MICROPROCESSOR & MICROCONTROLLER LABORATORY MANUAL 4TH SEMESTER ISE SUBJECT...

MICROPROCESSOR & MICROCONTROLLER

LABORATORY MANUAL

4TH SEMESTER ISE

SUBJECT CODE: 15CSL48

FACULTY: MRS DEEPTI C

DEPARTMENT OF

INFORMATION SCIENCE AND

ENGINEERING

PESIT – BANGALORE SOUTH CAMPUS

1KM BEFORE ELECTRONICS CITY,

HOSUR ROAD, BANGALORE-560100.

PESIT BANGALORE SOUTH CAMPUS DEPARTMENT OF ISE

4T H SEMESTER MICROPROCESSOR LAB 10CSL48

MICROPROCESSOR AND MICROCONTROLLER

LABORATORY [As per Choice Based Credit System (CBCS) scheme]

(Effective from the academic year 2016 -2017)

SEMESTER – IV

Subject Code 15CSL48 IA Marks 20

Number of Lecture Hours/Week 01I + 02 P Exam Marks 80

Total Number of Lecture Hours 40 Exam Hours 03 Credits – 02

Course objectives: This course will enable students to

· To provide practical exposure to the students on microprocessors, design and coding

knowledge on 80x86 family/ARM. To give the knowledge and practical exposure on

connectivity and execute of interfacing devices with 8086/ARM kit like LED displays,

Keyboards, DAC/ADC, and various other devices.

Description

Demonstration and Explanation hardware components and Faculty in-charge should

explain 8086architecture, pin diagram in one slot. The second slot, the Faculty in-charge

should explain instruction set types/category etc.

Students have to prepare a write-up on the same and include it in the Lab record and to be

evaluated.

Laboratory Session-1: Write-up on Microprocessors, 8086 Functional block diagram, Pin

diagram and description. The same information is also taught in theory class; this helps the

students to understand better.

Laboratory Session-2: Write-up on Instruction group, Timing diagrams, etc. The same

information is also taught in theory class; this helps the students to understand better.

Note: These TWO Laboratory sessions are used to fill the gap between theory classes and

practical sessions. Both sessions are evaluated as lab experiments for 20 marks.

Experiments

· Develop and execute the following programs using 8086 Assembly Language. Any

suitable assembler like MASM/TASM/8086 kit or any equivalent software may be used.

· Program should have suitable comments.

· The board layout and the circuit diagram of the interface are to be provided to the student

during the examination.

· Software Required: Open source ARM Development platform, KEIL IDE and Proteus for

Simulation



SOFTWARE PROGRAMS: PART A

1. Design and develop an assembly language program to search a

key element “X” in a list of „n‟16-bit numbers. Adopt Binary search

algorithm in your program for searching.

2. Design and develop an assembly program to sort a given set of

„n‟ 16-bit numbers in ascending order. Adopt Bubble Sort algorithm

to sort given elements.

3. Develop an assembly language program to reverse a given string

and verify whether it is a palindrome or not. Display the appropriate

message.

4. Develop an assembly language program to compute nCr using

recursive procedure. Assume that „n‟ and „r‟ are non-negative

integers.

5. Design and develop an assembly language program to read the

current time and Date from the system and display it in the standard

format on the screen.

6. To write and simulate ARM assembly language programs for data

transfer, arithmetic and logical operations (Demonstrate with the

help of a suitable program).

7. To write and simulate C Programs for ARM microprocessor

using KEIL (Demonstrate with the help of a suitable program)

Note: To use KEIL one may refer the book: Insider‟s Guide to the

ARM7 based microcontrollers, Hitex Ltd., 1st edition, 2005



HARDWARE PROGRAMS: PART B

8. a. Design and develop an assembly program to demonstrate BCD

Up-Down Counter (00-99) on the Logic Controller Interface.

b. Design and develop an assembly program to read the status of

two 8-bit inputs (X & Y) from the Logic Controller Interface and

display X*Y.

9. Design and develop an assembly program to display messages

“FIRE” and “HELP” alternately with flickering effects on a 7-

segment display interface for a suitable period of time. Ensure a

flashing rate that makes it easy to read both the messages (Examiner

does not specify these delay values nor is it necessary for the student

to compute these values).

10. Design and develop an assembly program to drive a Stepper

Motor interface and rotate the motor in specified direction

(clockwise or counter-clockwise) by N steps (Direction and N are

specified by the examiner). Introduce suitable delay between

successive steps. (Any arbitrary value for the delay may be assumed

by the student).

11. Design and develop an assembly language program to

a. Generate the Sine Wave using DAC interface (The output of the

DAC is to be displayed on the CRO).

b. Generate a Half Rectified Sine waveform using the DAC

interface. (The output of the DAC is to be displayed on the CRO).

12. To interface LCD with ARM processor--

ARM7TDMI/LPC2148. Write and execute programs in C language

for displaying text messages and numbers on LCD

13. To interface Stepper motor with ARM processor--

ARM7TDMI/LPC2148-Write a program to rotate stepper motor



Study Experiments:

1. Interfacing of temperature sensor with ARM freedom board (or any other

ARM microprocessor board) and display temperature on LCD

2. To design ARM cortex based automatic number plate recognition system

3. To design ARM based power saving system

Course Outcomes: The students should be able to:

· Learn 80x 86 instructions sets and gins the knowledge of how assembly

language works.

· Design and implement programs written in 80x86 assembly language

· Know functioning of hardware devices and interfacing them to x86 family

· Choose processors for various kinds of applications.

Graduate Attributes

· Engineering Knowledge

· Problem Analysis

· Modern Tool Usage

· Conduct Investigations of Complex Problems

· Design/Development of Solutions

Conduction of Practical Examination:

· All laboratory experiments (all 7 + 6 nos) are to be included for practical

examination.

· Students are allowed to pick one experiment from each of the lot.

· Strictly follow the instructions as printed on the cover page of answer script

for breakup of marks

· PART –A: Procedure + Conduction + Viva: 10 + 25 +05 (40)

· PART –B: Procedure + Conduction + Viva: 10 + 25 +05 (40)

· Change of experiment is allowed only once and marks allotted to the

procedure part to be made zero.



MASM COMMANDS:

C :/>cd foldername

C:/foldername>edit filename.asm

1. Type EDIT at the command prompt (C :\>\MASM\). A window will be opened with all the options

like File, Edit etc., In the workspace, type your program according to the assembly language syntax

and save the file with a .asm. extension. (say test.asm).After this command executed in command

prompt an editor window will open. Program should be typed in this window and saved. The

program structure is given below.

Structure of Program:

.model tiny/small/medium/large

.Stack <some number>

.data

; Initialize data

; which is used in program.

.code

; Program logic goes here.

;

end



To run the program, the following steps have to be followed:

C:/foldername>masm filename.asm

After this command is executed in command prompt if there are no errors in program

regarding to syntax the assembler will generates an object module as discuss above.

C:/foldername>link filename.obj

After verifying the program for correct syntax and the generated object files should be

linked together. For this the above link command should be executed and it will give an EXE file if

the model directive is small as discuss above.

C:/foldername>debug filename.exe

After generating EXE file by the assembler it’s the time to check the output. For this the

above command is used and the execution of the program can be done in different ways. It is as

shown below:

__ g ; complete execution of program in single step.

__ t ; Stepwise execution.

__d ds: starting address or ending address ; To see data in memory locations

__p ; Used to execute interrupt or procedure during stepwise execution of program

__ q ; To quit the execution.



Syntax of an instruction:

{label:} mnemonic {operand { , operand} } {; comment}

SIMPLIFIED SEGMENT DIRECTIVES

MASM version 5.0 and above, and TASM provide a simplified set of directives for declaring

segments called simplified segment directives. To use these directives, you must initialize a

memory model, using the .MODEL directive, before declaring any segment. The format of the

.MODEL directive is:

.MODEL memory-model

The memory-model may be TINY, SMALL, MEDIUM, COMPACT, LARGE, HUGE or

FLAT :

memory-

model

description

TINY One segment. Thus code and data together may not be greater than 64K

SMALL One code-segment. One data-segment. Thus neither code nor data may be

greater than 64K

MEDIUM More than one code-segment. One data-segment. Thus code may be greater than

64K

COMPACT One code-segment. More than one data-segment. Thus data may be greater than

64K

LARGE More than one code-segment. More than one data-segment. No array larger than

64K. Thus both code and data may be greater than 64K

HUGE More than one code-segment. More than one data-segment. Arrays may be

larger than 64K. Thus both code and data may be greater than 64K

FLAT One segment up to 4GB. All data and code (including system resources) are in a

single 32-bit segment.



Initialization of DS

Note: The instructions which initialize the DS register for an exe-format program with

simplified segment directives are:

MOV AX , @DATA

MOV DS , AX

At load time, @DATA is replaced with the 16-bit base address of the data segment.

Comments:

A semicolon starts a comment. A comment may follow a statement or it may be on a separate

line. Multiple-line comments can be written by using the COMMENT directive.

Numbers:

A binary number is suffixed by b or B.

e.g., 11010111B

A decimal number is suffixed by an optional d or D.

e.g., 42d -22D 3578

A hexadecimal number must begin with a decimal digit and it is suffixed by h or H

e.g., 20H 0bF2Ah

Characters:

A character is enclosed in a pair of single quotes or in a pair of double quotes.

e.g., ‘x’ “B”

Strings:

A string is enclosed in a pair of single quotes or in a pair of double quotes.

e.g., ‘ENTER YOUR NAME: ’

“THE MAXIMUM VALUE IS ”



Data definition

Each variable has a data type and is assigned a memory address by the program. The data-

defining directives are:

Directive Description of Initializers

BYTE, DB (byte) Allocates unsigned numbers from 0 to 255.

SBYTE (signed byte) Allocates signed numbers from –128 to +127.

WORD, DW (word = 2 bytes) Allocates unsigned numbers from0 to 65,535 (64K).

SWORD (signed word) Allocates signed numbers from–32,768 to +32,767.

DWORD, DD

(doubleword = 4 bytes), Allocates unsigned numbers from 0 to 4,294,967,295 (4

megabytes).

SDWORD (signed doubleword) Allocates signed numbers from –2,147,483,648 to +2,147,483,647.

A ? in place of an initializer indicates you do not require the assembler to initialize the

variable. The assembler allocates the space but does not write in it. Use ? for buffer areas or

variables your program will initialize at run time.

The DUP operator can be used to generate multiple bytes or words with known as well as un-

initialized values.

The EQU (equate) directive assigns a name to a constant expression

Syntax is:

name EQU constant_expression

Note: (i) No memory is allocated for EQU names

(ii) A name defined by EQU may not be redefined later in a program.

The LABEL directive, whose syntax is:

name LABEL type



1. a) Search a key element in a list of ‘n’ 16-bit numbers using the Binary search

algorithm.

ALGORITHM:

Step1: Initialize low=1,high=n;

Step2: while(low<=high)

{

/*Compare low > high if true the element is not found, exit*/

Find mid

else

mid=(low+high)/2;

Step3: if(key==a[mid])

found & exit;

else if(key>a[mid])

low=mid+1;

else

high=mid-1;

}

Step4: Not found.

Step5: Stop.

Algorithm:

Step 1 : Declare the array elements in the sorted order

Step 2 : Declare the search element

Step 4 : Assign low as 0 and high as n-1

Step 5 : Find mid index = (low +high)/2

Step 6 : Compare mid element with search element

Step 7 : If search element <mid element assign high as mid-1

Step 8 : If search element >mid element, assign low as mid+1

Step 9 : If search element = mid element , the search is successful so display the

location of the search element and go to step -12

Step 10 : Repeat step 5 to 9 until the search is successful

Step 11 : If search element is not available, display “element not available” Step 12 : Terminate the program



PROGRAM

.MODEL SMALL ; memory model

.DATA ; data segment area. Define all variables and messages

ARRAY DW 1111H,2222H,3333H,4444H,5555H ; declaring array of elements in asc order

LEN DW ($-ARRAY) ; finding length of array

KEY DW 3333H ; search element

MSG1 DB 10, 13,'ELEMENT FOUND AT POSITION $' ; message if key found

MSG2 DB 10, 13, „ELEMENT NOT FOUND$' ; message if key not found

RES DB ? 10,13,‟$‟

.CODE ; code segment starts

MOV AX, @DATA ; initializing data segment

MOV DS, AX

MOV BX, 01H ; set lower bound

MOV DX, LEN ; set upper bound

MOV CX, KEY ; move the key element to CX register

AGAIN: CMP BX, DX ; compare first and last element indexes

JA FAIL ; if lower bound>upper bound jump to FAIL

MOV AX, BX ; to calculate MID of array

ADD AX, DX ; find LOW+HIGH

SHR AX, 1 ; divide the value in AX by 2

MOV SI, AX

DEC SI

ADD SI, SI ; 2(AX-1) gives actual index of mid

CMP CX, ARRAY [SI] ; compare CX and the array element at MID

JAE BIGGER ; if equal jump to BIGGER

mailto:@data



DEC AX ; search elements below MID

MOV DX, AX ; HIGH=MID-1

JMP AGAIN ; jump to AGAIN

BIGGER: JE SUCCESS ; jump if equal to SUCCESS

INC AX ; search elements above MID

MOV BX, AX ; LOW=MID+1

JMP AGAIN ; unconditional jump to repeat the above

FAIL: LEA DX, MSG2 ;if not found display mesg2

MOV AH, 09H ;service no. to display message

INT 21H

JMP EXIT ; interrupt to terminate

SUCCESS: ADD AL,'0' ;get ASCII value of key position

MOV RES, AL

LEA DX, MSG1 ;if key element found display MESG1

MOV AH, 09H ;service no. to display message

INT 21H

MOV DL, RES ;display key position

MOV AH, 02H

INT 21H

EXIT: MOV AH, 4CH ;interrupt to return to DOS

INT 21H

END

*********************************



OUTPUT:

KEY=3333H ELEMENT FOUND

KEY=3331H ELEMENT NOT FOUND

RESULT: This program performs a search for a key element in an array. If the search

element is found a message „found‟ is displayed. If the search element (key in program) is

not present in the given array a message „not found‟ is displayed.



2. Sort a given set of ‘n’ numbers in ascending order using the Bubble Sort algorithm.

Algorithm:

Step 1: Initialize array elements in data segment.

Step 2: Initialize the number of elements in a array(N-array)

Number of elements in array (N)

=current address of memory pointer ($)– array of Array N(array name)

Step 3: Where List is array of ‘n’ elements

Bubble(int(list[],know)

{

int i,j;

for(i=1;i<=n;i++)

for(j=1;j<=n-i;j++)

if(list[j]>=list[j+1])

swap(list[j],list[j+1])

}

Step 4: Stop

ALGORITHM

Step 1 : Declare the array with the numbers that need to be sorted.

Step 2 : Initialize iteration count (n-1)

Step 3 : Initialize comparison counter

Step 4 : Compare num1 and num2

Step 5 : Num1<=num2 do not exchange

Step 6 : Num1>=num2 then exchange the positions of both numbers

Step 7 : Decrement iteration counter, comparison counter

Step 8 : Terminate the program



PROGRAM FOR ARRAY OF N 8-BIT NUMBERS

.MODEL SMALL

.STACK 64 ;stack segment

.DATA ;data segment

ARR DB 10H, 08H, 04H, 09H, 06H ;byte definition of array

LEN DB ($-ARR) ;length of the array

.CODE

MOV AX,@DATA ; initialise data segment

MOV DS, AX

MOV BL, LEN ;move the length into BL

DEC BL ;decrement BL by 01

OUTLOOP: MOV CL, BL ;move BL value into CL

MOV SI, 00H ;initialize SI to 00H

INLOOP: MOV AL, ARR[SI] ;move A[SI] into AL register

CMP AL, ARR[SI+1] ;compare values in AL and ARR[SI]

JB NEXT ;if AL<A[SI+1],jump to NEXT

XCHG AL,ARR[SI+1] ;swap values of AL and A[SI]

MOV ARR[SI],AL

NEXT: INC SI ;increment SI value by 01

LOOP INLOOP

DEC BL ;decrement BL value by 01

JNZ OUTLOOP ;jump to outloop till ZF=1

MOV AH,4CH ;end program

INT 21H

END



OUTPUT:

-G

-D DS:0000

13F7:0000 0A 00 46 E2 EB FE CB 75-E2 CC 04 06 08 09 10 00



PROGRAM FOR ARRAY OF N 16-BIT NUMBERS

.MODEL SMALL

.DATA

ARR DW 0020H,0070H,0040H,0010H,0050H ;word definition of array

CNT EQU 5

.CODE

MOV AX, @DATA ;initialise data segment

MOV DS, AX

MOV DX, CNT-1 ; no of iterations in DX reg

AGAIN0: MOV CX, DX ; no of comparisons

LEA SI, ARR ; SI = index of the array

AGAIN1: MOV AX, [SI] ; value of ARR [SI] moved to AX register

CMP AX, [SI+2] ; comparing elements ARR[SI] and ARR[SI+2]

JL LOOP1 ; exchange not required if ARR[SI] <ARR[SI+2]

XCHG [SI+2], AX ; exchange elements if bigger

XCHG [SI], AX

LOOP1: ADD SI, 02 ;repeat with every two numbers

LOOP AGAIN1 ;decrement number of comparisons

DEC DX ;decrement number of iterations each time

JNZ AGAIN0

MOV AH, 4CH ;terminate program

INT 21H

END

**************************

OUTPUT:

ASCENDING ORDER: 0010H, 0020H, 0040H, 0050H, 0070H

RESULT: This program performs an ascending order sort on given array. It will store the result in

data segment. The data segment memory location is observed for given array.

mailto:@data



3. Reverse a given string and check whether it is a palindrome or not.

Algorithm:

Step 1: Initialise a string (STR)

Step 2: SI points to the initial memory location and store the string in [SI].

Step 3: If [SI] is entered string then go to Step 4 to display the reversed string.

Step 4: Decrease SI by 1 location and move the content of SI to DI register. DI is assumed to point

at the reverse string.

Step 5: Compare the characters which is stored in SI and DI. If equal, display (“Palindrome”) using

‘09H’ service.

Else display (“Not a Palindrome”) using 09H service.

Step 6: Stop.

Algorithm:

Step 1 : Create display macro to display the message

Step 2: Declare the string

Step 3: Declare the message to display

Step 4: Find the reverse of string and store in string1

Step 5: Is string=string1, display it is a palindrome

Step 6: Else if display not a palindrome

Step 7: Terminate the program



PROGRAM

DISP MACRO MSG

LEA DX, MSG

MOV AH, 09H

INT 21H

ENDM

.MODEL SMALL

.DATA

STR DB "MALAYALAM$" ; array to store original string

LEN EQU ($-STR)

MSG1 DB 10, 13,"PALINDROME$"

MSG2 DB 10, 13,"NOT A PALINDROME$"

RSTR DB 40 DUP (0) ; array to store reverse string

.CODE

MOV AX, @DATA ; Initialisation of Data Segment

MOV DS, AX

MOV ES, AX ; Initialisation of Extra Segment

MOV CX,LEN

DEC CX

LEA SI, STR ;EA of STR is loaded to SI register

ADD SI,CX

DEC SI

LEA DI, RSTR ; EA of RSTR is loaded to DI register

AGAIN: MOV AL, [SI] ; Character in location pointed by SI is copied to AL

MOV [DI], AL ; Character in AL is copied to location pointed by DI

DEC SI ; decrements SI to point to next character of reversed string



INC DI ; increments DI to point to next character of entered string

DEC CX

CMP CX,0000H

JNZ AGAIN ;continue process till CL Register =Zero

MOV AL,"$"

MOV [DI],AL

LEA SI, STR

LEA DI, RSTR

MOV CX,LEN

REPE CMPSB ; compare original and reversed strings

JE PALIN ; if equal jump to PALIN

DISP MSG2 ; Display Message NOT PALINDROME

JMP EXIT

PALIN: DISP MSG1 ; Display Message PALINDROME

EXIT:MOV AH, 4CH

INT 21H

END

OUTPUT:

ABCDDCBA IS A PALINDROME

ABCDDC NOT A PALINDROME

*******************************************************************************

******

RESULT: This program reverses the string provided in data segment by keeping the original string

as it is and compares both the strings. It will check each and every character. If all the characters

are same then the given string is said to be as palindrome and it will display a message

“palindrome” on screen otherwise the given string is not palindrome and it will display a message

“not palindrome” on screen.



4. Compute nCr using recursive procedure. Assume that ‘n’ and ‘r’ are non-

negative integers.

Algorithm 1:

Step 1: if r=0 or n=r then nCr=1

Step 2: if r=1 or r=n-1 then nCr=n

Step 3: Recursive definition of nCr is nCr= (n-1)Cr + (n-1)C(r-1)

Algorithm 2:

nCr(r,n)

{

If(r==0 || n==r)

Return(1);

Else

If(r==1 || r==n-1)

Return(n);

Return(nCr (r,n-1)+ (n-1)Cr(r-1,n-1));

}



PROGRAM

.MODEL SMALL

.DATA

N DW 6H

R DW 3H

NCR DW 01 DUP (0) ; to store the result

MSG1 DB 10, 13, “Error $"

.CODE

MOV AX, @DATA ; initialize data segment

MOV DS, AX

MOV AX, N ; AX has value of N

MOV BX, R ; BX has value of R

CALL NCRPROC

MOV AH, 4CH

INT 21H

NCRPROC PROC

CMP AX, 0 ; if n=0, error condition

JZ Error

CMP BX, AX ; if n=r, ncr value is 1

JG Error

JE RES1

CMP BX, 0 ; if r=0, ncr value is 1

JE RES1

DEC AX

CMP BX, 1 ; if r=1, ncr value is n

JE RESN

CMP BX, AX ; if r=n-1, ncr value is n

JE RESN



PUSH AX

PUSH BX

CALL NCRPROC

POP BX

POP AX

DEC BX

PUSH AX

PUSH BX

CALL NCRPROC

POP BX

POP AX

RET

Error: LEA DX, MSG1 ; display message

MOV AH, 09H

INT 21H

RES1: INC NCR

RET

RESN: INC NCR ;initial ncr value=0 is incremented r=2 times(0+1+1)

ADD NCR, AX ;(1+5=6,6+4=10,10+3=13,13+2=15),i.e,[1+(n-1)+(n-2)+(n-3)+(n-4)]

RET ;addition is carried out (n-1) times

NCRPROC ENDP

END

*******************************************************************************

******



OUTPUT:

CASE1: N=0 R=2

Error

CASE 2: N=2 R=3

Error!

CASE3: N=3, R=3

NCR=1

CASE4: N=3, R=0

NCR=1

CASE5: N=3, R=1

NCR=3

CASE 6: N=3, R=2

NCR=3

CASE7: N=6, R=2

NCR=0FH

RESULT: This program performs nCr using recursive procedure. Output is stored in data

segment. The output is observed in data segment.

Note: From Pascal‟s triangular relation, nCr =

n-1Cr +

n-1Cr-1

In this formulation, one thing is to be noted that n and r keep decreasing, and sometimes is is

possible that n becomes smaller than r. So considering these cases the base conditions for the

recursive formula are,

nCn = 1,

nC1 = n,

nC0 = 1 and

nCr =

n-1Cr +

n-1Cr-1



5. Read the current time and date from the system and display it in the standard format

on the screen.

Algorithm

Step 1: Read the system time using ‘2CH’ service number.

Step 2: To display the hour, minute, seconds, convert from hexadecimal to BCD using ‘AAM’

instruction.

Step 3: Display using ‘02H’ service number.

Step 4: Stop.

Step 5: Read the system time using ‘2AH’ service number.

Step 6: To display the date, month, year, convert from hexadecimal to BCD using ‘AAM’ instruction.

Step 7: Display using ‘02H’ service number.

Step 8: Stop.



PROGRAM

CLRSCR MACRO ; macro definition to clear screen

MOV AL, 02H

MOV AH, 00H

INT 10H

ENDM

.MODEL SMALL

.DATA

msg1 DB 10, 13,'The system time is: $'

msg2 DB 10, 13,'The system date is: $'

.CODE

MOV AX,@DATA

MOV DS,AX

CLRSCR

LEA DX,msg1

MOV AH,09H

INT 21H

MOV AH,2CH ; DOS function to read system time

INT 21H

MOV AL, CH ; Hours in CH register

AAM ; unpack the digits

MOV BX, AX

CALL DISP ; use a procedure to convert to ASCII and display on screen

MOV DL,':' ; the format is hh: mm: ss

MOV AH, 02H

INT 21H

MOV AL, CL ; minutes in CL register

AAM



MOV BX,AX

CALL DISP

MOV DL,':'

MOV AH, 02H

INT 21H

MOV AL, DH ; seconds in DH register

AAM

MOV BX,AX

CALL DISP

LEA DX,msg2

MOV AH,09H

INT 21H

DAY: MOV AH, 2AH ; DOS Function to get System Date

INT 21H

MOV AL, DL ; Day is in DL

AAM

MOV BX, AX

CALL DISP

MOV DL,'/'

MOV AH, 02H ; To Print / in DOS

INT 21H

MONTH:MOV AH, 2AH ; To get System Date

INT 21H

MOV AL, DH ; Month is in DH

AAM

MOV BX, AX

CALL DISP



MOV DL,'/' ; To Print / in DOS

MOV AH, 02H

INT 21H

YEAR: MOV AH, 2AH ; To get System Date

INT 21H

ADD CX, 0F830H ; To negate the effects of 16bit value,

MOV AX, CX ; since AAM is applicable only for AL (YYYY -> YY)

AAM

MOV BX, AX

CALL DISP

MOV AH, 4CH

INT 21H

DISP PROC NEAR ; convert to ASCII and display

MOV DL, BH

ADD DL, 30h

MOV AH, 02h

INT 21h

MOV DL, BL

ADD DL, 30h

MOV AH, 02h

INT 21h

RET

DISP ENDP

END

*********************************************************************************

OUTPUT: The system time is: 12:02:06

The system date is: 03/12/16

RESULT: This program displays the present system time. The program displays the hours and

minutes and seconds in the format HH:MM: SS.

This program displays the present system date. The program displays the days, months and year in

the format DD/MM/YY.



8. Implement a BCD Up-Down Counter on the Logic Controller Interface.

D7 D6 D

5

D4 D3 D2 D1 D0

82H

1 0 0 0 0 0 1 0

I/O

Functio

n

Port A in

Mode 0

Port

A as

O/P

Port CU

As

O/P

Port B in

Mode 0

Port B

As I/P

Port

CL

As

O/P



Algorithm:

Step 1: Connect Logic Controller to PC.

Step 2: Set the 8255 using control word, initialize the control word i.e., 82H

Step 3: Store the control word in the Control register.

Step 4: Initialize the counter to 10(count to loop for displaying up counter), and AL with 00.

Step 5: Initialize the Port A address and display the contains of AL, call some delay and add 01h to

AL and again display it. Loop repeats until counter becomes 0.

Step 6: Again initialize counter to 10, and AL with 00, each time subtract 01 from AL and display

contents of AL, until counter becomes 0.

Step 7: Initialize the counter with 09H and AL, with 01H, display the contents of AL.

Step 8: Rotate the contains of AL once loop back and again display AL, the loop repeats until

counter becomes 0.

Step 9: Stop



PROGRAM:

.MODEL SMALL

.DATA

MSG1 DB “UP COUNTER STARTS $”

MSG2 DB “DOWN COUNTER STARTS $”

.CODE

MOV AX,@DATA ; Initialise data segment

MOV DS, AX

MOV DX, 9803H ; Control Word Register address

MOV AL, 82H ; set 8255 port B as input and port A and port C as output

OUT DX, AL

MOV CX, 0AH

LEA DX, MSG1

MOV AH, 09H

INT 21H

MOV AL, 00 ; AL holds initial count for BCD up counter

UP: MOV DX, 9800H

OUT DX, AL

INC AL ; increment count

DAA

CALL DELAY

LOOP UP

MOV CX, 0AH

LEA DX, MSG1

MOV AH, 09H

INT 21H

mailto:@DATA



DOWN: MOV DX, 9800H

OUT DX, AL

DEC AL ; decrement count

DAS

CALL DELAY

LOOP DOWN

QUIT: MOV AH, 4CH

INT 21H

DELAY PROC

PUSH BX

PUSH DX

MOV DX,0FFFFH

OUTER : MOV BX,0FFFFH

INNER : DEC BX

JNZ INNER

DEC DX

JNZ OUTER

POP DX

POP BX

RET

DELAY ENDP

END

*********************************************************************************

OUTPUT:

UPCOUNTER

0000 0001 0010 0011 0100 0101 0110 0111 1000 1001

DOWNCOUNTER

1001 1000 0111 0110 0101 0100 0011 0010 0001 0000



RESULT: The program performs the up-down counter based on the input data given on logic

controller read through port B. If the input is zero down counter starting from 99 down to 00 is

executed and if other than zero is the input then up counter starting from 00 down to 99 is executed.

The counting will continue until a key is pressed in the key board.



8b. Read the status of two 8-bit inputs (X & Y) from the Logic Controller Interface and

display X*Y.

D7 D6 D

5

D4 D3 D2 D1 D0

82H

1 0 0 0 0 0 1 0

I/O

Functio

n

Port A in

Mode 0

Port

A as

O/P

Port CU

As

O/P

Port B in

Mode 0

Port B

As I/P

Port

CL

As

O/P



ALGORITHM:

Step 1: Connect Logic Controller to PC.

Step 2: Set the 8255 using control word.

Step 3: Read the first data (Port B) from logic controller and store in X.

Step 4: Read the second data (Port B) from logic controller and store in Y.

Step 5: res=X*Y

Display(higher 8 bit data of res)

Step 6: Call delay procedure or wait some time.

Step 7: Stop



PROGRAM:

.MODEL SMALL

.DATA

MSG1 DB "ENTER VALUE OF X FROM THE INTERFACE: $"

MSG2 DB "ENTER VALUE OF Y FROM THE INTERFACE: $"

.CODE

MOV AX,@DATA

MOV DS,AX

MOV DX, 9803H ;initialize 8255 ports

MOV AL, 82H ;CW=82H Port B=Input Port,PortA=Output Port

OUT DX, AL

MOV DX , OFFSET MSG1

MOV AH , 09H

INT 21H

MOV DX, 9801H ;read first number X through port B of 8255

IN AL, DX

MOV BL, AL ; first number copied to BL reg

MOV DX ,OFFSET MSG2

MOV AH , 09H

INT 21H

MOV DX,9801H ;read second number through port B of 8255

IN AL,DX

MUL BL ; 16-bit product in AX reg

MOV DX,9800H ; send lower byte of the product to port A for display

OUT DX,AL

MOV AL,AH

OUT DX, AL ; send higher byte of the product to port A for display

MOV AH, 4CH ;terminate program

INT 21H

END



OUTPUT:

X00000011

Y00000010

PRODUCT: 00000110

RESULT: The program performs the multiplication between two bytes and gives the result. A

message appears on the screen to enter the first number. First byte is read from the port B of

logic controller and once enter key is pressed the second number is read from port B and

enter key is to be pressed. Result will be displayed on the port A of the logic controller.



9. Display messages FIRE and HELP alternately with flickering effects on a 7- segment

display interface for a suitable period of time. Ensure a flashing rate that makes it easy

to read both the messages (Examiner does not specify these delay values nor is it

necessary for the student to compute these values).

A seven segment display consists of seven LEDs arranged in the form of a squarish ‘8’ slightly

inclined to the right and a single LED as the dot character.

Different characters can be displayed by selectively glowing the required LED segments. Seven

segment displays are of two types, common cathode and common anode.

In common cathode type , the cathode of all LEDs are tied together to a single terminal which is

usually labelled as „com„ and the anode of all LEDs are left alone as individual pins labelled as

a, b, c, d, e, f, g & h (or dot) .

In common anode type, the anodes of all LEDs are tied together as a single terminal and cathodes

are left alone as individual pins. The pin out scheme and picture of a typical 7 segment LED

display is shown in the image below.

D7 D6 D5 D4 D3 D2 D1 D0

1 0 0 0 0 0 0 0

I/O

function

Port A in Mode 0 Port A as O/P Port CU

As O/P

Port B in

Mode 0

Port B

As O/P

Port

CL

As

O/P

80H

dp g f e D b c a

F 1 0 0 0 1 1 1 0 8EH

I 1 1 1 1 1 0 0 1 0F9H

R 1 0 0 0 1 0 0 0 88H

E 1 0 0 0 0 1 1 0 86H

H 1 0 0 0 1 0 0 1 89H

E 1 0 0 0 0 1 1 0 86H

L 1 1 0 0 0 1 1 1 0C7H

P 1 0 0 0 1 1 0 0 8CH



Algorithm:

Step 1: Connect Multiplex seven segment interface to PC.

Step 2: Set the 8255 using control word, initialize the control word i.e., 80H

Step 3: Store the control word in the Control register.

Step 4: Point the SI to FIRE and call display procedure.

Step 5: Display Procedure:

i) Initialize the counter to 4 to display 4 digit seven segment number.

ii) Move the contents of pointer to AL (which points to FIRE and HELP).

iii) Initialize Port A address and load the string byte to display on the seven segment display and the

pointer automatically increments to point to the next character (in the FIRE and HELP).

iv) Introduce some delay between each display and enable next digit D2 and so on until all the digits

from D1 to D4 is displayed.

Step 6: Point the SI to HELP and call display procedure

Step 7: Repeat from Step5 to Step8.

Step 8: To stop the display use service number 06H with DL content as 0FFH.

Step 9: Stop



PROGRAM

.MODEL SMALL

.DATA

FIRE DB 86H, 88H, 0F9H, 8EH ; initialize segment display value of FIRE

HELP DB 8CH, 0C7H, 86H, 89H ; initialize segment display value of HELP

.CODE

START : MOV AX,@DATA ;initialize data segment

MOV DS,AX

MOV DX,9803H ;initialize CWR

MOV AL,80H ; All ports are output ports

OUT DX,AL

MOV BH,0AH ;initialize number of times FIRE and HELP are displayed

AGAIN: MOV SI, OFFSET FIRE

CALL DISPLAY

CALL DELAY

MOV SI, OFFSET HELP

CALL DISPLAY

CALL DELAY

DEC BH ;decrement display count

CMP BH,00H

JE TERMINATE ;repeat till display count=0

JNZ AGAIN

DISPLAY PROC

MOV CX,04H ;initialize CX=4,as 4 displays are used

LOOP2 : MOV BL,08H ;initialize eight letters to BL register

MOV AL,[SI] ;copy first letter to AL

mailto:@DATA



NEXT : ROL AL,01H ;rotate contents of AL to the left

MOV DX,9801H ;initialize DX with port B address

OUT DX,AL ;send contents of AL to output port

PUSH AX

MOV AL,0FFH ; initialize AL=FFH,i/p =high

INC DX ; ;initialize DX with port C address

OUT DX,AL ;send high input to output port

MOV AL,00H ;initialize AL=00H,i/p =low

OUT DX,AL ;send low input to output port

DEC BL ; repeat 08 times

POP AX

JZ NEXT1

JMP NEXT

NEXT1 : INC SI ;get next letter

LOOP LOOP2 ;continue display till CX=0

RET

DISPLAY ENDP

DELAY PROC

PUSH AX

PUSH CX

MOV CX,0FFFH

LOOP1 : MOV AX,0FFFFH

LOOP3 : DEC AX

JNZ LOOP3

LOOP LOOP1

POP CX

POP AX



RET

DELAY ENDP

TERMINATE: MOV AH, 4CH

INT 21H

END START

*************************************************************************

Output: F I R E

H E L P

RESULT: This program displays “FIRE” and “HELP” on seven segment display interface

recursively one after the other with some delay till a key is pressed on key board. The data

which has to be displayed is provided in the program itself.



10. Drive a Stepper Motor interface to rotate the motor in specified direction

(clockwise or counter-clockwise) by N steps (Direction and N are specified by the

examiner). Introduce suitable delay between successive steps. (Any arbitrary value

for the delay may be assumed by the student).

A stepper motor is a widely used device that translates electrical pulses into mechanical movement. In

applications such as disk drives, dot matrix printers, and robotics, the stepper motor is used for

Position control. Every stepper motor has a permanent magnet rotor (also called the shaft.) surrounded

by a stator. The most common stepper motors have four common stator windings that are pairs with a

center-taped common. This type of stepper motor is commonly referred to as a four-phase stepper

motor.

A Stepper motor is stepped from one position to the next by changing the currents through the fields

in the motor. Common step sizes for stepper motors range from 0.9 degrees to 30 degrees.

Stepper motor is interfaced to 8086 with the help of 8255 IC (Programmable Peripheral Interface).

82C55A is used to provide the drive signals that are used to rotate the armature of the motor in either

the right-hand or left-hand direction. Stepper motor is IO device so CW should be (80) H.



The power circuit for one winding of the stepper motor is as shown in figure above. It is connected to

the port A (PA0) of 82C55A. Similar circuits are connected to the remaining lower bits of port A

(PA1, PA2, PA3). One winding is energized at a time. The coils are turned ON/OFF one at a time

successively.

The stepper motor showing full-step operation is shown below.

(A) 45-degrees. (B) 135-degrees

(C) 225-degrees (D) 315-degrees.



There are three commonly used excitation modes for step motors; these are full step, half step and

microstepping. In full step operation, the motor moves through its basic step angle, i.e., a 1.8° step motor takes

200 steps per motor revolution.

A typical stepper motor may have parameters like torque 3 kg-em, operating voltage 12V, current

rating 1.2A and a step angle 1.8 degree, i.e. 200 steps/revolution (number of rotor teeth).

Number of rotor teeth=200 steps/revolution

θR = 360degree/Number of rotor teeth

=360 degree/200=1.8 degree

Count N= Number of rotor teeth * θ =200*1.8 degree = 1 rotation

360 360

D7 D6 D5 D4 D3 D2 D1 D0

1 0 0 0 0 0 0 0

I/O

function

Port A in Mode

0

Port A as

O/P

Port

CU

As

O/P

Port B

in

Mode 0

Port B

As

O/P

Port

CL

As

O/P

80H



Algorithm:

Step 1: Initialize the control word 80H and store it in the control register.

Step 2: Initialize the number of steps (N) the stepper motor has to rotate.

Step 3: Coils are selected by storing 88H in AL register that is current is passed through coil.

Step 4: Send the data through Port C, hence initialize the Port C address.

Step 5: For clockwise rotation rotate the contents of AL register to right by 1 using ROR instruction

and decrease the steps. For anticlockwise rotation rotate the contents of AL register to left by 1

using ROL instruction and decrease the steps.

Step 6: Introduce delay counter after each steps.

Step 7: Repeat the steps 5 & 6 until n>0

Step 8: Stop.



PROGRAM:

.MODEL SMALL

.DATA

N DW 200 ; no of rotations. N=200 is one rotation

.CODE

MOV AX,@DATA

MOV DS, AX

MOV DX, 09803H

MOV AL, 80H

OUT DX,AL

MOV DX, 09800H

MOV AL,88H ; load pattern to energize the windings of motor into AL

MOV CX,N

AGAIN : OUT DX,AL ; energize winding of the motor

CALL DELAY

ROR AL,01 ; clockwise rotation

LOOP AGAIN

MOV CX,N

AGAIN1: OUT DX, AL

CALL DELAY

ROL AL,01

LOOP AGAIN1

MOV AH,4CH

INT 21H

mailto:@DATA



DELAY PROC ; delay procedure

PUSH BX

PUSH CX

MOV CX, 0FFFH

REPEAT: MOV BX, 0FFFFH

HERE1: DEC BX

JNZ HERE1

LOOP REPEAT

POP CX

POP BX

RET

DELAY ENDP

END

*********************************************************************************

***

RESULT: This program drives a stepper motor interface to rotate by N steps in clockwise and anti-

clockwise direction. After each rotation a delay is introduced to observe the rotation. After the

rotations are completed the execution is stopped.



11a. Generate the Sine Wave using DAC interface (The output of the DAC

is to be displayed on the CRO).

The pin details of DAC 0800 is given below and schematic diagram of the dual DAC

interface is given below.

The port A and port B of 82C55A peripheral are used as output ports. The digital inputs to

the DACs are provided through these ports. The analog outputs of the DACs are connected to

the inverting inputs of OP-amps 741 which acts as current to voltage converters. The outputs

from the OP-amps are connected to points marked X out and Y out at which the waveforms

are observed on a CRO. The power supplies of +12 and -12 are regulated for this interface.



D7 D6 D5 D4 D3 D2 D1 D0

1 0 0 0 0 0 0 0

I/O

function

Port A in Mode 0 Port A as O/P Port CU

As O/P

Port B in

Mode 0

Port B

As O/P

Port CL

As O/P

80H

Algorithm:

Calculate table values using the formula table[i]=127+127SinӨ (Equation to generate sine

wave) where Ө ranges from 0 to 180 in steps of 6 degree.


Step 2: Initialize the Port A address and initialize the counter to 31(total number of elements

in table).

Step 3: For (i=0 to31)

Out-port(Port A, Table[i]) ; /*will get +ve half cycle of sine wave

For (i=0 to 31)

Out-port(Port A, -table[i]) ;/*will get –ve half cycle of sine wave

Step 4: Stop.



Angle vs. Voltage Magnitude for Sine Wave

Angle θ

(degrees)

Sin θ θ Values

Sent to

DAC

(decimal)

=

127+127

Sin θ

θ Values

Sent to

DAC

Hexadecim

al

Angle θ

(degrees

)

Sin θ θ Values

Sent to

DAC

(decimal)

=

127+127

Sin θ

θ Values

Sent to DAC

(Hexadecim

al)

0 0 127 7F 186 -0.104 114 72H

6 0.104 140 8CH 192 -0.207 101 65H

12 0.207 153 99H 198 -0.309 88 58H

18 0.309 166 0A6H 204 -0.406 75 4CH

24 0.406 178 0B2H 210 -0.5 64 3FH

30 0.5 191 0BFH 216 -0.587 52 34H

36 0.587 201 0CAH 222 -0.66 42 2AH

42 0.66 212 0D4H 228 -0.743 33 21H

48 0.743 221 0DDH 234 -0.809 24 18H

54 0.809 230 0E6H 240 -0.866 17 11H

60 0.866 237 0EDH 246 -0.91 11 0BH

66 0.91 243 0F3H 252 -0.95 6 06H

72 0.95 248 0F8H 258 -0.978 3 03H

78 0.978 251 0FDH 264 -0.994 1 01H

84 0.994 253 0FCH 270 -1 0 00H

90 1 254 0FBH 276 -0.994 1 01H

96 0.994 253 0FCH 282 -0.978 3 03H

102 0.978 251 0FDH 288 -0.95 6 06H

108 0.95 248 0F8H 294 -0.91 11 0BH

114 0.91 243 0F3H 300 -0.866 17 11H

120 0.866 237 0EDH 306 -0.809 24 18H

126 0.809 230 0E6H 312 -0.743 33 21H

132 0.743 221 0DDH 318 -0.66 42 2AH

138 0.66 212 0D4H 324 -0.587 52 34H

144 0.587 201 0CAH 330 -0.5 64 3FH

150 0.5 191 0BFH 336 -0.406 75 4CH

156 0.406 178 0B2H 342 -0.309 88 58H

162 0.309 166 0A6H 348 -0.207 101 65H

168 0.207 153 99H 354 -0.104 114 72H

174 0.104 140 8CH 360 0 127 7FH

180 0 127 7FH



PROGRAM

.MODEL SMALL

.DATA

ARR DB 7FH,8CH,99H,0A6H,0B2H,0BFH,0CAH,0D4H,0DDH,0E6H,0EDH,0F3H,0F8H,0FDH,0FCH,

DB 0FBH,0F8H,0F3H,0EDH,0E6H,0DDH,0D4H,0CAH,0BFH,0B2H,0A6H,99H,08CH,7FH,72H,

DB 65H,58H,4CH,3FH,34H,2AH,21H,18H,11H,0BH,06H,03H,01H,00H,01H,

DB 03H, 06H,0BH,11H,18H,21H,2AH,34H,3FH,4CH,58H,65H,72H,7FH

MSG DB 10,13,”OBSERVE SINE WAVE ON CRO; PRESS ANY KEY TO EXIT$”

; array to store values of sinθ

.CODE

MOV AX,@DATA

MOV DS,AX

MOV DX,9803H ; make all ports as output

MOV AL,80H ; only port A of 8255 is used as output

OUT DX,AL

LEA DX,MSG

MOV AH,09H

INT 21H

UP1:LEA SI,ARR ; use SI as pointer to array

MOV CX,60 ; number of values in the array

the entire sine wave (1 cycle) is divided into 4 quadrants

UP: MOV AL,[SI] ; take sine value from array

MOV DX,9800H

OUT DX,AL ; and send it to port (CRO)

MOV DX,9801H

OUT DX,AL

CALL DELAY

INC SI

mailto:@DATA



LOOP UP

MOV AH,1

INT 16H

JNZ STOP

JMP UP1

MOV AH,4CH

INT 21H

DELAY PROC

PUSH BX

PUSH CX

MOV CX,0FFFH

REPEAT : MOV BX,0FFFFH

HERE1 : DEC BX

JNZ HERE1

LOOP REPEAT

POP CX

POP BX

RET

DELAY ENDP

END



RESULT: This program generates a sine wave of having amplitude of 5V. Output will be seen on a CRO. It will be continuous wave. It stops execution as soon as any key is pressed from the key board.

Output

Note :

Amplitude of sine wave vary from 0 to 5V.

F(x)= 127 +127 sinθ

(dc level) (amplitude)

where 0 <= θ <= 360 degree



11b) Generate a Half Rectified Sine wave form using the DAC interface.

(The output of the DAC is to be displayed on the CRO).

This program will display a half rectified sine wave on CRO device using DAC interface.

Algorithm: Calculate table values using the formula table[i]=127+127SinӨ (Equation to generate sine

wave) where Ө ranges from 0 to 180 in steps of 6 degree.


Step 2: Initialize the Port A address and initialize the counter to 31(total number of elements

in table).

Step 3: For (i=0 to31)

Out-port(Port A, Table[i]) ; /*will get +ve half cycle of sine wave

For (i=0 to 31)

Out-port(Port A, 127]) ;/*will get –ve half cycle of sine wave

Step 4: Stop.



PROGRAM

.MODEL SMALL

.DATA

ARR DB 7FH,8CH,99H,0A6H,0B2H,0BFH,0CAH,0D4H,0DDH,0E6H, 0EDH,0F3H, 0F8H,0FDH,0FCH,

DB 0FBH, 0F8H, 0F3H, 0EDH, 0E6H, 0DDH,0D4H, 0CAH,0BFH,0B2H,0A6H,99H,08CH,7FH,

DB 72H,65H, 58H,4CH,3FH,34H,00H,00H,00H,00H,00H,00H,00H,00H,00H,00H,00H,00H,00H,

DB 00H,00H,00H,00H,00H,00H,00H,00H,00H,00H,00H,00H

.CODE

MOV AX,@DATA

MOV DS,AX

MOV DX,9803H

MOV AL,80H

OUT DX,AL

LEA DX,MSG

MOV AH,9H

INT 21H

MOV DX,PORTA

UP1:LEA SI,ARR

MOV CX,60

UP: MOV AL,[SI]

MOV DX,9800H

OUT DX,AL

MOV DX,9801H

OUT DX,AL

CALL DELAY

INC SI

LOOP UP

mailto:@DATA



MOV AH,1

INT 16H

JNZ STOP

JMP UP1

MOV AH,4CH

INT 21H

DELAY PROC

PUSH BX

PUSH CX

MOV CX,09FFH

REPEAT : MOV BX,0FFFFH

HERE1 : DEC BX

JNZ HERE1

LOOP REPEAT

POP CX

POP BX

RET

DELAY ENDP

END

**************************************************************

Output



RESULT: This program generates a half - rectified sine wave of 5V amplitude. Output will

be seen in CRO. It stops execution as soon as any key is pressed from the key board.



6. To write and simulate ARM assembly language programs for data

transfer, arithmetic and logical operations (Demonstrate with the help

of a suitable program). Creating an Assembly project in Keil

1. Open the Keil IDE by clicking on its icon on the desktop.

2. From the Project menu Choose New µVision Project

3. Select a drive where you would like to create your project .Create a new folder and Name it

FirstProject. Type the name FirstProject for the project and click Save.Open that project

folder and give a name of your project executable file and save it.

4. In the Data base tree, choose the vendor and then the chip you want to use and then click

OK. For example, if you want to use the LPC2148, click on the ARM/NXP and then on the

LPC2148 and then press OK.

5. After selecting chip click on OK then it will display some window asking to add STARTUP

file. Click the No button when add the startup file to the project pop up window is displayed

A target is created.

6. To write your project code select a new file from FILE menu bar or Make a new file by

clicking on the New Icon. Press Ctrl+S to save the new file. (You can also save the file by

choosing Save from the File menu.).Name the file as program1.asm and save it in the

FirstProject directory.

7. Type the program in the file

8. Add the program1.asm file to the project. To do so: a. Right click on Source Group 1 and

choose Add Existing Files to Group. Then go to the FirstProject directory and choose

Program1.asm, press Add and then Close. The file will be added to target and it shows in the

project window.

Building

1. Now give a right click on target in the project window and select“Options for Target”.

2. It will show a window, in that go to output option and choose Create Hex file option by

selecting that box.In the same window go to Linker option and choose Use Memory Layout

from Target Dialog by selecting the box, and click OK.

3. To compile your project go to Project select Build Target option or press F7.

4. In the build OUT PUT window you can see the errors and warnings in your code. And here

Your project Hex file will be created

Debugging and Tracing

1. To start debugging click on Start/Stop Debug Session icon or choose Start/Stop Debug

Session from the Debug menu. (or simply press Ctrl+F5)

2. If it starts tracing successfully, a cursor appears in front of the next instruction to be

executed.

3. To trace the program use the Step Over button or click on Step Over from the Debug menu.

It executes the instructions of the program one after another.

4. To trace the program you can use the Step button, as well. The difference between the Step

Over and Step is in executing functions. While Step goes into the function and executes its

instructions one by one, Step Over executes the function completely and goes to the

instruction next to the function.

5. To exit from the debugging mode press Start/Stop Debug Session.



The Memory Window displays the memory area content. Several separate windows can be

used at a time. Open the windows from the toolbar or using the menu View – Memory

Windows.

Where

Address Enter an expression that evaluates to a memory address.

The Registers window shows and allows modifying the content of registers. The window also lists

all microcontroller operation modes. The currently active mode is highlighted in black. Currently

used registers are highlighted in blue.

Registers Window- Open this window through the toolbar button or the menu View - Registers

Window.

Where

Register Lists the operation modes and register names.

Value Shows the register value.

To modify a register value, double-click the value, or click on the value and press F2.

To refresh the values, halt program execution or click the button Update Windows of the Toolbox



DATA TRANSFER INSTRUCTIONS

1.32 bit Block Transfer from Location A to Location B

AREA Program, CODE, READONLY

ENTRY

start

MOV R0, #0x40000000

MOV R1, #0X4000003F

MOV R4, #0X05

loop LDR R2, [R0] ; load the value to be moved

STR R2, [R1] ; store it back in a different location

ADD R0, R0, #+4

ADD R1, R1, #+4

SUBS R4, R4, #0X01

BNE loop

SAME B SAME

END

Before execution

40000000: 11 22 33 44 55 66 77 88 AA BB CC DD EE FF 10 20 30 40 50 60

After execution

4000003C:11 22 33 44 55 66 77 88 AA BB CC DD EE FF 10 20 30 40 50 60



ARITHMETIC INSTRUCTIONS

2. Addition of two 32 bit numbers


ENTRY

start

MOV R0, #0x40000000

LDR R1, [R0]

ADD R0, R0, #04

LDR R2, [R0]

ADDS R1, R1, R2

ADD R0, R0, #04

STR R1, [R0]

SAME B SAME

END

Logical Instructions 3. ANDing of two 32 bit numbers


ENTRY

start

MOV R0, #0x40000000

LDR R1, [R0]

ADD R0, R0, #04

LDR R2, [R0]

AND R1, R1, R2

ADD R0, R0, #04

STR R1, [R0]

SAME B SAME

END

Before execution Value1 40000000: 11 22 33 44

Value2 40000004:55 66 77 88

After execution

40000008: CC AA 88 66



4. Ones complement of a given number

TTL ONESCOMP


ENTRY

MAIN

LDR R1, VALUE ; Load the number to be complimented

MVN R1, R1 ; NOT the contents of R1

MOV R3, #0X40000000

STR R1, [R3] ; Store the result

HERE B HERE

VALUE DCD &0123; Value to be complemented

END

Input Value: 0000C123

Output Result: FFFF3EDC



Creating a C project in Keil 1. Open the Keil IDE by clicking on its icon on the desktop.

2. From the Project menu Choose New µVision Project

3. Select a drive where you would like to create your project .Create a new folder and Name it

FirstProject. Type the name FirstProject for the project and click Save. Open that project

folder and give a name of your project executable file and save it.

4. In the Data base tree, choose the vendor and then the chip you want to use and then click

OK. For example, if you want to use the LPC2148, click on the ARM/NXP and then on the

LPC2148 and then press OK.

5. After selecting chip click on OK then it will display some window asking to add STARTUP

file. Click the Yes button when add the startup file to the project pop up window is

displayed. A target is created.

6. To write your project code select a new file from FILE menu bar or Make a new file by

clicking on the New Icon. Press Ctrl+S to save the new file. (You can also save the file by

choosing Save from the File menu.).Name the file as program1.asm and save it in the

FirstProject directory.

7. Type the program in the file

8. Add the program1.asm file to the project. To do so: a. Right click on Source Group 1 and

choose Add Files to Group. Then go to the FirstProject directory and choose Program1.asm,

press Add and then Close. The file will be added to target and it shows in the project

window.

Building

5. Now give a right click on target in the project window and select “Options for Target”.

6. It will show a window, in that go to output option and choose Create Hex file option by

selecting that box. In the same window go to Linker option and choose Use Memory Layout

from Target Dialog by selecting the box, and click OK.

7. To compile your project go to Project select Build Target option or press F7.

8. In the build Output window you can see the errors and warnings in your code. And here

Your project Hex file will be created.

Debugging and Tracing

1. To start debugging click on Start/Stop Debug Session icon or choose Start/Stop Debug

Session from the Debug menu. (or simply press Ctrl+F5)

2. To exit from the debugging mode press Start/Stop Debug Session.

Peripherals Menu

The menu Peripherals includes dialogs to view and change on-chip peripheral settings. The content

of this menu is tailored to show specific peripherals of the CPU selected for the application. This

menu is active only in Debug Mode.



LPC2148 C Programming

LPC2148 has 2 ports viz. Port 0 and Port 1. Each port can be used as I/P or O/P. Port 0 has 0-31

pins and Port1 has 16-31 pins.

Port 0 is a 32 bit wide I/O port (i.e. it can be used for max 32 pins where each pin refers to a

corresponding bit) and has dedicated direction bits for each of the pins present in the port. 28 out

of the 32 pins can be used as bi-directional I/O (digital) pins. Pins P0.24, P0.26 & P0.27 are

unavailable for use and Pin P0.30 can be used as output pin only. In port0, P0.24, P0.26, P0.27

are not physically available means you can‟t find them on chip but still you can write values to

them. & P0.31 is available for digital output only.

Port 1 is also a 32 bit wide I/0 port but Pins 0 to 15 i.e P1.0 – P1.15 are unavailable for use and

this port too has a dedicated direction bit for each of the usable pins. In port1, P1.0 to P1.15 is

not available physically. Only P1.16 to P1.31 is available. So out of 64 pins 45 pins are covered

by port0 & port1.

Each Port can be used as GPIO (general purpose I/p and O/p) or As SFR (Special Function).

Registers used for GPIO C programming.

1. IOxPIN (x=port number): GPIO Port Pin value register. This register can be used

to Read or Write values directly to the pins. Regardless of the direction set for the

particular pins it gives the current start of the GPIO pin when read. The current state

of the GPIO configured port pins can always be read from this register, regardless of

pin direction.

2. IOxDIR: This is the GPIO direction control register. Setting a bit to 0 in this

register will configure the corresponding pin to be used as an Input while setting it to

1 will configure it as Output. To set the direction of any Port as I/P or O/P we have

to SET and CLR a Register named as IODIR. This register individually controls the

direction of each port pin.

1---set as o/p

0---set as i/p

Example:

IO0DIR=0x00000000; // it means port 0 will act as i/p

IO1DIR=0x000000ff; // it means lower 8 bit of port 1 will act as o/p

3. IOxSET: GPIO Port Output Set registers. This register can be used to drive an

„output‟ configured pin to Logic 1 i.e HIGH. Writing Zero does NOT have any

effect and hence it cannot be used to drive a pin to Logic 0 i.e LOW. This register

controls the state of output pins in conjunction with the IOCLR register. Writing

ones produces highs at the corresponding port pins. Writing zeroes has no effect.

Example:

IO0SET=0x00000001; //it means 0th bit of Port 0 will be set

4. IOxCLR: This register can be used to drive an „output‟ configured pin to Logic 0

i.e LOW. Writing Zero does NOT have any effect and hence it cannot be used to

drive a pin to Logic 1. GPIO Port Output Clear registers. This Register controls the

state of output pins. Writing ones produces lows at the corresponding port pins and

clears the corresponding bits in the IOSET register.

Example:

IO0CLR=0x00000001; //it means 0th bit of Port 0 will be cleared

IOPIN Register is used to get the status of the pins.

IODIR Register is used to set the direction of the pins, when set to 1 means output and when 0

means input. IOSET Register is used to set the GPIO pins. IOCLR Register is used to clear the

GPIO pins.

Note:- When IOSET = 1 it will set the particular pin, but it doesn't means that to clear the pin

you will use IOSET = 0 or IOCLR = 0, this is different in case of ARM, to set the pin we have to

use IOSET=1 and to clear we have to use IOCLR=1.



PROGRAM 1-To calculate addition of two numbers #include <lpc214x.h>

void main (void)

{

unsigned char x, y, z ;

x=0x34;

y=0xA9;

IO0PIN=0xffffffff;

z=x+y;

IO0PIN=z;

}

PROGRAM 2-To turn on and off bit 10 of PORT 0

#include <lpc214x.h>

unsigned int delay;

int main (void)

{

IO0DIR = (1<<10) ; // Configure P0.10 as Output

While (1)

{

for (delay=0; delay<5000 ; delay++) ; // delay

IO0SET = (1<<10) ; // SET (1) P0.10 to turn LED ON

for (delay=0; delay<500000; delay++) ; // delay

IO0CLR = (1<<10) ; // CLEAR (0) P0.10 to turn LEDs OFF

}

}



To interface Stepper motor with ARM processor-- ARM7TDMI/LPC2148

A stepper motor is a brushless, synchronous electric motor that converts digital pulses into

mechanical rotation in steps. Every revolution of the stepper motor is divided into a discrete

number of steps, and for each pulse it receives the motor rotates through one step.

This program first configures the ARM Port0 as a GPIO and also as an out port. The

sequence code is sent to the driver IC using these port pins. A suitable delay is incorporated

between each step rotation. By applying the code in the reverse order, the stepper motor can

be rotated in the anticlockwise direction. Figure shows the interface of the Stepper Motor to

ARM 7 controller. The stepper motor is connected to Microcontroller using a ULN2003

driver IC. The ULN driver IC is connected to the Port0 pins P1.15 to P1.18 pins. So as the

microcontroller gives pulses with a particular frequency to ULN2003, the motor is rotated

either in clockwise or anticlockwise.

/* This program uses P0.15, P0.16,P0.17,P0.18 as output PINs connected to input of Driver

through which the Coil A,B,C,D of stepper motor driven */

/*.....................................................................

P0.15 P0.16 P0.17 P0.18

0x88- 1 0 0 0 1 0 0 0

0x44- 0 1 0 0 0 1 0 0

0x22- 0 0 1 0 0 0 1 0

0x11- 0 0 0 1 0 0 0 1

......................................................................*/



PROGRAM #include <LPC214x.h>

void delay(unsigned int value);

void gpio_init();

unsigned char cmd1[4]= {0x11,0x22,0x44,0x88};

//unsigned int j;

int main()

{ unsigned int i;

gpio_init();

while(1) //* // Loop forever

{

for (i=0;i<4;i++)

{

IO0PIN =cmd1[i]<<15;

delay(0X5000);

}

}

}

void gpio_init()

{

PINSEL0 = 0x00000000; // P1.15 - P1.18 as GPIO

IODIR0 |= 0x00078000; // Set Port0 as Output

IOCLR0 = 0x00078000;

}

void delay(unsigned int value)

{

unsigned int k;

for(k=0;k<value;k++);

{;

}

}



To interface LCD with ARM processor-- ARM7TDMI/LPC2148. Write

and execute programs in C language for displaying text messages and

numbers on LCD

APPARATUS REQUIRED:

1. PC

2. LPC2148 Kit

3. Serial port cable & power cable.

Theory: The ARM7 LPC21xx processor is interfaced to the 2x16 LCD module in 4-bit mode .The

interface diagram is shown below.The four data pins are connected with 4 data bits (P0.19 –

P0.22 pins to bits D4-D7), address bit (RS-P0.16), read/write bit (R/W-P0.17) and control

signal (E-P0.18) to make LCD display complete.The pins D0,D1,D2,D3 are left free with out

any connections.

16X 2 LCD is a 16 pin module . In which pins 1 &16 are grounded, 2 &15 are given to

VCC and 3rd

pin is given to potentiometer in order adjust the contrast of LCD. Pins 4, 5 & 6

corresponds to RS, R/W & EN respectively. Pins 7 to 14 are data lines from D0 to D7

respectively. Here the LCD is used in 4 bit mode i.e. upper 4 bits are used to transfer the data

with MSB first and LSB next. Port 0 pins i.e. from P0.16 to P0.23 are used for both data

and control signals. The interfacing diagram of 16X2 LCD is shown below

LCD looks almost like the one shown below. As you see that there are 8 data pins along with

3 control pins. One ground and two power pins are also there



/* This program is to interface 2x16 LCD display with LPC2148. It uses P0.28, P0.29, P0.30 of

LPC2148 connected to EN,RW & RS of LCD & P1.16 to P1.23 of LPC2148 to D0-D7 of LCD

Display

LCD Commands used

---------------------------------------------------------------------------------------------------------------------------

--

0x38 - Initialise LCD as 2X16 LCD Display

0x0E - Makes Display ON & Curser blink

0X06 - Shift curser RIGHT

0X01 - Clear the curser

0X80 - Initialise curser position as Line1 , Position1

---------------------------------------------------------------------------------------------------------------------------

--*/

ALGORITHM:

Step 1: Start

Step 2: Initialize LCD (Connect 3 pins of LCD RS, RW, EN to 3 pins of MCU and one port

of MCU to data lines of LCD. Give 5v supply to LCD)

Step 3: Send command signals [RS=0, R/W=0, En=1].

Step 4: Send data signals [RS=1, R/W=0, En=1].

Step 5: Display 2 messages in LCD display.

Step 6: Stop.



PROGRAM #include <LPC214x.h>

#define LCD_RS 0x40000000 // P0.30(0100 0000 0000 0000 0000 0000 0000 0000)

#define LCD_RW 0x20000000 // P0.29(0010 0000 0000 0000 0000 0000 0000 0000)

#define LCD_EN 0x10000000 // P0.28(0001 0000 0000 0000 0000 0000 0000 0000)

//........................... Functions defined..............................................................................

void lcd_cmd(unsigned char val) ; // Write Command

void lcd_data(unsigned char val) ; // Write LCD Display(ASCII)

void delay(unsigned int);

void gpio_init();

void lcd_clr();

//........................... ...............................................................................................

unsigned char cmd1[5]= {0x38,0x0e,0x06,0x01,0x80}; // LCD Commands

unsigned char msg1[]="PESIT-12345" ; // Commands to be displayed

int main()

{ unsigned int i;

gpio_init();

for(i=0;i<5;i++)

lcd_cmd(cmd1[i]);

for(i=0;msg1[i]!='\0';i++)

lcd_data(msg1[i]);

lcd_cmd(0xc0);

for(i=0;msg2[i]!='\0';i++) //#

lcd_data(msg2[i]); //#

while(1);

}

void gpio_init()

{

PINSEL1 = 0x00000000; // P0.16 to 31 as GPIO pins

PINSEL2 = 0x00000000; // P1.16 to 31 as GPIO pins

IODIR0 = 0x70000000;

// P0.28,P0.29,P0.30 as output pins i.e If corrosponding bit is 1:OUTPUT 0:INPUT

IODIR1 = 0x00FF0000; // P1.16 to P0,23 as OUTPUT

IOCLR0 = 0x70000000; // P0.28,P0.29,P0.30 is cleared

IOCLR1 = 0x00FF0000; // P1.16 to P0,23 is cleard

}



void lcd_cmd(unsigned char val)

{

IO1PIN=(val<<16);

IO0CLR=LCD_RS; // RS=0; Command register is enabled in LCD

IO0CLR=LCD_RW; // RW=0; Write Operation

IO0SET=LCD_EN ; // EN=1

delay(50);

IO0CLR=LCD_EN ; // EN=0 i.e HIGH to LOW transition at EN pin

delay(5000);

}

void lcd_data(unsigned char val)

{

IO1PIN=(val<<16);

IO0SET=LCD_RS ; // RS=1; Data register is enabled in LCD

IO0CLR=LCD_RW; // RW=0; Write Operation

IO0SET=LCD_EN ; // EN=1

delay(50);

IO0CLR=LCD_EN ; // EN=0 i.e HIGH to LOW transition at EN pin

delay(500000);

}

void delay(unsigned int value)

{ unsigned int j;

for(j=0;j<value;j++);

}

MICROPROCESSOR & MICROCONTROLLER LABORATORY MANUAL 4TH SEMESTER ISE SUBJECT...

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Transcript of MICROPROCESSOR & MICROCONTROLLER LABORATORY MANUAL 4TH SEMESTER ISE SUBJECT...