Upload
ajeyhareprasath
View
21.273
Download
1
Embed Size (px)
Citation preview
AARUPADAI VEEDU INSTITUTE OF TECHNOLOGY
PAIYANOOR
LAB MANUAL
MICROPROCESSORS AND MICROCONTROLLER
LABORATORY
DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
AARUPADAI VEEDU INSTITUTE OF TECHNOLOGY
OLD MAHABALIPURAM ROAD
LIST OF EXPERIMENTS
1. Programming 8085.
2. Programming 8086.
3. Programming 8051.
4. Pc based control systems.
5. Stepper Motor Controller.
6. PC interfacing.
7. AC & DC Motor Speed Control.
LIST OF EXPERIMENTS
A.8085 Programming
1. Study of 8085 Microprocessor
2. 16 bit Addition, Subtraction, Multiplication and Division.
3. BCD to HEX and HEX to BCD code conversion.
4. Largest and smallest of a given set of numbers.
5. Square Root of 8-bit number.
B.8086 Programming
1. Study of 8086 Microprocessor
2. 32 bit Addition and Addition of 3*3Matrices.
3. Ascending Order and Descending Order.
4. Reversal of a String.
C. Interfacing Experiments
1. Keyboard and Display Interface
2. ADC Interface.
3. DAC Interface.
4. Stepper Motor Interface.
5. Traffic Signal Modeling.
D.8051 Programming
Study of 8051.
1. 8 Bit Arithmetic Operations.
Addition, Subtraction, Multiplication and Division
2. Addition of two sixteen bit numbers.
3. Transferring a Block of data from
(i). Internal to External memory.
(ii). External to External memory.
(iii). External to Internal memory.
4. 8-bit Conversion
(i).ASCII to its equivalent Hexa decimal and
(ii). Hexa decimal to its equivalent ASCII
5. Arrange the given numbers in ascending and descending order.
6. Filling External and Internal Memory.
E. Interfacing Experiments
1. DAC Interfacing.
2. Stepper Motor Interfacing.
Experiment No: Date:
ADDITION OF TWO 8-BIT NUMBERS
AIM:
Write a program to add two 8 bit numbers and store the result in memory location
4200 and 4201.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard
(ii) Power cable.
ALGORITHM:
Step 1: Initialize register C to account for carry.
Step 2: The addend is brought to the accumulator.
Step 3: Move the augend to any one of the register.
Step 4: Perform the addition.
Step 5: Store the 8 bit and carry in two memory locations.
Step 6: Stop the process.
FLOWCHART
Start
Initialize C register to zero for carry
Get first data in Accumulator
Get second data in B register
ADD two numbers
Is carry NO
Flag
Set?
YES
Increment C register by one
Store the Sum & Carry in the desired memory locations
Stop
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100H MVI C,00H Initialize C reg. to store
carry.
4102H LXI H,4150H Initialize HL reg. pair to
point 1st data.
4105H MOV A,M Move 1st data to
accumulator.
4106H INX H Increment HL reg. pair to
point 2nd
data.
4107H MOV B,M Move 2nd
data to B reg.
4108H ADD B Add two numbers in A & B.
4109H JNC LOOP Check whether Carry Flag
is reset.
410CH INR C Increment C register.
LOOP 410DH STA 4200H Store 8 bit sum to memory
location 4200.
4110H MOV A,C Move carry to
Accumulator.
4111H STA 4201H Store carry to memory
4201.
4114H HLT Stop the process.
RESULT:
Thus the program for addition of two 8-bit numbers has been written and executed
and the sum is verified.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA- VOCE QUESTIONS
1. What is Microprocessor ?
2. What are the basic units of a microprocessor ?
3. What is a bus?
4. What is assembly language?
5. Why data bus is bi-directional?
Experiment No: Date:
SUBTRACTION OF TWO 8-BIT NUMBERS
AIM:
Write a program to subtract two 8 bit numbers and store the result in memory location
4502 and 4503.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard
(ii) Power cable
ALGORITHM:
Step 1: Initialize register C to account for Borrow.
Step 2: The Minuend is brought to the accumulator.
Step 3: Move the subtrahend to anyone register.
Step 4: Perform the subtraction.
Step 5: If the carry flag is reset go to step 7, else proceed to the next step.
Step 6: Take 2’s complement of the accumulator content and increment C register.
Step 7: Store the difference & borrow in memory location 4502 and 4503.
Step 8: Stop the process.
FLOWCHART
Start
Initialize C register to zero for Borrow
Get Minuend in Accumulator
Get Subtrahend in any other register
Subtract Subtrahend from Accumulator
Is carry NO
Flag
Set?
YES
Take 2’s Complement of Accumulator& increment C Reg.
Store the Results in the desired memory locations
Stop
PROGRAM :
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4300H MVI C,00H Initialize C reg to store carry.
4302H LXI H,4500H Initialize HL reg pair to point
1st data.
4305H MOV A,M Move 1st data to accumulator.
4306H INX H Increment HL reg pair to point
2nd
data.
4307H MOV B,M Move 2nd
to register B.
4308H SUB B Subtract the subtrahend from
minuend.
4309H JNC SKIP If Carry Flag reset, go to skip.
430CH CMA If the Subtraction result in
borrow take 2’s complement of
the accumulator 430DH ADI 01H
430FH INR C Increment register C
SKIP 4310H INX H Store the result in the desired
memory location 4311H MOV M,A
4312H INX H Store the carry flag to indicate
whether the result is positive or
negative 4313H MOV M,C
4314H HLT Stop the process.
RESULT:
Thus the subtraction program has been written & executed and the difference and
borrow were verified.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA-VOCE QUESTIONS
1. Why address bus is unidirectional?
2. What is the data and address size in 8086?
3. Define machine cycle.
4. Define T-State
5. What is the difference between CPU bus and system bus?
Experiment No: Date:
MULTIPLICATION OF TWO 8-BIT NUMBERS
AIM:
To write an Assembly Language Program for the multiplication of two 8-bit numbers.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard
(ii) Power cable.
ALGORITHM:
Step 1: Initialize the HL registers pair with the address 4200H.
Step 2: Clear C register.
Step 3: Get the multiplicand in accumulator and get the multiplier in B-reg.
Step 4: Move the Multiplicand to D register.
Step 5: Add the content of A register with the content of D register.
Step 6: Check the carry flag if CY=0 go to step 7 otherwise increment the C register.
Step 7: Decrement the B register.
Step 8: If the content of B reg is not equal to zero go to step 5, otherwise go to next
step.
Step 9: Store the content of accumulator in Next memory location.
Step 10: Store the higher order byte in next memory location.
Step 11; Stop the Process.
FLOWCHART
Start
Initialize the HL register pair and get the multiplicand in A & D
Get the multiplier in B-reg & decrement it by one
ADD accumulator (multiplicand) to D reg. content
NO
Is carry flag
Set?
YES
Increment carry register by one
Decrement Multiplier by one
Is multiplier NO
Zero
YES
Store the result in desired location
Stop
PROGRAM :
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100H LXI H,4200H Load the HL reg pair with the
address 4200H
4103H MVI C,00H Clear C reg
4105H MOV A,M Move the Multiplicand to the
accumulator.
4106H INX H Increment the HL reg pair.
4107H MOV B,M Move the Multiplier to B reg.
4108H DCR B Decrement B reg.
4109H MOV D,A Move the Multiplicand to D reg.
LOOP2 410AH ADD D Add the content of A reg with D
reg. content.
410BH JNC LOOP1 If CY=0, program control
jumps to label LOOP1.
410EH INR C Increment C reg.
LOOP1 410FH DCR B Decrement B reg.
4110H JNC LOOP2 If B is not equal to zero,
program control goes to
LOOP2; otherwise do next step
4113H INX H Increment the HL reg. pair.
4114H MOV M,A Store the least significant byte
of the product to the memory
location pointed by HL pair.
4115H INX H Increment the HL reg.
4116H MOV M,C Store the most significant byte
of the product to the memory
location pointed by HL pair.
4117H HLT Stop the Process
RESULT:
Thus the Assembly Language Program for the multiplication of two 8-bit data had
been written and executed & the results were stored in desired memory locations.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What does memory-mapping mean?
2. What is interrupt 1/0?
3. Why EPROM is mapped at the beginning of memory space in 8085
system?
4. What is the need for system clock and how it is generated in 8085?
5. What is the need for Port?
Experiment No: Date:
DIVISION OF TWO 8-BIT NUMBERS
AIM:
To write an Assembly Language Program for dividing two 8-bit numbers and store
the result in memory locations 4502 and 4503.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard
(ii) Power cable.
ALGORITHM:
Step 1: Initialize the HL register pair with address 4500H.
Step 2: Clear the C register.
Step 3: Get dividend in accumulator & get divisor in B register.
Step 4: Compare dividend and divisor.
Step 5: If carry occurs, go to step 8. Otherwise go to next step.
Step 6: Subtract divisor from dividend and increment C register.
Step 7: Go to step 4.
Step 8: Store the remainder in 4503H.
Step 9: store the quotient in 4502H.
Step 10: Stop the Process.
FLOWCHART
Start
Get dividend & divisor, Initialize C reg. for Quotient
Is
Dividend NO
>
Divisor
YES
Subtract Divisor from Dividend
Increment Quotient by one
Store the Quotient and Store the dividend as remainder
Stop
PROGRAM :
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100H LXI H,4500H Load the HL reg pair with the
address 4500H.
4103H MVI C,00H Clear C reg.
4105H MOV A,M Move the Dividend to the
accumulator.
4106H INX H Increment the HL reg pair.
4107H MOV B,M Move the Divisor to B reg.
REPEAT 4108H CMP B Compare B reg.
4109H JC SKIP If carry=1, jump to label
SKIP.
410CH SUB B Subtract the divisor from
dividend.
410DH INR C Increment C reg.
410EH JMP
REPEAT Jump to label REPEAT.
SKIP 4111H STA 4503H Store the remainder.
4114H MOV A,C Move the Quotient to
Accumulator.
4115H STA 4502H Store the Quotient.
4118H HLT Stop the Process
RESULT:
Thus the Assembly Language Program for the Division of two 8-bit data had been
written and executed & the results were stored in desired memory locations.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is a port?
2. Give some examples of port devices used in 8085 microprocessor based
system?
3. Write a short note on INTEL 8255
4. What is the drawback in memory mapped I/0?
5. How DMA is initiated?
Experiment No: Date:
LARGEST NUMBER IN GIVEN ARRAY
AIM:
To write an Assembly Language Program to find the largest number in the given
Array.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Initialize the HL register pair with the address 4700H.
Step 2: Get the first Data in accumulator.
Step 3: Get the count (i.e., Length of the array) in B register.
Step 4: Compare the next data from the memory address pointed by HL pair
to the accumulator.
Step 5: Check the Carry Flag. If the CF is reset go to step 7 otherwise go to
next step.
Step 6: Move the data from the memory address pointed by HL pair to the
accumulator.
Step 7: Decrement the count (content of B register)
Step 8: If the content of B register is not equal to zero, go to step 4,
otherwise go to next step.
Step 9: Store the content of accumulator in the address 5000H.
Step 10: Stop the Process.
FLOWCHART
Start
Get the length of the array as the count
Move the first data to Accumulator
Increment the memory pointer
Compare the second data with content of Accumulator
Is A content YES
Large?
NO
Move M to A
Decrement B reg
NO Is
Count=0?
YES
Store the Content of accumulator as largest number
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4100 LXI H, 4700H Initialize the memory
pointer with the address
4700H 4103 MOV B,M Get the length of the
array in B reg. 4104 INX H Increment the HL reg.
pair to point the next
memory location. 4105 MOV A,M Get the first data in
Accumulator. 4106 DCR B Decrement B reg.
content by one LOOP 4107 INX H Increment the HL reg.
pair to point the next
memory location. 4108 CMP M Compare the data in
accumulator with data in
memory. 410B JNC SKIP If CF is reset, the
program control goes to
label SKIP. 410C MOV A,M Move the data from
memory to accumulator. SKIP 410D DCR B Decrement B reg.
content
411E JNZ LOOP If ZF is reset go to
LOOP.
4111 STA 5000H Store the largest number
in 5000H
4114 HLT Stop the process
RESULT:
Thus the Assembly Language Program to find the largest of the given data had been
written and executed & the result was stored in the desired memory location
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is Instruction cycle?
2. What is fetch and execute cycle?
3. What is Block and Demand transfer mode DMA?
4. What is the need for timing diagram?
5. How many machine cycles constitute one instruction cycle in 8085?
Experiment No: Date:
SMALLEST NUMBER IN GIVEN ARRAY
AIM:
To write an Assembly Language Program to find the smallest number in the given
Array.
APPARATUS REQUIRED:
((i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Initialize the HL register pair with the address 4700H.
Step 2: Get the first Data in accumulator.
Step 3: Get the count (i.e., Length of the array) in B register.
Step 4: Compare the next data from the memory address pointed by HL pair
to the accumulator.
Step 5: Check the Carry Flag. If the CF is set, go to step 7 otherwise go to
next step.
Step 6: Move the data from the memory address pointed by HL pair to the
accumulator.
Step 7: Decrement the count (content of B register)
Step 8: If the content of B register is not equal to zero, go to step 4, otherwise
go to next step.
Step 9: Store the content of accumulator in the address 5000H.
Step 10: Stop the Process.
FLOWCHART
Start
Get the length of the array as the count
Move the first data to Accumulator
Increment the memory pointer
Compare the memory with content of Accumulator
Is A content YES
Small?
NO
Move M to A
Decrement B reg.
NO Is
Count=0?
YES
Store the Content of accumulator as smallest number
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4100 LXI H, 4700H Initialize the memory
pointer with the address
4700H 4103 MOV B,M Get the length of the
array in B reg. 4104 INX H Increment the HL reg.
pair to point the next
memory location. 4105 MOV A,M Get the first data in
Accumulator. 4106 DCR B Decrement B reg.
content by one LOOP 4107 INX H Increment the HL reg.
pair to point the next
memory location. 4108 CMP M Compare the data in
accumulator with data in
memory. 410B JC SKIP If CF is set the program
control goes to SKIP. 410C MOV A,M Move the data from
memory to accumulator. SKIP 410D DCR B Decrement B reg.
content.
411E JNZ LOOP If ZF is reset go to
LOOP.
4111 STA 5000H Store the smallest
number in 5000H
4114 HLT Stop the process
RESULT:
Thus the Assembly Language Program to find the smallest of the given data had
been written and executed & the result was stored in the desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. Define opcode and operand.
2. What is opcode fetch cycle?
3. What operation is performed during first T -state of every machine cycle
in 8085 ?
4. Why status signals are provided in microprocessor?
5 . How the 8085 processor differentiates a memory access and I/0
Access?
Experiment No: Date:
8085-ASCENDING ORDER
AIM:
To write an Assembly Language Program to arrange the given data in Ascending
Order.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Initialize the HL register pair with the address 4700H.
Step 2: Get the count 1, number of repetitions (length of array - 1) in B
register.
Step 3: Get the count 2, number of comparisons (length of array -1) in C
register.
Step 4: Save the count 2 in E register.
Step 5: Move the content of E to C register.
Step 6: Initialize the HL register pair with the address 4701H.
Step 7: Move the data from memory location pointed by HL pair to the
Accumulator.
Step 8: Compare the data from next memory location with the content of
Accumulator.
Step 9: Check CF. If carry flag is set (accumulator content is smaller than the
other number), go to step 12 otherwise do next step.
Step 10: Check ZF. If zero flag is set (accumulator content is equal to the other
number), go to step 12 otherwise do next step.
Step 11: Store the smallest number to the previous memory location & greatest
to the current memory location.
Step 12: Decrement the number of comparisons (count 2).
Step 13: If the number of comparison is not equal to zero, go to step 7,
otherwise do the next step.
Step 14: Decrement the number of repetitions (count 1).
Step 15: If the number of repetitions is not equal to zero, go to step 5,
otherwise do next step.
Step 16: Stop the Process.
FLOWCHART
Start
Initialize memory pointer to get N, the length of series
Initialize COUNT 1 for (N-1) repetitions
Initialize COUNT 2 for (N-1) Comparisons
Increment memory pointer and get the number from series
IS pointer < NO Interchange
(pointer+1)? Numbers
YES
Decrement COUNT 2
NO
Is COUNT 2 =
Zero?
YES
Decrement COUNT 1
NO
Is COUNT 1 =
zero?
YES
Stop
PROGRAM :
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100 LXI H, 4700H Initialize HL reg. pair
with 4700H
4103 MOV B,M
4104 DCR B B = COUNT 1 for
(N-1)repetitions
4105 MOV C,B C = COUNT 2 for
(N-1)Comparisons
4106 MOV E,C LOOP 4107 MOV C,E
4108 LXI H, 4701H REPEAT 410B MOV A,M
410C INX H
410D CMP M Compare consecutive
numbers
411E JC SKIP
4111 JZ SKIP
4114 MOV D,M Interchange numbers if
not in order 4115 MOV M,A
4116 DCX H
4117 MOV M,D
4118 INX H SKIP 4119 DCR C Decrement count 2
411A JNZ REPEAT
411D DCR B Decrement count 1
411E JNZ LOOP
4121 HLT Stop the Process
RESULT:
Thus the Assembly Language Program to arrange the given data in ascending order
had been written and executed.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. When the 8085 processor checks for an interrupt?
2. What is interrupt acknowledge cycle?
3. How the interrupts are affected by system reset?
4. What is Software interrupts?
5. What is Hardware interrupt?
Experiment No: Date:
8085-DESCENDING ORDER
AIM:
To write an Assembly Language Program to arrange the given data in Descending
Order.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with Keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Initialize the HL register pair with the address 4700H.
Step 2: Get the count 1, number of repetitions (length of array - 1) in B
register.
Step 3: Get the count 2, number of comparisons (length of array -1) in C
register.
Step 4: Save the count 2 in E register.
Step 5: Move the content of E to C register.
Step 6: Initialize the HL register pair with the address 4701H.
Step 7: Move the data from memory location pointed by HL pair to the
Accumulator.
Step 8: Compare the data from next memory location with the content of
Accumulator.
Step 9: Check CF. If carry flag is reset (accumulator content is larger than the
other number), go to step 12 otherwise do next step.
Step 10: Check ZF. If zero flag is set (accumulator content is equal to the other
number), go to step 12 otherwise do next step.
Step 11: Store the largest number to the previous memory location & smallest to
the current memory location.
Step 12: Decrement the number of comparisons (count 2).
Step 13: If the number of comparison is not equal to zero, go to step 7, otherwise
do the next step.
Step 14: Decrement the number of repetitions (count 1).
Step 15: If the number of repetitions is not equal to zero, go to step 5, otherwise
do next step.
Step 16: Stop the Process.
FLOWCHART
Start
Initialize memory pointer to get N, the length of series
Initialize COUNT 1 for (N-1) repetitions
Initialize COUNT 2 for (N-1) Comparisons
Increment memory pointer and get the number from series
IS pointer > NO Interchange
(pointer+1)? Numbers
YES
Decrement COUNT 2
NO
Is COUNT 2
= zero?
YES
Decrement COUNT 2
NO
Is COUNT 1
= zero?
YES
Stop
PROGRAM :
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100 LXI H, 4700H
4103 MOV B,M
4104 DCR B B = COUNT 1 for
(N-1)repetitions
4105 MOV C,B C = COUNT 2 for
(N-1)Comparisons
4106 MOV E,C LOOP 4107 MOV C,E
4108 LXI H, 4701H REPEAT 410B MOV A,M
410C INX H
410D CMP M Compare consecutive
numbers
411E JNC SKIP
4111 JZ SKIP
4114 MOV D,M Interchange numbers if
not in order 4115 MOV M,A
4116 DCX H
4117 MOV M,D
4118 INX H SKIP 4119 DCR C Decrement count 2
411A JNZ REPEAT
411D DCR B Decrement count 1
411E JNZ LOOP
4121 HLT Stop the Process
RESULT:
Thus the Assembly Language Program to arrange the given data in ascending order
had been written and executed.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is the difference between Hardware and Software interrupt?
2. What is Vectored and Non- Vectored interrupt?
3. List the Software and Hardware interrupts of 8085?
4. Whether HOLD has higher priority than TRAP or not?
5. What is masking and why it is required?
Experiment No: Date:
BLOCK OF DATA TRANSFER
AIM:
To write an Assembly Language Program to transfer the block of data starting from
memory location 4700H to the memory location 4800H. The total number of data (length of
block) is stored at 46FFH.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard
(ii) Power cord.
ALGORITHM:
Step 1: Initialize the HL register pair with the address 46FFH.
Step 2: Move the total number of data to the C register.
Step 3: Initialize the DE register pair with the address 4800H.
Step 4: Get byte from source memory block.
Step 5: Store the first data to the destination address.
Step 6: Increment the source block pointer.
Step 7: Increment the destination block pointer.
Step 8: Decrement the C register.
Step 9: If ZF=0, go to step 4.otherwise go to next step.
Step 10: Stop the Process.
FLOWCHART
Start
Initialize HL reg. pair as a source
Load the total number of input data In C reg.
Get the Data from source Block to Accumulator
Store data to destination block
Increment the source & destination block pointer
Decrement C reg.
NO Is
ZF=1?
YES
Stop
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100H LXI H,46FFH Initialize the source pointer
with the address 4200H.
4103H MVI C,M Get count of total number of
data bytes in C reg.
4104H INX H Point to first source block
byte.
4105H LXI D,4800H Initialize the destination
pointer with the address
4200H.
REPEAT 4108H MOV A,M Get the data byte from source
block.
4109H STAX D Store the data byte to the
destination block.
410AH INX H Increment source block
pointer.
410BH INX D Increment destination block
pointer.
410CH DCR C Decrement C register.
410DH JNZ
REPEAT If ZF=0, program control
jump to label REPEAT.
4110H HLT Stop the Process
RESULT:
Thus the assembly language program to transfer the block of data from memory
location 4700H to 4800H had been written and verified
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. When the 8085 processor accept hardware interrupt?
2. When the 8085 processor will disable the interrupt system?
3. What is the function performed by Dl instruction?
4. What is the function performed by El instruction?
5. How the vector address is generated for the INTR interrupt of 8085?
Experiment No:
Date:
DECIMAL TO HEXADECIMAL CONVERSION
AIM:
To write an Assembly Language Program to convert a Decimal number to a
Hexadecimal number. The Decimal data is stored in 4200H and Hexadecimal data is store in
4250H.
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Get the Decimal data in A reg. and save in E reg.
Step 2: Mask the lower nibble of the decimal data in A reg.
Step 3: Rotate the upper nibble to the lower position.
Step 4: Clear the Accumulator.
Step 5: Move the 0AH to C reg.
Step 6: Add B reg. content to the A reg. Content.
Step 7: Decrement the C reg. If ZF=0 go to step 6. I f ZF=1 go to next step.
Step 8: Save the product in B reg.
Step 9: Get the decimal in A reg. from E reg. and mask the upper
nibble(units).
Step 10: Add the units (A reg.) to product (B reg.).
Step 11: Stop the Process.
FLOWCHART: Start
Get the Decimal data in A reg. and save it in E reg.
Mask the lower nibble of the Decimal data in A reg.
Rotate the content of A reg. four times right & save it in B reg.
Clear A reg. and move 0AH to C reg.
ADD the content of B reg. to A reg.
Decrement C reg.
NO Check
if ZF=1?
YES
Move the content of A reg. to B reg.
Move the decimal data from E reg. to A reg. and mask the upper nibble
ADD the content of B reg. to A reg.
Store the Hexadecimal value from A reg. to memory
Stop
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100 LDA 4200H Get the data in A reg.
4103 MOV E,A Save in E reg.
4104 ANI F0H Mask the lower nibble
4106 RLC Rotate the upper nibble to
lower Nibble position save
in B reg. 4107 RLC 4108 RLC
4109 RLC 410A MOV B,A
410B XRA A Clear accumulator
410C MVI C,0AH Get the product of units digit
multiplied by 0AH in A reg. REPEAT 410E ADD B
410F DCR C
4110 JNZ REPEAT
4113 MOV B,A Save the product in B reg.
4114 MOVA,E Get the Decimal data in A
reg.
4115 ANI 0FH Mask the upper nibble
4117 ADD B Get the sum of units digit
and product in B reg. 4118 STA 4250H Store the Hexadecimal value
in memory
411B HLT Stop the Process
RESULT:
Thus the Assembly Language Program for Decimal to Hexadecimal conversion had
been written and executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. How clock signals are generated in 8085 and what is the frequency of the
internal clock?
2. What happens to the 8085 processor when it is resetted?
3. What are the operations performed by ALU of 8085?
4. What is a flag?
5. List the flags of 8085
Experiment No:
Date:
ADDITION OF TWO 16 BIT NUMBERS
AIM:
To write an Assembly Language Program to add two 16 bit numbers & store the
result
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Get the first data in DE pair
Step 2: Get the second data in HL pair
Step 3: Add the data in DE & HL pair
Step 4: Store the result in 4204.
Step 5: Stop the Process.
Stop
FLOWCHART: Start
Get the 1st data in DE register pair
Get the 2nd data in HLregister pair
Add DE & HL pair contents
Store the result in 4204 &4205
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100H LHLD 4200H Get first I6-bit number
4103H XCHG Save first I6-bit number in
DE
4104H LHLD 4202H Get second I6-bit number in
HL
4105H DAD D Add DE and HL
4106H SHLD 4204H Store I6-bit result in memory
locations 4004H and 4005H.
4109H HLT Stop the program
RESULT:
Thus the Assembly Language Program for 16 bit addition had been written and
executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. Which interrupt has highest priority in 8085? What is the priority of other
interrupts?
2. What is an ALE?
3. Explain the function of IO/M in 8085.
4. Where is the READY signal used?
5. What is HOLD and HLDA and how it is used?
Experiment No: Date:
SUBTRACTION OF TWO 16 BIT NUMBERS
AIM:
To write an Assembly Language Program to subtract two 16 bit numbers & store the
result
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Get the first data in DE pair
Step 2: Get the second data in HL pair
Step 3: Subtract the lower order byte of the 1st & 2
nd number
Step 4: Subtract the higher order byte of the 1st & 2
nd number
Step 5: Store the result in . in memory locations 4204H and 4205H.
Step6: Stop the process
Stop
FLOWCHART: Start
Get the 1st
& 2nd
data in DE & HL register pair
Subtract the lower order byte of the 1st & 2
nd number
Subtract the higher order byte of the 1st & 2
nd number
Store the result in 4204& 4205
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100H LHLD 4200H Get the 1st data in HL reg
pair
4103H XCHG Save in DE reg. pair
4104H LHLD 4202H Get second 16-bit number in
HL
4107H MOV A,E Get lower byte of the first
number
4108H SUB L Subtract lower byte of the
second number
410BH MOV L,A Store the result in L register
.
410CH MOV A,D Get higher byte of the first
number
410DH SBB H Subtract higher byte of
second number with borrow
410EH MOV H,A Store l6-bit result in
memory locations 4204H
and 4205H.
410FH SHLD 4204H Store l6-bit result in memory
locations 4004H and 4005H
4112H HLT Stop the Process
RESULT:
Thus the Assembly Language Program for 16 bit subtraction had been written and
executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is Polling?
2. What are the different types of Polling?
3. What is the need for interrupt controller?
4. List some of the features of INTEL 8259 (Programmable
Interrupt Controller
5. What is a programmable peripheral device ?
Experiment No: Date:
MULTIPLICATION OF TWO 16 BIT NUMBERS
AIM:
To write an Assembly Language Program to multiply two 16 bit numbers & store the
result
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Get the first data in HL pair
Step 2: Exchange the data with DE pair
Step 3:Get the second data in HL pair
Step 4: Copy the data in Stack Pointer(SP)
Step 5: DAD with SP
Step6: If there is carry, goto step7, else goto step8
Step7: Increment BC pair
Step8: Decrement DE pair
Step9: Move E-reg content to accumulator
Step10: OR DE pair content with accumulator. If the result is not zero, goto step5
Step10: Move BC register pair content to HL pair
Step11: Store the accumulator content
Step12: Stop the process
Stop
FLOWCHART:
Start
Load the HL pair with 1st data. Exchange the data with DE pair
Get the second data in HL pair. Exchange HL pair content with DE pair
Perform double addition
Is NO
there carry?
YES
Increment BC-reg content by 1
Decrement DE register pair by one
Move E-reg content to accumulator
DE reg pair. . OR accumulator content
With DE pair content
NO
Is the result
zero?
YES
Store the result
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START: 4100H LHLD 4500H Load the HL pair with 1st
data
4103H SPHL Exchange the data with DE
pair 4104H LHLD 4502 Get the second data in HL
pair
4107H XCHG Exchange HL pair content
with DE pair
4108H LXI H, 0000H
Initialise HL pair & DE
pair with00H
410BH LXI B, 0000H L2: 410EH DAD SP Perform double addition 410FH JNC L1 If there is no carry, goto L1 4112H INX B Increment the content of
BC reg pair by one L1: 4113H DCX D Decrement the content of
DE reg pair by one
4114H MOV A,E Move E-reg content to
accumulator
4115H ORA D OR accumulator content
with DE reg pair
4116H JNZ L2 Jump to L2 if not zero
4119H SHLD 4504H Store HL pair content in
4504H
411CH MOV A,C Move C-reg content to
accumulator
411DH STA 4506H Store HL pair content in
4506H
4120H MOV A,B Move B-reg content to
accumulator
4121H STA 4507H Store accumulator content
in 4507H
4124H HLT Stop the process
RESULT:
Thus the Assembly Language Program for 16 bit multiplication had been written and
executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1.
Experiment No: Date:
DIVISION OF TWO 16 BIT NUMBERS
AIM:
To write an Assembly Language Program to divide two 16 bit numbers & store the
quotient & remainder
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Get the first data in DE pair
Step 2: Get the second data in HL pair
Step 3: Subtract the lower order byte of the 1st & 2
nd number
Step 4: Subtract the higher order byte of the 1st & 2
nd number
Step 5: Store the result in . in memory locations 4204H and 4205H.
Step6: Stop the process
Stop
FLOWCHART:
Start
Dividend (2200H) & (2201H) , Divisor(2300H), Count=8, Quotient =0
Dividend = Dividend*2, Quotient = Quotient*2
YES
Is divisor
<= dividend Dividend =
? Dividend- Divisor
NO
Count = Count-1
NO Is count
= 0?
YES
Store the quotient & remainder
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START: 4100H MVI E, 00H Initialise Quotient = 0
4102H LHLD 4200H Get dividend
4105H LDA 4204H Get divisor 4108H MOV B, A Store divisor
4109H MVI C, 08 Initialise Count = 8
NEXT: 410BH DAD H Dividend = Dividend x 2
410CH MOV A, E
410DH RLC
410EH MOV E,A Quotient = Quotient x 2
410FH MOV A,H
4110H SUB B Is most significant byte of
Dividend > divisor 4111H JC SKIP No, go to Next step
4114H MOV H,A Subtract divisor
4115H INR E Quotient = Quotient + 1
SKIP: 4116H DCR C Count = Count - 1
4117H JNZ NEXT If count =0 repeat
411AH MOV A, E
411BH STA 4301H Store Quotient
411EH MOV A,H
411FH STA 4303H Store Remainder
4122H HLT Stop the process
RESULT:
Thus the Assembly Language Program for 16 bit division had been written and
executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is synchronous data transfer scheme?
2. What is asynchronous data transfer scheme?
3. What are the operating modes of 8212?
4. Explain the working of a handshake output port
5. What are the internal devices of 8255 ?
Experiment No:
Date:
SQUARE ROOT OF 8 BIT NUMBER
AIM:
To write an Assembly Language Program to find square root of a 8 bit number using
8085
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM:
Step 1: Move 01H to both C-reg & E-reg
Step 2: Load the accumulator with the content of 4200H
Step 3: Subtract accumulator content & C-reg content
Step 4: If accumulator content is zero, goto step7, else goto step5
Step 5: Increment C-reg content by two
Step6: Increment E-reg content by one & return to label
Step7: Move E- reg content to accumulator
Step8: Store the accumulator value in 4201H
Step9: Stop the process
FLOWCHART:
Start
Move 01H to both C-reg & E-reg. Load the accumulator
with the content of 4200H
: Subtract accumulator content & C-reg content
Is
Accumulator
content YES
zero?
NO
Increment C-reg content by two
Increment E-reg content by one &
return to label
Move E- reg content to accumulator
Store accumulator value in 4201H
Stop
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START: 4100H MVI C,01 Move 01 to C- reg
4102H MVI E,01 Move 01 to E- reg
4104H LDA 4200H Load the accumulator with
the content of 4200 LABEL: 4107H SUB C Subtract accumulator content
& C-reg content
4108H JZ LABEL1 If accumulator content is
zero, goto label1
410BH INR C Increment C-reg content by
two 410CH INR C
410DH INR E Increment E-reg content by
one
410EH JMP LABEL Return to label LABEL1: 4111H MOV A,E Move E- reg content to
accumulator
4112H STA 4201H Store the accumulator value
in 4201 4115H HLT Stop the process
RESULT:
Thus the Assembly Language Program for square root of a 8 bit number had been
written and executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is baud rate?
2. What is USART?
3. What are the control words of 8251A and what are its functions ?
4. Give some examples of input devices to microprocessor-based system
5. What are the tasks involved in keyboard interface?
Experiment No: Date:
BCD TO HEX CONVERSION
AIM:
To write an Assembly Language Program to convert a BCD number to a hexadecimal
number using 8085
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM
Step 1: Initialize memory pointer with 4300H
Step 2 Get the MSB & multiply it by 10 using repeated addition
Step 3: Add the LSB to the obtained result
Step4: Store the hexadecimal value in 4201H
Step5: Stop the process
Stop
FLOWCHART
Start
Initialize memory pointer with 4300H
Get the MSB & multiply it by 10 using repeated addition
Add the LSB to the obtained result
Store the hexadecimal result in the memory location
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START: 4100H LXI H,4300H
4103H MOV A,M Initialize memory pointer
4104H ADD A MSD *2 4105H MOV B,A Store MSB *2
4106H ADD A MSB *4
4107H ADD A MSB *8
4108H ADD B MSB *10
4109H INX H Point to LSB
410AH ADD M Add to form hex 410BH INX H
410CH MOV M,A Store the result 410DH HLT Stop the process
RESULT:
Thus the Assembly Language Program to convert BCD number to hexadecimal. had
been written and executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. How a keyboard matrix is formed in keyboard interface using 8279?
2. What is scanning in keyboard and what is scan time?
3. What are the internal devices of a typical DAC?
4. What is settling or conversion time in DAC?
5. What are the different types of ADC?
Experiment No: Date:
HEX TO BCD CONVERSION
AIM:
To write an Assembly Language Program to convert a hexadecimal number to BCD
using 8085
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM
Step 1: Initialize memory pointer with 4300H
Step 2 Get the hexadecimal number in C-reg
Step 3: Perform repeated addition for C number of times
Step4: Adjust for BCD in each step
Step5: Store the BCD number in the memory location
Step6: Stop the process
Stop
FLOWCHART
Start
Initialize memory pointer with 4300H
Get the hexadecimal number in C-reg
Adjust for BCD in each step & Perform repeated addition for C number of
times
Store the BCD number in the memory location
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START: 4100H LXI H,4300H
4103H MOV A,M Initialize memory pointer
4104H ADD A MSD *2 4105H MOV B,A Store MSB *2
4106H ADD A MSB *4
4107H ADD A MSB *8
4108H ADD B MSB *10
4109H INX H Point to LSB
410AH ADD M Add to form hex 410BH INX H
410CH MOV M,A Store the result 410DH HLT Stop the process
RESULT:
Thus the Assembly Language Program to convert hexadecimal number to BCD.
had been written and executed & the results were stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. Define stack
2. What is program counter? How is it useful in program execution?
3. How the microprocessor is synchronized with peripherals?
4. What is a minimum system and how it is formed in 8085?
5. Define bit, byte and word.
Experiment No: Date:
FACTORIAL OF A GIVEN NUMBER
AIM:
To write an Assembly Language Program to find factorial of a given
hexadecimal number
APPARATUS REQUIRED:
(i) 8085 Microprocessor kit with keyboard.
(ii) Power cable.
ALGORITHM
Step 1: Initialize memory pointer with 4200H
Step2: Get the multiplicand in E-reg & move it to the next memory location.
Decrement E-reg by 1
Step3: Move the multiplicand to accumulator
Step 4: Get the multiplier in B reg& move to the multiplicand to D-reg.
Clear accumulator contents
Step5: Add the contents of accumulator & D-reg.
Step6: Decrement B-reg by 1& if B-reg content is zero, goto step7, else
goto step5
Step7: Move accumulator contents to memory
Step8: Decrement E reg & move E-reg contents to C-reg & : Decrement C
reg
Step9: If C-reg content is zero goto step10, else goto step3
Step10: Stop the process
FLOWCHART
Start
Initialise the HL pair with the address 4200H
Get the multiplicand in E-reg & move it to the next memory location.
Decrement E-reg by 1
Move the multiplicand to accumulator.
Get the multiplier in Breg. Move to the multiplicand D-reg& clear
accumulator contents
Add the contents of accumulator & D-reg.
.
Decrement B-reg by one
Is B-reg
NO Content zero?
YES
Move accumulator toMemory
Decrement E reg & move E-reg contents
To C-reg
Decrement C-reg
.
Is C- reg NO
=0 ?
YES
Stop
PROGRAM:
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START: 4300H LXI H,4200H Initialize memory pointer
with 4200H
4303H MOV E,M Get the multiplicand in E-reg
& move it to the next
memory location.
Decrement E-reg by 1
4304H INX H 4305H MOV M,E
4306H DCR E LOOP1: 4307H MOV A,M Move the multiplicand to
accumulator
4308H MOV B,E Get the multiplier in B reg&
move to the multiplicand to
D-reg Clear accumulator
contents
4309H MOV D,A
430AH XRA A
LOOP2: 430BH ADD D Add the contents of
accumulator & D-reg.
430CH DCR B Decrement B-reg by 1 430DH JNZ LOOP2 Check whether B- reg
content is zero 4310H MOV M,A Move accumulator contents
to memory 4311H DCR E Decrement E-reg by 1
move E-reg contents to C-
reg
4312H MOV C,E
4313H DCR C Decrement C reg by 1 4314H JNZ LOOP1 Check whether C- reg
content is zero 4317H HLT Stop the process
RESULT:
Thus the Assembly Language Program to convert to find factorial of a given
hexadecimal number had been written and executed & the results were stored
in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is PSW?
2. What is the purpose of ALE?
3. What is the need for interrupt controller?
4. Give some examples of input devices to microprocessor-based system
5 Give some examples of port devices used in 8085 microprocessor
based system?
Experiment No: Date:
8086-ADDITION OF TWO 16-BIT NUMBERS
AIM:
To write an assembly language program to add two 16 bit numbers and store the sum
in the memory location 1200H and carry in 1202H.
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: Initialize the BL reg. with 00H to account for carry.
Step 2: Move one of the data to AX reg.
Step 3: Move next data to CX reg.
Step 4: Clear Carry Flag.
Step 5: ADD the CX reg. content with the AX reg. Jump to step 7 if no carry
occurs. Otherwise go to next step.
Step 6: Increment BL reg.
Step 7: Move the sum to 1200H.
Step 8: Move the carry to 1202H.
Step 9: Stop the Process
FLOWCHART:
Start
Initialize BL reg. with 00H
Move Data into AX and CX reg.
Clear Carry Flag
Add CX and AX reg.
Is Carry NO
Set?
` YES
Increment BL reg
Store Sum and Carry to Memory
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 1000H MOV BL,00H Move 00H to BL reg. to
account for carry.
1003H MOV
AX,FFFFH Move the data FFFFH to AX
reg.
1007H MOV CX,001FH Move the data 001FH to CX
reg.
100BH CLC Clear Carry Flag.
100CH ADD AX, CX Add the content of CX to AX.
100EH JNC LOOP If no carry occurs, jump to
LOOP.
1010H INC BL Increment BL reg.
LOOP 1012H MOV[1200H],AX Move the AX content (sum)to
1200H & 1201H.
1016H MOV[1202H],BL Move the BL content to
1202H.
101AH HLT Stop the Process.
RESULT:
Thus the Assembly Language Program for Addition of two 16-bit numbers had been
written and executed & the results were stored in desired memory locations.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is the name given to the register combination DX:AX?
2. What are the modes in which 8086 can operate?
3. What is the data and address size in 8086?
4. Explain the function of M/IO in 8086.
5. Write the flags of 8086
Experiment No: Date:
8086-SUBTRACTION OF TWO 16-BIT NUMBERS
AIM:
To write an assembly language program to Subtract two 16 bit numbers and store the
difference in the memory location 1200H and borrow in 1202H.
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: Initialize the BL reg with 00H to account for borrow.
Step 2: Move Minuend to AX reg.
Step 3: Move Subtrahend to CX reg.
Step 4: Clear Carry Flag.
Step 5: Subtract the CX reg. content from the AX reg. Jump to step 7 if no
carry occurs. Otherwise go to next step.
Step 6: Increment BL reg.
Step 7: Move the difference to 1200H.
Step 8: Move the borrow to 1202H.
Step 9: Stop the Process
FLOWCHART:
Start
Initialize BL reg. with 00H
Move Data into AX and CX reg.
Clear Carry Flag
Subtract CX from AX reg.
NO
Is Carry
Set?
YES
Increment BL reg
Store Difference and Borrow in memory
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 1000H MOV BL,00H Move 00H to BL reg. to
account for carry.
1003H MOV
AX,FFFFH Move the data FFFFH to AX
reg.
1007H MOV
CX,OO1FH Move the data 001FH to CX
reg.
100BH CLC Clear Carry Flag.
100CH SUB AX, CX Subtract the content of CX
from AX.
100EH JNC LOOP If no carry occurs, jump to
LOOP.
1010H INC BL Increment BL reg.
LOOP 1012H MOV[1200H],AX Move the AX content
(difference) to 1200H.
1016H MOV[1202H],BL Move the BL content to
1202H.
101AH HLT Stop the Process.
RESULT:
Thus the Assembly Language Program for Subtraction of two 16-bit numbers had
been written and executed & the results were stored in desired memory locations.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What are the interrupts of 8086?
2. How clock signal is generated in 8086?
3. What is the maximum internal clock frequency of 8086?
4. Write the special functions carried by the general purpose
registers of 8086
5. What is pipelined architecture?
Experiment No: Date:
8086-MULTIPLICATION OF TWO 16-BIT NUMBERS
AIM:
To write an assembly language program for Multiplication of two 16 bit numbers
using 8086 instruction set.
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: Initialize the BL reg. with 00H to account for carry.
Step 2: Move Multiplicand to AX reg.
Step 3: Move Multiplier to CX reg.
Step 4: Clear Carry Flag.
Step 5: Multiply the CX reg. with the AX reg. Jump to step 7 if no carry occurs.
Otherwise go to next step.
Step 6: Increment BL reg.
Step 7: Move the content of AX & BL to the specified memory locations.
Step 9: Stop the Process
FLOWCHART:
Start
Initialize BL reg. with 00H
Move Data into AX and CX reg.
Clear Carry Flag
Multiply CX and AX reg. content
Is Carry NO
Set?
YES
Increment BL reg
Store the result in memory location
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 1000H MOV BL,00H Move 00H to BL reg. to
account for carry.
1003H MOV AX,11A2H Move the data 11A2H to AX
reg.
1007H MOV CX,11B3H Move the data 11B3H to CX
reg.
100BH MUL CX Multiply the content of CX to
AX reg.
100DH JNC LOOP If no carry occurs, jump to
LOOP.
100FH INC BL Increment BL reg.
LOOP 1011H MOV[1200H],AX Move the AX content to
1200H & 1201H.
1015H MOV[1202],DX Move the DX content to
1202H & 1203H.
1019H MOV[1204H],BL Move the BL content to
1204H.
101DH HLT Stop the Process.
RESULT:
Thus the Assembly Language Program for Multiplication of two 16-bit numbers had
been written and executed & the results were stored in desired memory locations
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. List the segment registers of 8086
2. Illustrate the use of LEA, LDS, and LES instruction
3. What instructions are needed to add AL, 3L and DL together, and
Place the result in CL?
4. What is purpose served by CX register?
5. Which are pointers present in this 8086?
Experiment No: Date:
8086-DIVISION OF 32 BIT NUMBER BY 16-BIT NUMBER
AIM:
To write an assembly language program for Division of 32 bit number by 16 bit
number using 8086 microprocessor kit.
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: Move the lower order byte of dividend to AX reg. and higher order byte of
dividend to DX reg.
Step 2: Move Divisor to CX reg.
Step 3: Divide the content of AX&DX by CX reg.
Step 4: Move the content of AX & DX in the specified memory locations.
Step 5: Stop the Process.
FLOWCHART:
Start
Move the Dividend into AX and DX reg
Move the Divisor Value to CX reg
Divide DX & AX by CX contents
Store the Quotient value in desired memory
Store the remainder in desired memory
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 1000H MOV AX,FFFFH Move the data FFFFH to
AX reg
1004H MOV DX, 0012H Move the data 0012H to DX
reg
1008H MOV CX,01A1H Move the Divisor 01A1 to
CX reg
100CH DIV CX Divide the content of
AX&DX BY CX
100EH MOV[1200H],AX Move the AX content to
1200H.
1012H MOV[1202H],DX Move the DX content to
1202H.
1016H HLT Stop the Process
RESULT:
Thus the Assembly Language Program for the Division of 32 –bit number by 16-bit
number had been written and executed & the results were stored in desired memory location
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. How is register AL is used during execution of XLAT?
2. What is the use of PUSH and POP instruction?
3. What is the purpose of XCHG instruction?
4. What do square brackets means when they appear in an operand?
5. Write a routine to swap nibbles in AL.
Experiment No: Date:
8086 -REVERSAL OF A STRING
AIM:
To write an assembly language program to reverse a given string using 8086
microprocessor kit
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: AX is initialized with data & AX is moved into DS
Step 2: Initialize CX to 5
Step 3: Load the effective address in SI & DI. Add SI with 04
Step 4: Move SI to AL
Step 5: Decrement SI & Increment DI. Repeat this until an interrupt is raised
. Step 6: Stop the Process.
FLOWCHART:
Start
AX is initialized with data & AX is moved
into DS
CX is initialized to 5
Load the effective address in SI & DI. Add
SI with 04
SI is moved to AL
Decrement SI & Increment DI
Repeat this until an interrupt is raised
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START: MOV AX ,@
DATA AX is initialized with
data
MOV DS, AX AX is moved into DS
MOV CX, 0005H CX is initialized to 5
LEA SI, A1 Load the effective
address in SI & DI LEA DI, A2 ADD SI, 0004
Add SI content with 04
AGAIN: MOV AL,[SI]
SI is moved to AL
MOV [DI],AL AL is moved into DI
DEC SI
Decrement SI
INC DI
Increment DI
LOOP AGAIN
INT 3 Interrupt
END Stop the Process
RESULT:
Thus the Assembly Language Program for string reversal has been written and
executed & the results were stored in desired memory location
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. Mention the features of 8086?
2. Explain how physical address is formed in 8086.
3. What is stack and Subroutine?
4. Difference between Microprocessor & Microcontroller
5. What is the maximum memory addressing and I/O
addresing capabilities of 8086?
8086 –ADDITION OF TWO 32 BIT NUMBERS
AIM:
To write an assembly language program to add two 32 bit numbers using 8086
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: AX is initialized with data & AX is moved into DS
Step 2: Initialize CX to 5
Step 3: Load the effective address in SI & DI. Add SI with 04
Step 4: Move SI to AL
Step 5: Decrement SI & Increment DI. Repeat this until an interrupt is raised
. Step 6: Stop the Process.
FLOWCHART:
Start
AX is initialized with data & AX is moved
to DS
Data2 is moved to BX & added with Data4
Data3 is moved toCX &added with carry with data4
Data 4C is moved to AH & AL content is moved to DI
Interrupt is raised
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START: MOV AX ,@ DATA AX is initialized
with data
MOV DS, AX AX is moved into
DS
MOV BX, Data2 Move data2 to BX
ADD BX, Data4 Data4 is added with
BX content
MOV CX,Data1 Move data1 to CX
ADC CX, Data3
Data3 is added with
CX content
MOV AH,4CH Data 4C is moved
to AH
MOV [DI],AL AL content is
moved to DI
INT 21H Interrupt
END Stop the Process
RESULT:
Thus the Assembly Language Program for 32 –bit addition has been written and
executed & the results were stored in desired memory location
MODEL SMALL:
DATA:
Data1 DW
Data2 DW
Data3 DW
Data4 DW
VIVA VOCE QUESTIONS
1. What do square brackets means when they appear in an operand?
2. What is the difference between MOV AX, 0 and SUB AX, AX?
3. Write a routine to swap nibbles in AL.
4. Which are pointers present in this 8086?
5. What is meant by Maskable interrupts?
Experiment No: Date:
8086-ASCENDING ORDER
AIM:
To write an assembly language program to arrange the given data in ascending Order
using 8086 microprocessor kit.
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: Get the count 1, number of repetitions (length of array - 1) in B
register.
Step 2: Get the count 2, number of comparisons (length of array -1) in C
register.
.
Step 3: Move the data from memory location pointed to the Accumulator.
Step 4: Compare the data from next memory location with the content of
Accumulator.
Step5: Check CF. If carry flag is set (accumulator content is smaller than the
other number), decrement the repitition count
Step 6: Check ZF. If zero flag is set (accumulator content is equal to the other
number), decrement comparison count, otherwise do next step.
Step 7: Store the smallest number to the previous memory location & greatest
to the current memory location.
Step 8: Stop the Process.
FLOWCHART:
Start
Initialize memory pointer to get N, the length of series
Initialize COUNT 1 for (N-1) repetitions
Initialize COUNT 2 for (N-1) Comparisons
Increment memory pointer and get the number from series
IS pointer < NO Interchange
(pointer+1)? Numbers
YES
Decrement COUNT 2
NO
Is COUNT 2 =
Zero?
YES
Decrement COUNT 1
NO
Is COUNT 1 =
zero?
YES
Stop
DATA SEGMENT: A DB 5 DUP(0)
DATA ENDS
CODE SEGMENT: ASSUME CS: CODE, DS: DATA
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START: MOV CL,05H Move 05 to CL
MOV DL,05H Move 05 to DL
DEC CL Decrement CL
register
L1: MOV BX,00H Move 00H to BX
L2: MOV AL, A[BX] Move the 1st data to
AL
CMP AL, A[BX+1] Compare the 1st &
2nd
data
JB L3 If 1st data is lesser
than 2nd, goto L3
XCHG AL, A[BX+1] If 1st data is greater
than 2nd, exchange
the data
MOV A[BX], AL
Move AL content to
BX
L3: INC BX Increment BX by 1
LOOP L2 Goto L2
MOV DL,CL Move CL content to
DL
DEC DL Decrement DL
content
LOOP L1 If the count is not
zero, goto L1
END Stop the process
RESULT:
Thus the Assembly Language Program to arrange the given data in ascending Order
had been written and executed & the results were stored in desired memory location
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. From which address the 8086 starts execution after reset?
2. What are the modes in which 8086 can operate?
3. What is the data and address size in 8086?
4. What is purpose served by CX register?
5. Which are pointers present in this 8086
Experiment No: Date:
8086-DESCENDING ORDER
AIM:
To write an assembly language program to arrange the given data in descending
Order using 8086 microprocessor kit.
APPARATUS REQUIRED:
(i) 8086 Microprocessor kit with keyboard.
(ii) Power Cable.
ALGORITHM:
Step 1: Get the count 1, number of repetitions (length of array - 1) in B
register.
Step 2: Get the count 2, number of comparisons (length of array -1) in C
register.
.
Step 3: Move the data from memory location pointed to the Accumulator.
Step 4: Compare the data from next memory location with the content of
Accumulator.
Step5: Check CF. If carry flag is set (accumulator content is smaller than the
other number), decrement the repitition count
Step 6: Check ZF. If zero flag is set (accumulator content is equal to the other
number), decrement comparison count, otherwise do next step.
Step 7: Store the smallest number to the previous memory location & greatest
to the current memory location.
Step 8: Stop the Process.
FLOWCHART:
Start
Initialize memory pointer to get N, the length of series
Initialize COUNT 1 for (N-1) repetitions
Initialize COUNT 2 for (N-1) Comparisons
Increment memory pointer and get the number from series
IS pointer < NO Interchange
(pointer+1)? Numbers
YES
Decrement COUNT 2
NO
Is COUNT 2 =
Zero?
YES
Decrement COUNT 1
NO
Is COUNT 1 =
zero?
YES
Stop
DATA SEGMENT: A DB 5 DUP(0)
DATA ENDS
CODE SEGMENT: ASSUME CS: CODE, DS: DATA
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START: MOV CL,05H Move 05 to CL
MOV DL,05H Move 05 to DL
DEC CL Decrement CL
register
L1: MOV BX,00H Move 00H to BX
L2: MOV AL, A[BX] Move the 1st data to
AL
CMP AL, A[BX+1] Compare the 1st &
2nd
data
JA L3 If 1st data is lesser
than 2nd, goto L3
XCHG AL, A[BX+1] If 1st data is greater
than 2nd, exchange
the data
MOV A[BX], AL
Move AL content to
BX
L3: INC BX Increment BX by 1
LOOP L2 Goto L2
MOV DL,CL Move CL content to
DL
DEC DL Decrement DL
content
LOOP L1 If the count is not
zero, goto L1
END Stop the process
RESULT:
Thus the Assembly Language Program to arrange the given data in desscending
Order had been written and executed & the results were stored in desired memory location
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. Suppose that DS=1000h, SS=2000h, BP=1000h and DI=0100h. Determine
the memory address accessed by each of the following instruction.
MOV AL, [BP + DI]
2. Form a jump instruction that jumps to the address pointed by the BX register?
3. What is the difference between JUMP and LOOP instructions?
4. What is the instruction needed to count the number of 1’s found in AL?
5. What conditional jump instruction should be used after CMP AL, 30H to
jump when AL equals 30H?
Experiment No: Date:
8051-ADDITION OF TWO 8-BIT NUMBERS
AIM:
To perform the addition of two 8 bit numbers using immediate addressing and store
the result in memory.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Clear register Ro
Step 2: Get the data in accumulator.
Step 3: Add the second data with the first data.
Step 4: Make the data pointer to point the address 4500H.
Step 5: Check the carry flag. If no carry occur, go to step 7, otherwise do next
step.
Step 6: Increment the register Ro.
Step 7: Store the sum (content of accumulator)to the address pointed by the data
pointer.
Step 8: Make the data pointer to point the next address by increment the DPTR.
Step 9: Store the carry to that address.
Step 10: Stop the process.
FLOWCHART:
Start
Get first data in accumulator & clear Ro reg
Make DP to point 4500H
ADD second data with accumulator
Is NO
Carry
Set
YES
Increment Ro reg
Store Sum and Carry in memory using DP
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4100H MOV Ro,#00H Move the data 00H
to reg Ro. 4102H MOV A,#data1 Move the first data
in accumulator 4104H ADD A,#data2 Add the second data
with the
accumulator. 4106H MOV DPTR, #4500H Move the address
4500H to DPTR. 4109H JNC LOOP If no carry occurs go
to address 410C. 410BH INC Ro Increment the
DPTR. LOOP 410CH MOV X@DPTR,A Move the data from
accumulator to the
address pointed by
DPTR
410DH INC DPTR Increment the DPTR
410EH MOV A, Ro Move the content
from reg Ro to the
accumulator.
410FH MOV @ DPTR, A Move the data from
accumulator to the
address denoted by
DPTR
4110H NOP No operation.
RESULT:
Thus the addition of two 8-bit data using immediate addressing is performed and
executed.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is called microcontroller?
2. What is the difference between microprocessor and microcontroller?
3. How many flags are there in 8051
4. How many addressing modes are in 8051
5. What is called Datapointer?
Experiment No: Date:
8051-SUBTRACTION OF TWO 8-BIT NUMBERS
AIM:
To perform the subtraction of two 8 bit numbers using immediate addressing and
store the result in memory.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Clear register Ro
Step 2: Get the data in accumulator.
Step 3: Subtract the second data with the first data.
Step 4: Make the data pointer to point the address 4500H.
Step 5: Check the carry flag. If no carry occur, go to step 7, otherwise do next
step.
Step 6: Increment the register Ro.
Step 7: Store the difference (content of accumulator)to the address pointed by the
data pointer.
Step 8: Make the data pointer to point the next address by increment the DPTR.
Step 9: Store the borrow to that address.
Step 10: Stop the process.
FLOWCHART:
Start
Get first data in accumulator & clear Ro reg
Make DP to point 4500H
Subtract second data with accumulator
Is NO
Borrow
Obtained
YES
Increment Ro reg
Store Difference and Borrow in memory using DP
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4300H MOV Ro,#00H Move the data 00H
to reg Ro. 4302H MOV A,#data1 Move the first data
in accumulator 4304H SUB A,#data2 Subtract the second
data with the
accumulator. 4306H MOV DPTR, #4500H Move the address
4500H to DPTR. 4309H JNC LOOP If no carry occurs go
to address 410C. 430BH INC Ro Increment the
DPTR. LOOP 430CH MOV X@DPTR,A Move the data from
accumulator to the
address pointed by
DPTR
430DH INC DPTR Increment the DPTR
430EH MOV A, Ro Move the content
from reg Ro to the
accumulator.
430FH MOV @ DPTR, A Move the data from
accumulator to the
address denoted by
DPTR
4310H NOP No operation.
RESULT:
Thus the subtraction of two 8-bit data using immediate addressing is performed and
executed.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1 . How many registers are in 8051
2. What is called Immediate addressing mode?
3. What is called Indirect addressing mode?
4. How many flags are affected in SUB instructions
5. How many bytes is used for the instruction MOV DPTR, #4500H instruction
Experiment No: Date:
8051-MULTIPICATION OF TWO 8-BIT NUMBERS
AIM:
To write an 8051 assembly language program to multiply two 8-bit numbers and store
the result in memory.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Get the first data in accumulator.
Step 2: Get the second data in reg B.
Step 3: Multiply two 8-bit data.
Step 4: Get the data pointer to point the address 4500H.
Step 5: Store the LSB of the result from the accumulator to the address pointed by
DPTR.
Step 6: Increment the data pointer.
Step 7: Store the MSB of the result in the address pointed by DPTR.
Step 8: Stop the process.
FLOWCHART:
Start
Clear carry flag
Get multiplicand in accumulator
Get multiplier in B reg
Multiply accumulator with B
Store the result in memory
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4300H MOV A,#data1 Move the data 1 to
accumulator. 4302H MOV F0,#data2 Move the data 2 the
reg B. 4305H MUL B Multiply A & B. 4306H MOV DPTR, #4500H Move the address
4500H to DPTR. 4309H MOV X@DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 430AH INC DPTR
Increment the
DPTR. 430BH MOV A,B Move the content of
B reg to the
accumulator.
430DH MOV X@DPTR,A Move the data from
accumulator to the
address denoted by
DPTR.
430EH NOP No operation.
RESULT:
Thus the Assembly Language Program for the multiplication of two 8-bit data has
been written and executed & the results are stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is F0 in this instruction MOV F0,#data?
2. What is called Stack pointer?
3. Define Flag register
4. What is the use of this instruction MOVX@ DPTR, A?
5. What type of addressing mode is used for the instruction MUL AB?
Experiment No: Date:
8051-DIVISION OF TWO 8-BIT NUMBERS
AIM:
To write an 8051 assembly language program to divide two 8-bit numbers and store
the result in memory.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Get the first data in accumulator.
Step 2: Get the second data in reg B.
Step 3: Divide two 8-bit data.
Step 4: Get the data pointer to point the address 4500H.
Step 5: Store the Quotient of the result from the accumulator to the address pointed by
DPTR.
Step 6: Increment the data pointer.
Step 7: Store the remainder of the result in the address pointed by DPTR.
Step 8: Stop the process.
FLOWCHART:
Start
Clear carry flag
Get dividend in accumulator
Get divisor in B reg
Divide accumulator with B
Store the result in memory
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4300H MOV A,#data1 Move the data 1 to
accumulator. 4302H MOV F0,#data2 Move the data 2 the
reg B. 4305H DIV B Divide A by B. 4306H MOV DPTR, #4500H Move the address
4500H to DPTR. 4309H MOV X@DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 430AH INC DPTR
Increment the
DPTR. 430BH MOV A,B Move the content of
B reg to the
accumulator.
430DH MOV X@DPTR,A Move the data from
accumulator to the
address denoted by
DPTR.
430EH NOP No operation.
RESULT:
Thus the Assembly Language Program for the division of two 8-bit data has been
written and executed & the results are stored in desired memory location
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. The mnemonics used in writing a program is called
_______________________
2. What is called Fetch Cycle?
3. What is called Direct addressing mode?
4. What is called Implied addressing mode?
5. What is the use of POP instruction?
Experiment No: Date:
8051-ADDITION OF TWO 16-BIT NUMBERS
AIM:
To perform the addition of two 16 bit numbers using immediate addressing and store
the result in memory.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Clear register Ro
Step 2: Make the datapointer to point the address 4500 and move first data LB to R1
register.
Step 3: Increment the datapointer and move the firtdata HB to R2 register.
Step 4: Increment the datapointer and add the LB of 1st data with LB of 2
nd data and
store in R1 register.
Step 5: Increment the datapointer and add with carry HB of 2nd
data with HB of 1st
data.
Step 6 :Check the carry flag. If no carry occur, go to step8 , otherwise do next
step.
Step 7: Increment the register Ro.
Step 8: Store the sum (content of accumulator)to the address pointed by the data
pointer.
Step 9: Make the data pointer to point the next address by increment the DPTR.
Step 10: Store the carry to that address.
Step 11: Stop the process.
FLOWCHART:
Start
Clear R0 reg, Make DP to point 4500H
Add LB of 1st and 2
nd data
ADD with carry HB of2nd data with 1st
Is NO
Carry
Set
YES
Increment Ro reg
Store Sum and Carry in memory using DP
Stop
Move 1st data LB to R1 reg
.Increment DPTR move 1st data
HB to R2 reg.Increment DPTR
Increment DPTR
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4100H MOV Ro,#00H Move the data 00H
to reg Ro. 4102 MOV DPTR, #4500H Move the address
4500H to DPTR 4105 MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4106 MOV R1,A Move the content of
LB of 1st data from
accumulator to reg
R1 4107 INC DPTR Increment Data
pointer 4108 MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4109 MOV R2,A Move the content of
LB of 1st data from
accumulator to reg
R2 410A INC DPTR Increment Data
pointer 410B MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 410C ADD A,RI Add the content of
LB of 2nd
data from
reg R1with
accumulator 410D MOV R1,A Move the content of
accumulator to reg
R1 410E INC DPTR Increment Data
pointer 410F MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator
4110 ADDC A,R2 Add with carry the
content of HB of 2nd
data from reg R2
with accumulator 4111 MOV R2,A Move the content of
accumulator to the
reg R2. 4112 JNC LOOP If no carry occurs go
to address 4115 4114 INC Ro Increment reg R0 LOOP 4115 MOV A,R1 Move the content of
reg R1 to
accumulator 4116 INC DPTR .Increment the
DPTR. 4117 MOV X@DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 4118 MOV A,R2 Move the content of
reg R2 to
accumulator
4119 INC DPTR Increment the DPTR
411A MOVX @ DPTR, A Move the data from
accumulator to the
address pointed by
DPTR
411B MOV A,R0 Move the content of
reg R0 to
accumulator
411C INC DPTR Increment the DPTR
411D MOVX @ DPTR, A Move the data from
accumulator to the
address pointed by
DPTR
411E NOP No operation.
RESULT:
Thus the addition of two 8-bit data using immediate addressing is performed and
executed.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is the use of POP instruction?
2. What is called Vector Location?
3. Define Interrupt?
4. What is the length of Stack pointer?
5. What is the length of Instruction Register?
Experiment No: Date:
TRANSFERING A BLOCK OF DATA FROM INTERNAL TO EXTERNAL
MEMORY
AIM:
To write an Assembly Language Program to transfer the block of data starting from
Internal memory location 31 to the memory location 4401H. The total number of data is
stored at 4400H.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Make the data pointer to point the address 4400H.
Step 2: Move the total number of data to the R0 register.
Step 3: Move the internal RAM Memory location 31 to the register R1.
Step 4: Increment the destination block pointer.
Step 5: Move the indirect Ram to the accumulator.
Step 6: Store the first data to the destination address.
Step 7: Increment the source block pointer.
Step 8: Decrement the R0 register.
Step 9: If ZF=0, go to step 4.otherwise go to next step.
Step 10: Stop the Process.
FLOWCHART:
Start
Move DPTR to point to 4400
Load the total number of input data In R0 reg
Move direct byte 31 to R1 reg.
Store data byte to destination block
Increment the source & destination block pointer
Decrement R0 reg
NO
If
ZF=1
YES
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4200H MOV DPTR, #4400H Move the address
4400H to DPTR 4203 MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4204 MOV R0,A Get count of total
number of data bytes
in R0 reg. 4205 MOV R1,#31 Move immediate
data to the reg R1 LOOP 4207 INC DPTR Increment
datapointer 4208 MOV A,@R1 Movev indirect
RAM to accumulator 4209 MOV X @DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 420A INC R1 Increment the
register R1 420B DEC R0 Decrement the
register R0 420C JNZ LOOP If ZF=0,program
control jump to
label. 420E NOP No operation
RESULT:
Thus the assembly language program to transfer the block of data from Internal
memory location 31 to 4401H has been written and verified.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is the length of Memory Address Register?
2. What is the length of Data buffer Register?
3. What is the length of Program Counter?
4. What is the length of Temporary Register?
5. What is the length of Accumulator?
Experiment No: Date:
TRANSFERING A BLOCK OF DATA FROM EXTERNAL TO EXTERNAL
MEMORY
AIM:
To write an Assembly Language Program to transfer the block of data starting from
memory location 4501H to 4800H. The total number of data is stored at 4500H.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Make the data pointer to point the address 4500H.
Step 2: Move the total number of data to the R0 register.
Step 3: Move the Higher order byte of source to R1 register
Step 4: Move the Higher order byte of destination to R2 register
Step 5: Move the content of register R1 to datapointer.
Step 6: Increment the destination block pointer.
Step 7: Store the first data to the destination address.
Step 8: Decrement the R0 register.
Step 9: If ZF=0, go to step 5.otherwise go to next step.
Step 10: Stop the Process.
FLOWCHART:
Start
Move DPTR to point 4500
Move 45 to R1 and 48 to R2 reg
Move count to R0 and R1 reg content to DPH
Store data byte to destination block
Increment the source & destination block pointer
Decrement R0 reg
NO
If
ZF=1
YES
Stop
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4300H MOV DPTR, #4500H Move the address
4300H to DPTR 4303H MOV R1,#45H Move the higher
order byte of Data
pointer to R1
register. 4305H MOV R2,#48H Move the direct data
to R2 register. 4307H MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4308H MOV R0,A Get count of total
number of data bytes
in R0 reg. LOOP 4309H MOV DPH,R1 Move the content of
the reg R1 to higher
order byte of thev
Datapointer. 430B INC DPTR Increment
datapointer 430C MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 430D MOV DPH,R2 Move the content of
the register R2 to the
destination 430F MOV X @DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 4310 DEC R0 Decrement the
register R0 4311 JNZ LOOP If ZF=0,program
control jump to
label. 4313 NOP No operation
RESULT:
Thus the assembly language program to transfer the block of data from memory
location 4501H to 4800H has been written and verified.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. How many I/O ports can be accessed in Direct Method?
2. How many lines are there in address bus
3. What is called Program Counter?
4. What is called Instruction Register ?
5. What is the direction of databus?
Experiment No: Date:
ASCII TO ITS EQUIVALENT HEXADECIMAL CONVERSION
AIM:
To write an Assembly Language Program to convert a ASCII number to
Hexadecimal number. The ASCII data is stored in 4600H and Hexadecimal data is store in
4601H.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Make the data pointer to point the address 4600H.
Step 2: Move the ASCIIl number to accumulator.
Step 3: Move the direct byte 30 to the register R1.
Step 4: Subtract with borrow the content of the register R1 with the accumulator.
Step 5:Increment the Data pointer.
Step 6: Store the equivalent hexa decimal number to the datapointer.
Step 7: Stop the Process.
FLOWCHART:
Move Hexadecimal num to Acc
Mov 30 to R1 reg and Sub with Acc
Increment DPTR
Move the result to DPTR
STOP
Move DPTR to point 4600
START
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4300H MOV DPTR, #4600H Move the address
4600H to DPTR 4303 MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4304 MOV R1,#30 Move the direct byte
to the register R1 4306 SUBB A,R1 Subtract with borrow
the content of
register R1 with
accumulator 4307 INC DPTR Increment
datapointer 4308 MOV X @DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 4309 NOP No operation
RESULT:
Thus the assembly language program to convert a ASCII number to equivalent
Hexadecimal number has been written and verified.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. Which byte of an instruction is loaded into IR register
2. What is the length of Stackpointer?
3. What is the length of Status word?
4. How many interrupts are there in 8051
5. What is PSW?
Experiment No: Date:
HEXADECIMAL TO ITS EQUIVALENT ASCII CONVERSION
AIM:
To write an Assembly Language Program to convert a Hexadecimal number to its
equivalent ASCII number The Hexadecimal data is stored in 4600H and ASCII data is
stored in 4601H.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Make the data pointer to point the address 4600H.
Step 2: Move the content of datapointer to accumulator.
Step 3: Move the direct byte 30 to the register R1.
Step 4: Add the content of the register R1 with the accumulator.
Step 5:Increment the Data pointer.
Step 6: Store the equivalent ASCII number to the datapointer.
Step 7: Stop the Process.
FLOWCHART:
Move Hexadecimal num to Acc
Mov 30 to R1 reg and add with Acc
Increment DPTR
Move the result to DPTR
STOP
Move DPTR to point 4600
START
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4300H MOV DPTR, #4600H Move the address
4600H to DPTR 4303H MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4304H MOV R1,#30 Move the direct byte
to the register R1 4306H ADD A,R1 Add the content of
register R1 with
accumulator 4307H INC DPTR Increment
datapointer 4308H MOV X @DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 4309H NOP No operation
RESULT:
Thus the assembly language program to convert a ASCII number to equivalent
Hexadecimal number has been written and verified
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. How many I/o ports can be accessed by direct method
2. How many I/o ports can be accessed by Memory mapped method
3. What is called Vectored Interrupt?
4. What is called Maskable interrupt?
5. When interrupt service request is serviced
.
Experiment No:
Date:
8085-ASCENDING ORDER
AIM:
To write an Assembly Language Program to arrange the given data in Ascending
Order.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Move the datapointer to point to 4500H
Step 2: Get the count 1, number of repetitions (i.e., total number of data-1) in R0
register.
Step 3: Get the count 2, number of comparisons (i.e., total number of data-1) in A
register.
Step 4: Save the count 2 in R1 register.
Step 5: Move the datapointer to point to 4501H
Step 6: Move DPL to R2
Step 7: Move the 1st data to B register and increment DPTR and move the 2
nd data to
the
accumulator.
Step 8: Compare the data from next memory location with the content of
accumulator.
Step 9: Compare and jump if not equal goto step11 else do next step.
Step 10: Check ZF. If zero flag is set goto step else do next step
Step 11: check CF. If carry flag is not set (accumulator content is smaller than the
other number), go to step 13 otherwise do next step.
Step 12: Store the smallest number to the previous memory location & greatest to
the current memory location.
Step 13: Decrement the number of comparisons (count 2).
Step 14: If the number of comparison is not equal to zero, go to step 6, otherwise
do the next step.
Step 15: Decrement the number of repetitions (count 1).
Step 16: If the number of repetitions is not completed go to step 3, otherwise do
next step.
Step 17: Stop the Process.
FLOWCHART 8051-ASCENDING ORDER
Start
Initialize Datapointer to get N the length of series
Initialize COUNT 1 for (N-1) repetitions
Initialize COUNT 2 for (N-1) Comparisons
Increment memory pointer and get the number from series
IS pointer < NO Interchange
(pointer+1)? Numbers
YES
Decrement COUNT 2
NO
Is COUNT 2
Zero?
YES
Decrement COUNT 1
NO
Is COUNT 1
zero?
YES
Store sum as result
Stop
PROGRAM FOR SORTING A SERIES IN ASCENDING ORDER
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100 MOV DPTR,#4500
4103 MOVX A,@DPTR
4104 DEC A
4105 MOV R0,A R0 = COUNT 1
for (N-
1)repetitions
AGAIN 4106 MOV A,R0
4107 MOV R1,A R1= COUNT 2
for (N-
1)Comparisons
4108 MOV DPTR,#4501
BACK 410B MOV R2,DPL 410D MOVX A,@DPTR
410E MOV B,A
4110 INC DPTR
4111 MOVX A,@DPTR
4112 CJNE A,B LOOP1 Compare
consecutive
numbers
4115 JZ SKIP
LOOP1 4117 JNC SKIP
4119 MOV DPL,R2
411B MOVX @DPTR,A
411C INC DPTR
411D MOV A,B Interchange
numbers if not in
order 411F MOVX @DPTR,A
SKIP 4120 DJNZ R1,BACK Decrement
count2
4122 DJNZ R0,AGAIN Decrement count
1
4124 NOP Stop the Process
RESULT:
Thus the Assembly Language Program to arrange the given data in ascending order
has been written and executed & the results are stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. Which interrupt remains enabled even after a reset operation
2. What instruction is used to enable the interrupt?
3. What instruction is used to enable the TRAP interrupt?
4. What is the use of RIM instruction?
5. What is the purpose of READY signal?
Experiment No: Date:
8051-DESCENDING ORDER
AIM:
To write an Assembly Language Program to arrange the given data in Descending
Order.
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Move the datapointer to point to 4500H
Step 2: Get the count 1, number of repetitions (i.e., total number of data-1) in R0
register.
Step 3: Get the count 2, number of comparisons (i.e., total number of data-1) in A
register.
Step 4: Save the count 2 in R1 register.
Step 5: Move the datapointer to point to 4501H
Step 6: Move DPL to R2
Step 7: Move the 1st data to B register and increment DPTR and move the 2
nd data to
the
accumulator.
Step 8: Compare the data from next memory location with the content of
accumulator.
Step 9: Compare and jump if not equal goto step11 else do next step.
Step 10: Check ZF. If zero flag is set goto step else do next step
Step 11: check CF. If carry flag is set (accumulator content is smaller than the
other number), go to step 13 otherwise do next step.
Step 12: Store the smallest number to the previous memory location & greatest to
the current memory location.
Step 13: Decrement the number of comparisons (count 2).
Step 14: If the number of comparison is not equal to zero, go to step 6, otherwise
do the next step.
Step 15: Decrement the number of repetitions (count 1).
Step 16: If the number of repetitions is not completed go to step 3, otherwise do
next step.
Step 17: Stop the Process.
FLOWCHART 8051-DESCENDING ORDER
Start
Initialize Data pointer to get N the length of series
Initialize COUNT 1 for (N-1) repetitions
Initialize COUNT 2 for (N-1) Comparisons
Increment memory pointer and get the number from series
IS pointer < NO Interchange
(pointer+1)? Numbers
YES
Decrement COUNT 2
NO
Is COUNT 2
zero?
YES
Decrement COUNT 2
NO
Is COUNT 1
zero?
YES
Store sum as result
Stop
PROGRAM FOR SORTING A SERIES IN DESCENDING ORDER
LABEL ADDRESS MNEMONIC OPCODE OPERAND COMMENTS
START 4100 MOV DPTR,#4500
4103 MOVX A,@DPTR
4104 DEC A
4105 MOV R0,A R0 = COUNT 1
for (N-
1)repetitions
AGAIN 4106 MOV A,R0
4107 MOV R1,A R1= COUNT 2
for (N-
1)Comparisons
4108 MOV DPTR,#4501
BACK 410B MOV R2,DPL 410D MOVX A,@DPTR
410E MOV B,A
4110 INC DPTR
4111 MOVX A,@DPTR
4112 CJNE A,B LOOP1 Compare
consecutive
numbers
4115 JZ SKIP
LOOP1 4117 JC SKIP
4119 MOV DPL,R2
411B MOVX @DPTR,A
411C INC DPTR
411D MOV A,B Interchange
numbers if not in
order 411F MOVX @DPTR,A
SKIP 4120 DJNZ R1,BACK Decrement
count2
4122 DJNZ R0,AGAIN Decrement count
1
4124 NOP Stop the Process
RESULT:
Thus the Assembly Language Program to arrange the given data in descending order
has been written and executed & the results are stored in desired memory location.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is the purpose of ALE signal?
2. What addressing mode is used in the instruction MOV A,#data1?
3. What addressing mode is used in the instruction MOVX @DPTR, A?
4. What addressing mode is used in the instruction ADD A,R1?
5. What addressing mode is used in the instruction MOV A,@R1?
Experiment No: Date:
FILLING EXTERNAL MEMORY
AIM:
To write an Assembly Language Program to fill the data starting from memory
location 4200H to 4300H. The total number of data is stored at 4500H
APPARATUS REQUIRED:
(i) 8051 Microprocessor.
(ii) Power Cable.
ALGORITHM:
Step 1: Make the data pointer to point the address 4500H.
Step 2: Move the total number of data to the R0 register.
Step 3: Increment datapointer.
Step 4: Move the first data to Accumulator
Step 5: Make the data pointer to point the address 4200H.
Step 6: Store the first data to the destination address .
Step 7: Increment the destination block pointer.
Step 8:.Decrement the R0 register.
Step 9: If ZF=0, go to step 6.otherwise go to next step.
Step 10: Stop the Process.
FLOWCHART:
Start
Move DPTR to point 4500
Move count to R0 reg
Increment DPTR move 1st data to Accmulator
Move DPTR to point 4200
Store the data in the deastination address
II
Decrement R0 reg
NO
If
ZF=1
YES
Stop
Increment Data pointer`
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4100H MOV DPTR, #4500H Move the address
4500H to DPTR 4103H MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4104H MOV R0,A Move the content of
accumulator to R0
register. 4105H INC DPTR Increment
datapointer 4106H MOV X A,@DPTR Move the data from
address pointed by
DPTR to the
accumulator 4107H MOV DPTR, #4200H Move the address
4200H to DPTR LOOP1 410AH MOV X @DPTR,A Move the data from
accumulator to the
address pointed by
DPTR 410BH INC DPTR Increment
datapointer 410CH DEC R0 Decrement the
register R0 410DH JNZ LOOP1 If ZF=0,program
control jump to
label.
410FH NOP No operation
RESULT:
Thus the assembly language program to fill the data starting from memory location
4200H to 4300H and internal memory 30 to 40 has been written and verified.
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT
OUTPUT
OUTPUT
VIVA VOCE QUESTIONS
1. What is the use of JUMP instruction?
2. How many ports are used in 8051
3. The crystal oscillator is connected to pins__________
&________________
4. What is the use of RST pin?
5. “Pushing the stack” refers to___________ operation in a LIFO memory.
Experiment No: Date:
STEPPER MOTOR INTERFACE
AIM:
To run a stepper motor at different speed in two directions.
APPARATUS REQUIRED:
(i) 8085 Microprocessor.
(ii) Power Cable.
(iii)Stepper Motor Interface.
THEORY:
The hardware setup consists of a microprocessor motherboard and stepper motor
interface board. The motherboard consists of 8085 MPU, 8KB EPROM, 8KB RAM,
Keyboard and display controller 8279, 21-key Hex-keypad and six numbers of seven segment
LEDs and Bus Expansion connector. The stepper motor interface consists of driver transistors
for stepper motor windings and address decoding circuit. The microprocessor output the
binary sequence through data bus, which are converted to current pulses by the driver
transistors and used to drive stepper motor. The software for the system is developed in 8085
assembly language.
ALGORITHM:
Step 1: Initialize HL register pair with address 4150H.
Step 2: Initialize register B with total number of data for rotation.
Step 3: Get the data for rotation in accumulator.
Step 4: Send the data from accumulator to port address of stepper motor interface.
Step 5: Initialize DE register pair with data 030H.
Step 6: Decrement DE register pair.
Step 7: Check the value of DE reg pair. If it is not equal to zero go to step 3.
Otherwise go to step 1.
FLOWCHART: Start
Initialize HL reg with 411AH & Initialize B reg with 04H
Get data for rotation in accumulator
Send data to port address
Initialize DE reg pair with 0303H
Décrément DE reg
Is NO
DE=00
YES
Increment HL reg pair
Decrement B reg
Is NO
B=0
YES
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4100H LXI H,411AH 21 1A,41 Initialize HL reg with
411AH. 4103H MVI B,04H 06 04 Load data 04H to B reg. REPEAT 4105H MOV A,M 7E Send data to
accumulator. 4106H OUT C0H D3 C0 Send data to output port. 4108H LXI D,0303H 11 03,03 Initialize DE reg pair
with 0303H. DELAY 410BH NOP 00 No operation.
410CH DCX D 1B Decrement DE reg pair.
410DH MOV A,E 7B Move data from E to A.
410EH ORA D B2 OR the accumulator
content with D reg.
410FH JNZ DELAY C2 0B,41 If DE#0, go to DELAY.
4112H INX H 23 Increment HL reg pair.
4113H DCR B 05 Decrement B reg.
4114H JNZ REPEAT C2 05,41 If B#0, go to REPEAT.
4117H JMP START C3 00,41 Go to start.
Unipolar steeping scheme:
Two Phase Steeping Scheme:
RESULT:
Thus the program has been written and the stepper motor is rotated in two directions.
The stepper motor is also rotated at different speed.
clock wise Anticlock wise
Step A1 A2 B1 B2 A1 A2 B1 B2
1 1 0 0 0 1 0 0 0
2 0 0 0 1 0 0 1 0
3 0 1 0 0 0 1 0 0
4 0 0 1 0 0 0 0 1
clock wise Anticlock wise
Step A1 A2 B1 B2 A1 A2 B1 B2
1 1 0 0 1 1 0 1 0
2 0 1 0 1 0 1 1 0
3 0 1 1 0 0 1 0 1
4 1 0 1 0 1 0 0 1
OBSERVATION:
ADDRESS DATA COMMENTS
INPUT 411AH 0A 09 Data to rotate in
forward direction
411BH 06 05
411CH 05 06
411DH 09 0A
OUTPUT Stepper motor Rotated in forward Direction
VIVA VOCE QUESTIONS
1. What is the use of JUMP instruction?
2. How many machine cycles needed for the execution of ORA?
3. What is the difference between unipolar and two phase steeping
scheme?
4. A state during which nothing happens is known as
________________
5. microprocessor 8085 is the enhanced version of ____________ with
essentially the same construction set
Experiment No: Date:
ADC INTERFACING
AIM:
To write a program to initiate ADC and to store the digital data in memory
APPARATUS REQUIRED:
8085 Trainer Kit
ADC Interface board
THEORY:
The ADC0809 is an 8-bit successive approximation type ADC with inbuilt 8-channel
multiplexer. The ADC0809 is suitable for interface with 8086 microprocessor. The
ADC0809 is available as a 28 pin IC in DIP (Dual Inline Package). The ADC0809 has a
total unadjusted error of ±1 LSD (Least Significant Digit).The ADC0808 is also same as
ADC0809 except the error. The total unadjusted error in ADC0808 is ± 1/2 LSD.
ALGORITHM:
Step 1: Load data 10 to accumulator.
Step 2: Send the data from accumulator to output port address of ADC interface
Step 3: clear accumulator.
Step 4: Load data 10 to accumulator
Step 5: Send the data from accumulator to output port address of ADC interface
Step 6: Send the data from accumulator to input port address of ADC interface
Step 7: AND the contents 01 with accumulator and compare with it.
Step 8:check for zero flag if not set go to step 6,otherwise do the next step
Step 9: Send the data from accumulator to input port address of ADC interface
Step10:store the result and stop the operation
Decrement b-reg
START
TT
FLOWCHART:
Initialize HL reg with 411AH & Initialize B reg with 04H
Get data for rotation in accumulator
Send data to port address
Initialize DE reg pair with 0303H
Is NO
DE=00
YES
Increment HL reg pair
Decrement B reg
Is NO
B=0
YES
PROGRAM:
ADDRESS LABEL MNEMONICS OPCODE OPERAND COMMENTS
4100 START: MVI A,10 Load 10 to A-reg
4101 OUT C8 Send data to output
port
4102 MVI A,10 Load 10 to A-reg
4103 OUT C8 Send data to output
port
4104 MVI A,10 Load 10 to A-reg
4105 OUT D0 Send data to output
port
4106 XRA A Clear accumulator
4107 XRA A Clear accumulator
4108 XRA A Clear accumulator
4109 MVI A,00 Load 00 to A-reg
410A OUT D0 Send data to output
port
410B LOOP IN D8 Send data to input
port
410C ANI 01 And 01 with
accumulator
410D CPI 01 Compare 01 with
accumulator
410E JNZ LOOP If zero flag is not
set,go to specified
label
4111 IN C0 Send data to input
port
4112 STA 4150 Store accumulator
4115 HLT Stop operation
RESULT:
Thus the ADC was initiated and the digital data was stored at desired location
OBSERVATION:
Compare the data displayed at the LEDs with that stored at location 4150
VIVA VOCE QUESTIONS
1. What is the difference between LXI H,4100 and LHLD 4100?
2. What are the application of microprocessor?
3. What is called LIFO?
4. How many interrupts are there in 8085?
5. Which group of ports of 8255 PPI can be operated in 2modes
.Experiment No: Date
DAC INTERFACING
AIM:
To interface DAC with 8085 to demonstrate the generation of square,saw tooth and
rectangular wave.
APPARATUS REQUIRED:
8085 Trainer Kit
DAC Interface board
THEORY:
DAC 0800 is an 8-bit DAC and the output voltage variation is between -5V and + 5V. the
output voltage varied in steps of 10/256=0.04(appx). The digital data input and the
corresponding output voltages are presented in the table.1
Input Data in
HEX
Output Voltage
00 -5.00
01 -4.96
02 -4.92
… …
7F 0.00
… …
FD 4.92
FE 4.96
FF 5.00
Referring to Table.1, with 00H as input to DAC,the analog output is -5v. Similarly, with FFH
as input ,The output is +5v. Outputting digital data 00 and FF at regular intervals to
DAC,results in different waveforms namely square,triangular etc. the port address of DAC is
08H.
ALGORITHM:
1. SQUARE WAVE GENERATION:
(i) Load the initial value (00) to accumulator and move it to DAC
(ii) Call the delay program
(iii) Load the final value (FF) to accumulator and move it to DAC
(iv) Call the delay program
(v) Repeat steps 2 to 5
2. SAW TOOTH WAVE GENERATION:
(i) Load the initial value 00 to accumulator
(ii) Move the accumulator content to DAC
(iii) Increment the accumulator content by 1
(iv) Repeat steps 3 and 4.
3. TRIANGULAR WAVE GENERATION:
(i) Load the initial value (00) to accumulator
(ii) Move the accumulator content to DAC
(iii) Increment the accumulator content by1
(iv) If accumulator content is zero proceed to next step. Else go to step 3
(v) Load value (FF) to accumulator
(vi) Move the accumulator content to DAC
(vii) Decrement the accumulator content by 1
(viii) If accumulator content is zero go to step2. Else go to step 7.
FLOWCHART:
SQUARE WAVEFORM
Start
Move data to 00H to A
Call Delay subroutine
Move data FFH to A
Call Delay subroutine
Jump
DELAY:
Start
Move data FFH to C reg & 05 to B reg
Decrement C reg
YES Is
C#0
NO
Decrement B reg
Is YES
B#0
NO
Return
FLOWCHART:
SAWTOOTH WAVEFORM
Start
Move data to 00H to A
Increment A
If
YES There is
No zero
NO
YES
Jump
NO
Stop
TRIANGULAR WAVEFORM
Start
Move data 00H to L reg
Move L reg to Accumulator
Increment the reg L
If
YES There is
No zero
NO
Move data to FFH to L reg
Move L reg to Accumulator
Decrement the reg L
If
YES There is
No zero
NO
YES
Jump
NO
Stop
PROGRAM:
Square Wave Generation:
ADDRESS LABEL MNEMONICS OPCODE OPERAND COMMENTS
4100 START: MVI A,00 Load 00 to
accumulator
4101 OUT C8 Send data to output
port
4102 CALL DELAY Call operation
4103 OUT C8 Load 00 to
accumulator
4104 CALL DELAY Call operation
4105 JMP START Jump tothe label
specified
4106 DELAY: MVI B,05 Load 05 to B-reg
4107 L1: MVI C,FF Load FF to C-reg
4108 L2: DCR C Dec C-reg
4109 JNZ L2 Jump to the label
specified if zero
flag is not set
410A DCR B Decrement B-reg
410B JNZ L1 Jump to the label
specified if zero
flag is not set
410C RET Return to the
mainprogram
Saw tooth Wave Generation:
ADDRESS LABEL MNEMONICS OPCODE OPERAND COMMENTS
4100 START: MVI A,00 Load 00 to
accumulator
4101 L1 OUT C8 Load 00 to
accumulator
4102 INR A Increment A-reg
4103 JNZ L1 Jump to the label
specified if zero flag
is not set
4104 JMP START Jump to the label
specified
Triangular Wave Generation:
ADDRESS LABEL MNEMONICS OPCODE OPERAND COMMENTS
4100 START: MVI L,00 Load 00 to L-
reg
4101 L1: MOV A,L Load 00 to
accumulator
4102 OUT C8 Send data to
output port
4103 INR L incrementL-
reg
4104 JNZ L1 Jump to the
label
specified if
zero flag is
not set
4107 MVI L,FF Load FF to L-
reg
4108 L2: MOV A,L Load FF to
accumulator
4109 OUT C8 Send data to
output port
410A DCR L Decrement L-
reg
410B JNZ L2 Jump to the
label
specified if
zero flag is
not set
410E JMP START Jump to the
label
specified
RESULT:
Thus the square,triangular and saw tooth waveform were generated by interfacing DAC with
8085 trainer kit..
VIVA VOCE QUESTIONS
1. Write the two ways to initialize stack pointer at FFFFH?
2. Define instruction set?
3. is it possible to check AC flag status of 8085,how??
4. What is difference between PUSH and POP instruction?
5. How many flags are affected for the HALT?
Experiment No: Date:
TRAFFIC LIGHT CONTROLLER
AIM:
To control the traffic light system with an 8085 microcontroller program.
APPARATUS REQUIRED:
(i) 8085 Microprocessor.
(ii) Power Cable.
(iii)Traffic Light Interfacing.
(iii)Lights for traffic signaling.
THEORY:
The traffic lights are interfaced to Microprocessor system through buffer and ports of
programmable peripheral Interface 8255.So the traffic lights can be automatically switched
ON/OFF in desired sequence. The Interface board has been designed to work with parallel
port of Microprocessor system. The hardware of the system consists of two parts. The first
part is Microprocessor based system with 8085. Microprocessor as CPU and the peripheral
devices like EPROM, RAM, Keyboard & Display Controller 8279, Programmable as
Peripheral Interface 8255, 26 pin parallel port connector, 21 keys Hexa key pad and six
number of seven segment LED’s. The second part is the traffic light controller interface
board, which consist of 36 LED’s in which 20 LED’s are used for vehicle traffic and they are
connected to 20 port lines of 8255 through Buffer. Remaining LED’s are used for pedestrian
traffic. The traffic light interface board is connected to Main board using 26 core flat cables
to 26-pin Port connector. The LED’s can be switched ON/OFF in the specified sequence by
the Microprocessor.
ALGORITHM:
Step 1: Initialize HL pair with 4500H and C reg with 02H.
Step 2: Send control word to CNT register.
Step 3: Send data to port A & B.
Step 4: Call delay and increment HL pair.
Step 5: Decrement C reg.
Step 6: if C=0 go to step 1, otherwise go to step 3.
FLOWCHART:
Start
Initialize HL pair with 4500H & C reg with 02H
Send control word to CNT reg
Send data to port A & B
Call Delay
Increment HL pair
Decrement C register
If NO
C=0
YES
DELAY:
PUSH BC reg pair to stack
Move 05H to C reg
Load the DE with FFFFH
Décrément DE reg pair
Is NO
DE=0
YES
Décrément C reg
NO Is
C=0
YES
POP stack to BC rerg
Return
PROGRAM:
LABEL ADDRESS MNEMONICS OPCODE OPERAND COMMENTS
START 4100H LXI H,4500H 21 00,45 Initialize HL pair with
4500H ` 4103H MVI C,02H 0E 02 Move data 02H to C reg. 4105H MOV A,M 7E Get data from memory to
accumulator. 4106H OUT CNT D3 0F Send data to control reg. LOOP1 4108H INX H 232 Increment HL pair reg. 4109H MOV A,M 7E Get data from memory to
accumulator. 410AH OUT APRT D3 0C Send data to port A. 410CH INX H 23 Increment HL pair reg. 410DH MOV A,M 7E Get data from memory to
accumulator. 410EH OUT BPRT D3 0D Send data to port B. 4110H CALL
DELAY
CD 1B,41 Call delay routine.
4113H INX H 23 Increment HL pair reg. 4114H DCR C 0D Decrement HL pair. 4115H JNZ LOOP1 C2 09,41 If C reg value is non-zero
go to loop1. 4118H JMP START C3 00,41 Go to start. DELAY 411BH MVI B.05H 06 05 Move 05H to B reg. LOOP3 411DH LXI D,FFFFH 11 FF,FF Initialize DE reg pair with
FFFFH. LOOP2 4120H DCX D 1B Decrement DE reg pair. 4121H MOV A,D 7A Move data from D to A
reg. 4122H ORA E B3 OR reg with E reg. 4123H JNZ LOOP2 C2 20,41 Go to loop2. 4126H DCR B 05 Decrement B reg. 4127H JNZ LOOP3 C2 1D,41 Go to loop3.
412AH RET 09 Return.
RESULT:
Thus the Assembly Language Program to control the traffic light system has been
written and output is verified.
OBSERVATION:
ADDRESS DATA COMMENTS
4500H 80 Control Word
4501H 1A First Step Data
4502H A1 First Step Data
4503H 81 Second Step Data
4504H 5A Second Step Data
VIVA VOCE QUESTIONS
6. What is the use of latch?
7. What is the use of IN/OUT instruction?
8. What addressing mode is used for the instruction MOV A,M?
9. What is called buffer?
10. How many modes are there in 8255
Experiment No: Date
KEYBOARD / DISPLAY CONTROLLLER
AIM:
To interface 8279 Programmable Keyboard Display Controller to 8085 Microprocessor.
APPARATUS REQUIRED:
8085 Microprocessor toolkit
8279 Interface Board
VXT Parallel bus
Regulated D.C Power supply
THEORY:
& 310 A programmable keyboard and display interfacing chip. It Scans and encodes up to a 64-ey
keyboard and it controls up to a 16-digit numerical display. The keyboard section has a built-
in FIFO 8 character buffer. The display is controlled from an internal 16x8 RAM has stores
the coded display information.
PROGRAM:
RESULT:
Thus 8279 controller was interfaced with 8085 and program for rolling display was executed
successfully.
ADDRESS LABEL MNEMONICS OPCODE OPERAND COMMENTS
4100 START: LXI H,4130H Initialize HL pair
with 4130H
4101 MVI D,0FH Load 0F to D-reg
4102 MVI A,10H Load 10 to A-reg
4103 OUT C2H Send data to
outport port
4104 MVI A,CCH Load CC to A-reg
4107 OUT C2H Send data to
outport port
4108 MVI A,90H Load 90 to A-reg
4109 OUT C2H Send data to
outport port
410A LOOP: MOV A,M Move content of
memory to
accumulator
410B OUT C0H Send data to
outport port
410E CALL
DELAY
Call delay
4121 INX H Increment HL pair
4122 DCR D Decrement D-reg
4123 JNZ LOOP Jump to the label
specified if zero
flag is not set
4126 JMP START Jump to the label
specified
4129 DELAY: MVI B,A0H Load A0 to B-reg
412B LOOP2 MVI C,FFH Load FF to C-reg
412D LOOP1 DCR C Decrement C-reg
412E JNZ LOOP1 Jump to the label
specified if zero
flag is not set
4131 DCR B Decrement B-reg
4132 JNZ LOOP2 Jump to the label
specified if zero
flag is not set
4135 RET Return to the main
program
OBSERVATION:
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 -FF
4139 -FF
413A -FF
413B–FF
413C -FF
413D-FF
413E-FF
413F-FF
VIVA VOCE QUESTIONS
11. What is the use of CALL instruction?
12. What is the use of RETURN instruction?
13. What addressing mode is used for the instruction MVI C,FF?
14. What is the Jump range?
15. How many machine cycles are used for the instruction LXI H,4130H
Experiment No: Date
RAM ADDRESING
AIM:
To exhibit the RAM direct addressing and bit addressing schemes of 8051 microcontroller.
APPARATUS REQUIRED:
8051 Interfacing board
power supply
ALGORITHM:
For Bit addressing, select Bank 1 of RAM by setting 3rd
bit of PSW
Using Register 0 of Bank 1 and accumulator perform addition
For direct addressing provide the address directly(30 in this case)
Use the address and Accumulator to perform addition
Verify the results.
PROGRAM:
Bit Addressing:
SETB PSW.3
MOV R0,#data1
MOV A,#data2
ADD A,R0
MOV DPTR,#4500
MOVX @DPTR,A
HERE: SJMP HERE
Direct Addressing:
MOV 30,#data1
MOV A,#data2
ADD A,30
MOV DPTR,#4500
MOVX @DPTR,A
HERE: SJMP HERE
OBSERVATION:
Bit Addressing:
INPUT : 54
25
OUTPUT: 79(4500)
Direct Addressing:
INPUT : 54
25
OUTPUT: 79(4500)
RESULT:
Thus the program to exhibit different RAM addressing schemes of 8051 was executed.