Re: Determining amount of RAM on a board?

[david@uow.edu.au (David Wilson)]

"Ernest" <ernestls@home.com> writes:
>So if 8 chips equal 1 byte, then again, I would need 64 banks of 8 chips,
>right? Obviously that isn't right but I'm a little confused. If each chip
>has 8k bytes, or 65536 bits but you need a row of eight chips for the full
>64k bytes to work, then I think I understand. Each chip has lots of bits,
>but no bytes are formed unless all 8 chips offer one bit each, at any one
>time.

>Why did they set it up like this? Why not just use 8k bytes from each chip,
>and then move on to the next when you need more room? Wouldn't that make
>diagnostics easier, if you lose a chip.

Pin count. With DRAM, each address pin is used twice thus halving the number
off address pins. Have a look at the following:

Type	Addr	Data	A+D
64kx1	8	1	9
32kx2	8	2	10
16kx4	7	4	11
8kx8	7	8	15

Even if we used SRAM the table would be similar:

Type	Addr	Data	A+D
64kx1	16	1	17
32kx2	15	2	17
16kx4	14	4	18
8kx8	13	8	21

As you can see - the x1 format is most efficient pin wise. A 4164 is a 16pin
chip while a 4464 is an 18pin chip.
--
David Wilson  School of IT & CS, Uni of Wollongong, Australia