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Original question:

TPS7A63-Q1: Question for EN pin, WD short pulse and falling edge timing

TPS7A63-Q1: watchdog strategy using close window

user1751165
Prodigy 40 points
Part Number: TPS7A63-Q1

Hello Team：

I have 2 questions about this IC

1. From the figure 22, the watchdog must receive service only duting open windows, MCU/DSP and this IC have their oscilliation frequencys respectively, so how to create a relationship between the two frequencys, to ensure the high electrical level from MCU/DSP is exactly in the period of open windows？

2. why this IC use this strategy of watchdog? what about the strategy that ignoring the service during the close window?

thanks!

  • 0
    TI__Guru 50190 points

    Hi,

    For question #1, the window should be set up so that the MCU/DSP can service it in the middle of the open window. The window should be set wide enough to account for variations in the MCU/DSP's oscillator and the variations in the code to service the watchdog. 

    For question #2, I think you are asking why use a Window Watchdog vs a normal watchdog timer?  A normal watchdog timer can force a reset if the MCU/DSP is locked up and no longer able to generate the service signal. A Window watchdog does that and also resets if the MCU/DSP is resetting it too often. Meaning maybe the code is stuck in a loop where all it is doing is servicing the watchdog. A window watchdog is preferred for mission-critical or safety applications. 

    Does this answer your questions?

  • 0 in reply to JCHK
    Prodigy 40 points

    Hi JCHK:

    you are right, if the MCU/DSP can service in the middle of the open window, but my question is how to assume the MCU/DSP can service in the middle of the open window? as you know, width of open window is decided by the external resitor Rosc, and resistor has tolerance, even if only slight gap between this width and the time DSP counted, as time goes by, the MCU/DSP will serve beyond the middle of open window.

  • 0 in reply to user1751165
    TI__Guru 50190 points

    Hi,

    You are correct in that you need to consider the resistor tolerance as well. The resistor tolerances and oscillator tolerances need to be taken into account so statistically, you can guarantee that the window is wide enough. 

    I hope this helps.

  • 0 in reply to JCHK
    Prodigy 40 points

    Hi JCHK:

    in my opinion, even if the window is wide enough, there is still mistriggered risk, the only way to avoid the risk is this IC sends feeback to MCU/DSP, then MCU/DSP knows exactly when is the open window or close window.

    thanks a lot for your reply!

  • 0 in reply to user1751165
    TI__Guru 50190 points

    HI,

    There is always this risk if the DCP/MCU code is such that the service timing is not well controlled. For example, if the MCU is performing 8 different tasks  (with one of them being the task of servicing the watchdog) during its normal operation if each of the tasks has the same timing, then it is possible to set a window where the code can service it precisely every time.  If this is not possible, you may want to consider using a standard watchdog instead of a window watchdog.

    I hope this helps.

  • 0 in reply to JCHK
    Prodigy 40 points

    Hi JCHK

    GOT it， thanks！