MSPM0L1304: JTAG Communication after downloading

Hi,

I found a new phenomenon.

MSPM0 always goes into default_handler.

Hi SY,

Did you test based on the example code?

Which version of SDK are you using?

Which IDE and debugger are you using? 

Thanks

Yuhao

Hi, 

Did you test based on the example code?

-> Yes, I tested simply example like gpio toggle.

Which version of SDK are you using?

-> SDK version is 1.00.01.03

Which IDE and debugger are you using?

-> I used CCS and CCS theia with XDS110

The MSPM0 is production version on board.

I suspect that there is something wrong with your environment configuration.

Are you using the CCS12.3? Did you test the launchpad before?

I tested using the CCS12.3.

I don't have launchPad Rev A 

So I tested with launchPad Rev 1.0. 

This is an early version, but it worked.

How to determine which interrput triggered default_handler()?

We need to figure out what caused the default handler to run.

I am afraid that CCS12.3 and SDK1.00.01.03 is not suitable for the early samples.

Even if the programming does not report an error, any unknown error may occur on the chip.

I know that, I just test on early samples becuase I don't have prodcution sample. 

I am testing with a production sample and released SDK.

Do you know why MSPM0 run the Default_Handler()?

When is the default function executed?

How do I check the interrupt source?

As the comments indicate, when the processor receives an unexpected interrupt, it enters the default handler.

You can find the default handler in startup_mspm0g3507_ticlang.c.

If you want to check the interrupt, you can modified the startup_mspm0g3507_ticlang.c to add more handlers.

Or you can check the interrupt flag.

Where can I check the interrput flag?

I'm looking for interrupt flag register, but I couldn't find it.

You can refer to the register in TRM.

For example 2.6.1 SYSCTL Register

Hi,

As soon as mspm0 starts, it enters HardFault_Handler()

Is there any way to determine what is causing the hard fault?

Hi,

I solved this problem and thank you for your help.