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INA185: Regrading accuracy issue.

Hrishikesh Kamble
Prodigy 250 points
Part Number: INA185
Hello Sir/Ma'am,
We have built the prototype circuit of the INA185A3IDRLR current sense amplifier on the general-purpose board for our design testing & validation.
We design this circuit to work in low side sensing configuratiion according to that we built our test setup & tested it on the different currents. And we observed some accuracy-related issue on the higher currents.
Below mentioned is our setup specification.
  1. INA185A3IDRLR
  2. 1mohm current sense resistor.
  3. 3V3 input supply for the INA185A3IDRLR.
  4. Measured current range 1 to 20 Amp.
We connected this INA185A3IDRLR with a sense resistor in the low side configuration and measured the output of INA185.
For lower current, it is observed that there is a small difference between the measured value & calculated value but as we go on increasing the current, difference between mesured & calcaluted values are increasing.
Ex:
Sr.No Set current (A) Measured output voltage(V)  Theoretically calculated output voltage(V) Difference(V)
1 1 0.105 0.100 0.005
2 17 1.811 1.700 0.111
We used  "Imax X Rsense X gain < Vsp-Vref" formula for theoretical calculation.
VSp is 3.3V & Vref is 0 is  as we connected in the low side configuration.
If we see these readings we observed that, for 17Amp it is giving me an error of around 1 Amp.
For more detailed view, I am attaching our test setup block diagram and complete test readings on different currents.
Please have look and let us know what we can do to improve the accuracy at higher currents.
Thanks and Regards,
Hrishikesh Kamble
Acevin Solutions
  • 0
    Guru 140200 points

    Hi Hrishikesh,

    I heavily recommend to use a four terminal shunt supporting 4-wire connection. See section 10.1 of datasheet.

    Kai

  • 0
    TI__Mastermind 34200 points

    Hi,

    As Kai alluded to, this is most likely a system issue rather than simply a current sense amplifier issue.

    Specifically, it appears the delta in output voltage is dominated by gain error. Please refer to the last two columns in the spreadsheet, where gain error and output percentage error are calculated. As you can see, the two line up almost perfectly.

    I would suggest measure input differential voltage as well, and use it to divide output voltage. I think you’ll see improvement in accuracy.

    Regards, GuangNew_Current Sense amplifier test result _withGE.xlsx

  • 0 in reply to Guang Zhou
    Prodigy 250 points

    Hello Guang Zhou,

    Thank you for your reply.

    As per your suggestion, we did the testing and attached the test report for your reference.

    We divided the output voltage with differential voltage and readings are mentioned in the excel sheet.

    Thanks 

    Hrishikesh kamble

    Readings.xlsx

  • 0 in reply to Hrishikesh Kamble
    TI__Mastermind 34200 points

    Hi,

    Thank you making the additional measurements. As you can see, the INA185 output/input is exactly ~100. The 6% output error (compared with the product of ideal current and ideal shunt resistance) is due to the fact that the differential input voltage is 6% off from its ideal.

    This error can come from either the electronic load or the shunt resistor (and related wiring), or both. I would focus on retrieving the true current value and effective shunt resistance. Calibration of both will improve the system's accuracy.

    Regards, Guang

  • 0 in reply to Hrishikesh Kamble
    Guru 140200 points

    Hi Hrishikesh,

    the fact that your values are 6% too high means that your shunt resistance is 6% too big. The reason for this discrepancy is very probably the additional resistance of the soldering of shunt. That's why a four terminal shunt is heavily recommended whenever the shunt resistance is in the same range like the resistance of soldering, as in your case:

    Connect the wires running to the load and to power ground to the too fat terminals of above four terminal shunt and connect the input pins of INA185 to the two tiny terminals of above four terminal shunt. Don't confuse the different terminals and don't make any interconnections as this would destroy the perfomance.

    Having made the connections properly, the voltage drop across the solderings at the fat terminals would not be seen by the tiny terminals Relaxed

    Kai