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Memory Segmentation and Physical Address Calculation

Memory segmentation divides physical memory into smaller logical segments. The 8086 microprocessor divides memory into four segments: code, data, stack, and extra. Each segment is addressed by a 16-bit segment register that stores the segment's base address. To calculate the 20-bit physical address, the base address is appended with zeros and then combined with the 16-bit offset within the segment. For example, if the data segment register contains 2222H and the logical address is 2222H:0016H, the physical address would be 22220H + 0016H = 22236H.

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78 views18 pages

Memory Segmentation and Physical Address Calculation

Memory segmentation divides physical memory into smaller logical segments. The 8086 microprocessor divides memory into four segments: code, data, stack, and extra. Each segment is addressed by a 16-bit segment register that stores the segment's base address. To calculate the 20-bit physical address, the base address is appended with zeros and then combined with the 16-bit offset within the segment. For example, if the data segment register contains 2222H and the logical address is 2222H:0016H, the physical address would be 22220H + 0016H = 22236H.

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Mona Sayed
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© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Memory Segmentation

and
Physical address calculation
Introduction
Memory segmentation is nothing which is the
methods where whole memory is divided into
the smaller parts. in 8086 microprocessor
memory are divided into for parts which is
known as the segments. these segments are
data segment, code segment, stack segment
and extra segment.
Memory Segmentation

Thetotal memory size is divided into


segments of various sizes.

A segment is just an area in memory.


The process of dividing memory this

way is called Segmentation.


Memory Segmentation
In memory, data is stored as bytes.

Each byte has a specific address.

Intel 8086 has 20 lines address bus.

With 20 address lines, the memory that can be addressed is 2 power20


bytes.

2power20= 1,048,576 bytes (1 MB).

8086 can access memory with address

ranging from 00000 H to FFFFF H.


segments.
In 8086, memory has four different types
of segments.
These are:
Code Segment
Data Segment
Stack Segment
Extra Segment
Segment Register
Each of these segments are addressed by
an address stored in corresponding
segment register.
These registers are 16-bit in size.
Each register stores the base address
(starting address) of the corresponding
segment.
Because the segment registers cannot
store 20 bits, they only store the upper 16
bits.
Segment Register
How is a 20-bit address obtained if there are only 16-
bit registers?
The answer lies in the next few slides.
The 20-bit address of a byte is called its Physical
Address.
But, it is specified as a Logical Address.
Logical address is in the form of:
Base Address : Offset
Offset is the displacement of the memory location
from the starting location of the segment.
Example
The value of Data Segment Register
(DS) is 2222 H.
To convert this 16-bit address into 20-bit,
the BIU appends 0H to the LSBs of the
address.
After appending, the starting address of
the Data Segment becomes 22220H.
Contd.
If the data at any location has a logical
address specified as:
2222 H : 0016 H
Then, the number 0016 H is the offset.
2222 H is the value of DS.
Contd.
To calculate the effective address of the
memory, BIU uses the following formula:
Effective Address = Starting Address of
Segment + Offset
To find the starting address of the segment,
BIU appends the contents of Segment
Register with 0H.
Then, it adds offset to it.
Contd.
Therefore:
EA = 22220 H
+ 0016 H
------------
22236 H
Contd.
Maximum size of segment
All offsets are limited to 16-bits.
It means that the maximum size
possible for segment is 2power16= 65,535
bytes (64 KB).
The offset of the first location within the
segment is 0000 H.
The offset of the last location in the

segment is FFFF H
Where to look offset
Example
The contents of the following registers are:
CS = 1111 H
DS = 3333 H
SS = 2526 H
IP = 1232 H
SP = 1100 H
DI = 0020 H
Calculate the corresponding physical addresses for the

address bytes in CS, DS and SS.


Solution
1. CS = 1111 H
The base address of the code segment is 11110 H.
Effective address of memory is given by 11110H + 1232H =
12342H.
2. DS = 3333 H
The base address of the data segment is 33330 H.
Effective address of memory is given by 33330H + 0020H =
33350H.
3. SS = 2526 H
The base address of the stack segment is 25260 H.
Effective address of memory is given by 25260H + 1100H =
26350H.
Scope of Research
Herewe can various methods of memory
segmentation, or we can say that logical
segmentations. In logical segmentation we can
divide the memory in logically. And in
physical segmentation divide the memory
physically.

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