[参考译文] UCD3138064：使用 MFBALR17保护第二个32KB 程序闪存

https://e2e.ti.com/support/power-management-group/power-management/f/power-management-forum/1118591/ucd3138064-protecting-second-32kb-of-program-flash-using-mfbalr17

大家好、 我的程序将所有 64KB 用于代码。 我想为 两个闪存组启用 MFBALRx 只读保护

闪存组0很简单：将 TI 示例代码添加到 ASM 加载程序中

LDR		r4,c_mfbalr1_half0 ;point r4 at program flash base address register
MOV		r0,#0x62 ;make block size 32K, address 0, read only
STRH	r0,[r4]; store it there
.
.
.
c_mfbalr1_half0  .long 0xFFFFFE0E 	;low 16 bits of memory fine bank address low - FLASH#0

但我遇到了闪存组1的问题：当我在加载器中添加略微修改的代码时、CPU 会在执行期间复位

LDR		r4,c_mfbalr1_half0 ;point r4 at program flash base address register
MOV		r0,#0x62 ;make block size 32K, address 0, read only
STRH	r0,[r4]; store it there

LDR		r4,c_mfbalr17_half0 ;point r4 at program flash base address register
MOV		r0,#0x62 ;make block size 32K, address 0, read only
STRH	r0,[r4]; store it there
.
.
.
c_mfbalr1_half0  .long 0xFFFFFE0E 	;low 16 bits of memory fine bank address low - FLASH#0
c_mfbalr17_half0 .long 0xFFFFFE8E 	;low 16 bits of memory fine bank address low - FLASH#1

在从 函数 zero_out_integration_word 到 MFBALR17的任何写入过程中，也会重新启动 CPU (DecRegs.MFBALR17.all = MFBALRX_BYTE0_BLOCK_SIZE_32K；)

 MFBALR17是否 受到特别的写保护？