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
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
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).
The data segment register which is 16-bit is shifted left four _bits and then added to the offset register .tage@arbaminch
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.
The physical address in the 8086/8088 microprocessor is the offset (or effective address) plus 16 times the applicable segment register.
Physical address in the 8086/8088 is {Selected Segment Register} * 16 + {Effective Offset Address}. It is a 20-bit address .
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.
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....
for demultiplexing address/data bus
Effective address is the final address generated by offsetting and indexing which is sent to the virtual translation logic. It is the address of the operand in the virtual address space of the process, but not necessarily the address of the operand in the physical address space of the computer. In the 8085, efffective/virtual address is the same as physical address, because there is no virtual addressing logic in the 8085. In the 8086/8088, effective/virtual address is the same as physical address, but only in real mode. For example, in the 8086/8088, if the EBX register contains 1000000H, then the instruction MOV EAX,[EBX+1234H] has an effective address of 10001234H.
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.
Segmented memory in the 8086/8088 means that the effective address is added to 16 times the segment address. In the question, an effective address of 14A3H with DS of 7000H, means the physical address is 714A3H. However, the BIU in the 8086 only addresses words, not bytes, so the address generated is 714A2H. If the operation required more than one byte of data, subsequent even addresses of 714A4H, 714A6H, etc. would be generated.
The physical address is the final address that is presented to the bus, at the pins of the microprocessor chip, to form the address of the desired item in memory.In the 8085, physical and effective addresses are one and the same.In the 8086/8088, the physical address is the effective address plus 16 times one of the segment registers.In higher level processors, such as the 80386 and beyond, the physical address is formed by lookup of the effective address in a page table to convert from virtual/effective address to physical, or linear, address.The effective/virtual address is the address generated by the instruction and the programmer, without regard to any underlying addressing scheme. This is the address used when considering the "programming model", in "user mode".
The Instruction Pointer (IP) in an 8086 microprocessor contains the address of the next instruction to be executed. The processor uses IP to request memory data from the Bus Interface Unit, and then increments it by the size of the instruction.
queue of 8086 microprocessor is 6 bits
The 8086 Microprocessor operate to require frequency that is provided by clock generator to 8086 Microprocessor and also Synchronization various component of 8086.