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TPS92630-Q1: FAULT pin tied high?

Masatoshi Oguri
Expert 4780 points
Part Number: TPS92630-Q1

Hello,

In p.19 of the datasheet of TPS92630, a description goes:

When an entire string of LEDs is shorted (C as illustrated), the device pulls FAULT low to shut down all channels. With the FAULT pin tied high, only the faulted channel turns off.

But I wonder how a open-drain pin can be tied high. Even though FAULT pin is tied high, the device will pull it low anyway because it is open-drain. With a resister to pull FAULT to high, the device turns off all LED channels. How should I configure the pin to make the device only turns of the faulted channel?

Thank you.

Oguri (TIJ automotive FAE)

  • 0
    TI__Expert 5475 points

    Hi Oguri-san,

    The FAULT pin of TPS92630 is not an open-drain structure. It has internal strong pulldown current and weak pullup current. If connected to high with small pullup resistor, then the FAULT pin will still be high even with the pulldown current. In that case, only the faulty channel will be turned off.

    Best regards,

    Shirley He

  • 0 in reply to Shirley He
    TI__Expert 4780 points
    Hi Shirley,

    Thanks.
    Would you please suggest a recommended register value for the pullup?

    Please correct the wrong description in the p.15 of the datasheet as "Both fault pins are open-drain transistors
    with a weak internal pullup.".

    Regards,
    Oguri
  • 0 in reply to Masatoshi Oguri
    TI__Expert 5475 points
    Hi Oguri-san,

    In order to realize the one-fails-others-on, connect FAULT pin directly to a high voltage is ok. Or, when choosing the pullup resistance, ensure that when there's a fault, with the internal pulldown current, FAULT pin voltage is still higher the input high threshold (2V).

    You are right that the description of fault pin in the datasheet is not very proper. Thanks for pointing it out. We'll revise that in our next datasheet revision.

    Best Regards,
    Shirley He