https://www.ti.com/product/INA226

The nominal voltage and the nominal current through the load is 4 A and 5 V. The maximum current through the load is 5 A.

I started calculating shunt resistance by defining the peak power dissipation across the shunt resistor which is 200 mW.

Let's calculate the shunt voltage given the power is 200 mA and the max current is 5 A.

P = V x I

V = P / I

V = 200 mW / 5 A

V = 40 mV

This shunt voltage 40 mV is well within the input shunt voltage range according to datasheet INA226.

Now calculate the value of the shunt resistance given the maximum current (5 A) and the shunt voltage (40 mV)

V = I x R

R = V / I

R = 40 mV / 5 A

R = 8 m Ohm

The value of the shunt resistance is calculated as 8 m Ohm.

Now we need to calculate the Current LSB according to equation (2) on page 15/39 of the datasheet.

Current LSB = Maximum Current / 2^15

Current LSB = 5 A / 2^15

Current LSB = 152 uA/bit

Assuming the last two bits are noisy and not very stable. Or we do not want very high precision. Let's ignore the last two LSBs. We multiply by 4 to get new Current LSB.

Current LSB = 610 uA/bit

Is my calculation are ok until now ?

Now we calculate the Calibration Register value using equation (1) of the datasheet. The values of Current LSB and the R_Shunt need to be converted to A/bit and Ohm before using in equation (1)

CAL = 0.00512 / (Current_LSB x R_Shunt)

CAL = 0.00512 / (610 /1000/1000 x 8 /1000)

CAL = 1049 decimal

Is that number also correct to be written in Calibration Register ?

Two more questions.

The resolution of Shunt Voltage Register is 15 bit ? The LSB is 2.5 uV ?

Does it means that full scale reading will be 2^15 x 25 uV = 81 mV approx ?