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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.
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What does it mean for a computer to have a 16-bit processor?
When you say that a computer has a 16 bit processor, you mean that the fundamental data size of the accumulator and registers is 16 bits. Examples of 16 bit processors include the DEC PDP-11, the Intel 8086/8088, and the MODCOMP Classic, circa 1980.
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.
What size PVC for a washing machine drain?
2" minimum.
C program to copy one matrix to another matrix?
#include main() { int array[100], minimum, size, c, location = 1; printf("Enter the number of elements in array\n"); scanf("%d",&size); printf("Enter %d integers\n", size); for ( c = 0 ; c < size ; c++ ) scanf("%d", &array[c]); minimum = array[0]; for ( c = 1 ; c < size ; c++ ) { if ( array[c] < minimum ) { minimum = array[c]; location = c+1; } } printf("Minimum element is present at location number %d and it's value is %d.\n", location, minimum); return 0; }
What is the minimum number of stacks of size 10 required to implement a queue to size 10?
2 stacks required
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 the size of each memory segment of 8086 is?
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.
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 it the size bit address for 8085 and 8086?
128Kb
What is size of instruction queue of 8086?
6 bytes
Size of 8086 address bus?
The 8086/8088 has an internal 20-bit address bus and 16-bit data bus. Externally, the address bus is 20-bits, and the data bus is 16-bits for the 8086 and 8-bits for the 8088.The data bus in the 8086 is 16 bits in size, while the address bus is 20.
What is the word size of the Intel 80386?
The Intel 80386 is a 32-bit processor.
What does supported minimum size mean?
Supported minimum size
What is the size of address bus in 8086?
The address bus in the 8085 is 16 bits wide.
How physical address is generated in 8086 microprocessor?
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
Which group of Intel processors uses a 32-bit word size and a 64 bit word size?
Which group of Intel processors uses a 32-bit word size and a 64 bit word size?
What is the size of the instruction queue in 8086?
In 8086 the instruction queue is 6 byte long. This is because even the longest 8086 instruction is 6 byte long. Thus it is possible to prefetch even the longest instruction in the instruction set.
