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Flash Space calculations STM32L496RGT6

mm.51
Associate II

Looking into the available flash space on the chip:

For the 1Mbyte variant of the device the total number of bytes should be 1,024,000.  

2,000 bytes * 512page = 1,024,000 bytes of flash memory.

End of Address Space - Start of address space  = Amount of total space

 0x080F FFFF -  0x0800 0000 = 0xF FFFF bytes = 0d1,048,575 bytes

Am I incorrectly calculation incorrect ? If not where's the missing 24,575 ?

 

Thanks

1 ACCEPTED SOLUTION

Accepted Solutions
TDK
Guru

1MB = 1024^2 bytes

The flash length is 0x10 0000, which is 1024 * 1024 = 1048576 bytes = 1 MB.

 

If you feel a post has answered your question, please click "Accept as Solution".

View solution in original post

1 REPLY 1
TDK
Guru

1MB = 1024^2 bytes

The flash length is 0x10 0000, which is 1024 * 1024 = 1048576 bytes = 1 MB.

 

If you feel a post has answered your question, please click "Accept as Solution".