How physical address is generated in 8086 microprocessor?

Answer 1

For the formation of physical address we need Segment address and offset address

Consider an example

Segment Address : 1005H

Offset Address : 5555H

Segment address : 1005H 0001 0000 0000 0101

Shifted by 4 bit positions : 0001 0000 0000 0101 0000

Offset Address : + 0101 0101 0101 0101

Physical Address : 0001 0101 0101 1010 0101

1 5 5 A 5 H

Physical Address of given Segment Address : 155A5H

Answer 2

That is a port on the device that follows an ip address. The port is like a train station with multiple train tracks it has one place but there are multiple ways of sending information too and fro. There are 9999 ports on every computer if I am not mistaken. I hope this answered your question

Answer 3

The 8086/8088 calculates a physical address by taking the 16 bit offset address determined by the instruction and adding it to one of the four 16 bit segment registers after left shifting the segment register four places. This constructs a 20 bit address, giving a 1 MB address space, with a 64 KB segment size. The selection of segment register is by context or, in the case of operand, possibly by segment prefix override, and is one of Code Segment (CS), Data Segment (DS), Stack Segment (SS), or Extra Segment (ES).

Answer 4

The data segment register which is 16-bit is shifted left four _bits and then added to the offset register .tage@arbaminch

Answer 5

A physical address is generated in an 8086/8088 by adding the desired offset to the value of the appropriate segment register shifted left by 4 bits.

Answer 6

The physical address in the 8086/8088 microprocessor is the offset (or effective address) plus 16 times the applicable segment register.