mit

Micropro. & Interfacing Techniques (PU) L-1 Lab Manual
List of Experiments

Group A

Program 1 : Write X86/64 Assembly language program (ALP) to add array of N hexadecimal numbers
stored in the memory. Accept input from the user.
Explanation :
- Consider that a block of N bytes is present at source location.
- Let the number of bytes N = 10 for example.
- We have to add these N bytes.
- We will initialize this as count in the CX register.
- We know that source address is in the SI register. This SI register will act as
pointer.
- Clear the direction flag.
- Using ADD instruction add the contents, byte by byte of the block.
- Increment SI to point to next element.
- Decrement the counter and add the contents till all the contents are added.
- Result is stored in AL.
- Display the contents using display routine.
For example : Block Data : 01 02 03 04 05 06 07 08 09 0A
Result : 01 + 02 + 03 + 04 + 05 + 06 + 07 + 08 + 09 + 0A
= 37 H
Algorithm :
Step I : Initialise the data segment.
Step II : Initialise SI as pointer with source address.
Step III : Initialise CX register with count.
Step IV : Initialise direction flag to zero.
Step V : Add data, byte by byte.
Step VI : Increment pointer i.e. SI.
Step VII : Decrement counter CX.
Step VIII : Check for count in CX, if not zero goto step V else goto step IX.
Step IX : Display the result of addition.
Step X : Stop.
Flowchart : Refer flowchart A.1.
Micropro. & Interfacing Techniques (PU) L-2 Lab Manual
Program :
Label Instruction Comment
.model small
.data
blk1 db 01, 02,
03, 04, 05, 06,
07, 08, 09, 0AH

count dw 0AH
.code
mov ax, @data initialise data segment
mov ds, ax
mov ax, 0
mov si, offset
blk1
initialise pointer
mov cx, count initialise counter
cld df=0
l1: add al, [si] add numbers
inc si increment pointer
dec count decrement counter
jnz l1 check if all nos are added
mov ch, 02h Count of digits to be displayed
mov cl, 04h Count to roll by 4 bits
mov bl, al Result in reg bl
l2: rol bl, cl roll bl so that msb comes to lsb
mov dl, bl load dl with data to be displayed
and dl, 0fH get only lsb
cmp dl, 09 check if digit is 0-9 or letter A-F
jbe l4
add dl, 07 if letter add 37H else only add
30H
l4: add dl, 30H
mov ah, 02 INT 21H (Display character)
int 21H
dec ch Decrement Count
jnz l2
mov ah, 4cH Terminate Program
int 21H
end
Flowchart A.1

Micropro. & Interfacing Techniques (PU) L-3 Lab Manual
Result :
C:\programs>tasm blkadd.asm
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: blkadd.asm
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 437k
C:\programs>tlink blkadd
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>blkadd
37
Program 2 : Write X86/64 ALP to perform non-overlapped and overlapped block transfer (with and
without string specific instructions). Block containing data can be defined in the data
segment.
(A) To perform non-overlapped block transfer
Program statement :
- Write an ALP to move a block of N bytes of data from source to destination and
display the result. (Non-overlapped block transfer)
Explanation :
- Consider that a block of data of N bytes is present at source location. Now this block
of N bytes is to be moved from source location to a destination location.
- Let the number of bytes N = 10.
- We will have to initialize this as count in the CX register.
- We know that source address is in the SI register and destination address is in the
DI register.
- Clear the direction flag.
- Using the string instruction move the data from source location to the destination
location. It is assumed that data is moved within the same segment. Hence the DS
and ES are initialized to the same segment value.
- Display the contents using display routine.
Algorithm :
Step I : Initialise the data in the source memory and destination memory.
Step II : Initialise SI and DI with source and destination address.
Step III : Initialise CX register with the count.
Step IV : Initialise the direction flag to zero.
Step V : Transfer the data block byte by byte to destination.
Step VI : Decrement CX.
Step VII : Check for count in CX, if not zero goto step V else goto step VIII.
Step VIII : Display the bytes in destination location.
Step IX : Stop.
Micropro. & Interfacing Techniques (PU) L-4 Lab Manual
Flowchart : Refer flowchart A.2(a).
Program :
Label Instruction Comment
.model small
.data
src_blk db 01, 02, 03, 04,
05, 06, 07, 08, 09, 0AH

dest_blk db 10 dup(?)
count dw 0AH
.code
mov ax, @data initialize data & extra
segment
mov ds, ax
mov es, ax
mov si, offset src_blk si to point to source
block
mov di, offset dest_blk di to point to
destination block
mov cx, count initialize counter
cld df=0
again : rep movsb transfer contents
mov di, offset dest_blk di to point to
destination block
mov bh, 0Ah initialize counter
up: mov bl, [di] store result in bl
mov cx, 0204h Count of digits to be
displayed in ch and
digits to be mrolled
in cl
l1: rol bl, cl roll bl so that msb
comes to lsb
mov dl, bl load dl with data to be
displayed
and dl, 0fh get only lsb
cmp dl, 09h check if digit is 0-9 or
letter A-F
jbe l12
add dl, 07h if letter add 37H else
only add 30H
l12: add dl, 30h
Flowchart A.2(a)

Micropro. & Interfacing Techniques (PU) L-5 Lab Manual
Label Instruction Comment
mov ah, 02 Function 02 under
INT 21H
int 21h
dec ch Decrement Count
jnz l1
dec bh decrement counter
inc di
mov ah, 02h display space between
bytes
mov dl, ' '
int 21h
cmp bh, 00h repeat till all bytes are
displayed
jne up
mov ah, 4ch normal termination to
dos
int 21h
end
Result :
C:\programs>tlink revblock
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>tasm block
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: block.ASM
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 437k
C:\programs>tlink block
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>block
01 02 03 04 05 06 07 08 09 0A
C:\programs>
Micropro. & Interfacing Techniques (PU) L-6 Lab Manual
(B) To perform overlapped block transfer
Program Statement :
Write a program in the ALP of 8086 to move a block of N data bytes from source to
destination. The source begins at 2001 H and destination begins at location 2005 H.
(Overlapped block transfer)
Explanation :
- The source block is at address 2001 H and destination block is at address 2005 H.
Let the number of bytes in the block to be transferred be 10. Initialize this as count
in CX register. Now enter the data bytes in source block which you want to transfer
to the destination block. Here the destination block is overlapping the source block.
So first we will transfer the contents of last location, the second last location and so
on till all bytes are transferred. As SI and DI both are pointing to last location of
source and data block, once the data is transferred, we will use STD i.e. set direction
flag which autodecrements the SI and DI registers. Display the result.
Algorithm :
Step I : Initialize the data section with addresses of source and destination block.
Step II : Initialize SI = start of source block.
Step III : Enter data into source block.
Step IV : Initialize DI = start of destination block.
Step V : DI = DI + (count – 1) i.e. DI = last location of destination block.
Step VI : Initialize counter = 10.
Step VII : Autodecrement SI, DI.
Step VIII : Transfer contents from source location to destination location.
Step IX : Decrement counter.
Step X : Is counter = 0 ? If not go to step VIII.
Step XI : Display the contents.
Step XII : Stop.
Flowchart : Refer flowchart A.2(b).
Program :
Label Instruction Comment
.model small
.data
src_blk db 01, 02, 03,
04, 05, 06, 07, 08, 09,
0AH

dest_blk db 10 dup(?)
count dw 0AH
.code
mov ax, @data initialize data & extra
segment
mov ds, ax
mov es, ax
Micropro. & Interfacing Techniques (PU) L-7 Lab Manual
Label Instruction Comment
mov si, offset src_blk si to point to source block
mov di, offset dest_blk di to point to destination
block
mov cx, count initialize counter
cld df=0
again: rep movsb transfer contents
mov di, offset dest_blk di to point to destination
block
mov bh, 0Ah initialize counter
up: mov bl, [di] store result in bl
mov cx, 0204h Count of digits to be
displayed in ch and digits
to be rolled in cl
l1: rol bl, cl roll bl so that msb comes to
lsb
mov dl, bl load dl with data to be
displayed
and dl, 0fh get only lsb
cmp dl, 09h check if digit is 0-9 or letter
A-F
jbe l12
add dl, 07h if letter add 37H else only
add 30H
l12: add dl, 30h
mov ah, 02 Function 02 under
INT 21H
int 21h
dec ch Decrement Count
jnz l1
dec bh decrement counter
inc di
mov ah, 02h display space between bytes
mov dl, ' '
int 21h
cmp bh, 00h repeat till all bytes are
displayed
jne up
mov ah, 4ch normal termination to dos
int 21h
end
Flowchart A.2(b)

Micropro. & Interfacing Techniques (PU) L-8 Lab Manual
Result :
C:\programs>tlink revblock
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>tasm block
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: block.ASM
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 437k
C:\programs>tlink block
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>block
01 02 03 04 05 06 07 08 09 0A
C:\programs>
Program 3 : Write 64 bit ALP to convert 4-digit Hex number into its equivalent BCD number and 5-digit
BCD number into its equivalent HEX number. Make your program user friendly to accept
the choice from user for :
(a) HEX to BCD (b) BCD to HEX (c) EXIT.
Display proper strings to prompt the user while accepting the input and displaying the
result. (use of 64-bit registers is expected)
(A) HEX to BCD
Program statement :
- Write 8086 ALP to convert 4 digit Hex number to its equivalent BCD number.
Explanation :
- We have a 4 digit Hex number whose equivalent binary number is to be found.i.e.
FFFF H. Initially we compare FFFF H with decimal 10000 ( 2710 H in Hex ). If
number is greater than 10,000 we add it to DH register. Also, we subtract decimal
10,000 from FFFF H, each time comparision is made. Then we compare the number
obtained in AX by 1000 decimal. Each time we subtract 1000 decimal from AX and
add 1000 decimal to BX. Then we compare number obtained in AX by 100 decimals.
Each time we subtract 100 decimal from AX and add 100 decimal to BX to obtain
BCD equivalent. Then we compare number obtained in AX with 10 decimal. Each
time we subtract 10 decimal from AX and we add 10 decimal to BX. Finally we add
the result in BX with remainder in AX. The final result is present in register DH
with contains the 5
th
bit if present and register AX.
- Display the result.
Algorithm :
Step I : Initialize the data segment.
Step II : Initialize BX = 0000 H and DH = 00H.
Micropro. & Interfacing Techniques (PU) L-9 Lab Manual
Step III : Load the number in AX.
Step IV : Compare number with 10000 decimal. If below goto step VII else goto
step V.
Step V : Subtract 10,000 decimal from AX and add 1 decimal to DH
Step VI : Jump to step IV.
Step VII : Compare number in AX with 1000, if below goto step X else goto
step VIII.
Step VIII : Subtract 1000 decimal from AX and add 1000 decimal to BX.
Step IX : Jump to step VII.
Step X : Compare the number in AX with 100 decimal if below goto step XIII
Step XI : Subtract 100 decimal from AX and add 100 decimal to BX.
Step XII : Jump to step X
Step XIII : Compare number in AX with 10. If below goto step XVI
Step XIV : Subtract 10 decimal from AX and add 10 decimal to BX..
Step XV : Jump to step XIII.
Step XVI : Add remainder in AX with result in BX.
Step XVII : Display the result in DH and BX.
Step XVIII : Stop.
Flowchart : Refer flowchart A.3(a).
Program :
Label Instruction Comment
.model small
.stack 100
.code
mov ax, 0ffffh hex number to find it's bcd
mov bx, 0000
mov dh, 0
l9 : cmp ax, 10000 if ax>10000
jb l2
sub ax, 10000 subtract 10000
inc dh add 1 to dh
jmp l9
l2 : cmp ax, 1000 if ax>1000
jb l4
sub ax, 1000
add bx, 1000h add 1000h to result
jmp l2
l4 : cmp ax, 100 if ax>100
jb l6
sub ax, 100
Micropro. & Interfacing Techniques (PU) L-10 Lab Manual
Label Instruction Comment
add bx, 100h add 100h to result
jmp l4
l6 : cmp ax, 10 if ax>10
jb l8
sub ax, 10
add bx, 10h add 10h to result
jmp l6
l8 : add bx, ax add remainder to result
mov ah, 02
mov cx, 0204h Count to display 2 digits
go: rol dh, cl
mov dl, dh
and dl, 0fh
add dl, 30h display 2 msb digits
int 21h
dec ch
jnz go
mov ch, 04h Count of digits to be displayed
mov cl, 04h Count to roll by 4 bits
l12: rol bx, cl roll bl so that msb comes to lsb
mov dl, bl load dl with data to be
displayed
and dl, 0fH get only lsb
cmp dl, 09 check if digit is 0-9 or letter
A-F
jbe l14
add dl, 07 if letter add 37H else only add
30H
l14: add dl, 30H
mov ah, 02 Function 2 under INT 21H
(Display character)
int 21H
dec ch Decrement Count
jnz l12
mov ah, 4cH Terminate Program
int 21H
end

Flowchart A.3(a)

Micropro. & Interfacing Techniques (PU) L-11 Lab Manual
Result :
C:\programs>tasm hex2bcd.asm
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: hex2bcd.ASM
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 437k
C:\programs>tlink hex2bcd
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
C:\programs>hex2bcd
065535
(B) BCD to HEX
Program statement :
- Write 8086 ALP to convert 5 digit BCD number to its equivalent Hex number.
Explanation :
- We are given a five digit BCD number whose HEX equivalent is to be found i.e.
65535 whose HEX equivalent is to be found. First we will find the Hex equivalent of
60,000. We will compare 60,000H with 10,000H. Each time we compare a counter is
decremented by 10,000 and we add 10,000 decimal (2710 Hex). Then we will find the
equivalent of 5000. Now we will compare 5000H with 1000H. Each time we compare
the counter decrements by 1000 and we add 1000 decimal (3E8 H). Then we find the
equivalent of 500H by comparing it with 100H. Each time counter decrements by
100 and we add decimal 100 (64 H). Then we find equivalent of 30H by comparing it
with 10H. Each time counter decrements by 10 and we add 10 decimal (0A H). Then,
equivalent of 5 is 5H.
- Finally, all the equivalents obtained are added to get the equivalent of 65535.
Algorithm :
Step I : Initialize the data segment.
Step II : Load the MSB of word in register AX.
Step III : Compare it with 0, if zero goto step VII else goto step IV.
Step IV : decrement AX and initalize BX = 0000.
Step V : add 10000 decimal to BX.
Step VI : Jump to step III.
Step VII : Load LSB of word in register AX.
Step VIII : Compare it with 1000, if below ogto step XII else goto step IX.
Step IX : subtract 1000 H from AX.
Step X : Add 1000 decimal to BX.
Micropro. & Interfacing Techniques (PU) L-12 Lab Manual
Step XI : Jump to step VIII
Step XII : Compare number in AX now with 100 H, if below goto step XVI, else
goto step XIII.
Step XIII : Subtract 100 H from AX.
Step XIV : Add 100 decimal to BX.
Step XV : Jump to step XII.
Step XVI : Compare number in AX with 10H, if below goto step XX, else goto
step XVII.
Step XVII : Subtract 10 H from AX
Step XVIII : Add 10 decimal to BX
Step XIX : Jump to step XVI
Step XX : Add contents of AX and BX.
Step XXI : Display the result.
Step XXII : Stop.
Flowchart : Refer flowchart A.3(b).
Program :
Label Instruction Comment

.model small
.stack 100
.data
a dd 00065535h
.code
mov ax, @data Intialize data segment
mov ds, ax
mov ax, word ptr a+2 checking msb no
mov bx, 0000h intialize hex result
l11:
cmp ax, 0 cmp ax
jz l10
dec ax if ax=1 then it means
no>10000
add bx, 10000 so add 10000 to bx
jmp l11
l10: mov ax, word ptr a load lsb part in ax
l2 : cmp ax, 1000h if ax>1000h
jb l4
sub ax, 1000h
add bx, 1000 add 1000 to result
jmp l2
Micropro. & Interfacing Techniques (PU) L-13 Lab Manual
Label Instruction Comment
l4 : cmp ax, 100h if ax>100h
jb l6
sub ax, 100h
add bx, 100 add 100 to result
jmp l4
l6 : cmp ax, 10h if ax>10h
jb l8
sub ax, 10h
add bx, 10 add 10 to result
jmp l6
l8 : add bx, ax add remainder to result
mov ch, 04h Count of digits to be
displayed
mov cl, 04h Count to roll by 4 bits
mov bx, ax Result in reg bx
l2: rol bx, cl roll bl so that msb comes to
lsb
mov dl, bl load dl with data to be
displayed
and dl, 0fH get only lsb
cmp dl, 09 check if digit is 0-9 or
letter A-F
jbe l4
add dl, 07 if letter add 37H else only
add 30H
l4: add dl, 30H
mov ah, 02 Function 2 under INT 21H
(Display character)
int 21H
dec ch Decrement Count
jnz l2
mov ah, 4cH Terminate Program
int 21H
end
Flowchart A.3(b)
Micropro. & Interfacing Techniques (PU) L-14 Lab Manual
Result :
C:\programs>tasm bcd2hex.asm
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: bcd2hex.asm
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 437k
C:\programs>tlink bcd2hex
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
C:\programs>bcd2hex
FFFF
Program 4 : Write X86/64 ALP for the following operations on the string entered by the user. (use of
64-bit registers is expected)
(a) Calculate Length of the string (b) Reverse the string
(c) Check whether the string is palindrome or not
OR
Make your program user friendly by providing MENU like:
(a) Enter the string
(b) Calculate length of string
(c) Reverse string
(d) Check palindrome
(e) Exit
Display appropriate messages to prompt the user while accepting the input and
displaying the result.
Explanation :
- Using Macro display the Menu for entering string, calculate length, reverse,
palindrome and exit. Accept the choice from user using INT 21H function 01H.
- If choice = 1, call procedure for accepting string. Using interrupt INT 21H, function
0AH accept the string and end procedure. Return back to display Menu.
- If choice = 2, call procedure for finding length of the string. Display length and
return back to display Menu.
- If choice = 3, call procedure to reverse the string. Display the reverse string and
return back to display Menu.
- If choice = 4, call procedure to check if entered string is palindrome. If palindrome
displays, the string is a palindrome, otherwise display String is not a palindrome.
- If choice = 5, terminate the program. If any other key is pressed display invalid
choice.
Algorithm :
Step I : Initialize the data and stack memory.
Step II : Using Macro display Menu.
1. Accept 2. Length 3. Reverse
4. Palindrome 5. Exit.
Micropro. & Interfacing Techniques (PU) L-15 Lab Manual
Step III : Accept choice from user using INT 21H, function 01H.
Step IV : IS choice = 1 jump to step XI else goto step V.
Step V : IS choice = 2 jump to step XIV else goto step VI.
Step VI : IS choice = 3 jump to step XVII else goto step VII.
Step VII : IS choice = 4 jump to step XX else
goto step VIII.
Step VIII : IS choice = 5 jump to step
XXIII else goto step IX.
Step IX : Display Wrong choice.
Step X : Jump to step II.
Step XI : Call procedure accept.
Step XII : Accept string using INT
21H, function 0AH.
Step XIII : Return to main program
and goto step II.
Step XIV : Call procedure length.
Step XV : Calculate the length of
string and display it using
INT 21H, function 02H.
Step XVI : Return back to main
program and jump to
step II.
Step XVII : Call procedure reverse.
Step XVIII : Reverse the string and
display it.
Step XIX : Return back to main
program and jump to
step II.
Step XX : Call procedure palindrome.
Step XXI : Check if string is
palindrome. If yes display
string is palindrome else
string is not a palindrome.
Step XXII : Return back to main
program and jump to
step II.
Step XXIII : Terminate the program and
stop.
Flowchart : Refer flowchart A.4. Flowchart A.4
Micropro. & Interfacing Techniques (PU) L-16 Lab Manual
Program :
Label Instruction Comment
TITLE STRING OPERATIONS
MESS MACRO MSG DEFINITION OF MACRO MESS
MOV AH, 09H
LEA DX, MSG
INT 21H
ENDM
.MODEL SMALL
.STACK 100H
.DATA
STR1 DB 25 , ? , 25 DUP('$')
STR3 DB 25 , ? , 25 DUP('$')
MSG1 DB 0AH, 0DH, 'MENU $'
MSG21 DB 0AH, 0DH, '1.ACCEPT $'
MSG22 DB 0AH, 0DH, '2.LENGTH $'
MSG23 DB 0AH, 0DH, '3.REVERSE $'
MSG24 DB 0AH, 0DH, '4.PALINDROME $'
MSG25 DB 0AH, 0DH, '5.EXIT $'
MSG3 DB 0AH, 0DH, 'ENTER YOUR CHOICE : $'
MSG4 DB 0AH, 0DH, 'WRONG CHOICE $'
MSG5 DB 0AH, 0DH, 'ENTER THE STRING : $'
MSG6 DB 0AH, 0DH, 'STRING IS : $'
MSG7 DB 0AH, 0DH,'LENGTH IS : $'
MSG8 DB 0AH, 0DH, 'THE STRING IS A
PALINDROME $'

MSG9 DB 0AH, 0DH, 'THE STRING IS NOT A
PALINDROME $'

.CODE
mov ax, @data Intialize data and extra segment
mov ds, ax
mov es, ax
ak : mess msg1 display menu
mess msg21
mess msg22
mess msg23
mess msg24
mess msg25
mess msg3 accept choice
mov ah, 01h
Micropro. & Interfacing Techniques (PU) L-17 Lab Manual
Label Instruction Comment
int 21h
mov bl, al Choice BL
cmp bl, 31h if choice=1
je acc Accept string
cmp bl, 32h if choice=2
je len Find lenth of string
cmp bl, 33h if choice=3
je rev Reverse string
cmp bl, 34h if choice=4
je pal Check if string is palindrome
cmp bl, 35h if choice=5
je endd exit
mess msg4 Wrong Choice
jmp ak
acc : call accept
jmp ak
len : call lent
jmp ak
rev : call reverse
jmp ak
pal: call pall
jmp ak
endd: mov ah, 4ch
int 21h accept procedure
accept proc near
mess msg5
mov ah, 0ah Accept String
lea dx, str1
int 21h
RET
accept endp length procedure
lent proc near
mess msg7
mov dl, str1+1 Dl contains length of String
or dl, 30h
mov ah, 02h Display Length
int 21h
ret
lent endp reverse procedure
Micropro. & Interfacing Techniques (PU) L-18 Lab Manual
Label Instruction Comment
reverse proc near
mess msg6
mov ch, 00h
mov cl, str1+1 Cl has length of string
sub cl, 01h
lea si, str1+2 DESTINATION STRING
lea di, str1+2 DESTINATION STRING
repz movsb COPY TO TRAVERSE TILL END
OF FIRST STR
mov cl, str1+1
lea di, str3+2 DESTINATION STRING
loop1: mov dx, [si] dx contains rightmost character
mov ah, 02h
int 21h display character
mov [di], dx copy character to destination
dec si
inc di
dec cl
cmp cl, 00h
jne loop1
ret
reverse endp palindrome procedure
pall proc near
mess msg6
mov ah, 09h
lea dx, str1+2 str1 contains original string
int 21h
call reverse str3 has reversed string
lea di, str3+2
mov ah, 00h
mov dh, 00h
lea si , str1+2
mov cl, str1+1 CL contains Length of string
loop2 mov al, byte ptr[si]
mov bl, byte ptr[di]
dec cl Decrement count
cmp cl, 00h
je loopa
cmp al, bl Compare characters
Micropro. & Interfacing Techniques (PU) L-19 Lab Manual
Label Instruction Comment
je loop3 if same goto loop3
loopa cmp cl, 00h if checked all characters
je loop4
mess msg9 the strings are not same
jmp loop5
loop4 mess msg8 the strings are same
loop5 ret
loop3 inc si
inc di
jmp loop2 now check next character
pall endp
end end
Result :
C:\programs>tasm str
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: str.ASM
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 434k
C:\programs>tlink str
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
C:\programs>str
MENU
1. ACCEPT
2. LENGTH
3. REVERSE
4. PALINDROME
5. EXIT
ENTER YOUR CHOICE : 1
ENTER THE STRING : college
MENU
1. ACCEPT
2. LENGTH
3. REVERSE
4. PALINDROME
5. EXIT
ENTER YOUR CHOICE : 2
LENGTH IS : 7
Micropro. & Interfacing Techniques (PU) L-20 Lab Manual
MENU
1. ACCEPT
2. LENGTH
3. REVERSE
4. PALINDROME
5. EXIT
ENTER YOUR CHOICE : 3
STRING IS : egelloc
MENU
1. ACCEPT
2. LENGTH
3. REVERSE
4. PALINDROME
5. EXIT
ENTER YOUR CHOICE : 4
STRING IS : college
STRING IS : egelloc
THE STRING IS NOT A PALINDROME
MENU
1. ACCEPT
2. LENGTH
3. REVERSE
4. PALINDROME
5. EXIT
ENTER YOUR CHOICE : 1
ENTER THE STRING : madam
MENU
1. ACCEPT
2. LENGTH
3. REVERSE
4. PALINDROME
5. EXIT
ENTER YOUR CHOICE : 4
STRING IS : madam
STRING IS : madam
THE STRING IS A PALINDROME
Program 5 : Write 8086 ALP to perform string manipulation. The strings to be accepted from the user
is to be stored in data segment of program_l and write FAR PROCEDURES in code
segment program_2 for following operations on the string:
(a) Concatenation of two strings
(b) Number of occurrences of a sub-string in the given string Use PUBLIC and
EXTERN directive. Create .OBJ files of both the modules and link them to create an
EXE file.
Micropro. & Interfacing Techniques (PU) L-21 Lab Manual
(A) Concatenation of two strings
Program Statement :
Write a program in the assembly language of 8086, to concatenate two strings.
Explanation :
- Firstly, we will accept the two strings to be concatenated. Then, we will call
procedure CONCAT which will concatenate the two strings. Display the
concatenated strings.
Algorithm :
Step I : Start.
Step II : Accept string 1 from user.
Step III : Accept string 2 from user.
Step IV : Call procedure CONCAT.
Step V : Load length of string 1 in CX.
Step VI : Load the address of source string 1 in SI and DI.
Step VII : Copy the contents of string 1 to destination string.
Step VIII : Load SI with address of string 2.
Step IX : Copy the contents of string 2 to destination string.
Step X : Display the concatenated string.
Step XI : Procedure and return to calling program
Step XII : Stop.
Flowchart : Refer flowchart A.5(a).
Program :
Instruction Comment
page 100, 50
title string concatenatation
mess macro msg definition of macro mess
mov ah, 09h
lea dx, msg
int 21h
endm
.model small
.stack 100h
.data
str1 db 25 , ? , 25 dup('$')
str2 db 25 , ? , 25 dup('$')
msg1 db 0ah, 0dh, 'enter the string1: $'
msg2 db 0ah, 0dh, 'enter the string2: $'
msg3 db 0ah, 0dh, 'concatenated string
is : $'

.code
mov ax, @data data initialisation
Micropro. & Interfacing Techniques (PU) L-22 Lab Manual
Instruction Comment
mov ds, ax
mov es, ax
mess msg1 accept string1 from user
function 0ah
mov ah, 0ah under int 21h
lea dx, str1
int 21h
mess msg2 accept string1 from user
function 0ah
mov ah, 0ah under int 21h
lea dx, str2
int 21h
call concat call procedure
mov ah, 4ch normal termination to dos
int 21h
concat proc near begin procedure
mov ch, 00h
mov cl, str1+1 cl=length of string1
lea si, str1+2 destination string
lea di, str1+2 destination string
repz movsb copy to traverse till end of
string1
mov ch, 00h
mov cl, str2+1 cl=length of string2
lea si, str2+2 source string
cld df=0
repz movsb copy to traverse till end of
string2
mess msg3 display concatenated string
use function 09h
lea si, str1 under int 21h
mov ah, 09h
lea dx, str1+2
int 21h
ret
concat endp end procedure
end end program
Flowchart A.5(a)
Micropro. & Interfacing Techniques (PU) L-23 Lab Manual
Result :
C:\programs>TASM STR1
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: STR1.ASM
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 437k
C:\programs>TLINK STR1
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
C:\programs>STR1
ENTER THE STRING1: HELLO
ENTER THE STRING2: MONICA
CONCATENATED STRING IS : HELLOMONICA
C:\programs>
(B) Compare two strings
Program statement :
- Write a program in the ALP of 8086 to check the data in two strings are equal, if
equal display the message “equal strings”, and if not display the message “unequal
strings”.
Explanation :
- We will accept two strings from the user. After accepting the strings, the first step in
string comparison is to check whether their string lengths are equal. If the string
lengths are not equal, we print the message unequal strings. If the string lengths
are equal, we check if the contents of two strings are equal. The lengths of the two
stings are initialized in the CX register.
- The source and destination address are initialized in DS : SI and ES : DI registers.
- Using the string instruction REPE CMPSB, two data are compared character by
character.
- If all the characters are matching display the message “equal strings” otherwise
display “unequal strings”.
Algorithm :
Step I : Initialize the data memory.
Step II : Allocate data memory to save the strings.
Step III : Initialize DS and ES register.
Step IV : Accept the first string.
Step V : Accept the second string.
Step VI : Load the number of characters of first string in CL.
Step VII : Load the number of characters of second string in CH register.
Step VIII : Compare the lengths of the two strings. If not go to step XIII.
Step IX : Load number of characters to be compared in CX.
Micropro. & Interfacing Techniques (PU) L-24 Lab Manual
Step X : Compare the strings, character by character. If not same goto step XIII.
Step XI : Print “equal strings” using Macro.
Step XII : Jump to step XIV.
Step XIII : Print “unequal strings” using Macro.
Step XIV : Stop.
Flowchart : Refer flowchart A.5(b).
Program :
Label Instruction Comment
PRINT MACRO MES Macro to display
string
MOV AH, 09H
LEA DX, MES
INT 21H
ENDM
.MODEL SMALL
.DATA
MS1 DB 10, 13,
"ENTER FIRST
STRING : $"

MS2 DB 10, 13,
"ENTER SECOND
STRING : $"

MS3 DB 10, 13,
"EQUAL STRINGS $"

MS4 DB 10, 13,
"UNEQUAL
STRINGS $"

MS5 DB 10, 13, "$"
BUFF DB 25 , ? , 25
DUP('$')

BUFF1 DB 25 , ? ,
25 DUP('$')

.CODE
mov ax, @data Initialize DS and
ES
mov ds, ax
mov es, ax
print ms1
mov ah, 0ah ACCEPT first
STRING
lea dx, buff
int 21h
Flowchart A.5(b)
Micropro. & Interfacing Techniques (PU) L-25 Lab Manual
Label Instruction Comment
lea si, buff
print ms2
mov ah, 0ah ACCEPT OTHER
STRING
lea dx, buff1
int 21h
mov cl, buff+1 Number of
characters in str1
mov ch, buff1+1 Number of
characters in str2
cmp ch, cl check if length is
same
jnz para
mov ch, 00
mov cl, buff+1 Number of
characters
lea di, buff1
cld
repe cmpsb Compare string
char by char
jnz para if not same
goto PARA
print ms3 Strings are equal
jmp quit
para:
print ms4 Strings are
Unequal
quit:
mov ah, 4ch
int 21h
end
Result :
C:\programs>tasm COMPARE.ASM
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: COMPARE.ASM
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 437k
Micropro. & Interfacing Techniques (PU) L-26 Lab Manual
C:\programs>TLINK COMPARE.OBJ
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>COMPARE
ENTER FIRST STRING : hello
ENTER SECOND STRING : hell
UNEQUAL STRINGS
C:\programs>COMPARE
ENTER FIRST STRING : hello
ENTER SECOND STRING : hello
EQUAL STRINGS
(C) Number of occurrences of sub-string in the given string
Program Statement :
Write an assembly language program that determines if a given sub-string is
present or not in a main string of characters. The result is to be stored in a register
AL with FF : present, 00 : absent
Explanation :
- The substring is checked in the main string by comparing the first character of
substring with the characters of the main string. If there is a match of first
character, then it is required to check whether the length of the substring is less
than or equal to the remaining prortion of the main string. If the condition is
satisfied comparision continues. If string does not match, then first character of
substring is checked for remaining length of main string for the match and
process continues. If match is found, FF is stored in AL otherwise AL = 00H.
Algorithm :
Step I : Initialize data memory with main string, substring.
Step II : Initialize the DS and ES.
Step III : Accept the main string from user.
Step IV : Accept the substring from user.
Step V : CL = count of main string.
Step VI : DL = count of substring.
Step VII : Compare the first character of substring with main string till there is 0
match.
Step VIII : Subtract the length of substring to remaining length of main string till
match is found. If there is nothing goto step XIII.
Step IX : If result is negative, terminate program.
Step X : Compare all other substring with main string.
Step XI : If they match, store FF in the AL.
Step XII : If they don’t match store 00 in AL.
Step XIII : Stop.
Flowchart : Refer flowchart A.5(c).
Micropro. & Interfacing Techniques (PU) L-27 Lab Manual
Program :
Label
Instruction Comment

mess macro msg
definition of
macro mess
mov ah, 09h

lea dx, msg

int 21h
endm
.model small
.data
STR1 DB 25 , ? , 25
DUP('$')

Substr1 DB 25 , ? , 25
DUP('$')

msg1 db 10, 13, 'Enter the
string : $'

msg2 db 10, 13, 'Enter the
substring : $'

msg3 db 10, 13, '$'
count1 dw ?
count2 dw ?
res db 0
addr dw 0
buff db 25 , ? , 25 DUP('$')
.code








mov ax, @data Initialize data
section
mov ds, ax
mov es, ax
mess msg1
mov ah, 0ah
lea dx, str1
int 21h
mov cl, str1+1
mov ch, 00
mov count1, cx
mess msg2
mov ah, 0ah
lea dx, substr1
Micropro. & Interfacing Techniques (PU) L-28 Lab Manual
Label
Instruction Comment
int 21h
mov dl, substr1+1
mov dh, 00
mov count2, dx
mess msg3
mov si, offset substr1
mov di, offset str1
cld
mov al, [si]
next: repnz scasb

inc si
mov dx, cx
sub cx, count2
jb stop
mov cx, count2
dec cx
repz
cmpsb
cmp cx, 00
jnz stop
mov al, 0ffh
mov bh, al
jmp endd
stop: mov cx, 00
mov bh, cl
jmp endd
endd: mov ch, 02h Count of digits to
be displayed
mov cl, 04h Count to roll by 4
bits
l2: rol bh, cl roll bl so that
msb comes to lsb
mov dl, bh load dl with data
to be displayed
and dl, 0fH get only lsb
cmp dl, 09 check if digit is
0-9 or letter A-F
Flowchart A.5(c)
Micropro. & Interfacing Techniques (PU) L-29 Lab Manual
Label
Instruction Comment
jbe l4
add dl, 07 if letter add 37H
else only add
30H
l4: add dl, 30H







mov ah, 02 Function 2 under
INT 21H(Display
character)
int 21H
dec ch Decrement Count
jnz l2
mov ah, 4ch
int 21h
end
Result :
F:\C\PROGRAMS>TASM SUBSTR
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: SUBSTR.ASM
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 441k
F:\C\PROGRAMS>TLINK SUBSTR
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
F:\C\PROGRAMS>SUBSTR
Enter the string : TODAY IS FRIDAY
Enter the substring : IS
FF
F:\C\PROGRAMS>
Program 6 : Write X86/64 ALP to perform multiplication of two 8-bit hexadecimal numbers. Use
successive addition and add and shift method. Accept input from the user. (use of 64-bit
registers is expected)
(A) Successive Addition Method :
Program statement :
- Assuming that MUL instruction is not available in the instruction set of 8086, write
a program in assembly language of 8086 to simulate the MUL instruction.
Assuming that two digits are available in AL and BL registers. Use successive
addition method.
Micropro. & Interfacing Techniques (PU) L-30 Lab Manual
Explanation :
- Consider that a byte is present in the AL register and second byte is present in the
BL register.
- We have to multiply the byte in AL with the byte in BL.
- We will multiply the numbers using successive addition method.
- In successive addition method, one number is accepted and other number is taken as
a counter. The first number is added with itself, till the counter decrements to zero.
- Result is stored in DX register. Display the result, using display routine.
For example : AL = 12 H, BL = 10 H
Result = 12H + 12H + 12H + 12H + 12H + 12H + 12H + 12H + 12H + 12H
Result = 0120 H
Algorithm :
Step I : Initialise the data segment.
Step II : Get the first number.
Step III : Get the second number as counter.
Step IV : Initialize result = 0.
Step V : Result = Result + First number.
Step VI : Decrement counter
Step VII : If count ÷ 0, go to step V.
Step VIII : Display the result.
Step IX : Stop.
Flowchart : Refer flowchart A.6(a).
Program :
Label Instruction Comment
.model small
.data
a db 12H
b db 10H
.code
mov ax, @data Initialize data section
mov ds, ax
mov al, a Load number1 in al
mov bl, b Load number2 in bl
mov ah, 0
mov dx, 0 intialize result
ad: add dx, ax add numbers. Result in dx
dec bl dec number
cmp bl, 0
jnz ad
mov ch, 04h Count of digits to be displayed
Micropro. & Interfacing Techniques (PU) L-31 Lab Manual
Label Instruction Comment
mov cl, 04h Count to roll by 4 bits
mov bx, dx Result in reg bx
l2: rol bx, cl roll bl so that msb comes to lsb
mov dl, bl load dl with data to be displayed
and dl, 0fH get only lsb
cmp dl, 09 check if digit is 0-9 or letter A-F
jbe l4
add dl, 07 if letter add 37H else only add 30H
l4: add dl, 30H
mov ah, 02 Function 2 under INT 21H (Display
character)
int 21H
dec ch Decrement Count
jnz l2
mov ah, 4cH Terminate Program
int 21H
end
Result :
C:\programs>tasm succmul.asm
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: succmul.asm
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 438k
C:\programs>tlink succmul
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>succmul
0120
(B) Add and Shift Method :
Program statement :
- Assuming that MUL instruction is not available in the instruction set of 8086, write
a program in assembly language of 8086 to simulate the MUL instruction.
Assuming that two digits are available in AL and BL registers. Use add and shift
method.
Flowchart A.6(a)
Micropro. & Interfacing Techniques (PU) L-32 Lab Manual
Explanation :
- Consider that one byte is present in the AL register and another byte is present in
the BL register.
- We have to multiply the byte in AL with the byte in BL.
- We will multiply the numbers using add and shift method. In this method, you add
number with itself and rotate the other number each time and shift it by one bit to
left alongwith carry. If carry is present add the two numbers.
- Initialise the count to 4 as we are scanning for 4 digits. Decrement counter each
time the bits are added. The result is stored in AX. Display the result.
For example : AL = 11 H, BL = 10 H, Count = 4
Step I :
AX = 11
+ 11
22 H

Rotate BL by one bit to left along with carry.
BL = 10 H 0 ÷ 0 0 0 1 0 0 0 0
|
CY

BL = 0 0 0 1 0 0 0 0 0
CY

2


0

Step II : Now decrement counter count = 3.
Check for carry, carry is not there so add number with itself.
AX = 22
+ 22
44 H
Rotate BL to left,
BL = 0 ÷ 0 1 0 0 0 0 0 0
CY

4


0

Carry is not there.
Decrement count, count=2
Step III : Add number with itself
AX = 44
+ 44
88 H
Rotate BL to left,
BL = 0 1 0 0 0 0 0 0 0

CY


8

0

Carry is not there.
Micropro. & Interfacing Techniques (PU) L-33 Lab Manual
Step IV : Decrement counter count = 1.
Add number with itself as carry is not there.
AX = 88
+ 88
110 H
Rotate BL to left,
BL = 1 0 0 0 0 0 0 0 0
CY

0

0

Carry is there.
Step V : Decrement counter = 0.
Carry is present.
. add AX, BX
. 0110 i.e. 11 H
+ 0000 × 10 H
0110 H 0110 H
Algorithm :
Step I : Initialise the data segment.
Step II : Get the first number.
Step III : Get the second number.
Step IV : Initialize count = 04.
Step V : number 1 = number 1 × 2.
Step VI : Shift multiplier to left alongwith carry.
Step VII : Check for carry, if present goto step VIII else goto step IX.
Step VIII : number 1 = number1 + shifted number 2.
Step IX : Decrement counter.
Step X : If not zero, goto step V.
Step XI : Display the result.
Step XII : Stop.
Flowchart : Refer flowchart A.6(b).
Label Instruction Comment
.model small
.data
a db 11H
b db 10H
.code
mov ax, @data Initialize data section
mov ds, ax
mov al, a Load number1 in al
mov bl, b Load number2 in bl
mov ah, 0
mov dl, 04h initialize counter
Micropro. & Interfacing Techniques (PU) L-34 Lab Manual
Label Instruction Comment
ad: add ax, ax add numbers. Result in dx
rcl bl, 01
jnc skip
add ax, bx
skip: dec dl dec number
jnz ad
mov ch, 04h Count of digits to be displayed
mov cl, 04h Count to roll by 4 bits
mov bx, ax Result in reg bx
l2: rol bx, cl roll bl so that msb comes to lsb
mov dl, bl load dl with data to be
displayed
and dl, 0fH get only lsb
cmp dl, 09 check if digit is 0-9 or letter
A-F
jbe l4
add dl, 07 if letter add 37H else only add
30H
l4: add dl, 30H
mov ah, 02 Function 2 under INT 21H
(Display character)
int 21H
dec ch Decrement Count
jnz l2
mov ah, 4cH Terminate Program
int 21H
end
Result : Flowchart A.6 (b)
C:\programs>tasm shaddmul.asm
Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International
Assembling file: shaddmul.asm
Error messages: None
Warning messages: None
Passes: 1
Remaining memory: 438k
C:\programs>tlink shaddmul.obj
Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International
Warning: No stack
C:\programs>shaddmul
0110
C:\programs>
Micropro. & Interfacing Techniques (PU) L-35 Lab Manual
Program 7 : Write 8087 ALP to obtain :
(i) Mean (ii) Variance (iii) Standard Deviation
For a given set of data elements defined in data segment. Also display result.
Program :
Title Implement mathematical equation
Label Instruction Comment
.model small
.stack 100
.8087
.data
array_x DQ 10 DUP (112233H)
N DW 10
MEAN DQ ?
answer DQ ?
.code
MOV AX, @data Initialize data segment
MOV DS, AX
FINIT Initialize coprocessor
FLDZ accumulation = 0
XOR BX, BX clear pointer
MOV CX, N counter
loop_acc : FADD array_x [BX] add x(I) to TUS
ADD BX, 8 Point to next element
LOOP loop_acc Repeat N times
FIDIV N [Accumulation] N
FST MEAN Store mean
FLDZ Store 0 to TOS
XOR BX, BX clear pointer
MOV CX, N Counter
loop_final : FLD array_x[BX] Load data pointed by pointer
ADD BX, 8 offset
FSUB ST, ST(2) X(I) MEAN = TOS
FMUL ST, ST(0) = TOS
FADDP ST(1), ST Accumulate
LOOP loop_final loop till N = 0
FST answer store answer
end
Micropro. & Interfacing Techniques (PU) L-36 Lab Manual
Group B
Programs of 8255
Program 1 : Write 8086 ALP to convert an analog signal in the range of 0V to 5V to its corresponding
digital signal using successive approximation ADC and dual slope ADC. Find the
resolution used in both the ADC’S and compare the results.
Explanation :
ADC-0809 is an 8 bit successive approximation ADC. This chip has 8 channels
alongwith multiplexer. The channel select has address lines A, B, C. We will use channel 0
as input. Thus, address lines A, B, C will be grounded for channel 0.
The ALE pin is connected to the clock input. At the time of power on the valid
channel address is latched at the rising edge of the ALE singal. ADC 0809 has an SOC
(start of conversion) pin. A positive going pulse of short duration, is applied to this pin.
This pin starts the A/D conversion process. The OE should always be high, when data is to
be read. After the conversion, EOC is given through PC
7
indicating end of conversion. The
port A and C are defined in the input mode, whereas port B of 8255 is configured in output
mode. The data is read through port A of 8255. Positive (d.c.) and negative (d.c.) or (a.c.)
voltage is applied as the analog input at channel 0. Hence decoupling capacitors are used
to maintain minimum noise level. Internal oscillator can be enabled only when A/D
conversion is to be done. The oscillator oscillates till the INT.SOC enables pin PB
2
of the
8255.



Fig. 1
Micropro. & Interfacing Techniques (PU) L-37 Lab Manual
Algorithm :
Step I : Initialize the data section.
Step II : Initialize AL with control word.
Step III : Output contents of AL i.e. control word on control register.
Step IV : Load AL = 00H.
Step V : Output contents of AL on port B, to select channel 0.
Step VI : Send the ALE high.
Step VII : Call some delay i.e. wait for atleast 2.5 µs.
Step VIII : Make SOC high.
Step IX : Wait for atleast 0.5 µs.
Step X : Make SOC low.
Step XI : Check for EOC.
Step XII : If not wait.
Step XIII : Read the digital input availables on port A.
Step XIV : Go to step IV.
Control word :
I/O Mode A P
A
P
cu
Mode B P
B
P
CL

1 0 0 1 1 0 0 1 = 99H
Flowchart : Refer flowchart B.1(a).
Program :
Label Instruction Comment
.model small
.data
Port A EQU 0000 H
Port B EQU 0002 H
Port C EQU 0004 H
CWR EQU 0006 H
.code
MOV AX, @ Data
MOV DS, AX
MOV AL, 99H AL = Control
word.
MOV DX, CWR DX = address
of control word
register.
OUT DX, AL Output control
word on control
word register.
Micropro. & Interfacing Techniques (PU) L-38 Lab Manual
Label Instruction Comment
L1: MOV AL, 00H AL = 00 to
select channel
0.
MOV DX, Port B
OUT DX, AL Send address to
select channel
0.
MOV AL, 08H
OUT DX, AL Latch the given
address by
sending ALE
high.
Call Delay Wait for 2.5 µs.
MOV AL, 18H Make SOC high.
OUT DX, AL
NOP Wait for atleast
0.5 µs.
MOV AL, 08H
OUT DX, AL Make SOC low.
MOV DX, Port C
CHECK : IN DX, AL Check for EOC.
AND AL, 01H
JZ CHECK
MOV DX, Port A
IN DX, AL
JMP L1

The observation table will include.
Analog input
(V)
Digital
equivalent
Output =
255 × voltage
5V

0
1
2
3
4
5
Flowchart B.1(a)
Micropro. & Interfacing Techniques (PU) L-39 Lab Manual
Using Dual Slope ADC :
IC 7109 is used. It is a 12-bit dual slope A/D converter.
It has
––––––
RUN /Hold input and STATUS output, which monitors and control
conversion timing. It can operate with upto 30 conversions per second. Fig. 2 shows the
interfacing diagram.


Fig. 2 : Interfacing 8255, ADC 7109 and 8086
Algorithm :
Step I : Initialize the data section.
Step II : Make Run /
––––––
Hold signal high.
Step III : Check for status. Is it 0. If not wait till status = 0.
Step IV : Make Run /
––––––
Hold signal low.
Step V : Read low order byte on Port A.
Step VI : Read higher nibble on Port B.
Step VII : Stop.
Flowchart : Refer flowchart B.1(b).
Micropro. & Interfacing Techniques (PU) L-40 Lab Manual
Program :
Label Instruction Comment
.model small
.data
Port AEQU 0000 H
Port BEQU 0001 H
Port CEQU 0002 H
CWR EQU 0003 H
.Code
MOV AX, @Data
MOV DS, AX
MOV AL, 92H Initialize 8255 with control word.
MOV DX, CWR
OUT DX, AL Output control word on CWR.
MOV AL, 80H
MOV DX, Port C
OUT DX, AL Send the RUN signal to ADC.
MOV DX, Port B
BACK : IN DX, AL Stack for the status signal
AND AL, 40H
JNZ BACK If it is high check again.
MOV AL, 00H
Make RUN /
––––––––
HOLD signal low.
MOV DX, Port C
OUT DX, AL
MOV DX, Port A If low read lower byte.
IN DX, AL
MOV DX, Port B Get higher nibble
IN DX, AL
MOV AH, 4CH Terminate
INT 21 H
END
After execution of program 12 bit digital data is available on Port A and Port B of
the 8255. The higher nibble is on Port B and lower byte on Port A.
The resolution for both ADC’s.
ADC 0809 – 8 bit
ADC 7109 – 12 bit
Flowchart B.1(b)
Micropro. & Interfacing Techniques (PU) L-41 Lab Manual
Program 2 : Write 8086 ALP to interface DAC and generate following waveforms on oscilloscope.
(i) Square wave – variable duty cycle and frequency.
(ii) Sine wave – variable frequency.
(iii) Ramp wave – variable direction.
(iv) Trapezoidal wave.
(v) Stair case wave. ]
(i) Square wave – Variable duty cycle and frequency :
Explanation :
We are asked to generate a square wave using DAC interface. To generate square
wave we will output FFH and then 00H on port A of 8255. The output of 8255 (Port A) is
connected to the DAC 0808. We also have to vary duty cycle. Variation in duty cycle will
automatically change the frequency as,
duty cycle =
T
ON
T
ON
+ T
OFF
=
T
ON
Frequency

According to our duty cycle requirement, we can change the DELAY between the
two outputs FFH (for output high) and 00H (for output low).

Fig. 3
Algorithm :
Step I : Initialize 8255 Port A as output port.
Step II : Initialize AL = 00 H.
Step III : Ouput the contents of AL through
Step IV : Increment AL by one.
Step V : Check if AL = FF H ? If not, goto step III.
Step VI : Decrement AL by one.
Step VII : Output AL through port A.
Micropro. & Interfacing Techniques (PU) L-42 Lab Manual
Step VIII : Compare AL with 00. If AL = 00,
goto step III.
Step IX : If not, goto step VI.
Flowchart : Refer Flowchart B.2(a).
Program :
Label Instruction Comment
. MODEL SMALL
. CODE
MOV AL, 80 H Initialize Port A = Output Port.
MOV DX, 00 H Load DX with port address of port A.
MOV AL, 00 H
L1 : OUT DX, AL Ouput contents of AL through
port A.
INC AL Increment AL.
CMP AL, FF H Compare AL with FF H, if not
continue.
JNZ L1
L2 : DEC AL Decrement AL.
OUT DX, DL Output contents of AL to port A.
JNZ L2 Decrement till AL = 00.
JNZ L1 If AL = 00, goto L1 i.e. start from
beginning.
(ii) Sine wave – Variable frequency : Flowchart B.2(a)
Explanation :
In order to generate a sine wave, we have to output digital equivalent values which
represent the sine wave signal as shown in Fig. 4.


Fig. 4
Micropro. & Interfacing Techniques (PU) L-43 Lab Manual
Digital data 00H represents – 2.5 V, and FFH represents + 2.5 V.
The value of sin 0° = 0
sin 90° = 1
sin 270° = – 1
sin 360° = 0
The range of 0° to 90°, is distributed in 128 decimal steps.
Therefore, taking offset as 128.
The magnitude = 128 + 128 sin x
Where x : angle in degrees inorder to change, the frequency either increase or
decrease the steps.
The look up table shows the digital equivalent values, for the sine wave.
Degrees Equation Digital equivalent in
decimal
Digital Equivalent in
Hex
0° (128 + 128 sin 0°) 128 80H
10° (128 + 128 sin 10°) 150 96H
20° (128 + 128 sin 20°) 171 ABH
30° (128 + 128 sin 30°) 192 C0H
40° (128 + 128 sin 40°) 156 D2H
50° (128 + 128 sin 50°) 226 E2H
60° (128 + 128 sin 60°) 239 EFH
70° (128 + 128 sin 70°) 248 F8H
80° (128 + 128 sin 80°) 254 FEH
90° (128 + 128 sin 90°) 256 ÷ 255 FFH
100° (128 + 128 sin 100°) 254 FEH
110° (128 + 128 sin 110°) 248 F8H
120° (128 + 128 sin 120°) 239 EFH
130° (128 + 128 sin 130°) 226 E2H
140° (128 + 128 sin 140°) 156 D2H
150° (128 + 128 sin 150°) 192 C0H
160° (128 + 128 sin 160°) 171 ABH
170° (128 + 128 sin 170°) 150 96H
180° (128 + 128 sin 180°) 128 80H
190° (128 + 128 sin 190°) 106 6AH
200° (128 + 128 sin 200°) 84 54H
210° (128 + 128 sin 210°) 64 40H
220° (128 + 128 sin 220°) 46 2EH
230° (128 + 128 sin 230°) 30 1EH
240° (128 + 128 sin 240°) 17 11H
250° (128 + 128 sin 250°) 08 08H
Micropro. & Interfacing Techniques (PU) L-44 Lab Manual
Degrees Equation Digital equivalent in
decimal
Digital Equivalent in
Hex
260° (128 + 128 sin 260°) 02 02H
270° (128 + 128 sin 270°) 00 00H
280° (128 + 128 sin 280°) 02 02H
290° (128 + 128 sin 290°) 08 08H
300° (128 + 128 sin 300°) 17 11H
310° (128 + 128 sin 310°) 30 1EH
320° (128 + 128 sin 320°) 46 2EH
330° (128 + 128 sin 330°) 64 40H
340° (128 + 128 sin 340°) 84 54H
350° (128 + 128 sin 350°) 106 6AH
360° (128 + 128 sin 360°) 128 80H
Instead of storing all these values, we store the digital values from 0° to 90°, as
these values repeat again i.e. We store the first 10 values and generate the remaining
values at the time of execution.
Algorithm :
Step I : Initialize the date segment.
Step II : Initialize BX to point to start of look up table.
Step III : Initialize count in CL.
Step IV : Load the value from look up table.
Step V : Output that value on port A of 8255.
Step VI : Increment BX to point next value.
Step VII : Decrement count.
Step VIII : Check if count = 0 ? If not, goto step IV.
Step IX : Initialize CL with count.
Step X : Decrement BX to point value from look up table.
Step XI : Load AL with value from look up table.
Step XII : Output the value in AL on port A of 8255.
Step XIII : Decrement count.
Step XIV : Check if count = 0 ? If not, goto step X.
Step XV : Initialize CL = count
Step XVI : Initialize AH = FFH i.e. count for subtraction.
Step XVII : Load the value from look up table in AL.
Step XVIII : AL = AL – FFH i.e. compute the digitial equivalent value.
Step XIX : Output value in AL, to port A of 8255.
Step XX : Increment BX to next value in look up table.
Step XXI : Decrement count.
Step XXII : Is count = 0 ? If not, goto XVI.
Step XXIII : Initialize CL with count.
Step XXIV : Decrement BX with next value in the look up table.
Micropro. & Interfacing Techniques (PU) L-45 Lab Manual
Step XXV : Initialize AH = FFH.
Step XXVI : Load the value from look up table in AL.
Step XXVII : Compute the digital equaivalent AL = AL – FFH
Step XXVIII : Output the value in AL, on port A of 8255.
Step XXIX : Decrement count.
Step XXX : If not zero, goto step XXIV.
Step XXXI : Go back to start.
Program :
Label Instruction Comment
.model small
.data
PORTA EQU 80H
look_up dB 80H, 96H, 0ABH, 0COH,
0D2H, 0E2H, 0EFH,
0F8H, 0FEH, 0FFH
Count dB 0AH
.code
START : MOV AX, @Data Initialize data segment.
MOV DS, AX Generates sine wave from 0° to 90°
MOV BX, offset look_up Initialize BX to start of look up table.
MOV CL, count Initialize count.
L1 : MOV AL, [BX] AL = Value from look up table.
MOV DX, PORTA DX = address of port A.
OUT DX, AL Send data on port A.
INC BX Increment BX to next value from look up table.
DEC CL Decrement count.
JNZ L1 If count ÷ 0, continue till all values are sent on output
Generates sine wave from 90° to 180°.
MOV CL, Count Initialize counter.
L2 : DEC BX Decrement look up table pointer
MOV AL, [BX] AL = Value from look up table
MOV DX, Port A DX = address of Port A.
OUT DX, AL
DEC CL Decrement count.
JNZ L2 If count ÷ 0, goto L2.
Generates sine wave from 180° to 270°.
MOV CL, Count Initialize counter.
L3 : MOV AH, FFH Load count for subtraction.
MOV AL, [BX] AL = Value from look up table.
SUB AL, AH Calculate digital equivalent value.
MOV DX, Port A DX = address of port A.
OUT DX, AL
Micropro. & Interfacing Techniques (PU) L-46 Lab Manual
Label Instruction Comment
INC BX
DEC CL Decrement count.
JNZ L3 If count ÷ 0 goto L3
Generates sine wave from 270° to 360°.
MOV CL, Count Initialize counter.
L4 : DEC BX Decrement look up table pointer.
MOV AH, FFH Load count for subtraction.
MOV AL, [BX] AL = Value from look up table.
SUB AL, AH
MOV DX, Port A

OUT DX, AL
DEC CL Decrement count
JNZ L4
JMP START Continue
(iii) Ramp wave : Variable Direction :
Explanation :
We are asked to generate a ramp wave using DAC interface. To generate ramp wave
we will output 00 to FFH and FFH to 00H. If we want a ramp wave with reverse direction
then, we output in reverse manner. If we want ramp wave in forward direction, we will
initialize, AL = 00H, otherwise AL = FFH for reverse direction.

Fig. 5
Algorithm :
Step I : Start
Step II : Initialize AL.
Step III : Initialize BL = AL i.e. store AL value in register BL.
Step IV : AND AL with 80H, to check MSB.
Step V : If MSB = 1 goto step XIV.
Step VI : Load AL with value from BL.
Step VII : Output contents of AL through port A.
Step VIII : Increment AL by 1.
Micropro. & Interfacing Techniques (PU) L-47 Lab Manual
Step IX : Check if AL = FFH ? If not, goto step VI.
Step X : Decrement AL by one.
Step XI : Output contents of AL through port A.
Step XII : Compare AL with 00H ? If yes, goto step VII.
Step XIII : If not goto step X.
Step XIV : Load AL back.
Step XV : Jump to step X.
Flowchart : Refer Flowchart B.2(b).

Flowchart B.2(b)
Micropro. & Interfacing Techniques (PU) L-48 Lab Manual
Program :
Label Instruction Comment
.model small
.code
MOV AL, 80H Initialize Port A of 8255 as output port.
MOV DX, 00H Load DX with port address of port A.
MOV AL, FFH Initialize AL
MOV BL, AL
AND AL, 80H Check for MSB
JZ SKIP
MOV AL, BL Initialize AL
L1 : OUT DX, AL Output contents on port A.
INC AL
CMP AL, FFH Compare AL with FFH.
JNZ L1
L2 : DEC AL Decrement AL.
OUT DX, AL
JNZ L2.
JNZ L1.
SKIP : MOV AL, BL Initialize AL back.
JMP L2.
(iv) Trapezoidal wave :
Explanation :
We have to generate a trapezoidal wave using DAC interface. To generate
trapezoidal wave, first we will see the trapezoidal wave. Port A of 8255 is used as output
port.

Fig. 6 : Trapezoidal wave
Now here, we will output the Part A, Part B, Part C, Part D and the Part E. In Part
A, we will output 80 H to FFH. In Part B, we will output FFH, for a certain delay. In Part
C, we will output FFH to 00H. In Part D, we will output 00H, for a certain delay. In Part
E, we will output 00H to FFH.
Micropro. & Interfacing Techniques (PU) L-49 Lab Manual
Algorithm :
Step I : Initialize the data section.
Step II : Initialize port A as output port.
Step III : Initialize AL = 80H i.e. initial value of the trapezoidal wave.
Step IV : Output contents of AL on port A of 8255.
Step V : Increment AL.
Step VI : Check if AL = FFH ? If not, go to step IV.
Step VII : Output contents of AL i.e. FFH on port A.
Step VIII : Call delay 1.
Step IX : Decrement AL.
Step X : Output the contents of AL on port A of 8255.
Step XI : Is AL = 00 ? If not go to step IX.
Step XII : Output the contents of AL i.e. 00H on port A.
Step XIII : Call delay 2
Step XIV : Increment AL.
Step XV : Output the contents of AL on port A.
Step XVI : Check if AL = FFH ? If not, go to step XIV.
Step XVII : Jump to step VII.
Flowchart : Refer Flowchart B.2(c).


Flowchart B.2(c)
Micropro. & Interfacing Techniques (PU) L-50 Lab Manual
Program :
Label Instruction Comment
. model small
. data
PORTA EQU 0006H address of Port A.
Value db 80H initial value of trapezoidal wave.
. code
MOV AX, @ Data. Initialize the data segment.
MOV DS, AX
MOV AL, 80H Initialize port A as output port.
MOV DX, Port A DX = address of Port A of 8255.
OUT DX, AL
MOV AL, Value Initialize AL with initial value of trapezoidal wave.
L1 : MOV DX, AL Output the contents of AL on port A.
INC AL Increment AL.
CMP AL, FFH Compare AL with FFH.
JNZ L1
UP : OUT DX, AL Output FFH on port A.
CALL DELAY 1
L2 : DEC AL Decrement AL.
OUT DX, AL Output contents of AL on port A.
CMP AL, 00H Compare AL with 00H.
JNZ L2
OUT DX, AL Output 00H on port A.
CALL DELAY 2
L3 : INC AL Increment AL
OUT DX, AL Increment contents on AL on port A.
CMP AL, FFH Compare AL with FFH.
JNZ L3 If not zero, continue.
JMP UP
The delay 1 and delay 2 routines, will be different because at the output we want a trapezoidal wave.
Trapezoidal wave duration is not symmetric. Thus, delay routines are different.
DELAY 1 PROC
L4 : MOV BL, FFH Count for delay
DEC BL Decrement count.
JNZ L4
DELAY 1 ENDP
DELAY 2 PROC
L5 : MOV BH, 70H Count for delay
DEC BH Decrement count.
JNZ L5
DELAY 2 ENDP
Micropro. & Interfacing Techniques (PU) L-51 Lab Manual
(v) Staircase wave :
Explanation :
We have to generate a staircase wave using DAC interface. Fig. 7 shows a staircase
wave.



Fig. 7
We have considered 8 levels for drawing the stair case wave i.e. 2, 4, 8, 16, 32, 64,
128 and 256. Their hex equivalent are 02 H, 04 H, 08 H, 10 H, 20 H, 40 H, 80 H and Hex
equivalent for 255 is FFH. We will store these values in an array. You can also increase or
decrease the levels, for staircase waveform. Each level will be outputted on Port A of 8255
which is configured as output port.
Algorithm :
Step I : Initialize the data section.
Step II : Initialize BX to the start of array.
Step III : Initialize port A of 8255 as output port.
Step IV : Initialize count = 08H.
Step V : Load AL with first step of staircase.
Step VI : Output the contents of AL on port A.
Step VII : Call delay.
Step VIII : Increment BX to point next step value.
Step IX : Decrement count.
Step X : Check if count = 0 ? If not, go to step V.
Step XI : Initialize count = 08H in CL.
Step XII : Decrement BX to next value.
Micropro. & Interfacing Techniques (PU) L-52 Lab Manual
Step XIII : Load the value, pointed by BX in AL register.
Step XIV : Output the contents of AL on port A.
Step XV : Call delay.
Step XVI : Decrement count.
Step XVII : Is count = 00 ? If not, go to step XII.
Step XVIII : Jump step IV.
Flowchart : Refer Flowchart B.2(d).



Flowchart B.2(d)
Micropro. & Interfacing Techniques (PU) L-53 Lab Manual
Program :
Label Instruction Comment
. model small
. data
PORT A EQU 0050H
STEP DB 02H, 04H, 08H, 10H, 20H, 40H, 80H, FFH.
COUNT DB 08H.
. code
MOV AX, @ Data Initialize data segment.
MOV DS, AX
MOV AL, 80H Initialize 8255 Port A as output port.
MOV DX, Port A DX = address of Port A.
OUT DX, AL
MOV CL, COUNT Initialize CL = 08H.
MOV AL, 00H Initialize AL = 00H.
Call delay
MOV BX, OFFSET STEP BX = Start of STEP.
L1 : MOV AL, [BX] AL = Step value.
OUT DX, AL Output contents of AL on Port A.
CALL DELAY
INC BX Increment BX to point next step
value.
DEC CL Decrement count.
JNZ L1
MOV CL, COUNT Initialize CL = count.
L2 : DEC BX Decrement BX to point next step
value.
MOV AL, [BX]
OUT DX, AL Output contents of AL on Port A.
Call delay
DEC COUNT
JNZ L2
JMP L1
DELAY PROC
MOV CH, FFH
L3 : DEC CH
JNZ L3
DELAY ENDP
Program 3 : Write 8086 ALP to rotate a stepper motor for given number of steps at a given angle and
in the given direction of rotation based on the user choice such as,
i) If ‘C’ key is pressed - clockwise rotation.
ii) If ‘A’ key is pressed – anticlockwise rotation.
Micropro. & Interfacing Techniques (PU) L-54 Lab Manual
iii) If ‘B’ is pressed – ½ clockwise and Vz anticlockwise rotation.
iv) If ‘S’ key is pressed – stop rotation.
Also write routines to accelerate and deaccelerate the motor.
Explanation :
The block diagram of a microprocessor based control scheme for a stepper motor is
shown in Fig. 8. The motor has four windings A, B, C and D which are driven by a single
phase excitation scheme i.e. at a time only one winding is energized. The sequence in
which the windings are to be energized is stored in the memory of the microprocessor
system in the form of a look up table. The 8255 is configured such that port A acts as an
output port. The output of the 8255 are used to drive the four transistors. The
freewheeling diodes are connected across each phase of the phase windings to protect the
transistors against high forward voltage at the time of their turn off. When the look up
table is output on the output port in the sequence 0AH, 09H, 05H, 06H, the motor rotates
in clockwise direction. When the sequence of outputting the contents of look up table is
reversed i.e. 06H, 05H, 09H, 0AH the motor starts rotating in counter clockwise or
anticlockwise direction. If the sequence of rotation is 0AH, 09H, 06H, 05H the motor
rotates is ½ clockwise and ½ anticlockwise rotation.
The speed of motor can be changed by changing the rate at which look up table
contents are being sent out. The position control is possible by programming the processor
to keep a count of the number of pulses being sent out because each pulse corresponds to
rotation through a specific angle.


Fig. 8
Micropro. & Interfacing Techniques (PU) L-55 Lab Manual
Flowchart : Refer Flowchart B.3.
Program :
Label Instruction Comment
MESS MACRO MSG
MOV AH, 09W
LEA DX, MSG
INT 21H
ENDM
. MODEL SMALL
. DATA
Clockwise DB 0AH, 09H, 05H, 06H
Anticlockwise DB 06H, 05H, 09H, 0AH
Half-clk DB 0AH, 09H, 06H, 05H.
MSG 1 DB 0AH, 0DH, ‘Menu $’.
MSG 2 DB 0AH, 0DH, ‘C : clockwise rotation’
MSG 3 DB 0AH, 0DH, ‘A : Anticlockwise
rotation’.

MSG 4 DB 0AH, 0DH, ‘B : ½ clockwise and ½
anticlockwise rotation’

MSG 5 DB 0A, 0DH, ‘S : Stop rotation’
MSG 6 DB 0A, 0DH, ‘Accept choice’.
MSG 7 DB 0A, 0DH, ‘Wrong choice’.
. CODE
MOV AX, @Data
MOV DS, AX
L1 : MESS MSG 1 Display Menu.
MESS MSG 2
MESS MSG 3
MESS MSG 4
MESS MSG 5
MESS MSG 6 Accept choice from user.
MOV AH, 01H
INT 21H
MOV BL, AL Choice in BL
CMP BL, 41H If choice = A
Micropro. & Interfacing Techniques (PU) L-56 Lab Manual
Label Instruction Comment
JE clkrot
CMP BL, 42H Is choice = B. (42 is ASCII equivalent
for B)
JE half
CMP BL, 43H Is choice = ‘C’.
JE antirot
CMP BL, 53H Is choice = ‘S’.
JE endd.
MESS MSG 7 display ‘wrong choice’
JMP L1 display Menu again.
clkrot : CALL clockwiserot
JMP L1
half : CALL halfclockhalfanti
JMP L1
antirot : CALL anticlockwiserot.
JMP L1
endd: MOV AH, 4CH
INT 21H.
clockwiserot PROC NEAR
MOV AL, 80H Initialize port A as output port.
MOV DX, 00 Load port address of port A in DX.
OUT DX, AL Output contents of AL on port A.
MOV SI, Offset clockwise.
MOV BL, 04H Load sequence count.
L1 : MOV AL, [SI] AL = code
OUT DX, AL Output the excitation code on port A.
CALL DELAY Wait
INC SI Increment SI to point next code.
DEC BL decrement sequence count
JNZ L1
RET
clockwiserot ENDP.

Micropro. & Interfacing Techniques (PU) L-57 Lab Manual


Flowchart B.3
Program 4 : Write 8086 ALP to print a text message on printer using centronix parallel printer
interface.
Explanation :
Whenever data on computer is to be printed, the computer sends an INIT pulse
first, in order to initialize the printer. The computer then checks for a BUSY signal, in
order to know whether the printer is ready to receive data or not. If the BUSY signal is not
low, then the computer checks for the PE signal. If this signal is high, then display
message that the “Printer is OUT OF PAPER”. If the PE signal is low the computer checks
for the
––––––
ERROR signal. If this signal is low, it indicates that printer is in “Paper End”
state, “Offline” state and “Error” state. In such a case display message “PRINTER
OFFLINE” and stop.
Micropro. & Interfacing Techniques (PU) L-58 Lab Manual



Fig. 9


Fig. 10
If the BUSY signal is low, then the computer sends an ASCII code on eight parallel
data lines. The computer also sends a
–––
STB signal to the printer, to indicate that valid
data is available on the data bus. In response, the printer sends an
––––
ACK
(acknowledgement) signal, to indicate that data to be printed is received. The rising edge
of the
––––
ACK signal, resets the BUSY signal from the printer. When the computer sees that
BUSY signal is low, it sends the next character along with strobe and the sequence is
repeated till the last character to be printed is transferred. Fig. 10 shows the circuit for
interfacing centronix type parallel input printer to 8255 A. Port A is used as an output
port, to send 8-bit data to the printer. Port B is used an input port to check the
––––––
ERROR ,
PE and BUSY signals.
The port C signals, PC
7
is used as an
––––
ACK signal and PC
6
is used as
–––
STB signal.
The PC0 is used to send
––––
INIT signal to initialize the printer.
Micropro. & Interfacing Techniques (PU) L-59 Lab Manual
Algorithm :
Step I : Initialize the data section.
Step II : Initialize the pointer to point to the string.
Step III : Initialize the counter with number of characters in the string that are to
be printed.
Step IV : Initialize the 8255, port A with output port, portB as input port.
Step V : Send
––––
INIT signal to initialize the printer.
Step VI : Check for BUSY signal. If it is low go to step VII, else go to step XIII.
Step VII : Send the character to be printed.
Step VIII : Check for
––––
ACK . If not received, then wait till the
––––
ACK signal is received.
Step IX : Increment the string pointer, to point to the next character to be printed.
Step X : Decrement the counter.
Step XI : Is counter = 0 ? If yes, go to step XVII.
Step XII : If not goto step VI.
Step XIII : Check if PE signal is high. If not, go to step XV.
Step XIV : Display message “Paper out”.
Step XV : Check if
––––––
ERROR signal is low. If not goto step XVII.
Step XVI : Display message “Printer offline”.
Step XVII : Stop.
Flowchart : Refer flowchart B.4.
Program :
The control word for 8255 is
I/O Mode A PA PC
U
Mode B PB PC
L

1 0 1 0 X 0 1 0 = A2H

Label Instruction Comment
. model small
. data
PORTA EQU 0000
PORTB EQU 0002
PORTC EQU 0004
CWR EQU 0006
MSG 1 DB 10, 13, ‘Printer Out of Printer’.
MSG 2 DB 10, 13, ‘Printer offline’.
MSG 3 DB 10, 13, ‘Printing completed’.
MSG 4 DB 10, 13, ‘This matter is to be printed’.
COUNT DB 15.
. code
MOV AX, @ Data Initialize the data segment.
MOV DS, AX
LEA BX, MSG 4 Initialize pointer to start of string.
MOV DX, CWR DX = control word register address.
Micropro. & Interfacing Techniques (PU) L-60 Lab Manual
Label Instruction Comment
MOV AL, 0A2H Load control word in AL.
OUT DX, AL
MOV AL, 07H Make INTE
A
high to enable INTR
A
.
OUT DX, AL
MOV AL, 00H
OUT DX, AL
Make PC
0
low, so give
––––
INIT signal to
the printer for initialisation.
Back : MOV CX, 0FFFH Wait for 50 µs, so that printer will get
Initialized.
DEC CX
LOOP BACK
MOV AL, 01H

––––
INIT = 1
OUT DX, AX
UP : MOV DX, PORT B
IN AL, DX
MOV AH, AL Save the status in AH also.
AND AL, 01H
JNZ CHECK Check for BUSY signal. If high goto
CHECK.
MOV AL, [BX] Load the first character in AL.
MOV DX, Port A Load the address of Port A in DX.
OUT DX, AL Send the character to be printed.
MOV DX, Port C DX = address of Port C.
L2 : IN AL, DX
Check for
––––
ACK by checking status of
INTR
A
.
JNZ L2
MOV CL, COUNT Load count in CL.
DEC CL Decrement count.
JNZ UP Continue till all the characters are
printed.
JMP ENDD
CHECK : MOV AL, AH Load the printer status in AL.
AND AL, 02H Check for PE signal high.
MOV AL, AH Load status in AL.
JZ CHECK 1
LEA DX, MSG 1 Display Message “Printer out of
Paper”.
MOV AH, 09H
INT 21H
CHECK1 : AND AL, 04H
Check for
––––––
ERROR signal.
JNZ UP
LEA DX, MSG 2 Display Message “Printer offline”.
MOV AH, 09H
INT 21H
Micropro. & Interfacing Techniques (PU) L-61 Lab Manual
Label Instruction Comment
JMP UP
ENDD : LEA DX, MSG 3 Display Message “Printing
completed”.
MOV AH, 09H
INT 21H
MOV AH, 4CH Terminate Program.
INT 21H
END.



Flowchart B.4
Micropro. & Interfacing Techniques (PU) L-62 Lab Manual
Programs of 8253
Program 1 : Write 8086 ALP to program 8253 in mode 0, modify the program for hardware re-
triggerable monoshot mode. Generate a square wave with a pulse of 1 ms. Comment on
the difference between Hardware Triggered and software triggered strobe mode. Observe
the waveform at GATE and OUT pin of IC 8254 on CRO.
(i) To program 8254 in mode 0 means to initialize it.
Control word
Corel A

MOV AL, 31 H ; Counter 0, mode 0 CWR
MOV DX, 5006 ; CWR address
OUT DX, AL ; Loads control word in the CWR
(ii) The modification in the above program, for mode 5 i.e. Hardware retriggerable
monoshot mode is only to change the control word.
Control word
Corel B
MOV AL, 3BH ; Counter 0, mode 5 CWR
MOV DX, 5006 ; CWR address
OUT DX, AL ; Loads control word in CWR.
(iii) Program to generate a square wave with a pulse of 1 ms.
Step 1 : Assume the Counter 0 is used to generate square wave and addresses of
Counter 0 = 10 H and Control register = 13 H.
Step 2 : The output period required is 1 ms. The input frequency is 1 MHz, so
input period = 1 µs.
Count value =
Required period
Input period
=
1 ms
1 µs
= (1000)
10

Step 3 : The control word format to initialize counter 0, 16 bit counter, BCD
counting and square wave generator mode will be as follows :
Corel C
Micropro. & Interfacing Techniques (PU) L-63 Lab Manual
Step 4 : The 8253 initialisation program will be as follows :
MOV AL, 37H ; Initialise counter 0,
MOV DX, 0013 H ; mode 3 CWR address
OUT DX, AL ; 16 bit count and BCD counter
MOV AL, 00H ; Load LSB count value to counter 0
MOV DX, 10H
OUT DX, AL
MOV AL, 10H ; Load MSB count value to counter 0
MOV DX, 10H
OUT DX, AL
(iv) For the difference between Hardware triggered and software triggered strobe mode
please refer chapter 16.
(v) For the wave forms at GATE and OUT pin of IC 8254.
Programs of 8279
Program 1 : Write a program to initialize and operate 8279 for the following specifications. The
addresses of 8279 are A2 H and B2 H
(i) 12 digit display to display roll nos. from 1 to 12
(ii) Display scan time 15.24 msec. and external clock is 3 MHz.
(iii) Clear code is FF H
D
7
D
6
D
5
D
4
D
3
D
2
D
1
D
0

dp g f e d c b a
Soln. :
Step I : According to specification (i) the 8279 should be initialized in 16 digit right
entry mode.
(a) Keyboard /display mode set command :
DD = 11; KKK should be encoded KKK = 000 ;
The command word is,
D
7
D
6
D
5
D
4
D
3
D
2
D
1
D
0

0 0 0 0 0 0 0 0 = 18 H
Step II : Specification (ii) gives external clock frequency and display scan time.
. Scan time =
Display scan time
2
4
=
10.24
16
= 640 µs.
. Clock cycle time =
640 µs
64
= 10 µs
Internal clock = 100 kHz
PPPPP =
3 × 10
6
100 × 10
3
= 30
10
= 1E H
(b) Program clock :
PPPPP = 11110; The command word is, 0011 1110 = 3E H.
Step III : (c) Clear code command :
The command word is, 1101 1100 = DC H; CD
2
CD
1
CD
0
= 111, C
F
= 0, C
A
= 0
Micropro. & Interfacing Techniques (PU) L-64 Lab Manual
Step IV : According to specification (iii) the display is common anode type that is logic 0
corresponds to segment ‘ON’ and logic 1 corresponds to segment ‘OFF’. It also
gives the connection of segments with A
3
- A
0
and B
3
- B
0
lines.
.We will have look up Table B.4 as follows :
Table B.4 : Look-up Table
dp g f e d c b a Data Address
(D
7
D
6
D
5
D
4
D
3
D
2
D
1
D
0
)
1 1 1 1 1 1 0 0 1 F9 2000
2 1 0 1 0 0 1 0 0 A4 2001
3 1 0 1 1 0 0 0 0 B0 2002
4 1 0 0 1 1 0 0 1 99 2003
5 1 0 0 1 0 0 1 0 92 2004
6 1 0 0 0 0 0 1 0 82 2005
7 1 1 1 1 1 0 0 0 f8 2006
8 1 0 0 0 0 0 0 0 80 2007
9 1 0 0 1 0 0 0 0 90 2008
A 1 0 0 0 1 0 0 0 88 2009
B 1 0 0 0 0 0 1 1 83 200A
C 1 1 0 0 0 1 1 0 C6 200B
Step V : (d) Write display RAM command :
In right entry mode, the status of display is given as follows :

According to specification digit 12 to digit 15 are not connected. Hence the
first entry should be displayed on rightmost digit (digit 11). After first entry,
it shifts the address of digit and then displays contents of corresponding
location.
i.e. first entry will be displayed on digit 12
AI = 1, A
3
A
2
A
1
A
0
= 1100
Hence the command word is, 1001 1100 = 9C H
Step VI : Now we can find address of control/status and data register.
A
7
1
1
A
6
0
0
A
5

1
1
A
4

0
1
A
3
0
0
A
2
0
0
A
1

1
1
A
0

0
0
Here addresses show a change in A
4
bit. Hence A
4
of microprocessor is
connected to A
0
line of 8279. The address of control/status register is B2
while the address of data register is A2.
Step VII : Flowchart of the program is shown in Fig.10(a)
Micropro. & Interfacing Techniques (PU) L-65 Lab Manual
Title INITIALIZATION OF 8279
Label Instruction Comments
.model small
.code
MOV AL, 00H Keyboard/display mode
OUT B2 H, AL set command
MOV AL, 3E H Program CLK
Command
OUT B2, AL
MOV AL, DC H Clear code command
OUT B2, AL
MOV AL, 9C H 9CH for left entry mode
OUT B2, AL Write display RAM
command
L2: MOV BX, 2000 H
MOV CX, 000F H Character counter
L1: MOV AL, [BX] Read character code
from memory
OUT A2 H, AL
CALL DELAY (1 sec.)
INC BX
DEC CX
LOOPNE L1
LOOP L2
end

Programs of 8251
Program 1: Perform an experiment to establish communication between two 8251 systems A and B.
Program 8251 system A in asynchronous transmitter mode and 8251 system B in
asynchronous receiver mode. Write an ALP to transmit the data from system A and
receive the data at system B. The requirements are as follows:
Transmission :
- A message is stored as ASCII characters in the memory.
- A message specifies the number of characters to be transmitted as the first byte.
Reception :
- Message is retrieved and stored in the memory.
- Successful reception should be indicated.
(i) The interfacing schematic will be as Fig. 11.
(ii) The mode word required for asynchronous mode, 8 bit character. We assume baud
rate factor × 16, No parity check and 1 stop bit.
Fig. 10(a)
Micropro. & Interfacing Techniques (PU) L-66 Lab Manual




Fig. 11
The mode word format for transmitter and receiver will be as follows :
COREL 20
(iii) Command word format to enable Tx and Rx will be as follows :
COREL 21
(iv) Status word to check TxRDy will be 01H and RxRDy will be 02H.
(v) The flow chart for both systems will be as shown Flowchart B.5.
(vi) Program for both systems will be same.
Program :
Label Instruction Comment

MOV BX, address
1
Memory pointer for Tx
MOV DX, address
2
Memory pointer for Rx
Micropro. & Interfacing Techniques (PU) L-67 Lab Manual
Label Instruction Comment
MOV AL, 00 H Dummy 00's to 8251
OUT F3 H, AL
OUT F3 H, AL
OUT F3 H, AL
MOV AL, 40 H Internally reset 8251
OUT F3 H, AL
MOV AL, 4E H Mode word
OUT F3 H, AL
MOV AL, 15 H Command word
OUT F3 H, AL
UP : IN AL, F3 H Status word
AND AL, 01 H
TEST AL, 01 H Check TxRDy
CALL Tx data call subroutine 1
IN AL, F3 H status word
AND AL, 02 H
TEST AL, 02 H Check RxRDy
CALL Rx data Call subroutine 2
LOOP UP Go to up.
Subroutine 1
Tx data MOV AL, [BX] Take data from memory
OUT F2 H, AL Data ÷ 8251 Tx
INC BX memory pointer = memory pointer + 1
RET Go to main program.
Subroutine 2
Rx data IN F2 H, AL Take data from Rx
MOV [DX], AL Store data in memory
INC DX Memory pointer = Memory pointer +1
RET Go to main program

Micropro. & Interfacing Techniques (PU) L-68 Lab Manual



Flowchart B.5
8259 Programs
Program 1 : Write 8086 APL to interface 8259 in cascade mode (M/S) and demonstrate execution of
ISR in following manner :
Main program will display two digits up counter. When slave IRQ interrupt occurs, it
clears the counter and starts up counting again. When Master IR1 interrupt occurs, it
resets the counter to FFH and starts down counting.
Explanation :
Fig. 12 shows the interfacing of 8259 with 8086 in cascaded mode. Eight interrupt
lines IR0-IR7 are connected to the interrupt request register (IRR).A microprocessor uses
Micropro. & Interfacing Techniques (PU) L-69 Lab Manual
two PICs to provide 15 interrupt inputs (7 on the master PIC and 8 on the slave one). the
following table lists the interrupt sources on the PC .
Input on
8259A
Priority 80x86 INT Device
IRQ 0 Highest 08h Timer Chip
IRQ 1 09h Keyboard
IRQ 2 0Ah Cascade for controller 2 (IRQ 8-15)
IRQ 9/1 71h CGA vertical retrace (and other IRQ 2
devices)
IRQ 8/0 70h Real-time clock
IRQ 10/2 72h Reserved
IRQ 11/3 73h Reserved
IRQ 12/4 74h Reserved in AT, auxiliary device on PS/2
systems
IRQ 13/5 75h FPU interrupt
IRQ 14/6 76h Hard disk controller
IRQ 15/7 77h Reserved
IRQ 3 0Bh Serial Port 2
IRQ 4 0Ch Serial Port 1
IRQ 5 0Dh Parallel port 2 in AT, reserved in PS/2
systems
IRQ 6 0Eh Diskette drive
IRQ 7 Lowest 0Fh Parallel Port 1

For cascaded 8259, the INT pin of the slaves are connected to interrupt request pins
(IR0 – IR7) and
–––––
INTA to the
–––––
INTA of master 8259. The CAS2 – CAS0 lines work as
output for the master and input for the slave. The CAS2 – CAS0 lines act as select lines
for addressing the slaves. Fig. 12 shows cascaded 8259 interfaced with 8086 in maximum
mode.
Micropro. & Interfacing Techniques (PU) L-70 Lab Manual




m(12.15)Fig. 12 : Interfacing 8259 with 8086 (8259 – cascaded, 8086 – maximum mode)
Program :
Label Instruction Comment
START: MOV AX,1000H
MOV DS,AX
MOV AX,4000H
MOV SS,AX
MOV SP,0500H
MOV AX,0000H
MOV ES,AX
MOV AX,1000H
MOV [ES:0008H],AX
MOV AX,2000H
MOV [ES:0009H],AX
STI Enable interrupt
L1 : JMP L1
Micropro. & Interfacing Techniques (PU) L-71 Lab Manual
Interrupt service routine,
Instruction
MOV AL,[08H]
INC AL
DAA
POP AX
MOV AL,[09H]
DEC AL
POP AX
IRET

TSR Programs
Program 1 : Write a TSR program in 8086 ALP to implement Real Time Clock (RTC). Read the Real
Time from CMOS chip by suitable INT and FUNCTION and display the RTC at the bottom
right corner on the screen. Access the video RAM directly in your routine.
Program :
Label Instruction Comment
.model small
.stack 100h
.code
resi:
push ax
push bx
push cx
push dx
push es
push di
mov ax,0b800h
mov es,ax
mov di,140
mov ah,02
int 1Ah
mov bx,cx
mov ah,07
mov cx,0504h
cld
Micropro. & Interfacing Techniques (PU) L-72 Lab Manual
Label Instruction Comment
l2: cmp ch,03
jne l3
mov al,':'
jmp l4
l3: rol bx,cl
mov al,bl
and al,0fh
add al,30h
l4: stosw
dec ch
jnz l2
mov cx,0304h
cld
l5: mov al,':'
jmp l7
l6: rol dh,cl
mov al,dh
and al,0fh
add al,30h
l7: stosw
dec ch
jnz l6
pop di
pop es
pop dx
pop cx
pop bx
pop ax
jmp dword ptr cs: old_ip
data: old_ip dw 0
old_cs dw 0
init : mov ax,cs
mov ds,ax
mov ah,35h
mov al,08
int 21h
mov word ptr old_ip,bx
mov word ptr old_cs,es
mov ah,25h
Micropro. & Interfacing Techniques (PU) L-73 Lab Manual
Label Instruction Comment
mov al,08
lea dx,resi
int 21h
lea dx,init
add dx,100h
mov cl,4
shr dx,cl
inc dx
mov ah,31h
mov al,00
int 21h
end init

TSR Programs
Program 1 : Write a TSR program in 8086 ALP to implement Screen Saver. Screen Saver should get
activated if the keyboard is idle for 7 seconds. Access the video RAM directly in your
routine.
Program :
Label Instruction
.model small
.stack 500h
.code
resi_timer:
cmp cs:flag,1
je timer_2
dec cs:count
jnz timer_quit
push cx
push ax
push dx
push es
push di
push si
push ds
mov cx,0b800h
mov ds,cx
mov si,00
mov cx,cs
Micropro. & Interfacing Techniques (PU) L-74 Lab Manual
Label Instruction
mov es,cx
lea di,buf
cld
mov cx,2000
rep movsw
mov cs:flag,1
mov cx,0b800h
mov es,cx
mov di,0
mov cx,2000
mov ax,0720h
cld
rep stosw
pop ds
pop si
pop di
pop es
pop dx
pop ax
pop cx
timer_quit: jmp dword ptr cs:old_ip_timer
timer_2 :
dec cs:spd
jnz timer_2_quit
mov cs:spd,2
push cx
push ax
push dx
push es
push di
push si
push ds
mov cx,0b800h
mov es,cx
mov di,cs:pos
mov ax,0720h
cld
stosw
Micropro. & Interfacing Techniques (PU) L-75 Lab Manual
Label Instruction
inc cs: pos
inc cs: pos
cmp cs: pos,4000
jb t4
mov cs: pos,0
t4:
mov di,cs:pos

mov cx,cs
mov ds,cx
mov cx,5
lea si,string
mov ah,07
t8: lodsb
stosw
loop t8
pop ds
pop si
pop di
pop es
pop dx
pop ax
pop cx
timer_2_quit: jmp dword ptr cs: old_ip_timer
resi_kbd: mov cs:count,182
cmp cs:flag,0
je kbd_quit
push cx
push ax
push dx
push es
push di
push si
push ds
mov cx,0b800h
mov es,cx
mov di,00
mov cx,cs
mov ds,cx
lea si,buf
cld
Micropro. & Interfacing Techniques (PU) L-76 Lab Manual
Label Instruction
mov cx,2000
rep movsw
mov cs:flag,0
pop ds
pop si
pop di
pop es
pop dx
pop ax
pop cx
kbd_quit: jmp dword ptr cs:old_ip_kbd
old_ip_timer dw ?
old_cs_timer dw ?
old_ip_kbd dw ?
old_cs_kbd dw ?
flag db 0
count dw 182
buf db 4000 dup(0)
pos dw 0
spd db 2
string db 'HELLO'
init : mov ax,cs
mov ds,ax
mov ah,35h
mov al,8
int 21h
mov word ptr old_ip_timer,bx
mov word ptr old_cs_timer,es
mov ah,25h
mov al,8
lea dx,resi_timer
int 21h
mov ah,35h
mov al,9
int 21h
mov word ptr old_ip_kbd,bx
mov word ptr old_cs_kbd,es
mov ah,25h
mov al,9
Micropro. & Interfacing Techniques (PU) L-77 Lab Manual
Label Instruction
lea dx,resi_kbd
int 21h
lea dx,init
add dx,100h
mov cl,4
shr dx,cl
inc dx
mov ah,31h
mov al,00
int 21h
end init


Micropro. & Interfacing Techniques (PU) L-78 Lab Manual

Note