0
The 8086 was only capable of addressing 1Mbyte of memory. It was divided into segments of 65536 bytes (64 KB) each meaning about 16 segments.
Wiki User
∙ 13y ago
Wiki User
∙ 9y ago65536
Add your answer:
Earn +20 pts
What is minimum and maximum segment size in 8086?
In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment capacity is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient.
What is minimum size of a segment in Intel 8086 Why?
In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment capacity is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient. In 8086 microprocessor the total memory addressing capability is 1 mega bytes. For representing 1 mb there are minimum 4 hex digits are required i.e, 20 bits. but 8086 has fourteen 16-bit registers. That is there are no registers for representing 20 bit address. So,the total memory is divided into 16 logical segments and each segment size is 64 kb(kilo bytes). That is 16*64kb=1 mb.So,for representing 64 kb only 16 bit register is sufficient.
What is a segment used for?
A segment is a chunk (segment) of memory that is 64Kb in size. Due to the design of the 8086/8088 there are 64K possible segments, ecah overlapping the next by 16 bytes, for a total addressibility of 1 Mb. In the instruction model, a segment is the locus of addresses that can be reached in one instruction, without stopping to load a new value into a segment register. It is also called a near, or 16 bit address.
What is the size of RAM that 8086 can address?
8086 has 20 address lines. Therefore it can address 220 bits or 1,048,576 bits of memory, or roughly 1 MB (mega byte).
What is the difference between CPU working in real mode and protected mode?
In real mode, the effective memory address is the same as the linear, or physical, memory address. The segment register operates by specifying the base of a 64kb region of memory, each region overlapping the other by 16 bytes. In protected mode, the effective address is translated by the page table which provides a larger region size and more memory addressibility. The segment register, a/k/a selector, selects the particular page table to use.
What are the disadvantages of memory segmentation in 8086?
disadvantages of memory segmentation
What is the difference between near heap and far heap?
Near and far are obsolete terms used in the MSDOS and Windows 3.x platforms on the 8086/8088 processor. Near represents an area of memory that can be accessed using only a 16 bit offset and, as such, must lie within the default data segment, and is always less than 64kb in size. Far represents an area of memory that must be accessed using both a 16 bit offset and a 16 bit segment and, as such, can lie anywhere in memory and be larger than 64kb, at the expense of additional processing time and program size.
How does memory segmentation increase processors speed?
The segmentation function maintains a segment table that includes physical addresses of the segment, size, and other data. Segmentation speeds up a computer's information retrieval by assigning related data into a “segment table” between the CPU and the physical memory.
What is the need of Instruction PointerIP in 8086 microprocessor?
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.
What is segment overlapping?
Each segment register is multiplied by 16 before being added to the effective address during physical address computation. As a result, sequential values of segment registers result in physical addresses that are 16 bytes apart, yet, the maximum distance, based on effective address is 64 KB. That is why the segmented memory addressing approach in the 8086/8088 is overlapping - there are 64 K different combinations of segment register / effective address pair that will result is the same physical address. Well - sort of - its not wrap around - its a sliding window - call it a 16 byte paragraph size with an addressability range of 64 KB.
What is overlapping segment addresses?
Each segment register is multiplied by 16 before being added to the effective address during physical address computation. As a result, sequential values of segment registers result in physical addresses that are 16 bytes apart, yet, the maximum distance, based on effective address is 64 KB. That is why the segmented memory addressing approach in the 8086/8088 is overlapping - there are 64 K different combinations of segment register / effective address pair that will result is the same physical address. Well - sort of - its not wrap around - its a sliding window - call it a 16 byte paragraph size with an addressability range of 64 KB.
Difference between a page and a segment?
Paging is a virtual memory scheme which is transparent to the program at the application level and which divides memory into fixed-size blocks, such as 4 KBytes. The segmentation memory management scheme imposes a greater book-keeping burden on the application, and refers to memory using segments of variable size. segmentation is a logical unit visible to the user's program and id of arbitary size whereas paging is a physical unit invisible to the user's view and is of fixed size
