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INA3221: Conversion Time and Averaging Length Versus Max Sample Rate

Part Number: INA3221

Hi there,

I am trying to create a system which needs the highest possible sample rate.

In my understanding, the conversion time, averaging length & number of channels are all factors into increasing the INA3221 loop time.

INA3221 Loop time = (Bus Conversion Time * Averaging Length * # Channels) + (Shunt Conversion Time * Averaging Length * # Channels) 

We know that the INA chip has 3 channels, and lets assume we pick the lowest conversion times (144us) and lowest averaging length (1) to maximize the max sample rate of our system. Thus,

INA3221 Loop Time = (144us * 1 * 3 ) + (144us * 1 * 3)

INA3221 Loop Time = 864us

In my system, I am using 2 INA3221 chips,

thus the equation for max sample rate is as follows:

Max Sample Rate = (#1 INA3221 Loop Time) + ( #2 INA3221 Loop Time)

Using the example above using the lowest conversion times and averaging lengths,

Max Sample Rate = 864us + 864us = 1728us = 1.728ms

My question is : Are these equations right? Optimally, I would like to confirm these equations are right, then I can sample at 1.728ms and not worry about having duplicates.

Although, with these settings, I am finding that I have 25% duplicates at 50ms Sample Rate, so there must be some other factors that increase the loop time.

Thanks for reading!

  • Hello Zack,

    Thank you for using the TI forum.

    The equation you have for device loop time is correct, but the equation for Max Sample Rate is incorrect. You do not need to add the time from the two devices together since they can be converting data at the same time. This means that every 864µs you can read a sample from both devices. If the data on the line is not changing much, then you could still read the same value that was there before, which may account for your duplicates. 

    Regards,

    Mitch