Re: Determining amount of RAM on a board?

["Ernest" <ernestls@home.com>]

Thanks for explaining all this to me, folks. I was never much of a hardware
hack, so I never learned this stuff.  I never realized that each chip shares
one bit at a time. I thought it was just a fill up and move on sort of
thing.

I appreciate it.

E.



"Mark Cummings" <figjams@primus.com.au> wrote in message
3b811eb9@news.iprimus.com.au">news:3b811eb9@news.iprimus.com.au...
> >Wouldn't that mean that I would need a lot more than 8 chips then, if
each
> >chip is only one bit?
>
> They are refering to the bit width, not the total number of bits in the
> chip. 8x "1bit 64K" chips are needed to make "8bits of 64K" total.
>
> >> A byte requires eight bits being read simultaneously.  The 4164 only
> >> provides one bit, so you need eight of them to form a complete byte.
>
> correct sort-of. it would be 1 BYTE wide x 64K total for 8 chips. (or
65536
> bytes 8bits wide)
>
> >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.
>
> that makes two of us :)
>
> >If each chip
> >has 8k bytes,
>
> not really, each chip has 65536 bits in a 64K x 1bit array. To explain it
in
> other terms, each chip only has one Data pin, as most 8bit CPU's read
those
> bits in 8 at a time from the data bus you need 8 chips to fill up the full
8
> bits wide, so the CPU can deal with it in one read or write cycle.
>
> >or 65536 bits but you need a row of eight chips for the full
> >64k bytes to work, then I think I understand.
>
> I don't. :?
>
> >Each chip has lots of bits,
> >but no bytes are formed unless all 8 chips offer one bit each, at any one
> >time.
>
> we are getting there, I think.
>
> >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.
>
> Yes it would make diagnosis easier, but it is a trade off for total size
of
> the DRAM and the space it takes up on the circuit board..
> In fact I consider the inbuilt diagnosis on the IIe, Platinum IIe, ROM-1
GS
> and ROM-3 GS (and possibly the IIc) to be totally useless when it comes to
> inbuilt RAM testing. It only takes one stuck bit to stop the CPU in it's
> tracks.
> Generally if it had the full 8 bits the chip would be larger because of
the
> number of address and data lines. 3 less address lines and 7 more data
bits,
> making 4 more pins total. I have some "6264" static RAM chip which are 8K
> bytes (8bits x 8192), but uses 24 (or 28?) pins total. This is in a 600mil
> wide package which is twice the width of your typical 6164s and a bit
longer
> too. 8 of these on a board would make for a really large circuit board,
> compared to 8 of the 4164's. Circuit board space is generally considered
as
> expensive real estate, therefore the smaller you can make it the cheaper
it
> is to make.
>
> >> It may be useful to visualise the memory as a very tall and thin table,
> >> with each row corresponding to a memory location (65536 rows in total)
> >> and eight columns, each corresponding to one bit.  Each 4164 chip
> >> corresponds to a column of the table.  All eight chips (columns) are
> >> represented in every memory location (row), providing a single memory
> >> bit (row/column intersection).
> >
> >Yes, I get the picture (I think.) If you have 7 chips, then none of the
> >others will work because each chip needs to share one of it's bits to
form
> >the whole 1 byte, at any given time.
>
> yes, I think you have it now.
>
> Mark
>
>