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TPS27082L: TPS27082L Vout only 75% of Vin

Brian Han
Intellectual 330 points
Part Number: TPS27082L
Other Parts Discussed in Thread: TPS22918

Hello,

I'm using TPS27082 Load switch for one of my design and this is for both power sequence and reducing leakage.

My design is like the picture below.

When the load switch is on, everything works well. There is no problem.

But when the load switch is off and measure the Vout point where the BD4 is, it measures around 2.2~2.8V. This voltage varies by board.

If my understanding is correct, doesn't the Vout has to be near 0V when the switch is off?

+3.3V is always on suppy by PWM stepdown converter and I have checked the MCU's on/off pin and saw it works.

Or is it normal to be able to measure little voltage for load swtiches?

Thanks

  • 0
    TI__Genius 15740 points

    Hi Brian,

    When the TPS27082 is turned off, it does not have any circuitry for pulling the output down to 0V. Because the output is left floating, the load on the output of the switch would be responsible for pulling the voltage down.

    Can you use a load switch like the TPS22918? It is a more integrated solution and comes with a QOD pin that you can connect to the output to help drive the voltage down to 0V.

    Thanks,

    Alek Kaknevicius

  • 0 in reply to Aleksandras_Kaknevicius
    Intellectual 330 points

    Hello Alek,

    I missed the notification email, so I just saw this.

    I think we might be able to replace the load switch to TPS22918 for the future production with minor change, since the package is the same.

    But we have 400 SMT done boards that load switch can't be easily replaced.

    Is there a temporary work around using DIP type parts that I can do to make the output voltage 0V?

    Thank You

  • 0 in reply to Brian Han
    TI__Genius 15740 points

    Hi Brian,

    The simplest solution would be adding a resistor to the output. While this causes a constant current drain when the device is on, it will guarantee 0V on the output when the switch is off.

    Another solution would be to invert the ON signal and pass it to an NMOS transistor which would add the resistor to ground on the output.

    Thanks,

    Alek Kaknevicius

  • 0 in reply to Aleksandras_Kaknevicius
    Intellectual 330 points

    Hello Alek,

    Thanks for the suggestion I will check the best solution now.

    Best regards,

    - Brian