Ee6612 miroprocessor and microcontroller laboratory

Department of Electrical and Electronics Engineering - VEC
Valliammai Engineering College
SRM Nagar, Kattankulathur - 603203
Department of Electrical and Electronics Engineering
EE6612–Microprocessors and Microcontrollers Laboratory Manual
LAB MANUAL
VI Semester - Electrical and Electronics Engineering
Academic Year - 2016-2017
(2013 Regulation)
Prepared by,
Dr.G.Udhayakumar, Associate Professor
Mr.V.Sudhagar,AssistantProfessor
Mr.S.Balaji, AssistantProfessor


RANGANATHAN ENGINNERING COLLEGE
Coimbatore-641109
LABORATORY RECORD

EE6612 – Microprocessors and Microcontrollers Laboratory
SYLLABUS
EE6612- MICROPROCESSORS AND MICRO CONTROLLERS LABORATORY
OBJECTIVES:
To provide training on programming of microprocessors and microcontrollers and
understand the interface requirements.
LIST OF EXPERIMENTS:
1. Simple arithmetic operations: addition / subtraction / multiplication / division.
2. Programming with control instructions:
(i) Ascending / Descending order, Maximum / Minimum of numbers
(ii) Programs using Rotate instructions
(iii)Hex / ASCII / BCD code conversions.
3. Interface Experiments: with 8085
(i) A/D Interfacing. & D/A Interfacing.
4. Traffic light controller.
5. I/O Port / Serial communication
6. Programming Practices with Simulators/Emulators/open source
7. Read a key,interface display
8. Demonstration of basic instructions with 8051 Micro controller execution, including:
(i) Conditional jumps, looping
(ii) Calling subroutines.
9. Programming I/O Port 8051
(i) study on interface with A/D & D/A(ii
(ii) study on interface with DC & AC motor .
10. Mini project development with processors.
TOTAL: 45 PERIODS
OUTCOMES:
 Ability to understand and analyse, linear and digital electronic circuits.
 To understand and apply computing platform and software for engineering problems.

LIST OF EXPERIMENTS
CYCLE I
8-bit Microprocessor
1. Simple arithmetic operations:
(a) Addition (b) subtraction (c) multiplication (d) division.
2. Ascending / Descending order, Maximum / Minimum of numbers :
(a) Ascending order (b) Descending order (c) Maximum of numbers (d) Minimum of numbers.
3. Rotate instructions:
(a) Odd and Even Number (b) Positive and Negative Number
4.Code conversion.
(a) ASCII to hexadecimal number (b) hexadecimal to ASCII(c) hexadecimal to decimal
number (d) decimal to hexadecimal number and (e) hexadecimal to binary number
CYCLE II
8085 Interfacing Program
5. Interface Experiments:
(a) A/D Interfacing. (b) D/A Interfacing
6. Traffic light controller: I/O port - 8255
7. Serial Communication: 8251
8. Read a key, interface display: 8279
CYCLE III
8-bit Microcontroller
9. Demonstration of basic instructions with 8051 Micro controller execution, including:
(a)Addition (b) subtraction (c) multiplication (d) division.
10. (a) Conditional jumps, looping : Sum of elements in an array
(b) Calling subroutines : Check whether given number is Odd or Even using call option
11. Programming I/O Port :
(a) A/D Interfacing. (b) D/A Interfacing
12. 8085 Programming Practices with Simulators/ Jubin Open source Software
13. Interface with DC & AC motor
14. Mini project development with processors
ADDITIONAL EXPERIMENTS
15.Interfacing 8253 Timer With 8085
16. Stepper Motor Interfacing With 8051

S.no Date List Of Experiments Signature

8085 MICROPROCESSOR

Ex. No: 1 SIMPLE ARITHMETIC OPERATIONS
AIM:
To write an assembly language program to add, subtract, multiply and divide the given
data stored at two consecutive locations using 8085 microprocessor.
a. 8 BIT DATAADDITION:
ALGORITHM:
1. Initialize memory pointer to data location.
2. Get the first number from memory in accumulator.
3. Get the second number and add it to the accumulator.
4. Store the answer at another memory location.

FLOW CHART:
NO
YES
START
[HL] 4500H
[A] [M]
[A] [A]+[M]
[HL] [HL]+1
STOP
[HL] [HL]+1
[M] [A]
[C] 00H
[M] [C]
[HL] [HL]+1
Is there a
Carry ?
[C] [C]+1

PROGRAM:
ADDRESS OPCODE LABEL MNEMONICS OPERAND COMMENT
4100 START MVI C, 00 Clear C reg.
4101
4102 LXI H, 4500 Initialize HL reg. to
45004103
4104
4105 MOV A, M Transfer first data to
accumulator
4106 INX H Increment HL reg. to
point next memory
Location.
4107 ADD M Add first number to
acc. Content.
4108 JNC L1 Jump to location if
result does not yield
carry.
4109
410A
410B INR C Increment C reg.
410C L1 INX H Increment HL reg. to
point next memory
Location.
410D MOV M, A Transfer the result from
acc. to memory.
410E INX H Increment HL reg. to
point next memory
Location.
410F MOV M, C Move carry to memory
4110 HLT Stop the program

b. 8 BIT DATA SUBTRACTION
ALGORITHM:
2. Get the first number from memory in accumulator.
3. Get the second number and subtract from the accumulator.
4. If the result yields a borrow, the content of the acc. is complemented and 01H is added
to it (2’s complement). A register is cleared and the content of that reg. is incremented
in case there is a borrow. If there is no borrow the content of the acc. is directly taken as
the result.
5. Store the answer at next memory location.

FLOW CHART:
NO
YES
START
[HL] 4500H
[A] [M]
Is there a
Borrow?
[A] [A]-[M]
[HL] [HL]+1
[C] 00H
[C] [C]+1
STOP
[HL] [HL]+1
[M] [A]
[M] [C]
[HL] [HL]+1
Complement [A]
Add 01H to [A]

PROGRAM:
4100 START MVI C, 00 Clear C reg.
4101
4102 LXI H, 4500 Initialize HL reg. to
45004103
4104
4105 MOV A, M Transfer first data to
accumulator
point next mem.
Location.
4107 SUB M Subtract first number
from acc. Content.
4108 JNC L1 Jump to location if
borrow.
4109
410A
410B INR C Increment C reg.
410C CMA Complement the Acc.
content
410D ADI 01H Add 01H to content of
acc.410E
410F L1 INX H Increment HL reg. to
point next mem.
Location.
4110 MOV M, A Transfer the result from
acc. to memory.
point next mem.
Location.
4112 MOV M, C Move carry to mem.

c. 8 BIT DATA MULTIPLICATION:
ALGORITHM:
LOGIC: Multiplication can be done by repeated addition.
2. Move multiplicand to a register.
3. Move the multiplier to another register.
4. Clear the accumulator.
5. Add multiplicand to accumulator
6. Decrement multiplier
7. Repeat step 5 till multiplier comes to zero.
8. The result, which is in the accumulator, is stored in a memory location.

FLOW CHART:
NO
YES
NO
YES
[HL] 4500
B  M
A  00
C  00
Is there
any carry
C  C+1
B  B-1
[A]  [A] +[M]
[HL]  [HL]+1
IS B=0
A
START

A
STOP
[HL] [HL]+1
[M] [A]
[M] [C]
[HL] [HL]+1

PROGRAM:
4100 START LXI H, 4500 Initialize HL reg. to
4500
Transfer first data to
reg. B
4101
4102
4103 MOV B, M
point next mem.
Location.
4105 MVI A, 00H Clear the acc.
4106
4107 MVI C, 00H Clear C reg for carry
4108
4109 L1 ADD M Add multiplicand
multiplier times.
410A JNC NEXT Jump to NEXT if there
is no carry410B
410C
410D INR C Increment C reg
410E NEXT DCR B Decrement B reg
410F JNZ L1 Jump to L1 if B is not
zero.4110
4111
point next mem.
Location.
4113 MOV M, A Transfer the result from
acc. to memory.
point next mem.
Location.
4115 MOV M, C Transfer the result from
C reg. to memory.

d. 8 BIT DIVISION:
ALGORITHM:
LOGIC: Division is done using the method Repeated subtraction.
1. Load Divisor and Dividend
2. Subtract divisor from dividend
3. Count the number of times of subtraction which equals the quotient
4. Stop subtraction when the dividend is less than the divisor .The dividend now becomes
the remainder. Otherwise go to step 2.
5. Stop the program execution.

FLOWCHART:
YES
Yes
No
B  00
M  A-M
[B]  [B] +1
IS A<0
A A+ M
B  B-1
[HL] 4500
A  M
[HL]  [HL]+1
START
STOP
[HL] [HL]+1
[M] [A]
[M] [B]
[HL] [HL]+1

PROGRAM:
ADDRESS OPCODE LABEL MNEMONICS OPERAND COMMENTS
4100 MVI B,00 Clear B reg for quotient
4101
4102 LXI H,4500 Initialize HL reg. to
4500H4103
4104
4105 MOV A,M Transfer dividend to
acc.
point next mem.
Location.
4107 LOOP SUB M Subtract divisor from
dividend
4108 INR B Increment B reg
4109 JNC LOOP Jump to LOOP if
borrow
410A
410B
410C ADD M Add divisor to acc.
410D DCR B Decrement B reg
410E INX H Increment HL reg. to
point next mem.
Location.
410F MOV M,A Transfer the remainder
from acc. to memory.
point next mem.
Location.
4111 MOV M,B Transfer the quotient
from B reg. to memory.

OBSERVATION:
ADDITION:
S.NO INPUT OUTPUT
ADDRESS DATA ADDRESS DATA
1 4500 4502
4501 4503
2 4500 4502
4501 4503
SUBTRACTION:
S.NO INPUT OUTPUT
1 4500 4502
4501 4503
2 4500 4502
4501 4503
MULTIPLICATION:
S.NO INPUT OUTPUT
1 4500 4502
4501 4503
2 4500 4502
4501 4503
DIVISION:
S.NO INPUT OUTPUT
1 4500 4502
4501 4503
2 4500 4502
4501 4503

RESULT:
Thus the addition, subtraction, multiplication and division of two numbers was
performed using the 8085 microprocessor.

Ex. No: 2 Ascending / Descending order, Maximum / Minimum of numbers
AIM:
To write an assembly language program to arrange an array of data in ascending and
descending order and to find the smallest and largest data among the array.
a. ASCENDING ORDER
ALGORITHM:
1. Get the numbers to be sorted from the memory locations.
2. Compare the first two numbers and if the first number is larger than second then I
interchange the number.
3. If the first number is smaller, go to step 4
4. Repeat steps 2 and 3 until the numbers are in required order

FLOWCHART:
Yes
No
[HL]  4200H
[C]  [HL]
[C]  [C] -1
[HL] 4201H
[HL  [HL] + 1
Is
[A] < [HL]?
[B] [HL]
[HL]  [A]
[HL]  [HL] - 1
[HL]  [B]
[HL]  [HL] + 1
[D]  [D] – 01 H
A
[D]  [C]
START
[A]  [HL]

Yes
Yes
No
No
IS
[D] = 0?
A
[C]  [C]-1
IS
[C] = 0?
STOP

PROGRAM:
ADDRESS OPC
ODE
LABEL MNEMONICS OPERA
ND
COMMENTS
4100 LXI H,4200 Set pointer for array
4101
4102
4103 MOV C,M Load the Count
4104 DCR C Decrement Count
4105 LOOP 3 MOV D,C Transfer data from C to
D
4106 LXI H,4201 Load data from 4201
4107
4108
4109 LOOP2 MOV A,M Copy content of M to A
410A INX H Increment HL reg. to
point next memory
location
410B CMP M Compare M & A
410C JC LOOP1 If A is lesser than M
then go to loop1
410D
410E
410F MOV B,M Transfer data from M to
D reg
4110 MOV M,A Transfer data from acc
to M
4111 DCX H Decrement HL pair
4112 MOV M,B Transfer data from B to
M
4113 INX H Increment HL pair
4114 LOOP1 DCR D Decrement Dreg
4115 JNZ LOOP2 If D is not zero go to
loop24116
4117
4118 DCR C Decrement Creg
4119 JNZ LOOP3 If C is not Zero go to
loop3411A
411B
411C HLT Stop the program

b. DESCENDING ORDER
ALGORITHM:
1. Get the numbers to be sorted from the memory locations.
2. Compare the first two numbers and if the first number is smaller than second then I
interchange the number.
3. If the first number is larger, go to step 4
4. Repeat steps 2 and 3 until the numbers are in required order

FLOWCHART:
No
Yes
[HL]  4200H
[C]  [HL]
[C]  [C] -1
[HL] 4201H
[HL  [HL] + 1
Is
[A] < [HL]?
[B] [HL]
[HL]  [A]
[HL]  [HL] - 1
[HL]  [B]
[HL]  [HL] + 1
[D]  [D] – 01 H
A
[D]  [C]
START
[A]  [HL]

PROGRAM:
ADDRESS OPC
ODE
LABEL MNEMONICS OPERA
ND
COMMENTS
4100 LXI H,4200 Set pointer for array
4101
4102
4103 MOV C,M Load the Count
4104 DCR C Decrement Count
4105 LOOP 3 MOV D,C Transfer data from C to
D
4106 LXI H,4201 Load data from 4201
4107
4108
4109 LOOP2 MOV A,M Copy content of M to A
410A INX H Increment HL reg. to
point next memory
location
410B CMP M Compare M & A
410C JNC LOOP1 If A is lesser than M
then go to loop1
410D
410E
410F MOV B,M Transfer data from M to
D reg
4110 MOV M,A Transfer data from acc
to M
4111 DCX H Decrement HL pair
4112 MOV M,B Transfer data from B to
M
4113 INX H Increment HL pair
4114 LOOP1 DCR D Decrement D reg
4115 JNZ LOOP2 If D is not zero go to
loop24116
4117
4118 DCR C Decrement C reg
4119 JNZ LOOP3 If C is not Zero go to
loop3411A
411B
411C HLT Stop the program

c.LARGEST ELEMENT IN AN ARRAY
ALGORITHM:
1. Place all the elements of an array in the consecutive memory locations.
2. Fetch the first element from the memory location and load it in the accumulator.
3. Initialize a counter (register) with the total number of elements in an array.
4. Decrement the counter by 1.
5. Increment the memory pointer to point to the next element.
6. Compare the accumulator content with the memory content (next
element).
7. If the accumulator content is smaller, then move the memory content
(largest element) to the accumulator. Else continue.
9. Repeat steps 5 to 8 until the counter reaches zero
10. Store the result (accumulator content) in the specified memory location.

FLOW CHART:
NO
YES
NO
YES
[B] M
[HL]  [4200H]
[A]  [HL]
[HL  [HL] + 1
IS
[A] < [HL]?
[A] [HL]
[4300]  [A]
START
[B]  [B]-1
IS
[B] = 0?
STOP

PROGRAM:
ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPER
AND
COMMENTS
4200H4102
4103
4104 MOV B,M Initialize B reg with no. of
comparisons(n-1)4105 INX H
4106 MOV A,M Transfer first data to acc.
4107 LOOP1 INX H Increment HL reg. to point
next memory location
4108 CMP M Compare M & A
4109 JNC LOOP If A is greater than M then go
to loop410A
410B
410C MOV A,M Transfer data from M to A reg
410D LOOP DCR B Decrement B reg
410E JNZ LOOP1 If B is not Zero go to loop1
410F
4110
4111 STA 4300 Store the result in a memory
location.4112
4113

d.SMALLEST ELEMENT IN AN ARRAY
ALGORITHM:
1. Place all the elements of an array in the consecutive memory locations.
2. Fetch the first element from the memory location and load it in the accumulator.
3. Initialize a counter (register) with the total number of elements in an array.
5. Increment the memory pointer to point to the next element.
6. Compare the accumulator content with the memory content (next
element).
7. If the accumulator content is smaller, then move the memory content
(largest element) to the accumulator. Else continue.
9. Repeat steps 5 to 8 until the counter reaches zero
10. Store the result (accumulator content) in the specified memory location.

FLOW CHART:
YES
NO
NO
YES
[B] M
[HL]  [4200H]
[A]  [HL]
[HL  [HL] + 1
IS
[A] < [HL]?
[A] [HL]
[4300]  [A]
START
[B]  [B]-1
IS
[B] = 0?
STOP

PROGRAM:
ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPER
AND
COMMENTS
4200H4102
4103
4104 MOV B,M Initialize B reg with no. of
comparisons(n-1)4105 INX H
4106 MOV A,M Transfer first data to acc.
4107 LOOP1 INX H Increment HL reg. to point
next memory location
4108 CMP M Compare M & A
4109 JC LOOP If A is lesser than M then go
to loop410A
410B
410C MOV A,M Transfer data from M to A reg
410D LOOP DCR B Decrement B reg
410E JNZ LOOP1 If B is not Zero go to loop1
410F
4110
4111 STA 4300 Store the result in a memory
location.4112
4113
OBSERVATION:

A. ASCENDING ORDER
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4200(Array Size) 05
4201 4201
4202 4202
4203 4203
4204 4204
4205 4205
B. DESCENDING ORDER
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4200(Array Size) 05
4201 4201
4202 4202
4203 4203
4204 4204
4205 4205

C. SMALLEST ELEMENT
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4200(Array Size) 05
4201
4202 4300
4203
4204
4205
D. LARGEST ELEMENT
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4200(Array
Size)
05
4201
4202 4300
4203
4204
4205

RESULT:
Thus the sorting operations of arranging an array in ascending, descending order and
the largest and smallest element werefound using the 8085 microprocessor.

Ex. No: 3 FIND THE NUMBER OF EVEN AND ODD NUMBER IN A
BLOCK OF DATA
AIM:
To find the number of even and odd number, and number of positive and negative
numbers in a block of data.
a. Odd and Even Number
ALGORITHM:
1. Start.
2. Load the given data.
3. Clear the data in accumulator.
4. Move the content of accumulator to D and B register.
5. Increment HL register pair and move the content of memory to accumulator.
6. Rotate the obtain data in the form of RAR.
7. If carry occur increment B else increment D.
8. Decrement C, if C ≠0 go to step 5. Else increment HL pair.
9. Move the content of D to memory and increment HL pair and move the content B to
memory.
10. Stop.

Yes
No
NoYes
[A]  00H
[HL]  [4200H]
[C]  [HL]
[B]  [A]
[D]  [A]
[HL  [HL] + 1
[A]  [HL]
Is CY=1
START
[C]  [C]-1
Is
[C] = 0?
STOP
Rotate the A right
through carry
[B] [B]+1, for odd [D] [D]+1, for even
[HL] [HL]+1, [HL]  D
[HL] [HL]+1, [HL]  B

ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPER
AND
COMMENTS
4100 LXI H,4200 Load the number of data
4101
4102
4103 MOV C, M Move M to C
4104 XRA A Exclusive OR Reg A
4105 MOV D, A Move A to D
4106 MOV B, A Move A to B
4107 L3 INX H H Increment H
4108 MOV A,M Move M to A
4109
RAR
Rotate right through
carry
410A JNC L1 Jump no carry
410B
410C
410D INR D Increment D
410E JMP L2 Jump to L2
410F
4110
4111 L1 INR B Increment B
4112 L2 DCR C Decrement C
4113 JNZ L3 Jump non Zero
4114
4115
4116 INX H Increment H
4117 LOOP MOV M, D Move D to M
4118 INX H H Increment H
4119 MOV M, B Move B to M
411A HLT Halt the program

b. Positiveand Negative Number
ALGORITHM:
1. Start.
2. Load the given data.
3. Clear the data in accumulator.
4. Move the content of accumulator to D and B register.
5. Increment HL register pair and move the content of memory to accumulator.
6. Rotate the obtain data in the form of RAL.
7. If carry occur increment B else increment D.
8. Decrement C, if C ≠0 go to step 5. Else increment HL pair.
9. Move the content of D to memory and increment HL pair and move the content B to
memory.
10. Stop.

Yes
No
NoYes
[A]  00H
[HL]  [4200H]
[C]  [HL]
[B]  [A]
[D]  [A]
[HL  [HL] + 1
[A]  [HL]
Is CY=1
START
[C]  [C]-1
Is
[C] = 0?
STOP
Rotate the A left
through carry
[B] [B]+1, negative [D] [D]+1, positive
even
[HL] [HL]+1, [HL]  D
[HL] [HL]+1, [HL]  B

ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPER
AND
COMMENTS
4100 LXI H,4200 Load the number of data
4101
4102
4103 MOV C, M Move M to C
4104 XRA A Exclusive OR reg A
4105 MOV D, A Move A to D
4106 MOV B, A Move A to B
4107 L3 INX H H Increment H
4108 MOV A,M Move M to A
4109 RAL Rotate left through carry
410A JNC L1 Jump no carry
410B
410C
410D INR D Increment D
410E JMP L2 Jump to L2
410F
4110
4111 L1 INR B Increment B
4113 JNZ L3 Jump non Zero
4114
4115
4116 INX H Increment H
4117 LOOP MOV M, D Move D to M
4118 INX H H Increment H
4119 MOV M, B Move B to M
411A HLT Halt the program

Observation
a. Odd and even number
INPUT: OUTPUT:
4200 = 04 4205=03 (Odd numbers)
4201= 01 4206=01 (Even numbers)
4202 = 05
4203 = 03
4204 = 04
b. Positive and negative number
INPUT: OUTPUT:
4200 = 04 4205=01 (Negative numbers)
4201= 01 4206=03 (Positive numbers)
4202 = F5
4203 = 03
4204 = 74
RESULT:
Thus the number of even and odd number, and positive and negative number has been
found from the given numberof list.

Ex. No: 4CODE CONVERSIONS
AIM:
To write an assembly language program to perform the conversions of ASCII to
hexadecimal number, hexadecimal to ASCII, hexadecimal to decimal number, decimal number
to hexadecimal number and hexadecimal to binary number.
a.ASCII TO HEXADECIMAL
ALGORITHM:
1. Start the program
2. Load the data from address 4200 to A
3. Move data from accumulator to C
4. Move data from M to HL pair to accumulator
5. Subtract the data 30 from A
6. Decrement content of register
7. Stop the program if C is zero
8. Jump to Step 5
9. End the program

FLOWCHART:
YES
NO
Start
Set the ASCII value
Check
for
Carry?
Decrement the register content
Subtract 30 from A
Subtract 07 from A
Store the hex value
Stop

PROGRAM:
ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPER
AND
COMMENTS
4100 LDA 4200 Load data 4200 to A
4101
4102
4103 MOV C,A Move data from A to C
4104 LXI H,4201 Load address 4201 in HL
4105
4106
4107 LXI D,4301 Load address 4301 in DF
4108
4109
410A LOOP 1 MOV A,M Move data from M to A
410B SUI 30 Subtract 30 from A
410C
410D STAX D Store data from
accumulator to DE
410E DCR C Decrement from C
register
410F JZ LOOP Stop program if C is 0
4110
4111
4112 INX H Increment HL register
pair
4113 INX D Increment DE register
pair
4114 JMP LOOP 1 Jump to 410A
4115
4116
4117 LOOP HLT Stop

b.HEXADECIMAL TO ASCII
ALGORITHM:
2. Load the data from address 4200 to A
3. Move data from accumulator to C
4. Move data from M to HL pair to accumulator
5. Add the data 30 to A
6. Decrement content of register
7. Stop the program if C is zero
8. Jump to Step 5
9. End the program

FLOWCHART:
YES
NO
Set the ASCII value
Check
for
Carry?
Decrement the register content
Add 30 to A
Store the decimal value
Stop
Start

PROGRAM:
ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPER
AND
COMMENTS
4100 LDA 4200 Load data 4200 to A
4101
4102
4103 MOV C,A Move data from A to C
4104 LXI H,4201 Load address 4201 in HL
4105
4106
4107 LXI D,4301 Load address 4301 in DF
4108
4109
410A LOOP 1 MOV A,M Move data from M to A
410B ADI 30 Subtract 30 from A
410C
410D STAX D Store data from
accumulator to DE
410E DCR C Decrement from C
register
410F JZ LOOP Stop program if C is 0
4110
4111
4112 INX H Increment HL register
pair
4113 INX D Increment DE register
pair
4114 JMP LOOP 1 Jump to 410A
4115
4116
4117 LOOP HLT Stop

c.HEXADECIMAL TO BINARY
ALGORITHM:
2. Move the content of memory to accumulator
3. Move data 0B o register B
4. Increment the content of HL register pair
5. Rotate the accumulator right
6. Jump to the specified address if carry generated
7. Move 00 to memory
8. Jump to specified address if there is no zero
9. Move 01 to memory
10. Jump to specified address if there is no zero
11. End the program

FLOWCHART:
YES
NO
NO
YES
Load address in HL pair
Check for
Carry?
Initialize counter B to 08
Move data from M to A
Stop
Start
Increments HL register pair
Rotate accumulator right
Decrement B register
Move data from 01 to M
Move data from 00 to M
If B=0?

PROGRAM:
ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPERAND COMMENTS
4100 LXI H,4200 Load address in HL pair
4101
4102
4103 MOV A,M Move content of M to A
4104 MVI B, 08 Move 0B to register pair
4105
4106 L3 INX H Increment the content of
HL pair
4107 RRC Rotate accumulator right
4108 JC L1 Jump to specified address
if carry
4109
410A
410B MVI M, 00 Move 00 to M
410C
410D JMP L2 Decrement B register
410E
411F
4110 L1 MVI M, 01 Move 01 to M
4111
4112 L2 DCR B Decrement B by 1
4113 JNZ L3 Jump to the specified
address if no zero
4114
4115

d. BCD TO HEXADECIMAL
ALGORITHM:
1. Start the Program
2. Load the address in HL pair
3. Move the content of M to C Reg
4. AMove the content of M Reg to Accumulator
5. Extract Lower nibble of A Reg& copy to E Reg
6. Copy Value of CReg to A Reg
7. Move Values of Higher nibble to Lower nibble
8. Copy Value of A Reg to D Reg
9. Clear A Reg
10. ADD 0A with ‘D’ no of times
11. 4300 A (o/p- Hexadecimal)
12. A=A+E
13. Stop

FLOWCHART:
Start
HL 4200(Load BCD value)
C M
A M
Extract Lower nibble of A Reg
& copy to E Reg
Copy Value of CReg to A Reg
Copy Value of A Reg to D Reg
ADD 0A with ‘D’ no of times
Move values of Higher nibble to Lower nibble
Clear AReg
4300  A (o/p- Hexadecimal)
Stop
A=A+E

PROGRAM:
ADDRE
SS
OPCO
DE
LABEL MNEM
ONICS
OPERAND COMMENTS
4100 LXI H, 4200 Load address in HL pair
4101
4102
4103 MOV C,M Move content of M to C
4104 MOV A,C Move content of C to A
4105 ANI 0F And the data byte of
OFHwith the content of
aaccumulator
4106
4107 MOV E,A Move content of A to E
4108 MOV A,C Add B content with A
4109 ANI F0 And the data byte
FOHwith the content of
410A
410B RRC Rotate right through
carry the accumulator
410C RRC Rotate right through
410D RRC Rotate right through
410E RRC Rotate right through
410F MOV D,A Move content of D to A
4110 MVI A,00 Move immediate
00Hto the
accumulator
411A
411B L1 ADI 0A Add immediate data
OAHto the accumulator
411C
411D DCR D Decrement the D
register
411E JNZ L1 Jump if no zero to
4112H
411F
4120
4121 ADD E Add the content of E
register to Accumulator

4122 STA 4300 Store the accumulator
content at 4300h
4123
4124
4125 HLT Halt the execution
e. HEXADECIMAL TO DECIMAL
ALGORITHM:
2. Load the address in HL pair
3. Move the content from HL to A
4. Subtract 64 from A
5. Increment BC pair
6. Jump to address 4207
7. Subtract 0A from A
8. Increment HL pair
9. Rotate accumulator left
10. Increment HL pair
11. End the program

FLOWCHART:
YES
NO
NO
YES
Load address in HL pair
Check
Carry?
Clear accumulator
Initialize D register
Stop
Start
Move HL to C register
Add 01 with A
Store A in 4150 H
Move D to accumulator
Store A in 4151 H
Adjust A to BCD
Increment C register
Increment D register
Check
Carry?

PROGRAM:
ADDRE
SS
OPCODE LABEL MNEM
ONICS
OPER
AND
COMMENTS
4100 LXI H, 4150 Load data from 4150 to HL pair
4101
4102
4103 LXI B, 0000 Load data from address to BC
4104
4105
4106 MOV A, M Move the content from HL to A
4107 L4 SUI 64 Subtract 64 from A
4108
4109 JC L1 Stop if A has carry
410A
410B
410C INR B Increment BC
410D JMP L4 Jump to specified address
410E
410F
4110 L1 ADI 64 Add 64 to A
4111
4112 L3 SUI 0A Subtract 0A from A
4113
4114 JC L2 Stop if A has carry
4115
4116
4117 INR C Increment HL
4118 JMP L3 Stop if A has no carry
4119
411A
411B L2 ADI 0A Add 0A to A
411C
411D INX H Increment HL
411E MOV M, B Move B to M
411F MOV B, A Move A to B
4120 MOV A, C Move B to A
4121 RLC Rotate accumulator
4125 ADD B Add B to A

4126 INX H Increment H by 1
4127 MOV M, A Move content of A to M
OBSERVATION:
a. ASCII TO HEXADECIMAL
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4200(Array Size) 02 4301
4201 4302
4202
b. HEXADECIMAL TO ASCII
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4200(Array
Size)
02 4301
4201 4302
4202
c. HEXADECIMAL TO BINARY
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4200
4201 4205
4202 4206
4203 4207
4204 4208

d. BCD TO HEXADECIMAL
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4150
41524151
e. HEXADECIMAL TO DECIMAL
INPUT OUTPUT
MEMORY
LOCATION
DATA MEMORY
LOCATION
DATA
4150 4151
4152
RESULT:
Thus the assembly language programs for various code conversions are executed using
8085 microprocessor.

EX.No:5INTERFACING A/D AND D/A CONVERTER WITH 8085
AIM:
To write an assembly language program to convert an analog signal into a digital signal
and a digital signal into an analog signal using an ADC interfacing and DAC interfacing
respectively.
a. ADC INTERFACING WITH 8085
APPARATUS REQUIRED:
SL.NO ITEM SPECIFICATION QUANTITY
1 Microprocessor kit 8085,Vi Microsystems 1
2 Power supply +5 V dc 1
3 ADC Interface board Vi Microsystems 1
PROBLEM STATEMENT:
To program starts from memory location 4100H. The program is executed for various
values of analog voltage which are set with the help of a potentiometer. The LED display is
verified with the digital value that is stored in the memory location 4150H.
THEORY:
An ADC usually has two additional control lines: the SOC input to tell the ADC when
to start the conversion and the EOC output to announce when the conversion is complete. The
following program initiates the conversion process, checks the EOC pin of ADC 0419 as to
whether the conversion is over and then inputs the data to the processor. It also instructs the
processor to store the converted digital data at RAM 4200H.
ALGORITHM:
1. Select the channel and latch the address.
2. Send the start conversion pulse.
3. Read EOC signal.
4. If EOC =1 continue else go to step (3)
5. Read the digital output.
6. Store it in a memory location.
PROGRAM:

ADDRESS OPCO
DE
LABEL MNEMON ICS OPERA
ND
COMMENTS
4100 MVI A, 10 Select channel 0 and to
make accumulator low
4101
4102 OUT 0C8HC8 Output the data
4103
4104 MVI A, 18A, 18 Make accumulator high
4105
4106 OUT 0C8HC8 Display the data
4107
4108 MVI A, 01 Make 01 to accumulator
4109
410A OUT 0D0HD0 Display the data
410B
410C NOP
410D NOP
410E NOP
410F MVI A, 00 Make 00 to accumulator
4110
4111 OUT D0 Load D0 in output port
4112
4113 LOOP IN D8
4114
4115 ANI 0101 Do and operation directly
4116
4117 CPI 0101 Compare with accumulator
4118
4119 JNZ LOOP Jump to specified address
411A
411B
411C IN C0
411D
411E STA 4150 Store the data
411F
4120
4121 HLT End the program

ADC- CIRCUIT:
SOC JUMPER SELECTION:
J2:SOC Jumper selection
J5: Channel selection

OBSERVATION
ANALOG VOLTAGE DIGITAL DATA ON
LED DISPLAY
HEX CODE IN
LOCATION 4150

b. DAC INTERFACING WITH 8085
APPARATUS REQUIRED:
3 DAC Interface board Vi Microsystems 1
SOFTWARE EXAMPLES
The following examples illustrate how to control the DAC using 8085 and generate sine
wave, saw tooth wave by means of software.
(a) SQUARE WAVE GENERATION:
The basic idea behind the generation of waveforms is the continuous generation of
Analog output of DAC. With 00(HEX) as input to DAC2, the analog output is -5V.
Similarly, with FF (Hex) as input, the output is +5V. Outputting digital data 00 and FF at
regular intervals, to DAC2, results in a square wave of amplitude I5 Volts
ALGORITHM:
1. Load the initial value (00) to Accumulator and move it to DAC.
2. Call the delay program
3. Load the final value (FF) to accumulator and move it to DAC.
4. Call the delay program.
5. Repeat steps 2 to 5.
PROGRAM:
ADDRESS OPCO
DE
LABEL MNEMON ICS OPERAND COMMENT
4100 START MVI A, 00 Move 00 to A register
4101
4102 OUT C8 Load C8 to output port
4103
4104 CALL DELAY Call delay program
4107 MVI A, FF Load FF to B register
4109 OUT C8
410B CALL DELAY
410E JMP START START Jump to start of address
4111 DELAY MVI B, 05 Move 05 to B register

4113 L1 MVI C, FF Move FF to C register
4116 JNZ L2 Jump to L2 if no zero
4119 DCR B Decrement B register
411A JNZ L1 Jump to L1 if no zero
411D RET
Execute the program and using a CRO, verify that the waveform at the DAC2 output is a
square-wave. Modify the frequency of the square-wave, by varying the time delay.
(b) SAW TOOTH GENERATION:
ALGORITHM:
1. Load the initial value (00) to Accumulator
2. Move the accumulator content to DAC.
3. Increment the accumulator content by 1.
4. Repeat steps 3 and 4.
Output digital data from 00 to FF constant steps of 01 to DAC1 repeat this sequence again and
again. As a result a saw – tooth wave will be generated at DAC1 output.
PROGRAM:
ADDRESS OPCO
DE
LABEL MNEMON ICS OPERAN
D
COMMENT
4100 START MVI A, 00 Load 00 to accumulator
4102 L1 OUT C0 Load CO in output port
4104 INR A Increment A register
4108 JMP START Go to START
unconditionally
(c) TRIANGULAR WAVE GENERATION:
ALGORITHM:
1. Load the initial value (00) to Accumulator.
2. Move the accumulator content to DAC
4. If accumulator content is zero proceed to next step. Else go to step 3.
5. Load value (FF) to accumulator.
7. Decrement the accumulator content by 1.
8. If accumulator content is zero go to step 2. Else go to step 2.

The following program will generate a triangular wave at DAC2 output.
PROGRAM:
ADDRESS OPCOD
E
LABE
L
MNEMON ICS OPERA
ND
COMMENT
4100 START MVI L, 00 Move 00 to L register
4102 L1 MOV A, L Load L to a register
4103 OUT C8 Load c8 to output port
4105 INR L Increment L register
4109 MVI L, FF Load FF to L register
410B L2 MOV A, L Move L to a register
410C OUT C8 Load C8 to output port
410E DCR L Decrement L register
410F JNZ L2 Jump to L2 if no zero
4112 JMP START Go to START unconditionally
DAC - CIRCUIT:

WAEFORMS:
OBSERVATION:
WAVE FORMS AMPLITUDE TIME PERIOD
Square waveform
Saw tooth waveform
Triangular waveform

Result:
Thus the conversion of an analog signal into a digital signal and a digital signal into an
analog signal was done using interfacing of ADC and DAC respectively with 8085.

EX.No:6 TRAFFIC LIGHT CONTROLLERSWITH 8085
AIM
To write an assembly language program to simulate the traffic light at an intersection
using a traffic light interface.
APPARATUS REQUIRED:
3 Traffic light interface kit Vi Microsystems 1
ALGORITHM:
1. Initialize the ports.
2. Initialize the memory content, with some address to the data.
3. Read data for each sequence from the memory and display it through the ports.
4. After completing all the sequences, repeat from step2.
A SAMPLE SEQUENCE:
1. (a) Vehicles from south can go to straight or left.
(b) Vehicles from west can cross the road.
(c) Each pedestrian can cross the road.
(d) Vehicles from east no movement.
(e) Vehicles from north can go only straight.
2. All ambers are ON, indicating the change of sequence.
3. (a) Vehicles from east can go straight and left.
(b) Vehicles from south can go only left.
(c) North pedestrian can cross the road.
(d) Vehicles from north, no movement.
(e) Vehicles from west can go only straight.
5. (a) Vehicles from north can go straight and left.
(b) Vehicles from east can go only left.
(c) West pedestrian can cross the road.
(d) Vehicles from west, no movement.
(e) Vehicles from south can go only straight.

7. (a) Vehicles from west can go straight and left.
(b) Vehicles from north can go only left.
(c) South pedestrian can cross the road.
(d) Vehicles from south, no movement.
(e) Vehicles from east can go only straight.
9. (a) All vehicles from all directions no movement.
(b) All pedestrian can cross the road.
BIT ALLOCATION:
BIT LED BIT LED BIT LED
PA0 SOUTH LEFT PB0 NORTH LEFT PC0 WEST STRAIGHT
PA1 SOUTH RIGHT PB1 NORTH RIGHT PC1 NORTH STRAIGHT
PA2 SOUTH AMBER PB2 NORTH AMBER PC2 EAST STRAIGHT
PA3 SOUTH RED PB3 NORTH RED PC3 SOUTH STRAIGHT
PA4 EAST LEFT PB4 WEST LEFT PC4 NORTH PD
PA5 EAST RIGHT PB5 WEST RIGHT PC5 WEST PD
PA6 EAST AMBER PB6 WEST AMBER PC6 SOUTH PD
PA7 EAST RED PB7 WEST RED PC7 EAST PD

PATH REPRESENTATION:
CONTROL ----- 0F(FOR 8255 PPI)
PORT A----- 0C
PORT B ----- 0D
PORT C ----- 0E

PROGRAM:
ADD
RESS
OPCODE LABEL MNEMON
ICS
OPERAND COMMENT
4100 21 00 45 START LXI H,DATA Initialize HL reg.to Data
4103 0E 04 MVI A,41 A,41C,04 Initialize C reg with no of
comparisons
4105 7E MOV A,M Move M content to
accumulator
4106 D3 0F OUT CNT Load address 417B to HL
register
4108 23 INX H Increment H register
4109 7E LOOP 1: MOV A,M Move M content to
accumulator
410A D3 0C OUT APRT Output contents of
accumulator to OF port
410C 23 INX H Increment H register
410D 7E MOV A,M Move M content to
accumulator
410E D3 0D OUT BPRT Move M content to
accumulator
4110 CD 1B 41 CALL DELAY Call DELAY address
4113 23 INX H Increment H register
4114 0D DCR C Decrement C register
4115 C2 09 41 JNZ LOOP 1 Jump to LOOP 1 if no zero
4118 C3 00 41 JMP START Jump to start
411B C5 DELAY: PUSH B Push the register B
411C 0E 05 MVI C,05 Move 05 H immediately to
accumulator.
411E 7E LOOP3: LXI D,FFFF Load FF FF in DE register
pair
4121 D3 41 LOOP2 DCX D Decrement D register
4122 CD 66 41 MOV A,D Move D content to
accumulator
4123 EB ORA E OR content of E with
accumulator
4124 13 JNZ LOOP2 Jump To L2 IF no zero
4127 23 DCR C Decrement C register
4128 CD 42 41 JNZ LOOP3 Jump to LOOP 2 if no zero
412B EB POP B POP the register B

412C 7E RET Return to subroutine
4500 80 1A A1 64 DATA DB 80, 1A, A1,
64
4504 A4 81 5A 64 A4,81,5A,64
4508 54 8A B1 A8 DB 54,8A,B1,A8
450C B4 88 DA 68 B4,88,DA,68
4510 D8 1A E8 46 DB D8,1A,E8,46
4514 E8 83 78 86
74
E8,83,78,86,
74
RESULT:
Thus an assembly language program to simulate the traffic light at an intersection using a
traffic light interfaces was written and implemented.

EX.No:7INTERFACING 8251 WITH 8085
AIM:
To write a program to initiate 8251 and to check the transmission and reception
of character.
APPARATUS REQUIRED:
1. 8085 Microprocessor kit
2. 8251 Interface board
3. DC regulated power supply
THEORY:
The 8251 is used as a peripheral device for serial communication and is
programmed by the CPU to operate using virtually any serial data transmission technique.
The USART accepts data characters from the CPU in parallel format and the converts them
in a continuous serial data stream of transmission. Simultaneously, it can receive serial data
streams and convert them into parallel data characters for the CPU. The CPU can read the
status of USART at any time. These include data transmissions errors and control signals.
Prior to starting data transmission or reception, the 8251 must be loaded with a
set of control words generated by the CPU.These control signals define the complete
functional definition of the 8251 and must immediately follow a RESET operation. Control
words should be written in to the control register of 8251. Words should be written in to the
control register of 8251.words should be written in to the control register of
8251.Thesecontrol words are split into two formats.
1. MODE INSTRUCTION WORD
2. COMMAND INSTRUCTION WORD.
1. MODE INSTRUCTION WORD
This format defines the BAUD rate, character length, parity and stop bits required to
work with asynchronous data communication. By selecting the appropriate BAUD
factor synchronous mode, the 8251 can be operated in synchronous mode.
Initializing 8251 using the Mode instructions to the following conditions.
8 bit data
No parity
Baud rate factor (16X)
1 stop bit
Gives a mode command word of 01001110=4E(X)
ALGORITHM

1. Initialize timer (8253) IC
2. Move the Mode command word (4EH) to A reg.
3. Output it port address C2
4. Move the command instruction word (37H) to A reg.
5. Output it to port address C2
6. Move the data to be transfer to A reg.
7. Output it to port address C0.
8. Reset the system
9. Get the data through input port address C0.
10. Store the value in memory
11. Reset the system
PROGRAM:
ADDRESS OPCO
DE
L
A
BE
L
MNEM
ONICS
OPERA
ND
COMMENT
4100 MVI A, 36 Move 36 to A
4102 OUT CE Output contents of accumulator to
CE port
4104 MVI A, 0A Move 0A to accumulator
4106 OUT C8 Output contents of accumulator to
C8 port
4108 MVI A, 00 Move 00 to accumulator
410A OUT C8 Output contents of accumulator to
C8 port
410C MVI A, 4E Move 4E to accumulator
410E OUT C2 Output contents of accumulator to
C2 port
C2 port
C0 port
4118 RST1
4119 IN C0 Input the contents from port C0 to
accumulator
411B STA 4150 Store the output from accumulator
to 4150
411E RST1

SYNCHRONOUS MODE:
S2 S1 EP PEN L2 L1 B2 B1
0 1 0 1
0 0 1 1
5
BIT
6
BIT
7
BIT
8
BIT
EVEN PARITY GENERATION
0-Odd
1-Even
PARITY ENABLE
1-Enable
0-Disable
EXTERNAL SYNC DETECT
1-Sysdetect is an input
0-Sysdetect is an output
SINGLE CHARACTER SYNC
1-Single sync character
0-Double sync character

ASYNCHRONOUS MODE:
S2 S1 EP PEN L2 L1 B2 B1
0 1 0 1
0 0 1 1
Synch
mode
(1 X) (16 X) (64 X)
0 1 0 1
0 0 1 1
5
BIT
6
BIT
7
BIT
8
BIT
0 1 0 1
0 0 1 1
Invalid 61BIT 1.5BIT 2 BIT
EVEN PARITY GENERATION
0-Odd
1-Even
PARITY ENABLE
1-Enable
0-Disable

OBSERVATION:
MEMORY LOCATION INPUT DATA OUTPUT DATA
RESULT:
Thus the program to initiate 8251 was written and the transmission and reception of
character was checked by interfacing 8251 with 8085.

Ex No. 8 INTERFACING 8279 WITH 8085
AIM:
To interface 8279 Programmable Keyboard Display Controller to 8085 Microprocessor.
APPARATUS REQUIRED:
1.8085 Microprocessor toolkit.
3. Regulated D.C. power supply.
PROGRAM:
ADDRESS OPCO
DE
D
COMMENT
4100 START LXI
H, 4130H
H, 4130 Store the 16 bit address in
HL pair
4103 MVI D, 0FHD, 0F Move 0F to D register
4105 MVI A, 10 Move 10 to A
4107 OUT C2 Output the contents of A to
C2 output port
C2H
4109 MVI A, 90HA, CC Move CC to A
410B OUT C2 Output the contents of A to
C2 output port
410D MVI A, 90HA, 90 Move 90 to A
410F OUT
C2H
C2 Output the contents of A to
C2 output port
4111 LOOP MOV A, M Move content of M to A
4112 OUT C0HC0 Output the contents of M to
A
4114 CALL DELAY Call the delay address
4117 INX H Increment H register
4118 DCR
D
D Decrement D register
4119 JNZ LOOP Jump to specified address

411C JMP START STARTSTART Jump to START address
411F DELAY MVI B, A0 Move a to B register
4121 LOOP2 MVI C, FF Move FF to C register
4123 LOOP1 DCR C Decrement C register
4124 JNZ LOOP 1 Jump to LOOP 1 if no zero
4127 DCR B Decrement B register
4128 JNZ LOOP 2 Jump to LOOP 2 if no zero
412B RET
Pointer equal to 4130 .FF repeated eight times
4130 FF
4131 FF
4132 FF
4133 FF
4134 FF
4135 FF
4136 FF
4137 FF
4138 98
4139 68
413ª 7C
413B C8
413C 1C
413D 29
413E FF
413F FF
SEGMENT DEFINITION:

DATA BUS D7 D6 D5 D4 D3 D2 D1 D0
SEGMETS d c b a dp g f e
OBSERVATION:
LETTER 7
SEGMENT
DATA BUS
HEXADECIMAL
D7 D6 D5 D4 D3 D2 D1 D0

Accept a Key and display the same ( Two key lock out)
ADDRESS OPCO
DE
D
COMMENT
4100 MVI B, 08
4102 MVI A, 00
4104 OUT C2 Set mode and display
4106 MVI A, 0CC Clear Display
4108 OUT C2
410A MVI A, 90 Write display
410C OUT C2
410E MVI A, 0FF
4110 BACK OUT C0 Clear the display ram
4112 DCR B
4113 JNZ BACK
4116 LOP IN C2 Pressing of a key
4118 ANI 07
411A JZ LOP
411D MVI A, 40 Set to read FIFO ram
411F OUT C2
4121 IN C0
4123 ANI 0F get the corresponding code
from lookup table
4125 MOV L, A
4126 MVI H, 42
4128 MOV A,M
4129 OUT C0
412B JMP LOP
4200 0C 9F 4A 0B
4204 99 29 28 8F
4208 08 09 88 38
420C 6C 1A 68 E8

Accept a Key and display the same ( N Key rollover)
ADDRESS OPCO
DE
D
COMMENT
4100 MVI B, 08
4102 MVI A, 02
4104 OUT C2 Set mode and display
4106 MVI A, 0CC Clear Display
4108 OUT C2
410A MVI A, 90 Write display
410C OUT C2
410E MVI A, 0FF
4110 BACK OUT C0 Clear the display ram
4112 DCR B
4113 JNZ BACK
4116 LOP IN C2 Pressing of a key
4118 ANI 07
411A JZ LOP
411D MVI A, 40 Set to read FIFO ram
411F OUT C2
4121 IN C0
4123 ANI 0F get the corresponding code
from lookup table
4125 MOV L, A
4126 MVI H, 42
4128 MOV A,M
4129 OUT C0
412B JMP LOP
4200 0C 9F 4A 0B
4204 99 29 28 8F
4208 08 09 88 38
420C 6C 1A 68 E8
RESULT:
Thus 8279 controller was interfaced with 8085 and program for read a key and display and
rolling display were executed successfully.

MICROCONTROLLER

Ex.No: 98051 –BASIC SIMPLE PROGRAM
AIM:
To write an assembly language program to add, subtract, multiply and divide the given
data stored at two consecutive locations using 8051 microcontroller.
a. ADDITION OF TWO 8 BIT NUMBERS
Algorithm:
1. Set DPTR as pointer for data.
2. Move first data from external memory to accumulator and save it in R1 register.
3. Increment DPTR.
4. Move second data from external memory to accumulator
5. Clear R0 register to account for carry.
6. Add the content of R1 register to accumulator.
7. Check for carry. If carry is not set go to step 8. Otherwise go to next step.
8. Increment R0 register.
9. Increment DPTR and save the sum in external memory.
10. Increment DPTR, move carry to accumulator and save it in external memory.
11. Stop

LOAD THE ADDRESS IN DPTR
MOVE THE 1ST
DATA TO A – REG AND
SAVE IT IN R1 REG
INCREMENT DPTR AND MOVE THE 2ND
DATA TO A – REG
CLEAR R0 REGISTER
ADD A – REG WITH R1 REG TO GET SUM
If
CY=1
INCREMENT R0 REG
INCREMENT DPTR AND SAVE A – REG
CONTENT IN MEMORY
INCREMENT DPTR
Start
Stop
MOVE R0 TO A – REG AND THEN SAVE A – REG
CONTENT IN MEMORY
Yes
No

Label Program Comments
AHEAD:
HERE:
MOV DPTR,#4500
MOVX A,@DPTR
MOV R1,A
INC DPTR
MOVX A,@DPTR
MOV R0,#00
ADD A,R1
JNC AHEAD
INC R0
INC DPTR
MOVX @DPTR, A
INC DPTR
MOV A,R0
MOVX @DPTR, A
SJMP HERE
Load address of 1st
data in DPTR
Move the 1st
data to A
Save the first data in R1
Increment DPTR to point 2nd
data
Load the 2nd
data in A
Clear R0 for the account of carry
Get the sum in A reg
Check carry flag
If carry is set increment R0
Increment DPTR
Save the sum in external memory
Increment DPTR
Move carry to A reg
Save the carry in external memory
Remain idle in infinite loop
Observation:
Input:
4500: [Addend]
4501: [Augend]
Output:
4502: [Sum]
4503: [Carry]

MOVE THE MINUEND TO A – REG AND
SAVE IT IN R1 REG
INCREMENT DPTR AND MOVE
THE SUBTRAHEND TO A – REG
CLEAR R0 REGISTER AND CARRY FLAG
EXCHANGE R1 WITH A - REG
If
Cy=1
COMPLEMENT A AND
THEN INCREMENT
INCREMENT DPTR AND SAVE A – REG
CONTENT (DIFFERENCE) IN MEMORY
INCREMENT DPTR
Start
Stop
MOVE R0 TO A – REG AND THEN SAVE A – REG
CONTENT (SIGN) IN MEMORY
SUBTRACT R1 FROM A - REG
INCREMENT R0 REG
Yes
No

b. SUBTRACTION OF TWO 8 BIT NUMBERS
Algorithm:
1. Set DPTR as pointer for data.
2. Move the minuend from external memory to accumulator and save it in R1 register.
3. Increment DPTR.
4. Move subtrahend from external memory to accumulator
5. Exchange the contents of R1 and A such that minuend is in A and subtrahend is in
R1
6. Clear R0 register to account for sign.
7. Clear carry flag.
8. Subtract the content of R1 register from accumulator.
9. Check for carry. If carry is not set go to step 12. Otherwise go to next step.
10. Complement the content of A – reg and increment by 1 to get 2’s complement of
result in A – reg
11. Increment R0 register.
12. Increment DPTR and save the result in external memory.
13. Increment DPTR, move R0 (sign bit) to accumulator and then save it in external
memory.
14. Stop

AHEAD:
HERE:
MOV DPTR,#4500
MOVX A,@DPTR
MOV R1,A
INC DPTR
MOVX A,@DPTR
XCH A,R1
MOV R0,#00
CLR C
SUBB A,R1
JNC AHEAD
CPL A
INC A
INC R0
INC DPTR
MOVX @DPTR,A
INC DPTR
MOV A,R0
MOVX @DPTR,A
SJMP HERE
Load address of minuend in DPTR
Move the minuend to A
Save the minuend in R1
Increment DPTR to point subtrahend
Load the subtrahend in A
Get minuend in A and Subtrahend in R1
Clear R0 for the account of Sign
Clear carry
Subtract R1 from A
Check Carry flag. If carry is set then
Get 2’s complement of result in A
Set R0 to indicate negative sign
Increment DPTR
Save the result in external memory
Increment DPTR
Move sign bit to A reg
Save the sign in external memory
Observation:
Input:
4500: [Minuend]
4501: [Subtrahend]
Output:
4502: [Difference]
4503: [Sign Bit]

MOVE THE 1ST
DATA TO A – REG AND
SAVE IT IN B REG
INCREMENT DPTR AND MOVE
THE 2ND
DATA TO A – REG
MULTIPLY A AND B
INCREMENT DPTR
INCREMENT DPTR
MOVE B (HIGHER BYTE OF PRODUCT) TO A –
REG AND THEN SAVE A – REG CONTENT IN
MEMORY
SAVE A – REG CONTENT (LOWER BYTE OF
PRODUCT) IN MEMORY
Start
Stop

c. MULTIPLICATION OF TWO 8 BIT NUMBERS
Algorithm:
1. Load address of data in DPTR
2. Move the first data from external memory to A and save in B.
3. Increment DPTR and move second data from external memory to B.
4. Perform multiplication to get the product in A and B.
5. Increment DPTR and save A ( lower byte of product) in memory
6. Increment DPTR , move B ( lower byte of product) to A and save it in memory
7. Stop
HERE:
MOV DPTR,#4500
MOVX A,@DPTR
MOV B,A
INC DPTR
MOVX A,@DPTR
MUL AB
INC DPTR
MOVX @DPTR,A
INC DPTR
MOV A,B
MOVX @DPTR,A
SJMP HERE
Load address of 1st
data in DPTR
Move the 1st
data to A
Save the 1st
data in B
Increment DPTR to point 2nd
data
Load the 2nd
data in A
Get the product in A and B
Increment DPTR
Save the lower byte of result in external memory
Increment DPTR
Move the higher byte of product to A reg
Save it in external memory

Observation
Input:
4500: 22 [1st
data]
4501: 43 [2nd
data]
Output:
4502: E6 [Lower byte of product]
4503: 08 [Higher byte of product]

LOAD THE DIVIDEND TO A – REG AND
SAVE IT IN R0 REG
INCREMENT DPTR
LOAD THE DIVISOR IN A – REG AND SAVE IT IN B - REG
INCREMENT DPTR
INCREMENT DPTR
MOVE B (REMAINDER) TO A – REG AND THEN
SAVE A – REG CONTENT IN MEMORY
SAVE A – REG CONTENT (QUOTIENT) IN
MEMORY
MOVE THE DIVIDEND FROM R0 TO A - REG
DIVIDE A – REG CONTENT BY B – REG
Start
Stop

d. DIVISION OF TWO 8 BIT NUMBERS
Algorithm:
1. Load address of data in DPTR
2. Move the dividend from external memory to A and save it in R0 register.
3. Increment DPTR and move the divisor from external memory to A and save it in B
reg.
4. Move the dividend from R0 to A.
5. Perform division to get quotient in A and remainder in B.
6. Increment DPTR and save quotient (content of A - reg) in memory
7. Increment DPTR.
8. Move the remainder (Content of B – reg) to A and save in memory.
9. Stop
HERE:
MOV DPTR,#4500
MOVX A,@DPTR
MOV R0,A
INC DPTR
MOVX A,@DPTR
MOV B,A
MOV A,R0
DIV AB
INC DPTR
MOVX @DPTR,A
INC DPTR
MOV A,B
MOVX @DPTR,A
SJMP HERE
Load address of dividend in DPTR
Move the dividend to A
Save the dividend in R0
Increment DPTR to point divisor
Load the divisor in A
Move the divisor to B
Move the dividend to A
Divide the content of A by B
Increment DPTR
Save the quotient in external memory
Increment DPTR
Move the remainder to A reg
Save it in external memory

Observation:
Input:
4500: [Dividend]
4501: [Divisor]
Output:
4502: [Quotient]
4503: [Remainder]
Result:
Thus the addition, subtraction, multiplication and division of two numbers were
performed using the 8051 microcontroller.

Ex N0. 10a SUM OF ELEMENTS IN AN ARRAY
AIM:
To find the sum of elements in an array.
ALGORITHM:
1. Load the array in the consecutive memory location and initialize the
memory pointer with the starting address.
2. Load the total number of elements in a separate register as a counter.
3. Clear the accumulator.
4. Load the other register with the value of the memory pointer.
5. Add the register with the accumulator.
6. Check for carry, if exist, increment the carry register by 1. otherwise,
continue
7. Decrement the counter and if it reaches 0, stop. Otherwise increment the
memory pointer by 1 and go to step 4.

Yes
Yes
[A]  [DPTR]
[R0]  [A]
[DPTR]  [4200H],
Clear B, R1 and CY
[A]  [DPTR+1]
[A]  [A] +[B]
[A]  [B]
Is Cy=1
[R1] [R1]+1
[4500]  [A] [B]
[4501][A] [R1]
START
[R0]  [R0]-1
IS
[R0] = 0?
STOP
No
No

PROGRAM:
AGAIN
NC
HERE
MOV DPTR, #4200
MOVX A, @DPTR
MOV R0, A
MOV B, #00
MOV R1, B
CLR C
INC DPTR
MOVX A, @DPTR
ADD A, B
MOV B, A
JNC NC
INC R1
DJNZ R0,AGAIN
MOV DPTR, #4500
MOV A, B
MOVX @DPTR, A
INC DPTR
MOV A, R1
MOVX @DPTR, A
SJMP HERE
Load 4200 to DPTR, get array size
Copy array size value to A
Move contents of A to R0
Clear B, R1 & Carry Flag
Get value from datapointer
Sum of two numbers
If no carry jump to NC
Else increment R1
Decrement size of array, if not zero fetch data from
DPTR and add to the resultant value
Store the result in 4500 memory locations

OBSERVATION:
INPUT OUTPUT
4200(Array Size) 05 4500
4201
4202
45014203
4204
4205
RESULT:
The sum of elements in an array is calculated.

Exp. No 10(b)8051 –Check Odd or Even using Call Option
AIM:
To check whether given number is Odd or Even using call function.
ALGORITHM:
1. Start
2. Move the data to DPTR
3. Move the data to accumulator
4. Call check function whether given number is odd or even
5. Increment DPTR
6. Resultant value stored in DPTR-4201
7. Halt
Check Function
8. Rotate right through carry
9. If carry copy 00 to B reg (odd number)
10. Else copy EE to B reg (even number)
11. return

[DPTR]  [4200H]
[A]  [DPTR]
[DPTR]  [DPTR] +1
[A]  [B]
[DPTR]  [A]
START
STOP
Check Even &
odd
NoYes
Is CY=1
Rotate the A right
through carry
[B] 00H, for odd [B] EEH, for even
Check Even &
odd
Return

PROGRAM:
HERE
CHECK_EV_OD
L1
LAST
MOV DPTR, #4200
MOVX A, @DPTR
ACALL CHECK_EV_OD
INC DPTR
MOV A,B
MOVX @DPTR, A
SJMP HERE
RRC A
JC L1
MOV B,#0EEH
SJMP LAST
MOV B,#00H
RET
Load 4200 to DPTR, get a number
Copy value to A
Call check function whether given number is
odd or even
Resultant value stored in DPTR-4201
Rotate right through carry
If carry copy 00 to B reg (odd number)
Else copy EE to B reg (even number)
OBSERVATION:
INPUT OUTPUT
4200 4201
RESULT:
Thus the given number is either even or odd has been found using 8051 microcontroller.

11INTERFACING A/D & D/A CONVERTER WITH 8051
a. ADC INTERFACING WITH 8051
APPARATUS REQUIRED:
1 Microcontroller kit 8051,Vi Microsystems 1
3 ADC Interface board Vi Microsystems 1
PROBLEM STATEMENT:
To program starts from memory location 4100H. The program is executed for various
values of analog voltage which are set with the help of a potentiometer. The LED display is
verified with the digital value that is stored in the memory location 4150H.
THEORY:
An ADC usually has two additional control lines: the SOC input to tell the ADC when
to start the conversion and the EOC output to announce when the conversion is complete. The
following program initiates the conversion process, checks the EOC pin of ADC 0419 as to
whether the conversion is over and then inputs the data to the processor. It also instructs the
processor to store the converted digital data at RAM 4200H.
ALGORITHM:
1. Select the channel and latch the address.
2. Send the start conversion pulse.
3. Read EOC signal.
4. If EOC =1 continue else go to step (3)
5. Read the digital output.
6. Store it in a memory location.

PROGRAM:
WAIT
HERE
MOV DPTR, #FFC8
MOV A,#10
MOVX @DPTR, A
MOV A,#18
MOVX @DPTR, A
MOV DPTR, #FFD0
MOV A,#01
MOVX @DPTR, A
MOV A,#00
MOVX @DPTR, A
MOV DPTR, #FFD8
MOVX A,@DPTR
JNB E0,WAIT
MOV DPTR,#FFC0
MOVX A,@DPTR
MOV DPTR,#4150
MOVX @DPTR, A
SJMP HERE
Select Channel 0 and make ALE Low
make ALE High
SOC signal High
SOC signal low
Check for EOC
Read ADC data
Store the data in memory location

ADC- CIRCUIT:
SOC JUMPER SELECTION:
J2: SOC Jumper selection

J5: Channel selection
OBSERVATION
ANALOG VOLTAGE DIGITAL DATA ON
LED DISPLAY
HEX CODE IN
LOCATION 4150

(b)INTERFACING D/A CONVERTER WITH 8051
APPARATUS REQUIRED:
3 DAC Interface board Vi Microsystems 1
THEORY:
SOFTWARE EXAMPLES
After going through the software examples you can learn how to control the
DAC using 8051 and generate sine wave, saw tooth wave etc by means of software.
ALGORITHM:
(a) SQUARE WAVE GENERATION:
1. Load the initial value (00) to Accumulator and move it to DAC.
2. Call the delay program
3. Load the final value (FF) to accumulator and move it to DAC.
4. Call the delay program.
5. Repeat steps 2 to 5.

DAC - CIRCUIT:
WAVEFORMS:

OBSERVATION:
Square waveform
Saw tooth waveform
Triangular waveform
PROGRAM:
The basic idea behind the generation of waveforms is the continuous generation of
Analog output of DAC.
With 00(HEX) as input to DAC2, the analog output is -5V. Similarly, with FF (Hex) as
input, the output is +5V. Outputting digital data 00 and FF at regular intervals, to DAC2,
results in a square wave of amplitude I5 Volts.
ADDRESS LABEL MNEMON ICS OPCODE OPERAND COMMENT
MOV DPTR,#FFC8
START MOV A,#00
MOVX @DPTR,A
LCALL DELAY
MOV A,# FF
MOVX @DPTR,A
LCALL DELAY
LJMP START
DELAY MOV R1,#05
LOO[P MOV R2,#FF
DJNZ R2,HERE
DJNZ R1,LOOP
RET
SJMP START
Execute the program and using a CRO, verify that the waveform at the DAC2 output is
a square-wave. Modify the frequency of the square-wave, by varying the time delay.
(b) SAW TOOTH GENERATION
1. Load the initial value (00) to Accumulator
4. Repeat steps 3 and 4.
Output digital data from 00 to FF constant steps of 01 to DAC1 repeat this sequence again and
again. As a result a saw – tooth wave will be generated at DAC1 output.

PROGRAM:
MOV DPTR,#FFC0
MOV A,#00
LOOP MOVX @DPTR,A
INC A
SJMP LOOP
(c) TRIANGULAR WAVE GENERATION
1. Load the initial value (00) to Accumulator.
2. Move the accumulator content to DAC
4. If accumulator content is zero proceed to next step. Else go to step 3.
5. Load value (FF) to accumulator.
7. Decrement the accumulator content by 1.
8. If accumulator content is zero go to step 2. Else go to step 2.
The following program will generate a triangular wave at DAC2 output. The program is
self explanatory.
MOV DPTR,#FFC8
START MOV A,#00
LOOP1 MOVX @DPTR,A
INC A
JNZ LOOP1
MOV A,#FF
LOOP2 MOVX @DPTR,A
DEC A
JNZ LOOP2
LJMP START
OBSERVATION:
Square waveform
Saw tooth waveform
Triangular waveform

Result:
Thus the square, triangular and saw tooth wave form were generated by interfacing
DAC with 8051 trainer kit

Ex.No:14 INTERFACING 8253 TIMER WITH 8085
AIM:
To interface 8253 Interface board to 8085 microprocessor to demonstrate the generation of
square wave.
APPARATUS REQUIRED:
1. 8085 microprocessor kit
3. DC regulated power supply
4. CRO.
.
PROGRAM:
Address Opcodes Label Mnemonic Operands Comments
4100 3E 36 START: MVI A, 36 Channel 0 in mode 3
4102 D3 CE OUT CE Send Mode Control word
4104 3E 0A MVI A, 0A LSB of count
4106 D3 C8 OUT C8 Write count to register
4108 3E 00 MVI A, 00 MSB of count
410A D3 C8 OUT C8 Write count to register
410C 76 HLT
Set the jumper, so that the clock 0 of 8253 is given a square wave of frequency 1.5
MHz. This program divides this PCLK by 10 and thus the output at channel 0 is 150 KHz.
Vary the frequency by varying the count. Here the maximum count is FFFF H. So, the
square wave will remain high for 7FFF H counts and remain low for 7FFF H counts. Thus
with the input clock frequency of 1.5 MHz, which corresponds to a period of 0.067
microseconds, the resulting square wave has an ON time of 0.02184 microseconds and an OFF
time of 0.02184 microseconds.
To increase the time period of square wave, set the jumpers such that CLK2 of 8253 is
connected to OUT 0. Using the above-mentioned program, output a square wave of frequency
150 KHz at channel 0. Now this is the clock to channel 2.

CONTROL WORD:
SC1 SC2 RW1 RW0 M2 M1 M0 BCD
SC-SELECT COUNTER:
SC1 SC0 SELECT COUNTER
0 0 Select counter 0
1 1 Read back command
M-MODE:
M2 M1 M0 MODE
0 0 0 Mode 0
0 0 1 Mode 1
X 1 0 Mode 2
X 1 1 Mode 3
1 0 0 Mode 4
1 0 1 Mode 5
READ/WRITE:
RW1 RW0
0 0 Counter latch command
0 1 R/W least significant bit only
1 0 R/W most significant bit only
1 1 R/W least sig first and most sig byte

BCD:
0 Binary counter 16-bit
1 Binary coded decimal counter
Result:
Thus the 8253 has been interfaced to 8085 p and with different modes of 8253 have
been studied.

Ex.No:15STEPPER MOTOR INTERFACING WITH 8051
AIM:
To interface a stepper motor with 8051 microcontroller and operate it.
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 has 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.
2-PHASE SWITCHING SCHEME:
In this scheme, any two adjacent stator windings are energized. The switching scheme
is shown in the table given below. This scheme produces more torque.
ANTICLOCKWISE CLOCKWISE
STEP A1 A2 B1 B2 DATA STEP A1 A2 B1 B2 DATA
1 1 0 0 1 9h 1 1 0 1 0 Ah
2 0 1 0 1 5h 2 0 1 1 0 6h
3 0 1 1 0 6h 3 0 1 0 1 5h
4 1 0 1 0 Ah 4 1 0 0 1 9h
ADDRESS DECODING LOGIC:
The 74138 chip is used for generating the address decoding logic to generate the device
select pulses, CS1 & CS2 for selecting the IC 74175.The 74175 latches the data bus to the
stepper motor driving circuitry.
Stepper Motor requires logic signals of relatively high power. Therefore, the interface
circuitry that generates the driving pulses use silicon Darlington pair transistors. The inputs for
the interface circuit are TTL pulses generated under software control using the Microcontroller
Kit. The TTL levels of pulse sequence from the data bus are translated to high voltage output
pulses using a buffer 7407 with open collector.

BLOCK DIAGRAM:
REPRESENTATION:
8051
MICROCONTROLLER 8255
DRIVER CIRCUIT STEPPER MOTOR

PROGRAM :
Label
MNEM
ONICS
OPERAND Comments
ORG 4100h
START MOV DPTR, #TABLE Load the start address of
switching scheme data
TABLE into Data Pointer
(DPTR)
MOV R0, #04 Load the count in R0
LOOP: MOVX A, @DPTR Load the number in
TABLE into A
PUSH DPH Push DPTR value to
StackPUSH DPL
MOV DPTR, #0FFC0h Load the Motor port
address into DPTR
MOVX @DPTR, A Send the value in A to
stepper Motor port
address
MOV R4, #0FFh Delay loop to cause a
specific amount of time
delay before next data
item is sent to the Motor
DELAY: MOV R5, #0FFh
DELAY1: DJNZ R5, DELAY1
DJNZ R4, DELAY
POP DPL POP back DPTR value
from StackPOP DPH
INC DPTR Increment DPTR to point
to next item in the table
DJNZ R0, LOOP Decrement R0, if not zero
repeat the loop
SJMP START Short jump to Start of the
program to make the
motor rotate continuously
TABLE: DB 09 05 06 0Ah Values as per two-phase
switching scheme

PROCEDURE:
1. Enter the above program starting from location 4100.and execute the same.
2. The stepper motor rotates.
3. Varying the count at R4 and R5 can vary the speed.
4. Entering the data in the look-up TABLE in the reverse order can vary direction of
rotation.
RESULT:
Thus a stepper motor was interfaced with 8051 and run in forward and reverse
directions at various speeds.

Ex No. 12 8085 Programming Practices with Simulators/ Jubin Open source Software
Aim
To write an assembly language program to multiply the given data stored at two consecutive
locations using 8085 microprocessor simulator.
Motivation
Understanding of Intel 8085 microprocessor is fundamental to getting insight into the Von-
Neumann Architecture. It was first introduced in 1976, since then many generations of
computer architecture have come up, some still persists while others are lost in history. This
microprocessor still survives because it is still popular in university and training institutes to
get students acquainted with basic computer architecture. For this purpose 8085 trainer kit
are available on the market.
With this academic learning purpose in mind 8085 simulator software is designed. It helps
in get started easily with example codes, and to learn the architecture playfully. It also
provides a trainer kit as an appealing functional alternative to real hardware. The users can
write assembly code easily and get results quickly without even having the actual hardware.
Jubin 8085 simualor is open source software which is available at
https://8085simulator.codeplex.com/downloads/get/86234
Features
1. Assembler Editor
• Can load Programs written in other simulator
• Auto-correct and auto-indent features
• Supports assembler directives
• Number parameters can be given in binary, decimal and hexadecimal format
• Supports writing of comments
• Supports labeling of instructions, even in macros
• Has error checking facility
• Syntax Highlighting
2. Disassembler Editor
• Supports loading of Intel specific hex file format
• It can successfully reverse trace the original program from the assembly code, in
most of the cases
• Syntax Highlighting and Auto Spacing

3. Assembler Workspace
• Contains the Address field, Label, Mnemonics, Hex-code, Mnemonic Size, M-
Cycles and T-states
• Static Timing diagram of all instruction sets are supported
• Dynamic Timing diagram during step by step simulation
• It has error checking facility also
4. Memory Editor
• Can directly update data in a specified memory location
• It has 3 types of interface, user can choose from it according to his need.
– Show entire memory
content
– Show only loaded memory location
– Store directly to specified memory location
• Allows user to choose memory range
5. I/O Editor
• It is necessary for peripheral interfacing.
• Enables direct editing of content
6. Interrupt Editor
• All possible interrupts are supported. Interrupts are triggered by pressing the
appropriate column (INTR, TRAP, RST 7.5, RST 6.5, RST 5.5) on the interrupt
table. The simulation can be reset any time by pressing the clear memory in the
settings tab.
7. Debugger
• Support of breakpoints
• Step by step execution/debugging of program.
• It supports both forward and backward traversal of programs.
• Allows continuation of program from the breakpoint.
8. Simulator
• There are 3 level of speed for simulation:
– Step-by-step −→ Automatic line by line execution with each line
highlighting. The time to halt at each line is be decided by the user.
– Normal −→ Full execution with reflecting intermittent states periodically.

– Ultimate −→ Full execution with reflecting final state directly.
• There are 2 modes of simulator engine:
– Run all at a Time −→ It takes the current settings from the simulation speed
level and starts execution accordingly.
– Step by Step −→ It is manual mode of control of FORWARD and
BACKWARD traversal of instruction set. It also displays the in-line comment
if available for currently executed instruction.
• Allows setting of starting address for the simulator
• Users can choose the mnemonic where program execution should terminate
9. Helper
• Help on the mnemonics is integrated
• CODE WIZARD is a tool added to enable users with very little knowledge of
assembly code could also 8085 assembly programs.
• Already loaded with plenty SAMPLE programs
• Dynamic loading of user code if placed in user_code folder
• It also includes a user manual
10. Printing
• Assembler Content
• Workspace Content
11. Register Bank −→ Each register content is accompanied with its equivalent binary value
• Accumulator, Reg B, Reg C, Reg D, Reg E, Reg H, Reg L, Memory (M)
• Flag Register
• Stack Pointer (SP)
• Memory Pointer (HL)
• Program Status Word (PSW)
• Program Counter (PC)
• Clock Cycle Counter
• Instruction Counter
• Special blocks for monitoring Flag register and the usage of SIM and RIM
instruction
12. Crash Recovery
• Can recover programs lost due to sudden shutdown or crash of application

13. 8085 TRAINER KIT
• It simulates the kit as if the user is working in the lab. It basically uses the same
simulation engine at the back-end
14. TOOLS
• Insert DELAY Subroutine TOOL
– It is a powerful wizard to generate delay subroutine with user defined delay
using any sets of register for a particular operating frequency of 8085
microprocessor.
• Interrupt Service Subroutine TOOL
– It is a handy way to set memory values at corresponding vector interrupt
address
• Number Conversion Tool
– It is a portable interconversion tool for Hexadecimal, decimal and binary
numbers. So, that user do not need to open separate calculator for it.
Directives
Assembler
Directives Example
Description
ORG (Origin)
# ORG C000H The next block of instruction should be stored
in memory locations starting at C000H
BEGIN (Start) # BEGIN 2000H To start simulation from address 2000H
END (Stop) # END End of Assembly. It places the
mnemonic defined at "Settings → Stop
Simulation at Mnemonic"
EQU (Equal) # OUTBUF EQU
3945H
The value of the label OUTBUF
is 3945H. This may be used as
memory location.
DB (Define Byte) # DATA: DB F5H,12H Initializes an area byte by byte,in successive
memory locations until all values are stored. Label
DATA stores the initial address.

DW (Define Word) # LABEL: DW 2050H Initializes an area two bytes at a time.
DS (Define Storage) # STACK: DS 4 Reserves a specified number of memory
locations and set the initial address to label
STACK.
Debugging Mode
The debug mode allows the user to view and/or manipulate the program’s internal state for the
purpose of debugging. The software allows step wise or block wise line monitor with both
forward and backward traversal facility.
PROGRAM
#ORG 2000
#BEGIN 2000
LXI H, 4200
MOV B,M
INX H
MVI C,00
MVI A,00
L2: ADD M
JNC L1
INR C
L1: DCR B
JNZ L2
INX H
MOV M,A
INX H
MOV M,C
HLT
#ORG 4200
#DB 23,42
Click
ASSEMBLE, Run all at a Time and then Memory

Observation
Input Output
4200 4202
4201 4203
Result
Thus the multiplication of two numbers was performed using the 8085 microprocessor
simulator.

Ee6612 miroprocessor and microcontroller laboratory

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