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Current shunt resister of INA219AIDR

user6289930
Prodigy 40 points
Other Parts Discussed in Thread: INA219

Hi team,

Could you help me to check the current shunt resister on INA219AIDR?

My system will use 6A for INA219, what value of shunt resister should I choice?

I checked the voltage must less than 50mV on shunt resister in the datasheet. Than how small can that minmum be?

Thanks,

  • 0
    TI__Mastermind 32395 points

    Hello user6289930,

    Our apologies for the weekend delay. Thank you for posting to the forum.

    The shunt resistor value is dependent upon many system-level definitions such as:

    • What are the minimum and maximum currents you want to sense and at what allowable error?
    • Do you plan on changing the INA219 gain (PGA) in real time during normal system operation?
    • How much conversion time and averaging can you use during data acquisition?
    • What is the maximum allowed shunt power dissipation your system can allow?
    • Would or could you perform a one-time, offset calibration procedure in your system? By doing this for every system (or every INA219), you can remove the offset error; however, it requires taking a measurement at a known current level (most likely 0-A) and then storing the measurement in memory.

    Anyway let’s assume you won’t be performing any calibration and you will not be adjusting the INA219 gain in real time. Let’s say you can accept max 2% total error at the minimum current level you want to sense, 500mA. Let’s assume the INA219 offset (Vos) is +/-100uV max (PGA=/1). If your minimum shunt voltage (Vshunt) is 10mV, then a 100uV Vos will yield a 1% offset error. There will be an additional error from device gain error (1% over temperature) and shunt resistor tolerance.

    If we accept 10mV as the minimum shunt voltage, the Rshunt is 10mV/500mA = 20mOhm. This means that at 6 A of load the Vshunt = 6A*20mOhm = 120 mV. Unfortunately, this means you have to adjust the INA219 PGA because the full-scale range is +/-40mV when PGA=/1. Setting the PGA to /4 (FSR=160mV) will correct this so you can sense current at 500mA to 6A.

    Overall, the size of the shunt resistor depends on the error you can accept and error will be dependent upon the offset, gain error, shunt resistor tolerance, current noise, and any possible offset calibration employed.

    Hope this helps. Consider watching our training videos online detailing how to use our current sense amplifiers. https://training.ti.com/ti-precision-labs-current-sense-amplifiers?context=1139747-1139745-1138708

    Best,

    Peter

  • 0 in reply to Peter Iliya
    Prodigy 40 points

    Hi Peter,
    I will selected the 4m ohm of resister for Rshunt.
    My system is work on 10W~30W at 5V, So the current will between 2A~6A.


    1) Could you help me check if the Vsense is between 8mV~24mV, then I setting the PGA=/1 or not?
    2) Reference your information, I will have 3% summary tolerance, A 100uV Vos will yield a 1% offset error + an additional error from device gain error (1% over temperature) + shunt resistor tolerance, right?
    3) I checked the datasheet, there are three valures can be read immediate the voltage, current and power from table2, right?
    4) What is the calibartion can do?

    Thanks,

  • 0 in reply to user6289930
    TI__Mastermind 32395 points

    Hey user6289930,

    1. With a 4-Ω shunt, the Vsense will be 8mV to 24mV with currents from 2A to 6A. Thus setting PGA=/1 (+-40mV Vsense full-scale range) is the correct thing to do.

    2. The maximum total error will occur at the 8mV Vsense. If using INA219A at PGA=/1, then the Vos =±100uV. So offset error is 100uV/8mV *100 = 1.25%. Thus maximum total error is 3.25%. On average the RSS error will be SQRT(1.25^2 +1^2 +1^2) = 1.887%.

    3. You can immediately (without any programming/calibration) read shunt voltage and bus voltage values; however, in order to obtain current and power you must set the calibration register of the device.

    4. Programming the calibration register means giving the device the shunt resistor value and current range value so the device can perform mathematical processing and output current (Vshunt/Rshunt) and power (current*Vbus) in the current and power registers. See section 8.5.1 on how to calibrate the calibration register.

    Best,

    Peter