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SN74LVTH125: how to determine the value of pullup resister for /OE pin?

TSUTOMU HATTORI
Prodigy 90 points
Part Number: SN74LVTH125

Dear support team,

On the datasheet of the device, it's said that

 "/OE should be tied to VCC through a pullup resistor; the minimum value of the resistor is determined by the current-sinking capability of the driver."

 Does "the current-sinking capability of the driver" mean IOH(High-level output current) or IOL(Low-level output current)? or anythnig else?

When I determine the value of pullup resister for /OE pin, I guess the current-sinking capability of the driver means some parameter related to /OE pin,

but I couldn't find the appropriate value.

Could you please give me some advice for this issue?

  • 0
    TI__Mastermind 42600 points
    Hi,

    You can use a 10kohm for this and it'll work fine. The driver that is mentioned in the datasheet is for the drive controlling the OE pin or driving it High or Low. Whatever maximum current it can sink when it drives Low should help you determine the resistor size. For example: the device can only sink 4 mA and the pull up supply is 5 V, then you can use Ohms law to determine the value of the resistor needed: R = 5V / 4mA
  • 0 in reply to Dylan Hubbard
    Prodigy 90 points
    Dear Dylan,

    Thank you for your quick reply and good advice.
    I got it.

    I came to know that I hadn't understood that "the driver" mentioned in the datasheet is an external device of SN74LVTH125 to drive /OE pin.
    It was my misunderstanding.

    Thank you for your help.