nRF9160 SRAM Partition changes

[gonzalezjl]

Hello,

I would like to modify the current default SRAM partition from 64k (secure), 64k (bsd), 128k (non-secure).  I would like to change it as follows; 8k (secure), 8k (bsd), 240k (non-secure).  The secure_boot application is only using around 4k now, so 8k is more than enough.  Will the bsd be ok at 8k? What other potential problems should I look out for?

Thanks,

JL

[Martin Lesund]

Hi JL,

Update 20.01.2020:

Check the updated BSDLIB documentation for the latest information.

"The secure domain memory is divided into segments of 8 KB RAM blocks. Because the size of the BSD library block is 0xC020 bytes, it is not aligned to full 8 KB blocks. It occupies six blocks and a small part of a seventh block.

The leftover memory in that seventh block (8160 bytes, starting at address 0x2001 c020) is configured as non-secure, but it can be used by the application."

---

Old answer:
The BSD Library has an absolute position and is fixed at starting address 64K and size of 64K (0x20010000 – 0x2001FFFF).

This cannot be moved.

However, all other partitions might be moved. 

Note: The BSD memory always has to be Non-Secure configured.

---

You can find the "Default flash planning"  in the file: [/ncs/zephyr/boards/arm/nrf9160_pca10090/nrf9160_pca10090_partition_conf.dts]

There it's also documented that "64 kB SRAM reserved for and used by the BSD socket library."

Best regards,
Martin L.

[gonzalezjl]

Hi Martin,

Is it possible then to have discontiguous Non-Secure areas in SRAM to be used by the application?

&sram0 {

reg = <0x20000000 DT_SIZE_K(8)>;

};

&sram0_ns {

reg = <0x20002000 DT_SIZE_K(56)>;

};

&sram0_bsd {

reg = <0x20010000 DT_SIZE_K(64)>;

};

&sram1_ns {

reg = <0x20020000 DT_SIZE_K(128)>;

};

Thanks,

JL