This document discusses memory segmentation in the 8086 microprocessor. It contains the following key points: 1. The 8086 has a 20-bit address bus that can access 1MB of memory, but it only has 16-bit segment registers. To address the full 1MB, memory is divided into 64KB segments. 2. There are four segments - code, stack, data, and extra. Each segment is addressed using the corresponding 16-bit segment register containing the segment's base address. 3. To access a memory location within a segment requires a segment address, offset address, logical address, and physical address. The physical address is calculated by adding the segment base address to the offset.

1. Lecture 8
Memory segmentation
Iqra university

2. Contents
• Review: Registers Organization
• Memory segments
• Physical address, Offset address
• Memory configuration

3. 8086 REGISTER ORGANIZATION
ES
CS
SS
DS
IP
AH AL
BH BL
CH CL
DH DL
SP
BP
SI
DI
FLAGS
AX
BX
CX
DX
Extra Segment
Code Segment
Stack Segment
Data Segment
Instruction Pointer
Accumulator
Base Register
Count Register
Data Register
Stack Pointer
Base Pointer
Source Index
Destination Index
Type Register
size
Name of the
Register
General
purpose
registers
16 bit AX, BX, CX, DX
8 bit AL, AH, BL, BH,
CL, CH, DL, DH
16 bit SP, BPPointer
registers
Indexed
registers
16 bit SI, DI
Instruction
Pointer
16 bit IP
Segment
registers
16 bit CS, DS, SS, ES
Flags 16 bit Flag register

4. General Purpose Registers
8086 has 8 general purpose registers which:
Can be individually used for storing 8-bit data
Two registers can be combined to store 16-bit data
Valid register pairs are: AX, BX, CX, DX
AX is also called Accumulator having some special features
BX, also called Base Register is the only general purpose
register which can also be used in based-addressing mode
CX is also used as a default Counter Register in some
instructions (SHIFT, ROTATE & LOOP)
Acc. AH AL AX
Base
Counter
Data
BH BL BX
CH CL CX
DH DL DX
DX or Data Register is also
used by default to store upper
16-bit data in division and
multiplication operations

5. Register Name of the Register Special Function
AX 16-bit Accumulator Stores the 16-bit results of arithmetic and logic operations
AL 8-bit Accumulator Stores the 8-bit results of arithmetic and logic operations
BX Base register Used to hold base value in base addressing mode to access
memory data
CX Count Register Used to hold the count value in SHIFT, ROTATE and LOOP
instructions
DX Data Register Used to hold data for multiplication and division operations
SP Stack Pointer Used to hold the offset address of top stack memory location
BP Base Pointer Used to hold the offset or base value in based addressing
mode to access data from stack memory
SI Source Index Used to hold index value of source operand (data) for string
instructions
DI Data Index Used to hold the index value of destination operand (data) for
string operations

6. Memory Segmentation

7. Introduction
• Memory segmentation is nothing but a method where whole
memory is divided into the smaller parts. In 8086 microprocessor
memory are divided into four parts which is known as the segments.
these segments are
• Data Segment DS, Code Segment CS, Stack Segment SS and Extra
Segment ES

8. Memory Segmentation
• The total memory size is divided into its segments of various sizes.
• A segment is just an area in memory.
• The process of dividing memory is called Segmentation.

9. In memory, data is stored as bytes thatis 8bits.
Each byte has a specific address.
Intel 8086 has 20 lines address bus.
With 20 address lines it has 20 bit address, the memory that can be addressed is 2 ^20
bytes.
2^20= 1,048,576 bytes (1 MB).
8086 can access memory with address
ranging from 00000 H to FFFFF H.
Memory Segmentation

11. 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

12. Hence,
8086 has a 20-bit address bus & thus can address
a maximum of 1MB (2^20) of memory
But 8086 has 16-bit registers which can hold only
16-bit addresses
Solution is to divide 1MB memory in 64KB
segments
(2^16)(total sixteen) as they can be accessed by
using 16-bit
addresses. However, only 4 segments are
accessible by
8086 at a time. These 4 segments are:
Code segment: Used for storing the
codes or instructions
Stack segment: Used as a stack
•
•
•
•
Data segment: Used for storing data
Extra segment: Used for storing in-excess data

14. Advantages of Memory Segmentation:
Allows the memory capacity to be 1MB although actual addresses to be handled
are of 16-bit size
Allows the placing of code, data and stack portions of the same program in
different parts (segments) of the memory, for data and code protection.

15. 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Memory
00000H
FFFFFH
1MB
Address
Range
Each Segment is of 64KB
Memory= total 16 segments
Only 4 such segments can be
addressed (active) at a time
4
5
6
7
2^20= 1,048,576 bytes (1 MB).
8086 can access memory with address
ranging from 00000 H to FFFFF H

16. 1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
Memory
00000H
FFFFFH
1MB
Address
Range
Code Segment
Stack Segment
Data & Extra
Segments
•Segments may or may not be
consecutive
•Segments can also be contiguous
(adjacent), disjointed or overlapping

17. • Accessing to memory location there are four addresses that you all
have to understand
• Segment address
• Offset/Effective address
• Logical address
• Physical address
Accessing Addresses

18. Segment address
• Segment address relates to memory segments
• As memory is of 1MB= 2^20 bytes so segment address is of 20 bits as
00000 to FFFFFH
• Where, segment are DATA, Stack, Code, Extra
• Segment register hold only upper 16 bits that is 0000H to FFFFH and
0H is appended is the last while calculating physical address

19. Offset/Effective Address
• Offset address is displacement address required to move in 64KB
memory segment.
• Offset is of 16 bits 0000H to FFFFH

20. Logical Address
• Logical address is implemented practically by compiler through
different emulators
• Logical address it self comprises of two parts
• Segment Address : Offset

21. Physical Address
• Physical Address is calculated by adding segment address (appended
by 0H ) to the offset or effective address

23. • Thank you for listening
• Queries will be answered on
VLE