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Im measuring differential voltage over a 1 Ohm (shunt) resistor at max 1 Amp, load voltage is 12V DC using one of the channels on MCP3422. MCP3422 is powered from 3v3 DC.

The datasheet https://ww1.microchip.com/downloads/en/devicedoc/22088c.pdf on page 11, section 3.1 reads

"The maximum voltage range on each differential input pin is from VSS-0.3V to VDD+0.3V. Any voltage below or above this range will cause leakage currents through the Electrostatic Discharge (ESD) diodes at the input pins. This"

I want to be sure that I interpret this correctly - does this mean, that although the max voltage over that 1 Ohm resistor will be 1 V DC (1 Ohm * 1 Amp) the inputs will be saturated (ADC will fail), because the voltage across the CHAN1+ diff input pin and VSS is 12V (11V on CHAN1- input pin against VSS) - which is above the VDD+0.3V? Even though the voltage between differential inputs is below the max diff input threshold of 2.048V reference?

If this is correct - what is the best way to approach this on a basic level? Maybe have voltage dividers on both differential input pins?

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You are interpreting it correctly. The voltage referenced to ground needs to be between -0.3V and 3.6V on each pin.

You could achieve this by using a resistive 4:1 divider on each line, but this will limit your range to 0.25V. If you want the resolution for the maximum 1V range, the best option (if practicable) is to put the sense resistor on the low side. Note that this will mean that your circuit ground (for the circuit you're measuring) may differ from your supply ground by 1V. (In your original configuration, this 1V difference would be between the circuit +12V and your supply +12V.

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  • \$\begingroup\$ Thanks for the confirmation. I understand the limitations of the dividers. To the low side measurement - Im unsure how to implement this as Im unsure I understand this correctly. If measured on the low side would this mean I could actually just use single ended ADC channel - e.g. simply measure V difference between common ground and the voltage between the load and the shunt? \$\endgroup\$
    – MarcelH
    Commented Apr 9 at 19:44
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    \$\begingroup\$ That's what I mean. If your power input is on the left, you want the ADC ground on the left of the shunt (power input ground), as the power input ground may be up to 1V lower than the circuit ground. \$\endgroup\$
    – Cristobol Polychronopolis
    Commented Apr 9 at 20:01
  • \$\begingroup\$ Thanks!! That is perfect solution for my application. \$\endgroup\$
    – MarcelH
    Commented Apr 9 at 20:02

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