• State Resolved
  • Locked Locked
  • Replies 2 replies
  • Subscribers 48 subscribers
  • Views 395 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

INA219: Issue with the calibration register

Lucas Bagrowski
Prodigy 120 points
Part Number: INA219

Hi everybody,

I am currently working on a board with an INA219BIDR. And I have an issue with the calibration register, in the datasheet we can read that : (page 12)

So I defined my maximum current at 2 Amps, and I have Rshunt= 0.002 mOhms.

My calculations are :

Current_LSB = 2 / 2^15 = 0.000 061 035 156

Cal = trunc(0.04096 / (0.000 061 035 156 * 0.002)= 335 544

But the register length is 16 bits so the max value is 65535

Did I fail somewhere ?

Thanks 

Lucas

  • +1
    TI__Mastermind 23780 points

    Hello Lucas,

    Thank you for using the TI forum.  The math you showed is correct, however there is a problem with your setup.  If you have a maximum current of 2A, and use only a 2mΩ shunt resistor, then your full scale current measurement takes place over 2A*2mΩ=4mV.  4mV is a very small percentage of the device measurement range, so your measured value will be very noisy.  If you use the smallest measurement range device setting at PGA = /1, then the full scale range would be ±40mV. If you adjust your shunt to ~20mΩ then you will utilize the full 40mV range, giving you a much better measurement. This would also adjust your calibration value to 33554, which will be able to fit inside the calibration register.

  • 0 in reply to Mitch M
    Prodigy 120 points

    Thanks for your reply, I will modify this.

    Have a great day

    Lucas