I want to upload the device firmware to a file using dfu-util. How can I determine the correct size of flash memory?
After booting the device into DFU it can be found using:
dfu-util -l
For which I receive the following information:
Found DFU: [0483:df11] ver=2200, devnum=8, cfg=1, intf=0, alt=1, name="@Option Bytes /0x1FFFF800/01*016 e", serial="FFFFFFFEFFFF"
Found DFU: [0483:df11] ver=2200, devnum=8, cfg=1, intf=0, alt=0, name="@Internal Flash /0x08000000/064*0002Kg", serial="FFFFFFFEFFFF"
To upload the flash configuration to a file I need to determine the size of flash memory. Based on this article the size would be 64 x 1kB of flash memory.
What is the meaning of 'Kg' in 0002Kg?
The instructions I am following (elsewhere, for a different device, see above) is using 128 x 1kB
, instead which I believe is incorrect.
How can I calculate the size of flash memory and what will happen if I set the memory size too large to download an image?
The command is:
dfu-util -a 0 -s 0x08000000:131072 -U ./original.bin
I think it should be
dfu-util -a 0 -s 0x08000000:65536 -U ./original.bin