Re: Determining amount of RAM on a board?

["Mark Cummings" <figjams@primus.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