I was recently out to buy a 4 GB RAM chip for my laptop, and all they asked me was whether I wanted DDR-3? When I specified that I had a 64 bit system, they said that it's not important, my question is why? Shouldn't my system's word processing capabilities be a factor? Isn't memory designed into registers the size of the system's word length???

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  • I removed the tag "embedded" since it has nothing to do with that (OP says its for a laptop). Also voting to close for the same reason. – tcrosley May 6 '14 at 15:43
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    In fact, it is routine for memory interfaces to be wider than the processor word (for filling caches, or for using parallelism to deal with slow memory technologies), or also for them to be narrower (to simplify circuitry). Also, today 32 vs 64 bit is as likely to refer to the mode of operation utilized by an operating system as to the actual processor word width. – Chris Stratton May 6 '14 at 15:58
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    would it surprise you to find out that almost all hardware couldn't care less how long the data unit we define as a "word" is? that a single is 32bits on x86, and 64bits on x64? the CPU itself is pretty much all that cares. – Frank Thomas May 6 '14 at 17:03
  • @Frank Thomas: Yes it would surprise me, as I'm pretty new to system architecture, just moved here from digital electronics and transistor level designing, currently pursuing register transfer logic. Getting back on topic, what about the cache local to the processor? Or the other general purpose registers, special registers, accumulator? Are their word sizes chosen without any restrictions either, and if this is really so, how is a 64bit system any different from a 32 bit system? – ubuntu_noob May 6 '14 at 18:03
  • All those components are part of the CPU itself, so a CPU with an x64 instruction set would be written to accomodate 64b words. Ram however is grouped logically (not physically) so the Operating Systems memory management drivers handle all the abstraction involved in giving data in ram some kind of meaning. – Frank Thomas May 6 '14 at 18:22

No. Modern PC systems contain a memory interface chip or circuit that translates between the data bus width and the memory module width, the latter of which has been standardized by organizations such as JEDEC. This means that modules that use the same signaling and protocol are interchangeable regardless of the underlying architecture of the system.


The memory modules are standardized and have not been 32 bit in a while. even with a 32 bit processor there are layers of caching and on the dram side of the cache it doesnt matter what the processor prefers you want to read and write in multiples of that size so 64 bit is not a bad tradeoff. Likewise the (some, all, or latter) 32 bit processors adopted 64 bit busses anyway for similar reasons.

And if you look at the modules you either have a multiple of 8 chips (8 or 16) or a multiple of 9 (9 or 18) or 5,10 or 4,8

the chips often start 8 bits wide then over time can accomodate 16 bits wide for that speed. so you start off with 64 bits no ecc which is 64 bits per rank 8 parts per rank, one rank memories have one row of 8, two rank, 8 on one side 8 on the other. with ecc you get another 8 bits 72 bits, 9 parts per rank. for 16 bit wide memories 4 parts per rank no ecc and 5 with ecc.

But once you specify the connector/module type the choices become quite limited, how much ram and then depending on the brand/model you get whatever number of parts to implement that.

So for buying individual parts they need to know DDR, DDR2, DDR3 or DDR4, what speed, and density you want. The memory controller is specific to each and may not necessarily support anything but the one type, so a ddr2 may only support ddr2 and a ddr3 may only support ddr3, so you have to first start with that question, then move on into speed and with and density.

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