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TPS92611-Q1: TPS92611-Q1 - FAULT pin integrating with microcontroller logic (5V)

nikita alexandrov
Prodigy 60 points
Part Number: TPS92611-Q1

Hello,

I want to integrate TPS92611-Q1 with a microcontroller.  The absolute maximum voltage on any pin is 5.5V, whereas the voltage on the LED driver Fault pin is 6.5V in my application and up to 7V according to the datasheet.

What is the recommended way to interface the FAULT pin to a 5V logic if I am only using a single device and not a multi-device FAULT bus?

Apparently the 7V is due to a very weak pullup current, is it possible that this current is so weak I can interface it directly to a 5V logic?

Thank you for the help!

  • 0
    TI__Expert 5610 points

    Hi Nikita,

    You'd better not to interface it directly to a 5V logic. It will force the MCU pin to 7V when the FAULT presents high.

    If you just use MCU to monitor the FAULT pin, following is a solution. But please note that the logical is reversed.

    Thanks and best regards,

    Felix Wang

  • 0 in reply to Felix Wang
    Prodigy 60 points

    Thank you for the information.

    Since my MCU has an internal clamping diode to protect the pins, can I assume that the 7V pullup voltage on the FAULT pin will have a current lower than 1mA when it is in the high state? If that is the case then there should not be problems with damaging the pins even if the voltage exceeds 5V logic.   It is not clear to me the maximum expected current on a 7V high state, would it be I(FAULT_pullup) in the datasheet which would be 8uA typical?

    Would it be possible to use a diode to drop the voltage on FAULT going to the MCU logic? This will require a smaller component count for my needs.

    Would using a diode to drop 2V cause problems with the complex functionality of the FAULT pin? I am not using a FAULT bus, only using it to monitor open circuit and short conditions on the LED output.

    Another option might be to use a 10k resistor in series with the IO pin, therefore even if the voltage reached 7V and current flowed to the IO pin, the current would be below 1mA, which would not damage the IO pins due to the internal clamping diode?

    Could you please provide your thoughts on these two solutions for minimizing components?

    Thanks!

  • 0 in reply to nikita alexandrov
    TI__Expert 5610 points

    Hi Nikita,

    Yes, it would be I(FAULT_PULLUP) which would be 8uA typical.

    For your first solution, do you mean to use a 2V zener diode in series with the IO pin? If so, i think it would be ok, but you need to make sure the selected diode can provide 2V zener voltage with 5uA (FAULT minimum pullup current) zener current, and your MCU pin has the capability to source 1mA current when FAULT pulls down.

    For your second solution, there would be a problem when FAULT pulls down. The voltage on the MCU IO pin would equal to the voltage drop of the 10k resistor that is (pulldown current)*10k.

    As your MCU pin has an internal clamping diode, there would be another solution for your reference that is to connect MCU pin to FAULT directly just with a 10k pullup resistor to connect to MCU 5V logical.

    Thanks and best regards,

    Felix