**Part Number:**INA219

Dear Ladies and Gentlemen,

I am just getting into the workings of the INA219, but am having trouble understanding the function of the Calibration Register and how it works in general.

if I understand correctly, I can select a full scale range of 16 or 32 V. This would then be resolved with 12 bits to determine the bus voltage. This would then be resolved with 12 bits to determine the bus voltage. The lowest voltage resolution would therefore be approximately 7.8 mV or 3.9 mV. However, if the 16 or 32 V are the only reference potential, the shunt voltage must also be referred to this reference. A resolution of 3.9 mV seems extremely high to me, because with an assumed current of 2 A and a shunt resistor of 0.1 Ohm, the voltage drop at the shunt resistor would be only 200 mV. So I would have only 50 steps at my disposal? Am I making a mistake here?- the current in the current register is internally calculated by the formula: (shunt_voltage * calibration)/4096. But as I understand it, Calibration is a rather arbitrary value, because it depends on the estimated value of the maximum current. But how can a calibration value depend on an estimated value? For example, if I choose an assumed maximum current of 1 A and a shunt resistor of 0.1 Ohm, I get a calibration value of 13421... but if I assume a maximum current of 2 A, it is 6710... How can an assumed value have such a large influence on my measurement result. Again, the question: Where is my thinking error.

I would be very grateful for any assistance!

Best regards,

Werner Boecker