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LM26480: Value for nPOR pull-up resistor

Alexey Poleschuk
Prodigy 50 points
Part Number: LM26480

Hi!

I'm planning to use nPOR signal from this converter as system-wide reset on my board. The datasheet specifies value for a pull-up resistor on this pin as 100kOhm.

What is the smallest possible value of the pull-up resistor? I think that 100kOhm pull-up is too weak for my board.

--

Best regards,

Alex

  • 0
    TI__Genius 10885 points
    Hi Alex,

    It looks like the test conditions on VOL for nPOR indicate the open drain supporting a 500uA load. The minimum pull-up value would depend on the pull-up voltage.

    Best Regards,
    Rick S.
  • 0
    TI__Mastermind 30280 points
    Alex,

    I worked backwards through this question and essentially the VOL spec for nPOR is saying the RDS,on of the FET pulling low is worst-case 1kOhm (R = V/I = 0.5V/500uA)

    100kOhm will guarantee that the output voltage is very low, because the voltage divider of any logic-level voltage (1.8V, 2.5V, 3.3V) will produce a VOL of <50mV. But as you said, you think the pull-up is too weak (maybe because of a slow rise time?)

    I conclude that 10kOhm will work for any logic voltage, but the true minimum resistance would be:

    At 3.3V logic: (Vio - 0.5V) = IOL * Rpu,min --> Rpu,min = (Vio - 0.5V)/IOL = (3.3V - 0.5V)/500uA = 5.6kOhm
    At 1.8V logic: (Vio - 0.5V) = IOL * Rpu,min --> Rpu,min = (Vio - 0.5V)/IOL = (1.8V - 0.5V)/500uA = 2.6kOhm
  • 0 in reply to Brian Berner
    Prodigy 50 points
    Thank you very much, Brian!