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INA213: Error calculation

Part Number: INA213

Hi Team,

According to the tutorial in TI precision lab, current sensing amplifier's error(worst case) can be listed as below-

% ≈ + + + _ + + ℎ_ + _ + ℎ

However, voltage noise density error isn't listed above, is there any way to calculate the error caused by noise?

Thanks.

  • Repost my question, since some content isn't shown.

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    According to the tutorial in TI precision lab, current sensing amplifier's error(worst case) can be listed as below-

    However, voltage noise density error isn't listed above, is there any way to calculate the error caused by noise?

  • Charles,

    We don't list noise in the above equation, because the equation looks at accuracy based errors, where noise is a precision error. 

    For our analog devices, to calculate noise, you must find the integrated voltage noise density in the datasheet (shown below), and multiply this by the square root of the BW to achieve the RMS noise. Then, you may multiply this value by 6 (or 6.6 for 99% confidence) to achieve the peak to peak noise. 

    For our INA213, you can see from the datasheet that the BW is 80kHz. For the noise plot, I would suggest finding the additional area under the curve for a more accurate measurement, but a quick way to achieve theoretical results is to simply take the flat portion of the curve as the totality, as shown below:

    Then, this shows that, without any filtering on the output, the noise of the output signal will be 

    noise_RMS = 7.07uV_RMS

    noise_p2p = 46.67uVpp

    If you wanted to reduce these numbers, the equation shows that this nay be accomplished through reducing the BW of the device. 

    Let me know if you have any questions. 

  • Hi Charles,

    in opposite to your worst case total error summands most of the noise can easily be filtered out by low pass filtering or averaged out by multisampling techniques.

    Kai

  • Hi Kai,

    Thanks for your explanation, btw, I would like to confirm the gain error calculation, in INA213C datasheet, gain error 0.5% is based on -40C to 125C condition, so we don't need to consider gain error drift via temperature, am I correct? 

    Thanks.

  • Hi Kai,

    As for CMRR error, if my CM input volatge is 24V, should I calculate CMRR error by (24V-12V)*10^(-100(db)/20) or (24V-0V)*10^(-100(db)/20)?

    Thanks.

  • Charles,

    Regarding the gain error, this will be based on where you plan to operate the device. Note that the gain error test case is bidirectional, for a Vsense from -5mV to 5mV. If you plan to design to a sensing range outside of these test conditions, you may want to consider the gain error drift spec to establish a more conservative theoretical calculation for you system. 

    Regarding the CMRR, the correct method is the first you've listed. The device's common mode is characterized at 12V in the global test conditions, so this is the subtracted value for CM offset calculations.