i.mx28 u-boot dram configuration for dram replacement.

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i.mx28 u-boot dram configuration for dram replacement.

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RandyGraham
Contributor IV

Hello,

We have a custom i.mx28 board based on the mx28-evk reference design.

My hardware guy tells me that the dram chip used on the m28-evk board is no longer available and we need to use the micron equivalent, http://www.micron.com/~/media/Documents/Products/Data Sheet/DRAM/512MbDDR2.pdf, which is what we have on our custom board .

We are having issues getting u-boot to come up on our board, so I was wondering, should I need to adjust the dram configuration settings for this memory replacement ?

I'd assumed not since it should be functionally equivalent.

u-boot appears to initialize power, clocks, etc fine from SRAM but resets when relocating and trying to run from DRAM which is why I think we have a dram issue.

All of our voltages look correct.

Hopefully I am not asking a silly question :smileyhappy:

-Randy

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729 次查看
Yuri
NXP Employee
NXP Employee

I.MX28 DRAM system (under Linux) is initialized in bootlets, and strictly speaking

for new memory it is needed to provide some adaptation.

Source codes are located in

<ltib>/rpm/BUILD/imx-bootlets-src/boot_prep/init-mx28.c 

To get it, please use the next commands 

./ltib -m prep -p imx-bootlets-src

After editing :

./ltib -m scbuild -p imx-bootlets-src

./ltib -m scdeploy -p imx-bootlets-src

Next, please use the tool, linked below, to get optimal settings for specific DRAM :

Board_bring-up_and_DDR_initialization_tools :

https://community.freescale.com/docs/DOC-1455

https://community.freescale.com/wiki/images/1/15/MX28_DRAM_controller_register_programming_aid_v4.tar.gz

And finally, based on fact that both memory chips (old and new) are practically the same - perhaps
it makes sense to test memory (to avoid design or manufacture issues).

729 次查看
RandyGraham
Contributor IV

Thanks Yuri !

I am not using LTIB, but rather building the mainline u-boot directly.

Thanks for the doc and programming aid, I'll take a look.

-Randy

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