First, a similar question was asked before Understanding common mode vs output voltage graph for instrumentation amps, however, I failed to understand it with regards to my use case and need more clarification.
Essentially, I want to use an INA for high side average current sensing for power measurements. The system is powered using 3.0V with a current consumption max I estimate of 160 mA at peak and a low power state of <100 uA (what I am trying to find out). That means I need around 64 dB. However I do not necessarily need that full range, rather I am more interested in the low power current consumption, so 5 mA to 50 uA (40 dB) would be ideal. As I am rigging the circuit up from a previous board, my constraints are that I am using an INA823 with a Vs+ = 4.5 V, Vref = 0 V, and Vs- = 0 V. A note, I can modify the gain and sense resistor as needed, and the output will be fed into an oscilloscope. The 4.5 V reference comes from an LDO, while the 3.0 V comes from a buck converter. So my common mode voltage will be around 3V. See circuit below:
simulate this circuit – Schematic created using CircuitLab
The board has a 60 mΩ sense resistor, but that value can be easily changed. Now the downstream system doesn't have much headroom to function properly, min is 2.98 V.
So from what I understand, in this configuration I can set my INA823 gain to 10000 V/V to amplify 5 mA through a 60 mΩ sense resistor to output 3V. Now if I am reading the datasheet's common mode voltage vs output voltage graph correctly, this could be achievable at G = 10000 V/V? My confusion comes from understanding how the graph scales with gain. Furthermore, my supply is 4.5 V versus the 5.0 V used in the graph, but I assume if it scales with gain, this will not be an issue. Input common-mode voltage vs output voltage graph from INA823 datasheet:
So my underlying question is whether my assumption is correct, or my understanding and analysis of the system is wrong?
