Re: 8088 card and bank 15

Hello!

Ruud@Baltissen.org wrote:

> When studying the schematic I noticed a 283 adder. It genererates 
> the lines BP0..3 that are used to select the needed 64 KB DRAM bank 
> using the PLA.
> The adder adds one to the original address. Why? 

The adder is needed because 8088 memory must start at $00000 (there are
interrupt vectors there) whereas CBM-II memory starts at $10000. There
is nothing at $00000-$0FFFF; bank 0 is empty. Memory is present in banks
1 through 4. Therefore, to map 8088 memory to 6509 memory, 1 must be
added to the bank number.

> But when addressing 0Fxxxxh, BP0..3 are all zero. But what happens 
> if the 8088 addresses the 0Exxxxh area, BP0..3 are all one. Or in 
> other words: bank 15!
> 
> I studied the schematic and so far I don't see any reason why it 
> wouldn't work. So those who have a working CBM-II/8088 card 
> combination, please try it out.

Of course it will not work. Even though BP0-BP3 have the correct value,
memory access is still governed by RAS and CAS on the J2 port which go
to DRAM. The chips in bank 15 cannot see these signals.

Also, if the 8088 was allowed to touch bank 15, it would immediately
crash the 6509 which lives there. There is no possibility to stop the
6509; the AEC signal, though present on the chip, is not exposed on the
expansion connector (in fact it is not connected anywhere in the CBM-II
and it is quite likely that the motherboard hardware would not support
it even if ou piggybacked it).

Just to be sure, I fired up the computer and as expected, reads from
E000:0000 return garbage.

Regards,
Michau.

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