written 8.3 years ago by | • modified 8.3 years ago |
Banasthali University > Computer Science > Sem 5 > Microprocessor
Marks: 4 M
Year: Dec 2013
written 8.3 years ago by | • modified 8.3 years ago |
Banasthali University > Computer Science > Sem 5 > Microprocessor
Marks: 4 M
Year: Dec 2013
written 8.3 years ago by | modified 8.3 years ago by |
The 8088 is architecturally very similar to the 8086. The main difference is that there are only 8 data lines instead of the 8086's 16 lines. All of the other pins of the device perform the same function as they do with the 8086 with two exceptions.
First, pin 34 is no longer BHE (this is the high order byte select on the 8086 - the 8088 does not have a high order byte on its 8 bit data bus). Instead it outputs a maximum mode status, SSO. Combined with the IO/M and DT/R signals, the bus cycles can be decoded (It generally indicates when a write operation or an interrupt is in progress).
The second change is the pin that signals if a memory access or input/output access is being made has had it sense reversed. The pin on the 8088 is IO/M. On the 8086 part it is IO/M.
The reason for the reversal is that it makes the 8088 compatible with the 8085. The 8088 microprocessor can access data in any one out of 4 available segments, which limits the size of accessible memory to 256 KB (if all four segments point to different 64 KB blocks).
The 8088 is a version of the 8086 with an 8-bit data bus. The 8088 was used in the original IBM PC and its many clones. The 8088 has separate I/O and memory address spaces. Values in the I/O space are accessed with IN and OUT instructions. The port address is loaded into DX and the data is read/written to/from AL or AX:
`MOV DX,372H ;load DX with port address
OUT DX,AL ; output byte in AL to port
IN AX,DX ; input word to `