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University of Salahaddin College of Engineering Software & Informatics Dep

The document discusses memory segmentation in 8086 microprocessors. It contains the following key points: 1) The 8086 has 20 address lines allowing it to address up to 1MB of memory. This memory is divided into 64KB segments using segment registers. 2) The 8086 has four segment registers - code segment (CS), stack segment (SS), data segment (DS), and extra segment (ES). These point to the base addresses of 64KB segments. 3) A logical address consists of a segment and offset. The physical address is calculated by shifting the segment left by one bit and adding the offset. This allows accessing the full 1MB address space.

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Zayto Saeed
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98 views15 pages

University of Salahaddin College of Engineering Software & Informatics Dep

The document discusses memory segmentation in 8086 microprocessors. It contains the following key points: 1) The 8086 has 20 address lines allowing it to address up to 1MB of memory. This memory is divided into 64KB segments using segment registers. 2) The 8086 has four segment registers - code segment (CS), stack segment (SS), data segment (DS), and extra segment (ES). These point to the base addresses of 64KB segments. 3) A logical address consists of a segment and offset. The physical address is calculated by shifting the segment left by one bit and adding the offset. This allows accessing the full 1MB address space.

Uploaded by

Zayto Saeed
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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University of Salahaddin

College of Engineering
Software & Informatics Dep.

COMPUTER ARCHITECTURE
2 N D S TA G E

A S S I S TA N T L E C T U R E R : E S I L S E M I R K H U R S H E D

[Lec 3]
2020 – 2019
Memory Segmentation
• Total memory size is divided into segments of
various sizes.
• Memory data is stored as bytes.
• 8086 has 20 lines address bus.
• With 20 add line the memory can be
addressed as 220 bytes = 1MB
• Address ranging from 00000F to FFFFF H
Segment Registers of 8086 (cont..)
16-bit register containing address of 64 KB segment
1. Code segment (CS)
Processor use it for all accesses to instructions referenced by
instruction pointer (IP) .
The CS register is automatically updated during jump, call and
return instructions.
2. Stack segment (SS)
Used in the stack processing .
used two registers to referencing to it :
SP : Stack pointer
BP: Base pointer
Segment Registers of 8086 (cont..)

3. Data segment (DS)


The processor assumes that all data referenced by general
registers (AX, BX, CX, DX) and index register (SI, DI) is
located in the data segment.

4. Extra segment (ES)


The processor assumes that the DI register references the ES
segment in string manipulation instructions.

• It is possible to change default segments used by general and


index registers by prefixing instructions with a CS, SS, DS or
ES prefix.
Segment Register
• 8086 has 20-bit Address Bus. Thus, it can
address 1MB memory location.
• This 1MB memory can be accessed with the
help the above mentioned 4 Segment
Registers (CS, SS, DS & ES).
• This 1MB memory is divided into 16-bit logical
segments, each with a memory of 64KB.
• To locate any address in the memory, it needs
the Physical address of that memory location.

• It cannot directly get the 20-bit Physical


address using the 16-bit segment registers. It
needs to calculate the 20-bit Physical address.
Segment : Offset Address
• Logical Address is specified as segment: offset
• Physical address is obtained by shifting the
segment address one bit to the left and adding
the offset address.

• Thus the physical address of the logical address


2000:1000 is:
20000
+ 1000
21000
Segments, Segment Registers & Offset
Registers
• Segment Size = 64KB
• Maximum number of segments possible = 14
• Logical Address – 16 bits
• Physical Address – 20 bits
• 2 Logical Addresses for each Segments.
• Base Address (16 bits)
• Offset Address (16 bits)
• Segment registers are used to store the Base
address of the segment.
Segment and offsets

• Combination of Segment with Offset. Each


segment addressing 64 K byte.

Example :
let segment register=1000
Start Location = shift to left one hexadecimal digit
To be (10000 h)
End location= start location + FFFF h
So the end of this segment to be 1FFFF h.
Shift left
• 1000 H 10000 H

+ FFFF

1FFFF
More Examples:
Segment Register Start End
•
2000H

2001 H

2100 H

AB00 H

1234 H
Default Segment and Offset Registers

Segment Offset Special Purpose


CS IP Instruction addressing
SS SP, BP Stack addressing
DS BX, DI, SI, Data addressing
ES DI String Destination
addressing
Examples:

• Find the physical address if the logical address


SS:BP , SS=2000 H, BP=3000H

sol:
Physical Address = 20000+3000=23000 H to
addressing 1 M of memory from registers
consist of only 16 bit.
Home work:

If the Segment register was ( 100B h), with Offset (12 h )


Find the Physical Address?

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