programming of microprocessors

8/12/2019 Programming of MIcroprocessors

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Programming of Microprocessors

Slides Prepared By

Hardeep Singh

Lecturer/ECE

SBBSIET, Padhiana

Introduction A Computer can do only what the programmer asks

to do.

To perform a particular task the programmerprepares the sequence of instruction.

A set of programs for a particular computer is calledsoftware of the computer.

Program written in the form of 0s and 1s calledmachine language program.

Problems related with machine language

Difficult to understand or debug a program Programs are long

Program writing is difficult and tiresome

Chances of careless errors in writing the program

24-Oct-11 Hardeep Singh/Lecturer/SBBSIET 2


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To facilitate programmer machine codes canbe written in hexadecimal system.

Hexadecimal can easily be written inhexadecimal system compared to binarysystem.

Microprocessor kit converts hexadecimal intobinary as the microcomputer understands aprogram only in binary system

The hexadecimal is also not convenient tounderstand so we develop mnemonics for

each hexadecimal code which is in userfriendly language


Assembly Language

A programmer easily write a program inalphanumeric symbols instead of binary orhexadecimal.

Meaningful and easily rememberable symbols arechosen for the purpose. Example: ADD for addition

SUB for subtraction, etc.

Such symbols are called mnemonics. Program written in mnemonics are called

Assembly language program



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A microprocessor specific language is called low levellanguage.

A program which translate as assembly languageprogram into machine language program is calledassembler

Self Assembler

present on microcomputer which translates the code

Disadvantage is that it requires lot of memory space and time totranslates the code

Cross Assembler

Installed on desktop computer

Converts code into machine level objects and transfer tomicrocomputer

Disassembler Converts Machine code into assembly language program


Addition of two numbers placed in two consecutive memory

locations 2501H and 2502H

Mnemonics Operands Memory

Address

Machine Codes Comments

LXI H, 2501H 2000 21, 01, 25 Get the address of the 1st

number in HL pair

MOV A,M 2003 7E Get the 1st number in

accumulator

INX H 2004 23 Increment the content of

HL pair

ADD M 2005 86 Add the 1st and 2nd number

STA 2503H 2006 23, 03, 25 Store sum in 2503

HLT 2009 76 Stop



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Stack During the execution of a program sometimes it is

necessary to save the contents of the certain registersbecause the registers are required for some otheroperation in subsequent steps.

These contents moved to certain memory location tocertain memory locations by PUSH operations

After completing these operations those contents whichwere saved in the memory are transferred back to theregisters by POP operations.

Memory locations for this purpose is set aside by theprogrammer in the beginning.

The set of these memory locations are called stack

The last memory location of the occupied portion of thestack is called stacktop.

A 16-bit special register (Stack Pointer) holds the address ofstacktop



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The stack pointer is initialized in the beginning

of the program by LXI SP or SPHL instructions. Any of RAM can be used by stack.

There is no restriction on the location of the

stack in the memory.

Data stored in the stack on last-in-first-out

(LIFO) principle

Stack access is faster than memory access


Stack Related Instructions

PUSH Rp

POP Rp

SPHL

XTHL



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PUSH Rp

Push the contents of register pair on to stack

This instruction is used to write 16 bit data on the stack

When this instruction is executed the contents ofspecified register pair are copied on the stack in thefollowing sequence. The stack pointer in decremented by one and the contents

of higher order register of the specified register pair arecopied to the memory location pointed the stack pointer.

The stack pointer is again decremented by 1 and thecontents of lower order of register pair are copied to

memory location pointed by the stack pointer. Example: PUSH B


POP Rp

POP off stack contents to register pair

When this instruction is executed, the contents ofmemory location pointed by the stack pointer registerare copied to the lower order byte of register pair

The stack pointer in incremented by one and thecontents of that memory location are copied to thehigher order byte of register pair.

The stack pointer is again incremented by one

The process of POP is exactly opposite to PUSHoperation. So the contents stored by PUSH are takenback using POP instructions.

Example POP B



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SPHL

Load stack pointer with HL register paircontents

When this instruction is executed, thecontents of HL pair are transferred to the stackpointer

The contents of H register is copied to higherorder byte of stack pointer and the contents

of L register are copied to lower byte of stackpointer


XTHL

Exchange HL with top of stack

When this instruction is executed the contents

of L register are exchanged with stack location

pointed by the stack pointer. The contents of

H register are exchanged with the next stack

location

The contents of the stack pointer are not

altered.



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LXI SP

This instruction initialized the stack pointer

with 16 bit address

The stack point can be intialized by two

methods

Direct method: LXI SP, Data 16 Bit

Indirect method : LXI H, Data 16 Bit SPHL


Some other Commands

INX SP

This instruction increments the Stack Pointer by 1.

DAD SP

This instruction adds the contents of SP with the

contents of HL pair. The result of addition is stored

into the HL pair

DCX SP

This instruction decrements the stack pointer by 1



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2604

2605

2606

2607 SP

2608 6B 2608

2609

2610

Subroutines

While writing a program certain operations may occurseveral times and they are not available as individualinstructions.

Example is multiplication of numbers, as this programis used again & again

The concept of subroutines is developed to avoid therepetition of smaller programs.

The small program for a particular task is calledsubroutine.

Subroutine written separately & stored in the memory They are called at various points of the main program

by CALL instruction where they required.



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The last instruction of subroutine is RET

After completion of subroutine, the main programbegins from the instruction immediately followingthe CALL instruction

When program jumps from the main program tosubroutine , the counter of the program counter issaved in the stack so that the program executionmay come back to the main program as the proper

point after the completion of the subroutine.


Introduction to Programming

Step 1: Define the problem

Step 2: Solution Plan

Step 3: Flowchart

Step 4: Program

Step 5: Check the result



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Direction Arrow

Input & Output Unit

Process

Operation

Decision Unit

Predefined

Process/

Subrotines

Examples of Assembly Programming

Example 1: Place 05 in register B


Memory

Address

Machine

Codes

Mnemonics Operands Comments

FC00 06, 05 MVI B, 05 GET 05 IN

REGISTER B

FC02 76 HLT STOP


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Example 2: Get 05 in register A and

then move it to register BMemory

Address

Machine Codes Mnemonics Operands Comments

FC00 3E, 05 MVI A, 05 Get 05 in

register A

FC01 47 MOV B, A Transfer 05

from register A

to register B

FC02 76 HLT Stop


Place 05 in the accumulator. Increment it by one and

store the results in the memory location FC50 H

Memory

Address

Machine Codes Mnemonics Operands Comments

FC00 3E, 05 MVI A, 05 Get 05 in the

accumulator

FC01 3C INR A Increment the

contentof the

accumulator by

one

FC03 32, 50, FC STA FC05 H Store result in

FC05 H

FC06 76 HLT Halt



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8 Bit SubtractionMemory

Address

Machine

Codes

Mnemonics Operands Comments

2000 21, 01, 25 LXI H, 2501 H Get address of 1st number in H-

L pair

2003 7E MOV A, M 1st number in accumulator

2004 23 INX H Content of H-L pair increases

from 2501 H to 2501 H

2005 96 SUB M 1st Number 2nd Number

2006 23 INX H Content of H-L pair becomes

2503 H

2007 77 MOV M, A Store result in 2503 H

2008 76 HLT Halt


Problem

Write a program to provide the given traffic lights

(Green, Yellow, Red) and two pedestrian (Walk

and Dont Walk)

The signal lights and signs turned on/off by the

data bits of an output port as shown below


Lights Data Bits On Time (Seconds)

Green DO 15

Yellow D1

5

Red D2 20

Walk D3 15

Dont Walk D4 25


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Problem Analysis

The problem is primarily concerned with providing varioustime delays for a complete sequence of 40 seconds.


Timein

Sequence

Dont

Walk

Walk Red Yellow Green Hex

Code

D7 D6 D5 D4 D3 D2 D1 D0

0

15 0 1 0 0 0 0 0 1 41H

20 1 0 0 0 0 1 0 0 86H

40 1 0 0 1 0 0 0 0 90H


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