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INA236: How to convert shunt register voltage to current register and the behavior of Rshunt on GUI

Part Number: INA236

Hello Expert,

I have questions for INA 236 as follow.
First, We suspect that the formula of converting Current register from shunt register voltage as follows.

Current register = SHUT_VOLTAGE_REGISTER(binary) x 2.5[uV/LSB](uV) x 1/R_shunt[1/ohm](uA) x 1*10^-6(A) x 1/CURRENT_LSB[bit/A](binary)

But this formula has difference against  SHUT_CAL on datasheet.
Therefore, would you tell me how to convert shunt register voltage to current register?
We would like to know about whether we should consider quantization error at 2 time.

Next, I'd like to know about how Rshunt on GUI work for GUI's result of CURRENT.
Would you tell me it?

This is because there isn't CURRENT_LSB setting on GUI so we suspect this parameter has impact for CURRENT_LSB.
And, we want to know about what parameter has impact for error estimation. 

I'm looking forward to hearing back from you.


Best regards,
Kazuki Kuramochi

  • Hi Kuramochi-san,

    The equation you have is correct. There are two variables, R_shunt and CURRENT_LSB. Both must be defined before the device produces meaningful current/power readings. However the device has no way of knowing CURRENT_LSB directly.

    In the Shunt_Cal equation in the datasheet, the same two variables determine Shunt_Cal. In other words, these three quantities are inter-dependent, knowing any two is sufficient. Quantization error only needs to be considered once. Current is simply a scaled version of Vshunt.

    Regarding second question, you’ll need to specify the Rshunt value. CURRENT_LSB is something you choose based on maximum current. And it is usually chosen such that Shunt_Cal comes out to be a whole number to avoid truncation error.

    Here are a couple of examples – (1) Rshunt=2mOhm; I_LSB=1mA; then shunt_cal=0x0A00 (2) Rshunt=10mOhm; I_LSB=0.5mA; then shunt_cal=0x0400

    When using the GUI, simply program the Shunt_Cal register and set the RShunt, you’ll see correct current and power readings.

    Regards, Guang

  • Hello Guang-san,

    Thank you for your kindly explanation and sorry for my late reply.
    I understand.


    Thank you and best regards,
    Kazuki Kuramochi