Re: Determining amount of RAM on a board?

[dempson@actrix.gen.nz (David Empson)]

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".

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.

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).

If you had 4464 chips (64K x 4), then each chip would correspond to four
bits in each memory location, and you would need two of them to form a
complete byte (giving 64K bytes).

It is also possible to get 8-bit wide memory, such as static RAM chips
(e.g. used on Apple's SCSI cards, and for cache RAM on the Zip GS and
TransWarp GS).  A typical chip of this type would be a 6264 (8K x 8) or
a 62256 (32K x 8).  In these cases, a single chip is sufficient to
provide 8K bytes or 32K bytes of memory.  Using my table analogy, the
chip occupies all eight columns and an appropriate number of rows (8192
or 32768).


(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.)