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LM70860-Q1: PG Pin Current Capability

Evan Wang
Genius 9760 points
Part Number: LM70860-Q1

Tool/software:

Hi Team,

In LM708x0-Q1 datasheet, we didn't mark the IPG sink current.

Although we recommend to use 100k ohm PU, I believe 10k ohm should be fine.

May I know the limitation of the PU we add on PG pin?

Like from 10k~100k ohm?

Thank you.

  • 0
    TI__Guru 56345 points

    Hi Evan,

    The spec in the datasheet shows that when 2mA of current flows into the PG pin, the max VOL voltage of 400mV is produced. See below:

    Depending on the pull-up voltage will determine the current flowing into the PG pin. Using a 10Kohm resistor, the current flowing in the PG pin at:

    3.3V --> 290uA [(3.3V-0.4V)/10kohms]

    5V -->  460uA [(5V-0.4V)/10kohms]

    12V --> 1.16mA [(12V-0.4V)/10kohms]

    Therefore, a 10Kohm pull-up resistor should suffice at the pull-up voltage at 3.3V, 5V, and 12V.

    Ben

  • +1 in reply to Ben C
    TI__Genius 9760 points

    Hi Ben,

    Thank you for the sharing.

    I think we can assume the PD resistance based on 0.4V/2mA = 200 ohm.

    So depends on the PU we use, the VOL-PG could be higher but it also depends on the threshold of the cascaded rail, like a GPIO of a MCU.

    But is there a maximum restriction of that current, like we can't flow over 10mA into PG pin?

    Thank you.

  • 0 in reply to Evan Wang
    TI__Guru 56345 points

    Hi Even,

    Yes, it depends on the threshold of the cascaded rail.

    I would keep the current below 10mA..

    Ben

  • 0 in reply to Ben C
    TI__Genius 9760 points

    Hi Ben,

    Got it.

    Thank you for the comment.