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INA226: INA226: Choice of Shunt Resistance and configuration for 12V and 8-10A Max

Paolo Gaggero
Prodigy 80 points
Part Number: INA226
Other Parts Discussed in Thread: INA219

Good morning,
I have some questions about the INA226 sensor.

I also bought the INA219 but I read that as far as precision is concerned, the IN222 is better, so I bought that too.
The use of this module is for a design of a bench power supply with different fixed output voltages coming from 3 DC-DC converter stabilizers
(Vout1 = 12V
Vout2 = 5V
Vout3 = 5V (USB)
Vout4 = 3.3V).


Through 4 INA226 modules connected to Arduino I will read on a monitor the values ​​of CURRENT, VOLTAGE and POWER (V x I) used by the different loads connected on the outputs.


The INA226 modules I bought have a R100 shunt resistance or a 0.1 ohm.

The values ​​to read want them to be

- for VOLTAGES
12.00 V
5.00 V
3.30 V
with a precision of 2 decimal places (for example : 12,34 V ore 3,36 V )

-for the CURRENTS of each of the 3 I want that the values ​​are read at 3 decimal places with an error of 1 or 1.5 mA and maximum current of OUTPUT 8-10 A for each channel (for example : 7,685 A or 0,634 A )


1) Which configurations should I follow for the module?
2) Is the installed shunt resistance (0.1 ohm) ok for my purposes or will it be changed? If yes with which one is it and how many Watts?

Many thanks in advance
If you need more information or if you have any doubts ask me.

  • 0
    TI__Mastermind 30910 points

    Hi Paolo,

    Before recommending a resistor value, we need to know the current range for each of the rails. You can learn more about this topic from this video.

    By “configurations” I assume you’re referring to the ADC settings. This is an optimization process that takes into account accuracy and speed. You can read more about this topic in datasheet section 7.4.

    For further information on selecting shunt and setting Configuration register in order to receive valid Current and Power readings, you can refer to this video.

    Regards, Guang

  • 0 in reply to Guang Zhou
    Prodigy 80 points

    Hi Guang Zhou.

    In the meantime I tried to understand something about the INA226 sensor and I would like to see if what I intended is correct.

    All 4 INA226 sensors want them to have a maximum current of 8A; so :


    - the INA226 with 3.3V input voltage will have a maximum current of 8A
    - the INA226 with 5V input voltage will have a maximum current of 8A
    - the INA226 with 12V input voltage will have a maximum current of 8A
    - and for simplicity (even if for this it would be much less enough) the IN2222 with 5V USB input voltage will have a maximum current of 8A

    And the current of all the sensors I want to have a resolution of at least 1mA (if it is lower then 0.3mA or 0.2mA even better)

    My calculations are the following (correct me if I'm wrong at some point):

    MAX CURRENT CHOOSEN ( corresponding to the MAX input voltage ±81.92mV )                                  8 A            
    MIN CURRENT CHOOSEN ( i choose using it in UNIDIRECTIONAL MODE )                                                  0 A            
    MAX input voltage read on R Shunt                  81,92 mV            
    MIN voltage on Rshunt ( VShunt_LSB )                     2,5 uV            
    Rshunt ( =  MAX V Rshunt / MAX CURRENT Rshunt)                    10 mΩ        
    CURRENT Resolution Rshunt (= Vshunt_LSB / Rshunt)     0,000244 mA  ~  0,2 mA         
    Power Dissipation Rshunt                     0,65 W        

    So if I understand correctly, through the sensor I will be able to read up to 8 A with a resolution of 0,2 mA (so I would say that I am largely within my expectations that they were having precise mA current measurements) and with a dissipation at 8 A of 0,65 W.

    All this by replacing the resistance with a 2512 of 10 mΩ at 1% and 2 W power, correct?

    It can therefore also use a current of only 100 mA and in this case it will only change the power dissipated by the Shunt Resistance which will be

    P = R x I ^ 2 = 0.01 * (0.1 A) ^ 2 = 0.0001 W = 0.1 mW

    How do you think about it?

  • 0 in reply to Paolo Gaggero
    TI__Mastermind 30910 points

    Hi Paolo,

    Your analysis is pretty much spot-on, well done!

    I want to bring up a couple of points to supplement your analysis and for you to consider as you move along in the project:

    (1)   Accuracy is different from resolution – Yes based on current selection of Rshunt, the current resolution is 0.25mA. However the accuracy will be dominated by the 1% Rshunt tolerance. This error can be eliminated with calibration however.

    (2)   Looks like you’re going after accuracy, therefore you may want to configure the ADC to take advantage of averaging. This will improve noise performance. Please refer to figure 20 in section 7.4 for demonstration of the importance of averaging.

    Regards, Guang