How to convert voltage reading to current when using the INAx180 IC?

Question

I am using a INA4180-4181EVM to find a suitable solution for measuring 0-2A of current in a DC circuit. In the datasheet, there is no equation given for calculating the current based on the read output voltage. How can I calculate the current?

Datasheet

EDIT:

The maximum current that should be measured is 2A, the maximum heat dissipation is set to 1/3W. Here are my calculations based on the datasheet: $$R_{shunt} < \frac{PD_{max}}{I^2_{max}}$$ $$R_{shunt} < \frac{\frac{1}{3}W}{2^2A} = 0.0833\Omega $$ Meaning that the shunt resistor must not be larger than 0.0833 Ohms in order to satisfy the power dissipation requirement. To get a starting point for the gain selection, my calculations are as follows: $$ I_{max} \cdot R_{shunt} \cdot GAIN < V_{SP}\\ GAIN = \left(\frac{I_{max} \cdot R_{shunt}}{V_{SP}}\right)^{-1}\\ GAIN = \left(\frac{2A \cdot 0.075\Omega}{3.3V}\right)^{-1} = 22$$ Selecting a gain of 20 (and my ADC can read 0-3.3V): $$R_{shunt} = \left(\frac{2A \cdot 20}{3.3V}\right)^{-1} = 0.0825\Omega$$

However, (my understanding is that) by using this value the current is limited to exactly 2A, so it is better to select a value lower than 0.0825 to provide some headroom. I want to test the following values for my shunt resistor to find which one suits my circuit best: 0.070, 0.075 and 0.080.

The current calculation is then done as follows (using Rshunt = 0.08):

float Vshunt = V/20; // Gain reduction, V
float Rshunt = 0.08; // Ohms
float I = Vshunt/Rshunt; // A

Using the EVM board, this seems to match my Fluke multimeter pretty close (have not calculated exact difference yet since I dont have the correct Rshunt value at hand).

Answer 1

^{simulate this circuit – Schematic created using CircuitLab}

This circuit technically measures a voltage differential, rather than a current. Each pair of inputs goes into an amplifier with a gain, and so the voltage you read (OUT) is the result of the difference in voltage between IN+ and IN- multiplied by the gain of that device - looks like there are 20V/V, 50V/V, 100V/V, and 200V/V options.

Section 3.1 of the datasheet describes how to use this. You have to use an external shunt resistor, connecting IN+ to the positive terminal of the resistor and IN- to the negative side. The current you are trying to measure should be passed through that shunt resistor which should have a small value. Then, you can use the voltage you measure at OUT to calculate the voltage difference across the shunt resistor, which you can then plug into V=i*R to find i. I'm attaching a rough schematic to illustrate what this looks like, although your actual circuit will be different.