Re: Determining amount of RAM on a board?

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

"David Empson" <dempson@actrix.gen.nz> wrote in message
1eydwbq.1rmzydw1lx7l6oN%dempson@actrix.gen.nz">news:1eydwbq.1rmzydw1lx7l6oN%dempson@actrix.gen.nz...
> Ernest <ernestls@home.com> wrote:
>
> > "Paul Schlyter" <pausch@saaf.se> wrote in message
> > news:9ll3m7$b3l$1@merope.saaf.se...
> > > In article <JXmf7.31513$vW2.13941834@news1.sttln1.wa.home.com>,
> > > Ernest <ernestls@home.com> wrote:
> > > >
> > > >"william strutts" <wrstrutts1@nospam.home.com> wrote in message
> > > >news:EHmf7.38773$K6.15676876@news2...
> > > >
> > > >> Most chips will have something in the line of 44xx or 41xx
> > > >> The 44 series is 4 bits wide and 41 series is 1 bit wide.
> > > >> The older computers used for the most part 1 bit wide chips.
> > > >> 4104  - 4k x 1
> > > >> 4108 -  8k x 1
> > > >> 4116  -16k x 1
> > > >> 4164 - 64k x 1
> > > >
> > > >So what would a bank of eight 4164np-20 chips be?
> > >
> > > It would be 64K x 8 bits = 64 Kbytes !!!!
> >
> > 8 bits to a byte, right so...
> >
> > 64,000 divided by 8 = 8000, or 8k bytes per chip... times 8 chips = 64k
> > bytes.
>
> Except that it isn't "8K bytes per chip": it is "one bit from each of
> 64K bytes per chip".

Wouldn't that mean that I would need a lot more than 8 chips then, if each
chip is only one bit?

> 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.

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.

> 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.

> (Incidentally, when dealing with computer memory 8K is 8192, and 64K is
> 65536, not 8000 and 64000 respectively.  They are powers of two: "1K" is
> two to the power of ten, i.e. 1024.)

Yes, I know but I was trying to keep the image simple.