In the 8086 microprocessor, code segment addresses are generated using a segment:offset addressing scheme. The code segment (CS) register holds the starting address of the code segment, while the instruction pointer (IP) register holds the offset of the next instruction to be executed within that segment. The effective address of an instruction is calculated by adding the value in the CS register (shifted left by 4 bits) to the value in the IP register, allowing for a total addressable space of 1 MB. This segmentation allows for efficient memory management and organization of code.
Physical address in the 8086/8088 is {Selected Segment Register} * 16 + {Effective Offset Address}. It is a 20-bit address .
segment is for converting physical address to logical address , here on taking 8086 microprocessor as example, we have 20 address lines but it is capable of taking only 16 address lines.... so to convert that 20 into 16 segment is used....
It is mightily referring to Microprocessor 8086 . I think you saw "8086 microprocessor". The 8086 is nothing it indicates the number of microprocessor same as Digital or analog ic's . 8086 microprocessor has 20 Address buses and 8 data buses which has 1 Mb inbuilt memory for performing several type of airthmatical and logical operation.
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
for demultiplexing address/data bus
In the 8086 microprocessor, the physical address is calculated by shifting the segment address 4 bits to the left, which effectively multiplies it by 16. This operation aligns the segment address to the correct boundary in memory, allowing for a proper offset calculation. The combined segment and offset addresses then yield a 20-bit physical address, which can access a larger memory range than the 16-bit addresses used for segments and offsets individually. This addressing scheme enables the 8086 to utilize up to 1 MB of memory.
8086 has memory divided into segments.Each segment has its particular register like ES, DS, SS, CD (extra segment, data segment, stack segment and code segment).These registers hold the base address BA.Now, there two base registers(BX and BP) and two index registers(SI and DI) in 8086. These registers hols the effective address EA.Now Physical address PA is sum of EA and BA.That is,PA = BA+EA
Its 16bit microprocessor,and-> the 8086 has a 16bit databus 20bit address bus-> the intel 8086,is designed to operate in two modes namely(1) minimum mode(2) maximum mode
The 8086 microprocessor has 40 pins.
In the 8086 microprocessor, the Instruction Pointer (IP) register contains the address of the next instruction to be fetched and executed. It works in conjunction with the Segment Registers (such as CS - Code Segment) to form the complete address of the instruction in memory. The IP is automatically updated as instructions are executed, ensuring that the CPU always knows where to fetch the next instruction from.
the advantages of 8086 microprocessor over 8085A microprocessor is that 1)it is a 16 bit microprocessor 2)the disadvantages in 8085A microprocessor like low speed,limited number of registers,low memory addressing capability,less powerful instruction set have been overcome in 8086 registers
There are four segment registers in the 8086/8088, Code Segment (CS), Stack Segment (SS), Data Segment (DS), and Extra Segment (ES). As a result, there are four segments that can be directly addressed at a particular time, i.e. without an extra instruction to reload a segment register.