Microprocessor/Microcomputer

What is a Microcomputer
A complete computer based on a particular
microprocessor chip.
So the microprocessor is the most important
component in a microcomputer
So to study a microcomputer system, we
must first understand the microprocessor
What is a Microprocessor

Processor-on-a-chip can be described as

a microprocessor.
8051 series, 8086, Pentium series, etc

Block diagram of a generic

microcomputer system
RAM

Hard Disk
CD ROM

Data and address bus

Keyboard, mouse

Monitor, printer
2

Microprocessor based
system

sensor

Input

uP with
Control program

motor

Output

Structure of a modern
computer system

Model

Year

8086

Max. Clock
frequency at
introduction

Transis Register
tors
Sizes
per Die

Ext.
data
bus
size

Max.
Caches
external
address
space

1978 8 MHz

29K

16GP

16

1MB

None

486

1989 25MHz

1.2M

32GP
80FPU

32

4GB

L1:8KB

Pentium

1993 60MHz

3.1M

32GP
80FPU

64

4GB

L1:16K
B

P3

1999 500MHz

8.2M

32GP
80FPU
64MMX
128 XMM

64

64GB

L1:32K
B
L2:
512KB

64

64GB

L2:
1MB

Pentium 2007 1.6GHz to

Dual
2.4 GHz
Core
GP general purpose
FPU floating point unit

167 M

Register a device to store binary data

The Intel 8086

Microprocessor
The 8086
is a popular device used in

the early 70s and 80s and its

architecture is simple and suitable for
teaching computer architecture
Once we gain the basic concept of the
8086, we can then discuss the more
advanced microprocessors
Many features found in 8086 are still
being embedded in modern
microprocessors but enhanced!

8086 Microprocessor
This is a 16-bit microprocessor chip
manufactured by high-performance
metal-oxide semiconductor (HMOS)
technology
Circuitry on chip is approximately
29,000 transistors
Comes in a 40-pin package

Self test
Do you know what does it mean by
16-bit, 32-bit, or 64-bit processor?
How would you describe an Intel
Core2 Duo CPU ?

Basic 8086 features

True 16-bit microprocessor with 16-bit internal
and external data bus
The address bus and data bus are
multiplexed???
Multiplex address and data share the same
pin!!
A 20-bit address bus which allows access to 1
MB of memory.
Can address up to 64K byte-wide I/O ports
Or 32K word-wide ports (word = 16 bits)
Details regarding I/O ports will be discussed in
the I/O System

Pins layout for 8086

A/D address/data (address and
data share the pins - multiplexed)
Also pay attention to active high
and active low signals

10

8086 Features
The 8086 has two modes min. and
max.
Min. mode used as a typical
microprocessor
Max. mode use with multiple
processors, usually for floating-point
arithmetic)
The mode selection is via the MN/MX
input

11

Block diagram for a simple computer

system
Display unit
LCD

What are the

basic operations
performance by a computer?

memory

CPU

Get instruction from memory

Perform/Execute operation
Get next instruction

I/O

12

What are the basic operations

performed by a
microprocessor?
Get instruction from memory
Perform/Execute operation
Get next instruction
So inside the microprocessor, it is
organized into two units: Bus
Interface Unit (BIU) and Execution
Unit (EU). So that it can perform the
above operations effectively

13

Processor Model for 8086

14

The 8086 Internal

Architecture

The internal functions of the 8086 P are

divided between two separate processing
units. They are the Bus Interfacing Unit (BIU)
and the Execution Unit (EU).
The BIU is responsible for performing all bus
operations, such as instruction fetching,
reading and writing operands from/to memory,
and inputting and outputting of data for
peripherals.
The EU is responsible for executing instructions
The two units operate asynchronously so
overlapping instruction fetch and execution is
possible (whats the advantage of this???)
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Terminology
Program is stored in memory and consists of a
sequence of instructions and some data
To execute an instruction it may require some
operands
What is an operand?
Operand is the object that is being operated
upon!
Example, in an instruction ADD A, B (A = A+B)
ADD (addition is the operation)
A and B are the operands

16

Bus Interfacing Unit (BIU)

The BIU is the 8086s interface to the outside
world (external memory).
The major task of BIU is to get information
from the memory
Information includes data and instructions
How can we get data from memory?????
To access the memory, we need to issue an
address (via the address bus) and then read
the data (via the data bus) (Details of this
mechanism will be discussed when we discuss
the memory systems)

17

BIU
There is a full 16-bit bidirectional data bus
and 20-bit address bus
It has the following functions: instruction
fetch, instruction queueing, operand fetch
and storage, and bus control.
It contains the segment registers, internal
communication registers, instruction
pointer, instruction object code queue,
address summer (), and bus control logic.

18

How BIU and EU

collaborate

What does a program consists of ???

A program is a collection of instructions and
data
BIU fetch an instruction from memory and put
it in the queue and this is called instruction
queue (refer to the block diagram)
EU fetches the instruction from the queue and
executes
BIU and EU implements a pipeline (BIU->EU)
and pre-fetch to optimize the performance

19

BIU EU Pipeline
mechanism
Note: there are 3 components in the pipeline

Information
coming from memory

BIU

Control to access
the memory

EU
executes the instruction

queue that
can store 6 bytes
of instructions
EU requests BIU to get operands

20

Pre-fetch concept
Pre-fetching is similar to what you do when
youre having a buffet dinner.
You collect different kinds of food from the
buffet table, for example, you take the
sashimi, roast beef, soup, and salad etc.
When youre eating the salad, you have
already pre-fetched the sashimi and the
roast beef! If you do not pre-fetch then you
take the salad first, go back to the table, eat
your salad. When you finish the salad then
you go and get some other food.
Why pre-fetching your food??????
21

Pre-fetch
Is pre-fetching in a buffet dinner
exactly the same as the pre-fetching
mechanism in a microprocessor?
The plate is equivalent to which
component?
Is a big plate better than a small
plate?

22

Pre-fetch by BIU
Whats pre-fetch????
When the queue can store at least 2 bytes
EU is not requesting BIU to read or write operands
from memory
BIU will look ahead in the program by prefetching
the next sequential instruction
The prefetched instructions are held in the queue
which is a FIFO (First-in-first-out) device
Two bytes are fetched (16-bit data bus) in a single
memory cycle
EU will read one instruction byte from the output of
the queue

23

Pre-fetch
Int1a
Int1b
Int2a
Int2b
Int2c
Int3a
Int3b
Int4a
Memory

Queue

int1b int1a

int2b int2a int1b

EU

int1a
While EU is
processing int1a
Int2a and int2b have
already been
Pre-fetched
24

Instruction sequence

Fetch

Execute

Fetch

Execute

Fetch

Execute

25

Pre-fetching by BIU
If the instruction queue is full and EU is
not requesting access to operands in
memory, the BIU does not perform any
bus cycles this is called idle states
When BIU is in the process of fetching
an instruction when the EU requests its
services then BIU first completes the
instruction fetch bus cycle and then
serves the EU

26

Components in the BIU

BIU is to read/write the memory
What is needed to access the memory???
We need to generate an address and read/write the
data
BIU contains a dedicated adder () which is used to
generate the physical address of the memory location
Address is formed by adding an appended 16-bit
segment address and a 16-bit offset address
Example: the physical address of the next instruction
to be fetched is formed by combining the current
contents of the code segment (CS) register (16-bit)
and the current contents of the instruction pointer (IP)
register (16-bit)

27

Generating the physical

address

If CS (code segment) is 1005H

The IP is 5555H
What is the physical address? (how to
determine the physical address?)
Point to consider: the address bus of the
8086 is 20-bit, the registers are 16-bit.
Is it a problem????
Consider the sum of two 16-bit values,
what is the max. integer value
represented by 16-bit.

28

Segment concept

8086 can support up to 1M memory

Memory is divided into segments
Each segment is 64K
To access data inside a segment, we
need to know the base address of a
segment as well as the offset.
This is similar to a building, we can have
room 10B, 11B etc. The floor is the base
address and the letter B is the offset

29

Segment in 8086
Why segment mechanism is needed in
8086?
The address bus size (20-bit) > register size
(16-bit)
Example: if the address bus is 4 bits then you can
access 16 locations
If you can only output a 2-bit address from your
register then what will happen?

Save components can reduce the size of the

registers

A segment is a 64Kbyte memory block

30

Segment concept
Segment analogy
It is similar to an estate
Instead of building a very tall
building, we build smaller blocks
If you live in one of the smaller
blocks then your address will have
two components the block number
(segment) and the floor (offset)

31

Segment concept

A 64k segment

1M

How can we access locations

within a segment ??????
A 64k segment

32

Segment Registers
The segment registers are used for accessing the
memory
The 8086 address space is segmented into 64Kbyte segments and just four segments can be
active at a time. Because there are only 4
segment registers
In theory, how many segments can we have????
Total memory 1M and segment is 64K so 1M/64K number
of segment

33

The Segment concept

So the real address (physical
address) is =
Base address (20 bits) + offset
(16-bit)
Real address

Memory
(a segment)

Offset

The Base address must be

divisible by 16 so the last digit
is equal to 0 and the 0 is not
stored so a 16-bit register
can hold the rest of the address

Base address (segment address)

34

Segment concept
For example:
FFFFEH is not divisible by 16
FFFF0H is divisible by 16
12340H is also divisible by 16

35

Segment concept

The maximum value of a 16-bit value is FFFF (Hex), if two 16bit values added together, such as FFFF (segment) + FFFF
(offset), the result is 1FFFE (Hex) (physical) and it is only
a 17-bit value and values from 1FFFFH to FFFFFH
cannot be produced. As for a 20-bit pattern, it represents
values from 00000H - FFFFFH

So in 8086, you cannot randomly assign a segment. The

segment address must satisfy one condition, that is
the base address must be divisible by 16. If a value is
divisible by 16 and if we are using HEX (base 16) as the
number system then the last digit of the value must be a
0.

For example, the value in the segment register is 1234H

and the offset is 20H then the physical address is 12340H
+ 20H = 12360H.
36

Segment concept
The segment concept analogy
If you are design the elevator for a
very tall building, for example with
100 levels. How are you going to
arrange the buttons if the elevator
is able to reach all levels?

37

Execution Unit (EU)

The EU is responsible for decoding and
executing all instructions.
What is decoding ?
The EU will see data such as 8B C3 (10001011
11010011)
Decoding is to carry out the proper operation
according to the binary string (10001011
11010011)
8B C3 is (MOV AX, BX)
After decoding, EU will perform the move
(MOV) operation

38

Decoding

Instruction
10101100

Decoder

Control
signals

39

Execution Unit (EU)

(Contd)
EU consists of an ALU (Arithmetic and
Logic Unit), status and control flags, eight
general-purpose registers, temporary
registers, and queue control logic
The EU extracts instructions from the top
of the queue in the BIU, decodes them,
generates operand addresses if necessary,
passes them to the BIU and requests it to
perform the read or write bus cycles to
memory or I/O, and performs the operation
specified by the instruction on the
operands.
During execution of the instruction, the EU
tests the status and control flags and
updates them based on the results of
40
executing the instruction.

Functions of EU

ADD AX, 16 ; meaning add 16 to AX

Where AX is a register inside the CPU
If AX is 20 then after the operation it becomes 36
For the above operation, do we need to fetch
operand from memory?
16 in the above operation is called an immediate
Immediate values are stored as part of an
instruction and fetched together with the
instruction
Now if it is ADD AX, X ; X is a variable
Do we need to fetch the operand X from memory?
41

Functions of EU
If the instruction queue is empty, the EU
waits for the next instruction byte to be
fetched and shifted to the top of the
queue.
When the EU executes a branch or jump
instruction, it transfers control to a
location corresponding to another set of
sequential instructions. Whenever this
happens, the BIU automatically resets
the queue and then begins to fetch
instructions from this new location to
refill the queue.
42

Jump and branch

43

Summary
What is the pre-fetch concept?
What is a pipeline and its advantage?
What are functions performed by the
BIU and EU
What is a multiplexed address/data
bus
What is the segment concept
44

8086 Internal Registers

Registers are a very important component because
they are used as a temporary storage, as well as
storing the current status of the CPU. Contents of some
registers indicate the memory locations to be fetched.
Registers are internal components that we can control
with assembly language programming
4 groups of 16-bit register

Instruction Pointer (IP)

Data Registers (4)
Pointers and Index Registers (4)
Segment Registers (4)

The Flag Register

45

Instruction Pointer (IP)

Identifies the location of the next instruction
to be executed in the current code segment
IP contains an offset value not the physical
address of the next instruction
Physical address = IP+CS (code segment
register)
Every time an instruction word is fetched
from memory, the BIU updates the values in
IP (eg IP = IP+1) such that it points to the
next sequential instruction word in memory

46

Data Registers
4 general purpose data registers
and are used for temporary
storage of frequently used
intermediate results.
This can improve the speed
(why???)
Register can use either as 8-bit or
16-bit
Accumulator Register (AX: AH AL)
Base Register (BX: BH BL)
AX (16-bit)
Count Register (CX: CH CL)
AH (8-bit) AL (8-bit)
Data Register (DX: DH DL)
47

Data Registers
The general purpose data registers can be used
for arithmetic or logic operations
For example, to carry out an addition: add ax, bx
The result is stored in ax and it is equal to the
sum of values in ax and bx (in C, it is similar to
ax+=bx)
For string instruction, the CX register is used to
store a count value representing the number of
bytes to be moved
All I/O operations require data that are to be
input or output to be in the A register, while
register DX holds the address of the I/O port

48

Segment Registers
The segment registers are used for accessing the memory
The 8086 address space is segmented into 64K-byte
segments and just four segments can be active at a time.
In theory, how many segments can we have????
The segment registers are used to select the active segments

Code Segment (CS) Register

CS identifies the starting address of the 64-K byte segment
known as the code segment. Code segments of memory
contain instructions of the program.

Data Segment (DS) Register

DS register identifies the starting location of the current
data segment in memory. Data is stored in the data
segment.

49

Segment Registers
(Contd)
Stack Segment (SS) Register
SS register contains a logical address that identifies
the starting location of the current stack segment
in memory. Stack is used for temporary storage

Extra Segment (ES) Register

ES register identifies the extra segment usually
used for data storage.

The segment registers store the base address

of a segment. To determine the physical
address, an offset is required. The index
registers are used to store the offset value.

50

Pointer and Index

Registers
Stack Pointer (SP) permits easy access to
locations in the stack segment of memory

The value in SP represents the offset of the next stack

location which can be accessed relative to the current
address in the stack segment (SS) register, i.e., always
points to the top of the stack.

Base Pointer (BP)

BP represents an offset from the SS register. However,
it is used to access data within the stack segment.
Used in the based addressing mode
The applications of the various registers will be
discussed in details when we learn assembly language
programming

51

Pointer and Index

Registers
Index register are used to hold offset addresses for
instructions that access data stored in the data
segment of memory.

Source Index Register (SI)

SI is used to store an offset address for a source
operand under index addressing for string and
memory operation.

Destination Index Register (DI)

DI is used for storage of an offset that identifies the
location of destination operand also used in some
string operations.
Remarks: The offset value is always referenced to the
value in the data segment (DS) register.

52

Registers and pointers

Segment register

Pointer

CS (code segment)

IP (instruction pointer)

DS (data segment)

DI, SI

SS (Stack segment)

SP (stack pointer)
BP (base pointer)

ES (Extra segment)

DI

53

Flag Register
The flag register is a 16-bit register
within the execution unit. The
status flags in the register indicate
conditions that are produced as
the result of executing an
arithmetic or logic instruction.
What kind of conditions can you
think of???

54

8086 Flag Register (status

flag)

C - Carry Bit (set if there is a carryout or borrowin)

P - Parity Bit (set if lower byte of the result
contains even number of 1s) odd parity
Z - Zero Bit (set if result after an operation is
equal to zero)
S - Sign Bit (represent negative value produced
during an operation)
O - Overflow Bit (result is out of range). If the
result of a signed operation is not large enough to
be accommodated in a destination register.
The above are the most commonly used flag
registers, there are others but not discussed in
this subject!!!!!!

55

Example
If our data is only 8-bit then when we
do FFH + 1H = 1 0000 0000 this is a
9-bit value the 1 is the carry!!!!
Similarly when we do 00H 1H then
result is 1 1111 1111 the 1 is the
borrow bit.

56

Flags
consider using 8-bit values A and B,
determine flag status for C, S, Z and O
If
If
If
If
If
If

A = 0FH, B = 1; A+B
A = 0, B = 1; A-B
A = 7FH, B = 1; A+B
A = 80, B = 0F; A-B
A = FFH, B = 1; A+B
A = 2FH, B = 60, C = -1; (A+C) -B
57

Block diagram for a simple computer

system
Display unit
LCD

memory

CPU

Get instruction from memory

Perform operation
Get next instruction

I/O

58

Bus Cycle
Bus address and data
Bus cycle is used to access memory, I/O
devices, or the interrupt controller.
Bus cycle starts with an address being output
on the system bus followed by a read or write
data transfer.
A series of control signals are produced to
control the direction and timing of the bus
A standard bus cycle consists of 4 clock periods
Understand system bus timing will assist you to
choose the proper memory device
59

Bus cycle
T1 : BIU puts an address on the bus
T2: data are put on the bus (for write
cycle)
T2: bus in High Z mode (for read cycle)
T3: data on the bus
T4: data on the bus
For a 5MHz system, how long does it
take to complete 1 bus cycle????

60

Read cycle

61

Write cycle

62