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Editorial Team
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Asked: 2026-06-11T15:03:02+00:00

in 8086 microprocessor a 20 bit address is divided in 16bit+4bit address in which

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in 8086 microprocessor a 20 bit address is divided in 16bit+4bit address in which 4 bit binary is the segment address.when we convert a 4bit binary into hexadecimal it gets to 1bit hexadecimal.my question is when we encounter the problem of calculating the physical address from the logical, a 4bit hexadecimal segment address is given.why is it so?
Also in the calculation of physical address we append 0 in lsb to find the base address of the segment and then we add the offset into it. what is the logic behind appending 0?

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  1. Editorial Team

    One segment is equal to one paragraph. One paragraph is equal to 16 decimal bytes or 10 hexadecimal bytes. So a segment value of 89AB with zero offset is equal to 89AB x 10 or 89AB0 (note: all addresses are in hexadecimal for this context).

    For segment-offset to 20-bit absolute address conversion example, this is best represented like this:

    89AB:F012  ->  89AB   ->  89AB0   (paragraph to byte ->  89AB x 10 = 89AB0)
                F012  ->  0F012   (offset is already in byte unit)
                          ----- +
                          98AC2   (the absolute address)

    For absolute address to segment-offset conversion:

    98AC2  ->  9 8AC2  ->  9     ->  9000  ->  9000:8AC2
           (split)     8AC2      8AC2

    or…

    98AC2  ->  98AC 2  ->  98AC  ->  98AC  ->  98AC:0002
           (split)        2      0002

    or can be split at middle…

    98AC2  ->  98 AC2  ->  98    ->  9800  ->  9800:0AC2
           (split)      AC2      0AC2

    All above three segment-offset address including 89AB:F012 (the original address value) points to the same absolute address (same physical location).

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