3 ms·
Two explanations I read in the late 80's. When the decision was made you had the 8088 and the 68000. 8088 had an 8 bit bus and needed 8 DRAMs. 68000 had a 16 b
by Gibbon1 6y ago
Two explanations I read in the late 80's.
When the decision was made you had the 8088 and the 68000. 8088 had an 8 bit bus and needed 8 DRAMs. 68000 had a 16 bit buss and needed 16. And memory was about half the cost of a PC. So the 8088 minimum BOM was cheaper.
Second I think the 8088's instruction set was designed to make porting 8080/Z80 assembler easy. I can't say that was important in practice once things were bootstrapped. But might have got them to market faster.
I also think partly the dominant OS in the late 70's was CPM which ran on Intel/Zilog processors. 8088 was of course the manufacturer of the 8080. And my experience with Motorola in the 80's is they weren't exactly eager to sell into smaller markets. Of course now Intel is far worse and has been for the last 35 years.
- dboreham 6y ago68k was too new, too large a die (lots of articles at the time saying it might be unmanufacturable due to low yield) too costly.
- snvzz 6y ago68008 (1982) had 8bit bus, and some 68000 versions were made much later which supported both 8 and 16bit bus. But of course, this was way too late for IBM PC (released 1981).