AMC1200 has an input range of +-250mV, I use 300kOhm and 10kOhm for scaling the cell voltage to the I.A. range. That gives V.S.IN.DIV, and the maximum value of this is 240mV. This is fed into the VINP of the I.A. VINN of I.A. is connected to CGND.

The output of the I.A. is differential output. Therefore, I use an external opamp to implement a differential amplifier with a gain of 2 to get a single output. This output voltage is read by an MCU.

Questions

I am not sure what will output from the VOUTP and VOUTN of the I.A. According to the simplified schematic, assume my V.S.IN.DIV is 240mV and the gain is 8, my VOUTP will be 240mV*8 + 2.5V = 4.47V and my VOUTN will be 0 + 1.29V = 1.29V. So, the output of my external opamp will be (4.47-1.29)*2 = 6.36V. Is my calculation correct? Also, is this a reasonable solution to measure the cell's voltage?

Answer 1

Just a couple of issues with your calculations: VOUTP and VOUTN are centered at 2.55 volts if VDD2 is 5 volts, so the gain of eight (1.92 volts) means that VOUTP will be 3.51 (2.55 + 0.96) and VOUTN will be 1.59 (2.55 - 0.96).

Your gain of two will give you 1.92 x 2 or 3.84 volts at your op amp output. (You could never have achieved 6.36 V since the op amp power rail is only 5V.)

Your circuit approach is fine as is.