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INA213: How to set the current sense resistor for low current application?

Navendu Shekhar
Prodigy 20 points
Part Number: INA213


Hi TI Team, 

We are using INA213`AIDCKR part in our application. Our requirement is to measure current till 0.35A. what shunt value resistor should I use since the datasheet mentions min drop allowed should be>= 10mV.

  • 0
    Prodigy 20 points

    What will be the minimum current it can measure and produce the Output Voltage?
    Calculating the offset error(%) at 10mA of load current with 100uV Vos(INA213) and Rshunt as 10mE will give the offset error of 100%. 
    So how the offset error is related to the output voltage.?

  • 0 in reply to Navendu Shekhar
    TI__Intellectual 1235 points

    Hi Navendu, 

    To answer your question:

    Navendu Shekhar said:
    What will be the minimum current it can measure and produce the Output Voltage?

    You can calculate the minimum and maximum current for the scenario you provided using these equations:

    Please note, Imin refers to measuring current in the opposite direction of Imax, thus is actually -1.4Amps. So with 10mOhm shunt and 10mA current, the measurable current range is -1.4A to 8.1A. This will change once you have determined a new shunt resistance value using the resources linked below.

    We have a online video training series that will help you, please see: 2.2 How to choose a shunt resistor, 3.2 Offset error, and 3.8 Output swing. 

    Cierra

  • 0 in reply to Cierra Cowley
    Prodigy 20 points

    Hi Cierra
    Please help in getting below answers for below queries-
    1. Can you please share the doc which shows the Imin and Imax calcualtion?
    2. With the equation which you have provided it looks like it will not fit our requirement since our requirement is to measure current from 10mA to 350mA. Am I right?
    3. What is the VSP and VSN in this current calcualtion equation?
    4. In this statement which current(10mA) you are talking about "So with 10mOhm shunt and 10mA current, the measurable current range is -1.4A to 8.1A".
    5. How the VOS error affects the output Voltage? WHat is the equation for that?

  • 0 in reply to Navendu Shekhar
    TI__Intellectual 1235 points

    Hi Navendu,

    Navendu Shekhar said:
    what shunt value resistor should I use since the datasheet mentions min drop allowed should be>= 10mV

    First, I forgot to address this statement in my previous reply. The INA21X products are capable of sensing across a minimum drop of 10mV, but you do not need to limit your Rshunt drop to 10mV. You should choose an Rshunt based on the current range you want to measure and power dissipation requirements, as mentioned in this training: 2.2 How to choose a shunt resistor

    1. These equations are not from a document, I am finding the input current range based on the output voltage limitations, using Vin = Iin * Rshunt, where Vin = Vout/Gain. Vout is limited to 0.05V to 4.8V due to output swing limitations, as explained below and in the 3.8 Output swing training. Vout is also affected by the reference voltage (Vref = 0.75V from the NCP114AMX075TCG).

    2. No, you will be able to measure 10mA to 350mA, which is within the range -1.4A to 8.1A, but this range will change once you choose a new resistor. You will want to choose a larger resistor to reduce the offset voltage error %. First, determine the measurement range desired and power dissipation requirements, then choose a resistor for that range: 2.2 How to choose a shunt resistor

    3. Sorry for leaving these definitions out, VSP: Swing to Vs power supply rail, VSN: Swing to GND. These terms are explained in this video training: 3.8 Output swing

    4. I meant to refer to this statement. The current range is -1.4A to 8.1A. I crossed out '10mA' in the reply above.

    Navendu Shekhar said:
    Calculating the offset error(%) at 10mA of load current with 100uV Vos(INA213) and Rshunt as 10mE will give the offset error of 100%.

    5. The offset voltage on the datasheet is referred to input (RTI), so you will need to multiple that value by the gain to determine how it will impact the output voltage. For the INA213, the max offset voltage referred to input is 100uV. The gain is 50V/V, therefore the max offset voltage you will see at the output will be .005V or 5mV.

    Cierra

  • 0 in reply to Cierra Cowley
    Prodigy 20 points

    Hi Cierra

    Some query related to the statement-
    "you will be able to measure 10mA to 350mA, which is within the range -1.4A to 8.1A"

    Going through the INA213 datasheet the output voltage swing will be from 50mV to 4.8V
    considering the 10mA current and vref=0v case and calculating the output voltage swing
    (vinmin-vos)*Gain-->(0.1mv-0.1mv)*50=0v, so we cannot observe the output voltage swing. so how we will be able to measure the current?
    In order to see the output voltage swing the current should be greater than 100mA to see the output voltage swing. Am I correct?

  • 0 in reply to Navendu Shekhar
    Guru 140200 points

    Hi Navendra,

    a minimum voltage drop of 10mV across the shunt is recommended to keep the error caused by the maximum input offset voltage of INA213A of 100µV below 1%. 10mV is 100 times bigger than 100µV. That's all.

    I would increase the shunt so much that you have a maximum wide output voltage swing which is from 0.75V to about 4.8V. Something like this:

    The additional input offset error voltage due to the input bias current is in the 10µV range then (35µA x 0.22R = 8µV) which is much less than the maximum input offset voltage of IN213A. (But keep in mind that the input bias current depends on the common mode input volatge which you didn't mention so far.)

    The minimum current you can measure up to an error of 1% would be 10mV / 0.22R = 45mA in the above example.

    Of course, if the voltage drop is too high at 350mA with a 0.22R shunt, which is 77mV, then you can decrease the shunt. The minimum current for 1% error changes accordingly then. (For a 100mR shunt the minimum current for 1% error would increase to 10mV / 0.1R = 100mA.)

    Kai

  • 0 in reply to Navendu Shekhar
    TI__Intellectual 1235 points

    Hi Navendu,

    Navendu Shekhar said:
    considering the 10mA current and vref=0v case and calculating the output voltage swing
    (vinmin-vos)*Gain-->(0.1mv-0.1mv)*50=0v, so we cannot observe the output voltage swing. so how we will be able to measure the current?

    From your schematic, we see that the NCP114AMX075TCG output sets the Vref to 0.75V, Vref=0.75V is assumed in the equations above. When Vref=0V, the output swing limitation is violated as you mention. If you were to change Vref to 0V, you would need to increase Rshunt > 10mOhm to avoid 'Swing to GND' limitation.