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LM26480: NPOR maximum current

Fred Archer1
Prodigy 10 points
Part Number: LM26480

Hello,

We have a design which requires a much stronger pull-up on the NPOR pin than the recommended 100k. We currently have a maximum of 6mA being sunk by this NPOR pin. I cannot see any maximum rated current for this pin in the datasheet. Can you advise what the maximum rated is?

Thanks,

Fred

  • 0
    TI__Genius 10885 points

    Hi Fred,

    This is likely not a characterized specification. 100k is likely a typical recommendation to ensure VOL is within expectations. Looping in the product expert to comment.

    Best Regards,

    Rick S.

  • 0
    TI__Mastermind 30310 points

    Fred,

    The VOL of the nPOR pin is specified with a load (IOL) of 500uA, on page 9 of the LM26480 datasheet.

    Using 0.5V as the maximum voltage at the nPOR pin, RDS-ON,max = VOL,max/IOL = 1kOhm. 

    Assuming 3.3V is the pull-up voltage, VPU, then RPU,min = (VPU - VOL,max)/IOL.

    The minimum resistance that will satisfy the equation is 5.6kOhm.

    For every IOL value, there will be a different VOL max value. The VOL spec gives a reference point, so you can ensure that nPOR will pull the line down below the VIL requirement of the input it is connected to at the other end. RDS-ON should be relatively constant, meaning you could plug in other values for IOL in and see what the VOL value will be. If this RDS-ON value holds true, then there is no way you are going to get 6mA through the open-drain FET and hold a low value of <0.5*VIO.

    I cannot visualize where this 6mA load is coming from, unless you are driving an LED directly with this pull-up resistor to indicate the buck regulators have turned on. If this is the case, then you might want to consider driving the gate of a FET to turn on the LED instead so that the 6mA comes from the input voltage (5V, for example) or the I/O voltage (3.3V, for example) but the current does not sink through the nPOR pin directly.