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Each segment in the 8086/8088 is 64KB because that is how Intel designed the microprocessor. The offset address is 16 bits, making the allowable range 64KB. See the related questions link for a further discussion of segmented architecture.
Wiki User
∙ 13y ago
Wiki User
∙ 12y agoBecause that's the way Intel designed it.
A segment register is 16 bits. That is 65536, or 64K, different values. Also, the processor is a 16 bit processor, and offset values can only contain 65536, or 64K, different values.
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8086 micro processor memory segmentation and advantage and rules?
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Why you Use Memory Segmentation In 8086 Microprocessor?
The 8086/8088 is a 16 bit processor running on a 16 bit (8086) or 8 bit (8088) bus with a 20 bit address bus. In order to obtain the extra 4 bits of addressibility, Intel designed segment registers that are effectively multiplied by four and then added to the 16 bit offset address generated by the instruction. This yields 64K segments of 64KB each, although they overlap each other at a distance of 16 bytes.
How many bytes are addressable the 8086?
The 8086 can address 1,114,080 bytes. (One Mb + 64Kb - 16) That does not count I/O space, it only counts memory space.
What is memory interfacing in 8086 microprocessor?
The 8086/8088 is a 16 bit computer running on a 20 bit address bus. Processes use a segmented memory architecture to access one of four 64kb memory segments from a physical space of 1mb.
Why 4 bit address bus is grounded?
The data bus in the 8086 is 16 bits in size, while the address bus is 20 (16bits would only address 64KB of memory, an extra 4 bits allows to address the total of 1MB, this is done trough segmentation of the memory). To form a multiplexed of data bus and address bus, four bits of 8086 address bus are grounded.
Why 8086 requires address bus of 20 bits while using 16 bit address for segement register?
This is due to the fact that 16bits would only address 64KB of memory, which even then was very little. The answer for this was to come up with an extra 4 bits to address the total 1MB, this is done trough segmentation of the memory. Google it.
What are the disadvantages of memory segmentation in 8086?
disadvantages of memory segmentation
How can the memory other than four 64KB segments be accessed by 8086?
You can access any location in memory. You need only to load its segment address and then refer to its offset address, using the appropriate segment register.
What is the difference between microprocessors 85 and 86?
85 is a 8 bit processor,number of flags are 5 and memory capacity is 64KB while 86 is a 16 bit processor ,number of flags are 9 and memory capacity is 1 MB.The main difference between 8085 and 8086 is that 8086 uses pipelining.
How segmentation and pipe lining in 8086 are related to each other?
The process of dividing total memory sizes to the segment of various sizes is called segmentation. The device which is used to fetch,decode and execute is called pipe lining.
How do you address 1GB memory with 8085 or 8086?
You cannot address 1GB memory with the 8085 or the 8086/8088 without some kind of external demultiplexor that is software controlled. The address bus on the 8085 is 16 bits, giving addressibility of 64KB; while the address bus on the 8086/8088 is 20 bits, giving addressibility of 1MB. To address 1GB, you need a 30 bit address bus.
Define the extra segment and stack segment?
The extra segment in the 8086/8088 is a 64kb region of memory that is indexed by the displacement address of the destination of certain string operations, relative to DI. Contrast that with the data segment, which is a 64kb region of memory that is indexed by the displacment address of the displacement of most operand addresses. The stack segment is similar, but it is used for stack oriented data, relative to SP or BP.
