S32M244 WDOG fast test question

[WendyWang]

Hi ,

      I done a WDOG fast test, The Watchdog Control and Status Register (CS)'s value be setting 0x25a8(TST=01), But after watchdog reset(RCM->SRS shows 0x20) it alway shows 0x2520(TST=00). Seems the TST be cleared.

      In the S32M24x Reference Manual, Any other reset except POR does not affect the value of the TST field.  

Q：Why it different as the reference manual?   

[danielmartynek]

Hi @WendyWang,

Are you certain that the startup code is not reconfiguring the WDOG?

Here is an example for your reference:
https://community.nxp.com/t5/S32K-Knowledge-Base/Example-S32K116-WDOG-Fast-Test/ta-p/1103420

Regards,

Daniel

[WendyWang]

Hi Daniel,

My project base on the RTD( SW32K1_S32M24X_RTD_R21-11_3.0.0_D2503), and I have reference the example code, just change it to RTD version(let WDOG_IP_ENABLE_TEST_MODE to STD_ON) and before the test init the WDOG (because i found after power on the WDOG is disable, but  I read the CS value before the init code then base it to do the test). I'm sure i did not add the startup code to reconfiguring the WDOG.

And I try read the CS's value in the "void SystemInit(void)" , it always show 0x2520 too. According the S32M24x Reference Manual, it should be 0x2580.  But I cannot found any code reconfiguring the WDOG before the "void SystemInit(void)" .

[danielmartynek · Accepted solution]

HI @WendyWang,

In the startup code of the project, you might have this:

That would explain the behavior.

BR, Daniel

[WendyWang]

Ok, I see .

Is that means for the WDOG fast test I need del this code?

Thanks

[danielmartynek]

Yes, but remember that the WDOG is enabled by default and needs to be serviced or reconfigured before its timeout period expires.

Regards,

Daniel