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INA190: CMRR dB calculation for current-sense amplifiers

Vic L
Prodigy 10 points
Part Number: INA190

Tool/software:

I have not able to find any methods of calculating CMRR for current-sense amplifiers online, so please tell me if my way is incorrect.

To my understanding, a current-sense amplifier operates measures the voltage drop across a shunt resistor between the two input terminals, like so.

Also to my understanding, CMRR is defined as (Differential gain/Common-mode gain) and measured with the amplifier terminals connected to a common-mode voltage. As you can see, the shunt resistor is bypassed.

DC transfer characteristics for IS1 = 0 to 4A

V_CM = VIN/2 = 36/2 = 18V

A_CM = VOUT/VCM = 369.15u/18 = 20.51u

CMRR = 20log(A_d/A_CM) = 20log(50/20.51u) = 127.74 dB

which is rather close to the minimum CMRR of 132 dB specified for INA190.

  • 0
    TI__Genius 17260 points

    Hello, 

    Thank you for your post. 

    Just so you have it, I am providing the link to the video explaining how CMRR is measured for CSAs: Precision labs series: Current sense amplifiers | TI.com

    You did manage to simulate it correctly, although you can use a voltage signal generator without the current source and shunt to simplify the simulation.  

    Here are some equations for CMRR below that I hope you find useful, but you already got the AD/ACM!

    I hope this helps, 

    Joe

  • 0 in reply to Joe Vanacore
    Prodigy 10 points

    thanks for answering, Joe!

    My follow up question: In the actual application where the input terminals are not shorted, does the shunt resistor being there affect the actual CMRR? Otherwise, it seems like the CMRR does not taken into account at all the resistance between the Vin+ and Vin- terminal.

  • 0 in reply to Vic L
    TI__Genius 17260 points

    Hi Vic, 

    In the actual application, the common mode voltage of each pin will differ slightly due to the current flowing through the shunt. 

    This primarily has an impact on the DC component of the common mode voltage. 

    Whatever AC component each input sees, you can apply the CMRR attenuation to each pin as the AC components will be slightly different due to this voltage drop.

    In reality, you can consider the CMRR for each pin independently. 

    I hope this makes sense, 

    Joe