Hi,
I am using TI INA219 test board connecting to a DC power supply. Power supply plus --> Vin+. My shunt resistor is 0.025 ohm between Vin+ and Vin-. Vin- -->.10 ohm resistor --> power supply minus. I was able to read the bus voltage correctly with bus_voltage_LSB of 0.0004. I calculated the calibration value for the calculation base on 4096 / 0.025 (shunt resistor). I played with the calibration value and got the current value looks correct. However, the power value is way off. Please help.
My example is the DC power voltage is 6.0V, the power supply current reading is 0.59A. I have a SW calibrate value of 1.01833 to give an exact bus voltage 6.0V. I changed the voltage on the DC power supply to many points from 0 to 13V and I got the adjusted bus voltage right on the number. This is a great start.
For the calibration value for the calibration register, I can't figure out how do you come up with the current LSB. My calculation was done at the datasheet example 1 on page 17. V_bux_max=15, V_shunt_max=0.25, R_shunt=0.025, MaxPossible_I=10, Minimum_LSB=0.0003, Maximum_LSB=0.0025 and Current_LSB=0.0004. The calculation using in equation 4 is 4096 for the calibration value. This value can't be right since Current=ShuntVoltage*Calibration/4096 therefore Current=ShuntVoltage. Please let me know what is wrong in my calibration value. Thanks a lot.
Dennis
