80286

I heard that the 80286 had 20 bits in for each address in memory but an
address bus of only 16 bits!

How did I access an address in memory with this limitation?

On 31 Oct 2006 08:07:59 -0800, "Gaijinco" wrote:

I heard that the 80286 had 20 bits in for each address in memory but an
address bus of only 16 bits!

How did I access an address in memory with this limitation?

segment register

How did I access an address in memory with this limitation?

segment register

Can you explain a little bit of what segment register means?

Thanks!

Externally, the 80286 has 24 address lines, but the internal address
register is only 16 bits. This is where the segment register comes
into play. In the 286 the segment register adds 8 significant bits
or 256 segments to the address register. Each segment is 64Kb,
which multiplied by 256 gives the 16Mb address space of the
80286. In a 32 bit processor the segment register is 16 bits which
gives 4Gb of addressing space, (64Kb x 64Kb).

In a 80286 memory dump the address will display like this example:
0043:0100
To the left of the colon is the hex offset value, and to the right
is the hex value of the 16 bit address register - called the offset.