How you calculate the physical address in 8086 microprocessor with example?

Answer 1

The physical address in the 8086/8088 is calculated by adding the effective address with the contents of one of the segment registers left shifted by 4 bit positions. This results in a 20 bit address bus.

As an example, if the CS register contains 1234H, and the IP register contains 5678H, then the next instruction is fetched from physical address 179B8H, which is 1234H times 16 (12340H) plus 5678H.

The segment register used is selected by context, or by using a segment override prefix, however, the code segment register (CS) can not be overidden during instruction fetch, nor can the stack segment register (SS) be overidden during stack pushes and pops.

Answer 2

To calculate the physical address in the 8086 microprocessor, you combine the contents of the segment register with the offset address. The physical address is obtained by shifting the segment register contents left by 4 bits and adding the offset address. For example, if the segment register (let's say DS) contains 1000H and the offset address is 200H, the physical address would be 12000H (1000H << 4 + 200H).

Answer 3

if DS=7FA2H AND THE OFFSET IS 438EHthen physical address=7FA20+438E

logical address:

7FA2:438E