This document provides a list of experiments to be conducted using microprocessors and microcontrollers for two cycles. The first cycle involves programs written for the 8086 assembler using TASM software. The second cycle involves programs written for the 8051 assembler using TOP VIEW SIMULATOR software for interfacing experiments. A total of minimum 10 programs must be conducted between the two cycles.

1. MICROPROCESSORS AND MICROCONTROLLERS
LABORATORY (ECE) MANUAL
Prepared by
S.Sreedhar Babu Assoc.prof[ece],
(COURSE COORDINATOR)
SCHOOL OF ELECTRICAL SCIENCES
KONERU LAKSHMAIAH
UNIVERSITY
2012-13.

2. List of Experiments
Cycle - I: The following Programs/Experiments are to be written for the assembler
Using TASM software.
1. Programs on Data Transfer in forward and reverse direction using 8086.
2. Programs on Arithmetic Instructions with 16-bit data using 8086.
3. Programs on logical Instructions using 8086.
4. Programs on String manipulation using 8086.
5. Programs on Sorting and searching an array using 8086.
6. Programs on Procedures and Macros for BCD to Binary conversion, factorial
using 8086.
7. Programs on Interrupts for using 8086.
Cycle – II: Interfaced with 8051 using Machine language program and some
Programs/Experiments are to be written for the assembler using TOP VIEW SIMULATOR
software for the interfacing.
8. Programs on Arithmetic, Logic& Bit manipulation Instructions 8051.
9. Interfacing 7-Segment Display 8051.
10. LCD Interfacing to 8051.
11. Interfacing of Binary Counters 8051.
12. Interfacing Stepper Motor 8051.
13. Interfacing A/D & D/A to 8051
14. Keyboard Interface 8051.
15. Data Transfer between two PCs using RS.232 C Serial Port.
Note: Minimum of 10 programs to be conducted.
Lab Incharge HOD

3. INTRODUCTION
EDITOR
An editor is a program, which allows you to create a file containing the assembly language Statements for
your program. As you type in your program, the editor stores the ASCII codes for the letters and numbers in
successive RAM locations. When you have typed in all of your programs, you then save the file on a floppy
of hard disk. This file is called source file. The next step is to process the source file with an assembler. In
the TASM assembler, you should give your source file name the extension, ASM.
ASSEMBLER
An assembler program is used to translate the assembly language mnemonics for instructions to the
corresponding binary codes. When you run the assembler, it reads the source file of your Program the disk,
where you saved it after editing on the first pass thro ugh the source program the assembler determines the
displacement of named data items, the offset of labels and pails this information in a symbol table. On the
second pass through the source program, the assembler produces the binary code for each instruction and
inserts the offset etc that is calculated during the first pass. The assembler generates two files on floppy or
hard disk. The first file called the object file is given the extension. OBJ. The object file contains the binary
codes for the instructions and information about the addresses of the instructions. The second file generated
by the assembler is called assembler list file. The list file contains your assembly language statements, the
binary codes for each instructions and the offset for each instruction. In TASM assembler, TASM source file
name ASM is used to assemble the file. Edit source file name LST is used to view the list file, which is
generated, when you assemble the file.
LINKER
A linker is a program used to join several object files into one large object file and convert to an exe file.
The linker produces a link file, which contains the binary codes for all the combined modules. The linker
however doesn’t assign absolute addresses to the program, it assigns is said to be relocatable because it can
be put anywhere in memory to be run. In TASM, TLINK source filename is used to link the file.
DEBUGGER
A debugger is a program which allows you to load your object code program into system memory, execute
the program and troubleshoot are debug it the debugger allows you to look at the contents of registers and
memory locations after your program runs. It allows you to change the contents of register and memory
locations after your program runs. It allows you to change the contents of register and memory locations and
return the program. A debugger also allows you to set break point at any point in the program. If you inset a
breakpoint the debugger will run the program upto the instruction where the breakpoint is set and stop

4. execution. You can then examine register and memory contents to see whether the results are correct at that
point. In TASM, td filename is issued to debug the file.
DEBUGGER FUNCTIONS
1. Debugger allows to look at the contents of registers and memory locations.
2. We can extend 8-bit register to 16-bit register which the help of extended register option.
3. Debugger allows to set breakpoints at any point with the program.
4. The debugger will run the program upto the instruction where the breakpoint is set and then stop
execution of program. At this point, we can examine registry and memory contents at that point.
5. With the help of dump we can view register contents.
6. We can trace the program step by step with the help of F7.
7. We can execute the program completely at a time using F8.
DEBUGGER COMMANDS
ASSEMBLE:
To write assembly language program from the given address.
A starting address <cr>
Eg: a 1000H <cr>
Starts program at an offset of 1000H.
DUMP:
To see the specified memory contents
D memory location first address last address
(While displays the set of values stored in the specified range, which is given above)
Eg: d 0100 0105 <cr>
Display the contents of memory locations from 100 to 105(including).
ENTER:
To enter data into the specified memory locations(s).
E memory location data data data data data …<cr>
Eg: e 1200 10 20 30 40 ….
Enters the above values starting from memory locations 1200 to 1203, by loading 10 into
1200, 20 into 1201 and soon.
GO:
To execute the program
G: one instruction executes (address specified by IP)
G address <cr>: executes from current IP to the address specified
G first address last addresses <cr>: executes a set of instructions specified between the given address.
MOVE:
Moves a set of data from source location to destination location

5. M first address last address destination address
Eg: m100 104 200
Transfers block of data (from 100 to 104) to destination address 200.
QUIT:
To exit from the debugger.
Q <cr>
REGISTER:
Shows the contents of Registers
R register name
Eg: r ax
Shows the contents of register.
TRACE:
To trace the program instruction by instruction.
T = 0100 <cr>: traces only the current instruction. (Instruction specified by IP)
T = 0100 02 <cr>: Traces instructions from 100 to 101, here the second argument spec ifies the number of
instructions to be traced.
UNASSEMBLE:
To unassembled the program. Shows the opcodes along with the assembly language program.
U 100 <cr>: unassembled 32 instructions starting from 100th location.
U 0100 0109 <cr>: unassembles the lines from 100 to 104.

6. Using Turbo – Assembler – Linker – Debugger
(TASM, TLINK, TD)
1. Open an MSDOS window.
2. Set the PATH so that the TASM programs are available. The TASM programs are on the C drive; set the
path so that DOS can find them. This only needs to be done once each time you open an MSDOS prompt.
set PATH=%PATH%; C:TASMBIN.
3. Use a Text Editor to Edit the .ASM File.
Create your file using one of the following programs:
notepad proj.asm
WordPad proj.asm
edit proj.asm
4. Compile the source code to create an object module.
tasm/z/zi proj.asm
The /z switch causes TASM to display the lines that generate compilation errors. The /zi switch enables
information needed by the debugger to be included in the .OBJ file. Note that you should use "real mode"
assembler, TASM.EXE. Do not use the "protected mode" assembler TASM32.EXE for the assignments that
will be given in class
5. Run Linke r TLINK.EXE- generate .EXE file from the .OBJ file
tlink/v proj.
6. Run the Program
Your final program (if there were no errors in the previous step) will have an .EXE ending. To just run it,
type:
proj
If you want to use the debugger to examine the instructions, registers, etc., type:
td proj
This brings up the regular full-screen version of the Turbo debugger.
1. Tracing the Program's Execution
The Turbo debugger first starts, a Module Window which displays the Executable lines of program code,
marked with a bullet in the left column of the window. You can set breakpoints or step to any of these lines
of code. An arrow in the first column of the window indicates the location of the instruction pointer. This
always points to the next statement to be executed. To execute just that instruction use one of the two
methods listed under the Run menu item:
o Trace into (can use F7 key): executes one instruction; traces "into" procedures.

7. o Step over (can use F8 key): executes one instruction; skips (does not trace into) procedures.
Hitting either of these executes the instruction, and moves the arrow to the next instruction. As each
instruction executes, the effects might be visible in the Registers Window and Watches Window.
2. Setting and Removing Breakpoints
To set a breakpoint, position the cursor on the desired line of source code and press F2. The line containing
the breakpoint will turn red. Pressing F2 again removes the breakpoint. To execute all instructions from the
current instruction pointer up to the next encountered breakpoint, choose Run (can use F9 key) from the Run
menu item.
3. Examining Registers
Another window, the Registers Window, can be opened to examine the current value of the CPU registers
and flags. The View menu can be used to open this Registers Window. The registers and flags might change
as each instruction is executed.
4. Examining Memory
To examine memory, you will need to open an Inspector window. An Inspector window shows the contents
of a data structure (or simple variable) in the program you are debugging. It also allows you to modify the
contents of the data structure or variable. To open an Inspector window, place the cursor on what you want
to inspect and press CTRL-I. After you've examined the data item, press the ESC key to remove the
Inspector window.
5. Vie wing the Program's Output
Output written to the screen by a program is not immediately visible, since the main purpose of using a
debugger is to examine the internal operation of the program. To observe what the user would see, press
ALT-F5. The entire screen will change to a user-view showing the program's input and output (and possibly
that of previous programs as well). Press any key to return to the debugger screen.

8. AN INDRODUCTION TO 8086, SIMPE
PROGRAMS USING 8086 TRAINER KIT
(a) AIM : Addition of two 8-bit numbers using IMMEDIATE addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic comment
2000 B0 05 Mov AL,05 Data 05 is stored in reg AL
Data 04 is stored in reg BL
2002 B3 04 Mov BL,04
Data of AL and BL are
2004 02 C3 Add AL,BL
added and stored in AL
2006 CC Int 03 Stop
RESULT: AL: 09
b) AIM : Subtraction of two 8-bit numbers using immediate addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic comment
2000 B0 05 Mov AL,05 Data 05 is stored in reg AL
Data 04 is stored in reg BL
2002 B3 04 Mov BL,04
Data of BL is subtracted
2004 2B C3 Sub AL,BL
from AL and stored in AL
2006 CC Int 03 Stop
RESULT: AL: 01
(c) AIM : Multiplication of two 8-bit numbers using immediate addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic comment
2000 B0 05 Mov AL,05 Data 05 is stored in reg AL
Data 04 is stored in reg BL
2002 B3 04 Mov BL,04
Data of AL and BL are
2004 F7 E3 Mul BL multiplied and RESULT is
stored in AL
2006 CC Int 03 Stop
RESULT : AL

9. d) AIM : Division of two 8-bit numbers using immediate addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic comment
2000 B0 05 Mov AL,05 Data 05 is stored in reg AL
Data 04 is stored in reg BL
2002 B3 04 Mov BL,04
Data of AL is divided by
2004 F6 F3 Div BL BL and RESULT stored in
AL
2006 CC Int 03 Stop
RESULT : AL: 01Remainder
AH: 01Quotient
E) AIM : Addition of two 16-bit numbers using direct addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic comment
Move contents of 1700
2000 8B060017 Mov AX,[1700]
in reg AX
Move contents of 1702
2004 8B1E0217 Mov BX,[1702]
in reg BX
Data of AX and BX are
2008 01D8 Add AX,BX added and RESULT
stored in AX
200A CC Int 03 Stop
RESULT : AX=
Input:-
Location Data
1700 88
1701 00
1702 44
1703 00

10. f) AIM : Subtraction of two 16-bit numbers using direct addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic comment
Move contents of 1700
2000 8B 06 00 17 Mov AX,[1700]
in reg AX
Move contents of 1702
2004 8B 1E 02 17 Mov BX,[1702]
in reg BX
Data of BX is
subtracted from AX
2008 29 D8 Sub AX,BX
and RESULT stored in
AX
200A CC Int 03 Stop
RESULT : AX=
Input:-
Location Data Out put :
1700 88
1701 00
1702 44
1703 00
g) AIM : Addition of two 8-bit numbers using indirect addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic comment
Move Data from 2000
2000 BE0017 Mov SI,1700
to SI
Move contents of SI to
2003 8A04 Mov AL,[SI]
reg AL
2005 46 Inc SI Incrementing SI
Moving data from SI to
2006 8A1C Mov BL,[SI]
BL
Adding data of AL and
2008 02C3 Add AL,BL
BL
Move contents of AL
200A 89060018 Mov [1800],AL
to 1800
200E
CC Int 03 Stop

11. RESULT :
Input:-
Location Data
1700 04
1701 04
Output :-
1800 08
b) AIM : Subtraction of two 8-bit numbers using indirect addressing mode.
APPARATUS : 8086 Trainer kit.
Program:
Offset Address Opcode Label Mnemonic Comment
Move Data from 2000
2000 BE0017 Mov SI,1700
to SI
Move contents of SI to
2003 8A04 Mov AL,[SI]
reg AL
2005 46 Inc SI Incrementing SI
Moving data from SI to
2006 8A1C Mov BL,[SI]
BL
Subtract data of BL
2008 2BC3 Sub AL,BL
from data of AL
Move contents of AL
200A 89060018 Mov [1800],AL
to 1800
200E
CC Int 03 Stop
RESULT :
Input:-
Location Data
1700 09
1701 04
Output :-
1800 05

12. Ex.No.1 DATA TRANSFER Date:
Aim:
A. Write an ALP to transfer the data stored in consecutive memory locations, in the forward direction.
B. Write an ALP to transfer the data stored in consecutive memory locations, in the reverse direction.
C. Write an ALP to transfer the data stored in consecutive memory locations, in the forward direction
overlapping.
Software Used:
Computer system with TASM.
PROGRAM:
A. Forward Direction
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
ARR1 DB 0H,1H,2H,3H,4H,5H,6H,7H,8H,9H
COUNT EQU 10D
ORG 3000H
ARR2 DB 10D DUP (0H)
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV SI, 2000H
MOV DI, 3000H
MOV CX, COUNT
BACK: MOV AH, [SI]
MOV [DI], AH
INC SI

13. INC DI
LOOP BACK
MOV AH, 4CH
INT21H
CODE ENDS
END START
B.Reverse Direction
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
ARR1 DB 0H,1H,2H,3H,4H,5H,6H,7H,8H,9H
COUNT EQU 10D
ORG 3000H
ARR2 DB 10D DUP (0H)
DATA ENDS
CODE SEGMENT
START: MOV AX, DATA
MOV DS, AX
MOV SI, 2000H
MOV DI, 3000H
MOV CX, COUNT
ADD DI, COUNT-1
BACK: MOV AH, [SI]
MOV [DI], AH
INC SI
DEC DI
LOOP BACK

14. MOV AH, 4CH
INT21H
CODE ENDS
END START
C.Forward Direction with Overlapping
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
ARR1 DB 0H,1H,2H,3H,4H,5H,6H,7H,8H,9H
COUNT EQU $-ARR1
OVERLAP EQU 06D
ORG 3000H
ARR2 DB 10D DUP (0H)
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV SI, 2000H
MOV DI, 3000H
MOV CX, COUNT-OVERLAP
NXTP: MOV AH, [SI]
MOV [DI], AH
INC SI
INC DI
LOOP NXTP
AGAIN: LEA SI, ARR1

15. MOV CX, COUNT
MOV AH, [SI]
MOV [DI], AH
INC SI
INC DI
LOOP AGAIN
MOV AH, 4CH
INT21H
CODE ENDS
END START
Result: Data stored in consecutive memory locations is transferred from 2000h memory
location to 3000h memory location (a) in the forward direction (i.e. in the same order), (b) in
the reverse direction and (c) with overlapping in the forward direction.

16. Ex.No.2 ARITHMETIC OPERATIONS Date:
Aim:
A. Write an ALP to perform addition on 16-bit data stored in consecutive memory
locations and store the result from the next location onwards.
B. Write an ALP to perform subtraction on 16-bit data stored in consecutive memory
locations and store the result from the next location onwards.
C. Write an ALP to perform multiplication on 16-bit data stored in consecutive memory
locations and store the result from the next locations onwards.
D. Write an ALP to perform division on 16-bit data stored in consecutive memory
locations and store the result from the next location onwards.
Software Used:
Computer system with TASM.
PROGRAM:
A. Addition
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
ADDEND DW 8765H
ADDER DW 9876H
SUM DW 0H
CARRY DB 0H
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV AX, ADDEND
ADD AX, ADDER
JNC SKIP
INC CARRY
SKIP: MOV SUM, AX
MOV AH, 4CH
INT 21H
CODE ENDS
END START

17. B. Subtraction
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
SUBTRAHEND DW 8765H
SUBTRACTOR DW 9876H
DIFFERENCE DW 0H
BARROW DB 0H
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV AX,SUBTRAHEND
SUB AX, SUBTRACTOR
JNC SKIP
INC BARROW
SKIP: MOV DIFFERENCE, AX
MOV AH, 4CH
INT 21H
CODE ENDS
END START
C.Multiplication
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
MULTIPLICANT DW 0FFFFH
MULTIPLIER DW 123AH
RES DD 0H
DATA ENDS
CODE SEGMENT

18. ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV AX, MULTIPLICANT
MOV BX, MULTIPLIER
MUL BX
MOV WORD PTR [RES], AX
MOV WORD PTR [RES+2], DX
MOV AH, 4CH
INT21H
CODE ENDS
END START
D.Division
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
DIVIDEND DW 8765H
DIVISOR DW 1234H
QUOTIENT DW 0H
REMAINDER DW 0H
DATA ENDS
CODE SEGMENT
START: MOV AX,DATA
MOV DS, AX
MOV AX, DIVIDEND
MOV BX, DIVISOR
DIV BX
MOV QUOTIENT, AX
MOV REMAINDER, DX
MOV AH, 4CH

19. INT 21H
CODE ENDS
END STAR T
Result:- An ALP is written to perform a) addition, (b) subtraction, (c) multiplication and (d)
division operations using arithmetic instructions and the same is verified.

20. Ex.No.3 LOGICAL OPERATIONS Date:
Aim:
A. Write an ALP to find number of 1’s in a given word.
B. Write an ALP to find the number of even and odd numbers in the given array.
C. Write an ALP to find the number of elements in the array having “1” in their 5th bit position.
Software Used:
Computer system with TASM.
PROGRAM:
A. Number of 1’s in a word
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
NUM DW 5464
COUNT EQU 16D
BITCOUNT DB 0H
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV CL, COUNT
MOV AX, NUM
NXTP: ROR AX, 01H
JNC GO
INC BITCOUNT
GO: DEC CL
JNZ NXTP
MOV AH, 4CH

21. INT 21H
CODE ENDS
END START
B.Number of even and odd numbers
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
SERIES DW 3456H,4533H,1234H,1567H,0FFFFH,145AH,56D7,4E34H,3421H,
89C5H
COUNT EQU 0AH
ODDCOUNT DB 00H
EVENCOUNT DB 00H
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
LEA SI, SERIES
MOV CL, COUNT
NXTP: MOV AX, [SI]
ROR AX, 01H
JC ODD
INC EVENCOUNT
JMP OTHER
ODD: INC ODDCOUNT
OTHER: INC SI
DEC CL
JNZ NXTP
MOV AH, 4CH

22. INT 21H
CODE ENDS
END START
C.Number of elements having 1’s in their 5th bit position
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
SERIES DB 21H,54H,05H,34H,32H,14H,18H,17H,53H,58H
COUNT EQU 0AH
BITCOUNT DB 00H
DATA ENDA
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
LEA SI, SERIES
MOV CL, COUNT
NXTP: MOV AX, [SI]
TEST AX, 10H
JZ GO
INC BITCOUNT
GO: INC SI
DEC CL
JNZ NXTP
MOV AH, 4CH
INT 21H
CODE ENDS

23. END START, Result: Logical operations such as Shift, rotate and test are used to find the a) no. of
1’s in the given byte, b) no. of even and add numbers, c)no. of positive and negative numbers and d)
no. of elements having 1’s in their 5th bit position.
Ex.No.4 STRING MANIPULATIONS Date:
Aim:
A. Write an ALP to transfer the data in forward direction using string instructions.
B. Write an ALP to transfer the data in reverse direction using string instructions.
Software Used:
Computer system with TASM.
PROGRAM:
A. Forward Direction using String Instructions
ASSUME DS: DATA, CS: CODE, ES: EXTRA
DATA SEGMENT
ORG 2000H
STRING1 DB 'MICROPROCESSOR'
COUNT EQU $-STRING1
DATA ENDS
EXTRA SEGMENT
ORG 3000H
STRING2 DB 14D DUP (00H)
EXTRA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV AX, EXTRA
MOV ES, AX
MOV CX, COUNT
LEA SI, STRING1
LEA DI, STRING2

24. CLD
REP MOVSB
MOV AH, 4CH
INT 21H
CODE ENDS
END START
B.Reverse Direction using String Instructions
ASSUME DS: DATA, CS: CODE, ES: EXTRA
DATA SEGMENT
ORG 2000H
STRING1 DB 'MICROPROCESSOR'
LENGTH_STRING DW $-STRING1
DATA ENDS
EXTRA SEGMENT
ORG 3000H
STRING2 DB 14D DUP (00H)
EXTRA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV AX, EXTRA
MOV ES, AX
LEA SI, STRING1
LEA DI, STRING2
MOV CX, LENGTH_STRING
ADD DI, CX
DEC DI
BACK: MOVSB

25. SUB DI, 02H
LOOP BACK
MOV AH, 4CH
INT 21H
CODE ENDS
END START
Result: String manipulations such as a) forward string, b) reverse string.

26. Ex.No.5 SORTING Date:
Aim:
A. Write an ALP to sort the given array in signed ascending order.
B. Write an ALP to sort the given array in unsigned descending order.
C. Write an ALP to find the maximum and the minimum element in the given array.
Software Used:
Computer system with TASM.
PROGRAM:
A. Signed ascending order
ASSUME DS: DATA, CS: CODE
DATA SEGMENT
ORG 3000H
ARRAY DB 03H,07H,05H,01H,09H,04H,06H,02H,08H
COUNT EQU 09H
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV DX, COUNT
DEC DX
BEGIN: MOV CX, DX
LEA SI, ARRAY
BACK: MOV AL, [SI]
CMP AL,[SI+01]
JNZ SKIP
XCHG AL, [SI+01]

27. XCHG AL, [SI]
SKIP: INC SI
LOOP BACK
DEC DX
JNZ BEGIN
MOV AH, 4CH
INT 21H
CODE ENDS
END START
B.Unsigned descending order
ASSUME DS: DATA, CS: CODE
DATA SEGMENT
ORG 3000H
ARRAY DB 03H,07H,05H,01H,09H,04H,06H,02H,08H
COUNT EQU 09H
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV DX, COUNT
DEC DX
BEGIN: MOV CX, DX
LEA SI, ARRAY
BACK: MOV AL, [SI]
CMP AL,[SI+01]
JNB SKIP
XCHG AL, [SI+01]

28. XCHG AL, [SI]
SKIP: INC SI
LOOP BACK
DEC DX
JNZ BEGIN
MOV AH, 4CH
INT 21H
CODE ENDS
END START
C.Maximum and Minimum elements
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
ARRAY DW 5555H,9999H, 7777H, 2222H, 1111H, 8888H, 6666H
A_LENTH EQU ($-ARRAY)/2
MAX_NO DW 0H
MIN_NO DW 0H
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV CX, A_LENTH-1
LEA SI, ARRAY
MOV AX, [SI]
MOV BX, AX
BACK: INC SI
INC SI
CMP AX, [SI]

29. JNC SKIP
MOV AX, [SI]
JMP NEXT
SKIP: CMP BX, [SI]
JC NEXT
MOV BX, [SI]
NEXT: LOOP BACK
MOV MAX_NO,AX
MOV MIN_NO, BX
MOV AH, 4CH
INT 21H
CODE ENDS
END START
Result: Sorting the array in a) signed ascending order, b) unsigned descending order and c)
searching for the maximum and minimum elements in the unsigned array, are implemented.

30. Ex.No.6 PROCEDURE AND MACROS Date:
Aim:
A. Write an ALP to convert Hexadecimal numbers to BCD numbers.
B. Write an ALP to find the factorial of given number.
Software Used:
Computer system with TASM.
Flow Chart:
PROGRAM
A. Hexadecimal to BCD conversion
ASSUME DS: DATA, CS: CODE
DATA SEGMENT
ORG 2000H
HEXA DB 25H, 57H, 89H, 0A4H
COUNT EQU 04
DECI DB 12 DUP (00H)
BCD DW 04 DUP (00H)
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
MOV SI, OFFSET HEXA
MOV DI, OFFSET DECI
MOV CX, COUNT
BACK: XOR AX, AX
MOV AL, [SI]
MOV BL, 64H

31. DIV BL
MOV [DI], AL
INC DI
MOV AL, AH
XOR AH, AH
MOV BH, 0AH
DIV BH
MOV [DI], AL
INC DI
MOV [DI], AH
INC SI
INC DI
LOOP BACK
CALL UP2P
MOV AH, 4CH
INT 21H
UP2P PROC NEAR
MOV CH,COUNT
MOV DI,OFFSET DECI
MOV BP,OFFSET BCD
CONTINUE: XOR AX,AX
MOV AH,[DI]
INC DI
MOV AL,[DI]
MOV CL,4
SHL AL,CL
INC DI
ADD AL,[DI]
MOV DS:BP,AX
INC DI

32. INC BP
INC BP
DEC CH
JNZ CONTINUE
RET
UP2P ENDP
CODE ENDS
END START
B.Factorial of a given number
FACTORIAL MACRO
XOR CX, CX
XOR AX, AX
INC AX
MOV CL, NUMBER
CMP CL, 0
JE GO
REPEAT: MUL CX
LOOP REPEAT
GO: MOV FACT, AX
ENDM
ASSUME CS: CODE, DS: DATA
DATA SEGMENT
ORG 2000H
NUMBER DB 08H
FACT DW 0
DATA ENDS
CODE SEGMENT
START: MOV AX, DATA
MOV DS, AX

33. FACTORIAL
MOV AH, 4CH
INT 21H
CODE ENDS
END START
Result: Programs on a) Procedure for converting Hexadecimal numbers to packed BCD and b)
Macro for finding the factorial of a given number are implemented.

34. Ex.No.7 INTERRUP TS Date:
Aim:
A. Write an ALP to find whether the given string is a palindrome or not.
B. Write an ALP to enter the string through keyboard and display it.
Software Used:
Computer system with TASM.
PROGRAM:
A. String is palindrome or not
ASSUME DS: DATA, CS: CODE
DATA SEGMENT
ORG 2000H
STRING DB 'RADAR'
COUNT EQU $-STRING-1
MSG1 DB "STRING IS PALINDROME $"
MSG2 DB "STRING IS NOT PALINDROME $"
DATA ENDS
CODE SEGMENT
ORG 1000H
START: MOV AX, DATA
MOV DS, AX
LEA SI, STRING
LEA DX, MSG2
MOV CX, COUNT
MOV DI, SI
ADD DI, COUNT-1
SHR CX, 1

35. BACK: MOV AL, [SI]
CMP AL, [DI]
JNZ NEXT
INC SI
DEC DI
LOOP BACK
LEA DX, MSG1
NEXT: MOV AH, 09H
INT 21H
MOV AH, 4CH
INT 21H
CODE ENDS
END START
B.Enter and Display the string
ASSUME DS: DATA, CS: CODE
DATA SEGMENT
ORG 2000H
INPUT DB “ENTER THE STRING”, 0DH, 0AH, “$”
OUTPUT DB “THE ENTERED STRING IS:” 0DH, 0AH, “$”
S_LENTH DB 0
BUFFER DB 80 DUP (0)
DATA ENDS
CODE SEGMENT
START: MOV AX, DATA
MOV DS, AX
XOR CL, CL
LEA DX, INPUT
MOV AH, 09H
INT 21H
LEA BX, BUFFER
REPEAT: MOV AH, 01H

36. INT 21H
CMP AL, 0DH
JZ EXIT
INC CL
MOV [BX], AL
INC BX
JMP REPEAT
EXIT: MOV S_LENTH, CL
LEA BX, BUFFER
ADD BL, CL
MOV AL,’$’
MOV [BX], AL
LEA DX, OUTPUT
MOV AH, 09H
INT 21H
LEA DX, BUFFER
MOV AH, 09H
INT 21H
CODE ENDS
ENDF START
Result: Interrupts on a) whether the given string is a palindrome or not and b) entering the string through
keyboard and displaying the same, are verified.

37. 8051 INTRODUCTION AND
INTERFACING WITH
PERIPHERALDEVICES
USING TOP VIEWIMULATOR

38. Ex.No.8 Arithmetic, Logic and Bit manipulation Instructions Date:
Aim: Write an ALP to perform Addition, Subtraction, Multiplication, Division, Swap,Reset,
Set, Bit, Byte manipulation Instructions.
Software Used:
Computer system with TOPVIEW SIMULATOR.
PROCEDURE:
In Top view simulator:
 Select device 89C51 with 12000000Hz frequency.
 Open the hex file (Ctrl + O or file->Load program).
 Run the program at full speed. ( Run->go)
 To stop the execution Run->stop execution.
A.ADDITION
$MOD51
ORG 0000H
MOV DPTR, #0101H
MOVX A,@DPTR
MOV B, A
DEC DPL
MOV A,@DPTR
ADD A, B
INC DPTR
INC DPTR
MOVX @DPTR, A
CLR A
ADDC A, #00H
INC DPTR
MOVX @DPTR, A

39. STOP: SJMP STOP
END
B.SUBTRACTION
$MOD51
ORG 0000H
MOV DPTR, #0101H
MOVX A,@DPTR
MOV B, A
DEC DPL
MOVX @DPTR, A
CLR C
SUBB A,B
INC DPTR
INC DPTR
MOVX @DPTR, A
CLR A
ADDC A, #00H
INC DPTR.
MOVX @DPTR,A
STOP: SJMP STOP
END
C.MULTIPLICATION
$MOD51
ORG 0000H
MOV DPTR, # 0101H
MOVX A,@DPTR
MOV B, A
DEC DPL

40. MOVX A, @DPTR
MUL A B
INC DPTR
INC DPTR
MOVX @DPTR, A
MOV A, B
INC DPTR
MOVX @DPTR, A
STOP: SJMP STOP
END
D.DIVISION
$MOD51
ORG 0000H
MOV DPTR, # 0101H
MOV X A,@DPTR
MOV B, A
DEC DPL
MOVX A, @DPTR
DIV AB
INC DPTR
INC DPTR
MOVX @DPTR, A
MOV A, B
INC DPTR
MOVX @DPTR, A
STOP: SJMP STOP
END
D.SWAP

41. $MOD51
ORG 0000H
MOV DPTR, # 0101H
MOVX A, @ DPTR
SWAP A
INC DPTR
MOVX @ DPTR, A
STOP: SJMP STOP
END
E.RESET
$MOD51
ORG 0000H
MOV DPTR, # 0101H
CLR C
MOV A, ODOH
MOV DPTR, # 0101H
MOVX @ DPTR, A
STOP: SJMP STOP
END
F.SET
$MOD51
ORG 0000H
MOV DPTR, # 0101H
SETB C
MOV A,0D0H
MOV DPTR, # 0101H
MOVX @ DPTR, A
STOP: SJMP STOP

42. END
G.BIT
$MOD51
ORG 0000H
MOV DPTR, # 0101H
MOV A, ODOH
MOVX @ DPTR, A
CPL C
MOV A, ODOH
INC DPTR
MOVX @DPTR, A
STOP: SJMP STOP
END
H.BYTE
$MOD51
ORG 0000H
MOV DPTR, # 0101H
MOVX A @ DPTR
CPL A
INC DPTR
MOVX @ DPTR, A
STOP: SJMP STOP
END
I.COMPLIMENT
$MOD51
ORG 0000H
MOV DPTR,#0101H
MOVX A,@DPTR

43. CPL A
INC DPTR
MOVX @DPTR,A
STOP: SJMP STOP
END
Result: All the Arithmetic and special instruction operations are performed by using 8051.

44. Ex. No 9 Interfacing of 7 – Segment Displays Date:
Problem statement:
Interface 7-segment displays (using BCD to 7-seg decoder) to the 8051
microcontroller.
Aim: To display two Digit Decimal counter on 7- segment displays.
Software Used:
Computer system with TOPVIEW SIMULATOR.
PROCEDURE:
In Top view simulator:-
 Select device 89C51 with 12000000Hz frequency.
 Open the hex file (Ctrl + O or file->Load program).
 Open LED module settings to configure the LED connections.( File-> External
modules settings-> LED)
 In seven segment display select
Interface selection-> non-multiplexed.
Display type->common cathode.
Data input selection -> BCD.
 Click on selection of port line and number of digits.
 Select no. of digits as 2
 Select control lines and port lines as shown in the below table.
 Check with summary window for correct connections.
 Open the LED module. (View->External modules->LED).
 Run the program at full speed. ( Run->go)
 The two digit decimal counter will shown up.
 To stop the execution Run->stop execution.

45. Table for Port line selection
7-Segment Display PIN on 8051
Digit2 A P1.0
Digit2 B P1.1
Digit2 C P1.2
Digit2 D P1.3
Digit2 Dp GND
Digit1 A P1.4
Digit1 B P1.5
Digit1 C P1.6
Digit1 D P1.7
Digit1 Dp GND
Circuit:
Digit2
P1.0 A 7
P1.1 B BCD to 7 SEG
SEG
P1.2 C
8051
P1.3 D |_|
Digit1
P1.4 A A
BCD to 7 7
P1.5 BB SEG SEG
P1.6 C C
P1.7 D

46. PROGRAM
$MOD51
ORG 0000H
SJMP MAIN
ORG 0030H
MAIN:MOV A,#00H
AG AIN:MOV P1,A
LCALL DELAY
ADD A,#01
DA A
SJMP AGAIN
ORG 0040H
DELAY:MOV R2,0FFH
UP1:MOV R1,0FFH
UP2:NOP
NOP
NOP
DJNZ R1,UP2
DJNZ R2,UP1
RET
END

47. Simulated output:
Result: Decimal counter is displayed on 7-Segment displays.

48. Ex.No.10 Interfacing of 2 – line 16 – character LCD display Date:
Problem statement:
Interface LCD controller to the 8051 microcontroller.
Aim: To display strings in two lines on a LCD module.
Software Used:
Computer system with TOPVIEW SIMULATOR.
Procedure:
In Top view simulator
 Select device 89C51 with 12000000Hz frequency.
 Open the hex file ( Ctrl + O or file->Load program).
 Open LCD module settings to configure the LCD connections.( File-> External
modules settings-> LCD) Check with the summary window for correct
connections.
 Select LCD as 2lines 16 characters and Data bus width as 8 bit.
 In port line selection configure the port lines as per requirement. ( see the table for
port selection).
 Open the LCD module. (View->External modules->LCD).
 Run the program at full speed. ( Run->go)
 The strings are displayed on the LCD.
 To stop the execution Run->stop execution.
Table for Port line selection
PIN on 8051 PIN on LCD
P1.0 … P1.7 D0…D7(Data lines)
P2.0 RS
P2.1 RW
P2.2 E

49. Circuit:
P1.0 D0
LCD
controller
P1.7 D7
RS RW E
8051
P2.0
P2.1
P2.2
Flow Chart:
PROGRAM
$MOD51
RS EQU P3.1
EN EQU P3.2
RW EQU P3.3
ORG 0000H
MOV A,#38H
LCALL LCD_COMD
MOV A,#0EH
LCALL LCD_COMD
MOV A,#06H
LCALL LCD_COMD
MOV A,#01H
LCALL LCD_COMD
MOV A,#'E'
LCALL LCD_TEXT
MOV A,#'C'

50. LCALL LCD_TEXT
MOV A,#'E'
LCALL LCD_TEXT
LOOP:SJMP LOOP
LCD_COMD:CLR C
LCALL WRITE
RET
LCD_TEXT:SETB C
LCALL WRITE
RET
WRITE:SETB EN
CLR RW
MOV RS,C
MOV P1,A
CLR EN
LCALL DELAY
RET
DELAY:MOV R0,#60
LOOP2:MOV R1,#255
LOOP1:DJNZ R1,LOOP1
DJNZ R0,LOOP2
RET
END

51. Simulator Output:
Result:
The strings in two lines are displayed on LCD module.

52. Ex.No.11 LIGHT EMITTING DIODE (LED) Date:
Problem statement:
Interface LED controller to the 8051 microcontroller.
Aim:
Write an ALP for LED using 8051.
Software Used:
Computer system with TOPVIEW SIMULATOR.
PROCEDURE:
In Top view simulator:
 Select device 89C51 with 12000000Hz frequency.
 Open the hex file (Ctrl + O or file->Load program).
 Open LED module settings to configure the LED connections.( File-> External
modules settings-> LED)
 In seven segment display select
Interface selection-> non-multiplexed.
Display type->common cathode.
Data input selection -> BCD.
 Click on selection of port line and number of digits.
 Select no. of digits as 2
 Select control lines and port lines as shown in the below table.
 Check with summary window for correct connections.
 Open the LED module. (View->External modules->LED).
 Run the program at full speed. ( Run->go)
 The two digit decimal counter will shown up.
 To stop the execution Run->stop execution.
Flow Chart:

53. PROGRAM
$MOD51
CLR A
REPEAT:MOV P1,A
LCALL DELAY
INC A
SJMP REPEAT
DELAY:MOV R2,#0FFH
BACK:MOV R1,#0FFH
UP1:NOP
NOP
DJNZ R1,UP1
DJNZ R2,BACK
RET
END
Simulator Output:
Result: Decimal counter is displayed on LED displays.

54. Exp.12 Interfacing of Stepper Motor Date:
Aim: Interface the stepper motor to 8051 target system and write program to operate in
different speeds both clock wise and anti clock wise directions.
STEPPER MOTOR CONTROLLER INTERFACING MODULE
CIRCUIT DESCRIPTION
The four winding of the motor is connected with PA0 to PA3 through buffer and driving
circuit. So, the Port A of 8255 will have to initializes in output mode.
HARDWARE INSTALLATION
1. Connect Stepper Motor interfacing module to 8255 – I of 8051 / 8085 Trainer Kit
through 26 pin FRC Cable.
2. Be sure about the direction of the cable i.e. Pin No.1 of module should be
connected to Pin No.1 of 8255 connector.
3. Connect +5V, GND from the trainer kit (+5V & GND signals are available in the
25 & 26 pin of FRC 8255 – I Connector).
4. Connect Motor Supply to +12V from the Trainer Kit.
8255 Port Address:
Port A – FF00H
Port B –FF01H
Port C – FF02H
Control Word – FF03H
Procedure:
1. Create a source file in Assembly language.
2. Assemble the source file using ASM51 cross Assembler and create LST & HEX files.
3.Configure the windows hyper terminal with a baud rate 9600 , 8 data bits, no parity, 1
stop bit and flow control XON / XOFF.

55. 4. Establish a serial communication link between Host and Target system using hyper
terminal.
5. Down load the HEX file from the host to the target system.
6. Connect stepper motor interface module to the target system and run the program.
Program:
This program will move the motor in Clock wise direction. For Anti Clock wise
direction change F9, F5, F6, FA in place of FA, F6, F5, F9.
Address Code Label Mnemonic Operand Comments
2000 90 FF 03 MOV DPTR, #0FF03H
2003 74 80 MOV A,#80H
2005 F0 MOVX @DPTR,A Init ports of 8255 as
all
2006 90 FF 00 START: MOV DPTR,#0FF00H output ports
2009 74 FA MOV A,#0FAH output code for step
1
200B F0 MOVX @DPTR,A
200C 11 1F ACALL delay between two
DELAY steps
200E 74 F6 MOV A,#0F6H output code for step
2
2010 F0 MOVX @DPTR,A
2011 11 1F ACALL
DELAY
2013 74 F5 MOV A,#0F5H output code for step
3
2015 F0 MOVX @DPTR,A

56. 2016 11 1F ACALL
DELAY
2018 74 F9 MOV A,#0F9H output code for step
4
201A F0 MOVX @DPTR,A
201B 11 1F ACALL delay between two
DELAY steps
201D 80 E7 SJMP START repeat for next
cycle
201F 7F 3F DELAY: MOV R7,#03FH delay count for
controlling speed
2021 7E 3F DELA: MOV R6,#03FH
2023 00 DELA1: NOP
2024 00 NOP
2025 00 NOP
2026 DE FB DJNZ R6,DELA1
2028 DF F7 DJNZ R7,DELA
202A 22 RET
Result:
The Stepper Motor is interfaced to 8051 and operated with different speeds and
directions.

57. 8051 INTERFACING WITH
PERIPHERAL DEVICES
USING ASSEMBLY LANGUAGE
PROGRAM

58. BINARY COUNTER INTERFACING WITH 8051
AIM:
To interface Binary Counter 8051 parallel port to demonstrate the generation of convert a binary number to
a decimal.
APPARATUS REQUIRED;
8051 Trainer Kit
LED’S
Resistors -330k ohm
THEORY:
Before starting with counters there is some vital information that needs to be understood. The most
important is the fact that since the outputs of a digital chip can only be in one of two states, it must use a
different counting system than you are accustomed to. Normally we use a decimal counting system; meaning
each digit in a number is represented by one of 10 characters (0-9). In a binary system, there can only be two
characters, 0 and 1.
A computer does not recognize 0 or 1. It only works o n voltage changes. What we call logic 0 to a
computer is zero volts. What we call logic 1 is +5 volts. When a logic state changes from a zero to a one the
voltage at the pin in question goes from zero volts to +5 volts. Likewise, when a logic state changes from a
one to a zero the voltage is changing from +5 volts to zero volts.
When counting up in a decimal system, we start with the first digit. When that digit ‘overflows’, i.e. gets
above 9, we set it to 0 and add one to the next digit over. The same goes for a binary system. When the
count goes above 1 we add one to the next digit over and set the first digit to 0. Here is an example
DECIMAL TO BINARY CONVERSION
Decimal Number (base 10) Binary Number (base 2)
0 0
1 1
2 10
3 11
4 100
5 101
6 110
Decimal Number (base 10) Binary Number (base 2)
7 111
8 1000
9 1001

59. BINARY COUNTING
To convert a binary number to a decimal, we use a simple system. Each digit or ‘bit’ of the binary
number represents a power of two. All you need to do to convert from binary to decimal is add up the
applicable powers of 2. In the example below, we find that the binary number 10110111 is equal to 183. The
diagram also shows that eight bits make up what is called a byte. Nibbles are the upper or lower four bits of
that byte. Referring to nibbles and bytes are useful when dealing with other number systems such as
hexadecimal, which is base 16.
fig1. Interfacing
PROGRAM:
MOV DPTR, # 2023; Control register of 8255.
MOV A, #80; 8255 in I/O mode operation
MOVX @DPTR, A
MOV A, #00H
START: MOV DPTR, # 2020 (port A address )
INC A
MOVX @DPTR, A
LCALL DELAY

60. CJNE A, #FFH, START
XX: SJMP XX
DELAY: MOV R1, # FFh
CC: MOV R2, # FFh
AA: DJNZ R2, AA
MOV R3, # FFH
BB: DJNZ R3, BB
DJNZ R1, CC
RET
Result:

61. Seven Segment Display Interfacing With 8051
AIM:
To interface Seven Segment Display to 8051 to generate a digit from 0 to 9 using common anode and
Common cathode technique.
APPARATUS REQUIRED
8051 TRAINER
Common cathode Seven Segment Display
Common anode Seven Segment Display
INTRODUCTION
For the seven segment display you can use the LT-541 or LSD5061-11 chip. Each of the segments of the
display is connected to a pin on the 8051 (the schematic shows how to do this). In order to light up a
segment on the the pin must be set to 0V. To turn a segment off the corresponding pin must be set to 5V.
This is simply done by setting the pins on the 8051 to '1' or '0'.
LED displays are
Power-hungry (10ma per LED)
Pin- hungry (8 pins per 7-seg display)
But they are cheaper than LCD display
7-SEG Display is available in two types -1. Common anode & 2. Common cathode, but command anode
display is most suitable for interfacing with 8051 since 8051 port pins can sink current better than sourcing
it.

62. CREATING DIGIT PATTERN
For displaying Digit say 7 we need to light segments -a ,b, c. Since we are using Common anode display , to
do so we have to to provide Logic -0 (0 v) at anode of these segments.so need to clear pins- P1.0
,P1.1,P1.2. that is 1 1 1 1 1 0 0 0 -->F8h .
Connection Hex Code
Segment Seg.
8051 pin number Digit Seg. h Seg. g Seg. f Seg. e Seg. d Seg. c Seg. a HEX
number b
a P1.0 0 1 1 0 0 0 0 0 0 C0
b P1.1 1 0 0 0 0 0 1 1 0 06
c P1.2 2 1 0 1 0 0 1 0 0 A4
d P1.3 3 1 0 1 1 0 0 0 0 B0
e P1.4 4 1 0 0 1 1 0 0 1 99
f P1.5
g p1.6 COMMON ANODE
h(dp) P1.7
Segment Seg.
8051 pin number Digit Seg. h Seg. g Seg. f Seg. e Seg. d Seg. c Seg. a HEX
number b
a P1.0 0 0 0 1 1 1 1 1 1 3f
b P1.1 1 0 0 0 0 0 1 1 0 06
c P1.2 2 0 1 0 1 1 0 1 1 5b
d P1.3 3 0 1 0 0 1 1 1 1 4f
e P1.4 4 0 1 1 0 0 1 1 0 66
f P1.5
g p1.6
h(dp) P1.7 COMMON CATHODE

63. You can also do this for some characters like A, E. But not for D or B because it will be same
as that of 0 & 8 . So this is one of limitation of 7-seg display.
Since we can enable only one 7-seg display at a time, we need to scan these display at fast rate .The
scanning frequency should be high enough to be flicker- free. At least 30HZ .Therefore – time one digit is
ON is 1/30 seconds
INTERFACING
Note that I am using Common Anode display. So the common Anode pin is tied to 5v .The cathode pins are
connected to port 1 through 330 Ohm resistance (current limiting).
SOURCE CODE:
8000: MOV DPTR, #2023; Control register of 8255
MOV A, # 80;
MOVX @DPTR, A; select I/O mode in 8255, out at Control register of 8255
MOV R0, # 00;
MOV A, R0;
XX: MOV DPTR, # 8500; Look up table address
MOVC A, @A+DPTR;
MOV DPTR, #2020; Port “A” Address
MOVX @DPTR, A;
LCALL DELAY
INC R0

64. CJMP R0, # 00FH, XX;
YY: SJMP YY
DELAY: MOV R1, # FF
CC: MOV R2, # FF
AA: DJNZ R2, AA
MOV R3, # FF
BB: DJNZ R3, BB
DJNZ R1, CC
RET
LOOK UP TABLE
8500: C0, 06, A4, B0, 99, -----------------
RESULT: To generate digits from 0 to 9 using common anode and common cathode technique.

65. INTERFACING DAC WITH 8051
AIM:
To interface DAC with 8051 parallel port to demonstrate the generation of square, saw tooth and
Triangular wave.
APPARATUS REQUIRED;
8051 Trainer Kit
DAC Interface Board
THEORY:
DAC 0809 is an 8 – bit DAC and the output voltage variation is between – 5V and +5V.The output voltage
varies in steps of 10/256 = 0.04 (appx.). The digital data input and the corresponding output voltages are
presented in the Table below
INPUT DATA IN HEX OUTPUT VOLTAGE(V)
00 -5.00
01 -4.96
02 -4.92
7F 00
FD 4.92
FE 4.96
FF 5.00
Table 1.
Referring to Table1, with 00 H as input to DAC, the a nalog output is – 5V. Similarly, With FF H as input,
the output is +5V. Outputting digital data 00 and FF at regular intervals, to DAC, results in different wave
forms namely square, triangular, etc,
Two methods of creating a DAC : binary weighted and R/2R ladder. The vast majority of integrated
circuit DACs, including the MC1408 (DAC 0809) used in this section use the R/2R method since it can
achieve a much higher degree of precision. The first criterion for judging a DAC is its resolution, which is a
function of the number of binary inputs. The common ones are 8, 10, and 12 bits. The number of data bit
inputs decides the resolution of the DAC since the number of analog output levels is equal to 2 n , where n is
the number of data bit inputs. Therefore, an 8-input DAC.
Such as the DAC 0809 provides 256 discrete voltage (or current) levels of output.
Similarly, the 12-bit DAC provides 4096 discrete voltage levels. There are also
16-bit DACs, but they are more expensive.

66. Figure 1. 8051Connection to DAC 0809
ALGORITHM:
(a) Square Wave Generation
1. Move the port address of DAC to DPTR.
2. Load the initial value (00) to Accumulator and move it to DAC.
3. Call the delay program.
4. Load the final value (FF) to accumulator and move it to DAC.
5. Call the delay program.
6. Repeat the steps 2 to 5.
(b) Saw tooth Wave Generation
1. Move the port address of DAC to DPTR.
2. Load the initial value (00) to Accumulator.
3. Move the accumulator content to DAC.
4. Increment the accumulator content by 1.
5. Repeat Steps 3 and 4.
(c) Triangular Wave Generation
1. Move the port address of DAC to DPTR.
2. Load the initial value (00) to Accumulator.
3. Move the accumulator content to DAC

67. 4. Increment the accumulator content by 1.
5. If accumulator content is zero proceed to next step. Else go to step 3.
6. Load value (FF) to Accumulator
7. Move the accumulator content to DAC
8. Decrement the accumulator content by 1.
9. If accumulator content is zero go to step2. Else go to step 7.
PROGRAM: (a) Square Wave Generation
MOV DPTR, # 2023; Control register of 8255.
MOV A, #80; 8255 in I/O mode operation
MOVX @DPTR, A
MOV DPTR, # 2021 (port B address of DAC start)
START: MOV A, #00H
MOVX @DPTR, A
LCALL DELAY
MOV A, #FFH
MOVX @DPTR, A
LCALL DELAY
SJMP START
DELAY: MOV R1, #05H
LOOP: MOV R2, #FFH
HERE: DJNZ R2, HERE
DJNZ R1, LOOP
RET
PROGRAM: (b) Saw Tooth Wave Generation
MOV DPTR, # 2023; Control register of 8255.
MOV A, #80; 8255 in I/O mode operation

68. MOVX @DPTR,
START: MOV A, #00H
MOV DPTR, # 2021 (port B address of DAC start)
GO: MOVX @DPTR, A
INC A
CJNE A, #FFH, GO
SJMP START
PROGRAM: (c) Triangular Wave Generation
MOV DPTR, # 2023; Control register of 8255.
MOV A, #80; 8255 in I/O mode operation
MOVX @DPTR, A
START: MOV A, #00H
MOV DPTR, # 2021 (port B address of DAC start)
GO: MOVX @DPTR, A
INC A
CJNE A, #FFH, GO
GO1: MOVX @DPTR, A
DEC A
CJNE A, #00H, GO1
SJMP START
Result : Thus the square, triangular and saw tooth wave form were generated by interfacing DAC with
8051 trainer kit.

69. ADC 0808/0809 INTERFACING WITH 8051
AIM:
To interface ADC with 8051 to generate digital output by giving an analog input voltage.
APPARATUS REQUIRED;
8051 Trainer Kit
ADC Interface Board
THEORY:
One of the most commonly used ADC is ADC0808/0809. ADC 0808/0809 is a Successive
approximation type with 8 channels i.e. it can directly access 8 single ended analog signals.
I/O Pins
ADDRESS LINE A, B, and C: The device contains 8-channels. A particular channel is selected by using the
address decoder line. The TABLE 1 shows the input states for address lines to select any channel.
Address Latch Enable ALE: The address is latched on the Low – High transition of ALE.
START: The ADC’s Successive Approximation Register (SAR) is reset on the positive edge i.e. Low- High
of the Start Conversion pulse. Whereas the conversion is begun on the falling edge i.e. high – Low of the
pulse.
Output Enable: Whenever data has to be read from the ADC, Output Enable pin has to be pulled high thus
enabling the TRI-STATE outputs, allowing data to be read from the data pins D0-D7.
End of Conversion (EOC): This Pin becomes high when the conversion has ended, so the controller comes
to know that the data can now be read from the data pins.
Clock: External clock pulses are to be given to the ADC; this can be given either from LM 555 in Astable
mode or the controller can also be used to give the pulses.
ALGORITHM:
1. Start. 2. Select the channel. 3. A Low – High transition on ALE to latch in the address. 4. A Low –
High transition on Start to reset the ADC’s SAR. 5. A High – Low transition on ALE. 6. A High – Low
transition on start to start the conversion. 7. Wait for End of cycle (EOC) pin to become high. 8. Make

70. Output Enable pin High. 9. Take Data from the ADC’s output 10. Make Output Enable pin Low. 11.
Stop
The total numbers of lines required are: Datalines:8,ALE:1,START:1,EOC:1,Output Enable:1
I.e. total 12 lines. You can directly connect the OE pin to Vcc. Moreover instead of polling for EOC just put
some delay so instead of 12 lines you will require 10 lines. We Can also provide the clock through the
controller thus eliminating the need of external circuit for clock.
Calculating Step Size
ADC 0808 is an 8 bit ADC i.e. it divides the voltage applied at Vref+ & Vref- into 28 i.e. 256 steps.
Step Size = (Vre f+ - Vref-)/256
Suppose Vref+ is connected to Vcc i.e. 5V & Vref- is connected to the Gnd then the step size will be
Step size= (5 - 0)/256= 19.53 mv.
Calculating D out.
The data we get at the D0 - D7 depends upon the step size & the Input voltage i.e. Vin.
Dout = Vin /step Size.
If you want to interface sensors like LM35 which has output 10mv/°C then I would suggest that you set the
Vref+ to 2.56v so that the step size will be
Step size= (2.56 - 0)/256= 10 mv.
So now whatever reading that you get from the ADC will be equal to the actual temperature.
Here is a program for interfacing the ADC to microcontroller, as stated above have assumed that the
OE pin is connected to Vcc & the clock is given by the controller.
This program selects channel 0 as input channel reads from it & saves in the accumulator.
NOTE: In the LAB, ADC is interfaced with 8051 Microcontroller through 8255 PPI.
Port 2 of 8051 is connected to Port C of 8255, Port 1 of 8051 is connected to Port A of 8255,
Port C is for interfacing signals and Port A is for Reading data from ADC.

71. Fig.1 interfacing circu it
PROGRAM:
MOV DPTR, #2023
MOV A, #90: PA-I/P PORT, PORT B, C-O/P PORTS
MOVX @DPTR, A
BEGIN: MOV DPTR, #2022; Select CH-0
MOV A, #00
MOVX @DPTR, A
MOV A, #0DH; Set PC6 (OE)
MOV DPTR, #2023
MOVX @DPTR, A
MOV A, #0FH; SET SOC
MOVX @DPTR, A
LCALL DELAY
MOV A, #0EH
MOVX @DPTR, A
MOV A, #0CH
MOVX @DPTR, A

72. MOV DPTR, #2020
BACK: MOVX A,@DPTR
JB 0E7, BACK
MOV DPTR, #2020
REP: MOVX A,@DPTR
ANL A, #80H
JNB 0E7, REP
MOV A, #0DH
MOV DPTR, #2023
MOVX @DPTR, A
MOV DPTR, #2020
MOVX A,@DPTR
MOV DPTR, #9000H; Read the digital output from this address
MOVX @DPTR, A
SJMP BEGIN
DELAY: MOV R3, #30H
AG AIN: MOV R4, #FFH
BACK: NOP
NOP
DJNZ R4, BACK
DJNZ R3, AGAIN
RET
RESULT: Given input voltage varying from 5v to 0v at Channel 0 and observe the digital output value at
the address at 9000h, which is given in code.

73. STEPPER MOTOR INTERFACING WITH 8051
AIM:
To interface a stepper motor with 8051 microcontroller and operate it.
APPARATUS:
8051
Stepper motor interfacing board
THEORY:
A motor in which the rotor is able to assume only discrete stationary angular position is a stepper
motor. The rotary motion occurs in a step-wise manner from one equilibrium position to the next. Stepper
Motors are used very wisely in position control systems like printers, disk drives, process control machine
tools, etc.
The basic two-phase stepper motor consists of two pairs of stator poles. Each of the four poles ha s its
own winding. The excitation of any one winding generates a North Pole. A South Pole gets induced at the
diametrically opposite side. The rotor magnetic system has two end faces. It is a permanent magnet with one
face as South Pole and the other as North Pole.
The Stepper Motor windings A1, A2, B1, B2 are cyclically excited with a DC current to run the
motor in clockwise direction. By reversing the phase sequence as A1, B2, A2, B1, anticlockwise stepping
can be obtained.
8051 Interfacing with stepper motor

74. Stepper motors can be driven in two different patterns or sqeunces. Namely,
1. Full Step Sequence 2. Half Step Sequence
A normal 4 step sequence is like below.
Step # Winding A Winding B Winding C Winding D
1 1 0 0 1
2 1 1 0 0
3 0 1 1 0
4 0 0 1 1
Going from step 1 to 4 we rotate motor clockwise, going from step 4 to 1 we rotate motor counter clockwise. The
stepper motor discussed here has total 6 leads. 4 leads representing 4 windings and 2 commons for centre tapped leads.
Here we have used 2 phase 4 step sequence. It must also be noted we can start from any step. Be it step 1, 2, 3 or 4.
But we must continue in proper order for proper rotation.
In programs on blog I have mostly begun from step 3. Other step sequences are as below
Half Step 8 Step Sequence
Step # Winding A Winding B Winding C Winding D
1 1 0 0 1
2 1 0 0 0
3 1 1 0 0
4 0 1 0 0
5 0 1 1 0
6 0 0 1 0
7 0 0 1 1
8 0 0 0 1
Wave drive 4 step sequence
Step # Winding A Winding B Winding C Winding D
1 1 0 0 0
2 0 1 0 0
3 0 0 1 0
4 0 0 0 1

75. The Movement is associated with a step angle:
Step angle Steps per revolution
.72 500
1.8 200
2 180
2.5 144
5 72
7.5 48
15 24
4 Pins Of Stepper Motor Controlled By 4 Bits Of 8051 P1.0 - P1.3; It Is Interfaced Using Darlington
Arrays Such As Uln2003 As; 8051 Lacks Current To Run The Motor. So Uln2003 Is Used For
Energizing Stator.; Program Below Shows Just Rotating The Motor Step Wise; But Step Width Is
Unknown
MOV A, #66H ; Load The Step Sequence
BACK :MOV P1, A ; Load Sequence To Port
RR A ; Change Sequence Rotate Clockwise
ACALL DELAY : Wait For It
SJMP BACK ; Now Keep Going
DELAY :MOV R2, #100
H1 :MOV R3, #255
H2 :DJNZ R3, H2
DJNZ R2, H1
RET
Step Angle:
Step angle of the stepper motor is defined as the angle traversed by the motor in one step. To
calculate step angle, simply divide 360 by number of steps a motor takes to complete one revolution. As
we have seen that in half mode, the number of steps taken by the motor to complete one revolution gets
doubled, so step angle reduces to half.
As in above examples, Stepper Motor rotating in full mode takes 4 steps to complete a revolution, So
step angle can be calculated as...Step Angle ø = 360° / 4 = 90°.And in case of half mode step angle gets half

76. so 45°.So this way we can calculate step angle for any stepper motor. Usually step angle is given in the spec sheet
of the stepper motor you are using.
Knowing stepper motor's step angle helps you calibrate the rotation of motor also to helps you move the motor
to correct angular position.
; This Other Program Code How In Same Interfacing To Rotate A Stepper Motor
; 64 Degrees In Clockwise Direction Using 4 Step Sequence
; It Takes Some Calculations
; Consider A Motor With Step Angle 2 Degree
; So Steps Per Revolution = 180
; No Of Rotor Teeth = 45
; Movement Per 4 Step Sequence is 8 Degrees
; For 64 Degree Movement We Need To Send 8 Consecutive 4 Step Sequence
; That Is It Will Cover 32 Steps
ORG 0000H
MOV A, #66H
MOV R1, #32H ; 32 Steps To Be Taken
BACK :RR A ; Rotate Clockwise
MOV P1, A ;
ACALL DELAY ;
DJNZ R0, BACK ;
END
DELAY: MOV R2, #100
H1: MOV R3, #255
H2: DJNZ R3, H2
DJNZ R2, H1
RET
Result: By applying step sequence in program rotate the motor in clock wise and anti clock wise direction.