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The 8086 forms a 20 bit address by adding the effective address (a 16 bit value) to a segment register (another 16 bit value) which is left shifted by 4. That gives a 20 bit address in the range of 00000H to FFFFFH.

cs register holds the base address (16 bit) and the IP has the offset.

(ex):

CS --->348A

IP --->4214(offset)

generation of 20 bit:

CS*10+IP

(ie)

348A0

04214 +

----------------------

38AB4(20 BIT)

----------------------

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11y ago
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13y ago

Complete effective (or external or Physical) address is calculated with: 1) Segment resistors 2) offset resistors. Let us take a example and solve the complicated issue.

Segment address ----> 1005h (h stands of hexadecimal)

Offset address -------> 5555h

Segment address (In Binary) --------------------> 0001 0000 0000 0101

Shifted by 4 bit positions Left with carry ------> 0001 0000 0000 0101 0000

+

(Add) offset address (in Binary)------------------> 0000 0101 0101 0101 0101

Therefore, we get Physical Address-------------> 0001 0101 0101 1010 0101

Physical address (in hexadecimal) --------------> 155A5h

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Q: How the 20 bit effective address is calculated in 8086?
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What is physical address in 8086 microprocessor?

Physical address in the 8086/8088 is {Selected Segment Register} * 16 + {Effective Offset Address}. It is a 20-bit address .


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 it the size bit address for 8085 and 8086?

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In the 8085, the high order address is A8-A15. In the 8086/8088, the high order address is A8-A19. (For a 16-bit address, the answer is A8-A15, but the answer above reflects the chosen categories, 8085 and 8086/8088, with the 8086/8088 running in 20-bit mode.) In Windows XP, running in 32-bit mode, the high order address is A8-A31, a 32 bit address.


How many address lines are required to access 1MB RAM using microprocessor 8086?

The 8086/8088 microprocessor has a 20 bit address bus, so the number of memory locations it can address is 220 or 1,048,576.


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The difference between the 8086 and the 8088 is that the 8086 has a 16 bit data bus and that the 8088 has an 8 bit data bus. Both processors are the same 16 bit processor, and both have a 20 bit address bus. The 8086 is twice as fast as the 8088 in terms of data transfer rate on the bus for the same bus clock speed.


Why memory is divided in even and odd banks in 8086 microprocessor?

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What is the Comparison between 8086 and 80286 in micro processor?

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Difference between 8086 and 80386?

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What is the difference between the Intel 8085 and the Intel 8086?

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