3 Jul
2016
3 Jul
'16
5:32 a.m.
On 2016-Jun-30 22:21:27 +0300, Diomidis Spinellis <dds(a)aueb.gr> wrote:
First, the 8088 was a 16-bit CPU with an 8-bit data bus
in a cheap
40-pin package. This halved the number DRAM chips required and allowed
the IBM PC to be easily designed along existing easily-available 8-bit
peripherals. In contrast the 68000 had a 16-bit data bus in a more
expensive 64-pin package. Remember that in the 1980s glue logic was
implemented through simple TTL chips, so adopting the 68000 might have
doubled the number of chips on the motherboard.
My understanding was that the 8-bit bus was a requirement so IBM could
have a 64KB base model using the then new 64k×1 chips. IBM also
emasculated the PC so it didn't compete with their existing minis. The
68008 wasn't available until later (and this would explain why Motorola
pushed the 6809 as a solution).
Both the 8086 and M68k could relatively easily use 8-bit peripherals
(both Intel and Motorola had a range of 8-bit peripherals that they
didn't want to make instantly obsolete).
In addition, the 8086 architecture was an extension of
the 8080 one,
which made it easier to make the MS-DOS API compatible with the CP/M
Since IBM was buying the software, I'm not sure how much of a driver
this was. Definitely, porting from 8080 to 8086 was easier but writing
from scratch would be far easier on M68k.
--
Peter Jeremy