AnsweredAssumed Answered

STM32F427VI boot from bank 2

Question asked by brown.larry on Jan 24, 2016
Latest reply on Jan 24, 2016 by Clive One

     I've read the documentation and searched the forum and internet, but still having trouble with booting from the flash bank2. I have BOOT0 = 0, BOOT1 = 0, BFB2 = 0. I have programmed an image at 0x8000000 and a slightly different image at 0x8100000. I have read back the binary and confirmed that they are loaded correctly. Reading AN2606 Fig 37 indicates that this should work, the application should test bank2 and then boot from there. But the application always boots to the version at bank1, 0x8000000.

     If I set BFB2 to 1 or Boot0 to a 1, the device does not boot at all. 

     I am confused on a few fronts. AN2606 discusses remapping the flash from 0x8100000 to 0x0, indicating that the image for bank2 is built exactly the same was as for bank1. However, RM0090 in the discussion on BFB2 discusses changing the interrupt vector table. Seems contradictory. I am also confused how BFB2 and BOOT0/1 interact. And finally, how does SYSCFG_MEMRMP play into all of this?

     Any advice would be appreciated.

Thanks, Larry