7 Feb
2008
7 Feb
'08
2:59 p.m.
Date: Fri, 08 Feb 2008 01:43:37 +1030
From: Alexis
I'm attaching a hard drive to my 8080 computer and
there seems to be
something about the way the sector translation works that's caught my
attention. Before I go wasting my time trying to figure it out myself I'll
ask here because there's going to be someone who knows (it's 1:30am and
I'm feeling a little lazy).
Another thing is that the total sectors per track in
the Disk Parameter
Block is a 16-bits, not 8-bits.
Apologies if this has been discussed and I've missed it. (Perhaps
someone can provide a link if it has been)
I'll confine my answer to CP/M 2.2, since that's probably what you're
working with.
There's lots of 16-bit arithmetic in CP/M 2.2--in particular, the 128-
byte block number is maintained as a 16-bit value. This limits the
size of a disk to 65536 128-byte records or 8MB. Yes, you can have a
single track with 65535 sectors on it or 65535 tracks with a a single
sector on each. In theory, the maximum disk size should be about 1GB
(DRM=65535 and BSH=7).
Check it out for yourself in the source at:
http://www.cpm.z80.de/source.html
Note that CP/M 2.2 contains no division routines in its sector and
track computation--the block-to-track-and-sector translation is
handled by repetitive subtraction. There is a little bit of code to
reduce the burden of this by using the last translated block number
to compute the next, though I wonder how many cycles are saved in
practice.
Hope this helps.
Cheers,
Chuck