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INA3221: I2C Level Shift?

Part Number: INA3221
Other Parts Discussed in Thread: PCA9306, TXS0102, , INA233

Hello team, 

Can INA3221 be directly tied to a 1.8V I2C bus if the supply voltage for INA3221 is 3.3V? I would think not and a level shifter like TXS0102 or PCA9306 is needed, but wanted to make sure for customer design.

Thanks!

  • Hello Sepeedah,

    In order to determine the feasibility of this scenario, I would direct you to the "Digital Input/Output" section of the Electrical Characteristics specification table. You will see that it shows the high-level input voltage to be 0.7*Vs and low-level input voltage to be 0.3*Vs, which equate to 2.31V and 0.99V when Vs is 3.3V.

    So this would not work since the 1.8V bus won't exceed 2.31V and clock edges won't be detected. We do have power monitors that can operate on bus voltage levels independent of the supply such as the INA233, but INA3221 is the only three channel device.

    Anyway using the level shift will be necessary for the device to operate on the 1.8V rail with 3.3V supply.

    Sincerely,

    Peter

  • Thank you Peter for the confirmation! Much appreciated. 

    A follow on question, I see the datasheet calls out a hysteresis voltage of 500mV on the same table, how is that to be interpreted? Is it a 500mV window on VIH and VIL thresholds basically?

    Thanks!

  • Hey Sepeedah,

    I don't believe so. As long as the digital input voltage satisfies the VIH and VIL specification ranges, then the device will always read a HIGH or LOW respectively.

    The hysteresis specification should apply to situations where the digital signal is leaving its current state. So if voltage is HIGH at 3V, then as it is going low the device won't stop seeing HIGH until voltage reaches 2.5V, potentially as low as (VIH_minimum - 0.5V). It always more robust to consider the worst-case which would be the minimum of VIH and maximum of VIL. So state might not change from HIGH until (0.7*Vs - 0.5V) and might not change from LOW until (0.3*Vs + 0.5V).

    Hope this helps.

    Best,

    Peter