MPC5748G question about boot vector

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MPC5748G question about boot vector

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kmh48301
Contributor IV

Hello.

I need to know the meaning of Boot vector, BAF code entry/exit point.

kmh48301_1-1610605875903.png

 

So I Dumped the memory of that address

kmh48301_2-1610606470058.png

kmh48301_5-1610606740282.png

0040_4100 ~ 0040_57D0 is reserved space in memory map.

So I guess this dumped contents is generated by internally

for boot process. Is my thought correct?

-------------------------------------------------------------------------------

But 400b_fd80 is SRAM region.

kmh48301_4-1610606559139.png

I have no idea for this..

 

==========================================

kmh48301_0-1610605793467.png

And when "Exit from STANDBY/LPU_STANDBY to DRUN" ,

Does this mean that i can execute my own boot code?

 

Thanks

Best regards

phillip

 

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petervlna
NXP TechSupport
NXP TechSupport

Hello,

But, what is code exit point in it..?

it says code exit point is 400B_FD80.

But as you can see in my dump, the last point of BAF code is 0x0040_57D0.

And 400B_FD80 is SRAM region.

It means if NXP programmed something in here, It can't be reserved (SRAM is volatile).

1. Is my understanding about 'BAF code entry point' is correct?

BAF entry point is 0040_4100 as you can see after destructive reset.

 

2. Please explain more about 'BAF code exit point'

once BAF finish execution the device will boot at the address specified by user in boot header.

petervlna_0-1610961139461.png

petervlna_1-1610961182187.png

please refer to the reference manual :

57.1.3 Search boot header and boot options

 

best regards,

Peter

 

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petervlna
NXP TechSupport
NXP TechSupport

Hello,

0040_4100 ~ 0040_57D0 is reserved space in memory map.

So I guess this dumped contents is generated by internally for boot process. Is my thought correct?

The BAF is programmed in NXP factory, and as I know for this device is OTP in latest cut.

petervlna_0-1610708398067.png

I do not understand your question. Could you be more specific?

 

And when "Exit from STANDBY/LPU_STANDBY to DRUN" ,

Does this mean that i can execute my own boot code?

are you referring to SRAM boot after standby exit? Or the flash boot?

In these cases the BAF is not executed as it was already executed after power on reset and the device is configured. After wake up it will start from boot vector in flash or SRAM.

best regards,

Peter

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kmh48301
Contributor IV

Hello. Thank you for explanation.

 

Sorry for my lack of good English.

 

"The BAF is programmed in NXP factory, and as I know for this device is OTP in latest cut."

 

The Dump memory : 0x0040_4100 ~ 0x0040_57D0 => This contents are Programmed in NXP factory.

=> this contents are maybe boot code or something for boot.

So In RM(page 2839), it says 0040_4100 is BAF code entry point.

 

But, what is code exit point in it..?

it says code exit point is 400B_FD80.

 

But as you can see in my dump, the last point of BAF code is 0x0040_57D0.

 

And 400B_FD80 is SRAM region.

It means if NXP programmed something in here, It can't be reserved (SRAM is volatile).

 

1. Is my understanding about 'BAF code entry point' is correct?

2. Please explain more about 'BAF code exit point'

 

 

Thanks

Best regards

Phillip

 

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734 Views
petervlna
NXP TechSupport
NXP TechSupport

Hello,

But, what is code exit point in it..?

it says code exit point is 400B_FD80.

But as you can see in my dump, the last point of BAF code is 0x0040_57D0.

And 400B_FD80 is SRAM region.

It means if NXP programmed something in here, It can't be reserved (SRAM is volatile).

1. Is my understanding about 'BAF code entry point' is correct?

BAF entry point is 0040_4100 as you can see after destructive reset.

 

2. Please explain more about 'BAF code exit point'

once BAF finish execution the device will boot at the address specified by user in boot header.

petervlna_0-1610961139461.png

petervlna_1-1610961182187.png

please refer to the reference manual :

57.1.3 Search boot header and boot options

 

best regards,

Peter

 

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