Chris Zuhars <
czu...@gmail.com> wrote:
> I'm wanting to buy new DRAM for my Ramfactor III card. The existing 2
> chips it has say 41256-15 which I know are 256k 150ns chips.
>
> However, my understanding of the board is: fully populated,it is 1mb,
> which means 4 banks of 8 256k DRAM chips. Correct so far?
Sounds right.
http://apple2.info/wiki/index.php?title=RAM_Cards
> So when I spec 32 new DRAM chips from Jameco or someplace, what comes up
> is often 256x1 chips for that model (41256). So those wouldn't be the same
> as 256x8 chips, are they?
Correct. The number after the 'x' is data bus width of the memory chip.
A 256Kx1 chip has 1 data line (and enough address pins for 256K
locations). Each address produces a single bit of data from that chip,
so you need a set of 8 matching chips to produce 8 bits of data.
A 256Kx8 chip has 8 data lines (and enough address pins for 256K
locations). A single chip produces 8 bits of data.
The wider data bus means the 256Kx8 chip has more pins and will be a
physically larger package. It won't even fit in the socket of a board
which expects to use 256Kx1 chips.
> Do I need to specifically choose 256x8 chips? Or would these 256x1s work?
> I think the former, but I just wanted to make sure.
256Kx8 chips will NOT work. Look specifically for the "41256" part
number to get 256Kx1, and you will need 32 of them to fully populate the
card.
--
David Empson
dem...@actrix.gen.nz