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I have some ARM processor that have 72-bit width (8 for ECC) RAM controller.

  1. Can I buy eg. two 32bit chips and combine them? If yes then how?
  2. Can I buy just one 32bit and pull down rest of data pins on processor?

Processor brief if needed: https://docs.broadcom.com/docs/1211168571391

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Can I buy eg. two 32bit chips and combine them?

I imagine so. The product brief describes ECC as "optional", which indicates to me that it can operate with non-ECC memory:

The memory system supports the latest DDR4 memories in addition to DDR3 and DDR3L. Optional ECC support is provided as well.

In practice, though, this part appears to be designed to operate with very large quantities of memory. (For instance, it advertises "address space beyond 4 GB".) Using it with the bare minimum of memory seems like it'd be a waste of its capabilities.

If yes then how?

Contact the vendor for a full datasheet. Keep in mind that Broadcom typically only works with large customers; if you just have a few of these parts from a gray-market source, you are probably out of luck.

Can I buy just one 32bit and pull down rest of data pins on processor?

You'd have to ask the vendor, but it seems unlikely.

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  • \$\begingroup\$ Thanks for Your response, I found retail seller for these processors, so I won't talk with Broadcom (excluding Datasheet). \$\endgroup\$ Commented Oct 25, 2018 at 20:35
  • \$\begingroup\$ That's not going to work. You will need the datasheet to do anything with this part, and Broadcom will not provide that datasheet to an individual, or to a company which does not intend to buy their parts. \$\endgroup\$ Commented Oct 25, 2018 at 20:39
  • \$\begingroup\$ Actually we are small firm, and w have some investors, so we will see how Broadcom will behave. \$\endgroup\$ Commented Oct 25, 2018 at 20:46
  • \$\begingroup\$ Many ARM SOC chips will work with narrower memory than the chip supports (e.g. 32 bit memory on a 64 bit bus) but this requires that the SOC memory controller be so configured. The Broadcom documents should tell you. \$\endgroup\$ Commented Oct 25, 2018 at 23:29

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