Need some help decoding "module organization" and "chip organization"

[uOpt]

I want to buy 2x 2 GB modules that play nice with an existing set of 2x 2 GB that I have.

The existing modules are Aeneon AET860UD00-370A08X, 533 MHz 4-4-4-12.

Organization is given as "256Mx64" and "128M x 4".

If I decode that correctly this means:

"module organization": 256Mx64, which means 64 bit width (non-ECC,
only other option for DDRx would be 72, which would be ECC), and 256
Mega-words of that width = 2 GByte.
"chip organization" or "DRAM organization": is given as "128M x 4",
which means "4 chips, 128 M per chip" but omits rank and/or sides.


Can somebody help me decode the "chip organization"? Does that mean 128 Mbyte per chip, 4 chips per rank/side, 2 ranks, 2 sides == 16 * 128Mbyte == 2048 MByte?

I'm confused, the math doesn't play nice.

[RyderOCZ]

Your Aeneon modules are 2GB and say they are 128x4? A 128x4 IC is used to make a 1GB module based on info I have. We used them on a couple products, but some boards have issues with them. AFAIK, they were 1GB modules, although they may have been single sided, not sure.

Regardless of the math and whether I could do it right, all our 2GB modules (4GB kits) are 128x8 ICs and there are 16 IC's on a module (8 per side)

[uOpt]

Thanks, Ryder.

Now I actually found other information (the manufacturer doesn't have their datasheets online) that say 128Mx8, which makes more sense.

[RyderOCZ]

I asked a couple questions and if I understand what I was told correctly:

128x4 is 128Mbit density, but with only 4 internal ranks, so therefore, its the same as a 64x8 which is 64Mbit density, but with 8 internal ranks, so they are essentially the same thing. We make our 1GB modules out of 64x8 IC's for the most part, so a 128x4 should net the same module size.

Glad you found that other data, that does indeed make more sense since 128x8 should be twice the size of 64x8 and you will have a 2GB module.