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Original question:

INA210: Measurement of voltage and current

INA210: power off and on for testing

Silvia Luevano Cerda
Prodigy 60 points
Part Number: INA210

Hello, 

The configuration I am reading is a sleepmode for a PCB circuit, but the normal operation is like 700 mA , I only want to use it on sleep current so the 100u that I am interested, so my question is if I dont power on the INA210 I should use the same configuration without damaging the component?

thanks in advance.

  • 0
    TI__Mastermind 26660 points
    Hello Silvia,

    Thank you very much for posting to the forum.

    If the INA210 is not powered then it should not break as long as the Absolute Maximum conditions (Table 6.1) are not violated. I am unsure about what your schematic values are, but if the load current is 700mA and your shunt resistor is 20Ω, then the shunt voltage drop is 14V and this will be the input differential voltage for INA210 when system is not in sleep mode. This will not break the part whether it is powered or not since 14V < 26V. However, this means your shunt resistor needs a power rating of 14V*0.7A = 9.8W.

    Hope this helps and post back if you have more questions.

    Sincerely,
    Peter Iliya
    Current Sensing Applications