I would like to know the Erase Block size for the Intel SSD 320 series.
Of course I need it to be able properly align partitions, in OS X.
Is it 128k or 512k or something else?
Pls. only confirmed sources, not guessing, and not saying, it does not need to be aligned.
For aligned partitions, 4KByte (4096 bytes) alignment is correct and is what you should use for partition alignment. Many docs online recommend skipping the first 128 "sectors", which is 512 * 128 = 65536 bytes (64KBytes), for your first/primary partition. 65536 is also divisible by 4096 in case you're worried about the "4KB sector size" concern with SSDs (though most still SSDs report 512-byte "sectors" to the OS and merge multiple reads/writes into 4KByte segments internal to the drive). 64KBytes is enough for bootloaders to fit into as well, so all is good there.
Actual NAND flash pages on the drive itself are of a size that's unknown to me (Intel hasn't stated what the flash page size is), but this doesn't matter to the underlying OS (for any reason; performance or otherwise), as the FTL takes care of the translation.
I don't know what utilities OS X uses for partition creation, but ultimately it doesn't matter, the same rules above apply.
According to Anandtech's reporting on IMTF's anouncement for the 25nm flash used in the 320 the flash has a page size of 8K, and a erase block is 256 pages, which gives a block size of 2MB.
The drive cannot erase a full NAND "block" if there's evidence of something still using it (the FTL handles this); that's completely unrelated to page alignment (more specifically, the important of partition alignment). Align your partition to a boundary evenly divisible by 8KBytes (8192 bytes) and you should be fine.
I'm very curious why you're so concerned about "erase block size", however.
Why am I concerned? There's a ton of info on forums regarding partiotion alignment on SSDs and how this effects performance.
Ideally partition boundaries match Erase Block Size boundaries.
I would like an official answer... On other, competitor's products (SSDs), this is publicly available info.
If Intel has a secret technology and does not want to make the answer public, it is fine, just help us by telling us what proper partition boundaries are.
You, like most people on forums (so don't feel alone), are misunderstanding the technology. The erase block size doesn't matter nor does it have anything to do with performance -- it's the NAND flash page ("cell") size and alignment that matters.
Please take the time to read the following Wikipedia page, which explains how all of this works. Again: NAND flash page ("cell") size is what you need to be confirmed about, not the erase page size.
You'll want to read sections titled "Basic SSD operation", "Garbage collection", "Background garbage collection", and "Over-provisioning". Don't just skim and look at the pretty pictures -- you need to read the text to understand fully.
Any benefits if I make NTFS page size 8K instead of default 4K in this case?
One SSD cell will hold only one piece of data - or something like that