rwessel (robertwessel.delete@this.yahoo.com) on January 13, 2020 9:40 pm wrote:
>
> It depends on what you mean by seemlessly.
>
> The 8086 bus was similar enough that pretty much all the 8-bit 8080 peripherals
> could be trivially connected to it. The one gotcha is that the I/O ports (or memory
> mapped locations), would be on alternate bytes as viewed from the CPU.
>
> If you wanted them on consecutive bytes, as you would if you were attempting to
> emulate the IBM PC, you need some logic to generate that 8-bit bus from the 16-bit
> bus. It's not hard, and you really only need to do it once in the system.
>
> You can see the required circuit on page 192 of the follow (page 5-8, figure 5-6). This happens to be the
> Intel recommendation for the 286, but doing this for the 8086 would have been essentially identical.
>
> http://bitsavers.trailing-edge.com/components/intel/80286/210760-002_80286_Hardware_Reference_Manual_1987.pdf

The low and high byte selection is irrelevant because the 8-bit operation will only read from the low or high byte anyway and this points to the flaw. If a 16-bit operation is performed, only one bus cycle occurs which can only return the same low and high byte for both. Offhand I do not remember if any of the 8-bit peripherals could be accessed usefully like that but it would explain why 8086 PCs had some compatibility problems.

286 PCs must have used some state machine logic to divide 16-bit accesses into two 8-bit accesses because the 16-bit ISA bus included provisions for returning the access width from the expansion card.