Chapter 4

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CHAPTER 4: MICROCONTROLLER/MICROPROCESSOR

INSTRUCTION

SALTIHIE BIN [email protected]

ELECTRICAL ENGINEERING DEPARTMENT

POLITEKNIK MUKAH SARAWAK

Micrprocessor and Microcontroller System - Saltihie Zeni

mailto:[email protected]


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INTRODUCTION

• Instruction set – A list of instruction that can be executed by a microprocessor.

• Instruction are categorized according to basic operation performed:– Data transfer– Arithmetic– Logic– Shift and Rotates– Bit manipulation– BCD– Program control– System control

Instruction is written using mnemonic

(easy to remember code) because it is not practical to use

binary code for programming


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BASIC INSTRUCTION SET


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68000 REGISTER SET


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68000 REGISTER


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ADDRESSING MODE


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1. REGISTER DIRECT ADDRESSING


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2. IMMEDIATE ADDRESSING MODE


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IMMEDIATE ADDRESSING MODE


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3. ABSOLUTE ADDRESSING MODE


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4. ADDRESS REGISTER INDIRECT ADDRESSING MODE


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5. RELATIVE ADDRESSING MODE


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6. INHERENT ADDRESSING MODE


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Example 4.1

• State the addressing mode for the following instruction:

i. MOVE.W D2,D3

ii. JSR

iii. MOVE.B (A0), D3

iv. CLR.W $2003

v.MOVE.W #$1234,D0

vi.BRA DELAY

vii. SWAP D0,D1

viii. MOVE.B #%10010011,D3


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BUILDING A PROGRAM

• Program design is a process to provide an executable program in microprocessor.

• A task that need to be done by microprocessor need a conversion to machine language that can be understood by microprocessor.

• 5 Steps to build a program:– Definition of Problem– Logic Design– Programming– Load, Test and Debug– Documentation


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Problem analysis is a phase to convert a task that need to be done to an abstract shape and not particular to a microprocessor.This phase include a design using pseudo code or flow chart (will be discuss later).6 mini steps in problem analysis:– Clarify objectives and users– Clarify desired outputs– Clarify desired inputs– Clarify desired process– Double-check feasibility of implementing the program– Document the analysis.

1. DEFINITION OF PROBLEMS


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2. LOGIC DESIGN


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2. LOGIC DESIGN

• Algorithm can be represented in flow chart or pseudo code.

• Flow chart – A graphical representation of algorithm

• Pseudo – Artificial and informal language– Similar to everyday English


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2. LOGIC DESIGNFlow chart – Basic blocks


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• This phase is where a programmer write a task using editor(discussed in chapter 3).

• Editor is any software that can provide ASCII text like notepad or particular software like SciTE or programmers notepad.

• After source code provided, it then need to compile to produce a program that can be understood by microprocessor.

3. PROGRAMMING


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• In program writing, there will be a mistake or technically called error.

• 2 types of error:– Syntax error– Semantic error

3. PROGRAMMING


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• Syntax error– Mistake in program writing for example more or

less comma than needed. This error can be detect early and easy to do a correction.

• Semantic error– Logic mistake for example programmer done a

operation which is not appropriate and not suitable in the program.

– This error cannot detected by compiler and only can be seen when the program already execute.

3. PROGRAMMING


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3. PROGRAMMING


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4. LOAD, TESTING AND DEBUGGING


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5. DOCUMENTATION


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PROGRAM EXECUTION STRUCTURE

SEQUENCE SELECTION LOOPING


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JUMP Instruction

JMP = JumpEffect : Program Counter Effective AddressSyntax:

JMP <ea>Example:

JMP START ;START = $1000

JMP $2000

JMP (A0) ; A0 = $1000Useful in selection structure especially to go to sub-routine or sub-program.


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BRANCH ALWAYS Instruction

BRA = Branch AlwaysEffect : PC > PC +offsetSyntax:

BRA <ea>Example:

BRA LAST


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if-then STRUCTURE

• Basic control structure – conditional statement• Syntax in high level language programming

if <condition> then <statement>

• In 68000, this structure is:

<condition> ; testing condition and set CCRBCC SKIP ; select subroutine<Statement> ;execute statement

SKIP ….. ;continue


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CONDITIONAL BRANCH (Bcc)• Syntax:

Bcc <ea>• cc represent the condition that will be tested.• If condition TRUE = Subroutine executed• If condition FALSE = Subroutine not executed, program go to next instruction.


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Example Using Conditional Branch

1. Clear data in D7 if operation D6-1 equal to zero.

2. Add 1 to D0 if addition of D1+D2 produce carry.


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Compare Instruction

Compare = CMP• Function : Compare 2 data. Subtract data and set CCR. Destination must

data register. It look like subtract instruction but data register not change.


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Variation of CMP Instruction• CMPI (Compare immediate)

– To compare immediate data– Operation size : B,W,L– Destination : Data or memory register

• CMPA(Compare Address)– To compare address register– Operation size : W,L . W will be extended to L before comparison– Destination : memory register

• CMPM(Compare Memory)– To compare data inside memory – Operation size : B,W,L– Destination : (An)+ only


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Example of CMP and Bcc1. Subtract D1 from D0. Jump if result is 0.

2. Jump if value in D0 equal to 10.

CMPI #10, D0 ; Compare D0 with 10BEQ SAMA ; Jump if D0=10


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if-then-else Structure

• Syntax in high level language:if <condition> then<statement 1> else <statement 2>

• In 68000, this structure is:


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do-while Structure

• Syntax in high level programmingdo <statement>while <condition>

• In 68000, the structure is:


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while Structure

• Syntax in high level programmingwhile <condition> do <statement>

• In 68000, the structure is:


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SIMPLE PROGRAMMING EXAMPLE(ADDITION OF 2 NUMBERS)

1. Define the problemInput : 2 numbers (for example decimal 12 and 34) – save in data register D0 and D1Process : Addition of 2 numbersOutput : Add from Data register D0 and D1 and save result in D1.

2. Draw a flow chart or pseudo code.3. Using the assembly language, write the program.


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Enter first number to data register D0

Enter second number to register

D1

Calculate:D1 = D0 +D1ADD D0, D1

START

END

Flow chart to add 2 number saved in

data register


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PROGRAM

ORG $1000MOVE.B #12,D0MOVE.B #34,D1ADD.B D0,D1END

Write the following program in easy68k and observe the change in

register


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Example 4.2

Case study

1. A robot will move forward when the sensor 1 (S1) enabled. When the sensor 2 (S2)

is activated,) it will make a round of U (right and stop for 20 seconds, then

continue to move forward. Until the sensor 3 (S3) is activated, then the robot will

stop. Draw a flow chart to illustrate the process.

2. In a classroom, when switch 1 is ON it will ON an air-conditioner and all light

respectively. After 3 minutes, LCD screen move down and 2 minutes after that LCD

projector also ON. When switch 1 is turn OFF, 5 minutes after that, air-conditioner,

light and LCD also turn OFF and LCD screen will move up. Draw a flow chart to

illustrate the process.


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Example 4.3

Writing a program

Write a program for the flow chart given. Assume the starting address of your program

is $1000.

D0>5

D1=D0+7D0=D0-1

D7=D0/7

FALSETRUE

END

START


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Note ….

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