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MaxVer
Level 1
Level 1
First like received 5 sign-ins First reply posted

Dear,

I don't understand the evaluation sequence in "3.1.1.7.7 Debug System handling".

If there is no external tool request debug access by writing CMD_KEY_EXCHANGE into COMDATA register, the SSW will write into COMDATA a 32-bit value UNIQUE_CHIP_ID_32BIT, so what's the purpose of "write UNIQUE_CHIP_ID_32BIT into COMDATA" ?

And if the flash read protection is not activated, the debug interface will be unlocked!!! I don't understand why the debug interface will be unlocked even without any authentication, and from my point, although the UCB_PFlash is set with flash read protection, the read protection will also be not activated if the boot mode is internal start.

1 Solution
Yuva
Moderator
Moderator
Moderator
250 replies posted 250 sign-ins 100 solutions authored

Hello,

Though the external tool didn't request debug access by writing CMD_KEY_EXCHANGE into the COMDATA register, it can be connected at a later stage.

There are also other conditions evaluated when flash read protection is not activated before unlocking as mentioned in the note.

I suggest to refer more details in the AP32399 application note, you can access this by requesting in MyICP.

Thanks.

 

View solution in original post

1 Reply
Yuva
Moderator
Moderator
Moderator
250 replies posted 250 sign-ins 100 solutions authored

Hello,

Though the external tool didn't request debug access by writing CMD_KEY_EXCHANGE into the COMDATA register, it can be connected at a later stage.

There are also other conditions evaluated when flash read protection is not activated before unlocking as mentioned in the note.

I suggest to refer more details in the AP32399 application note, you can access this by requesting in MyICP.

Thanks.