MPD

MICROPROCESSOR & MICROCONTROLLER LAB MANUAL
8085 MICROPROCESSOR
PROGRAMS
C.SARAVANAKUMAR. M.E.,
LECTURER, DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING
1
ADDITION OF TWO 8 BIT NUMBERS
AIM:
To perform addition of two 8 bit numbers using 8085.
ALGORITHM:
1) Start the program by loading the first data into Accumulator.

2) Move the data to a register (B register).
3) Get the second data and load into Accumulator.
4) Add the two register contents.
5) Check for carry.
6) Store the value of sum and carry in memory location.
7) Terminate the program.
PROGRAM:
MVI C, 00 Initialize C register to 00

LDA 4150 Load the value to Accumulator.
MOV B, A Move the content of Accumulator to B register.
ADD B Add the value of register B to A
JNC LOOP Jump on no carry.
INR C Increment value of register C
LOOP: STA 4152 Store the value of Accumulator (SUM).
MOV A, C Move content of register C to Acc.
STA 4153 Store the value of Accumulator (CARRY)
HLT Halt the program.
OBSERVATION:
Input: 80 (4150)
80 (4251)
Output: 00 (4152)
01 (4153)
RESULT:
Thus the program to add two 8-bit numbers was executed.
2
SUBTRACTION OF TWO 8 BIT NUMBERS
AIM:
To perform the subtraction of two 8 bit numbers using 8085.
ALGORITHM:
1. Start the program by loading the first data into Accumulator.

2. Move the data to a register (B register).
3. Get the second data and load into Accumulator.
4. Subtract the two register contents.
5. Check for carry.
6. If carry is present take 2’s complement of Accumulator.
7. Store the value of borrow in memory location.
8. Store the difference value (present in Accumulator) to a memory
9. location and terminate the program.
PROGRAM:
MVI C, 00 Initialize C to 00
LDA 4150 Load the value to Acc.
MOV B, A Move the content of Acc to B register.
SUB B
CMA Complement Accumulator contents.
INR A Increment value in Accumulator.
INR C Increment value in register C
LOOP: STA 4152 Store the value of A-reg to memory address.
MOV A, C Move contents of register C to Accumulator.
STA 4153 Store the value of Accumulator memory address.
HLT Terminate the program.
3
OBSERVATION:
Input: 06 (4150)
02 (4251)
Output: 04 (4152)
01 (4153)
RESULT:
Thus the program to subtract two 8-bit numbers was executed.
4
MULTIPLICATION OF TWO 8 BIT NUMBERS
AIM:
To perform the multiplication of two 8 bit numbers using 8085.
ALGORITHM:
1) Start the program by loading HL register pair with address of memory location.
5) Check for carry.
6) Increment the value of carry.
7) Check whether repeated addition is over and store the value of product and carry
in memory location.
PROGRAM:
MVI D, 00 Initialize register D to 00

MVI A, 00 Initialize Accumulator content to 00
LXI H, 4150
MOV B, M Get the first number in B - reg
INX H
MOV C, M Get the second number in C- reg.
LOOP: ADD B Add content of A - reg to register B.
JNC NEXT Jump on no carry to NEXT.
INR D Increment content of register D
NEXT: DCR C Decrement content of register C.
JNZ LOOP Jump on no zero to address
STA 4152 Store the result in Memory
MOV A, D
STA 4153 Store the MSB of result in Memory
5
OBSERVATION:
Input: FF (4150)
FF (4151)
Output: 01 (4152)
FE (4153)
RESULT:
Thus the program to multiply two 8-bit numbers was executed.
6
DIVISION OF TWO 8 BIT NUMBERS
AIM:
To perform the division of two 8 bit numbers using 8085.
ALGORITHM:
2) Move the data to a register(B register).
4) Compare the two numbers to check for carry.
5) Subtract the two numbers.
6) Increment the value of carry .
7) Check whether repeated subtraction is over and store the value of product and
carry in memory location.
PROGRAM:
LXI H, 4150
MOV B, M Get the dividend in B – reg.
MVI C, 00 Clear C – reg for qoutient
INX H
MOV A, M Get the divisor in A – reg.
NEXT: CMP B Compare A - reg with register B.
JC LOOP Jump on carry to LOOP
SUB B Subtract A – reg from B- reg.
INR C Increment content of register C.
JMP NEXT Jump to NEXT
LOOP: STA 4152 Store the remainder in Memory
MOV A, C
STA 4153 Store the quotient in memory
7
OBSERVATION:
Input: FF (4150)
FF (4251)
Output: 01 (4152) ---- Remainder

FE (4153) ---- Quotient
RESULT:
Thus the program to divide two 8-bit numbers was executed.
8
LARGEST NUMBER IN AN ARRAY OF DATA
AIM:
To find the largest number in an array of data using 8085 instruction set.
ALGORITHM:
1) Load the address of the first element of the array in HL pair

2) Move the count to B – reg.
3) Increment the pointer
4) Get the first data in A – reg.
5) Decrement the count.
7) Compare the content of memory addressed by HL pair with that of A - reg.
8) If Carry = 0, go to step 10 or if Carry = 1 go to step 9
9) Move the content of memory addressed by HL to A – reg.
10) Decrement the count
11) Check for Zero of the count. If ZF = 0, go to step 6, or if ZF = 1 go to next step.
12) Store the largest data in memory.
PROGRAM:
LXI H,4200 Set pointer for array

MOV B,M Load the Count
INX H
MOV A,M Set 1st element as largest data
DCR B Decrement the count
LOOP: INX H
CMP M If A- reg > M go to AHEAD
JNC AHEAD
MOV A,M Set the new value as largest
AHEAD: DCR B
JNZ LOOP Repeat comparisons till count = 0
STA 4300 Store the largest value at 4300
HLT
9
OBSERVATION:
Input: 05 (4200) ----- Array Size

0A (4201)
F1 (4202)
1F (4203)
26 (4204)
FE (4205)
Output: FE (4300)
RESULT:
Thus the program to find the largest number in an array of data was executed
10
SMALLEST NUMBER IN AN ARRAY OF DATA
AIM:
To find the smallest number in an array of data using 8085 instruction set.
ALGORITHM:

8) If carry = 1, go to step 10 or if Carry = 0 go to step 9
12) Store the smallest data in memory.
PROGRAM:

INX H
LOOP: INX H
CMP M If A- reg < M go to AHEAD
JC AHEAD
MOV A,M Set the new value as smallest
AHEAD: DCR B
HLT
11
OBSERVATION:

0A (4201)
F1 (4202)
1F (4203)
26 (4204)
FE (4205)
Output: 0A (4300)
RESULT:
Thus the program to find the smallest number in an array of data was executed
12
ARRANGE AN ARRAY OF DATA IN ASCENDING ORDER
AIM:
To write a program to arrange an array of data in ascending order
ALGORITHM:
1. Initialize HL pair as memory pointer

2. Get the count at 4200 into C – register
3. Copy it in D – register (for bubble sort (N-1) times required)
4. Get the first value in A – register
5. Compare it with the value at next location.
6. If they are out of order, exchange the contents of A –register and Memory
7. Decrement D –register content by 1
8. Repeat steps 5 and 7 till the value in D- register become zero
9. Decrement C –register content by 1
10. Repeat steps 3 to 9 till the value in C – register becomes zero
PROGRAM:
LXI H,4200
MOV C,M
DCR C
REPEAT: MOV D,C
LXI H,4201
LOOP: MOV A,M
INX H
CMP M
JC SKIP
MOV B,M
MOV M,A
DCX H
MOV M,B
INX H
SKIP: DCR D
JNZ LOOP
DCR C
JNZ REPEAT
HLT
13
OBSERVATION:
Input: 4200 05 (Array Size)

4201 05
4202 04
4203 03
4204 02
4205 01
Output: 4200 05(Array Size)

4201 01
4202 02
4203 03
4204 04
4205 05
RESULT:
Thus the given array of data was arranged in ascending order.
14
ARRANGE AN ARRAY OF DATA IN DESCENDING ORDER
AIM:
To write a program to arrange an array of data in descending order
ALGORITHM:

PROGRAM:
LXI H,4200
MOV C,M
DCR C
REPEAT: MOV D,C
LXI H,4201
LOOP: MOV A,M
INX H
CMP M
JNC SKIP
MOV B,M
MOV M,A
DCX H
MOV M,B
INX H
SKIP: DCR D
JNZ LOOP
DCR C
JNZ REPEAT
HLT
15
OBSERVATION:

4201 01
4202 02
4203 03
4204 04
4205 05

4201 05
4202 04
4203 03
4204 02
4205 01
RESULT:
Thus the given array of data was arranged in descending order.
16
BCD TO HEX CONVERSION

AIM:
To convert two BCD numbers in memory to the equivalent HEX number using 8085
instruction set
ALGORITHM:
1) Initialize memory pointer to 4150 H

2) Get the Most Significant Digit (MSD)
3) Multiply the MSD by ten using repeated addition
4) Add the Least Significant Digit (LSD) to the result obtained in previous step
5) Store the HEX data in Memory
PROGRAM:
LXI H,4150
MOV A,M Initialize memory pointer
ADD A MSD X 2
MOV B,A Store MSD X 2
ADD A MSD X 4
ADD A MSD X 8
ADD B MSD X 10
INX H Point to LSD
ADD M Add to form HEX
INX H
MOV M,A Store the result
HLT
OBSERVATION:
Input: 4150 : 02 (MSD)

4151 : 09 (LSD)
Output: 4152 : 1D H
RESULT:
Thus the program to convert BCD data to HEX data was executed.
17
HEX TO BCD CONVERSION

AIM:
To convert given Hexa decimal number into its equivalent BCD number using 8085
instruction set
ALGORITHM:

2) Get the Hexa decimal number in C - register
3) Perform repeated addition for C number of times
4) Adjust for BCD in each step
5) Store the BCD data in Memory
PROGRAM:
LXI H,4150 Initialize memory pointer

MVI D,00 Clear D- reg for Most significant Byte
XRA A Clear Accumulator
MOV C,M Get HEX data
LOOP2: ADI 01 Count the number one by one
DAA Adjust for BCD count
JNC LOOP1
INR D
LOOP1: DCR C
JNZ LOOP2
STA 4151 Store the Least Significant Byte
MOV A,D
STA 4152 Store the Most Significant Byte
HLT
OBSERVATION:
Input: 4150 : FF
Output: 4151 : 55 (LSB)

4152 : 02 (MSB)
RESULT:
Thus the program to convert HEX data to BCD data was executed.
18
HEX TO ASCII CONVERSION

AIM:
To convert given Hexa decimal number into its equivalent ASCII number using
8085 instruction set.
ALGORITHM:
1. Load the given data in A- register and move to B – register

2. Mask the upper nibble of the Hexa decimal number in A – register
3. Call subroutine to get ASCII of lower nibble
4. Store it in memory
5. Move B –register to A – register and mask the lower nibble
6. Rotate the upper nibble to lower nibble position
7. Call subroutine to get ASCII of upper nibble
9. Terminate the program.
PROGRAM:
LDA 4200 Get Hexa Data

MOV B,A
ANI 0F Mask Upper Nibble
CALL SUB1 Get ASCII code for upper nibble
STA 4201
MOV A,B
ANI F0 Mask Lower Nibble
RLC
RLC
RLC
RLC
CALL SUB1 Get ASCII code for lower nibble
STA 4202
HLT
SUB1: CPI 0A
JC SKIP
ADI 07
SKIP: ADI 30
RET
19
OBSERVATION:
Input: 4200 E4(Hexa data)
Output: 4201 34(ASCII Code for 4)

4202 45(ASCII Code for E)
RESULT:
Thus the given Hexa decimal number was converted into its equivalent ASCII Code.
20
ASCII TO HEX CONVERSION

AIM:
To convert given ASCII Character into its equivalent Hexa Decimal number using
ALGORITHM:
1. Load the given data in A- register

2. Subtract 30 H from A – register
3. Compare the content of A – register with 0A H
4. If A < 0A H, jump to step6. Else proceed to next step.
6. Store the result
7. Terminate the program
PROGRAM:
LDA 4500
SUI 30
CPI 0A
JC SKIP
SUI 07
SKIP: STA 4501
HLT
OBSERVATION:
Input: 4500 31
Output: 4501 0B
RESULT:
Thus the given ASCII character was converted into its equivalent Hexa Value.
21
SQUARE OF A NUMBER USING LOOK UP TABLE

AIM:
To find the square of the number from 0 to 9 using a Table of Square.
ALGORITHM:
1. Initialize HL pair to point Look up table

2. Get the data .
3. Check whether the given input is less than 9.
4. If yes go to next step else halt the program
5. Add the desired address with the accumulator content
6. Store the result
PROGRAM:
LXI H,4125 Initialsie Look up table address

LDA 4150 Get the data
CPI 0A Check input > 9
JC AFTER if yes error
MVI A,FF Error Indication
STA 4151
HLT
AFTER: MOV C,A Add the desired Address
MVI B,00
DAD B
MOV A,M
STA 4151 Store the result
HLT Terminate the program
LOOKUP TABLE:
4125 01
4126 04
4127 09
4128 16
4129 25
4130 36
4131 49
4132 64
4133 81
22
OBSERVATION:
Input: 4150: 05
Output: 4151 25 (Square)
Input : 4150: 11
Output: 4151: FF (Error Indication)
RESULT:
Thus the program to find the square of the number from 0 to 9 using a Look up table
was executed.
23
INTERFACING WITH 8085
24
INTERFACING 8251 (USART) WITH 8085 PROCESSOR
AIM:
To write a program to initiate 8251 and to check the transmission and reception of
character
THEORY:
The 8251 is used as a peripheral device for serial communication and is programmed
by the CPU to operate using virtually any serial data transmission technique. The USART
accepts data characters from the CPU in parallel format and then converts them into a
continuous serial data stream for transmission. Simultaneously, it can receive serial data
streams and convert them into parallel data characters for the CPU. The CPU can read the
status of USART ant any time. These include data transmission errors and control signals.
Prior to starting data transmission or reception, the 8251 must be loaded with a set
of control words generated by the CPU. These control signals define the complete
functional definition of the 8251 and must immediately follow a RESET operation. Control
words should be written into the control register of 8251. These control words are split into
two formats:
1. MODE INSTRUCTION WORD

2. COMMAND INSTRUCTION WORD
This format defines the Baud rate, Character length, Parity and Stop bits required to
work with asynchronous data communication. By selecting the appropriate baud factor sync
mode, the 8251 can be operated in Synchronous mode.
Initializing 8251 using the mode instruction to the following conditions
8 Bit data
No Parity
Baud rate Factor (16X)
1 Stop Bit
gives a mode command word of 01001110 = 4E (HEX)
25
MODE INSTRUCTION - SYNCHRONOUS MODE
S2 S1 EP PEN L2 L1 B2 B1
BAUD RATE FACTOR

0 1 0 1
0 0 1 1
SYNC MODE (1X) (16X) (64X)
CHARACTR LENGTH
0 1 0 1
0 0 1 1
6 7
5 BITS BITS BITS 8 BITS
PARITY ENABLE
1= ENABLE 0 = DISABLE
EVEN PARITY GEN/CHECK

0 =ODD 1 = EVEN
NUMBER OF STOP BITS

0 1 0 1
0 0 1 1
1.5
INVALID 1 BIT BIT 2 BIT
26
MODE INSTRUCTION - ASYNCHRONOUS MODE
CHARACTER LENGTH
0 1 0 1
0 0 1 1
5 BITS 6 BITS 7 BITS 8 BITS
PARITY ENABLE

0 =ODD 1 = EVEN
EXTERNAL SYNC DETECTS

1 = SYSDET IS AN INPUT
0 = SYSDET IS AN IOUTPUT
SINGLE CHARACTER SYNC

1 = SINGLE SYNC CHARACTER
0 = DOUBLE SYNC CHARACTER
This format defines a status word that is used to control the actual operation of
8251. All control words written into 8251 after the mode instruction will load the command
instruction.
The command instructions can be written into 8251 at any time in the data block
during the operation of the 8251. to return to the mode instruction format, the master reset
bit in the command instruction word can be set to initiate an internal reset operation which
automatically places the 8251 back into the mode instruction format. Command instructions
must follow the mode instructions or sync characters.
Thus the control word 37 (HEX) enables the transmit enable and receive enable bits,
forces DTR output to zero, resets the error flags, and forces RTS output to zero.
27
EH IR RTS ER SBRK RXE DTR TXEN
TRANSMIT ENABLE
1=Enable 0 = Disable
DATA TERMINAL READY

HIGH will force DTR
Output to Zero
RECEIVE ENABLE
SEND BREAK CHARACTER

1 = Forces TXD LOW
0 = Normal Operation
ERROR RESET
1=Reset Error Flags
PE,OE,FE
REQUEST TO SEND
HIGH will force RTS
Output to Zero
INTERNAL RESET
HIGH Returns 8251 to
Mode Instruction Format
ENTER HUNT MODE

1= Enable a Search for
Sync Characters( Has
No Effect in Async mode)
COMMAND INSTRUCTION FORMAT
28
ALGORITHM:
1. Initialise timer (8253) IC

2. Move the mode command word (4E H) to A -reg
3. Output it to port address C2
4. Move the command instruction word (37 H) to A -reg
6. Move the the data to be transferred to A -reg
8. Reset the system
9. Get the data through input port address C0
10. Store the value in memory
11. Reset the system
PROGRAM:
MVI A,36H
OUT CEH
MVI A,0AH
OUT C8H
MVI A,00
OUT C8H
LXI H,4200
MVI A,4E
OUT C2
MVI A,37
OUT C2
MVI A,41
OUT C0
RST 1
ORG 4200
IN C0
STA 4500
RST 1
OBSERVATION:
Output: 4500 41
RESULT:
Thus the 8251 was initiated and the transmission and reception of character was
done successfully.
29
INTERFACING ADC WITH 8085 PROCESSOR
AIM:
To write a program to initiate ADC and to store the digital data in memory
PROGRAM:
MVI A,10
OUT C8
MVI A,18
OUT C8
MVI A,10
OUT D0
XRA A
XRA A
XRA A
MVI A,00
OUT D0
LOOP: IN D8
ANI 01
CPI 01
JNZ LOOP
IN C0
STA 4150
HLT
OBSERVATION:
Compare the data displayed at the LEDs with that stored at location 4150
RESULT:
Thus the ADC was initiated and the digital data was stored at desired location
30
INTERFACING DAC WITH 8085

AIM:
To interface DAC with 8085 to demonstrate the generation of square, saw tooth and
triangular 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 varies in steps of 10/256 = 0.04 (appx.). The digital data input and
the corresponding output voltages are presented in the Table1.
Input Output
Data in Voltage
HEX
00 - 5.00
01 - 4.96
02 - 4.92
… …
7F 0.00
… …
FD 4.92
FE 4.96
FF 5.00
Referring to Table1, with 00 H as input to DAC, the analog output is – 5V. Similarly,
with FF H as input, the output is +5V. Outputting digital data 00 and FF at regular intervals,
to DAC, results in different wave forms namely square, triangular, etc,. The port address of
DAC is 08 H.
31
ALGORITHM:
(a) Square Wave Generation
1. Load the initial value (00) to Accumulator and move it to DAC

2. Call the delay program
3. Load the final value(FF) to accumulator and move it to DAC
4. Call the delay program.
5. Repeat Steps 2 to 5
(b) Saw tooth Wave Generation
1. Load the initial value (00) to Accumulator

2. Move the accumulator content to DAC
3. Increment the accumulator content by 1.
4. Repeat Steps 3 and 4.
(c) Triangular Wave Generation

5. If accumulator content is zero proceed to next step. Else go to step 3.
6. Load value (FF) to Accumulator
8. Decrement the accumulator content by 1.
9. If accumulator content is zero go to step2. Else go to step 7.
32
PROGRAM:
START: MVI A,00

OUT Port address of DAC
CALL DELAY
MVI A,FF
CALL DELAY
JMP START
DELAY: MVI B,05

L1: MVI C,FF
L2: DCR C
JNZ L2
DCR B
JNZ L1
RET
START: MVI A,00

L1: OUT Port address of DAC
INR A
JNZ L1
JMP START
33
START: MVI L,00

L1: MOV A,L
INR L
JNZ L1
MVI L,FF
L2: MOV A,L
DCR L
JNZ L2
JMP START
RESULT:
Thus the square, triangular and saw tooth wave form were generated by interfacing
DAC with 8085 trainer kit.
34
INTERFACING 8253 (TIMER IC) WITH 8085 PROCESSOR
AIM:
To interface 8253 Programmable Interval Timer to 8085 and verify the operation
of 8253 in six different modes.
APPARATUS REQUIRED:
1) 8085 Microprocessor toolkit.

2) 8253 Interface board.
3) VXT parallel bus.
4) Regulated D.C power supply.
5) CRO.
MODE 0-Interrupt On Terminal Count:-
The output will be initially low after mode set operation. After loading the counter, the
output will remain low while counting and on terminal count, the output will become high
until reloaded again.
Let us see the channel in mode0. Connect the CLK 0 to the debounce circuit and
execute the following program.
PROGRAM:
MVI A, 30H ;Channel 0 in mode 0.
OUT CEH
MVI A, 05H ;LSB of count.
OUT C8H
MVI A, 00H ;MSB of count.
OUT C8H
HLT
It is observed in CRO that the output of channel 0 is initially low. After giving ‘x’ clock
pulses, we may notice that the output goes high.
MODE 1-Programmable One Shot:-
After loading the count, the output will remain low following the rising edge of the
gate input. The output will go high on the terminal count. It is retriggerable; hence the
output will remain low for the full count after any rising edge of the gate input.
35
The following program initializes channel 0 of 8253 in Mode 1 and also initializes triggering
of gate. OUT 0 goes low as clock pulses and after triggering It goes back to high level after
five clock pulses. Execute the program and give clock pulses through the debounce logic and
verify using CRO.
PROGRAM:
OUT CEH ;
OUT C8H
OUT C8H
OUT DOH ;Trigger Gate 0.
HLT
MODE 2-Rate Generator:
It is a simple divide by N counter. The output will be low for one period of the input
clock. The period from one output pulse to next equals the number of input count in the
count register. If the count register is reloaded between output pulses, the present period will
not be affected, but the subsequent period will reflect a new value.
MODE 3-Square Generator:
It is similar to mode 2 except that the output will remain high until one half of the
count and goes low for the other half provided the count is an even number. If the count is
odd the output will be high for (count +1)/2 counts. This mode is used for generating baud
rate of 8251.
PROGRAM:

OUT CEH ;
MVI A, 0AH ;LSB of count.
OUT C8H
OUT C8H
HLT
We utilize mode 3 to generate a square wave of frequency 150 kHz at Channel 0.Set the
jumper so that the clock of 8253 is given a square wave of Frequency 1.5 MHz. This
program divides the program clock by 10 and thus the Output at channel 0 is 150 KHz.
36
MODE 4-Software Triggered Strobe:
The output is high after the mode is set and also during counting. On Terminal
count, the output will go low for one clock period and becomes high again. This mode can
be used for interrupt generation.
MODE 5-Hardware Triggered Strobe:
Counter starts counting after rising edge of trigger input and the output goes low for
one clock period. When the terminal count is reached, the counter is retrigerrable. On
terminal count, the output will go low for one clock period and becomes high again. This
mode can be used for interrupt generation.
RESULT:
Thus the 8253 PIT was interfaced to 8085 and the operations for mode 0, Mode 1
and mode 3 was verified.
37
INTERFACING 8279 KEYBOARD/DISPLAY CONTROLLER

WITH 8085 MICROPROCESSOR
AIM:
To interface 8279 Programmable Keyboard Display Controller to 8085

Microprocessor.
APPARATUS REQUIRED:

PROGRAM:
START: LXI H,4130H

MVI D,0FH ;Initialize counter.
MVI A,10H
OUT C2H ;Set Mode and Display.
MVI A,CCH ;Clear display.
OUT C2H
MVI A,90H ;Write Display
OUT C2H
LOOP: MOV A,M
OUT C0H
CALL DELAY
INX H
DCR D
JNZ LOOP
JMP START
DELAY: MVI B, A0H

LOOP2: MVI C, FFH
LOOP1: DCR C
JNZ LOOP1
DCR B
JNZ LOOP2
RET
38
Pointer equal to 4130 .FF repeated eight times.
4130 - FF
4131 –FF
4132 –FF
4133 –FF
4134 –FF
4135 –FF
4136 –FF
4137 –FF
4138 –98
4139 –68
413A -7C
413B -C8
413C -1C
413D -29
413E -FF
413F -FF
RESULT:
Thus 8279 controller was interfaced with 8085 and program for rolling display was
executed successfully.
39
8051 MICROCONTROLLER
PROGRAMS
40
ADDITION OF TWO 8 – BIT NUMBERS
AIM:
To perform addition of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:
1. Clear C – register for Carry

2. Get the data immediately .
3. Add the two data
4. Store the result in memory pointed by DPTR
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
ADD A,#data2
MOV DPTR,#4500
MOVX @DPTR,A
HERE: SJMP HERE
OBSERVATION:
Input: 66
23
Output: 89 (4500)
RESULT:
Thus the program to perform addition of two 8 – bit numbers using 8051 instruction set
was executed.
41
SUBTRACTION OF TWO 8 – BIT NUMBERS
AIM:
To perform Subtraction of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:

3. Subtract the two data
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
SUBB A,#data2
MOV DPTR,#4500
MOVX @DPTR,A
HERE: SJMP HERE
OBSERVATION:
Input: 66
23
Output: 43 (4500)
RESULT:
Thus the program to perform subtraction of two 8 – bit numbers using 8051 instruction
set was executed.
42
MULTIPLICATION OF TWO 8 – BIT NUMBERS
AIM:
To perform multiplication of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:
1. Get the data in A – reg.

2. Get the value to be multiplied in B – reg.
3. Multiply the two data
4. The higher order of the result is in B – reg.
5. The lower order of the result is in A – reg.
6. Store the results.
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
MOV B,#data2
MUL AB
MOV DPTR,#4500
MOVX @DPTR,A
INC DPTR
MOV A,B
MOVX @DPTR,A
HERE: SJMP HERE
43
OBSERVATION:
Input: 80
80
Output: 00 (4500)
19 (4501)
RESULT:
Thus the program to perform multiplication of two 8 – bit numbers using 8051
instruction set was executed.
44
DIVISION OF TWO 8 – BIT NUMBERS
AIM:
To perform division of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:

2. Get the value to be divided in B – reg.
3. Divide the two data
4. The quotient is in A – reg.
5. The remainder is in B – reg.
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
MOV B,#data2
DIV AB
MOV DPTR,#4500
MOVX @DPTR,A
INC DPTR
MOV A,B
MOVX @DPTR,A
HERE: SJMP HERE
45
OBSERVATION:
Input: 05
03
Output: 01 (4500)
02 (4501)
RESULT:
46
RAM ADDRESSING
AIM:
To exhibit the RAM direct addressing and bit addressing schemes of 8051
microcontroller.
ALGORITHM:
1. For Bit addressing, Select Bank 1 of RAM by setting 3rd bit of PSW
2. Using Register 0 of Bank 1 and accumulator perform addition
3. For direct addressing provide the address directly (30 in this case)
4. Use the address and Accumulator to perform addition
5. 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
47
OBSERVATION:
Bit addressing:
Input: 54
25
Output: 79 (4500)
Direct addressing:
Input: 54
25
Output: 79 (4500)
RESULT:
Thus the program to exhibit the different RAM addressing schemes of 8051 was
executed.
48
INTERFACING STEPPER MOTOR WITH 8051
AIM:
To interface stepper motor with 8051 parallel port and to vary speed of motor, direction of
motor.
APPARATUS REQUIRED:

• Stepper Motor Interface Board
THEORY:
A motor in which the rotor is able to assume only discrete stationary angular position is a
stepper motor. The rotor motion occurs in a stepwise manner from one equilibrium
position to next.
The motor under our consideration uses 2 – phase scheme of operation. In this scheme,
any two adjacent stator windings are energized. The switching condition for the above said
scheme is shown in Table.
Clockwise Anti - Clockwise

A1 B1 A2 B2 A1 B1 A2 B2
1 0 0 1 1 0 1 0
0 1 0 0 0 1 0 1
0 1 0 1 0 1 0 0
1 0 1 0 1 0 0 1
In order to vary the speed of the motor, the values stored in the registers R1, R2, R3 can
be changed appropriately.
49
ALGORITHM:
1. Store the look up table address in DPTR

2. Move the count value (04) to one of the register (R0)
3. Load the control word for motor rotation in accumulator
4. Push the address in DPTR into stack
5. Load FFC0 in to DPTR.
7. Send the control word for motor rotation to the external device.
8. Pop up the values in stack and increment it.
9. Decrement the count in R0. If zero go to next step else proceed to step 3.
10. Perform steps 1 to 9 repeatedly.
PROGRAM:
ORG 4100
START: MOV DPTR,#4500H
MOV R0,#04
AGAIN: MOVX A,@DPTR
PUSH DPH
PUSH PDL
MOV DPTR,#FFC0H
MOV R2, 04H
MOV R1,#FFH
DLY1: MOV R3, #FFH
DLY: DJNZ R3,DLY
DJNZ R1,DLY1
DJNZ R2,DLY1
MOVX @DPTR,A
POP DPL
POP DPH
INC DPTR
DJNZ R0,AGAIN
SJMP START
50
DATA:
4500: 09, 05, 06, 0A
RESULT:
Thus the speed and direction of motor were controlled using 8051 trainer kit.
51
AIM:
To interface DAC with 8051 parallel port to demonstrate the generation of square,
saw tooth and triangular wave.
APPARATUS REQUIRED:

THEORY:
the corresponding output voltages are presented in the Table below
.
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
to DAC, results in different wave forms namely square, triangular, etc,.
52
ALGORITHM:
1. Move the port address of DAC to DPTR



53
PROGRAM:
ORG 4100
MOV DPTR,PORT ADDRESS OF DAC
START: MOV A,#00
MOVX @DPTR,A
LCALL DELAY
MOV A,#FF
MOVX @DPTR,A
LCALL DELAY
LJUMP START
DELAY: MOV R1,#05

LOOP: MOV R2,#FF
HERE: DJNZ R2,HERE
DJNZ R1,LOOP
RET
SJMP START
ORG 4100
MOV A,#00
LOOP: MOVX @DPTR,A
INC A
SJMP LOOP
54
ORG 4100
START: MOV A,#00
LOOP1: MOVX @DPTR,A
INC A
JNZ LOOP1
MOV A,#FF
LOOP2: MOVX @DPTR,A
DEC A
JNZ LOOP2
LJMP START
RESULT:
55
PROGRAMMING 8051 USING KEIL SOFTWARE
AIM:
To perform arithmetic operations in 8051 using keil software.
PROCEDURE:
1. Click KeilµVision2 icon in the desktop

2. From Project Menu open New project
3. Select the target device as ATMEL 89C51
4. From File Menu open New File
5. Type the program in Text Editor
6. Save the file with extension “.asm”
7. In project window click the tree showing TARGET
8. A source group will open.
9. Right Click the Source group and click “Add files to Source group”
10. A new window will open. Select our file with extension “.asm”
11. Click Add.
12. Go to project window and right click Source group again
13. Click Build Target (F7).
14. Errors if any will be displayed.
15. From Debug menu, select START/STOP Debug option.
16. In project window the status of all the registers will be displayed.
17. Click Go from Debug Menu.
18. The results stored in registers will be displayed in Project window.
19. Stop the Debug process before closing the application.
PROGRAM:
ORG 4100
CLR C
MOV A,#05H
MOV B,#02H
DIV AB
56
OBSERVATION:
A: 02
B: 01
SP:07
Note that Stack pointer is initiated to 07H
RESULT:
Thus the arithmetic operation for 8051 was done using Keil Software.
57
SYSTEM DESIGN USING MICROCONTROLLER
AIM:
To Design a microcontroller based system for simple applications like security

systems combination lock etc.
PROCEDURE:
1. Read number of bytes in the password

2. Initialize the password
3. Initialize the Keyboard Display IC (8279) to get key and Display
4. Blank the display
5. Read the key from user
6. Compare with the initialized password
7. If it is not equal, Display ‘E’ to indicate Error.
8. Repeat the steps 6 and 7 to read next key
9. If entered password equal to initialized password, Display ‘O’ to indicate open.
PROGRAM:
MOV 51H,#
MOV 52H,#
MOV 53H,#
MOV 54H,#
MOV R1,#51
MOV R0,#50
MOV R3,#04
MOV R2,#08
MOV DPTR,#FFC2
MOV A,#00
58
MOVX @DPTR,A
MOV A,#CC
MOVX @DPTR,A
MOV A,#90
MOVX @DPTR,A
MOV A,#FF
MOV DPTR,#FFCO
LOOP: MOVX @DPTR,A
DJNZ R2,LOOP
AGAIN: MOV DPTR,#FFC2
WAIT: MOVX A,@DPTR
ANL A,#07
JZ WAIT
MOV A,#40
MOVX @DPTR,A
MOV DPTR,#FFCO
MOVX A,@DPTR
MOV @R0,A
MOV A,@R1
CJNE A,50H,NEQ
INC R1
DJNZ R3,AGAIN
MOV DPTR,#FFCO
MOV A,#OC
MOVX @DPTR,A
XX: SJMP XX
NEQ: MOV DPTR,#FFCO

MOV A,#68
MOVX @DPTR,A
YY: SJMP YY
RESULT:
Thus the program for security lock system was executed
59
Tutorial
On
Introduction to 8085 Architecture and Programming

Contents
1. Internal architecture of 8085 microprocessor

2. 8085 system bus
3. 8085 pin description.
4. 8085 functional description.
5. Programming model of 8085 microprocessor
6. Addressing modes.
7. Instruction set classification.
8. Instruction format.
9. Sample programs.
1. Internal Architecture of 8085 Microprocessor
Control Unit
Generates signals within uP to carry out the instruction, which has been decoded. In
reality causes certain connections between blocks of the uP to be opened or closed, so
that data goes where it is required, and so that ALU operations occur.
Arithmetic Logic Unit

The ALU performs the actual numerical and logic operation such as ‘add’, ‘subtract’,
‘AND’, ‘OR’, etc. Uses data from memory and from Accumulator to perform
arithmetic. Always stores result of operation in Accumulator.
Registers
The 8085/8080A-programming model includes six registers, one accumulator, and

one flag register, as shown in Figure. In addition, it has two 16-bit registers: the stack
pointer and the program counter. They are described briefly as follows.
The 8085/8080A has six general-purpose registers to store 8-bit data; these are
identified as B,C,D,E,H, and L as shown in the figure. They can be combined as
register pairs - BC, DE, and HL - to perform some 16-bit operations. The
programmer can use these registers to store or copy data into the registers by using
data copy instructions.
Accumulator
The accumulator is an 8-bit register that is a part of arithmetic/logic unit (ALU). This
register is used to store 8-bit data and to perform arithmetic and logical operations.
The result of an operation is stored in the accumulator. The accumulator is also
identified as register A.
Flags
The ALU includes five flip-flops, which are set or reset after an operation according
to data conditions of the result in the accumulator and other registers. They are called
Zero(Z), Carry (CY), Sign (S), Parity (P), and Auxiliary Carry (AC) flags; they are
listed in the Table and their bit positions in the flag register are shown in the Figure
below. The most commonly used flags are Zero, Carry, and Sign. The microprocessor
uses these flags to test data conditions.
For example, after an addition of two numbers, if the sum in the accumulator id larger
than eight bits, the flip-flop uses to indicate a carry -- called the Carry flag (CY) -- is
set to one. When an arithmetic operation results in zero, the flip-flop called the
Zero(Z) flag is set to one. The first Figure shows an 8-bit register, called the flag
register, adjacent to the accumulator. However, it is not used as a register; five bit
positions out of eight are used to store the outputs of the five flip-flops. The flags are
stored in the 8-bit register so that the programmer can examine these flags (data
conditions) by accessing the register through an instruction.
These flags have critical importance in the decision-making process of the micro-
processor. The conditions (set or reset) of the flags are tested through the software
instructions. For example, the instruction JC (Jump on Carry) is implemented to
change the sequence of a program when CY flag is set. The thorough understanding
of flag is essential in writing assembly language programs.
Program Counter (PC)
This 16-bit register deals with sequencing the execution of instructions. This register
is a memory pointer. Memory locations have 16-bit addresses, and that is why this is a
16-bit register.
The microprocessor uses this register to sequence the execution of the instructions.
The function of the program counter is to point to the memory address from which the
next byte is to be fetched. When a byte (machine code) is being fetched, the program
counter is incremented by one to point to the next memory location
Stack Pointer (SP)
The stack pointer is also a 16-bit register used as a memory pointer. It points to a
memory location in R/W memory, called the stack. The beginning of the stack is
defined by loading 16-bit address in the stack pointer. The stack concept is explained
in the chapter "Stack and Subroutines."
Instruction Register/Decoder
Temporary store for the current instruction of a program. Latest instruction sent here
from memory prior to execution. Decoder then takes instruction and ‘decodes’ or
interprets the instruction. Decoded instruction then passed to next stage.
Memory Address Register

Holds address, received from PC, of next program instruction. Feeds the address bus
with addresses of location of the program under execution.
Control Generator
Generates signals within uP to carry out the instruction which has been decoded. In
reality causes certain connections between blocks of the uP to be opened or closed, so
that data goes where it is required, and so that ALU operations occur.
Register Selector
This block controls the use of the register stack in the example. Just a logic circuit
which switches between different registers in the set will receive instructions from
Control Unit.
General Purpose Registers
uP requires extra registers for versatility. Can be used to store additional data during a
program. More complex processors may have a variety of differently named registers.
Microprogramming
How does the µP knows what an instruction means, especially when it is only a
binary number? The microprogram in a uP/uC is written by the chip designer and tells
the uP/uC the meaning of each instruction uP/uC can then carry out operation.
2. 8085 System Bus

Typical system uses a number of busses, collection of wires, which transmit binary
numbers, one bit per wire. A typical microprocessor communicates with memory and
other devices (input and output) using three busses: Address Bus, Data Bus and
Control Bus.
Address Bus
One wire for each bit, therefore 16 bits = 16 wires. Binary number carried alerts
memory to ‘open’ the designated box. Data (binary) can then be put in or taken
out.The Address Bus consists of 16 wires, therefore 16 bits. Its "width" is 16 bits. A
16 bit binary number allows 216 different numbers, or 32000 different numbers, ie
0000000000000000 up to 1111111111111111. Because memory consists of boxes,
each with a unique address, the size of the address bus determines the size of memory,
which can be used. To communicate with memory the microprocessor sends an
address on the address bus, eg 0000000000000011 (3 in decimal), to the memory. The
memory the selects box number 3 for reading or writing data. Address bus is
unidirectional, ie numbers only sent from microprocessor to memory, not other way.
Question?: If you have a memory chip of size 256 kilobytes (256 x 1024 x 8 bits),
how many wires does the address bus need, in order to be able to specify an address in
this memory? Note: the memory is organized in groups of 8 bits per location,
therefore, how many locations must you be able to specify?
Data Bus
Data Bus: carries ‘data’, in binary form, between µP and other external units, such as
memory. Typical size is 8 or 16 bits. Size determined by size of boxes in memory and
µP size helps determine performance of µP. The Data Bus typically consists of 8
wires. Therefore, 28 combinations of binary digits. Data bus used to transmit "data",
ie information, results of arithmetic, etc, between memory and the microprocessor.
Bus is bi-directional. Size of the data bus determines what arithmetic can be done. If
only 8 bits wide then largest number is 11111111 (255 in decimal). Therefore, larger
number have to be broken down into chunks of 255. This slows microprocessor. Data
Bus also carries instructions from memory to the microprocessor. Size of the bus
therefore limits the number of possible instructions to 256, each specified by a
separate number.
Control Bus
Control Bus are various lines which have specific functions for coordinating and
controlling uP operations. Eg: Read/NotWrite line, single binary digit. Control
whether memory is being ‘written to’ (data stored in mem) or ‘read from’ (data taken
out of mem) 1 = Read, 0 = Write. May also include clock line(s) for
timing/synchronising, ‘interrupts’, ‘reset’ etc. Typically µP has 10 control lines.
Cannot function correctly without these vital control signals.
The Control Bus carries control signals partly unidirectional, partly bi-directional.
Control signals are things like "read or write". This tells memory that we are either
reading from a location, specified on the address bus, or writing to a location
specified. Various other signals to control and coordinate the operation of the system.
Modern day microprocessors, like 80386, 80486 have much larger busses. Typically
16 or 32 bit busses, which allow larger number of instructions, more memory
location, and faster arithmetic. Microcontrollers organized along same lines, except:
because microcontrollers have memory etc inside the chip, the busses may all be
internal. In the microprocessor the three busses are external to the chip (except for the
internal data bus). In case of external busses, the chip connects to the busses via
buffers, which are simply an electronic connection between external bus and the
internal data bus.
3. 8085 Pin description.
Properties
Single + 5V Supply
4 Vectored Interrupts (One is Non Maskable)
Serial In/Serial Out Port
Decimal, Binary, and Double Precision Arithmetic
Direct Addressing Capability to 64K bytes of memory
The Intel 8085A is a new generation, complete 8 bit parallel central processing unit
(CPU). The 8085A uses a multiplexed data bus. The address is split between the 8bit
address bus and the 8bit data bus. Figures are at the end of the document.
Pin Description
The following describes the function of each pin:
A6 - A1s (Output 3 State)
Address Bus; The most significant 8 bits of the memory address or the 8 bits of the I/0
address,3 stated during Hold and Halt modes.
AD0 - 7 (Input/Output 3state)
Multiplexed Address/Data Bus; Lower 8 bits of the memory address (or I/0 address)
appear on the bus during the first clock cycle of a machine state. It then becomes the
data bus during the second and third clock cycles. 3 stated during Hold and Halt
modes.
ALE (Output)
Address Latch Enable: It occurs during the first clock cycle of a machine state and
enables the address to get latched into the on chip latch of peripherals. The falling
edge of ALE is set to guarantee setup and hold times for the address information.
ALE can also be used to strobe the status information. ALE is never 3stated.
SO, S1 (Output)
Data Bus Status. Encoded status of the bus cycle:
S1 S0
O O HALT
0 1 WRITE
1 0 READ
1 1 FETCH
S1 can be used as an advanced R/W status.
RD (Output 3state)
READ; indicates the selected memory or 1/0 device is to be read and that the Data
Bus is available for the data transfer.
WR (Output 3state)
WRITE; indicates the data on the Data Bus is to be written into the selected memory
or 1/0 location. Data is set up at the trailing edge of WR. 3stated during Hold and Halt
modes.
READY (Input)
If Ready is high during a read or write cycle, it indicates that the memory or
peripheral is ready to send or receive data. If Ready is low, the CPU will wait for
Ready to go high before completing the read or write cycle.
HOLD (Input)
HOLD; indicates that another Master is requesting the use of the Address and Data
Buses. The CPU, upon receiving the Hold request. will relinquish the use of buses as
soon as the completion of the current machine cycle. Internal processing can continue.
The processor can regain the buses only after the Hold is removed. When the Hold is
acknowledged, the Address, Data, RD, WR, and IO/M lines are 3stated.
HLDA (Output)
HOLD ACKNOWLEDGE; indicates that the CPU has received the Hold request and
that it will relinquish the buses in the next clock cycle. HLDA goes low after the Hold
request is removed. The CPU takes the buses one half clock cycle after HLDA goes
low.
INTR (Input)
INTERRUPT REQUEST; is used as a general purpose interrupt. It is sampled only

during the next to the last clock cycle of the instruction. If it is active, the Program
Counter (PC) will be inhibited from incrementing and an INTA will be issued. During
this cycle a RESTART or CALL instruction can be inserted to jump to the interrupt
service routine. The INTR is enabled and disabled by software. It is disabled by Reset
and immediately after an interrupt is accepted.
INTA (Output)
INTERRUPT ACKNOWLEDGE; is used instead of (and has the same timing as) RD
during the Instruction cycle after an INTR is accepted. It can be used to activate the
8259 Interrupt chip or some other interrupt port.
RST 5.5
RST 6.5 - (Inputs)
RST 7.5
RESTART INTERRUPTS; These three inputs have the same timing as I NTR except
they cause an internal RESTART to be automatically inserted.
RST 7.5 ~~ Highest Priority
RST 6.5
RST 5.5 o Lowest Priority
The priority of these interrupts is ordered as shown above. These interrupts have a
higher priority than the INTR.
TRAP (Input)
Trap interrupt is a nonmaskable restart interrupt. It is recognized at the same time as

INTR. It is unaffected by any mask or Interrupt Enable. It has the highest priority of
any interrupt.
RESET IN (Input)
Reset sets the Program Counter to zero and resets the Interrupt Enable and HLDA
flipflops. None of the other flags or registers (except the instruction register) are
affected The CPU is held in the reset condition as long as Reset is applied.
RESET OUT (Output)
Indicates CPlJ is being reset. Can be used as a system RESET. The signal is
synchronized to the processor clock.
X1, X2 (Input)
Crystal or R/C network connections to set the internal clock generator X1 can also be
an external clock input instead of a crystal. The input frequency is divided by 2 to
give the internal operating frequency.
CLK (Output)
Clock Output for use as a system clock when a crystal or R/ C network is used as an
input to the CPU. The period of CLK is twice the X1, X2 input period.
IO/M (Output)
IO/M indicates whether the Read/Write is to memory or l/O Tristated during Hold and
Halt modes.
SID (Input)
Serial input data line The data on this line is loaded into accumulator bit 7 whenever a
RIM instruction is executed.
SOD (output)
Serial output data line. The output SOD is set or reset as specified by the SIM
instruction.
Vcc
+5 volt supply.
Vss
Ground Reference.
4. 8085 Functional Description
The 8085A is a complete 8 bit parallel central processor. It requires a single +5 volt
supply. Its basic clock speed is 3 MHz thus improving on the present 8080's
performance with higher system speed. Also it is designed to fit into a minimum
system of three IC's: The CPU, a RAM/ IO, and a ROM or PROM/IO chip.
The 8085A uses a multiplexed Data Bus. The address is split between the higher 8bit
Address Bus and the lower 8bit Address/Data Bus. During the first cycle the address
is sent out. The lower 8bits are latched into the peripherals by the Address Latch
Enable (ALE). During the rest of the machine cycle the Data Bus is used for memory
or l/O data.
The 8085A provides RD, WR, and lO/Memory signals for bus control. An Interrupt
Acknowledge signal (INTA) is also provided. Hold, Ready, and all Interrupts are
synchronized. The 8085A also provides serial input data (SID) and serial output data
(SOD) lines for simple serial interface.
In addition to these features, the 8085A has three maskable, restart interrupts and one
non-maskable trap interrupt. The 8085A provides RD, WR and IO/M signals for Bus
control.
Status Information
Status information is directly available from the 8085A. ALE serves as a status strobe.
The status is partially encoded, and provides the user with advanced timing of the
type of bus transfer being done. IO/M cycle status signal is provided directly also.
Decoded So, S1 Carries the following status information:
HALT, WRITE, READ, FETCH
S1 can be interpreted as R/W in all bus transfers. In the 8085A the 8 LSB of address
are multiplexed with the data instead of status. The ALE line is used as a strobe to
enter the lower half of the address into the memory or peripheral address latch. This
also frees extra pins for expanded interrupt capability.
Interrupt and Serial l/O
The8085A has5 interrupt inputs: INTR, RST5.5, RST6.5, RST 7.5, and TRAP. INTR
is identical in function to the 8080 INT. Each of the three RESTART inputs, 5.5, 6.5.
7.5, has a programmable mask. TRAP is also a RESTART interrupt except it is non-
maskable.
The three RESTART interrupts cause the internal execution of RST (saving the
program counter in the stack and branching to the RESTART address) if the interrupts
are enabled and if the interrupt mask is not set. The non-maskable TRAP causes the
internal execution of a RST independent of the state of the interrupt enable or masks.
The interrupts are arranged in a fixed priority that determines which interrupt is to be
recognized if more than one is pending as follows: TRAP highest priority, RST 7.5,
RST 6.5, RST 5.5, INTR lowest priority This priority scheme does not take into
account the priority of a routine that was started by a higher priority interrupt. RST
5.5 can interrupt a RST 7.5 routine if the interrupts were re-enabled before the end of
the RST 7.5 routine. The TRAP interrupt is useful for catastrophic errors such as
power failure or bus error. The TRAP input is recognized just as any other interrupt
but has the highest priority. It is not affected by any flag or mask. The TRAP input is
both edge and level sensitive.
Basic System Timing
The 8085A has a multiplexed Data Bus. ALE is used as a strobe to sample the lower
8bits of address on the Data Bus. Figure 2 shows an instruction fetch, memory read
and l/ O write cycle (OUT). Note that during the l/O write and read cycle that the l/O
port address is copied on both the upper and lower half of the address. As in the 8080,
the READY line is used to extend the read and write pulse lengths so that the 8085A
can be used with slow memory. Hold causes the CPU to relingkuish the bus when it is
through with it by floating the Address and Data Buses.
System Interface
8085A family includes memory components, which are directly compatible to the
8085A CPU. For example, a system consisting of the three chips, 8085A, 8156, and
8355 will have the following features:
· 2K Bytes ROM
· 256 Bytes RAM
· 1 Timer/Counter
· 4 8bit l/O Ports
· 1 6bit l/O Port
· 4 Interrupt Levels
· Serial In/Serial Out Ports
In addition to standard l/O, the memory mapped I/O offers an efficient l/O addressing
technique. With this technique, an area of memory address space is assigned for l/O
address, thereby, using the memory address for I/O manipulation. The 8085A CPU
can also interface with the standard memory that does not have the multiplexed
address/data bus.
5. The 8085 Programming Model
In the previous tutorial we described the 8085 microprocessor registers in reference to

the internal data operations. The same information is repeated here briefly to provide
the continuity and the context to the instruction set and to enable the readers who
prefer to focus initially on the programming aspect of the microprocessor.
The 8085 programming model includes six registers, one accumulator, and one flag
register, as shown in Figure. In addition, it has two 16-bit registers: the stack pointer
and the program counter. They are described briefly as follows.
ACCUMULATOR A (8)
FLAG REGISTER
B (8) C (8)
D (8) E (8)
H (8) L (8)
Stack Pointer (SP) (16)
Program Counter (PC) (16)
Data Bus Address Bus
8 Lines Bidirectional 16 Lines unidirectional
Registers
The 8085 has six general-purpose registers to store 8-bit data; these are identified as
B,C,D,E,H, and L as shown in the figure. They can be combined as register pairs -
BC, DE, and HL - to perform some 16-bit operations. The programmer can use these
registers to store or copy data into the registers by using data copy instructions.
Accumulator
The accumulator is an 8-bit register that is a part of arithmetic/logic unit (ALU). This
register is used to store 8-bit data and to perform arithmetic and logical operations.
The result of an operation is stored in the accumulator. The accumulator is also
identified as register A.
Flags
The ALU includes five flip-flops, which are set or reset after an operation according
to data conditions of the result in the accumulator and other registers. They are called
Zero(Z), Carry (CY), Sign (S), Parity (P), and Auxiliary Carry (AC) flags; their bit
positions in the flag register are shown in the Figure below. The most commonly used
flags are Zero, Carry, and Sign. The microprocessor uses these flags to test data
conditions.
D7 D6 D5 D4 D3 D2 D1 D0
S Z AC P CY
For example, after an addition of two numbers, if the sum in the accumulator id larger
than eight bits, the flip-flop uses to indicate a carry -- called the Carry flag (CY) -- is
set to one. When an arithmetic operation results in zero, the flip-flop called the
Zero(Z) flag is set to one. The first Figure shows an 8-bit register, called the flag
register, adjacent to the accumulator. However, it is not used as a register; five bit
positions out of eight are used to store the outputs of the five flip-flops. The flags are
stored in the 8-bit register so that the programmer can examine these flags (data
conditions) by accessing the register through an instruction.
These flags have critical importance in the decision-making process of the micro-
processor. The conditions (set or reset) of the flags are tested through the software
instructions. For example, the instruction JC (Jump on Carry) is implemented to
change the sequence of a program when CY flag is set. The thorough understanding
of flag is essential in writing assembly language programs.
Program Counter (PC)
This 16-bit register deals with sequencing the execution of instructions. This register
is a memory pointer. Memory locations have 16-bit addresses, and that is why this is a
16-bit register.
The microprocessor uses this register to sequence the execution of the instructions.
The function of the program counter is to point to the memory address from which the
next byte is to be fetched. When a byte (machine code) is being fetched, the program
counter is incremented by one to point to the next memory location
Stack Pointer (SP)
The stack pointer is also a 16-bit register used as a memory pointer. It points to a
memory location in R/W memory, called the stack. The beginning of the stack is
defined by loading 16-bit address in the stack pointer.
This programming model will be used in subsequent tutorials to examine how these
registers are affected after the execution of an instruction.
6. The 8085 Addressing Modes
The instructions MOV B, A or MVI A, 82H are to copy data from a source into a
destination. In these instructions the source can be a register, an input port, or an 8-bit
number (00H to FFH). Similarly, a destination can be a register or an output port. The
sources and destination are operands. The various formats for specifying operands are
called the ADDRESSING MODES. For 8085, they are:
1. Immediate addressing.
2. Register addressing.
3. Direct addressing.
4. Indirect addressing.
Immediate addressing
Data is present in the instruction. Load the immediate data to the destination provided.
Example: MVI R,data
Register addressing
Data is provided through the registers.
Example: MOV Rd, Rs
Direct addressing
Used to accept data from outside devices to store in the accumulator or send the data
stored in the accumulator to the outside device. Accept the data from the port 00H and
store them into the accumulator or Send the data from the accumulator to the port
01H.
Example: IN 00H or OUT 01H
Indirect Addressing
This means that the Effective Address is calculated by the processor. And the
contents of the address (and the one following) is used to form a second address. The
second address is where the data is stored. Note that this requires several memory
accesses; two accesses to retrieve the 16-bit address and a further access (or accesses)
to retrieve the data which is to be loaded into the register.
7. Instruction Set Classification

An instruction is a binary pattern designed inside a microprocessor to perform a
specific function. The entire group of instructions, called the instruction set,
determines what functions the microprocessor can perform. These instructions can be
classified into the following five functional categories: data transfer (copy)
operations, arithmetic operations, logical operations, branching operations, and
machine-control operations.
Data Transfer (Copy) Operations
This group of instructions copy data from a location called a source to another
location called a destination, without modifying the contents of the source. In
technical manuals, the term data transfer is used for this copying function. However,
the term transfer is misleading; it creates the impression that the contents of the
source are destroyed when, in fact, the contents are retained without any modification.
The various types of data transfer (copy) are listed below together with examples of
each type:
Types Examples
1. Between Registers. 1. Copy the contents of the register B into

register D.
2. Specific data byte to a register or a 2. Load register B with the data byte 32H.
memory location.
3. Between a memory location and a 3. From a memory location 2000H to register

register. B.
4. Between an I/O device and the 4.From an input keyboard to the

accumulator. accumulator.
Arithmetic Operations
These instructions perform arithmetic operations such as addition, subtraction,

increment, and decrement.
Addition - Any 8-bit number, or the contents of a register or the contents of a

memory location can be added to the contents of the accumulator and the sum is
stored in the accumulator. No two other 8-bit registers can be added directly (e.g., the
contents of register B cannot be added directly to the contents of the register C). The
instruction DAD is an exception; it adds 16-bit data directly in register pairs.
Subtraction - Any 8-bit number, or the contents of a register, or the contents of a

memory location can be subtracted from the contents of the accumulator and the
results stored in the accumulator. The subtraction is performed in 2's compliment, and
the results if negative, are expressed in 2's complement. No two other registers can be
subtracted directly.
Increment/Decrement - The 8-bit contents of a register or a memory location can be

incremented or decrement by 1. Similarly, the 16-bit contents of a register pair (such
as BC) can be incremented or decrement by 1. These increment and decrement
operations differ from addition and subtraction in an important way; i.e., they can be
performed in any one of the registers or in a memory location.
Logical Operations
These instructions perform various logical operations with the contents of the
accumulator.
AND, OR Exclusive-OR - Any 8-bit number, or the contents of a register, or of

a memory location can be logically ANDed, Ored, or Exclusive-ORed with the
contents of the accumulator. The results are stored in the accumulator.
Rotate- Each bit in the accumulator can be shifted either left or right to the next
position.
Compare- Any 8-bit number, or the contents of a register, or a memory location can
be compared for equality, greater than, or less than, with the contents of the
accumulator.
Complement - The contents of the accumulator can be complemented. All 0s are

replaced by 1s and all 1s are replaced by 0s.
Branching Operations
This group of instructions alters the sequence of program execution either

conditionally or unconditionally.
Jump - Conditional jumps are an important aspect of the decision-making process in

the programming. These instructions test for a certain conditions (e.g., Zero or Carry
flag) and alter the program sequence when the condition is met. In addition, the
instruction set includes an instruction called unconditional jump.
Call, Return, and Restart - These instructions change the sequence of a program
either by calling a subroutine or returning from a subroutine. The conditional Call and
Return instructions also can test condition flags.
Machine Control Operations
These instructions control machine functions such as Halt, Interrupt, or do nothing.
The microprocessor operations related to data manipulation can be summarized in

four functions:
1. copying data
2. performing arithmetic operations
3. performing logical operations
4. testing for a given condition and alerting the program sequence
Some important aspects of the instruction set are noted below:

1. In data transfer, the contents of the source are not destroyed; only the contents of
the destination are changed. The data copy instructions do not affect the flags.
2. Arithmetic and Logical operations are performed with the contents of the
accumulator, and the results are stored in the accumulator (with some
expectations). The flags are affected according to the results.
3. Any register including the memory can be used for increment and decrement.
4. A program sequence can be changed either conditionally or by testing for a given
data condition.
8. Instruction Format
An instruction is a command to the microprocessor to perform a given task on a
specified data. Each instruction has two parts: one is task to be performed, called the
operation code (opcode), and the second is the data to be operated on, called the
operand. The operand (or data) can be specified in various ways. It may include 8-bit
(or 16-bit ) data, an internal register, a memory location, or 8-bit (or 16-bit) address.
In some instructions, the operand is implicit.
Instruction word size
The 8085 instruction set is classified into the following three groups according to
word size:
1. One-word or 1-byte instructions

2. Two-word or 2-byte instructions
3. Three-word or 3-byte instructions
In the 8085, "byte" and "word" are synonymous because it is an 8-bit microprocessor.
However, instructions are commonly referred to in terms of bytes rather than words.
One-Byte Instructions
A 1-byte instruction includes the opcode and operand in the same byte. Operand(s)
are internal register and are coded into the instruction.
For example:
Task Op Operand Binary Hex

code Code Code
Copy the contents of the accumulator in MOV C,A 0100 1111 4FH
the register C.
Add the contents of register B to the ADD B 1000 0000 80H
contents of the accumulator.
Invert (compliment) each bit in the CMA 0010 1111 2FH
accumulator.
These instructions are 1-byte instructions performing three different tasks. In the first
instruction, both operand registers are specified. In the second instruction, the operand
B is specified and the accumulator is assumed. Similarly, in the third instruction, the
accumulator is assumed to be the implicit operand. These instructions are stored in 8-
bit binary format in memory; each requires one memory location.
MOV rd, rs
rd <-- rs copies contents of rs into rd.
Coded as 01 ddd sss where ddd is a code for one of the 7 general registers which is
the destination of the data, sss is the code of the source register.
Example: MOV A,B

Coded as 01111000 = 78H = 170 octal (octal was used extensively in instruction
design of such processors).
ADD r
A <-- A + r
Two-Byte Instructions
In a two-byte instruction, the first byte specifies the operation code and the second
byte specifies the operand. Source operand is a data byte immediately following the
opcode. For example:
Task Opcode Operand Binary Hex Code

Code
Load an 8-bit data MVI A, Data 3E First Byte
0011 1110
byte in the
accumulator. Data Second Byte
DATA
Assume that the data byte is 32H. The assembly language instruction is written as
Mnemonics Hex code
MVI A, 32H 3E 32H
The instruction would require two memory locations to store in memory.
MVI r,data
r <-- data
Example: MVI A,30H coded as 3EH 30H as two contiguous bytes. This is an
example of immediate addressing.
ADI data
A <-- A + data
OUT port
where port is an 8-bit device address. (Port) <-- A. Since the byte is not the data but
points directly to where it is located this is called direct addressing.
Three-Byte Instructions
In a three-byte instruction, the first byte specifies the opcode, and the following two
bytes specify the 16-bit address. Note that the second byte is the low-order address
and the third byte is the high-order address.
opcode + data byte + data byte
For example:
Task Opcode Operand Binary code Hex Code

Transfer the JMP 2085H C3 First byte
program 1100 0011
sequence to 85 Second Byte
1000 0101
the memory
location 0010 0000 20 Third Byte
2085H.
This instruction would require three memory locations to store in memory.
Three byte instructions - opcode + data byte + data byte
LXI rp, data16
rp is one of the pairs of registers BC, DE, HL used as 16-bit registers. The two data
bytes are 16-bit data in L H order of significance.
rp <-- data16
Example:
LXI H,0520H coded as 21H 20H 50H in three bytes. This is also immediate
addressing.
LDA addr
A <-- (addr) Addr is a 16-bit address in L H order. Example: LDA 2134H coded as
3AH 34H 21H. This is also an example of direct addressing.
9. Sample Programs
Write an assembly program to add two numbers
Program
MVI D, 8BH
MVI C, 6FH
MOV A, C
ADD D
OUT PORT1
HLT
Write an assembly program to multiply a number by 8
Program
MVI A, 30H
RRC
RRC
RRC
OUT PORT1
HLT
Write an assembly program to find greatest between two numbers
Program
MVI B, 30H
MVI C, 40H
MOV A, B
CMP C
JZ EQU
JC GRT
OUT PORT1
HLT
EQU: MVI A, 01H

OUT PORT1
HLT
GRT: MOV A, C
OUT PORT1
HLT
Instruction Set Design
• One goal of instruction set design is to minimize instruction length
• Many instructions were designed with compilers in mind.
• Determining how operands are addressed is a key component of instruction
set design
Instruction Format
• Defines the layout of bits in an instruction
• Includes opcode and includes implicit or explicit operand(s)
• Usually there are several instruction formats in an instruction set
• Huge variety of instruction formats have been designed; they vary widely from
processor to processor
Instruction Length
• The most basic issue
• Affected by and affects:
― Memory size
― Memory organization
― Bus structure
― CPU complexity
― CPU speed
• Trade off between a powerful instruction repertoire and saving space with
shorter instructions
Instruction format trade-offs
• Large instruction set => small programs
• Small instruction set => large programs
• Large memory => longer instructions
• Fixed length instructions same size or multiple of bus width => fast fetch
• Variable length instructions may need extra bus cycles
• Processor may execute faster than fetch
― Use cache memory or use shorter instructions
• Note complex relationship between word size, character size, instruction size
and bus transfer width
― In almost all modern computers these are all multiples of 8 and related to
each other by powers of 2
Allocation of bits
• Determines several important factors
• Number of addressing modes
― Implicit operands don’t need bits
― X86 uses 2-bit mode field to specify interpretation of 3-bit operand fields
• Number of operands
―3 operand formats are rare
―For two operand instructions we can use one or two operand mode
indicators
―X86 uses only one 2-bit indicator
• Register versus memory
―Tradeoff between # of registers and program size
―Studies suggest optimal number between 8 and 32
―Most newer architectures have 32 or more
―X86 architecture allows some computation in memory
Allocation of bits
• Number of register sets
– RISC architectures tend to have larger sets of uniform registers
– Small register sets require fewer opcode bits
– Specialized register sets can reduce opcode bits further by implicit
reference (address vs. data registers)
• Address range
– Large address space requires large instructions for direct addressing
– Many architectures have some restricted or short forms of displacement
addressing
Ex: x86 short jumps and loops, PowerPC 16-bit displacement addressing
• Address granularity
– Size of object addressed.
– Typically 8,16, 32 and 64 instruction variants
Addressing Modes
Addressing Modes
• For a given instruction set architecture, addressing modes define how
machine language instructions identify the operand (or operands) of each
instruction.
• An addressing mode specifies how to calculate the effective memory address
of an operand by using information held in registers and/or constants
contained within a machine instruction or elsewhere.
• Different types of addresses involve tradeoffs between instruction length,
addressing flexibility, and complexity of address calculation
• Common addressing modes
―Direct
―Immediate
―Indirect
―Register
―Register indirect
―Displacement
―Implied (stack)
Direct Addressing
• The instruction tells where the value can be found, but the value itself is out in
memory.
• The address field contains the address of the operand
• Effective address (EA) = address field (A)
• In a high level language, direct addressing is frequently used for things like global
variables.
• Advantage
– Single memory reference to access data
– More flexible than immediate.
Instruction
A
Memory
operand
Direct Addressing
for the following examples, assume an accumulator machine structure and that an add
instruction is stored in memory, beginning at location 12
memory
assembly lang. addr contents hardware actions
------------------- ------- ------------ -----------------------
... ...
add(one) 12 | 40 | acc <- acc + memory[24]
13 | 24 | = acc + 1
... ...
word(one,1) 24 |1| effective address = 24
... ...
so, when the PC points to 12:
40 (that is, the contents of location 12) is interpreted as an opcode
24 (that is, the contents of location 13) is interpreted as an address
1 (that is, the contents of location 24) is interpreted as data
note that there are no tags or other indicators that the number 40 in location 12 has
to be an opcode; it could just as well be used as an address or as data
Immediate Addressing
• the instruction itself contains the value to be used; located in the addresss field of
the instruction
• the value is stored in memory immediately after the instruction opcode in
memory
• Similar to using a constant in a high level language
• Advantage
– fast since the value is included in the instruction; no memory reference to
fetch data
• Disadvantage
– not flexible, since the value is fixed at translation-time
– can have limited range in machines with fixed length instructions
Instruction
operand
Immediate Addressing
for the following example, assume an accumulator machine structure and that an add
memory
------------------- ------- ------------ -----------------------
... ...
add_immediate(1) 12 | 41 | acc <- acc + 1
13 |1|
... ... no additional memory
fetch for data beyond the
instruction fetch (since the
instruction contains the
data being used)
since an add must have different hardware actions than an add_immediate,
add_immediate has to be a different opcode (or there has to be an extra type-of-
addressing-mode code in the instruction format to go along with the opcode)
Example of direct and immediate addressing
Suppose we have a statement in C like
b = a + 10;
a and b are variables, so they are out in memory.
To execute this statement, we will need to fetch a from memory, and write
our result to b.
That means the instructions we generate need to have the addresses of a and
b, and need to read and write those addresses as appropriate.
The number 10 is an actual value appearing in the statement. So, our code
needs to include 10 itself.
Memory-Indirect Addressing
The memory cell pointed to by the address field contains the address of (pointer to)
the operand
• EA = (A)
Instruction
A
Memory
operand
Indirect Addressing
memory
------------------- ------- ------------ -----------------------
... ...
add_indirect(ptr) 12 | 42 | acc <- acc + memory[memory[36]]
13 | 36 | = acc + memory[24]
... ... = acc + 1
word(one,1) 24 |1|
... ... effective address = 24
word(ptr,one) 36 | 24 |
... ...
the address included in the instruction is that of a pointer, that is, a word that holds
another address
Register Addressing
• Operand(s) is (are) registers
• EA = R
– Register R is EA (not contents of R)
Instruction
R
operand
Registers
Register Addressing
• There is a limited number of registers
– A very small address field is needed
– Shorter instructions
– Faster instruction fetch
– X86: 3 bits used to specify one of 8 registers
• No memory access needed to fetch EA
• Very fast execution
• Very limited address space
• Multiple registers can help performance
• Requires good assembly programming or compiler writing
Note: in C you can specify register variables
register int a;
– This is only advisory to the compiler; no guarantees
Register-Indirect Addressing
• Similar to memory-indirect addressing
• EA = (R)
• Operand is in memory cell pointed to by contents of register R
• Large address space (2n)
• One fewer memory address than memory-indirect
Instruction
R
Memory
operand
Registers
Displacement Addressing
• Combines register-indirect addressing and direct addressing
• EA = A + (R)
• Address field holds two values
– A = base value
– R = register that holds displacement
– Or visa versa
Instruction
R A
Memory
operand
Registers
Types of Displacement Addressing
• Relative Addressing
• Base-register addressing
• Indexing
Relative Addressing
• EA = A + (PC)
• Address field A is treated as 2’s complement integer to allow backward references
• Fetch operand from PC+A
• Can be very efficient because of locality of reference & cache usage
– But in large programs code and data may be widely separated in memory
Base-Register Addressing
• A holds displacement
• R holds pointer to base address
• R may be explicit or implicit
– E.g. segment registers in 80x86 are base registers and are involved in all EA
computations
– X86 processors have a wide variety of base addressing
Indexed Addressing
• A = Base
• R = displacement
• EA = A + R
• Good for accessing arrays
– EA = A + R
– R++
• Iterative access to sequential memory locations is very common
• Some architectures provide auto-increment or auto-decrement
• Preindex EA = A + (R++)
• Postindex EA = A + (++R)
Indexed Addressing
memory
------------------- ------- ------------ -----------------------
... ...
add_indexed(b0,x) 12 | 43 | acc <- acc + memory[20+memory[36]]
13 | 20 | = acc + memory[20+4]
14 | 36 | = acc + memory[24]
... ... = acc + 1
word(b0,5) 20 |5|
word(b1,-2) 21 | -2 | effective address = 24
word(b2,3) 22 |3|
word(b3,9) 23 |9|
word(b4,1) 24 |1|
... ...
word(x,4) 36 |4|
... ...
Addressing modes using registers
on machines with multiple registers, addresses and index values can be held in registers,
for example:
Direct load(x,r1) // r1 <- memory[ x ]
immediate load_imm(3,r2) // r2 <- 3
indexed for array access load_ind(a,r3,r4) // r4 <- memory[ a + r3 ]
(fixed array base address
and index in a register)
register indirect as part load_ind(0,r5,r6) // r6 <- memory[ 0 + r5 ]
of indexed (i.e., a pointer
is in a register)
base plus displacemennt as load_ind(2,r7,r8) // r8 <- memory[ 2 + r7 ]
part of indexed (i.e., // accesses 3rd word of
structure access w/ ptr. in // a structure
reg. and constant offset)
Branch addressing modes
direct addressing, such as the accumulator machine

Assembly lang addr contents hardware actions
------------------ ---------- ----------- -----------------------
... ...
ba(target) 20 | 70 | pc <- 30
21 | 30 |
... ...
label(target) 30 | | next instruction after branch
... ...
Branch addressing modes
pc-relative addressing, such as the JVM Memory

Assembly lang addr contents hardware actions
------------------ ---------- ----------- -----------------------
... ...
goto(target) 20 | 167 | pc <- 21 + 9
21 | 9 | = 30
... ...
label(target) 30 | | next instruction after goto
... ...
note that other machines may make the offset relative to the address of the branch (e.g.,
20 above) or the fully-updated pc (e.g., 22 above)
Stack Addressing
• Operand is implicitly on top of stack
– PUSH
– POP
Alternate Addressing
• Offset addressing – offset is added or subtracted from value in base

register
• Preindex addressing
– Memory address is formed the same way as offset addressing, but the
memory address is written back to the base register after adding or
subtracting the displacement
– The writeback occurs before the store to memory
• Postindex addressing
– Similar to preindex addressing, but the writeback of the effective
address occurs after the store to memory
DTEL (Department for Technology Enhanced Learning)
The Centre for Technology enabled Teaching & Learning
Teaching Innovation - Entrepreneurial - Global 1

NAGAR YUWAK SHIKSHAN SANSTHA’S
SHRI DATTA MEGHE POLYTECHNIC
DEPARTMENT OF COMPUTER TECHNOLOGY
Microprocessor and Programming

AUTHORS
MANOJ JETHWA
2
CONTENT: MICROPROCESSOR AND PROGRAMMING
1 CHAPTER 1: Basics of Microprocessor

.
2 CHAPTER 2: 16 Bit Microprocessor: 8086
3 CHAPTER 3: Instruction Set of 8086 Microprocessor
The Art of Assembly Language

4 CHAPTER 4: Programming
8086 Assembly Language

5 CHAPTER 5: Programming.
Procedure and Macro in Assembly

6 CHAPTER 6: Language Program
DTEL 3
SYLLABUS GENERAL OBJECTIVE
The student will be able to:
1 Understand What is microprocessor Architecture.
Understand the execution of instructions in

2 pipelining and address generation.
Apply instructions in Assembly Language Program for

3 different problem statements.
Use the procedures and macros in assembly

4 language programming.
DTEL 4
CHAPTER-1 Basics of Microprocessor
1 .
Topic 1: Evolution of Microprocessor and types
2 Topic 2: 8085 Microprocessor,
3 Topic 3: Salient features of 8085
4 Topic 4: Architecture of 8085 - Functional Block diagram,
5 Topic 5: Pin description,
DTEL 5
CHAPTER-1 SPECIFIC OBJECTIVE / COURSE OUTCOME
Draw the architecture of 8085 and understand the functions of

1 different pins of 8085
Identify status of different flags and understand register

2 organisation of 8085
DTEL 6
LECTURE 1: BASIC BLOCK OF COMPUTER
CPU OR MICROPROCESSOR
ALU
OUTPUT
INPUT Devices
Devices
Control Unit
DATA PATH CONTROL SIGNALS
MEMORY (Primary) BASIC BLOCK OF COMPUTERS

CONSISTS
ü ALU
MEMORY ü INPUT DEVICE
(Secondary) ü OUTPUT DEVICE
ü MEMORY
ü CONTROL UNIT
DTEL 7
LECTURE 1:- BASIC BLOCK OF COMPUTER
The typical Computer system consists of:
§ CPU (central processing unit)
ü ALU (arithmetic-logic unit)
ü Control Logic
ü Registers, etc…
§ Memory
§ Input / Output interfaces
Interconnections between these units are through 3 basic buses:

§ Address Bus
§ Data Bus
§ Control Bus
DTEL 8
LECTURE 1:- CPU
ü The main function of ALU is to perform arithmetic and
logical operations on binary numbers.
ü The Input Device is used to feed data and command for

the CPU.
ü The output device is used for display of result /data

/program etc.
ü The memory is used for storing information.
ü The control unit Synchronizes operation of ALU with IO

and Memory.
DTEL 9
LECTURE 1:- BUS
The interconnections (known as Interfacing) between the 5 units
of computer system is carried by 3 basic buses i) Address Bus ii)
Data Bus iii) Control Bus. A bus(from the Latin omnibus, meaning
"for all") is essentially a set of wires which is used in computer
system to carry information of the same logical functionality. The
function of the 3 buses is
ü The address bus selects memory location or an I/O device for
the CPU.
ü The data bus transfers information between the
microprocessor and its memory or I/O device. Data transfer
can vary in size, from 8-bits wide to 64 bits wide in various
members of microprocessors.
ü The Control bus generates command signals to synchronise
the CPU operation with IO and Memory devices.
DTEL 10
LECTURE 1:- Evolution of Microprocessor
Processo Date of Clock Data Bus Adress Bus Addressable Memory Size
r Launch speed Width
4004 1971 740 khz 4 bit 12 4 KB
8-BIT PROCESSOR
8008 1972 800 Khz 8 bit 14 16 Kb
8080 1974 2 Mhz 8 bit 16 64 kb
8085 1976 3 Mhz 8 bit 16 64 kb
16-BIT PROCESSOR
8086 1978 5 Mhz 16 20 1M
80286 1982 16 Mhz 16 24 16 M
DTEL 11
LECTURE 1:- Evolution of Microprocessor
Processor Date of Clock Data Bus Adress Addressable Memory Size
Launch speed Width Bus
32-BIT PROCESSOR
80386 1985 33 Mhz 32 32 4G
80486 1989 40 Mhz 32 32 4G+ 8k cache
Petium I 1993 100 Mhz 32 32 4G+16k cache
Petium II 1997 233 Mhz 32 32 4G+16k cache + L2 256 Cache
Petium III 1999 1.4 Ghz 32 32 4G+32k cache + L2 256 Cache
Petium IV 2000 2.66 Ghz 32 Internal 32 4G+32k cache + L2 256 Cache
64 External
64-BIT PROCESSOR
2.66 Ghz 64G+Independent L1 64 Kb+
Dual Core 2006 64 36 Common L2 256 kb Cache
Core 2 64G+Independent L1 128 Kb+
Duo 2006 3 Ghz 64 36 Common L2 4 Mb Cache
64G+Independent L1 64 Kb+
Common L2 256 kb Cache + 8 Mb
I7 2008 3.33 Ghz 64 36 L3 Cache
DTEL 12
LECTURE 2:- 8085 ARCHITECTURE
INTA INTR RST 5.5 RST 6.5 RST 7.5 TRAP SOD SID
INTERRUPT CONTROL SERIAL CONTROL
INTERNAL BUS
ACMULATOR
TEMP
INSTRUCTION B C
.REGISTER
REGISTER (IR) D E
FLAG 8-Bit H L
code
STACK POINTER (SP)
INSTRUCTION
DECODER and
PROGRAM .COUNTER (PC)
ALU MACHINE
CYCLE
ADDRESS
INCREMENTER / DECREMENTER
ENCODING
256-Bit
MULTIPLEXER
X1
TIMMING & CONTROL ADRESS /
Clock WAIT ADRESS
GEN DATA
RESET DMA STATUS CONTROL STATES BUFFER BUFFER
X2
Clock out Reset in HOLD ALE

A15-8 AD7-0
S0 S1 WR READY
Reset out IO / M RD
HLDA
DTEL 13
LECTURE 2:- Processing Unit
üArithmetic and Logic Unit

üAccumulator
üStatus Flags
üTemporary Register
DTEL 14
LECTURE 2:- Processing Unit
Processing INTERRUPT CONTROL SERIAL CONTROL
Unit
INTERNAL BUS
8-Bit
code
ACMULATOR
TEMP
INSTRUCTION B C
.REGISTER
REGISTER (IR) D E
FLAG 8-Bit H L
code
STACK POINTER (SP)
INSTRUCTION
DECODER and
ALU MACHINE
CYCLE
ADDRESS
ENCODING
256-Bit
MULTIPLEXER
X1
Clock WAIT ADRESS
GEN DATA
X2

A15-8 AD7-0
S0 S1 WR READY
Reset out IO / M RD
HLDA
DTEL 15
LECTURE 2:- Arithmetic & Logic Unit (ALU)
ü It performs Arithmetic and logic operations on binary nos.
ü The result is stored in accumulator in most cases, hence A

is known as accumulator.
ü Arithmetic Operations:
ü Addition, Subtraction, Increment, Decrement .
ü Logic Operations:
ü AND, OR, X-OR, Complement .
DTEL 16
ü It the main register of microprocessor directly connected

with the ALU.
ü It is also called register ‘A’.
ü It is an 8-bit register.
ü It is used in the arithmetic and logic operations.
ü It always contains one of the operands on which

arithmetic/logic has to be performed.
ü After the arithmetic/logic operation, the contents of

accumulator are replaced by the result.
DTEL 17
Status Flag
F7 F0
SF ZF AF PF CF
The 5 Status Flags are Ø Status

Flags
CF: Carry flag
affected immediately after an
PF: Parity flag
arithmetic or logical operation AF: Auxiliary carry flag
performed by the ALU. The ZF: Zero flag
SF: Sign flag
SET or RESET condition of OF: Overflow flag
each flag is used to indicate
the status of the result
generated by the ALU.
DTEL 18
Status Flag
ü Sign Flag: It is used to indicate whether the result is positive or negative. It will
set (SF=1) if the result is –ve and if the result +ve then SF=0.
ü Zero Flag: It is used to indicate whether the result is a Zero or non-zero. It will set
(ZF=1) if the result is zero else ZF=0.
ü Auxiliary carry Flag: It is used to indicate whether or not the ALU has generated a
carry/Borrow from D3 bit position to D4 bit. It will set if there was a carry out
from bit 3 to bit 4 of the result else AF=0. The auxiliary carry flag is used for
binary coded decimal (BCD) operations.
ü Parity Flag: It is used to indicate parity ( Even or Odd) of the result. It will set if the
parity is even else PF =0.
ü Carry Flag: It is used to indicate whether a carry/Borrow has been generated

/occurred during addition/subtraction It will set if there was a carry is generated
from the MS-bit during addition, or borrow during subtraction/comparison else
CF=0.
DTEL 19
Program Status Word (PSW)
• The Accumulator and Status Flags clubbed
together is known as Program Status Word (PSW).
• It is a 16-bit word.
Accumulator (8) FLAGs (8)
DTEL 20
LECTURE 3:- Register sets & pointer
Storage ,
Pointer and
Interface
INTERNAL BUS
ACMULATOR
TEMP
INSTRUCTION B C
.REGISTER
REGISTER (IR) D E
FLAG 8-Bit H L
code
STACK POINTER (SP)
INSTRUCTION
DECODER and
ALU MACHINE
CYCLE
ADDRESS
ENCODING
256.
MULTIPLEXER
X1
Clock WAIT ADRESS
GEN DATA
X2

A15-8 AD7-0
S0 S1 WR READY
Reset out IO / M RD
HLDA
DTEL 21
LECTURE 3:- Register sets & pointer
ü The 8085 has set of 8 register (of 8-bit) and 2 memory pointers (of
16-bit) . The register A and flag are directly connected with ALU,While
B,C,D,E,H,& L are indirectly connected through internal bus. The
register A is used to store data as well as result of an operation
performed by the ALU.The Flag is used to store status of result. The
register B,C,D,E,H,L are used to store 8-bit data. It can also be paired
to store 16-bit data. The pairing combination can be,
B-C D-E H-L
The register pairs can also be used to generate 16-bit address.
ü The pointers are used to generate 16-bit address for selection of

memory location. The PC generates address during execution of a
program. It contains the memory address (16 bits) of the instruction
that will be executed in the next step.
While SP generates address during stack operation.
DTEL 22
LECTURE 3:- Timming-Control and IR -Decoder
INTERNAL BUS
ACMULATOR
TEMP
INSTRUCTION B C
.REGISTER
REGISTER (IR) D E
FLAG 8-Bit H L
code
STACK POINTER (SP)
INSTRUCTION
DECODER and
ALU MACHINE
CYCLE
ADDRESS
ENCODING
Instruction
Unit
256-Bit
MULTIPLEXER
X1
Clock WAIT ADRESS
GEN DATA
X2

A15-8 AD7-0
S0 S1 WR READY
Reset out IO / M RD
HLDA
DTEL 23
LECTURE 3:- IR-DECODER-TIMMING & CONTROL
• Instruction Register : It is used to store the current

instruction code which is fetched from the memory It
is an 8-bit register.
• Instruction Decoder : Instruction is decoded and the

meaning in the form of signal is given to TC.
• Timing and Control Unit : It generates internal and

external signals to execute the code
DTEL 24
PIN DIAGRAM OF 8085
25
LECTURE 4:- Pin diagram
DTEL 26
LECTURE 4:- Pin diagram
ü It was introduced in 1977 by Intel.
ü It is 8-bit microprocessor.
ü It is NMOS device consisting of 6200
transistors .
ü Its data bus is 8-bit and address bus is 16-
bit.
ü Its clock speed was 3 MHz. Could
execute 7,69,230 instructions per second.
ü Its data bus is 8-bit and address bus is 16-
bit.
ü It had 6,500 transistors.
ü It is 40 pins Dual-Inline-Package (DIP).
DTEL 27
LECTURE 4:- X1 & X2 Pin 1 and Pin 2
— It is also called Oscillator

Pins to which crystal of
max 6.14 Mhz should be
connected so that 8085
can generate clock signals
internally.
— The internal Clock pulses
will be ½ times crystal
value i.e, 3.07 Mhz.
DTEL 28
LECTURE 4:- RESET IN and RESET OUT Pin 36 and Pin 3
◦ RESET IN is used to reset the

microprocessor by making the pin Low.
When the signal on this pin is low for at
least 3 clock cycles, it forces the
microprocessor to reset . Thereby ,
1. Clear the PC and IR.
2. Disable all the interrupts (except
TRAP) and the SOD pin.
3. HIGH output pulse to RESET OUT pin.
◦ Reset OUT is used to reset the external

peripheral devices and ICs on the
circuit. It is active high signal. The
output on this pin goes high whenever
RESET IN pin is made low .
DTEL 29
LECTURE 4:- SID and SOD Pin 4 and Pin 5
— SID (Serial Input Data): It
receives 1-bit from external
device and Stores the bit at the
MSB of the Accumulator. RIM
(Read Interrupt Mask)
instruction is used to transfer
the bit from SID MS Bit of Acc.
— SOD (Serial Output Data): It

transmits MSB of Accumulator
through this pin. SIM (Set
Interrupt Mask) instruction is
used to transfer the MS bit of
Acc through SOD.
DTEL 30
LECTURE 4:- Interrupt Pins 6 to 11
— Interrupt: (INTR,RST5.5,RST6.6,RST7.5
and TRAP pins)
• It allows external devices to
interrupt the normal program
execution of the microprocessor.
• When microprocessor receives
interrupt signal, it temporarily
stops current program and starts
executing new program indicated
by the interrupt signal.
• Interrupt signals are generated by
external peripheral devices like
keyboard , sensors, printers etc.
• After execution of the new
program, microprocessor returns
back to the previous program.
DTEL 31
— The 5 Hardware Interrupt Pins are TRAP , RST 7.5 , RST

6.5 , RST 5.5 and INTR. Interrupts can be classified as,
1. Maskable and Non-Maskable
2. Vectored and Non-Vectored
3. Edge Triggered and Level Triggered
4. Priority Based Interrupts
DTEL 32
Maskable and Non-Maskable
Maskable interrupts are those interrupts which can be enabled or
disabled. Enabling and Disabling can be done by software
instructions like EI , DI and SIM. The interrupt pins RST7.5, RST6.5
,RST5.5 and INTR are Maskable.
The interrupts which cannot be disabled are called non-maskable

interrupts. These interrupts can never be disabled by any
software instruction. TRAP is a non-maskable interrupt.Such pins
are normally used for emergency cases like fire alarming , fire
extinguisher system ,intruder detector etc.
DTEL 33
Vectored and Non-Vectored
• Vectored interrupts which have particular memory location where program
control is transferred when interrupt occur. Each vectored interrupt points to the
particular location in memory. RST 7.5 , RST 6.5 ,RST 5.5, TRAP are vectored
Interrupts.
— The addresses to which program control is transferred are :
Name Vectored Address
RST 7.5 003C H (7.5 x 0008 H)
RST 6.5 0034 H (6.5 x 0008 H)
RST 5.5 002C H (5.5 x 0008 H)
TRAP 0024 H (4.5 x 0008 H)
— Absolute address is calculated by multiplying the RST no with 0008 H.
• Non-Vectored interrupts don't have fixed memory location for transfer of

program control. The address of the memory location is given by interrupting
device to the processor along with the interrupt. INTR is a non-vectored
interrupt.
DTEL 34
Edge Triggered and Level Triggered
The interrupts that are triggered at leading or trailing edge are
called edge triggered interrupts. RST 7.5 is an edge triggered
interrupt. It is triggered during the leading (positive) edge.
The interrupts which are triggered at high or low level are called
level triggered interrupts. RST 6.5,RST 5.5, INTR, are level triggered
…..
interrupt.
The TRAP is edge and level triggered interrupt.
DTEL 35
Priority Based Interrupts
When there is a simultaneous interrupt request at two or more
interrupt pins then the microprocessor will execute program of
that pin that has higher priority. To avoid confusion in such cases
all microprocessor assigns priority level to each interrupt pins.
Priority is considered by microprocessor only when there are
simultaneous requests. The priority of 8085 pins are:
Interrupt Priority
TRAP 1
RST 7.5 2
RST 6.5 3
RST 5.5 4
INTR 5
DTEL 36
LECTURE 4:- Address and Data Pins
• Address Bus pins A15–A8 and

AD7–AD0 : The address bus are
used to send address for memory
or IO device. It selects one of the
many locations in memory or a
particular IO port. 8085 has 16-bit
bus.
• Data Bus pins AD7–AD0 : It is
used to transfer data between
microprocessor and memory /IO
Device. 8085 Data bus is of 8-bit.
DTEL 37
LECTURE 4:- ALE Pin 30
• On this pin 8085 generates a

pulse in the T1 clock of each
Machine cycle to latch lower
byte adress from AD7 – AD0
pins.From mid of T2 to T3 the
bus can transfer data.
• It indicates whether the bus has

address or data .Since the bus
AD7 – AD0 is used for Adress as
well as data therefore it is
known as Multiplexed bus. Due
to multiplexing 8085 has less
no of pins.
DTEL 38
LECTURE 4:- Status S1 ,So Pin 31 ,29
• S0 and S1 are called Status
Pins.They indicate the
status of current operation
which is in progress by
8085.The 4 status
indicated by 8085 are
S0 S1 Operation
0 0 Halt
0 1 Write
1 0 Read
1 1 Opcode
Fetch
DTEL 39
LECTURE 4:- IO/M Pin 34
• This pin indicates whether I/O

or memory operation is being
performed by microprocessor.
• If IO/M = 1 then
– I/O operation is being
performed.
• If IO/M = 0 then
– Memory operation is being
performed.
DTEL 40
LECTURE 4:- RD Pin 32
• It is a control signal used to

perform Read operation
from memory or from
Input device. It is active
low signal. A low signal
indicates that data on the
data bus must be placed by
the selected memory
location or input device.
DTEL 41
LECTURE 4:- WR Pin 31
• It is a control signal used to
perform Write operation
into memory location or to
output device. It is also
active low signal. A low
signal indicates that data on
the data bus must be
written into selected
memory location or to
output device.
DTEL 42
LECTURE 4:- READY Pin 35
• This pin is used to
synchronize slower peripheral
devices with high speed of
microprocessor.
• A low pulse in T2 causes the

microprocessor to enter into
wait state.
• The microprocessor remains

in wait state until the input at
this pin goes high.
DTEL 43
LECTURE 4:- HOLD Pin 38
• HOLD pin is used to request the
microprocessor for DMA
transfer.
• A high signal on this pin is a

request to microprocessor, by
external device, to relinquish
the hold on buses.
• The request is sent by external

device through DMA controller.
DTEL 44
LECTURE 4:- HLDA Pin 39
• The HLDA signal is send to DMA
Controller as acknowledgement
to DMA controller to indicate
that microprocessor has
relinquished the system bus.
• After data transfer When HOLD

is made low by DMA controller,
HLDA is also made low by 8085
so that the microprocessor
takes control of the buses again.
DTEL 45
LECTURE 4:- VSS and VCC Pin 20 and Pin 40
• +5V DC power supply

is connected to VCC to
bias internal circuit.
• Ground signal is
connected to VSS.
DTEL 46
LECTURE 5:- Chapter 1 Question Bank
• What are the technical features of 8085?
• Explain the function of ALU section of 8085.
• Describe the function of the following blocks of 8085
• ALU ii) Timming & control iii) Instruction Decoder
• Explain the function of various registers of 8085.
• Draw the Block (Architecture) of 8085 and explain IR, stack pointer and
programme counter.
• What are the various Flag of 8085?
• What are the pointers of 8085.Explain the function of Pointers of 8085?
• Explain the function of Interrupt section of 8085.
• List Maskable and non-maskable Interrupts of 8085.
• Explain the function of SID & SOD of 8085.
• Describe microprocessor evolution with suitable example?
• Differentiate, any six ,between 8085 & 8086.
DTEL 47
LECTURE 5:- Summary
1.The typical Computer system consists of:
ü ALU (arithmetic-logic unit)
ü Control Logic
ü Memory
ü Input devices
ü Output devices
2. A bus(from the Latin omnibus, meaning "for all") is essentially a set of

wires which is used in computer system to carry information of the same
logical functionality. The interconnections (known as Interfacing) between the
5 units of computer system is carried by 3 basic buses
i) Address Bus ii) Data Bus iii) Control Bus.
3. 8085 can be divided into sections like i) Processing unit ii) Register &
pointers iii) instruction register-decoder-timming & control.
4.The 8085 has 8-bit flag but 5 are affected by Arithmetic / logical operation.
DTEL 48
CHAPTER-2 16 Bit Microprocessor: 8086
Topic 1: Salient features of 8086
1 .
Architecture of 8086 - Functional Block

2 Topic 2: diagram,
3 Topic 3: Register organization,
4 Topic 4: Concepts of pipelining, Memory segmentation
5 Topic 5: Operating Modes of 8086
DTEL 49
Draw the architecture of 8086 and understand the functions of

1 different pins of 8086
2 Understand the operating modes of 8086
DTEL 50
Lecture 1: Intel 8086 Microprocessor
Key Features:
üIntroduction date: March 1978
üIt is 16-bit HMOS microprocessor implemented
with 29,000 transistors
üIt can be operated with clock Frequency of 5MHz
üTechnology: HMOS
üNumber of Pins : 40
ü It has 20-bit Address lines and hence it can address 220 = 1 Mbytes
memory location.
üIt can generate 16-bit address for IO devices and can address 216 = 64K IO
ports.
üIt can be operated in two Modes : Maximum and Minimum
üIt has two stage pipeline architecture.
üNumber of instructions: 135 instructions with eight 8-bit registers and eight
16-bit registers
üDC Power Supply +5v
DTEL 51
LECTURE 1 Architecture of 8086
MEMORY
Σ 6 6-Byte Q
5
4
3
CS
2
DS
1
ES
SS
IP
Control
BIU
system
EU
AH AL
BH BL ALU
CH CL
DH DL
SP
BP FLAGS
SI
DI
DTEL 52
LECTURE 2 Architecture of 8086
• The architecture of 8086 provides a number of
improvements over 8085 architecture. It supports a 16-bit
ALU, a set of 16-bit registers and provides segmented
memory addressing capability, a rich instruction set,
powerful interrupt structure, fetched instruction queue
for overlapped fetching and execution.
• The complete architecture of 8086 can be logically
divided into two units a) Bus Interface Unit (BIU) and (b)
Execution Unit (EU). Both units operate asynchronously to
provide the 8086 to overlap instruction fetch and
execution operation , which is called as parallel
processing. This results in efficient use of the system bus
and enhance system performance.
DTEL 53
LECTURE 3 Pipelining and parallel processor
An instruction pipeline is a technique used in the design of
microprocessors to increase the number of instructions that can
be executed in a unit of time. Pipeline technique is used in
advanced microprocessors where the microprocessor begins
operation on next instruction before it has completed operation
on the previous. That is, several instructions are simultaneously
in the pipeline at a different stage of processing. The pipeline is
divided into different Stages and each Stage can perform its
particular operation simultaneously with the other stages.
When a stage completes an operation, it passes the result to
the next stage in the pipeline and fetches the next operation
from the preceding stage. The final results of each instruction
emerge at the end of the pipeline in rapid succession. Since all
units perform operation concurrently on different instructions ,
it is known as parallel processor.
DTEL 54
LECTURE 3 Pipelining and parallel processor
Pipelining of 8086
INSTRUCTION
NO. EXECUTION PHASES
1 Fetch-1 Decode-1 Execute-1
Machine
cycle
1 2 3 4 5 6 7 8
Non-Pipelining Process of 8085
Inst ruction-1 Inst ruction-2 Inst ruction-3
Fetch-1 Decode-1 Execute-1 Fetch-2 Decode-2 Execute-2 Fetch-3 Decode-3 Execute-3
M.
cycle 1 2 3 4 5 6 7 8 9
DTEL 55
LECTURE 4 Register of 8086
8086 has a powerful set of registers that can be grouped as
Ø General Data register
Ø Segment registers
Ø Pointers & Index registers
Ø FLAG
Ø Only GPRs can be accesses as
8/16-bit while others as 16-bit only
AX AH AL CS IP FLAGS /
BH BL DS PSW
BX SI
CX CH CL ES DI
DX DH DL SS SP
General Data Segment BP
registers registers
Pointers and
Index registers
DTEL 56
LECTURE 4 Special Purpose Registers:
• Special Purpose Registers: The special purpose registers are
Ø Segment registers
Ø Pointers and index registers
• Segment Registers : Unlike 8085, the 8086 addresses a

segmented memory of 1MB, which the 8086 is able to address.
The 1 MB is divided into 16 logical segments (16 X 64 KB = 1024
KB = 1 MB). Each segment thus contains 64 Kbytes of memory.
There are four segment registers, viz. Code Segment Register
(CS), Data Segment Register (DS), Extra Segment Register (ES) and
Stack Segment Register (SS).
DTEL 57
LECTURE 4 Special Purpose Registers:
Pointers and Index Registers
The pointers contain offset within the particular
segments. The pointers IP, BP and SP usually contain
offsets within the code, data and stack segments
respectively. The index registers are used as general
purpose registers as well as for offset storage in case of
indexed, based indexed and relative based indexed
addressing modes. The register SI is generally used to
store the offset of source data in DMS while the
register DI is used to store the offset of destination in
DMS or EMS. The index registers are particularly useful
for string manipulations.
DTEL 58
LECTURE 5 Flag Register
• The FLAG is nothing but group of flip-flops which are affected
(SET or RESET) immediately after an arithmetic or logical
operation performed by the ALU.
• The flags of 8086 can be divided into two types: Conditional
Flags and Control Flags
• Conditional Flags are affected immediately after an arithmetic or
logical operation performed by the ALU. The SET or RESET
condition of each flag is used to indicate the status of the result
generated by the ALU.The 8086 has 6 conditional flags, out of
which 5 are similar to the 8085 while Overflow flag is the
additional flag.
• Control Flag are not affected by Arithmetic or logical operation
performed by the ALU but programmer can SET or RESET these
Flags to Control certain operation/Instructions.
DTEL 59
LECTURE 5 Flag Register
Same as 8085
Additional Flags
XX XX XX XX OF DF IF TF SF ZF XX AF XX PF XX CF
F15 F8 F7 F0
Ø Control Flags Ø Status Flags

IF: Interrupt enable flag CF: Carry flag
DF: Direction flag PF: Parity flag
TF: Trap flag AF: Auxiliary carry flag
ZF: Zero flag
SF: Sign flag
XX: Don’t Care OF: Overflow flag
DTEL 60
LECTURE 5 Conditional Flag
The 6 Status or Conditional Flags are affected immediately after an
arithmetic or logical operation performed by the ALU. The SET or
RESET condition of each flag is used to indicate the status of the
result generated by the ALU.
• Sign Flag: It is used to indicate whether the result is positive or

negative. It will set (SF=1) if the result is –ve and if the result +ve then
SF=0.
• Zero Flag: It is used to indicate whether the result is a Zero or non-
zero. It will set (ZF=1) if the result is zero else ZF=0.
•Auxiliary carry Flag: It is used to indicate whether or not the ALU has
generated a carry/Borrow from D3 bit position to D4 bit. It will set if
there was a carry out from bit 3 to bit 4 of the result else AF=0. The
auxiliary carry flag is used for binary coded decimal (BCD) operations.
DTEL 61
LECTURE 5 Conditional Flag
•Parity Flag: It is used to indicate parity ( Even or Odd) of the result.
It will set if the parity is even else PF =0.
•Carry Flag: It is used to indicate whether a carry/Borrow has been

generated /occurred during addition/subtraction It will set if there
was a carry is generated from the MS-bit during addition, or borrow
during subtraction/comparison else CF=0.
•Overflow Flag: The OF indicates a signed arithmetic result

overflow. If result of an operation is too large a positive number or
too small a negative number to fit in the destination then OF will
SET, else it will RESET.
DTEL 62
LECTURE 5 Control Flag
TF (Trap Flag) : It is used for Single step operation .If TF=1 then 8086
executes single instruction at a time and stop momentarily. If TF=0
then 8086 executes the given programme in natural sequence.
IF (Interrupt-enable flag) : When IF=1 then maskable Interrupt INTR

will cause the CPU to transfer control to an interrupt vector location.
DF (Direction flag) : Causes string instructions to auto decrement/

increment the index registers (SI/DI) by 1 ( for byte operation) or 2
by word operation). If DF=1 will decrement and DF=0 will increment
index registers.
DTEL 63
LECTURE 6 20-bit Physical address generation
ü Since the 8086 can generate 20-bit physical address therefore
it can access 2 20= 1048576 locations or 1024 Kbytes location
or 1 Mbytes locations addressed from 00000h TO FFFFFh .
ü For programme flexibility the 1Mbytes location is logically
segmented (divided or organized) into
Ø Code Memory Segment (CMS),
Ø Data Memory Segment (DMS),
Ø Extra Memory Segment (EMS) and
Ø Stack Memory Segment (SMS).
DTEL 64
LECTURE 6 20-bit Physical address generation
ü Each memory segment can be maximum of 64 Kbytes.
ü To access a particular location of memory segment the 20-bit

physical address is generated by the addition of Base Address
(BA) provided by the segment register and 16/8 bit offset
address/displacements (OA) is provided by Pointers/index
registers.
The selection of segment registers and

pointers/index registers is according to the rule
given in the table.
DTEL 65
LECTURE 6 Default combination of seg reg & pointer
Segment Default
Name of register Pointers/ Index Memory
Memory used for register used segment
segment base value for offset used for Segment selection rule
address
CMS Automatic during
(Code memory CS IP Instructions execution of a
Segment) programme to prefetch
code.
DMS During execution of a
(Data memory DS BX/SI/16/8bit Local data string instruction or
Segment) displacement data transfer.
During execution of a
(Data memory ES DI/16/8bit External string instruction or
Segment) displacement data data transfer from IO.
SMS During execution of a
(Data memory SS SP/BP Stack stack instruction.
Segment)
DTEL 66
LECTURE 7 Memory Address generation
4-bit
Inserted
0’s
16-BIT SEGMENT VALUE 0000
+ 16-BIT OFFSET
20-BIT PHYSICAL ADRESS
DTEL 67
LECTURE 7 Memory Address generation Example
Inserted 0’s
Segment
Converted to 20-bit
2500 Base
2 5 0 0 0 Logical Address
Added with
95F3 Offset
9 5 F 3
+
2 E 5 F 3 FFFFF H
20-bit Physical Address to

select Memory location
.
External Memory consisting of
. 1024 K location with address
. range from 0000H to FFFFFH
.
.
00000 H
DTEL 68
LECTURE 8 RANGE OF CMS-DMS-EMS-SMS
FFFFF H
Maximum range of one 16 –BIT NUMBER IN
Memory segment is 64 K SEGMENT REGISTER
Physical Address
range from 00000H
to FFFFFH is Inserted 0’s by ∑ of BIU
1024 KB = 1 MB.
Hence we can create
4 set of 4FFFF H
CMS-DMS-EMS-SMS SMS
40000 H 4000 0
20-bit Base Address
3FFFF H
( for SMS)
EMS
30000 H 3000 0
2FFFF H 20-bit Base Address
( for EMS)
DMS
20000 H 2000 0
1FFFF H 20-bit Base Address
CMS ( for DMS)
Total capacity of 1
set of Memory 10000 H 1000 0
segment will be 64 20-bit Base Address
K + 64 K + 64 K + 64 ( for CMS)
K = 256 K 00000 H + 16-bit Offset (0000 to FFFF )
Hence CMS will range from
10000 h to 1FFFFh
DTEL 69
LECTURE 8 Advantages of memory segmentation
ü Allows the memory capacity to be 1Mb even though the

addresses associated with the individual register / instructions are
only 16 bits wide.
ü Facilitate the use of separate memory areas for the program, its
data and the stack and allows a program and/or its data to be put
into different areas of memory each time the program is
executed. Due to which relocatibility of information becomes
efficient.
DTEL 70
LECTURE 8 Advantages of memory segmentation
üThe greatest advantage of segmented memory is that programs
that reference logical addresses only can be loaded and run
anywhere in memory. This is because the logical addresses always
range from 00000h to 0FFFFh, independent of the code segment
base. Such programs are said to be relocatable, meaning that they
can be executed at any location in memory. The requirements for
writing relocatable programs are
1. No reference should be made to physical addresses, and
2. No changes to the segment registers be allowed once
initialised.
ü Since more than 1 set of CMS-DMS-EMS-SMS can be created
therefore multiprogramming can be implemented easily. Also
sharing of segments by different process is also possible.
DTEL 71
LECTURE 9 MINIMUM & MAXIMUM MODE
8086 works in two modes: GND 1
8086
40 +5V
AD14 2 39 AD15
1) Minimum Mode and 2) Maximum AD13 3 38 A16/S3
Mode AD12
AD11
4
5
37 A17/S4
36 A18/S5
If pin 33 MN/MX is high, it works in

AD10 6 35 A19/S6
AD9 7 34 BHE/ S7
minimum mode and If pin 33 MN/MX AD8 8 33 MN/MX MN/MX =1 MN/MX =0

MINIMUM MAXIMUM
is low, it works in maximum mode. AD7
AD6 10
9 32 RD
31
MODE
SIGNALS
HOLD
MODE
SIGNALS
RG/Gto
AD5 11 30 HLDA RQ/GT1
AD4 12 29 WR LOCK
Pins 24 to 31 generates two different

AD3 13 28 M/IO S2
AD2 14 27 DT / R S1
sets of signals. One set of signals is AD4 15 26 DEN So
generated in minimum mode. Other AD1 16
AD0 17
25
24
ALE
INTA
Qso
QS1
set of signals is generated in maximum NMI 18 23 TEST
mode. CLK 19 22 READY
GND 20 21 RESET
DTEL 72
LECTURE 9 Pin Description for Minimum Mode
• Pin 24 is an interrupt acknowledge. When microprocessor
receives INTR signal, it uses this pin to send acknowledgment by
generating 3 active low signal.
• Pin 25 is an Address Latch Enable signal. It indicates that valid
address is generated on bus AD15 – AD0.It generates a pulse
during T1 state.It is connected to enable external latch .
• Pin 26 is a Data Enable signal. This signal is used to enable the
external transceiver like 8286. Transceiver is used to separate
the data from the address/data bus AD15 – AD0.It is an active low
signal.
• Pin 27 is a Data Transmit/Receive signal. It controls the direction
of data flow through the transceiver. When it is high, data is
transmitted out.When it is low, data is received in.
DTEL 73
LECTURE 9 Pin Description for Minimum Mode
• Pin 28 is issued by the microprocessor to distinguish whether
memory or /O access. When it is high, memory can be
accessed. When it is low, I/O devices can be accessed.
• Pin 29 is a Write signal. It is used to write data in memory or
output device depending on the status of M/IO signal. It is an
active low signal.
• Pin 30 is a Hold Acknowledgement signal.It is issued after
receiving the HOLD signal.It is an active high signal.
• Pin 31 During DMA operation microprocessor receives HOLD
signal from DMA controller.
DTEL 74
LECTURE 9 Pin Description for Maximum Mode
QS1 and QS0
Pin 24 and 25
• These pins provide the status of

internal instruction queue.
QS1 QS0 Status
0 0 No operation
0 1 1st byte of opcode from
queue
1 0 Empty queue
1 1 Subsequent byte from queue
DTEL 75
S0, S1, S2
Pin 26, 27, 28
• These status signals
S2 S1 S0 Status
indicate the operation
0 0 0 Interrupt
being to be performed Acknowledge
by the microprocessor.
0 0 1 I/O Read
• These information 0 1 0 I/O Write
decoded by the Bus 0 1 1 Halt
Controller 8288 which 1 0 0 Opcode Fetch
generates all memory 1 0 1 Memory Read
and I/O control signals. 1 1 0 Memory Write
1 1 1 Passive
DTEL 76
LOCK
Pin 29
• This signal indicates that external processors like
8087 should not request CPU to relinquish the
system bus as it is locked with important
operation. This pin is activated by using LOCK
prefix before any instruction.
• When it goes low, all interrupts are masked and
HOLD request is not granted.
DTEL 77
RQ/GT1 and RQ/GT0
Pin 30 and 31 (Bi-directional)
• These are Request/Grant pins.
• External processors like 8087 can request the CPU
through these lines to release the system bus.
• After receiving the request, CPU sends acknowledge
signal on the same lines.
• RQ/GT0 has higher priority than RQ/GT1.
DTEL 78
LECTURE 10 De-multiplexing Address/Data Pin Description
WR RD A[19:16] FFFFF H
A[19:8]
Buffer
ALE CLK
External memory
8086 A[15:0]
with
AD[15:0] 1024 KB
D Q (i,e. 10,48,576)
locations
D latches
00000 H
DT/ R DEN
Data Bus
D[15:0]
Trans-receiver
DTEL 79
LECTURE 11 Memory Read Timing Diagrams
T1 T2 T3 T4
CLK A[19:6]
ALE Buffer
A[19:0]
A[19:16] A[19:16] S3-S6

8086
AD[15:0]
D latch
AD[15:0] A[15:0] D[15:0]

Memory
IO/M D[15:0]
Trans
DT/R
DT/R -ceiver
DEN
DEN
IO/M
RD WR
RD
WR
DTEL 80
LECTURE 11 Memory Write Timing Diagrams
T1 T2 T3 T4
CLK A[19:6]
ALE Buffer
A[19:0]
A[19:16] A[19:16] S3-S6

8086
AD[15:0]
D latch
AD[15:0] A[15:0] D[15:0]

Memory
IO/M D[15:0]
Trans
DT/R
DT/R -ceiver
DEN
DEN
IO/M
WR WR
RD
RD
DTEL 81
LECTURE 12 Chapter 2 Question Bank
• What is pipeline?
• Explain the function of Q 8086.
• Describe the function EU & BIU.
• Explain the function of various registers of 8086.
• Explain function of segment register & pointer .
• What are the various Flag of 8086?
• How 20-bit address is generated in 8086?
• Explain the Minimum and Maximum mode of 8086.
• Explain the timming diagram of memory read 8086.
• Explain the timming diagram of memory write 8086.
DTEL 82
LECTURE 12 Summary
1.The 8086 logically divided into:
ü BIU &
ü EU
Both units operate asynchronously to give the 8086 an overlapping instruction
fetch and execution mechanism which is called as Pipelining.
2. For programme flexibility the 1Mbytes location is logically segmented (divided or

organized) into Code Memory Segment (CMS), Data Memory Segment (DMS), Extra
Memory Segment (EMS) and Stack Memory Segment (SMS).
3.To access a particular location of memory segment the 20-bit physical address
is generated by the addition of Base Address (BA) provided by the segment
register and 16/8 bit offset address/displacements (OA) is provided by
Pointers/index registers.
4. The flags of 8086 can be divided into two types: Conditional Flags and Control Flags
5. 8086 works in two modes: 1) Minimum Mode 2) Maximum Mode.
DTEL 83
CHAPTER-3 Instruction Set of 8086 Microprocessor
Topic 1: Machine Language Instruction format,
1 . addressing modes
2 Topic 2: Instruction set,
3 Topic 3: Groups of Instructions,
DTEL 84
1 Understand the different types of instructions
Identify the addressing modes of instruction and the operation of

2 an instructions
DTEL 85
LECTURE 1 INTRODUCTION TO ASSEMBLY PROGRAMMING
A programme is nothing but set of Instructions written sequentially one below
the other and stored in computers memory for execution by microprocessor.
Instruction consists of a mnemonic and one or two operands (data).
ØMachine Language: In this Programs consist of 0s and 1s.
ØAssembly Languages : It uses short form notations , called , mnemonics ,

to write a programme .The Mnemonics are like MOV , ADD , SUB, etc.
Ø High level languages: It uses English like sentences with proper syntax to
write a programme.
ØAssembler translates Assembly language program into machine code.
Ø Compilers like Pascal, Basic, C etc translate the HLL program into
machine code. The programmer does not have to be concerned with internal
details of the CPU.
DTEL 86
Conversion from HLL to ML is

Done by Compiler / Interpreter
HIGH LEVEL
ASSEMBLY LEVEL
MACHINE
LEVEL
Conversion from ALL to ML

is Done by Assembler like
MASM
DTEL 87
LECTURE 2 Instruction Format
q General Format of Instructions is
Label: Opcode Operands ; Comment
Ø Label: It provides a symbolic address that can be used in branch instructions

Ø Opcode: It specifies the type of instructions
Ø Operands: Instructions of 8086 family can have one, two, or zero operand
Ø Comments: programmers can use for effective reference
q Machine Code Format
Opcode Mode Operand1 Operand2
MOV AL, BL 1000100011000011
binary code of MOV
DTEL 88
LECTURE 2 What is Addressing Modes?
Ø Addressing modes is define as the way in which data is addressed in the
operand part of the instruction. It indicates the CPU finds from where to get data
and where to store results
Ø When a CPU executes an instruction, it needs to know where to get data
and where to store results. Such information is specified in the operand
fields of the instruction.
1000100011000011 MOV AL, BL
Opcode Mode Operand1 Operand2
Ø An operand can be:

— A datum
— A register
— A memory location
DTEL 89
LECTURE 2 Types of Addressing Modes
Addressing Modes Examples
1] Immediate addressing MOV AL, 1FH

2] Register addressing MOV AL, BH
3] Direct addressing MOV [0200H], AH
4] Register Indirect addressing MOV DX, [SI]
5] Relative Based addressing MOV BX, [SI+4]
6] Relative Indexed addressing MOV [DI+8], BL
7] Based indexed addressing MOV [BP+SI], AL
8] Relative Based indexed with displacement addressing MOV AL, [BX+SI+2]
DTEL 90
LECTURE 2 1] Immediate Addressing Mode
In this AM 8/16-bit Data is specified in the operand part of instruction
immediately
• MOV AL,1Fh
• MOV AX,0FC8h
• MOV AH,4Eh
• MOV DX,1F00h
DTEL 91
LECTURE 2 2] Register Addressing
Ø In this data is specified through register in operand part of instruction
Ø Operands are the names of internal register. The processor gets data
from the register specified by instruction .
For Example: move the value of register BL to register AL
MOV AL, BL AH AL
BH BL
q If AX = 1F00H and BX=8086H, after the execution of MOV AL, BL

what are the new values of AX and BX?
DTEL 92
LECTURE 2 3] Direct Addressing
Ø In this AM 16-bit OFFSET address is specified , with symbol [ ] , in the
operand part of the instruction.
ØThe processor will access memory location by adding this OFFSET with
Base address given by DS.
DS × 10H + OFFSET = Memory location
— Example: If DS = 1000H, then explain the operation
MOV AX, 8085 H

DS: 1 0 0 0 _0
MOV [8000H], AX
+ Disp: 8 0 0 0
AH AL 18000
80 85
85 18000H
80 18001H
DTEL 93
LECTURE 2 3] Direct Addressing example
— Example: If DS = 1000H, then explain the operation
1 MOV AX, F2A5 H

MOV [8000H], AH
2 MOV AX, F2A5 H

MOV [8000H], AL
DTEL 94
LECTURE 2 4] Register Indirect Addressing
Ø In this AM OFFSET address is specified indirectly through one of the
registers BX, SI, DI in the instruction operand.
Ø The index register is specified using symbol [ ].
ØThis value is added with DS to generate 20-bit Physical address
For Example: MOV DL, [SI]
Ø Memory address is calculated as following:
BX
DS × 10H + SI = 20-bit Memory address
DI
DTEL 95
LECTURE 2 4] Register Indirect Addressing example
Ø Example 1: assume DS = 0800H then explain the operation

MOV DL, 12H
MOV SI, 2000H DH DL
MOV DL, [SI] 12
0B000H 12
DS: 0 8 0 0 0_
+ SI: 300 0
memory 0 B000
Ø Example 2: assume DS = 0900H, BX=3000H
MOV DL,E7H
MOV [BX], DL
DTEL 96
LECTURE 2 5] Relative Based Addressing
Ø In this AM OFFSET address is specified indirectly by adding an 8-bit (or 16-bit)
constant (displacement) with one of the registers BX, BP in the instruction operand.
Ø If BX appears in the instruction operand field, segment register DS is used in
address calculation and If BP appears in the instruction operand field, segment
register SS is used in address calculation
Ø Calculation of memory address
DS BX
× 10H + + 8/16-bit Displacement = 20-bit Memory address
SS BP
For Example: If DS = 2000H then explain MOV AX, [ BX+4 ]
DTEL 97
LECTURE 2 5]Relative Based Addressing example
Ø Example 1: assume DS = 0100H, BX=0700H
MOV AX, F4E0H
MOV AX, [ BX+4 ] AH AL
C0 B0
DS: 0 1 0 0 0 01705H C0
+ BX: 070 0 01704H B0
+ Disp.: 0 0 0 4
01704 memory
Ø Example 2: assume SS = 0A00H, BP=0012H, CH = ABH
MOV [BP +7], CH
DTEL 98
LECTURE 2 6] Relative Indexed Addressing
Ø In this AM OFFSET address is specified indirectly by adding an 8-bit (or 16-bit)
constant (displacement) with one of the Index registers SI, DI in the instruction
operand. This value is then added with DS to generate 20-bit Physical address
Ø Calculation for memory address
SI
DS × 10H + + 8 / 16 bit Displacement = Memory address
DI
For Example: MOV BL,17H

MOV DI,0030H
Ø Example: assume DS = 0200H MOV [DI+9],BL
MOV [DI+9], BL BH BL
DS: 0 2 0 0 _0 17
+ DI: 003 0 17 02039H
- Disp.: 0 0 0 9
0 203 9
DTEL 99
LECTURE 2 7] Based Indexed Addressing
Ø In this AM OFFSET address is specified indirectly by adding one of the Index
registers SI /DI with based register BX / BP in the instruction operand. This value is
added with DS to generate 20-bit Physical address.
Ø Calculation for memory address
DS BX SI
× 10H + + = 20-bit Memory address
SS BP DI
For Example: MOV [BX] [SI], AH

or
MOV [BX+SI], AH
q If BX appears in the instruction operand field, then segment register DS is used

in address calculation
q If BP appears in the instruction operand field, segment register SS is used in
address calculation
DTEL 100
LECTURE 2 7] Based Indexed Addressing example
Ø Example 1: assume SS = 2000H explain the operation
MOV AH,07H
MOV SI, 0800H
MOV BP,4000H AH AL
MOV [BP] [SI], AH 07
SS: 2 0 0 0 0_ 24800H 07
+ BP: 4 0 0 0
+ SI.: 080 0
24800 memory
Ø Example 2: assume DS = 0B00H, BX=0112H, DI = 0003H, CH=ABH
MOV [BX+DI], CH
DTEL 101
LECTURE 2 8] Relative Based Indexed Addressing
Ø In this AM OFFSET address is specified indirectly by adding 8/16-bit
displacement with one of the Index registers SI /DI with based register BX / BP in
the instruction operand. This value is added with DS to generate 20-bit Physical
address.
Ø Calculate memory address
DS BX SI 8/16-bit
× 10H + + +
displacement
SS BP DI
For Example: MOV CL, [BX+DI+2080H]
DTEL 102
LECTURE 2 8] Relative Based Indexed Addressing example
Ø Example 1: assume DS = 0300H, BX=1000H, DI=0010H

MOV BX,1000H
MOV DI, 0010H
MOV CL, [BX+DI+2080H] CH CL
20
DS: 0 3 0 0 0_
+ BX: 1 0 0 0 06090H 20
+ DI.: 0010
+ Disp. 2 0 8 0
memory
06090
Ø Example 2: assume SS = 1100H, BP=0110H, SI = 000AH, CH=ABH
MOV [BP+SI+0010H], CH
DTEL 103
LECTURE 2 Instruction Types
1] Data transfer instructions
2] Arithmetic instructions
3] String instructions
4] Bit manipulation instructions
5] Loop and jump instructions
6] Subroutine and interrupt instructions
7] Processor control instructions
DTEL 104
LECTURE 2 1] Data transfer instructions
Various Data transfer Instructions are

a) Memory/Register Transfers
MOV Move byte or word to register or memory
XCHG Exchange byte or word
XLAT Translate byte using look-up table
b) Stack Transfers
PUSH Push data onto stack
PUSHF Push flags onto stack
POP Pop data from stack
POPF Pop flags off stack
c) AH/Flags Transfers
LAHF Load AH from flags
SAHF Store AH into flags
DTEL 105
LECTURE 2 1] Data transfer instructions
d) Address Translation
LEA Load effective address
LDS Load pointer using data segment
LES Load pointer using extra segment
e) Port I/O Instructions

IN Input byte or word from port
OUT Output word to port
DTEL 106
LECTURE 4 2] Arithmetic Instructions
Addition
ADD Add byte or word
ADC Add byte or word with carry
INC Increment byte or word by 1
AAA ASCII adjust for addition
DAA Decimal adjust for addition
Subtraction
SUB Subtract byte or word
SBB Subtract byte or word with borrow
DEC Decrement byte or word by 1
NEG Negate byte or word
AAS ASCII adjust for subtraction
DAS Decimal adjust for subtraction
Multiplication
MUL Multiply byte or word unsigned
IMUL Integer multiply byte or word
AAM ASCII adjust for multiplication
Division
DIV Divide byte or word unsigned
IDIV Integer divide byte or word
AAD ASCII adjust for division
CBW Convert byte to word
CWD Convert word to double word
DTEL 107
LECTURE 4 3] Bit Manipulation Instructions
a) Logical Instructions
q NOT Destination
§ Inverts each bit of the destination
§ Destination can be a register or a memory location
q AND Destination, Source
§ Performs logic AND operation for each bit of the destination with corresponding
source bit and stores result into destination
§ Source can be immediate no while destination can be register or memory
§ Destination and source can not be both memory locations at the same time
q OR Destination, Source
§ Performs logic OR operation for each bit of the destination with source; stores
result into destination
DTEL 108
a) Logical Instructions
q XOR Destination, Source

§ Performs logic XOR operation for each bit of the destination with source; stores
result into destination
q TEST Destination, Source

§ Performs logic AND operation for each bit of the destination with source
§ Updates Flags depending on the result of AND operation
§ Do not store the result of AND operation anywhere
DTEL 109
b) Shift Instruction
q SHL Destination, Count
§ SHift LEFT destination bits; the number of times bits shifted is given by CL
§ During the shift operation, the MSB of the destination is shifted into CF and
zero is shifted into the LSB of the destination
CF Destination 0
MSB LSB
q SHR Destination, Count

§ SHift RIGHT destination bits; the number of times bits shifted is given by CL
§ During the shift operation, the LSB of the destination is shifted into CF and
zero is shifted into the MSB of the destination
0 Destination CF
MSB LSB
DTEL 110
b) Shift Instructions
q SAR Destination, Count
§ Shift RIGHT destination bits; the number of times bits shifted is given by CL
§ The LSB of the destination is shifted into CF and the MSB of the destination
is copied in the MSB itself i.e, it remains the same
Destination CF
MSB LSB
DTEL 111
c) Rotate Instructions
q ROL Destination, Count
§ Left shift destination bits; the number of times bits shifted is given by CL
§ The MSB of the destination is shifted into CF, it is also rotated into the LSB .
MSB LSB
CF Destination
q ROR Destination, Count

§ Right shift destination bits; the number of times bits shifted is given by CL
§ The LSB of the destination is shifted into CF, it is also rotated into the MSB .
MSB LSB
Destination CF
DTEL 112
c) Rotate Instructions
q RCL Destination, Count
§ Left shift destination bits; the number of times bits shifted is given by CL
§ The MSB of the destination is shifted into CF; the old CF value is rotated into the LSB.
MSB LSB
CF Destination
q RCR Destination, Count

§ Right shift destination bits; the number of times bits shifted is given by CL
§ The LSB of the destination is shifted into CF, the old CF value is rotated into the MSB.
MSB LSB
Destination CF
DTEL 113
LECTURE 5 4] String Instructions
q String is a collection of bytes or words stored in successive memory locations of
DMS or EMS that can be up to 64KB in length .
q String instructions can have two operands. One is source string and the second is
destination string .
§ Source string is located in Data Segment and SI register points to the current
element of the source string
§ Destination string is located in Extra Segment and DI register points to the
current element of the destination string
DS : SI ES : DI
0510:0000 5F 02A8:2000 5F
0510:0001 4E 02A8:2001 4E
0510:0002 4A 02A8:2002 4A
0510:0003 5B 02A8:2003 5B
0510:0004 D0 02A8:2004 D0
0510:0005 CA 02A8:2005 CA
0510:0006 55 02A8:2006 55
Source String Destination String
DTEL 114
Repeat Prefix Instructions
q REP String Instruction
— The prefix instruction repeatedly execute the instruction until CX AUTO-decrements to 0
(During the execution, CX is decremented by one after execution of the string instruction ).
— For Example: MOV CX, 09

REP MOVSB
By the above two instructions, the microprocessor will execute MOVSB 9 times.
— Execution flow of REP MOVSB is as below:
While (CX!=0) Check_CX: If CX!=0 Then

{ CX = CX –1;
CX = CX –1; OR MOVSB;
MOVSB; goto Check_CX;
} next instruction
DTEL 115
Repeat Prefix Instructions
q REPZ String Instruction

§ Repeatedly execute the string instruction until CX=0 OR zero flag is clear
q REPNZ String Instruction

§ Repeatedly execute the string instruction until CX=0 OR zero flag is set
q REPE String Instruction

§ Repeatedly execute the string instruction until CX=0 OR zero flag is clear
q REPNE String Instruction

§ Repeatedly execute the string instruction until CX=0 OR zero flag is set
DTEL 116
q MOVSB (MOVSW)
§ Move a byte (word) at source memory location of DMS (DS:SI) to destination
memory location (ES:DI) and update SI and DI according to status of DF.
§ After transfer Increment SI/DI by 1 ( or 2) if DF=0 and Decrement SI/DI if
DF=1.
Example: DS : SI ES : DI
MOV AX, 0510H 0510:0000 5E 0300:0100 5E
MOV DS, AX 0510:0001 48 0300:0101 ?
MOV SI, 0 0510:0002 4F 0300:0102 ?
MOV AX, 0300H 0510:0003 50 0300:0103 ?
MOV ES, AX 0300:0104
0510:0004 50 ?
MOV DI, 100H
0510:0005 45
CLD
MOV CX, 5
0510:0006 52
REP MOVSB Source String Destination String
INT 21
DTEL 117
q MOVSB (MOVSW)
§ Move a byte (word) at source memory location of DMS (DS:SI) to destination
memory location (ES:DI) and update SI and DI according to status of DF.
§ After transfer Increment SI/DI by 1 ( or 2) if DF=0 and Decrement SI/DI if
DF=1.
§ Example:
DS : SI ES : DI
MOV AX, 0510H 0510:0000 5E 0300:0100 5E
MOV DS, AX 0510:0001 48 0300:0101 48
MOV SI, 0 0510:0002 4F 0300:0102 4F
MOV AX, 0300H 50 0300:0103 50
0510:0003
MOV ES, AX
0510:0004 50 0300:0104 50
MOV DI, 100H
0510:0005 45
CLD
MOV CX, 5 0510:0006 52
REP MOVSB Source String Destination String
INT 21
DTEL 118
q CMPSB (CMPSW)
§ Compare bytes (words) at memory locations DS:SI and ES:DI;

update SI and DI according to DF and the width of the data being compared
§ Example:
Assume: ES = 02A8H
DI = 2000H
DS = 0510H DS : SI ES : DI
SI = 0000H 0510:0000 4D M 02A8:2000 4D M
0510:0001 4A J 02A8:2001 4A J
0510:0002 45 E 02A8:2002 45 E
CLD 0510:0003 54 54
T 02A8:2003 T
MOV CX, 7 48 H H
0510:0004 02A8:2004 48
REPZ CMPSB 0510:0005 57 W 02A8:2005 57 W
INT 21
0510:0006 41 A 02A8:2006 4E N
What will be the values of CX after Source String Destination String
The execution?
DTEL 119
q SCASB (SCASW)
§ Compare byte in AL (or word in AX) with data at memory location ES:DI;
It updates DI depending status of DF and the length of the data being compare
q LODSB (LODSW)
§ Load byte (word) at memory location DS:SI to AL (AX);
It updates SI depending status of DF and the length of the data being transferred
q STOSB (STOSW)
§ Store byte (word) at in AL (AX) to memory location ES:DI;
It updates DI depending status of DF and the length of the data being transferred
DTEL 120
LECTURE 7 5] Program Transfer Instructions unconditional
q JMP Label
§ Unconditionally Jump to specified Label or address location.
§ Label can be represented by a word or Alphabet with no.
MOV CX, 0007h

MOV AX,F2FEh
ADD AH,CL
JMP L1 Current
SUB AH,CL instruction
JMP
L1: MOV [0200],AH
INT21
Jump
Next
instruction
DTEL 121
LECTURE 7 5] Program Transfer Instructions conditional
Ø Conditional Jumps
q JZ: Label_1
§ If ZF =1, jump to the target address labeled by Label_1; else do not jump
q JNZ: Label_1
§ If ZF =0, jump to the target address labeled by Label_1; else do not jump
Ø Other Conditional Jumps
JNC JAE JNB JC JB JNAE JNG

JNE JE JNS JS JNO JO JNP
JPO JP JPE JA JBNE JBE JNA
JGE JNL JL JNGE JG JNLE JLE
DTEL 122
LECTURE 8 5] Program Transfer Instructions conditional
Mnemonic Meaning Jump Condition

JA Jump if Above CF = 0 and ZF = 0
JAE Jump if Above or Equal CF = 0
JB Jump if Below CF = 1
JBE Jump if Below or Equal CF = 1 or ZF = 1
JC Jump if Carry CF = 1
JE Jump if Equal ZF = 1
JNC Jump if Not Carry CF = 0
JNE Jump if Not Equal ZF = 0
JNZ Jump if Not Zero ZF = 0
JPE Jump if Parity Even PF = 1
JPO Jump if Parity Odd PF = 0
JZ Jump if Zero ZF = 1
DTEL 123
LECTURE 9 5] Program Transfer Instructions (Looping)
q LOOP Label CX = CX –1
If CX != 0 Then
JMP Label else
Next Instruction
CX = CX –1
q LOOPE/LOOPZ Label If CX != 0 & ZF=1 Then
JMP Label else
Next Instruction
q LOOPNE/LOOPNZ Label
CX = CX –1
If CX != 0 & ZF=0 Then
JMP Label else
Next Instruction
DTEL 124
LECTURE 10 6] Processor Control Instructions
q CLC Clear carry flag
q STC Set carry flag
q CMC Complement carry flag
q CLD Clear direction flag
q STD Set direction flag
q CLI Clear interrupt-enable flag
q STI Set interrupt-enable flag
q HLT Halt microprocessor operation

q NOP No operation
q LOCK Lock Bus During Next Instruction
DTEL 125
LECTURE 10 7] Subroutine Instructions
Ø A subroutine or procedure is a collection of instructions, written
separately from main program, and can be called from a program.
Ø Instruction used is CALL Procedure-Name
Ø RET instruction lets the microprocessor to return from a subroutine to the called
program.
Example
•••
MOV AL, 1
CALL M1
MOV BL, 3 The order of execution will be :
••• MOV AL, 1
MOV CL, 2
MOV BL, 3
M1 PROC
MOV CL, 2
RET
M1 ENDP
DTEL 126
• What Opcode & Operand?
• List various Instructions of 8086.
• Describe the Data transfer Instruction.
• Describe the Arithmetic Instruction.
• Describe the Data Bit manipilation Instruction.
• Explain Various Program control Instruction.
• Describe the Processor Control Instruction.
• What Adressing mode ?
• Explain various Adressing modes of 8086.
DTEL 127
LECTURE 12 Summary
1. Addressing modes is define as the way in which data is addressed in the
operand part of the instruction. It indicates the CPU finds from where to get data
and where to store results.
2. 8086 has 8 Adressing modes a] Immediate addressing b] Register addressing

c] Direct addressing d] Register Indirect addressing e] Relative Based
f] Relative Indexed addressing g] Based indexed addressing h] Relative Based
indexed with displacement addressing
3. 8086 Instructions cab be grouped as ,

1] Data transfer instructions
2] Arithmetic instructions
3] String instructions
4] Bit manipulation instructions
5] Loop and jump instructions
6] Subroutine and interrupt instructions
7] Processor control instructions
DTEL 128
CHAPTER-4 The Art of Assembly Language Programming
1 .
Topic 1: Program development steps
2 Topic 2: Assembly Language Programming Tools
3 Topic 3: Assembler directives and Operators
DTEL 129
Know the program development steps

1
Use the different program development tools

2
DTEL 130
A programme is nothing but set of Instructions written sequentially one

below the other and stored in computers memory for execution by
microprocessor.
Instruction consists of a mnemonic and one or two operands (data).
Ø Machine Language: In this Programs is written in 0s and 1s.
Ø Assembly Languages : It uses short form notations , called ,

mnemonics , to write a programme .The Mnemonics are like MOV , ADD
, SUB, etc.
Ø High level languages: It uses English like sentences with proper

syntax to write a programme.
Ø Assembler translates Assembly language program into machine

code.
DTEL 131
Conversion from HLL to ML is

Done by Compiler / Interpreter
HIGH LEVEL
ASSEMBLY LEVEL
MACHINE
LEVEL
Conversion from ALL to ML

is Done by Assembler like
MASM
DTEL 132
LECTURE 2 Assemble, Link and execute Program
Step & Input Software Output

operation
1.Editing / Note pad or Note pad / Filename.asm
writing Word MS-Word
programme
2.Assemble Filename.asm MASM Filename.obj
3.Link Filename.obj LINK Filename.exe
DTEL 133
LECTURE 2 How to Build Executable Programs
Filename.obj
Assembler
checks syntax
Executable
Editor and translate
File
programme like source code Linker filename.exe
note-pad / Word into machine
code (like
MASM)
Other Debug or
OBJ
files & code view
Filename.asm Library if any error
DTEL 134
LECTURE 3 How to Build Executable Programs
Editor
A text editor is required in order to create assembly language
source files, where you’ll be writing your code. You can use
Notepad, DOS editor, or any other editor of your choice that
produces plain ASCII text files.
Debugger
A debugger program allows tracing of program execution and
examination of registers and memory content.
For 16-bit programs, MASM’s debugger named CodeView can be
used to debug .
DTEL 135
LECTURE 3 Assembler
Source files Assembler

Syntax check OBJ
Translate source files
files into
machine code
An assembler is a program that converts source-code programs

written in assembly language into object files in machine
language. Popular assemblers include MASM (Macro Assembler
from Microsoft), TASM (Turbo Assembler from Borland), NASM
(Netwide Assembler for both Windows and Linux), and GNU
assembler distributed by the free software foundation.
DTEL 136
LECTURE 3 Linker
library Executable
Source files Assembler files
Syntax check
Translate source
OBJ
files
Linker
files into
machine code
OBJ
files
A linker program combines your program's object file created by the assembler
with other object files and link libraries, producing a single executable
program. You need a linker utility to produce executable files.
Two linkers: LINK.EXE and LINK32.EXE are provided with the MASM 6.15
distribution to link 16-bit real-address mode and 32-bit protected-address
mode programs respectively.
DTEL 137
LECTURE 3 Assembler Directives
It provides information to assist the assembler in producing

executable code. It creates storage for a variable and initialize it.
Assembler directives (pseudo-instructions) give directions to the
assembler about how it should translate the Assembly language
instructions into machine code, Where as other instructions
discussed in the above section give command to the 8086
microprocessor. Assembler directives are specific for a particular
assembler. However all the popular assemblers like the Intel 8086
macro assembler, the turbo assembler and the IBM macro
assembler use common assembler directives. The basic structure
of a program in ASM will look like this
DTEL 138
Lecture 4 Program format
; Program for addition of two 8-bit nos. Comments / Remark
ASSUME CS: Code DS: Data

Data SEGMENT
N1 DB 2FH
N2 DB 0EH Assembler directives
SUM DB 1 DUP(?)
Data ENDS
Code SEGMENT
MOV AL,N1
MOV BL,N2 Program
ADD AL,BL
MOV SUM,AL
Code ENDS Assembler directives

END
DTEL 139
LECTURE 4 ASSUME
1) The ASSUME directive tells the assembler the name of the

logical segment it should use for a specified segment. It
associates segment names with segment registers. For example
ASSUME CS: Code tells assembler that the
instructions for programme are in the logical segment named
Code.
Similarly for
ASSUME DS: Data tells assembler that for
any program if instruction refers to the data segment then it
should use the logical segment named Data. For example in the
instruction MOV AX, [BX] the memory segment referred to by
[BX] is in logical segment Data.
DTEL 140
LECTURE 4 ASSUME
The 8086 contains a segment register (DS) that is dedicated to a
data memory segment. This register is the default segment register
used for all memory references used for data. The user is
responsible loading the DS register with the appropriate value and
telling the assembler where the DS register points so that it can
calculate the offsets correctly. The standard is to define a segment
to be a data segment. This is a convenient way of keeping data and
code separate. The most common way of doing this is:
Data SEGMENT
...
Data ENDS ;indicates the end of the data segment
Hence the Assume directive is required to inform assembler of

location of DS pointer by : ASSUME DS: Data;
DTEL 141
LECTURE 4 Data directive
2) Data storage directive
Each variable has a data type and is assigned a memory address by
the program. Data directives are used to reserve and provide name
for memory location in data segments. The symbols used for data
types are:
Data type Symbol
Byte B
Word W
Double word D
Quad Word Q
Ten Bytes T
DTEL 142
LECTURE 4 Data directive Example
For byte variable we should use
ü DB for declaration
N1 DB 4 ; initialise variable N1 with value 4 in decimal

N2 DB AFH ; initialise variable N2 with value AF in Hex
Name DB “JETHWA” ; allocates 6 byte with variable Name
J 4A Name
E 45
T 54
H 48
W 57
A 41
DTEL 143
LECTURE 4 DUP Operator
DUP Operator is used to create arrays of elements whose initialize
value is same.
The basic syntax is count DUP ( initial value)

Example :
N1 DB 100 DUP( 0) ; create 100 bytes arrays with value 0 in each
N2 DW 5 DUP (?) ; CREATE 5 arrays of uninitialized words
L1 DB 5, 4, 3 DUP( 2,3 DUP(0),1) is same as

L1 DB 5,4,2,0,0,0,1,2,0,0,0,1,2,0,0,0,1
DTEL 144
• What is Machine language?
• What is Assembly language?
• What is High level language?
• Describe the function of Linker.
• Describe the function of Assembler & debugger.
• What is Assemble directives?
• Explain various Assemble directives.
DTEL 145
LECTURE 5 Summary
1. A programme is nothing but set of Instructions written sequentially one below the other
and stored in computers memory for execution by microprocessor. Program can be written
in 3 levels a) Machine Language b) Assembly Languages c) High level languages
2. Assembler translates Assembly language program into machine code.
3. Compilers like Pascal, Basic, C etc translate the HLL program into machine code.
The programmer does not have to be concerned with internal details of the CPU.
4. A text editor is required in order to create assembly language source files, where you’ll be writing
your code. You can use Notepad, DOS editor, or any other editor of your choice that produces lain
ASCII text files.
5. A debugger program allows tracing of program execution and examination of registers and
memory content.
6. An assembler is a program that converts source-code programs written in assembly
language into object files in machine language.
7. A linker program combines your program's object file created by the assembler with
other object files and link libraries, producing a single executable program. You need
a linker utility to produce executable files.
8. Assembler directives provides information to assist the assembler in producing
executable code. It creates storage for a variable and initialize it. Assembler directives
(pseudo-instructions) give directions to the assembler about how it should translate the
Assembly language instructions into machine code
DTEL 146
CHAPTER-5 8086 Assembly Language Programming.
Topic 1: Model of 8086 assembly language programs

1 .
Topic 2: Programming using assembler -

2
DTEL 147
Write a appropriate programs using editor

1
Run program using assembler & linker and Debug program

2 using debugger
DTEL 148
Lecture 1 Program for 8-bit addition

Data SEGMENT
N1 DB 2FH
N2 DB 0EH
SUM DB 1 DUP(?)
Data ENDS
Code SEGMENT
MOV AX, Data
MOV DS,AX
MOV AL,N1
MOV BL,N2
ADD AL,BL
MOV SUM,AL
Code ENDS
END
DTEL 149
Lecture 2 Program for 16-bit addition

Data SEGMENT
N1 DW 002FH
N2 DW 000EH
SUM DW 1 DUP (?)
Data ENDS
Code SEGMENT
MOV AX, Data
MOV DS,AX
MOV AX,N1
MOV BX,N2
ADD AX,BX
MOV SUM,AX
Code ENDS
END
DTEL 150
Lecture 3 Program for 16-bit Multiplication
; Program for MULTIPLICATION of two 16-bit nos. Comments / Remark

Data SEGMENT
N1 DW 002FH
N2 DW 000EH
RES DW 2 DUP (?)
Data ENDS
Code SEGMENT
MOV AX, Data
MOV DS,AX
MOV AX,N1
MOV BX,N2
MUL BX
MOV RES,AX
MOV RES+2,BX
Code ENDS
END
DTEL 151
Lecture 4 Program for 16-bit Division
; Program for DIVISION of 16-bit no. Comments / Remark

Data SEGMENT
N1 DW 0F2FH
N2 DB 0EH
RESQ DB 2 DUP (?)
Data ENDS
Code SEGMENT
MOV AX, Data
MOV DS,AX
MOV AX,N1
MOV BL,N2
DIV BL
MOV RESQ,AL
MOV RESQ+1,AH
Code ENDS
END
DTEL 152
Lecture 5 Program for 32-bit Division
; Program for DIVISION of 32-bit no. Comments / Remark
Data SEGMENT
N1 DD FE000F2FH
N2 DW E40EH
RESQ DW 2 DUP (?)
Data ENDS
Code SEGMENT
MOV AX, Data
MOV DS,AX
MOV AX,N1
MOV DX,N1+2
MOV BX,N2
DIV BX
MOV RESQ,AX
MOV RESQ+2,DX
Code ENDS
END
DTEL 153
• Write program to transfer a block of 50 bytes B1 to another
block B2.The block B1 begins with offset address 1000h and
block B2 from 2000h?
• Write program to exchange data of block of 10 bytes B1 to with
another block B2.The block B1 begins with offset address
0200h and block B2 from 0300h?
• Write program to arrange a block of 50 bytes in ascending
order. The block begins with offset address 1000h?
• Write program to arrange a block of 50 bytes in descending
order. The block begins with offset address 1000h?
DTEL 154
CHAPTER-6 Procedure and Macro in Assembly Language Program
1 .
Topic 1: Procedure
2 Topic 2: Defining Macros.
DTEL 155
Understand the purpose of procedure and macros

1
Use procedure and macros

2
DTEL 156
LECTURE 1 Procedures
Procedure is a part of code that can be called from your
program in order to make some specific task. Procedures make
program more structural and easier to understand. Generally
procedure returns to the same point from where it was called.
The syntax for procedure declaration:

name PROC
; here goes the code

; of the procedure ...
RET
name ENDP
DTEL 157
LECTURE 1 Procedures
name - is the procedure name, the same name should be in the top and the
bottom, this is used to check correct closing of procedures.
RET instruction is used from procedure
PROC and ENDP are compiler directives, so they are not assembled into any
real machine code. Compiler just remembers the address of procedure.
CALL instruction is used to call a procedure.
DTEL 158
LECTURE 1 Procedures example
ORG 100h
MOV AL,2FH
MOV BL,F2H
CALL m1
MOV [SI] , AX
RET ; return to operating system.
Procedure for multiplication

m1 PROC
MUL BL
RET ; return to caller.
m1 ENDP
END
DTEL 159
LECTURE 1 Procedures example
There are several ways to pass parameters to procedure, the easiest way to pass
parameters is by using registers, here is another example of a procedure that
receives two parameters in AL and BL registers, multiplies these parameters and
returns the result in AX register:
ORG 100h
MOV AL, 1
MOV BL, 2
CALL m2 In this example value of AL register
CALL m2 is update every time the procedure is
called, BL register stays unchanged,
CALL m2
so this algorithm calculates 2 in
CALL m2 power of 4,
RET ; return to operating
system. so final result in AX register is 16 (or
10h).
m2 PROC
MUL BL ; AX = AL * BL.
RET ; return to caller.
m2 ENDP
END
DTEL 160
LECTURE 1 NEAR CALL & FAR CALL
A near CALL is a call to a procedure which is in the same code

memory segment as that of CALL instruction in this the 8086
decrements Stack pointer by 2 and copies the IP on the
STACK.
A far CALL is a call to a procedure which is in different code

segment as that of CALL instruction. . In this the 8086
decrements Stack pointer by 2 and copies the CS first on the
STACK and then again decrement SP by 2 to copy IP on the
STACK.
DTEL 161
LECTURE 2 Recursive Procedure
Recursive Procedure: A recursive procedure is procedure which calls itself. It is
used to work with complex data structures called trees. If the procedure is called
with N (known as recursion depth) =3 then the n is decremented by 1 after each
procedure CALL and the procedure is called until n=0 as shown in the diagram
below:
Main P Proc Fact Proc Fact Proc Fact
CALL CALL CALL
RET RET RET

With 3! With 2! With 1!
DTEL 162
LECTURE 2 To find out factorial of number
;Program to find out factorial of number Using Recursion
Data SEGMENT
NUMBER DB 03H
FACTORIAL DW 1DUP(?)
ENDS
Stack SEGMENT
DW 128 DUP(0)
ENDS
Code SEGMENT
ASSUME CS:Code, DS:Data, SS:Stack
; INITIALISE SEGMENT REGISTERS:
MOV AX, Data
MOV DS, AX
MOV AX, Stack
MOV SS, AX
MOV CX,NUMBER
CALL FACT
RET
DTEL 163
LECTURE 2 To find out factorial of number
;PROCEDURE FOR FACTORIAL PROGRAM

;CX CONTAINS INPUT NUMBER
;DX CONTAINS RESULT
FACT PROC NEAR
CMP CX, 01H
JNE CONT
MOV DX,01H
RET
CONT: PUSH CX ; FOR BACKUP
DEC CX
CALL FACT
POP AX ; BACKUP OF CX IE N
MUL DX ; N*(N-1)!
MOV DX, AX ; RESULT INTO DX
RET
FACT ENDP
ENDS
DTEL 164
LECTURE 3 Reentrant Procedure
Reentrant Procedure: A program or subroutine is called reentrant
if it can be interrupted in the mid (i.e. the control flow is
transferred outside of the subroutine, either due to an internal
action such as a jump or call, or by an external action such as a
hardware interrupt or signal), and then can then safely be called
again before its previous invocation has been completed, and once
the reentered invocation completes, the previous invocations
should be able to resume execution correctly.
If procedure1 is called from main program and procedure2 is
called from procedure1 and procedure1 again from procedure2
then such is called as reentrant procedure as shown below:
DTEL 165
LECTURE 3 Macros
Macros
Macros are just like procedures, but not really. Macros look
like procedures, but they exist only until your code is
compiled, after compilation all macros are replaced with real
instructions. For Macro assembler generates the code in
program each time where the macro is “called”. If you
declared a macro and never used it in your code, compiler will
simply ignore it.
Macro definition:
name MACRO
[parameters,...]
<instructions>
ENDM
DTEL 166
LECTURE 3 Macros example
MyMacro MACRO p1, p2, p3
MOV AX, p1
MOV BX, p2
MOV CX, p3
ENDM
ORG 100h
MyMacro 1, 2, 3
The code is expanded into:
MyMacro 4, 5, DX
RET MOV AX, 00001h
MOV BX, 00002h
MOV CX, 00003h
MOV AX, 00004h
MOV BX, 00005h
MOV CX, DX
DTEL 167
LECTURE 4 Compare Procedure & Macro
Procedure Macro
Accessed during assembly with
Accessed by CALL & RET
name given during program
instruction
execution to macro when defined
Machine code for
Machine code is generated for
instruction is put only
instruction each time when macro is
once in the memory
called.
With procedures
With macro more memory is
less memory is required
required
Parameters can be passed

Parameters passed as part of
in registers,
statement which calls macro
memory locations or stack
DTEL 168
LECTURE 5 Advantages and Disadvantages of MACRO
Advantages of MACRO
üProgram written with MACRO is more readable
üMACRO can be called by just writing its name along with its
parameters;
hence no extra code is required like CALL & RET.
üExecution time is less as compared to Procedure
üFinding errors is easy
Disadvantages of MACRO
üObject code is generated every time Macro is called, hence
object file
becomes lengthy
üFor large group of instruction macro is not preferred
DTEL 169
• What is Procedures?
• What are the instructions to implement Procedures?
• What is Re-entrant Procedures?
• Describe the function MACROS.
• What are the differences between Procedures & MACROS.
• List various Advantages and disadvantages of MACROS.
DTEL 170
LECTURE 5 Summary
1. Procedure is a part of code that can be called from your program in order to make some specific
task. Procedures make program more structural and easier to understand. Generally procedure
returns to the same point from where it was called. CALL instruction is used to call a procedure.
2. A near CALL is a call to a procedure which is in the same code memory segment as that of CALL
instruction in this the 8086 decrements Stack pointer by 2 and copies the IP on the STACK.
3. A far CALL is a call to a procedure which is in different code segment as that of CALL instruction. . In
this the 8086 decrements Stack pointer by 2 and copies the CS first on the STACK and then again
decrement SP by 2 to copy IP on the STACK.
4. Reentrant Procedure: A program or subroutine is called reentrant if it can be interrupted in the

middle (i.e. the control flow is transferred outside of the subroutine, either due to an internal action
such as a jump or call, or by an external action such as a hardware interrupt or signal), can then
safely be called again before its previous invocations have been completed, and once the reentered
invocation completes, the previous invocations should be able to resume execution correctly.
5. Macros are just like procedures, but not really. Macros look like procedures, but they exist only
until your code is compiled, after compilation all macros are replaced with real instructions. For
Macro assembler generates the code in program each time where the macro is “called”. If you
declared a macro and never used it in your code, compiler will simply ignore it.
DTEL 171
Recommended Books:
1. Advanced Microprocessor and Peripherals (Architecture, Programming
& Interfacing) by A.K. Roy & K.M. Bhurchandi, Tata Mcgraw Hill
2. Fundamentals of MIcroprocessors by B RAM, Dhanpat Rai Publications
3. Microprocessors by A.P.Godse, Technical Publications
4. 8085 Microprocessor: Programming And Interfacing 1st Edition
Author: N.K.Srinath, PHI Learning Private Limited
5. Microprocessor 8085 And Its Interfacing, Author A.P.Mathur , PHI
Learning Pvt. Ltd.
6. The 8088 and 8086 Microprocessors: , Author: Walter A. Triebel, Avtar
Singh,
7. Microprocessor 8085, 8086 , by Abhishek Yadav
8. Microprocessors: Theory and Applications : Intel and Motorola
by Mohamed Rafiquzzaman
9. “Microcomputer Systems: The 8086/88 Family”, Liu, Gibson, 2nd
Edition, PHI Learning Private Limited
DTEL 172
References Books:
1. Microprocessor Architecture,Programming and Applications with the 8085
by Ramesh S. Gaonkar , Penram International Publishing (India)
2. Microprocessor & interfacing (programming & hardware) Revised Second
Edition by Douglas V. Hall , Tata McGraw Hill
3. The Intel Microprocessors: 8086/8088, 80186/80188, 80286, 80386
Barry Bray , Pearson Education; Eighth edition (2011)
4. The 8086/8088 Family: Design, Programming And Interfacing 1st Edition,
John Uffenbeck, Prentice-Hall
5. Microcomputer Systems the 8086/8088 Family : Architecture,
Programming and Design 2nd Edition Author: Gleen A.Gibson , Prentice-
Hall
6. 8085 Microprocessor: Programming And Interfacing 1st Edition
Author: N.K.Srinath, PHI Learning Private Limited
7. The 8086 Microprocessor :Programming & Interfacing the PC with CD
Kenneth Ayala, Publisher: Cengage Learning
8. Assembly programming and the 8086 microprocessor, Douglas Samuel
Jones ,Oxford University Press
DTEL 173
References Web:
1. www.intel.com
2. www.pcguide.com/ref/CPU
3. www.CPU-World.com /Arch /
4. www.techsource .com / Engineering parts/ microprocessor.html
5. www.slideshare.net
6. www.powershow.com
7. www.authorstream.com
8. www.youtube.com
9. www.scribd.com
10. www.eazynotes.com
11. www.electronicstutorialsblog.com
12. ece.uprm.edu
DTEL 174
8085 MICROPROCESSOR
PROGRAMS
1
ADDITION OF TWO 8 BIT NUMBERS
AIM:
To perform addition of two 8 bit numbers using 8085.
ALGORITHM:
1) Start the program by loading the first data into Accumulator.

5) Check for carry.
6) Store the value of sum and carry in memory location.
PROGRAM:
MVI C, 00 Initialize C register to 00

MOV B, A Move the content of Accumulator to B register.
ADD B Add the value of register B to A
INR C Increment value of register C
LOOP: STA 4152 Store the value of Accumulator (SUM).
MOV A, C Move content of register C to Acc.
STA 4153 Store the value of Accumulator (CARRY)
HLT Halt the program.
OBSERVATION:
Input: 80 (4150)
80 (4251)
Output: 00 (4152)
01 (4153)
RESULT:
Thus the program to add two 8-bit numbers was executed.
2
AIM:
To perform the subtraction of two 8 bit numbers using 8085.
ALGORITHM:
1. Start the program by loading the first data into Accumulator.

2. Move the data to a register (B register).
3. Get the second data and load into Accumulator.
4. Subtract the two register contents.
5. Check for carry.
6. If carry is present take 2’s complement of Accumulator.
7. Store the value of borrow in memory location.
8. Store the difference value (present in Accumulator) to a memory
9. location and terminate the program.
PROGRAM:
MVI C, 00 Initialize C to 00
MOV B, A Move the content of Acc to B register.
SUB B
CMA Complement Accumulator contents.
INR A Increment value in Accumulator.
INR C Increment value in register C
LOOP: STA 4152 Store the value of A-reg to memory address.
MOV A, C Move contents of register C to Accumulator.
STA 4153 Store the value of Accumulator memory address.
3
OBSERVATION:
Input: 06 (4150)
02 (4251)
Output: 04 (4152)
01 (4153)
RESULT:
Thus the program to subtract two 8-bit numbers was executed.
4
MULTIPLICATION OF TWO 8 BIT NUMBERS
AIM:
To perform the multiplication of two 8 bit numbers using 8085.
ALGORITHM:
5) Check for carry.
6) Increment the value of carry.
7) Check whether repeated addition is over and store the value of product and carry
in memory location.
PROGRAM:
MVI D, 00 Initialize register D to 00

MVI A, 00 Initialize Accumulator content to 00
LXI H, 4150
MOV B, M Get the first number in B - reg
INX H
MOV C, M Get the second number in C- reg.
LOOP: ADD B Add content of A - reg to register B.
JNC NEXT Jump on no carry to NEXT.
INR D Increment content of register D
NEXT: DCR C Decrement content of register C.
JNZ LOOP Jump on no zero to address
STA 4152 Store the result in Memory
MOV A, D
STA 4153 Store the MSB of result in Memory
5
OBSERVATION:
Input: FF (4150)
FF (4151)
Output: 01 (4152)
FE (4153)
RESULT:
Thus the program to multiply two 8-bit numbers was executed.
6
DIVISION OF TWO 8 BIT NUMBERS
AIM:
To perform the division of two 8 bit numbers using 8085.
ALGORITHM:
2) Move the data to a register(B register).
4) Compare the two numbers to check for carry.
5) Subtract the two numbers.
6) Increment the value of carry .
7) Check whether repeated subtraction is over and store the value of product and
carry in memory location.
PROGRAM:
LXI H, 4150
MOV B, M Get the dividend in B – reg.
MVI C, 00 Clear C – reg for qoutient
INX H
MOV A, M Get the divisor in A – reg.
NEXT: CMP B Compare A - reg with register B.
JC LOOP Jump on carry to LOOP
SUB B Subtract A – reg from B- reg.
INR C Increment content of register C.
JMP NEXT Jump to NEXT
LOOP: STA 4152 Store the remainder in Memory
MOV A, C
STA 4153 Store the quotient in memory
7
OBSERVATION:
Input: FF (4150)
FF (4251)
Output: 01 (4152) ---- Remainder

FE (4153) ---- Quotient
RESULT:
Thus the program to divide two 8-bit numbers was executed.
8
LARGEST NUMBER IN AN ARRAY OF DATA
AIM:
To find the largest number in an array of data using 8085 instruction set.
ALGORITHM:

8) If Carry = 0, go to step 10 or if Carry = 1 go to step 9
12) Store the largest data in memory.
PROGRAM:

INX H
LOOP: INX H
CMP M If A- reg > M go to AHEAD
JNC AHEAD
MOV A,M Set the new value as largest
AHEAD: DCR B
HLT
9
OBSERVATION:

0A (4201)
F1 (4202)
1F (4203)
26 (4204)
FE (4205)
Output: FE (4300)
RESULT:
Thus the program to find the largest number in an array of data was executed
10
SMALLEST NUMBER IN AN ARRAY OF DATA
AIM:
To find the smallest number in an array of data using 8085 instruction set.
ALGORITHM:

8) If carry = 1, go to step 10 or if Carry = 0 go to step 9
12) Store the smallest data in memory.
PROGRAM:

INX H
LOOP: INX H
CMP M If A- reg < M go to AHEAD
JC AHEAD
MOV A,M Set the new value as smallest
AHEAD: DCR B
HLT
11
OBSERVATION:

0A (4201)
F1 (4202)
1F (4203)
26 (4204)
FE (4205)
Output: 0A (4300)
RESULT:
Thus the program to find the smallest number in an array of data was executed
12
ARRANGE AN ARRAY OF DATA IN ASCENDING ORDER
AIM:
To write a program to arrange an array of data in ascending order
ALGORITHM:

PROGRAM:
LXI H,4200
MOV C,M
DCR C
REPEAT: MOV D,C
LXI H,4201
LOOP: MOV A,M
INX H
CMP M
JC SKIP
MOV B,M
MOV M,A
DCX H
MOV M,B
INX H
SKIP: DCR D
JNZ LOOP
DCR C
JNZ REPEAT
HLT
13
OBSERVATION:

4201 05
4202 04
4203 03
4204 02
4205 01

4201 01
4202 02
4203 03
4204 04
4205 05
RESULT:
Thus the given array of data was arranged in ascending order.
14
ARRANGE AN ARRAY OF DATA IN DESCENDING ORDER
AIM:
To write a program to arrange an array of data in descending order
ALGORITHM:

PROGRAM:
LXI H,4200
MOV C,M
DCR C
REPEAT: MOV D,C
LXI H,4201
LOOP: MOV A,M
INX H
CMP M
JNC SKIP
MOV B,M
MOV M,A
DCX H
MOV M,B
INX H
SKIP: DCR D
JNZ LOOP
DCR C
JNZ REPEAT
HLT
15
OBSERVATION:

4201 01
4202 02
4203 03
4204 04
4205 05

4201 05
4202 04
4203 03
4204 02
4205 01
RESULT:
Thus the given array of data was arranged in descending order.
16
BCD TO HEX CONVERSION

AIM:
To convert two BCD numbers in memory to the equivalent HEX number using 8085
instruction set
ALGORITHM:

2) Get the Most Significant Digit (MSD)
3) Multiply the MSD by ten using repeated addition
4) Add the Least Significant Digit (LSD) to the result obtained in previous step
5) Store the HEX data in Memory
PROGRAM:
LXI H,4150
MOV A,M Initialize memory pointer
ADD A MSD X 2
MOV B,A Store MSD X 2
ADD A MSD X 4
ADD A MSD X 8
ADD B MSD X 10
INX H Point to LSD
ADD M Add to form HEX
INX H
MOV M,A Store the result
HLT
OBSERVATION:
Input: 4150 : 02 (MSD)

4151 : 09 (LSD)
Output: 4152 : 1D H
RESULT:
Thus the program to convert BCD data to HEX data was executed.
17
HEX TO BCD CONVERSION

AIM:
To convert given Hexa decimal number into its equivalent BCD number using 8085
instruction set
ALGORITHM:

2) Get the Hexa decimal number in C - register
3) Perform repeated addition for C number of times
4) Adjust for BCD in each step
5) Store the BCD data in Memory
PROGRAM:
LXI H,4150 Initialize memory pointer

MVI D,00 Clear D- reg for Most significant Byte
XRA A Clear Accumulator
MOV C,M Get HEX data
LOOP2: ADI 01 Count the number one by one
DAA Adjust for BCD count
JNC LOOP1
INR D
LOOP1: DCR C
JNZ LOOP2
STA 4151 Store the Least Significant Byte
MOV A,D
STA 4152 Store the Most Significant Byte
HLT
OBSERVATION:
Input: 4150 : FF
Output: 4151 : 55 (LSB)

4152 : 02 (MSB)
RESULT:
Thus the program to convert HEX data to BCD data was executed.
18
HEX TO ASCII CONVERSION

AIM:
To convert given Hexa decimal number into its equivalent ASCII number using
ALGORITHM:
1. Load the given data in A- register and move to B – register

2. Mask the upper nibble of the Hexa decimal number in A – register
3. Call subroutine to get ASCII of lower nibble
5. Move B –register to A – register and mask the lower nibble
6. Rotate the upper nibble to lower nibble position
7. Call subroutine to get ASCII of upper nibble
9. Terminate the program.
PROGRAM:
LDA 4200 Get Hexa Data

MOV B,A
ANI 0F Mask Upper Nibble
CALL SUB1 Get ASCII code for upper nibble
STA 4201
MOV A,B
ANI F0 Mask Lower Nibble
RLC
RLC
RLC
RLC
CALL SUB1 Get ASCII code for lower nibble
STA 4202
HLT
SUB1: CPI 0A
JC SKIP
ADI 07
SKIP: ADI 30
RET
19
OBSERVATION:
Input: 4200 E4(Hexa data)
Output: 4201 34(ASCII Code for 4)

4202 45(ASCII Code for E)
RESULT:
Thus the given Hexa decimal number was converted into its equivalent ASCII Code.
20
ASCII TO HEX CONVERSION

AIM:
To convert given ASCII Character into its equivalent Hexa Decimal number using
ALGORITHM:
1. Load the given data in A- register

3. Compare the content of A – register with 0A H
4. If A < 0A H, jump to step6. Else proceed to next step.
6. Store the result
7. Terminate the program
PROGRAM:
LDA 4500
SUI 30
CPI 0A
JC SKIP
SUI 07
SKIP: STA 4501
HLT
OBSERVATION:
Input: 4500 31
Output: 4501 0B
RESULT:
Thus the given ASCII character was converted into its equivalent Hexa Value.
21
SQUARE OF A NUMBER USING LOOK UP TABLE

AIM:
To find the square of the number from 0 to 9 using a Table of Square.
ALGORITHM:
1. Initialize HL pair to point Look up table

2. Get the data .
3. Check whether the given input is less than 9.
4. If yes go to next step else halt the program
5. Add the desired address with the accumulator content
6. Store the result
PROGRAM:
LXI H,4125 Initialsie Look up table address

LDA 4150 Get the data
CPI 0A Check input > 9
JC AFTER if yes error
MVI A,FF Error Indication
STA 4151
HLT
AFTER: MOV C,A Add the desired Address
MVI B,00
DAD B
MOV A,M
STA 4151 Store the result
HLT Terminate the program
LOOKUP TABLE:
4125 01
4126 04
4127 09
4128 16
4129 25
4130 36
4131 49
4132 64
4133 81
22
OBSERVATION:
Input: 4150: 05
Output: 4151 25 (Square)
Input : 4150: 11
Output: 4151: FF (Error Indication)
RESULT:
Thus the program to find the square of the number from 0 to 9 using a Look up table
was executed.
23
INTERFACING WITH 8085
24
INTERFACING 8251 (USART) WITH 8085 PROCESSOR
AIM:
To write a program to initiate 8251 and to check the transmission and reception of
character
THEORY:
The 8251 is used as a peripheral device for serial communication and is programmed
by the CPU to operate using virtually any serial data transmission technique. The USART
accepts data characters from the CPU in parallel format and then converts them into a
continuous serial data stream for transmission. Simultaneously, it can receive serial data
streams and convert them into parallel data characters for the CPU. The CPU can read the
status of USART ant any time. These include data transmission errors and control signals.
Prior to starting data transmission or reception, the 8251 must be loaded with a set
of control words generated by the CPU. These control signals define the complete
functional definition of the 8251 and must immediately follow a RESET operation. Control
words should be written into the control register of 8251. These control words are split into
two formats:

This format defines the Baud rate, Character length, Parity and Stop bits required to
work with asynchronous data communication. By selecting the appropriate baud factor sync
mode, the 8251 can be operated in Synchronous mode.
Initializing 8251 using the mode instruction to the following conditions
8 Bit data
No Parity
Baud rate Factor (16X)
1 Stop Bit
gives a mode command word of 01001110 = 4E (HEX)
25
MODE INSTRUCTION - SYNCHRONOUS MODE
BAUD RATE FACTOR

0 1 0 1
0 0 1 1
SYNC MODE (1X) (16X) (64X)
CHARACTR LENGTH
0 1 0 1
0 0 1 1
6 7
5 BITS BITS BITS 8 BITS
PARITY ENABLE

0 =ODD 1 = EVEN
NUMBER OF STOP BITS

0 1 0 1
0 0 1 1
1.5
INVALID 1 BIT BIT 2 BIT
26
MODE INSTRUCTION - ASYNCHRONOUS MODE
CHARACTER LENGTH
0 1 0 1
0 0 1 1
5 BITS 6 BITS 7 BITS 8 BITS
PARITY ENABLE

0 =ODD 1 = EVEN
EXTERNAL SYNC DETECTS

1 = SYSDET IS AN INPUT
0 = SYSDET IS AN IOUTPUT
SINGLE CHARACTER SYNC

1 = SINGLE SYNC CHARACTER
0 = DOUBLE SYNC CHARACTER
This format defines a status word that is used to control the actual operation of
8251. All control words written into 8251 after the mode instruction will load the command
instruction.
The command instructions can be written into 8251 at any time in the data block
during the operation of the 8251. to return to the mode instruction format, the master reset
bit in the command instruction word can be set to initiate an internal reset operation which
automatically places the 8251 back into the mode instruction format. Command instructions
must follow the mode instructions or sync characters.
Thus the control word 37 (HEX) enables the transmit enable and receive enable bits,
forces DTR output to zero, resets the error flags, and forces RTS output to zero.
27
EH IR RTS ER SBRK RXE DTR TXEN
TRANSMIT ENABLE
DATA TERMINAL READY

HIGH will force DTR
Output to Zero
RECEIVE ENABLE
SEND BREAK CHARACTER

1 = Forces TXD LOW
0 = Normal Operation
ERROR RESET
1=Reset Error Flags
PE,OE,FE
REQUEST TO SEND
HIGH will force RTS
Output to Zero
INTERNAL RESET
HIGH Returns 8251 to
Mode Instruction Format
ENTER HUNT MODE

1= Enable a Search for
Sync Characters( Has
No Effect in Async mode)
COMMAND INSTRUCTION FORMAT
28
ALGORITHM:
1. Initialise timer (8253) IC

2. Move the mode command word (4E H) to A -reg
4. Move the command instruction word (37 H) to A -reg
6. Move the the data to be transferred to A -reg
8. Reset the system
9. Get the data through input port address C0
10. Store the value in memory
11. Reset the system
PROGRAM:
MVI A,36H
OUT CEH
MVI A,0AH
OUT C8H
MVI A,00
OUT C8H
LXI H,4200
MVI A,4E
OUT C2
MVI A,37
OUT C2
MVI A,41
OUT C0
RST 1
ORG 4200
IN C0
STA 4500
RST 1
OBSERVATION:
Output: 4500 41
RESULT:
Thus the 8251 was initiated and the transmission and reception of character was
done successfully.
29
INTERFACING ADC WITH 8085 PROCESSOR
AIM:
To write a program to initiate ADC and to store the digital data in memory
PROGRAM:
MVI A,10
OUT C8
MVI A,18
OUT C8
MVI A,10
OUT D0
XRA A
XRA A
XRA A
MVI A,00
OUT D0
LOOP: IN D8
ANI 01
CPI 01
JNZ LOOP
IN C0
STA 4150
HLT
OBSERVATION:
Compare the data displayed at the LEDs with that stored at location 4150
RESULT:
Thus the ADC was initiated and the digital data was stored at desired location
30

AIM:
To interface DAC with 8085 to demonstrate the generation of square, saw tooth and
triangular wave.
APPARATUS REQUIRED:

THEORY:
the corresponding output voltages are presented in the Table1.
Input Output
Data in Voltage
HEX
00 - 5.00
01 - 4.96
02 - 4.92
… …
7F 0.00
… …
FD 4.92
FE 4.96
FF 5.00
to DAC, results in different wave forms namely square, triangular, etc,. The port address of
DAC is 08 H.
31
ALGORITHM:



32
PROGRAM:
START: MVI A,00

CALL DELAY
MVI A,FF
CALL DELAY
JMP START
DELAY: MVI B,05

L1: MVI C,FF
L2: DCR C
JNZ L2
DCR B
JNZ L1
RET
START: MVI A,00

L1: OUT Port address of DAC
INR A
JNZ L1
JMP START
33
START: MVI L,00

L1: MOV A,L
INR L
JNZ L1
MVI L,FF
L2: MOV A,L
DCR L
JNZ L2
JMP START
RESULT:
34
INTERFACING 8253 (TIMER IC) WITH 8085 PROCESSOR
AIM:
To interface 8253 Programmable Interval Timer to 8085 and verify the operation
of 8253 in six different modes.
APPARATUS REQUIRED:

5) CRO.
MODE 0-Interrupt On Terminal Count:-
The output will be initially low after mode set operation. After loading the counter, the
output will remain low while counting and on terminal count, the output will become high
until reloaded again.
Let us see the channel in mode0. Connect the CLK 0 to the debounce circuit and
execute the following program.
PROGRAM:
OUT CEH
OUT C8H
OUT C8H
HLT
It is observed in CRO that the output of channel 0 is initially low. After giving ‘x’ clock
pulses, we may notice that the output goes high.
MODE 1-Programmable One Shot:-
After loading the count, the output will remain low following the rising edge of the
gate input. The output will go high on the terminal count. It is retriggerable; hence the
output will remain low for the full count after any rising edge of the gate input.
35
The following program initializes channel 0 of 8253 in Mode 1 and also initializes triggering
of gate. OUT 0 goes low as clock pulses and after triggering It goes back to high level after
five clock pulses. Execute the program and give clock pulses through the debounce logic and
verify using CRO.
PROGRAM:
OUT CEH ;
OUT C8H
OUT C8H
OUT DOH ;Trigger Gate 0.
HLT
MODE 2-Rate Generator:
It is a simple divide by N counter. The output will be low for one period of the input
clock. The period from one output pulse to next equals the number of input count in the
count register. If the count register is reloaded between output pulses, the present period will
not be affected, but the subsequent period will reflect a new value.
MODE 3-Square Generator:
It is similar to mode 2 except that the output will remain high until one half of the
count and goes low for the other half provided the count is an even number. If the count is
odd the output will be high for (count +1)/2 counts. This mode is used for generating baud
rate of 8251.
PROGRAM:

OUT CEH ;
MVI A, 0AH ;LSB of count.
OUT C8H
OUT C8H
HLT
We utilize mode 3 to generate a square wave of frequency 150 kHz at Channel 0.Set the
jumper so that the clock of 8253 is given a square wave of Frequency 1.5 MHz. This
program divides the program clock by 10 and thus the Output at channel 0 is 150 KHz.
36
MODE 4-Software Triggered Strobe:
The output is high after the mode is set and also during counting. On Terminal
count, the output will go low for one clock period and becomes high again. This mode can
be used for interrupt generation.
MODE 5-Hardware Triggered Strobe:
Counter starts counting after rising edge of trigger input and the output goes low for
one clock period. When the terminal count is reached, the counter is retrigerrable. On
terminal count, the output will go low for one clock period and becomes high again. This
mode can be used for interrupt generation.
RESULT:
Thus the 8253 PIT was interfaced to 8085 and the operations for mode 0, Mode 1
and mode 3 was verified.
37
INTERFACING 8279 KEYBOARD/DISPLAY CONTROLLER

WITH 8085 MICROPROCESSOR
AIM:
To interface 8279 Programmable Keyboard Display Controller to 8085

Microprocessor.
APPARATUS REQUIRED:

PROGRAM:
START: LXI H,4130H

MVI D,0FH ;Initialize counter.
MVI A,10H
OUT C2H ;Set Mode and Display.
MVI A,CCH ;Clear display.
OUT C2H
MVI A,90H ;Write Display
OUT C2H
LOOP: MOV A,M
OUT C0H
CALL DELAY
INX H
DCR D
JNZ LOOP
JMP START
DELAY: MVI B, A0H

LOOP2: MVI C, FFH
LOOP1: DCR C
JNZ LOOP1
DCR B
JNZ LOOP2
RET
38
Pointer equal to 4130 .FF repeated eight times.
4130 - FF
4131 –FF
4132 –FF
4133 –FF
4134 –FF
4135 –FF
4136 –FF
4137 –FF
4138 –98
4139 –68
413A -7C
413B -C8
413C -1C
413D -29
413E -FF
413F -FF
RESULT:
Thus 8279 controller was interfaced with 8085 and program for rolling display was
executed successfully.
39
8051 MICROCONTROLLER
PROGRAMS
40
AIM:
To perform addition of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:

3. Add the two data
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
ADD A,#data2
MOV DPTR,#4500
MOVX @DPTR,A
HERE: SJMP HERE
OBSERVATION:
Input: 66
23
Output: 89 (4500)
RESULT:
Thus the program to perform addition of two 8 – bit numbers using 8051 instruction set
was executed.
41
AIM:
To perform Subtraction of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:

3. Subtract the two data
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
SUBB A,#data2
MOV DPTR,#4500
MOVX @DPTR,A
HERE: SJMP HERE
OBSERVATION:
Input: 66
23
Output: 43 (4500)
RESULT:
Thus the program to perform subtraction of two 8 – bit numbers using 8051 instruction
set was executed.
42
AIM:
To perform multiplication of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:

2. Get the value to be multiplied in B – reg.
3. Multiply the two data
4. The higher order of the result is in B – reg.
5. The lower order of the result is in A – reg.
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
MOV B,#data2
MUL AB
MOV DPTR,#4500
MOVX @DPTR,A
INC DPTR
MOV A,B
MOVX @DPTR,A
HERE: SJMP HERE
43
OBSERVATION:
Input: 80
80
Output: 00 (4500)
19 (4501)
RESULT:
44
AIM:
To perform division of two 8 – bit numbers using 8051 instruction set.
ALGORITHM:

2. Get the value to be divided in B – reg.
3. Divide the two data
4. The quotient is in A – reg.
5. The remainder is in B – reg.
PROGRAM:
ORG 4100
CLR C
MOV A,#data1
MOV B,#data2
DIV AB
MOV DPTR,#4500
MOVX @DPTR,A
INC DPTR
MOV A,B
MOVX @DPTR,A
HERE: SJMP HERE
45
OBSERVATION:
Input: 05
03
Output: 01 (4500)
02 (4501)
RESULT:
46
RAM ADDRESSING
AIM:
To exhibit the RAM direct addressing and bit addressing schemes of 8051
microcontroller.
ALGORITHM:
1. For Bit addressing, Select Bank 1 of RAM by setting 3rd bit of PSW
2. Using Register 0 of Bank 1 and accumulator perform addition
3. For direct addressing provide the address directly (30 in this case)
4. Use the address and Accumulator to perform addition
5. 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
47
OBSERVATION:
Bit addressing:
Input: 54
25
Output: 79 (4500)
Direct addressing:
Input: 54
25
Output: 79 (4500)
RESULT:
Thus the program to exhibit the different RAM addressing schemes of 8051 was
executed.
48
INTERFACING STEPPER MOTOR WITH 8051
AIM:
To interface stepper motor with 8051 parallel port and to vary speed of motor, direction of
motor.
APPARATUS REQUIRED:

• Stepper Motor Interface Board
THEORY:
A motor in which the rotor is able to assume only discrete stationary angular position is a
stepper motor. The rotor motion occurs in a stepwise manner from one equilibrium
position to next.
The motor under our consideration uses 2 – phase scheme of operation. In this scheme,
any two adjacent stator windings are energized. The switching condition for the above said
scheme is shown in Table.
Clockwise Anti - Clockwise

A1 B1 A2 B2 A1 B1 A2 B2
1 0 0 1 1 0 1 0
0 1 0 0 0 1 0 1
0 1 0 1 0 1 0 0
1 0 1 0 1 0 0 1
In order to vary the speed of the motor, the values stored in the registers R1, R2, R3 can
be changed appropriately.
49
ALGORITHM:
1. Store the look up table address in DPTR

2. Move the count value (04) to one of the register (R0)
3. Load the control word for motor rotation in accumulator
4. Push the address in DPTR into stack
5. Load FFC0 in to DPTR.
7. Send the control word for motor rotation to the external device.
8. Pop up the values in stack and increment it.
9. Decrement the count in R0. If zero go to next step else proceed to step 3.
10. Perform steps 1 to 9 repeatedly.
PROGRAM:
ORG 4100
START: MOV DPTR,#4500H
MOV R0,#04
AGAIN: MOVX A,@DPTR
PUSH DPH
PUSH PDL
MOV DPTR,#FFC0H
MOV R2, 04H
MOV R1,#FFH
DLY1: MOV R3, #FFH
DLY: DJNZ R3,DLY
DJNZ R1,DLY1
DJNZ R2,DLY1
MOVX @DPTR,A
POP DPL
POP DPH
INC DPTR
DJNZ R0,AGAIN
SJMP START
50
DATA:
4500: 09, 05, 06, 0A
RESULT:
Thus the speed and direction of motor were controlled using 8051 trainer kit.
51
AIM:
To interface DAC with 8051 parallel port to demonstrate the generation of square,
saw tooth and triangular wave.
APPARATUS REQUIRED:

THEORY:
the corresponding output voltages are presented in the Table below
.
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
to DAC, results in different wave forms namely square, triangular, etc,.
52
ALGORITHM:



53
PROGRAM:
ORG 4100
START: MOV A,#00
MOVX @DPTR,A
LCALL DELAY
MOV A,#FF
MOVX @DPTR,A
LCALL DELAY
LJUMP START
DELAY: MOV R1,#05

LOOP: MOV R2,#FF
HERE: DJNZ R2,HERE
DJNZ R1,LOOP
RET
SJMP START
ORG 4100
MOV A,#00
LOOP: MOVX @DPTR,A
INC A
SJMP LOOP
54
ORG 4100
START: MOV A,#00
LOOP1: MOVX @DPTR,A
INC A
JNZ LOOP1
MOV A,#FF
LOOP2: MOVX @DPTR,A
DEC A
JNZ LOOP2
LJMP START
RESULT:
55
PROGRAMMING 8051 USING KEIL SOFTWARE
AIM:
To perform arithmetic operations in 8051 using keil software.
PROCEDURE:
1. Click KeilµVision2 icon in the desktop

2. From Project Menu open New project
3. Select the target device as ATMEL 89C51
4. From File Menu open New File
5. Type the program in Text Editor
6. Save the file with extension “.asm”
7. In project window click the tree showing TARGET
8. A source group will open.
9. Right Click the Source group and click “Add files to Source group”
10. A new window will open. Select our file with extension “.asm”
11. Click Add.
12. Go to project window and right click Source group again
13. Click Build Target (F7).
14. Errors if any will be displayed.
15. From Debug menu, select START/STOP Debug option.
16. In project window the status of all the registers will be displayed.
17. Click Go from Debug Menu.
18. The results stored in registers will be displayed in Project window.
19. Stop the Debug process before closing the application.
PROGRAM:
ORG 4100
CLR C
MOV A,#05H
MOV B,#02H
DIV AB
56
OBSERVATION:
A: 02
B: 01
SP:07
Note that Stack pointer is initiated to 07H
RESULT:
Thus the arithmetic operation for 8051 was done using Keil Software.
57
SYSTEM DESIGN USING MICROCONTROLLER
AIM:
To Design a microcontroller based system for simple applications like security

systems combination lock etc.
PROCEDURE:
1. Read number of bytes in the password

2. Initialize the password
3. Initialize the Keyboard Display IC (8279) to get key and Display
4. Blank the display
5. Read the key from user
6. Compare with the initialized password
7. If it is not equal, Display ‘E’ to indicate Error.
8. Repeat the steps 6 and 7 to read next key
9. If entered password equal to initialized password, Display ‘O’ to indicate open.
PROGRAM:
MOV 51H,#
MOV 52H,#
MOV 53H,#
MOV 54H,#
MOV R1,#51
MOV R0,#50
MOV R3,#04
MOV R2,#08
MOV DPTR,#FFC2
MOV A,#00
58
MOVX @DPTR,A
MOV A,#CC
MOVX @DPTR,A
MOV A,#90
MOVX @DPTR,A
MOV A,#FF
MOV DPTR,#FFCO
LOOP: MOVX @DPTR,A
DJNZ R2,LOOP
AGAIN: MOV DPTR,#FFC2
WAIT: MOVX A,@DPTR
ANL A,#07
JZ WAIT
MOV A,#40
MOVX @DPTR,A
MOV DPTR,#FFCO
MOVX A,@DPTR
MOV @R0,A
MOV A,@R1
CJNE A,50H,NEQ
INC R1
DJNZ R3,AGAIN
MOV DPTR,#FFCO
MOV A,#OC
MOVX @DPTR,A
XX: SJMP XX
NEQ: MOV DPTR,#FFCO

MOV A,#68
MOVX @DPTR,A
YY: SJMP YY
RESULT:
Thus the program for security lock system was executed
59
Microprocessors
About the Tutorial

A microprocessor is a controlling unit of a micro-computer, fabricated on a small chip
capable of performing Arithmetic Logical Unit (ALU) operations and communicating with
the other devices connected to it.
In this tutorial, we will discuss the architecture, pin diagram and other key concepts of
microprocessors.
Audience
This tutorial is designed for all those readers pursing either Bachelor’s or Master’s degree
in Computer Science. It will help them understand the basic concepts related to
Microprocessors.
Prerequisites
In this tutorial, all the topics have been explained from elementary level. Therefore, a
beginner can understand this tutorial very easily. However if you have a prior knowledge
of computer architecture in general, then it will be quite easy to grasp the concepts
explained here.
Disclaimer & Copyright

 Copyright 2016 by Tutorials Point (I) Pvt. Ltd.
All the content and graphics published in this e-book are the property of Tutorials Point (I)
Pvt. Ltd. The user of this e-book is prohibited to reuse, retain, copy, distribute or republish
any contents or a part of contents of this e-book in any manner without written consent
of the publisher.
We strive to update the contents of our website and tutorials as timely and as precisely as
possible, however, the contents may contain inaccuracies or errors. Tutorials Point (I) Pvt.
Ltd. provides no guarantee regarding the accuracy, timeliness or completeness of our
website or its contents including this tutorial. If you discover any errors on our website or
in this tutorial, please notify us at contact@tutorialspoint.com.
i
Microprocessors
Table of Contents
About the Tutorial ............................................................................................................................................ i
Audience ........................................................................................................................................................... i
Prerequisites ..................................................................................................................................................... i
Disclaimer & Copyright ..................................................................................................................................... i
Table of Contents ............................................................................................................................................ ii
MICROPROCESSOR...................................................................................................................... 1
1. Microprocessor − Overview ...................................................................................................................... 2

How does a Microprocessor Work? ................................................................................................................ 2
Features of a Microprocessor .......................................................................................................................... 3
2. Microprocessor − Classification ................................................................................................................. 4

RISC Processor ................................................................................................................................................. 4
CISC Processor ................................................................................................................................................. 6
Special Processors ........................................................................................................................................... 7
8085 MICROPROCESSOR ............................................................................................................. 9
3. 8085 – Architecture................................................................................................................................. 10
8085 Microprocessor – Functional Units ....................................................................................................... 10
8085 Architecture .......................................................................................................................................... 12
4. 8085 − Pin Configuration ......................................................................................................................... 13
5. 8085 − Addressing Modes & Interrupts ................................................................................................... 16

Addressing Modes in 8085 ............................................................................................................................ 16
Interrupts in 8085 .......................................................................................................................................... 16
6. 8085 – Instruction Sets ............................................................................................................................ 19

Control Instructions ....................................................................................................................................... 19
Branching Instructions ................................................................................................................................... 21
Arithmetic Instructions .................................................................................................................................. 24
Data Transfer Instructions ............................................................................................................................. 26
8085 – Demo Programs ................................................................................................................................. 29
8086 MICROPROCESSOR ........................................................................................................... 32
7. 8086 − Overview ..................................................................................................................................... 33

Features of 8086 ............................................................................................................................................ 33
Comparison between 8085 &8086 Microprocessor...................................................................................... 33
Architecture of 8086...................................................................................................................................... 34
8. 8086 – Functional Units........................................................................................................................... 35

EU (Execution Unit) ....................................................................................................................................... 35
BIU (Bus Interface Unit) ................................................................................................................................. 36
ii
Microprocessors
9. 8086 − Pin Configuration ......................................................................................................................... 38
10. 8086 − Instruction Sets ............................................................................................................................ 43

Data Transfer Instructions ............................................................................................................................. 43
Arithmetic Instructions .................................................................................................................................. 44
Bit Manipulation Instructions ........................................................................................................................ 45
String Instructions ......................................................................................................................................... 46
Program Execution Transfer Instructions (Branch & Loop Instructions) ....................................................... 46
Processor Control Instructions ...................................................................................................................... 47
Iteration Control Instructions ........................................................................................................................ 47
Interrupt Instructions .................................................................................................................................... 48
11. 8086 − Interrupts .................................................................................................................................... 49

Hardware Interrupts ...................................................................................................................................... 49
Software Interrupts ....................................................................................................................................... 50
12. 8086 − Addressing Modes ....................................................................................................................... 53
MULTIPROCESSOR CONFIGURATION ........................................................................................ 55
13. Multiprocessor Configuration − Overview............................................................................................... 56

Coprocessor Configuration ............................................................................................................................ 56
Closely Coupled Configuration ...................................................................................................................... 57
Loosely Coupled Configuration ..................................................................................................................... 58
14. 8087 Numeric Data Processor ................................................................................................................. 60

8087 Architecture .......................................................................................................................................... 60
8087 Pin Description ..................................................................................................................................... 61
I/O INTERFACING ...................................................................................................................... 63
15. I/O Interfacing − Overview ...................................................................................................................... 64
16. 8279 − Programmable Keyboard ............................................................................................................. 66

Operational Modes of 8279 .......................................................................................................................... 70
17. 8257 − DMA Controller ........................................................................................................................... 71

How DMA Operations are Performed? ......................................................................................................... 71
Features of 8257 ............................................................................................................................................ 71
8257 Architecture .......................................................................................................................................... 72
MICROCONTROLLERS................................................................................................................ 76
18. Microcontrollers − Overview ................................................................................................................... 77

Difference between Microprocessor and Microcontroller............................................................................ 77
Types of Microcontrollers.............................................................................................................................. 77
Applications of Microcontrollers ................................................................................................................... 78
iii
Microprocessors
19. 8051 ─ Architecture................................................................................................................................. 79
20. 8051 − Pin Description ............................................................................................................................ 80
21. 8051 − Input Output Ports....................................................................................................................... 82

Pins Current Limitations ................................................................................................................................ 83
22. 8051 − Interrupts .................................................................................................................................... 84
PERIPHERAL DEVICES ................................................................................................................ 86
23. Intel 8255A − Programmable Peripheral Interface .................................................................................. 87

Ports of 8255A ............................................................................................................................................... 87
Operating Modes ........................................................................................................................................... 87
Features of 8255A ......................................................................................................................................... 88
8255 ............................................................................................................................................................... 88
Architecture ................................................................................................................................................... 88
24. Intel 8255A − Pin Description .................................................................................................................. 89
25. Intel 8253 ─ Programmable Interval Timer .............................................................................................. 91

Difference between 8253 and 8254 .............................................................................................................. 91
Features of 8253 / 54 .................................................................................................................................... 91
8254 Architecture .......................................................................................................................................... 92
8254 Pin Description ..................................................................................................................................... 92
26. Intel 8253/54 − Operational Modes ........................................................................................................ 95
iv
Microprocessors
Microprocessor
5
1. Microprocessor − Overview Microprocessors
Microprocessor is a controlling unit of a micro-computer, fabricated on a small chip capable

of performing ALU (Arithmetic Logical Unit) operations and communicating with the other
devices connected to it.
Microprocessor consists of an ALU, register array, and a control unit. ALU performs
arithmetical and logical operations on the data received from the memory or an input device.
Register array consists of registers identified by letters like B, C, D, E, H, L and accumulator.
The control unit controls the flow of data and instructions within the computer.
Block Diagram of a Basic Microcomputer
Microprocessor
Input Output
Device (ALU +Register array + Device
Control unit)
Memory
How does a Microprocessor Work?

The microprocessor follows a sequence: Fetch, Decode, and then Execute.
Initially, the instructions are stored in the memory in a sequential order. The microprocessor
fetches those instructions from the memory, then decodes it and executes those instructions
till STOP instruction is reached. Later, it sends the result in binary to the output port. Between
these processes, the register stores the temporarily data and ALU performs the computing
functions.
List of Terms Used in a Microprocessor

Here is a list of some of the frequently used terms in a microprocessor:
6
Microprocessors
 Instruction Set: It is the set of instructions that the microprocessor can understand.
 Bandwidth: It is the number of bits processed in a single instruction.
 Clock Speed: It determines the number of operations per second the processor can
perform. It is expressed in megahertz (MHz) or gigahertz (GHz).It is also known as
Clock Rate.
 Word Length: It depends upon the width of internal data bus, registers, ALU, etc. An
8-bit microprocessor can process 8-bit data at a time. The word length ranges from 4
bits to 64 bits depending upon the type of the microcomputer.
 Data Types: The microprocessor has multiple data type formats like binary, BCD,
ASCII, signed and unsigned numbers.
Features of a Microprocessor
Here is a list of some of the most prominent features of any microprocessor:
 Cost-effective: The microprocessor chips are available at low prices and results its
low cost.
 Size: The microprocessor is of small size chip, hence is portable.
 Low Power Consumption: Microprocessors are manufactured by using metal-oxide

semiconductor technology, which has low power consumption.
 Versatility: The microprocessors are versatile as we can use the same chip in a
number of applications by configuring the software program.
 Reliability: The failure rate of an IC in microprocessors is very low, hence it is reliable.
7
2. Microprocessor − Classification Microprocessors
A microprocessor can be classified into three categories:
RISC Processor
RISC stands for Reduced Instruction Set Computer. It is designed to reduce the execution
time by simplifying the instruction set of the computer. Using RISC processors, each
instruction requires only one clock cycle to execute results in uniform execution time. This
reduces the efficiency as there are more lines of code, hence more RAM is needed to store
the instructions. The compiler also has to work more to convert high-level language
instructions into machine code.
Some of the RISC processors are:
 Power PC: 601, 604, 615, 620

 DEC Alpha: 210642, 211066, 21068, 21164
 MIPS: TS (R10000) RISC Processor
 PA-RISC: HP 7100LC
8
Microprocessors
Architecture of RISC
RISC microprocessor architecture uses highly-optimized set of instructions. It is used in
portable devices like Apple iPod due to its power efficiency.
Hardwired Data Path

Control Unit
Instruction
Data cache
cache
(Instruction)
+ (Data)
Main memory
Characteristics of RISC
The major characteristics of a RISC processor are as follows:
 It consists of simple instructions.

 It supports various data-type formats.
 It utilizes simple addressing modes and fixed length instructions for pipelining.
 It supports register to use in any context.
 One cycle execution time.
 “LOAD” and “STORE” instructions are used to access the memory location.
 It consists of larger number of registers.
 It consists of less number of transistors.
9
Microprocessors
CISC Processor
CISC stands for Complex Instruction Set Computer. It is designed to minimize the number
of instructions per program, ignoring the number of cycles per instruction. The emphasis is
on building complex instructions directly into the hardware.
The compiler has to do very little work to translate a high-level language into assembly level
language/machine code because the length of the code is relatively short, so very little RAM
is required to store the instructions.
Some of the CISC Processors are:
 IBM 370/168
 VAX 11/780
 Intel 80486
Architecture of CISC
Its architecture is designed to decrease the memory cost because more storage is needed in
larger programs resulting in higher memory cost. To resolve this, the number of instructions
per program can be reduced by embedding the number of operations in a single instruction.
Characteristics of CISC
 Variety of addressing modes.
 Larger number of instructions.
10
Microprocessors
 Variable length of instruction formats.

 Several cycles may be required to execute one instruction.
 Instruction-decoding logic is complex.
 One instruction is required to support multiple addressing modes.
Special Processors
These are the processors which are designed for some special purposes. Few of the special
processors are briefly discussed:
Coprocessor
A coprocessor is a specially designed microprocessor, which can handle its particular function
many times faster than the ordinary microprocessor.
For example: Math Coprocessor.
Some Intel math-coprocessors are:
 8087-used with 8086

 80287-used with 80286
 80387-used with 80386
Input/Output Processor
It is a specially designed microprocessor having a local memory of its own, which is used to
control I/O devices with minimum CPU involvement.
For example:
 DMA (direct Memory Access) controller

 Keyboard/mouse controller
 Graphic display controller
 SCSI port controller
Transputer (Transistor Computer)

A transputer is a specially designed microprocessor with its own local memory and having
links to connect one transputer to another transputer for inter-processor communications. It
was first designed in 1980 by Inmos and is targeted to the utilization of VLSI technology.
A transputer can be used as a single processor system or can be connected to external links,
which reduces the construction cost and increases the performance.
11
Microprocessors
For example:16-bit T212, 32-bit T425, the floating point (T800, T805 & T9000) processors.
DSP (Digital Signal Processor)

This processor is specially designed to process the analog signals into a digital form. This is
done by sampling the voltage level at regular time intervals and converting the voltage at
that instant into a digital form. This process is performed by a circuit called an analogue to
digital converter, A to D converter or ADC.
A DSP contains the following components:
 Program Memory: It stores the programs that DSP will use to process data.
 Data Memory: It stores the information to be processed.
 Compute Engine: It performs the mathematical processing, accessing the program

from the program memory and the data from the data memory.
 Input/Output: It connects to the outside world.
Its applications are:
 Sound and music synthesis

 Audio and video compression
 Video signal processing
 2D and 3d graphics acceleration.
For example: Texas Instrument’s TMS 320 series, e.g., TMS 320C40, TMS320C50.
12
Microprocessors
8085 Microprocessor
13
3. 8085 – Architecture Microprocessors
8085 is pronounced as "eighty-eighty-five" microprocessor. It is an 8-bit microprocessor

designed by Intel in 1977 using NMOS technology.
It has the following configuration:
 8-bit data bus

 16-bit address bus, which can address upto 64KB
 A 16-bit program counter
 A 16-bit stack pointer
 Six 8-bit registers arranged in pairs: BC, DE, HL
 Requires +5V supply to operate at 3.2 MHZ single phase clock
It is used in washing machines, microwave ovens, mobile phones, etc.
8085 Microprocessor – Functional Units

8085 consists of the following functional units:
Accumulator
It is an 8-bit register used to perform arithmetic, logical, I/O & LOAD/STORE operations. It
is connected to internal data bus & ALU.
Arithmetic and logic unit

As the name suggests, it performs arithmetic and logical operations like Addition, Subtraction,
AND, OR, etc. on 8-bit data.
General purpose register

There are 6 general purpose registers in 8085 processor, i.e. B, C, D, E, H &
L. Each register can hold 8-bit data.
These registers can work in pair to hold 16-bit data and their pairing combination is like B-C,
D-E & H-L.
Program counter
It is a 16-bit register used to store the memory address location of the next instruction to be
executed. Microprocessor increments the program whenever an instruction is being executed,
so that the program counter points to the memory address of the next instruction that is
going to be executed.
14
Microprocessors
Stack pointer
It is also a 16-bit register works like stack, which is always incremented/decremented by 2
during push & pop operations.
Temporary register
It is an 8-bit register, which holds the temporary data of arithmetic and logical operations.
Flag register
It is an 8-bit register having five 1-bit flip-flops, which holds either 0 or 1 depending upon the
result stored in the accumulator.
These are the set of 5 flip-flops:
 Sign (S)
 Zero (Z)
 Auxiliary Carry (AC)
 Parity (P)
 Carry (C)
Its bit position is shown in the following diagram:
D7 D6 D5 D4 D3 D2 D1 D0
S Z AC P CY
Instruction register and decoder

It is an 8-bit register. When an instruction is fetched from memory then it is stored in the
Instruction register. Instruction decoder decodes the information present in the Instruction
register.
Timing and control unit

It provides timing and control signal to the microprocessor to perform operations. Following
are the timing and control signals, which control external and internal circuits:
 Control Signals: READY, RD’, WR’, ALE

 Status Signals: S0, S1, IO/M’
 DMA Signals: HOLD, HLDA
 RESET Signals: RESET IN, RESET OUT
15
Microprocessors
Interrupt control
As the name suggests it controls the interrupts during a process. When a microprocessor is
executing a main program and whenever an interrupt occurs, the microprocessor shifts the
control from the main program to process the incoming request. After the request is
completed, the control goes back to the main program.
There are 5 interrupt signals in 8085 microprocessor: INTR, RST 7.5, RST 6.5, RST 5.5, TRAP.
Serial Input/output control

It controls the serial data communication by using these two instructions: SID (Serial input
data) and SOD (Serial output data).
Address buffer and address-data buffer

The content stored in the stack pointer and program counter is loaded into the address buffer
and address-data buffer to communicate with the CPU. The memory and I/O chips are
connected to these buses; the CPU can exchange the desired data with the memory and I/O
chips.
Address bus and data bus

Data bus carries the data to be stored. It is bidirectional, whereas address bus carries the
location to where it should be stored and it is unidirectional. It is used to transfer the data &
Address I/O devices.
8085 Architecture
We have tried to depict the architecture of 8085 with this following image:
16
Microprocessors
17
4. 8085 − Pin Configuration Microprocessors
The following image depicts the pin diagram of 8085 Microprocessor:
The pins of a 8085 microprocessor can be classified into seven groups:
18
Microprocessors
Address bus
A15-A8, it carries the most significant 8-bits of memory/IO address.
Data bus
AD7-AD0, it carries the least significant 8-bit address and data bus.
Control and status signals

These signals are used to identify the nature of operation. There are 3 control signal and 3
status signals.
Three control signals are RD, WR & ALE.
 RD: This signal indicates that the selected IO or memory device is to be read and is
ready for accepting data available on the data bus.
 WR: This signal indicates that the data on the data bus is to be written into a selected
memory or IO location.
 ALE: It is a positive going pulse generated when a new operation is started by the
microprocessor. When the pulse goes high, it indicates address. When the pulse goes
down it indicates data.
Three status signals are IO/M, S0 & S1.
IO/M
This signal is used to differentiate between IO and Memory operations, i.e. when it is high
indicates IO operation and when it is low then it indicates memory operation.
S1 & S0
These signals are used to identify the type of current operation.
Power supply
There are 2 power supply signals: VCC & VSS. VCC indicates +5v power supply and VSS
indicates ground signal.
Clock signals
There are 3 clock signals, i.e. X1, X2, CLK OUT.
 X1, X2: A crystal (RC, LC N/W) is connected at these two pins and is used to set
frequency of the internal clock generator. This frequency is internally divided by 2.
19
Microprocessors
 CLK OUT: This signal is used as the system clock for devices connected with the
microprocessor.
Interrupts & externally initiated signals

Interrupts are the signals generated by external devices to request the microprocessor to
perform a task. There are 5 interrupt signals, i.e. TRAP, RST 7.5, RST 6.5, RST 5.5, and INTR.
We will discuss interrupts in detail in interrupts section.
 INTA: It is an interrupt acknowledgment signal.
 RESET IN: This signal is used to reset the microprocessor by setting the program
counter to zero.
 RESET OUT: This signal is used to reset all the connected devices when the
microprocessor is reset.
 READY: This signal indicates that the device is ready to send or receive data. If READY
is low, then the CPU has to wait for READY to go high.
 HOLD: This signal indicates that another master is requesting the use of the address
and data buses.
 HLDA (HOLD Acknowledge): It indicates that the CPU has received the HOLD request
and it will relinquish the bus in the next clock cycle. HLDA is set to low after the HOLD
signal is removed.
Serial I/O signals

There are 2 serial signals, i.e. SID and SOD and these signals are used for serial
communication.
 SOD (Serial output data line): The output SOD is set/reset as specified by the SIM
instruction.
 SID (Serial input data line): The data on this line is loaded into accumulator whenever
a RIM instruction is executed.
20
Microprocessors
End of ebook preview

If you liked what you saw…
Buy it from our store @ https://store.tutorialspoint.com
21
EE0310-Microprocessor & Microcontroller Lab
LABORATORY MANUAL
EE0310 – MICROPROCESSOR &

MICROCONTROLLER LAB
DEPARTMENT OF ELECTRICAL & ELECTRONICS ENGINEERING

FACULTY OF ENGINEERING & TECHNOLOGY
SRM UNIVERSITY, Kattankulathur – 603 203
1
LIST OF EXEPRIMENTS
Sl.No. Name of the Experiments Page No.

1 Induction to 8085 Microprocessor
2 a) Addition of 2 - 8 bit numbers
b) Subtraction of 2 - 8 bit numbers
3 a) Addition of 2 - 16 bit numbers
b) Subtraction of 2 – 16 bit numbers
4 a) Multiplication of 2 - 8 numbers
b) Division of 2 - 8 bit numbers
5 a) Ascending order
b) Descending order
6 a) Fibonaci Series
b) Sum of finite series
7 Factorial of Given Numbers
8 a) Multiplication of 2 - 16 bit numbers
b) Division of 2 - 16 bit numbers
9 a) Binary to BCD code conversions
b) BCD to Binary code conversions
10 a) Rolling Display
b) Flashing Display
11 Stepper motor rotate forward and reverse direction
12 Digital analog conversion
13 Analog digital conversion
14 Microcontroller
a) Addition
b) Subtraction
c) Multiplication
d) Division
2
INTRODUCTION TO MICROPROCESSOR 8085
Aim
To study the microprocessor 8085
Architecture of 8085 Microprocessor
a) General purpose register

It is an 8 bit register i.e. B,C,D,E,H,L. The combination of 8 bit register is known as
register pair, which can hold 16 bit data. The HL pair is used to act as memory pointer is
accessible to program.
b) Accumulator
It is an 8 bit register which hold one of the data to be processed by ALU and stored
the result of the operation.
c) Program counter (PC)

It is a 16 bit pointer which maintain the address of a byte entered to line stack.
d) Stack pointer (Sp)

It is a 16 bit special purpose register which is used to hold line memory address for
line next instruction to be executed.
e) Arithmetic and logical unit

It carries out arithmetic and logical operation by 8 bit address it uses the accumulator
content as input the ALU result is stored back into accumulator.
f) Temporary register
It is an 8 bit register associated with ALU hold data, entering an operation, used by
the microprocessor and not accessible to programs.
g) Flags
Flag register is a group of fire, individual flip flops line content of line flag register
will change after execution of arithmetic and logic operation. The line states flags are
i) Carry flag (C)
ii) Parity flag (P)
iii) Zero flag (Z)
iv) Auxiliary carry flag (AC)
v) Sign flag (S)
h) Timing and control unit

Synchronous all microprocessor, operation with the clock and generator and control
signal from it necessary to communicate between controller and peripherals.
i) Instruction register and decoder

Instruction is fetched from line memory and stored in line instruction register decoder
the stored information.
j) Register Array
These are used to store 8 bit data during execution of some instruction.
3
PIN Description
Address Bus
1. The pins Ao – A15 denote the address bus.
2. They are used for most significant bit
Address / Data Bus

1. AD0 – AD7 constitutes the address / Data bus
2. These pins are used for least significant bit
ALE : (Address Latch Enable)

1. The signal goes high during the first clock cycle and enables the lower order
address bits.
IO / M
1. This distinguishes whether the address is for memory or input.
2. When this pins go high, the address is for an I/O device.
S0 – S1
S0 and S1 are status signal which provides different status and functions.
RD
1. This is an active low signal
2. This signal is used to control READ operation of the microprocessor.
WR
1. WR is also an active low signal
2. Controls the write operation of the microprocessor.
HOLD
1. This indicates if any other device is requesting the use of address and data bus.
HLDA
1. HLDA is the acknowledgement signal for HOLD
2. It indicates whether the hold signal is received or not.
INTR
1. INTE is an interrupt request signal
2. IT can be enabled or disabled by using software
INTA
1. Whenever the microprocessor receives interrupt signal
2. It has to be acknowledged.
RST 5.5, 6.5, 7.5

1. These are nothing but the restart interrupts
2. They insert an internal restart junction automatically.
4
TRAP
1. Trap is the only non-maskable interrupt
2. It cannot be enabled (or) disabled using program.
RESET IN
1. This pin resets the program counter to 0 to 1 and results interrupt enable and
HLDA flip flops.
X1, X2
These are the terminals which are connected to external oscillator to produce the
necessary and suitable clock operation.
SID
This pin provides serial input data
SOD
This pin provides serial output data
VCC and VSS

1. VCC is +5V supply pin
2. VSS is ground pin
Specifications
1. Processors
Intel 8085 at E144 MHz clock
2. Memory
Monitor RAM: 0000 – IFFF
EPROM Expansion: 2000 – 3FFF’s
0000 – FFF
System RAM: 4000 – 5FFF
Monitor data area 4100 – 5FFF
RAM Expansion 6000 – BFFF
3. Input / Output
Parallel: A8 TTL input timer with 2 number of 32-55 only input timer available in -85 EBI.
Serial: Only one number RS 232-C, Compatible, crucial interface using 8281A
Timer: 3 channel -16 bit programmable units, using 8253 channel ‘0’ used for no band late.
Clock generator. Channel ‘1’ is used for single stopping used program.
Display: 6 digit – 7 segment LED display with filter 4 digit for adder display and 2 digit for
data display.
Key board: 21 keys, soft keyboard including common keys and hexa decimal keys.
RES: Reset keys allow to terminate any present activity and retain to  - 85 its on initialize
state.
5
INT: Maskable interrupt connect to CPU’s RST 7.5 interrupt
DEC: Decrement the adder by 1
EXEC: Execute line particular value after selecting address through go command.
NEXT: Increment the address by 1 and then display its content.
Key Functions:
E i. Hex entry key ‘0’

0 ii. Substituting memory content where “next” key is paused immediately
SUB after 1, take used to st cutting address.
iii. Register key ‘E’
RD i) Hex code entry (1)

1 ii) Register key ‘D’
REG
i) Hex code entry ‘2’

C ii) Retricre data from data ‘memory’ to data top
2 iii) Register key ‘C’
TN
B i) Hex code entry ‘3’

3 ii) Retricre data from memory to top
TR iii) Register key ‘B’
F i) Hex key entry ‘C’

4 ii) Block search from byte
BLOC iii) Register key ‘F’
A i) Hex key entry ‘5’

5 ii) Fill block of RAM memory with desired data
FILL iii) Register key ‘A’
L i) Hex key entry ‘6’

6 ii) TN/Tl used for sending (or) receiving
SER iii) Register key ‘H’
H
7 i) Hex key entry ‘7’
F2 ii) Register key ‘H’
6
I i) Register key ‘S’

8 ii) Register key ‘I’
G0
PL i) Hex key entry ‘A’

9 ii) Function key F3
SNG iii) Register key “ph”
A i) Hex key entry “y”

PH ii) Signal step program (instruction by instruction)
F3
C i) Hex key entry “c”

SH ii) Much a block of memory from a linear block
MOV iii) Register key “SH”
D i) Hex key D
ii) Compare 2 memory block
CMP
B i) Hex key entry ‘B’

SL ii) Check a block from flame
BC iii) Register key “SPL”
E
i) Hex key ‘E’
ii) Insert by test into memory (RAM)
INS
F i) Hex key ‘F’

ii) Delete byte from memory RAM
DEL
System Power Consumption

Micro BSEB2 MICRO SSEB
+5V @ 1Amp +5V@ 800 mA
+12V @ 200 mA
- 12V @ 100 mA
7
Power Supply Specification

MICRO SSEM
230V, AC @ 80 Hz
+5V @ 600 mA
Key Function
8
IC’s Used
8085 - 8 bit p
8253 - programmable internal timer
8255 - programmable peripheral interface
8279 - programmable key boards / display interface
8251 - programmable communication interface
2764 - 8 KV VV EPROM
6264 - 8K STATIC PROM
7414 - Hex inverter
7432 - Quad 21/p OR GATE
7409 - Quad 21/p AND GATE
7400 - NAND Gate
7404 - Dual D-FF
74373 - Octal ‘D’ Latch
74139 - Dual 2 to 4 line decoder
74138 - 3 to 8 line decoder
9
In Enter Program into Trainer Kit

1. Press ‘RESET’ key
2. Sub (key processor represent address field)
3. Enter the address (16 bit) and digit in hex
4. Press ‘NEXT’ key
5. Enter the data
6. Again press “NEXT”
7. Again after taking the program, are use HLT instruction
its Hex code
8. Press “NEXT”
How to executive program

1. Press “RESET”
2. Press “GO”
3. Enter the address location in which line program was executed
4. Press “Execute” key
Result:
Thus 8085 microprocessor was studied successfully.
10
ADDITION OF TWO 8-BIT NUMBERS
Aim:
To write an assembly language for adding two 8 bit numbers by using micro
processor kit.
Apparatus required:
8085 micro processor kit
(0-5V) DC battery
Algorithm:
Step 1 : Start the microprocessor
Step 2 : Intialize the carry as ‘Zero’
Step 3 : Load the first 8 bit data into the accumulator
Step 4 : Copy the contents of accumulator into the register ‘B’
Step 5 : Load the second 8 bit data into the accumulator.
Step 6 : Add the 2 - 8 bit datas and check for carry.
Step 7 : Jump on if no carry
Step 8 : Increment carry if there is
Step 9 : Store the added request in accumulator
Step 10 : More the carry value to accumulator
Step 11 : Store the carry value in accumulator
Step 12 : Stop the program execution.
11
START
Intialise the carry as zero
Load the 1st 8 bit number
Transfer the 1st number to register ‘B’
Load the 2nd 8 bit number
Transfer and Add the contents of A and B
No
Check for
carry?
Yes
Increment carry by one
Store the added value in accumulator
Move the contents of carry into accumulator
Store the value of carry in accumulator
END
12
Address Label Mnemonics Hex Code Comments

4100 MVI C,00 OE, 00 Initialize the carry as zero
4102 LDA 4300 3A, (00, 43) Load the first 8 bit data
4105 MOV, B,A 47 Copy the value of 8 bit data
into register B
4106 LDA 4301 3A, (01, 43) Load the second 8 bit data
into the accumulator
4109 ADD B 80 Add the hoo values
410A JNC D2, 0E, 41 Jump on if no carry
410D INR C OC If carry is there increment it
by one
410E Loop STA 4302 32 (02, 43) Stone the added value in the
accumulator
4111 MOV A,C 79 More the value of carry to
the accumulator from
register C
4112 STA 4303 32 (03, 43) Store the value of carry in
the accumulator
4115 HLT 76 Stop the program execution
Input
Without carry
Input Address Value
4300 04
4301 02
Output
Output Address Value
4302 06
4303 00 (carry)
With carry
Input Address Value
4300 FF
4301 FF

4302 FE
4303 01 (carry)
Calculation 1111 1111

1111 1111
---------------
(1) 1111 1110
=========
F E
Result:
The assembly language program for 8 bit addition of two numbers was executed
successfully by using 8085 micro processing kit.
13
Aim:
To write a assembly language program for subtracting 2 bit (8) numbers by using-
8085 micro processor kit.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Intialize the carry as ‘Zero’
Step 3 : Load the first 8 bit data into the accumulator
Step 4 : Copy the contents of contents into the register ‘B’
Step 5 : Load the second 8 bit data into the accumulator.
Step 6 : Subtract the 2 8 bit datas and check for borrow.
Step 7 : Jump on if no borrow
Step 8 : Increment borrow if there is
Step 9 : 2’s compliment of accumulator is found out
Step 10 : Store the result in the accumulator
Step 11 : More the borrow value from ‘c’ to accumulator
Step 12 : Store the borrow value in the accumulator
Step 13 : Stop program execution
14
START
Intialise the borrow as zero
Load the 1st 8 bit number
Move the 1st 8 bit data to register ‘B’
Subtract the two values
No
Check for
carry?
Yes
Increment carry by one
1’s compliment of 2nd value
Add 1 to 1’s compliment for 2’s compliment
Store the value of result in accumulator
Move the carry into the accumulator
Store the value of carry in accumulator
END
15

4100 MVI C,00 OE, 00 Initialize the carry as zero
4102 LDA 4300 3A, (00, 43) Load the first 8 bit data into the
accumulator
4105 MOV, B,A 47 Copy the value into register ‘B’
4106 LDA 4301 3A, (01, 43) Load the 2nd 8 bit data into the
accumulator
4109 SUB B 90 Subtract both the values
410A Loop INC D2, 0E, 41 Jump on if no borrow
410D INR C OC If borrow is there, increment it by
one
410E Loop CMA 2F Compliment of 2nd data
410F ADI, 01 6, 01 Add one to 1’s compliment of 2nd
data
4111 STA 4302 32,02,43 Store the result in accumulator
4114 MOV A,C 79 Moul the value of borrow into the
accumulator
4115 STA 4303 32,03,43 Store the result in accumulator
4118 HLT 76 Stop Program execution
Input
Without borrow
Input Address Value
4300 05
4301 07
Output
4302 02
4303 00 (borrow)
With carry borrow
Input Address Value
4300 07
4301 05

4302 02
4303 01 (borrow)
Calculation 05 – 07
07 – 0111
CMA 1000
ADJ 0.1 0001
------
1001
05 - 0101
------
1110 (-2)
16
Result:
The assembly language program subtraction of two 8 bit numbers was executed
successfully by using 8085 micro processing kit.
17
Aim:
To write an assembly language program for adding two 16 bit numbers using 8085
micro processor kit.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Get the 1st 8 bit in ‘C’ register (LSB) and 2nd 8 bit in ‘H’
register (MSB) of 16 bit number.
Step 3 : Save the 1st 16 bit in ‘DE’ register pair
Step 4 : Similarly get the 2nd 16 bit number and store it in ‘HL’ register
pair.
Step 5 : Get the lower byte of 1st number into ‘L’ register
Step 6 : Add it with lower byte of 2nd number
Step 7 : tore the result in ‘L’ register
Step 8 : Get the higher byte of 1st number into accumulator
Step 9 : Add it with higher byte of 2nd number and carry of the lower bit
addition.
Step 10 : Store the result in ‘H’ register
Step 11 : Store 16 bit addition value in ‘HL’ register pair
Step 12 : Stop program execution
18
START
C = 00H
Load ‘HL’ with 1st Data
Transfer HL - DE
Load ‘HL’ with 2nd Data
DE + HL = HL
If
Cy =0
C = C + 01
Store ‘HL’ in memory (SUM)
Transfer C - A
Store ‘A’ in memory (Cy)
STOP
19

4500 MVI C,00 0E C = 00H
4501 00
4502 LHLD 4800 2A HL – 1st No.
4503 00
4504 48
4505 XCHG EB HL – DE
4506 LHLD 4802 2A HL – 2nd No.
4507 02
4508 48
4509 DAD D 19 Double addition DE +
HL
450A JNC Ahead D2 If Cy = 0, G0 to 450E
450E
450B 0E
450C 45
450D INR C 0C C = C + 01
450E AHEAD SHLD 4804 22 HL – 4804 (sum)
450F 04
4510 48
4511 MOV C,A 79 Cy – A
4512 STA 4806 32 Cy – 4806
4513 06
4514 48
4515 HLT 76 Stop excution
Input
Without
Input Address Value
4800 01 (addend)
4801 04
4802 02 (augend)
4803 03 (augend)
Output
4804 03 (sum)
4805 07 (sum)
4806 00 (carry)
Calculation 0000 0100 0000 0001

0000 0011 0000 0010
---------------------------------
0000 0111 0000 0011
0 7 0 3
20
With carry
Input Address Value
4800 FF (addend)
4801 DE (addend)
4802 96 (augend)
4803 DF (augend)

4804 95 (sum)
4805 BE (sum)
4806 01 (carry)
Calculation 1101 1110 1111 1111

1101 1111 1001 0101
---------------------------------
1011 1110 1001 0101
B E 9 5
Result:
The assembly language program for addition of two 16 bit numbers was executed
using 8085 micro processing kit.
21
Aim:
To write an assembly language program for subtracting two 16 bit numbers using
8085 microprocessor kit.
Apparatus required:
8085 microprocessor kit
(0-5V) DC battery
Algorithm:
Step 2 : Get the 1st 16 bit in ‘HL’ register pair
Step 3 : Save the 1st 16 bit in ‘DE’ register pair
Step 4 : Get the 2nd 16 bit number in ‘HL’ register pair
Step 5 : Get the lower byte of 1st number
Step 6 : Get the subtracted value of 2nd number of lower byte by
subtracting it with lower byte of 1st number
Step 7 : Store the result in ‘L’ register
Step 8 : Get the higher byte of 2nd number
Step 9 : Subtract the higher byte of 1st number from 2nd number with
borrow
Step 10 : Store the result in ‘HL’ register
Step 11 : Stop the program execution
22
START
C = 00H
Load ‘HL’ with 1st Data
Transfer HL - DE
Load ‘HL’ with 2nd Data
Transfer E – A (LSB)
A = A – L (LSB)
Store ‘A’ in memory (LSB)
Transfer D – A (MSB)
A – A – H – Borrow (MSB)
Store ‘A’ in memory (MSB)
STOP
23

4500 MVI C,00 0E C = 00H
4501 00
4502 LHLD 4800 2A L – 1st No.
4503 00
4504 48
4505 XLHG EB HL – DE
4506 LHLD 4802 2A HL – 2nd No.
4507 02
4508 48
4509 MOV A,E 7B LSB of ‘1’ to ‘A’
450A SUB L 95 A–A–L
450B STA 4804 32 A – memory
450C 04
450D 48
450E MOV A,D 7A MSB of 1 to A
450F SBB H 9C A- A – H
4510 STA 4805 32 A – memory
4511 05
4512 48
4513 HLT 76 Stop execution
Input
Without borrow
Input Address Value
4800 07
4801 08
4802 05
4803 06
Output
4804 02
4805 02
4807 00
24
With borrow
Input Address Value
4800 05
4801 06
4802 07
4803 08

4804 02
4805 02
4806 01
Calculation
05 06 - 07 08
05 06 0101 0110 07 08 0111 1000

CMA 1010 1001 CMA 1000 0111
ADI 0000 0001 ACI 0000 0001
--------------- --------------
1010 1010 1000 1000
05 06 + 07 08
1010 1010
1000 1000
---------------
(1) 0010 0010
02 02
Result:
The assembly language program for subtraction of two 16 bit numbers was executed
by using 8085 micro processing kit.
25
Aim:
To write an assembly language for multiplying two 8 bit numbers by using 8085
micro processor kit.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Get the 1st 8 bit numbers
Step 3 : Move the 1st 8it number to register ‘B’
Step 4 : Get the 2nd 8 bit number
Step 5 : Move the 2nd 8 bit number to register ‘C’
Step 6 : Intialise the accumulator as zero
Step 7 : Intialise the carry as zero
Step 8 : Add both register ‘B’ value as accumulator
Step 9 : Jump on if no carry
Step 10 : Increment carry by 1 if there is
Step 11 : Decrement the 2nd value and repeat from step 8, till the 2nd
value becomes zero.
Step 12 : Store the multiplied value in accumulator
Step 13 : Move the carry value to accumulator
Step 14 : Store the carry value in accumulator
26
START
Get the 1st 8 bit number
Move it to register ‘B’
Get the 2nd 8 bit number
Intialize the accumulator as zero & carry as zero
Add the accumulator with 1st number
No
Check for
carry?
Yes
Increment carry
Decrement 2nd number
No 2nd Number
Yes
Store the value f carry in accumulator
END
27

4100 LDA 4500 3A, 00, 45 Load the first 8 bit number
4103 MOV B,A 47 Move the 1st 8 bit data to
register ‘B’
4104 LDA 4501 3A, 01, 45 Load the 2nd 16 it number
4107 MOV C,A 4F Move the 2nd 8 bit data to
register ‘C’
4108 MVI A, 00 3E, 00 Intialise the accumulator as
zero
410A MVI D, 00 16, 00 Intialise the carry as zero
410C ADD B 80 Add the contents of ‘B’ and
accumulator
410D INC D2 11, 41 Jump if no carry
4110 INR D 14 Increment carry if there is
4111 DCR C OD Decrement the value ‘C’
4112 JNZ C2 0C, 41 Jump if number zero
4115 STA 4502 32 02, 45 Store the result in
accumulator
4118 MOV A,D 7A Move the carry into
accumulator
4119 STA 4503 32,03,45 Store the result in
accumulator
411C HLT 76 Stop the program execution
Input
Input Address Value
4500 04
4501 02
Output
4502 08
4503 00
Result:
The assembly language program for multiplication of two 8 bit numbers was executed
using 8085 micro processing kit.
28
Aim:
To write an assembly language program for dividing two 8 bit numbers using
microprocessor kit.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Intialise the Quotient as zero
Step 3 : Load the 1st 8 bit data
Step 4 : Copy the contents of accumulator into register ‘B’
Step 5 : Load the 2nd 8 bit data
Step 6 : Compare both the values
Step 7 : Jump if divisor is greater than dividend
Step 8 : Subtract the dividend value by divisor value
Step 9 : Increment Quotient
Step 10 : Jump to step 7, till the dividend becomes zero
Step 11 : Store the result (Quotient) value in accumulator
Step 12 : Move the remainder value to accumulator
Step 13 : Store the result in accumulator
29
START
Get the divided
Intialise the Quotient as zero
Get the divisor
Compare the dividend & divisor
Add the accumulator with 1st number
No
Check for
carry?
Increment carry
Decrement 2nd number
Dividend
Yes
Store the Quotient in accumulator
Move the remainder to accumulator
Store the remainder in accumulator
END
30

4100 MVI C, 00 0E, 00 Intialise Quotient as zero
4102 LDA, 4500 3A 00, 45 Get the 1st data
4105 MOV B,A 47 Copy the 1st data into
register ‘B’
4106 LDA, 4501 3A 01, 45 Get the 2nd data
4109 CMP B B8 Compare the 2 values
410A JC (LDP) DA 12,41 Jump if dividend lesser than
divisor
410D Loop 2 SUB B 90 Subtract the 1st value by 2nd
value
410E INR C 0C Increment Quotient (410D)
410F JMP (LDP, 41) C3, 0D, 41 Jump to Loop 1 till the value
of dividend becomes zero
4112 Loop 1 STA 4502 32 02,45 Store the value in
accumulator
4115 MOV A,C 79 Move the value of remainder
to accumulator
4116 STA 4503 32 03,45 Store the remainder value in
accumulator
4119 HLT 76 Stop the program execution
Input
Input Address Value
4500 09
4501 02
Output
4502 04 (quotient)
4503 01 (reminder)
1001
0010 – I
------
0111
0010 – II
------
0101
0010 – III
------
0011
0010 – IV
------
0001 – carry
Quotient - 04
Carry - 01
31
Result:
The assembly language program for division of two 8 bit numbers was executed using
8085 micro processing kit.
32
ASCENDING ORDER
Aim:
To write a program to sort given ‘n’ numbers in ascending order
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Accumulator is loaded with number of values to sorted and it is
saved
Step 3 : Decrement 8 register (N-1) Repetitions)
Step 4 : Set ‘HL’ register pair as data array
Step 5 : Set ‘C’ register as counter for (N-1) repetitions
Step 6 : Load a data of the array in accumulator
Step 7 : Compare the data pointed in ‘HL’ pair
Step 8 : If the value of accumulator is smaller than memory, then jump
to step 10.
Step 9 : Otherwise exchange the contents of ‘HL’ pair and accumulator
Step 10 : Decrement ‘C’ register, if the of ‘C’ is not zero go to step 6
Step 11 : Decrement ‘B’ register, if value of ‘B’ is not zero, go step 3
33
START
Load the count value in accumulator and save it in register ‘B’
Decrement ‘B’ register (N-1)
Load the starting address of array in ‘HL’
Decrement ‘C’ register of array in ‘HL’
Increment the data pointer
Compare the data pointed by ‘HL’ with ‘A’
Compare No
the datas
A<M Exchange the contents of
Yes memory pointer by ‘HL’
C=0? Increment carry by one

No
Yes
Decrement register ‘B’
No
B=0?
END
34

4100 LDA 4500 3A, 00,45 Load the number of values
4103 MOV B,A 47 Move it ‘B’ register
4104 DCR B 05 For (N-1) comparisons
4105 Loop 3 LXI H, 4500 21, 00,45 Set the pointer for array
4108 MOV C,M 4E Count for (N-1) comparisons
4109 DCR C 0D For (N-1) comparisons
410A INX H 23 Increment pointer
410B Loop 2 MOV A,M 7E Get one data in array ‘A’
410C INX H 23 Increment pointer
410D CMP M BE Compare next with
accumulator
410E JC DA, 16, 41 If content less memory go
ahead
4111 MOV D,M 56 If it is greater than
interchange it
4112 MOV M,A 77 Memory content
4113 DCX H 2B Exchange the content of
memory pointed by ‘HL’ by
previous location
4114 MOV M,D 72 One in by ‘HL’ and previous
location
4115 INX H 23 Increment pointer
4116 Loop 1 DCR C 0D Decrement ‘C’ register
4117 JNZ Loop 1 C2, 0B, 41 Repeat until ‘C’ is zero
411A DCR B 05 Decrement in ‘B’ values
411B JNZ Loop 2 C2, 05, 41 Repeat till ‘B’ is zero
411E HLT 76 Stop the program execution
Input
Input Address Value
4500 04
4501 AB
4502 BC
4503 01
4504 0A
Output Address & Value

4500 04
4501 01
4502 0A
4503 AB
4504 BC
Result:
The assembly language program for sorting numbers in ascending order was executed
by microprocessor kit.
35
DESCENDING ORDER
Aim:
To write a program to sort given ‘n’ numbers in descending order
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Load the number of values into accumulator and save the
number of values in register ‘B’
Step 3 : Decrement register ‘B’ for (N-1) Repetitions
Step 4 : Set ‘HL’ register pair as data array address pointer and load the
data of array in accumulator
Step 5 : Set ‘C’ register as counter for (N-1) repetitions
Step 6 : Increment ‘HL’ pair (data address pointer)
Step 7 : Compare the data pointed by ‘HL’ with accumulator
Step 8 : If the value of accumulator is larger than memory, then jump
to step 10, otherwise next step.
Step 9 : Exchange the contents of memory pointed by ‘HL’ and
accumulator
Step 10 : Decrement ‘C’ register, if the of ‘C’ is not zero go to step 6,
otherwise next step.
Step 11 : Decrement ‘B’ register, if ‘B’ is not zero, go step 3, otherwise
next step.
36
START
Load the count value to accumulator
Save the value in ‘B’ register
Decrement ‘B’ register (N-1)
Load the starting address of data array in ‘HL’ pair
Using data pointer, load the count value in ‘C’
Decrement ‘C’ register (counter for N-1)
Increment the data pointer ‘HL’ with ‘A’
Compare the data pointed at ‘HL’ with ‘A’
A<M
No
Exchange the contents
C=0? Decrement register C
Yes
Decrement register ‘C’

Decrement
No
B=0?
END
No Yes
37

4100 LDA 4500 3A, 00,45 Load the number of values
in accumulator
4103 MOV B,A 47 Move it to ‘B’ register
4104 DCR B 05 For (N-1) comparisons
4105 Loop 3 LXI H, 4500 21, 00,45 Set the pointer for array
4108 MOV C,M 4E Count for (N-1) comparisons
4109 DCR C 0D For (N-1) comparisons
410A INX H 23 Increment pointer
410B Loop 2 MOV A,M 7E Get one data from array
410C INX H 23 Increment pointer
410D CMP M BE Compare next with number
410E ICE, Loop 1 D2, 16,41 If content ‘A’ is greater than
content of ‘HL’ pair
4111 MOV D,M 56 If it is greater than
interchange the datas
4112 MOV M,A 77 Accumulator to memory
value
4113 DCX H 2B Decrement memory pointer
4114 MOV M,D 72 Move the old to ‘HL’ and
previous location
4116 Loop 1 DCR C 0D Decrement ‘C’ register
4117 JNZ Loop 2 C2, 0B, 41 Repeat till ‘C’ is zero
411A DCR B 05 Decrement in ‘B’ values
411B JNZ Loop 3 C2, 05, 41 Jump to loop till the value of
‘B’ be
411E HLT 76 Stop the program execution
Input
Input Address Value
4500 04
4501 AB
4502 BC
4503 01
4504 0A
Output Address & Value

4500 04
4501 BC
4502 AB
4503 0A
4504 01
Result:
The assembly language program for sorting ‘4’ numbers in descending order was
executed successfully using microprocessor kit.
38
SUM OF DATAS
Aim:
To write an assembly language program to calculate the sum of datas using 8085
microprocessor kit.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Load the number of values in series in accumulator and move it
to register C and load the starting address of array
Step 3 : Intialize the value of A as ‘00’
Step 4 : Move the value of ‘A’ to ‘B’ register
Step 5 : Add the content of accumulator with the data pointed by ‘HL’
pair
Step 6 : If there exists a carry, increment ‘B’ by 1, if not continue
Step 7 : Increment the pointer to next data
Step 8 : Decrement the value of ‘C’ by 1, which is used as counter
Step 9 : If ‘C’ is equal to zero, go to step 10 if not go to step 5.
Step 10 : Store the value of ‘A’ to memory, it shows the result
Step 11 : Move the content of B to A
Step 12 : Store the value of A to memory
Step 13 : Stop the program
39
START
Load the number of values in ‘A’
Move it to register ‘C’
Load the starting value of array
Move the value from A to B
Add the accumulator date with memory
No
Check for
carry?
Yes
Increment ‘B’ register
Decrement ‘C’ register by 1
Is
C=0?
Store the result in accumulator
Move the value from B to A
Store the value in A
END
40

4100 LDA 4200 3A 00,42 Load the accumulator with
number of values
4103 MOV B,A 4F Move it from A to C
4104 LXI H, 4201 21,01,42 Load the starting address of
data array
4107 SUB A 97 Intialise ‘A’ as 00
4108 MOV B,A 47 Intialise ‘B’ as 00
4109 Loop ADD M 86 Add the previous sum with
next data
410A JNC Skip D2, 0E, 41 Jump on if no carry
410D INR B 04 Increment carry by one
410E Skip INX H 23 Increment pointer for next
data
410F DCR C 0D Decrement ‘C’ by one
4110 JNZ Loop C2, 09, 41 Jump if not zero
4113 STA 4400 32,00,44 Store the sum in
accumulator
4116 MOV A,B 78 Move the value of carry to A
from B
4117 STA 4401 32,01,44 Store the carry in memory
411A HLT 76 End of program
Input
Input Address Value
4200 04
4201 07
4202 09
4203 03
4204 04
Output

4400 17
4401 00
07 + 09 + 03 + 04 = 23
= 17 (in Hexa decimal)

(0F + 8 = 233)
0F = 0000 1111
08 = 0000 1000
---------------
0001 0111
1 7
41
Result:
The assembly language program for sum of datas was executed successfully using
42
FACTORIAL OF 8 BIT NUMBER
Aim:
To write an program to calculate the factorial of a number (between 0 to 8)
Apparatus required:
(0-5V) power supply
Algorithm:
Step 1 : Intialize the stack pointer
Step 2 : Get the number in accumulator
Step 3 : Check for if the number is greater than 1. If no store the result
otherwise go to next step.
Step 4 : Load the counter and initialize result
Step 5 : Now factorial program in sub-routine is called.
Step 6 : In factorial,
initialize HL RP with 0.
Move the count value to B
Add HL content with Rp.
Decrement count (for multiplication)
Step 7 : Exchange content of Rp (DE) with HL.
Step 8 : Decrement counter (for factorial) till zero flag is set.
Step 9 : Store the result
Step 10 : Hault
Memory address Content

4250 05
4251 (12010)
43
START
Intialize stack pointer
Get the number
If
Number < 2 ?
Result = 1
Load counter
Initialize result
CALL facto
END
44
Facto
Result = Result X no
No = No -1
No
If
No = 0 ?
Yes
RET
45
Memory Hex Code Label Mnemonics Comments

Location Op code Operand
4200 3A LDA 4250 Get the number in
4201 50 accumulator
4202 42
4203 FE CPI 02H Compare data with 2
4204 02 and check it is greater
than 1
4205 DA JC Loop 1 If cy =1 jump to loop 1
4206 17 If cy = 0 proceed
4207 42
4208 5F MOV E,A Move content of A to E
4209 16 MVI D,00 Load this term as a
420A 00 result
420B 3D DCR A Decrement
accumulator by 1
420C 4F MOV C,A Move ‘A’ content to
‘C’ (counter 1 less than
A)
420D CD CALL Facto Call sub routine
420E 00 programe Facto
420F 46
4210 EB XCHG Exchange (DE) – (HL)
4211 22 SHLD 4251 Store content of HL in
4212 51 specified memory
4213 42 location
4214 C3 JMP Loop 3 Jump to Loop 3
4215 1D
4216 42
4217 21 Loop 1 LXI H,0001H HL is loaded with data
4218 00 01
4219 01
421A 22 SHLD 4251 Store the result in
421B 51 memory
421C 42
421D 76 Loop 3 HLT Terminate the program
Sub Routine
4600 21 Facto LXI H,0000 Initialize HL pair
4601 00
4602 00
4603 41 MOV B,C Content of ‘C’ is
moved to B
4604 19 Loop 2 DAD D Content of DE is added
with HL
4605 05 DCR B ‘B’ is decremented
4606 C2 JNZ Loop 2 Multiply by successive
4607 04 addition till zero flag is
4608 46 set
46
4609 EB XCHG [DE] – [HL]

460A 0D DCR C Decrement counter
value
460B C4 CNZ Facto Call on no zero to facto
460C 00 (i.e repeat process till
460D 46 zero flag for c = 1)
460E C9 RET Return to main
program
Memory address Content

4250 04
4251 18
1 x 2 x 3 x 4 = 24
Hexadecimal
16 24
1-8
Result:
Thus, factorial program was done successfully
47
FIBANOCCI SERIES
Aim:
To write an assembly language program to displace Fibanocci Series.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Load the length of series in the accumulator and decrement it
by 2
Step 3 : Move the value to register ‘D’
Step 4 : Load the starting value of data value address
Step 5 : Intialise the 1st number as 00
Step 6 : Move the pointer to 2nd data and intialise them as ‘01’
Step 7 : Move the pointer to next position for next data
Step 8 : Intialise B as ‘00’ and C as ‘01’ for calculations
Step 9 : Copy the contents of ‘B’ to accumulator
Step 10 : Add the content of ‘C’ register to accumulator
Step 11 : Move the content ‘C’ to ‘B’ and ‘A’ to C
Step 12 : Now store the result to memory pointed by ‘HL’ pair
Step 13 : Move the pointer to next pointer
Step 14 : Decrement 0 by 1 for counter
Step 15 : If ‘D’ is not zero, go to step 9
Step 16 : if ‘D’ is zero, end the program
48
START
Load the lenth of series
Decrement it by 2
Move it to ‘D’ register
Load the starting value of data
Intialise their values as ‘00’
Move the pointer to next position
Intialise the value as ‘01’
Intialise ‘B’ as ‘00’ & ‘C’ as ‘01’
Move it from ‘B’ to ‘A’
Store the result in accumulator
Decrement D by 1
No
Check
D=0?
Yes
END
49

4200 LDA 4300 3A, 00, 43 Store the length of series in
‘A’
4203 SUI 02 D6, 02 Decrement ‘A’ by 02
4205 MOV D,A 57 Move ‘A’ to ‘D’ (counter)
4206 LXI H, 4301 21,01,43 Load the starting address of
array
4209 MVI M,00 36,00 Intialise 4301 as ‘00’
420B INX H 23 Increment pointer
420C MVI M, 01 36,01 Initialize 2nd as ‘01’
420E INX H 23 Increment pointer
420F MVI B,00 06,00 Intialise ‘B’ as ‘00’
4211 MVI, C, 01 0E, 01 Intialise ‘C’ as ‘01’
4213 Loop MOV A,B 78 Move B to A
4214 ADD C 81 Add ‘A’ and ‘C’
4215 MOV B,C 41 Move C to B
4216 MOV C,A 4F Move A to C
4217 MOV M,A 77 Move the result to memory
4219 DCR D 15 Decrement counter
421A JNZ loop C2, 13,42 If D = 0, jump to loop
421D HLT 76 Stop the program
Input
Input Address Value
4300 05
Output

4301 00
4302 01
4303 01
4304 02
4305 03
00 + 01 = 01
01+ 01 = 02
02 + 01 = 03
Result:
The assembly language for Fibonaci series was executed successfully using 8085
microprocessor kit.
50
16 – BIT MULTIPLICATION
Aim:
To write an assembly language program for 16 bit multiplication by using 8085
microprocessor kit.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Load the 1st data in ‘HL’ register pair
Step 3 : Move content of ‘HL’ pair to stack pointer
Step 4 : Load the 2nd data in ‘HL’ and move it to ‘DE’
Step 5 : Make ‘HL’ pair as ‘00’ and ‘00’
Step 6 : Add ‘HL’ pair and ‘SP’
Step 7 : Check for carry condition, if carry is present increment it by
one else move to next step.
Step 8 : Decrement DE register
Step 9 : Then move E to ‘A’ and perform ‘OR’ operation with ‘a’ and
‘D’
Step 10 : The value of operation is zero, then store the value else go to
step 3
51
START
Load the lenth of series
Move the value ‘HL’ to ‘SP’
Intialise both ‘BC’ & ‘HL’ as ‘0000’
Add the content of ‘HL’ & ‘SP’
No
If
Carry=0?
Yes
Increment BC pair
Decrement DE pair
Move ‘E’ to ‘a’ & “or” operation with ‘D’
Result or ≥1?
No
Yes
Store the result in ‘HL’
END
52

4100 2A,00,42 LHLD 4200 Get the 1st data in HL
4103 F9 SP HL Save it in stack
pointer4106
4106 2A,02,42 LHLD 4202 Get the 2nd data in HL
4107 EB XCHG Exchange ‘HL’ and
‘DC’
4108 21,00,00 LXI H 0000 Make HL – 0000
410B 01,00,00 LXI B 0000 Make BC – 0000
410E 39 Next DAD SP Add ‘SP’ and ‘HL’
410F D2, 13, 41 JNC Loop Jump to loop if no
carry
4112 03 INX B Increment ‘BC’ by one
4113 1B Loop DCX D Decrement ‘DE’ by
one
4114 7B MOV A,E Make E – A
4115 B2 ORA D ‘OR’ gate between A
&D
4116 C2,0E,41 JNZ Next Jump on if number
zero
4119 22,04,42 SHLD 4204 Store the LSB in
memory
411C 69 MOV L,C Make C to L
411D 60 MOV H,B Make B to H
411E 22,06,42 SHLD 4206 Store the MSB in
memory
4121 76 HLT Stop the program
Input
Input Address Value
4200 04
4201 07
4202 02
4203 01
Output
4204 08
4205 12
4206 01
4207 00
Result:
Thus the assembly language program for 16 bit multiplication was executed
successfully.
53
16 – BIT DIVISION
Aim:
To write an assembly language program for 16 bit division in 8085 microprocessor.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 2 : Intialise ‘BC’ as ‘0000’ for Quotient
Step 3 : Load the divisor in ‘HL’ pair and save it in ‘DE’ register pair
Step 4 : Load the dividend in ‘HL’ pair
Step 5 : Move the value of ‘a’ to register ‘E’
Step 6 : Subtract the content of accumulator with ‘E’ register
Step 7 : Move the content ‘A’ to ‘C’ & ‘H’ to ‘A’
Step 8 : Subtract with borrow, the content of ‘A’ with ‘D’
Step 9 : Move the value of ‘a’ to ‘H’
Step 10 : If cy = 1, go to step 12, otherwise next step
Step 11 : Increment ‘B’ register & jump to step ‘4’
Step 12 : Add both contents of ‘DC’ and ‘HL’
Step 13 : Store the remainder in memory
Step 14 : Move the content of ‘C’ to ‘L’ & ‘B’ to ‘H’
Step 15 : Store the Quotient in memory
54
START
Intialise the Quotient as zero
Load the divisor in ‘HL’ & move it ‘DE’
Load the dividend in ‘HL’ pair
Intialise A with compare ‘E’ & ‘L’
Yes
Check for
if cy = 1
Subtract ‘HL’ from
‘DE’ increment BC pair
Store te remainder in HL in memory
Move the content of BC to ‘HL’ pair
Store the Quotient in HL to memory
END
55

4500 LXI B,0000 0,00,00 Intialise Quotient as ‘0000’
4503 LHLD 4802 2A,02,48 Load the divisor in ‘HL’
4506 XCHG EB Exchange ‘HL’ and ‘DE’
4507 LHLD 4800 2A,00,48 Load the dividend
450A Loop 2 MOV A,L 7D Move the ‘L’ value to ‘A’
450B SUB E 93 (A-E) – A
450C MOV L,A 6F A- L (A value is move t L)
450D MOV A,H 7C H – A (a is stored with H)
450E SBB D 9A Subtract ‘D’ from ‘A’
450F MOV H,A 67 Then A is moved to ‘H’
4510 JC loop 1 DA,17,45 If cy is present go to loop 1
4513 INX B 03 Increment BC pair by 1
4514 JMP loop 2 C3, 0A, 45 Jump to loop 2
4517 Loop 1 DAD ‘D’ 19 ‘DE’ and ‘HL’ pair all added
4518 SHLD 4806 22,06,48 HL is stored in memory
451B MOV L,C 69 Move ‘C’ register data to ‘L’
451C MOV H,B 60 Move ‘B’ register data to
‘H’
451D SHLD 4804 22,04,48 Store the result in ‘HL’ pair
4520 HLT 76 Stop the program
Input
Input Address Value
4800 04
4801 00
4802 02
4803 00
Output

4804 02
4805 00
4806 FE
4807 FF
Result:
Thus the assembly language program for 16 bit division was executed successfully.
56
BINARY TO BCD CONVERSION
Aim:
To write an assembly language program to convert an 8 bit binary data to BCD using
Apparatus required:
(0-5V) power supply
Algorithm:
Step 2 : Clear ‘D’ and ‘E’ register to account for hundred’s and ten’s
load the binary data in accumulator
Step 3 : Compare ‘A’ with 64 if cy = 01, go step C otherwise next step
Step 4 : Subtract 64 from (64+1) ‘A’ register
Step 5 : Increment ‘E’ register
Step 6 : Compare the register ‘A’ with ‘0A’, if cy=1, go to step 11,
otherwise next step
Step 7 : Subtract (0AH) from ‘A’ register
Step 8 : Increment D register
Step 9 : Go to step 7
Step 10 : Combine the units and tens to from 8 bit result
Step 11 : Save the units, tens and hundred’s in memory
57
START
Intialise ‘D’ as ‘00’ for hundreds and ‘e’ as ‘00’ for tens
Load the binary data in ‘A’
Compare the accumulator with ‘64’
Cy = ?
Subratct 64 from accumulator
Increment ‘E’ register
Compare with 0A
Cy = ?
Subtract ‘0A’ from accumulator
Increment register ‘D’
Combine the units & tens to form 8 bit

cCom
Save the units, tens & hundreds in memory
END
58

4100 0E,00 MVI E,00 Clear ‘E’ register
(Hund)
4102 53 MOV D,E Clear ‘D’ register
(tens)
4103 3A,00,42 LDA 4200 Get the data in ‘A’
4106 C3,06,41 HUND CPI 64 Compare the data with
64
4108 DA,11,41 JC TEN If content is less jump
to ten
410B D6, 64 SUI 64 Subtract data by 64
410D IC INR E Increment carry each
time
410E C3,06,41 JMP HUND Jump to hundred &
repeat
4111 C3, 0A TEN CPI 0A Compare the data with
0A
4113 DA,1C,41 JC UNIT If data is less jump to
unit
4116 D6, 0A SUI 0A Subtract the data by 0A
4118 14 INR D Increment ‘D’ each
time
4119 C3,11,41 JMP TEN Jump to ten & repeat
411C 4F UNIT MOV 4A Move the value ‘A’ to
‘C’
411D 7A MOV A,D Move the value ‘D’ to
‘A’
411E 07 RLC Rotate the value of ‘A’
411F 07 RLC Of ‘A’ so that
4120 07 RLC Lower and upper
niddle
4121 07 RLC Gets exchanged
4122 81 ADD C Add ‘A’ and ‘C’
4123 32,50,42 STA 42,50 Save ten’ & units in
‘M’
4126 7B MOV A,E Move to E to A
4127 32,51,42 STA 4251 Save hundreds unit in
‘A’
412A 76 HLT Stop the program
execution
59
Input
Input Address Value
4200 54
Output

4250 84
4251 00
Result:
Thus the binary to BCD conversion was executed successfully
60
BCD TO BINARY
Aim:
To write an assembly language program to convert BCD data to Binary data using
Apparatus required:
(0-5V) power supply
Algorithm:
Step 2 : Get the BCD data in accumulator and save it in register ‘E’
Step 3 : Mark the lower nibble of BCD data in accumulator
Step 4 : Rotate upper nibble to lower nibble and save it in register ‘B’
Step 5 : Clear the accumulator
Step 6 : Move 0AH to ‘C’ register
Step 7 : Add ‘A’ and ‘B’ register
Step 8 : Decrement ‘C’ register. If zf = 0, go to step 7
Step 9 : Save the product in ‘B’
Step 10 : Get the BCD data in accumulator from ‘E’ register and mark
the upper nibble
Step 11 : Add the units (A-ug) to product (B-ug)
Step 12 : Store the binary value in memory
Step 13 : End the program
61
START
Get the BCD data in accumulator
Move ‘A’ to register ‘E’
Lower nibble of BCD data in ‘A’
Rotate upper -> lower & save it in ‘B’
Clear the accumulator & move ‘0A’ to ‘C’
Yes
Zf = 0 ?
Save the product in ‘B’ register
Get the BCD data in A, move it to E and mark it
Add the unit A and (B-ug) product
Store the binary value in ‘M’
END
62

4100 3A,00,42 LDA 4200 Get the data in ‘A’
4103 5E MOV E,A Save in ‘E’ register
4104 E6, F0 ANI F0 Mark the lower nibble
4106 07 RLC Rotate the upper
4107 07 RLC To lower nibble
4108 07 RLC And save in
4109 07 RLC Register B
410A 47 MOV B,A Move it from ‘A’ to
‘B’
410B AF XRA A Clear the accumulator
410C 0E,0A MVI C,0A Intialise ‘C’ as ‘0A’
410E 08 REP
410F 0D DCR C Decrement ‘C’ register
4110 C2,0E,41 JNZ Jump till value ‘C’ is 0
4113 47 MOV B,A Move the value A to B
4114 7B MOV A,E Get the BCD in ‘A’
4115 E6, 0F ANI 0F Mark the upper nibble
4117 80 ADD B Add ‘A’ and ‘B’
4118 32,01,42 STA 4201 Save the binary data
411B 76 HLT Stop the program
execution
Input
Input Address Value
4200 68
Output

4201 44
16 68
4-4
Result:
Thus the BCD to binary conversion was executed successfully
63
SPEED CONTROL OF STEPPER MOTOR
Aim:
To write an assembly program to make the stepper motor run in forward and reverse
direction.
Apparatus required:
Stepper motor
(0-5V) power supply
Algorithm:
Step 1 : Load the ‘HL’ pair wit value from table
Step 2 : Move it to ‘B’ register for setting the counter
Step 3 : Move the memory value to accumulator and display it by
control word
Step 4 : Load ‘DE’ register pair with FFFF for starting delay subroutine
Step 5 : Run the delay loop control D-register becomes zero.
Step 6 : Increment ‘H’ address for next value from table
Step 7 : Jump on no zero
Step 8 : When B = 0, go to start and restart the program
64
START
Load ‘HL’ register pair with data
Load ‘E’ with ‘04’ (count)
Move memory to accumulator
Display the accumulator content (8 bit port)
Load the ‘DE’ pair with ‘FFFF’
Start delay subroutine
Decrement ‘D’ by one, check ‘OR’ gate

between ‘D’ and E
D=0?
Increment ‘HL’ pair by one (count)
Decrement ‘B’ count by one
Cy = ?
65

4100 Start LXI H,Look up 21,1A,41 Load the ‘HL’ with
data
4103 MVI B,04 06,04 B = 04
4105 Repeat MOV A,M 7E Memory value to ‘A’
4106 OUT C0 D3, C0 Display it
4108 LXI D,03,03 11 Load ‘DE’ with FFFF
410B Delay NOP 00 Start delay loop
410C DCX D 1B Decrement DE by 1
410D MOV A,E 7B Move ‘E’ to ‘A’
410E ORA D B2 Check De = 0 or not
410F JNZ DELAY C2, 0B,41 Jump on zero
4112 INX H 23 Increment HL by 1
4113 DCR B 05 Decrement B by 1
4114 JNZ Repeat C2,05,41 Jump on no zero
4117 JMP START C3,00,41 Jump to start
Input
Input Address Value
411A 0A
411B 06
411C 05
411D 09
Reverse Direction

411A 09
411B 05
411C 06
411D 0A
Result:
Thus, an assembly language program to control of stepper motor was written using
66
FLASHING DISPLAY
Aim:
To write an assembly language program to obtain the following flashing display of a
particular data.
Apparatus required:
8085 micro processing kit
(0-5V) power supply
Algorithm:
Step 1 : Get the control words in accumulator and output words through
8 bit address
Step 2 : Load ‘HL’ register pair with memory address
Step 3 : Get the count value in ‘C’ register
Step 4 : Increment the register pair by one and display the character and
call for delay.
Step 5 : Clear the display and call delay routine to step 7
Step 7 : Load ‘DE’ register pair with memory address
Step 8 : Decrement ‘DE’ pair with memory address
Step 9 : If the content is not equal to zero, go to step 8
Step 10 : Return to main program
67
START
Load the control loads for flashing display
Load the register pair ‘HL’ with specified address
Transfer count from memory to ‘C’ register
Increment memory address
Transfer memory count in accumulator
Output accumulator content to a data register
Increment count in register ‘c’
No
Check for
carry?
Yes
Call delay
Load control word for linear screen
Call delay
Load ‘DE’ with memory address
No
If
Content ?
Yes
Return
68

4300 MVI A,00 3E,00 Intialise ‘A’ as ‘00’
4302 OUT 01 DE,01 Out the control word
trough 8 bit
4304 MVI A,90 3E,90 Intialise ‘a’ with cw for
RAM
4306 OUT 01 D3,01 Out the cw
4308 MVI A,CC 3E,CC A = CC
430A OUT 01 0D,01 Out the cw
430C Loop 2 LXI H,5000 21,00,50 Load ‘HL’ with
430F MOV C,M 4E M to C
4310 Loop 1 INX H 23 Increment ‘H’ by
4311 MOV A,M 7E Move M to A
4312 OUT 00 D3, 00 Out the character
4314 DCR C 0D Decrement ‘C’ by 1
4315 JNZ Loop 1 C2,10,43 Check for zero
4318 CALL Delay C0,00,46 Call subroutine
431B MVI A,DC 3E,DC A <- 0C
431D OUT 01 D3, 01 A<-01
431F CALL Delay CD,00,46 Call subroutine
4322 JMP Loop 2 C3 0C,43 Check for zf
4600 Delay LXI D,FFFF 11,FF,FF Intialise DE=FFFF
4603 Loop 3 DCX D 1B Decrement DE by 1
4604 MOV A,E 7B Move ‘E’ to ‘A’
4605 ORA D B2 Check ‘De’ = ‘00’
4606 JNZ Loop 3 C2,03,46 Jump on no zero
4609 RET C9 C9 Return to main
program
Input
Input Address Value
5000 05
5001 68
5002 68
5003 68
5004 FD
5005 88
Output
EEE – A
Result:
Thus, an assembly language program to obtain flashing display of a particular data
was written using 8085 microprocessor kit.
69
ROLLING DISPLAY
Aim:
To write an assembly language program to obtain a rolling display of a particular data
by using 8085 microprocessor
Apparatus required:
8085 micro processing kit
(0-5V) power supply
Algorithm:
Step 1 : Get the control words in accumulator and output the control
words through 8 bit port address
Step 2 : Load ‘HL’ register pair with memory address and transfer
memory content to ‘C’ register
Step 3 : Increment ‘HL’ pair with one and transfer the particular bit
pattern through 8 bit port address
Step 4 : Call subroutine delay at step 6
Step 5 : If the count value in ‘C’ is not equal to zero then go to step 3
else go to step 2
Step 6 : Load ‘DE’ register pair by memory address
Step 7 : Decrement ‘DE’ register pair by one
Step 8 : If DE is not equal to zero, go to step 7 else main program
70
Delay
Intialise register ‘D’ with maximum count ‘EE’
Intialise register ‘c’ with maximum count ‘EE’
Decrement register ‘C’ count by one
If No
Reg ‘C’
Count ?
Yes
Decrement register ‘D’ content by one
No
If reg ‘D’
Count ?
Yes
Return
71

4500 MVI A,00 3E,00 Initialise A 00
4502 OUT 01 DE, 01 Control word through 8
bit
4504 MVI A,90 3E, 90 A = RAM cw
4506 OUT 01 DE,01 Output cw through 8
bit port
4508 MVI A,CC 3E,CC A = CC
450A OUT 01 DE,01 Output cw through 8
bit port
450C Loop 2 LXI H,5000 21,00,50 Memory -> HL
location
450F MOV C,M 4E M -> C
4510 Loop 1 INX H 23 Increment ‘HL’
4511 MOV A,M 7E Move ‘H’ to ‘A’
4512 OUT 00 DE, 00 Output the character
4514 CALL Loop CD,00,46 Call the subroutine
4517 DCR C 0D Decrement ‘C’ by one
451B JMP Loop 2 C3,0C,45 Jump to L2
4600 Loop LXI D,FFFF 11,FFFF Load DE-FFFF
4603 Loop 3 DCX D 1B Decrement ‘DE’ by 1
4604 MOV A,D 7A Move ‘D’ to ‘A’
4605 ORA E B3 (A) = (A) check
4609 RET C9 Return to main
program
Input
Input Address Value
5000 06
5001 98
5002 68
5003 7A
5004 C8
5005 1A
5006 2C
Output
HELPUS
Result:
Thus, an assembly language program to obtain rolling display of a particular value
written using 8085 microprocessor kit.
72
SQUARE WAVE GENERATOR
Aim:
To write a program and to generate square generator using DAC.
Apparatus required:
(0-5V) power supply
Algorithm:
Step 1 : Intialise ‘A’ as ‘00’ and take data pointer to port C8
Step 2 : Call delay
Step 3 : Move FF to A and take port ‘C8’
Step 4 : Call delay
Delay Subtroutine
Step 1 : Counter 1 = 05
Step 2 : Counter 2 = FF
Step 3 : Decrement counter 2
Step 4 : Check if c= 0, if no jump to step 3
Step 5 : Decrement counter 1
Step 6 : Check if B = 0, if no jump to step 2
Step 7 : Return to main program
73
START
Load the control words as for displaying
Call the delay subtroutine program
Intialise the accumulator as ‘EE’
Output the accumulator contents register
Call the delay subroutine program
Jump to the start of program
Intialise ‘B’ as 05 and ‘C’ as ‘FF’ and decrement ‘c’ by one
Check for
c=?
No
Yes
Decrement the value of ‘B’
No
Check for
B=0?
Yes
Return
74

4100 3E,00 Start MVI A,00 Intialise ‘A’ with ‘00’
4102 D3,C8 OUT C8 Load the control words
4104 CD,11,41 CALL Delay Call delay sutroutine
4107 3E,FF MVI A,FF Intialise ‘A’ with ‘FF
4109 D3,C8 OUT C8 A -> C8
410B CD,11,41 CALL Delay Call delay subroutine
410E C3,00,41 JMP Start Jump to start
4111 06,05 Delay MVI B,05 B -> 05
4113 0E Loop 1 MVI C,FF [C] => FF
4115 OD Loop 2 DCR C Decrement ‘C’ register
4116 C2,15,41 JNZ Loop 2 Jump on no zero
4119 05 DCR B Decrement ‘B’ register
411A C2,13,41 JNZ Loop 1 Jump on n zero
411D C9 RET Return to main
program
Result:
Thus square wave was generated using 8085 microprocessor kit.
75
TRIANGULAR WAVE GENERATOR
Aim:
To write an assembly language program for generating triangular wave using DAC.
Apparatus required:
(0-5V) DC battery
Algorithm:
Step 1 : Move content of ‘C’ to ‘A’ where ‘L’ is intialised to ‘00’
Step 2 : Output content of C8
Step 3 : Increment L till zf = 0
Step 4 : Intialise ‘L’ register with FF
Step 5 : Move content of ‘L’ to accumulator and output to port
Step 6 : Decrement ‘L’ if not equal to zero jump else go to next step
Step 7 : Jump on next step
76
Delay
Intialise the value of ‘1’ as ‘00’
Move the values of ‘1’ to ‘A’
Output the control word for control signal
Increment the value of ‘L’
L=0?
No Yes
Set the value of ‘1’ as ‘FF’
Move the ‘1’ value to accumulator
Output the control word
L=0?
No
Yes
Jump to start
77

4300 2E,00 Start MVI L,00 Intialise ‘L’ as ‘00’
4302 7D Loop 1 MOV A,L [L] -> [A]
4303 D3,C8 OUT C8 Load the control words
4305 2C INR L Increment register ‘L’
4306 C2,02,43 JNZ Loop 1 Jump on no zero to
loop 1
4309 2E, FF MVI L,FF L = FF
430B 70 Loop 2 MOV A,L L -> A
430C D3,C8 OUT C8 [C8] -> [A]
430E 2D DCR L Decrement L by one
430F C2,0B,43 JNZ Loop 2 Jump on no zero to
430B
4312 C3,00.43 JMP Start Repeat process
Result:
Thus the triangular wave was generated using 8085 microprocessor kit.
78
SAWTOOTH WAVE GENERATOR
Aim:
To write an assembly language program for generating Sawtooth waveform by using
microprocessor 8085.
Apparatus required:
(0-5V) power supply
Algorithm:
Step 1 : Intialise accumulator with ‘00’
Step 2 : Output current address specified
Step 3 : Increment accumulator by one
Step 4 : Jump to step one
Delay
Intialise the accumulator as ‘00’
Display the output port using cw
Increment the accumulator
Jump to loop 1

4500 3E,00 Start MVI A,00 Intialise ‘A’ as ‘00’
4502 D3, C8 Loop 1 OUT C8 A = [C8]
4504 3C INR A Increment ‘A’ by one
4505 C3,02,45 JMP Loop 1 Jump to loop one
TRIANGULAR WAVE
79
SAW TOOTH WAVE
SQUARE WAVE
Result:
Thus the Sawtooth wave was generated using 8085 microprocessor kit.
80
ANALOG TO DIGITAL CONVERTER
Aim:
To write an assembly language program to convert analog to digital signal and to
display it in 7 segment LED display
Apparatus required:
(0-5V) power supply
Algorithm:
Step 1 : Access the channel of ADC
Step 2 : Intialise the accumulator with start of conversion signal &
output it to the ADC
Step 3 : Send ‘0’ signal for ending the conversion for ADC
Step 4 : Get the analog value converted to display from ADC
Step 5 : The digital signal is separated into two nibbles and displayed in
hexadecimal from by calling service subroutine.
81
START
Load the control word necessary for generation of ALE signal to control register
Load the control word necessary to start the conversion to control register
Get the output port specified
If the 1st
LSB bit=1?
Get accumulator output in port specified
Transfer accumulator content to ‘B’
Mask all MSR & Store memory in location
Mask all 4 LSB & store memory in location
Load accumulator with 03 & with 08
Load the pair with address of MSR
Call the service subroutine
82

5000 3E,10 MVI A,10 Intialise ‘a’ with 10
5002 D3,C OUT C8 Output channel through
5004 3E,18 MVI A,18 Intialise ‘A’ with 18
5006 D3, C8 OUT C8 Output channel through
8 bit port
5008 00 NOP No operation
5009 00 NOP No operation
500A 3E,10 MVI A,10 Intialise ‘A’ with 2nd
signal
500C D3,C8 OUT C8 Output channel through
8 bit port
500E 3E,01 L2 MVI A,01 Intialise ‘A’ with 2nd
5010 D3,D0 OUT D0 Output through 8 bit
5012 00 NOP
5013 00 NOP
5014 00 NOP
5015 3E,00 MVI A,00
5017 D3,D0 OUT D0
5019 DB,D8 L1 IN D8
501B E6,01 ANI 01
501D CA,19,50 JZ L1
5020 DB,C0 IN C0 Get input from
5022 47 MOV B,A B -> A
5023 E6,0F ANI 0F And of with ‘A’
5025 32,51,51 STA 5151 Store in 5151
5028 78 MOV A,B B -> A
5029 E6,F0 ANI F0 And F0 with A
502B 0F RRC Rotate content ‘A’
502C 0F RRC
502E 0F RRC
502F 32,50,51 STA 550 Store MSB in 5150
5032 3E,03 MVI A,03 03 -> A
5034 0E,08 MVI C,08 08 -> C
5036 21,50,51 LXI H 5150 Load ‘HL’ pair with
5150
5039 CD,05,00 CALL 0005 Call device subroutine
503C C3,0E,50 JMP 500E Jump to 500E
Result:
Thus the analog to digital conversion was done microprocessor.
83
ARTHMETIC OPERATIONS USING 8051
Aim:
To do the arithmetic operations using 8051 microprocessor
Apparatus required:
DAC interface kit
Keyboard
Algorithm:
Addition / Subtraction
Step 1 : Move 1H data to memory
Step 2 : Add or subtract 1H data with 2nd data
Step 3 : Initialize data pointer.
Step 4 : Move result to memory pointed by DPTR.
START
Out 1H data in memory
Add or subtract 1H and 1st data
Initialize DPTR
Move result to memory preset by DPTR
Stop
84
Program: 8-bit Addition:
Memory Label Opcode Mnemonics Comments

Location
4100 Start C3 CLR C Clear the carry flat
4101 74DA MOV A, # data 1 Moves data 1 to
register A
4103 24DA ADD A, # data 2 Add content of A and
data 2 and store in A
4105 464500 MOV DPTR, # 4500 Moves data 4500 to
DPTR
4108 F0 MOV A @ DPTR, A Moves control of A to
location pointed DTPR
4109 80 FE SJMP 4109 Short jump to 4109
Execution:
Addition:
ML Input ML Output
4103 0L 4500 05
4109 03
Program: 8-bit Subtraction:

Location
4100 Start C3 CLR C Clear the carry flat
4101 74DA MOV A, # data 1 Moves data 1 to
register A
4103 24DA SUB B, # data 2 Subtract data 2 from
content of A and store
result in A
4105 464500 MOV DPTR, # 4500 Moves 4500 to DPTR
4108 F0 MOV X @ DPTR, A Moves result by
location by DTPR
4109 80 FE SJMP 4109 Short jump to 4109
Execution:
Subtraction:
ML Input ML Output
4101 05 4500 03
4103 02
Result:
Thus 8-bit addition, subtraction is performed using 8051.
85
ARTHMETIC OPERATIONS USING 8051
Aim:
To do the arithmetic operations using 8051 microprocessor
Apparatus required:
DAC interface kit
Keyboard
Algorithm:
Multiplication / Division
Step 1 : Get 1H data and 2nd data to memory
Step 2 : Multiply or divide 1H data with 2nd data
Step 3 : Initialize data pointer.
Step 4 : Move result to memory pointed by DPTR (first port)
Step 5 : Increment DPTR
Step 6 : Move 2nd part of result to register A
Step 7 : Move result to 2nd memory location pointer by DPTR
START
Get data into the register
Complement the data
Move the data to pointer by DPTR
Increment data
Increment DPTR
Move data into paste location
Short jump to preset location
Stop
86
Yes
Program: 8-bit Multiplication:

Location
4100 Start 7403 MOV A, # data 1 Move immediate data
to accumulator
4101 75F003 MOV B, # data 2 Move 2nd data to B
register
4105 A4 MUL A B Get the product in A &
B
4106 904500 MOV DPTR, # 4500 Load data in 4500
location
4109 F0 MOV X @DPTR, A Move A t ext RAM
410B E5F0 MOV A,B Move 2nd data in A
410D F0 MOV A @ DPTR Same the ext RAM
410E 80FE SJMP 410E Remain idle in infinite
loop
Execution:
Multiplication:
ML Input Output Address Value
4101 0L 4500 08
4103 04
Program: 8-bit Division:

Location
4100 Start 7408 MOV A, # data 1 Move immediate data
to accumulator
4102 75F002 MOV B, @ data 2 DIV Move immediate to B
AB reg.
4105 84 DIV AB Divide content of A &
B
4106 904500 MOV DPTR, # 4500 Load data pointer with
4500 location
4109 F0 MOV X @ DPTR, A Move A to ext RAM
410A A3 INC DPTR Increment data pointer
410B ESF0 MOV A,B Move remainder to A
410D F0 MOV @ DPTR, A Move A to ext RAM
410E 80FE SJMP 410E Remain idle in infinite
loop
Execution:
Division:
ML Input Output Address Value
4101 08 4500 02
4103 04
Result:
Thus 8-bit multiplication & division is performed using 8051.
87
2
The 8085 Microprocessor
1. Draw the pin configuration and functional pin diagram of μP 8085.
Ans. The pin configuration and functional pin diagram of μP 8085 are shown below:
+5V GND
1 2 40 20
x1 1 40 VCC Serial SID 5 x1 x2 vcc
I/O SOD 4
x2 2 39 HOLD ports A15 28
High order
RESET OUT 3 38 HLDA address bus
A8
SOD 4 37 CLK(OUT) 21
TRAP 6
SID 5 36 RESET IN
initiated signals
RST7.5 7
Externally
TRAP 6 35 READY RST6.5 8 AD0 19 Multiplexed

RST5.5 9 address/data
RST 7.5 7 34 IO/M
INTR 10 AD7 12 bus
RST 6.5 8 33 S1
RST 5.5 9 8085A 32 RD READY 35

HOLD 39
INTR 10 31 WR 8085A
RESET IN 36
INTA 11 30 ALE
acknowledgment
INTA 11
External signal
AD0 12 29 S0
HLDA 38 30
AD1 13 28 A15 ALE
29 S0
AD2 14 27 A14 33
S1 Control
AD3 15 26 34 IO/M and
A13
32 status
AD4 25 A12 RD signals
16
31
WR
AD5 17 24 A11
AD6 18 23 A10
AD7 19 22 A9
VSS 20 21 A8
3 37
Pin Configuration
RESET OUT CLK OUT
Fig. 2.1: Functional pin diagram
12 Understanding 8085/8086 Microprocessors and Peripheral ICs through Questions and Answers
2. In how many groups can the signals of 8085 be classified?

Ans. The signals of 8085 can be classified into seven groups according to their functions. These
are:
(1) Power supply and frequency signals (2) Data and Address buses (3) Control bus
(4) Interrupt signals (5) Serial I/O signals (6) DMA signals (7) Reset signals.
3. Draw the architecture of 8085 and mention its various functional blocks.
Ans. The architecture of 8085 is shown below:
INTA RST 6.5 TRAP
INTR SID SOD
RST 5.5 RST 7.5
Interrupt control Serial I/O control
8-Bit Internal data bus
B C
REG REG
Temp Flag Instruction D E
Accumulator reg flip-flops register REG REG
H L
REG REG
Stack pointer
Instruction Program counter

Arithmetic
logic decoder and
unit machine Incrementer/
(ALU) cycle decrementer
encoding address latch
x2 x1 GND +5V
Address/
Address
CLK Data
DMA Reset buffer
GEN Control Status buffer
CLK GEN READY RD WR ALE S0 S1 IO/M HOLD HLDA RESET IN RESET OUT A15 – A8 AD7 – AD0
Fig. 2.2: Architecture of 8085
The various functional blocks of 8085 are as follows:

z Registers
z Arithmetic logic unit
z Address buffer
z Incrementer/decrementer address latch
z Interrupt control
z Serial I/O control
z Timing and control circuitry
z Instructions decoder and machine cycle encoder.
The 8085 Microprocessor 13
4. What is the technology used in the manufacture of 8085?

Ans. It is an NMOS device having around 6200 transistors contained in a 40 pin DIP package.
5. What is meant by the statement that 8085 is a 8-bit microprocessor?
Ans. A microprocessor which has n data lines is called an n-bit microprocessor i.e., the width
of the data bus determines the size of the microprocessor. Hence, an 8-bit microprocessor
like 8085 can handle 8-bits of data at a time.
6. What is the operating frequency of 8085?
Ans. 8085 operates at a frequency of 3 MHz, and the minimum frequency of operation is 500
kHz.
The version 8085 A-2 operates at a maximum frequency of 5 MHz.
7. Draw the block diagram of the built-in clock generator of 8085.

Ans. The built-in clock generator of 8085, in block schematic, is shown below:
VCC
CLK (out)
T Q Q1
x1(Pin1) CLK
x1(Pin2) Q Q2
Fig. 2.3: Block diagram of built-in clock generator
The internal built-in clock generator, LC or RC tuned circuits, piezo-electric crystal

or external clock source acts as an input to generate the clock. The T F/F, shown in Fig.
2.3 divides the input frequency by 2. Thus the output frequency of 8085 (obtained from
pin 37) is half the input frequency.
8. What is the purpose of CLK signal of 8085?
Ans. The CLK (out) signal obtained from pin 37 of 8085 is used for synchronizing external
devices.
9. Draw the different clock circuits which can be connected to pins 1 and 2 of 8085.
Ans. The different external clock circuits which can be connected to pins 1 and 2 of 8085 are
shown below in Fig. 2.4:
+VCC
Pull-up resistance
CLK (out) x1(1)
x1(1) 37 CLK (out)
R XTAL
x1(1) 37 CLK (out)
C x2
(2) 8085
x2(2) 8085
C
x2(2)
(a) R C circuit NC
(b) Crystal clock circuit (c) External frequency source
Fig. 2.4: The different external clock circuits

The output frequency obtained from pin 37 of Fig. 2.4(b) is more stable than the RC
circuit of Fig. 2.4(a).
10. What are the widths of data bus (DB) and address bus (AB) of 8085?
Ans. The width of DB and AB of 8085 are 8-bits (1 byte) and 16-bits (2 bytes) respectively.
11. What is the distinguishing feature of DB and AB?
Ans. While the data bus is bidirectional in nature, the address bus is unidirectional.
Since the µP can input or output data from within it, hence DB is bidirectional. Again
the microprocessor addresses/communicates with peripheral ICs through the address bus,
hence it is unidirectional, the address comes out via the AB of µP.
12. The address capability of 8085 is 64 KB. Explain.
Ans. Microprocessor 8085 communicates via its address bus of 2-bytes width – the lower byte
AD0 – AD7 (pins 12-19) and upper byte D8 – D15 (pins 21–28). Thus it can address a
maximum of 216 different address locations. Again each address (memory location) can
hold 1 byte of data/instruction. Hence the maximum address capability of 8085 is
= 216 × 1 Byte
= 65, 536 × 1 Byte
= 64 KB (where 1 K = 1024 bytes)
13. Does 8085 have serial I/O control?
Ans. 8085 has serial I/O control via its SOD and SID pins (pins 4 and 5) which allows it to
communicate serially with external devices.
14. How many instructions 8085 can support?
Ans. 8085 supports 74 different instructions.
15. Mention the addressing modes of 8085.
Ans. 8085 has the following addressing modes: Immediate, Register, Direct, Indirect and
Implied.
16. What jobs ALU of 8085 can perform?
Ans. The Arithmetic Logic Unit (ALU) of 8085 can perform the following jobs:
z 8-bit binary addition with or without carry.
z 16-bit binary addition.
z 2-digit BCD addition.
z 8-bit binary subtraction with or without borrow.
z 8-bit logical OR, AND, EXOR, complement (NOT function).
z bit shift operation.
17. How many hardware interrupts 8085 supports?
Ans. It supports five (5) hardware interrupts—TRAP, RST 7.5, RST 6.5, RST 5.5 and INTR.
18. How many I/O ports can 8085 access?
Ans. It provides 8-bit I/O addresses. Thus it can access 28 = 256 I/O ports.
19. Why the lower byte address bus (A0 – A7) and data bus (D0 – D7) are multiplexed?
Ans. This is done to reduce the number of pins of 8085, which otherwise would have been a
48 pin chip. But because of multiplexing, external hardware is required to demultiplex
the lower byte address cum data bus.
20. List the various registers of 8085.
Ans. The various registers of 8085, their respective quantities and capacities are tabulated
below:
Table 2.1: List of Various Registers in 8085
S. No. Name of the Register Quantity Capacity

1. Accumulator (or) Register A 1 8-bit
2. Temporary register 1 8-bit
3. General purpose registers (B, C, D, E, H and L) 6 8-bit each
4. Stack pointer (SP) 1 16-bit
5. Program counter (PC) 1 16-bit
6. Instruction register 1 8-bit
7. Incrementer/Decrementer address latch 1 16-bit
8. Status flags register 1 8-bit
21. Describe the accumulator register of 8085.
Ans. This 8-bit register is the most important one amongst all the registers of 8085. Any data
input/output to/from the microprocessor takes place via the accumulator (register). It is
generally used for temporary storage of data and for the placement of final result of
arithmetic/logical operations.
Accumulator (ACC or A) register is extensively used for arithmetic, logical, store and
rotate operations.
22. What are the temporary registers of 8085?
Ans. The temporary registers of 8085 are temporary data register and W and Z registers. These
registers are not available to the programmer, but 8085 uses them internally to hold
temporary data during execution of some instructions.
23. Describe W and Z registers of 8085.
Ans. W and Z are two 8-bit temporary registers, used to hold 8-bit data/address during
execution of some instructions.
CALL-RET instructions are used in subroutine operations. On getting a CALL in the
main program, the current program counter content is pushed into the stack and loads
the PC with the first memory location of the subroutine. The address of the first memory
location of the subroutine is temporarily stored in W and Z registers.
Again, XCHG instruction exchanges the contents H and L with D and E respectively.
W and Z registers are used for temporary storage of such data.
24. Describe the temporary data register of 8085.
Ans. The temporary data register of 8085 is an 8-bit register, which is not available to the
programmer, but is used internally for execution of most of the arithmetic and logical
operations.
ADD D instruction adds the contents of accumulator with the content of D. The
content of D is temporarily brought into the temporary data register. Thus the two inputs
to the ALU are—one from the accumulator and the other from the temporary data
register. The result is stored in the accumulator.
25. Describe the general purpose registers of 8085?
Ans. The general purpose registers of 8085 are: B, C, D, E, H and L. They are all 8-bit registers
but can also be used as 16-bit register pairs—BC, DE and HL. These registers are also
known as scratch pad registers.
26. In what other way HL pair can be used?
Ans. HL register pair can be used as a data pointer or memory pointer.
27. Mention the utility of the general purpose registers.
Ans. General purpose registers store temporary data during program execution, which can also
be stored in different accessible memory locations. But storing temporary data in memory
requires bus access—hence more time is needed to store. Thus it is always advisable to
store data in general purpose registers.
The more the number of general purpose registers, the more is flexibility in
programming—so a microprocessor having more such registers is always advantageous.
28. Which are the sixteen bit registers of 8085.
Ans. 8085 has three (3) sixteen bit registers—Program Counter (PC), Stack Pointer (SP) and
Incrementer/Decrementer address latch register.
29. Discuss the two registers program counter and stack pointer.
Ans. Program counter (PC) is a sixteen bit register which contains the address of the
instruction to be executed just next. PC acts as a address pointer (also known as memory
pointer) to the next instruction. As the processor executes instructions one after another,
the PC is incremented—the number by which
the PC increments depends on the nature of
R/W memory
PC
the instruction. For example, for a 1-byte
instruction, PC is incremented by one, while

Auto-increment
for a 3-byte instruction, the processor
facility
increments PC by three address locations.
Stack pointer (SP) is a sixteen bit register
which points to the ‘stack’. The stack is an

User program
area in the R/W memory where temporary
ends here
data or return addresses (in cases of
Gap
subroutine CALL) are stored. Stack is a auto-
decrement facility provided in the system. The

Auto-decrement
facility
stack top is initialised by the SP by using the
Stack
area
instruction LXI SP, memory address.
SP
In the memory map, the program should Fig. 2.5: Auto-increment and
be written at one end and stack should be auto-decrement facility for PC
initialised at the other end of the map—this is and SP respectively
done to avoid crashing of program. If sufficient
gap is not maintained between program memory location and stack, then when the stack
gets filled up by PUSH or subroutine calls, the stack top may run into the memory area
where program has been written. This is shown in Fig. 2.5.
30. Describe the instruction register of 8085.
Ans. Program written by the programmer resides in the R/W memory. When an instruction
is being executed by the system, the opcode of the instruction is fetched from the memory
and stored in the instruction register. The opcode is loaded into the instruction register
during opcode fetch cycle. It is then sent to the instruction decoder.
31. Describe the (status) flag register of 8085.
Ans. It is an 8-bit register in which five bit positions contain the status of five condition flags
which are Zero (Z), Sign (S), Carry (CY), Parity (P) and Auxiliary carry (AC). Each of these
five flags is a 1 bit F/F. The flag register format is shown in Fig. 2.6:
D7 D6 D5 D4 D3 D2 D1 D0
S Z X AC X P X CY
Fig. 2.6: The flag register format
z Sign (S) flag: – If the MSB of the result of an operation is 1, this flag is set, otherwise
it is reset.
z Zero (Z) flag:– If the result of an instruction is zero, this flag is set, otherwise reset.
z Auxiliary Carry (AC ) flag:– If there is a carry out of bit 3 and into bit 4 resulting from
the execution of an arithmetic operation, it is set otherwise reset.
This flag is used for BCD operation and is not available to the programmer to change
the sequence of an instruction.
z Carry (CY) flag:– If an instruction results in a carry (for addition operation) or borrow
(for subtraction or comparison) out of bit D7, then this flag is set, otherwise reset.
z Parity (P) flag:– This flag is set when the result of an operation contains an even
number of 1’s and is reset otherwise.
32. State the characteristics of the flag register.
Ans. The following are the characteristics of flag register:
z It is an 8-bit register.
z It contains five flags—each of one bit.
z The flag register can’t be written into.
33. What is the purpose of incrementer/decrementer address latch register?

Ans. This 16-bit register increments/decrements the contents of PC or SP when instructions
related to them are executed.
34. Mention the blocks on which ALU operates?
Ans. The ALU functions as a part which includes arithmetic logic group of circuits. This
includes accumulator, flags F/Fs and temporary register blocks.
35. What is the function of the internal data bus?

Ans. The width of the internal data bus is 8-bit and carries instructions/data between the CPU
registers. This is totally separate from the external data bus which is connected to
memory chips, I/O, etc.
The internal and external data bus are connected together by a logic called a
bidirectional bus (transreceiver).
36. Describe in brief the timing and control circuitry of 8085.
Ans. The T&C section is a part of CPU and generates timing and control signals for execution
of instructions. This section includes Clock signals, Control signals, Status signals, DMA
signals as also the Reset section. This section controls fetching and decoding operations.
It also generates appropriate control signals for instruction execution as also the signals
required to interface external devices.
37. Mention the following:
(a) Control and Status signals
(b) Interrupt signals
(c) Serial I/O signals
(d) DMA signals
(e) Reset signals.
 ææ æ
Ans. The control and status signals are ALE, RD , WR , IO/ M , S0, S1 and READY.
ææææ
The interrupt signals are TRAP, RST 7.5, RST 6.5, RST 5.5, INTR. INTA is an
interrupt acknowledgement signal indicating that the processor has acknowledged an
INTR interrupt.
Serial I/O signals are SID and SOD
DMA signals are HOLD and HLDA
æææææææ
Reset signals are RESET IN and RESET OUT.
38. What is the function of ALE and how does it function?
Ans. Pin 30 of 8085 is the ALE pin which stands for ‘Address Latch Enable’. ALE signal is used
to demultiplex the lower order address bus (AD0 – AD7).
Pins 12 to 19 of 8085 are AD0 – AD7 which is the multiplexed address-data bus.
Multiplexing is done to reduce the number of pins of 8085.
Lower byte of address (A0 – A7) are available from AD0 – AD7 (pins 12 to 19) during
T1 of machine cycle. But the lower byte of address (A0 – A7), along with the upper byte
A8 – A15 (pins 21 to 28) must be available during T2 and rest of the machine cycle to access
memory location or I/O ports.
Now ALE signal goes high at the beginning of T1 of each machine cycle and goes low
at the end of T1 and remains low during the rest of the machine cycle. This high to low
transition of ALE signal at the end of T1 is used to latch the lower order address byte
(A0 – A7) by the latch IC 74LS373, so that the lower byte A0 – A7 is continued to be
available till the end of the machine cycle. The situation is explained in the following
figure:
19 74LS373
AD7
8 A7 – A0
AD0 (Latch)
0 12 (Lower byte of address bus,
8 available from T2 state of each
5 G machine cycle)
ALE
30
ALE signal
Fig. 2.7: Lower byte of address latching achieved by the H to L transition of ALE signal,
which occurs at the end of T1 of each machine cycle
39. Explain the function of the two DMA signals HOLD and HLDA.
Ans. DMA mode of data transfer is fastest and pins 39 and 38 (HOLD and HLDA) become active
only in this mode.
When DMA is required, the DMA controller IC (8257) sends a 1 to pin 39 of 8085. At
the end of the current instruction cycle of the microprocessor it issues a 1 to pin 38 of
the controller. After this the bus control is totally taken over by the controller.
When 8085 is active and 8257 is idle, then the former is MASTER and the latter is
SLAVE, while the roles of 8085 and 8257 are reversed when 8085 is idle and 8257 becomes
active.
40. Discuss the three signals IO/ M , S0 and S1.
Ans. IO/ M signal indicates whether I/O or memory operation is being carried out. A high on
this signal indicates I/O operation while a low indicates memory operation. S0 and S1
indicate the type of machine cycle in progress.
æææææææ
41. What happens when RESET IN signal goes low?
æææææææ
Ans. RESET IN is an input signal which is active when its status is low. When this pin is
low, the following occurs:
z The program counter is set to zero (0000H).
z Interrupt enable and HLDA F/Fs are resetted.
z All the buses are tri-stated.
z Internal registers of 8085 are affected in a random manner.
æææææææ
42. Is there any minimum time required for the effective RESET IN signal?
æææææææ
Ans. For proper resetting to take place, the reset signal RESET IN must be held low for at
least 3 clock cycles.
43. Indicate the function of RESET OUT signal.
Ans. When this signal is high, the processor is being reset. This signal is synchronised to the
processor clock and is used to reset other devices which need resetting.
44. Write the advantages/disadvantages of having more number of general purpose

registers in a microprocessor.
Ans. Writing of a program becomes more convenient and flexible by having more number of
general purpose registers.
But there are certain disadvantages of having more GPRs. These are as follows:
The more the number of GPRs in a microprocessor, more number of bits would be
required to identify individual registers. This would reduce the number of operations that
can be provided by the microprocessor.
In programs involving subroutine CALL, if more GPRs are involved, then their
status are to be saved in stack and on return from the subroutine, they are to be restored
from the stack. This will thus put considerable overhead on the microprocessor.
If more number of GPRs are used in a microprocessor, considerable area of the chip
is used up in accommodating the GPRs. Thus there may be some problem in
implementing other functions on the chip.
45. Draw the lower and higher order address bus during the machine cycles.
Ans. The lower byte of address (AD0 – AD7) is available on the multiplexed address/data bus
during T1 state of each machine cycle, except during the bus idle machine cycle, shown
in Fig. 2.8.
The higher byte of address (A8 – A15) is available during T1 to T3 states of each
machine cycle, except during the bus idle machine cycle, shown in Fig. 2.9.
Machine cycle 1 Machine cycle 2
T1 T2 T3 T4 T1 T2 T3
AD0 – AD7
A0 – A7 A0 – A7
Fig. 2.8: Lower byte address on the multiplexed bus
T1 T2 T3 T4 T1 T2 T3
A8 – A15 A8 – A15 A8 – A15
Fig. 2.9: Higher byte address on A8 – A15
46. Draw the appearance of data in the read and write machine cycles.
Ans. Data transfer from memory or I/O device to microprocessor or the reverse takes place
during T2 and T3 states of the machine cycles.
In the read machine cycle, data appears at the beginning of T3 state, whereas in the
write machine cycle, it appears at the beginning of T2, shown in Fig. 2.10.
T1 T2 T3 T1 T2 T3
AD0 – AD7 Address Data Address Data
(a) Read machine cycle (b) Write machine cycle
Fig. 2.10: Data bus
47. Draw the status signals during opcode fetch and memory read machine cycles.
Ans. The status signals are IO/ M , S0 and S1. Their conditions indicate the type of machine
cycle that the system is currently passing through. These three status signals remain
active right from the beginning till the end of each machine cycle, shown in Fig. 2.11.
IO/M, S0, S1:
T1 T2 T3 T4 T1 T2 T3
AD0 – AD7
IO/M = 0, S0 = 1, S1 = 1 IO/M = 0, S0 = 0, S1 = 1
Opcode Memory read
Fig. 2.11: Status signals

ææ ææ
48. Show the RD and WR signals during the Read cycle and Write cycle.
ææ
Ans. When RD is active, microprocessor reads data from either memory or I/O device while
ææ
when WR is active, it writes data into either memory or I/O device.
Read cycle Write cycle
T1 T2 T3 T1 T2 T3
RD
WR
Fig. 2.12: RD and WR signals

Data transfer (reading/writing) takes place during T2 and T3 states of read cycle or
write cycle and is shown in Fig. 2.12.
49. Indicate the different machine cycles of 8085.
Ans. 8085 has seven different machine cycles. These are:
(1) Opcode Fetch (2) Memory Read (3) Memory Write (4) I/O Read (5) I/O Write
(6) Interrupt Acknowledge (7) Bus Idle.
50. Draw the Opcode Fetch machine cycle of 8085 and discuss.
Ans. The first machine cycle of every instruction is the Opcode Fetch. This indicates the kind
of instruction to be executed by the system. The length of this machine cycle varies
between 4T to 6T states—it depends on the type of instruction. In this, the processor
places the contents of the PC on the address lines, identifies the nature of machine cycle
æ
(by IO/ M , S0, S1) and activates the ALE signal. All these occur in T1 state.
Opcode fetch
T1 T2 T3 T4
CLK
A15
High order memory address Unspecified
A8
A7
Low order Opcode
A0
Memory address
ALE
IO/M Status IO/M = 0, S0 = 1, S1 = 1 Opcode fetch
RD
Fig. 2.13: Opcode fetch machine cycle

ææ
In T2 state, RD signal is activated so that the identified memory location is read from
and places the content on the data bus (D0 – D7).
In T3, data on the data bus is put into the instruction register (IR) and also raises
ææ
the RD signal thereby disabling the memory.
In T4, the processor takes the decision, on the basis of decoding the IR, whether to
enter into T5 and T6 or to enter T1 of the next machine cycle.
One byte instructions that operate on eight bit data are executed in T4. Examples are
ADD B, MOV C, B, RRC, DCR C, etc.
51. Briefly describe Memory Read and Write machine cycles and show the wave-
forms.
Memory read Opcode write
T1 T2 T3 T1 T2 T3
CLK CLK
A15 – A8 Memory address A15 – A8 Memory address
ALE ALE
A7 – AD0 A7 – A0 Data from memory A7 – AD0 A7 – A 0 Data from CPU
IO/M, S1, S0 IO/M = 0, S1 = 1, S0 = 0 IO/M IO/M = 0, S1 = 0, S0 = 1
RD WR
(a) Memory read machine cycle (b) Memory write machine cycle
Fig. 2.14: Memory read and write machine cycle
Ans. Both the Memory Read and Memory Write machine cycles are 3T states in length. In
Memory Read the contents of R/W memory (including stack also) or ROM are read while
in Memory Write, it stores data into data memory (including stack memory).
As is evident from Fig. 2.14 during T2 and T3 states data from either memory or CPU
are made available in Memory Read or Memory Write machine cycles respectively. The
status signal (IO/ M , S0, S1) states are complementary in nature in Memory Read and
Memory Write cycles. Reading or writing operations are performed in T2.
In T3 of Memory Read, data from data bus are placed into the specified register (A,
ææ
B, C, etc.) and raises RD so that memory is disabled while in T3 of Memory Write
æææ
WR signal is raised which disables the memory.
52. Draw the I/O Read and I/O Write machine cycles and discuss.
Ans. I/O Read and Write machine cycles are almost similar to Memory Read and Write
machine cycles respectively. The difference here is in the IO/ M signal status which
remains 1 indicating that these machine cycles are related to I/O operations. These
machine cycles take 3T states.
In I/O read, data are available in T2 and T3 states, while during the same time (T2
and T3) data from CPU are made available in I/O write.
The I/O read and write machine cycles are shown in Fig. 2.15.
I/O Read I/O Write
T1 T2 T3 T1 T2 T3
CLK CLK
ALE ALE
A15 – A8 I/O Addr A15 – A8 I/O Addr
A7 – AD0 I/O Addr I/O Data A7 – AD0 I/O Addr I/O Data
RD WR
IO/M,S1,S0 IO/M = 1, S1 = 1, S0 = 0 IO/M,S1,S0 IO/M = 1, S1 = 0, S0 = 1
(a) I/O read machine cycle (b) I/O write machine cycle
Fig. 2.15: I/O read and write machine cycles

53. Draw the Interrupt Acknowledge cycles for (a) RST instruction (b) CALL
instruction.
Ans. The following figure shows the Interrupt Acknowledge cycle for RST instruction.
Restart instruction
M1 M2 M3
T1 T2 T3 T4 T5 T6 T1 T2 T3 T1 T2 T3
CLOCK
A8 –A15 PCH (SP-1)H (SP-2)H
AD0 –AD7 PCL RST (SP-1)L D0–D7(PCH) (SP-2)L D0–D7(PCL)
ALE
INTA
INTA
IO/M,S1,S0 (1,1,1) (0,0,1) (0,0,1)
RD
WR
Fig. 2.16: Restart instruction

In M1, RST is decoded. This initiates a CALL to the specific vector location. Contents
of the PC are stored in stack in machine cycles M2 and M3.
M1 M2 M3 M4 M5
T1 T2 T3 T4 T5 T6 T1 T2 T3 T1 T2 T3 T1 T2 T3 T1 T2 T3
Clock
Higher order Higher order Higher order PCH PCH

A8–A15 Unspecified address byte address byte
address byte
Address Opcode Data Data PCL Data PCL Data
ALE
IO/M,S1,S0 IO/M =1, S1 =1,S0 = 1 IO/M =1, S1 =1,S0 = 1 IO/M =1, S1 =1,S0 = 1 IO/M =1, S1 =0,S0 = 1 IO/M =1, S1 =0,S0 = 1
INTA
WR
Fig. 2.17: Timing diagram of INTA machine cycle and execution of call instruction
The above figure shows an Interrupt Acknowledge cycle for CALL instruction. M2 and
M3 machine cycles are required to call the 2 bytes of the address following the CALL.
Memory write are done in machine cycles M4 and M5 in which contents of PC are stored
in stack and then a new instruction cycle begins.
54. What is meant by Bus Idle Machine cycle?
Ans. There are a few situations in which machine cycles are neither Read or Written into.
These are called Bus Idle Machine cycle.
Such situations arise when the system executes a DAD or during the internal opcode
generation for the RST or TRAP interrupts.
The ALE signal changes state during T1 of each machine cycle, but in Bus Idle
Machine cycles, ALE does not change state.
55. Explain the DAD instruction and draw its timing diagram.
Ans. DAD instruction adds the contents of a specified register pair to the contents of H and L.
For execution of DAD, 10 T-states are needed. Instead of having a single machine
cycle having 10 T-states, it consists of the Opcode Fetch machine cycle (4T states) and 6
extra T-states divided into two machine cycles. These two extra machine cycles are Bus
Idle Machine cycles which do not involve either memory or I/O.
The timing diagram for DAD instruction is shown below:

Instruction cycle of DAD Instruction
Opcode Fetch Bus Idle

T1 T2 T3 T4 T1 T2 T3 T4 T5 T6
CLOCK
ALE
A15 – A8 A15 – A8 Unspecified Unspecified Unspecified

Opcode
AD7 – AD0 A15 – A8
for DAD
IO/M, S1 , S0 IO/M =0, S1 = 1, S0 =1 IO/M = 0, S1 = 0, S0 = 0
RD
WR
INTA
Fig. 2.18: Timing diagram for DAD instruction
56. Discuss the concept of WAIT states in microprocessors.

Ans. So many times it may happen that there is speed incompatibility between microprocessor
and its memory and I/O systems. Mostly the microprocessor is having higher speed.
So in a given situation, if the microprocessor is ready to accept data from a peripheral
device while there is no valid data in the device (e.g. an ADC), then the system enters
into WAIT states and the READY pin (an input pin to the microprocessor, pin no. 35 for
8085) is put to a low state by the device.
Once the device becomes ready with some valid data, it withdraws the low state on
the READY pin of 8085. Then 8085 accepts the data from the peripheral by software
instructions.
57. Does 8085 have multiplication and division instructions?
Ans. No, 8085 does not have the above two instructions. It can neither multiply nor divide two
8-bit numbers. The same are executed by the processor following the process of repetitive
addition or subtraction respectively.
58. Indicate the bus drive capability of 8085.
Ans. 8085 buses can source up to 400 mA and sink 2 mA of current. Hence 8085 buses can drive
a maximum of one TTL load.
Thus the buses need bus drivers/buffers to enhance the driving capability of the buses
to ensure that the voltage levels are maintained at appropriate levels and malfunctioning
is avoided.
59. What are the buffers needed with the buses of 8085?
Ans. An 8-bit unidirectional buffer 74LS244 is used to buffer the higher order address bus
(A8 – A15). It consists of eight non-inverting buffers with tri-state outputs. Each pin can
sink 24 mA and source 15 mA of current.
A bidirectional buffer 74LS245 (also called octal bus transreceivers) can be used to
drive the bidirectional data bus (D0 – D7) after its demultiplexing. The DIR pin of the IC
controls the direction of flow of data through it.
60. Explain the instruction cycle of a microprocessor.
Ans. When a processor executes a program, the instructions (1 or 2 or 3 bytes in length) are
executed sequentially by the system. The time taken
Instruction cycle
by the processor to complete one instruction is called
Execute
the Instruction Cycle (IC).
Fetch cycle
cycle
Clock
An IC consists of Fetch Cycle (FC) and an
Execute Cycle (EC). Thus IC = FC + EC. It is shown

FC EC
in Fig. 2.19. Depending on the type of instruction, IC
IC
time varies.
61. Explain a typical fetch cycle (FC).

Fig. 2.19: Instruction cycle showing
Ans. The time required to fetch an opcode from a memory FC, EC and IC
location is called Fetch Cycle.
A typical FC may consist of 3T states. In the first T-state, the memory address,
residing in the PC, is sent to the memory. The content of the addressed memory (i.e.,
the opcode residing in that memory location) is read in the second T-state, while in the
third T-state this opcode is sent via the data bus to the instruction register (IR). For slow
memories, it may take more time in which case the processor goes into ‘wait cycles’. Most
microprocessors have provision of wait cycles to cope with slow memories.
A typical FC may look like the following:
Send Reading Transferring
address the opcode opcode to
to memory from memory µP via DB
T1 T2 T3
Fig. 2.20: A typical FC
62. Draw the block schematic of a typical Instruction Word flow diagram and
explain the same.
Ans. There are two kinds of words—instruction word and data word. The 2 byte content of the
PC is transferred to a special register–called memory address register (MAR) or simply
address register (AR) at the beginning of the fetch cycle. Since the content of MAR is an
address, it is thus sent to memory via the address bus. The content of the addressed
memory is then read under ‘Read control’ generated by T&C section of the
microprocessor. This is then sent via the data bus to the memory data register (MDR)
or simply data register (DR) existing in CPU. This is placed in the instruction register
(IR) and is decoded by the instruction decoder and subsequently executed. The PC is then
incremented if the subsequent memory location is to be accessed.
Instruction
register MDR
Data bus
Instruction
CPU Memory
decoder
Program
Control MAR
counter
Address
bus
Fig. 2.21: Flow of instruction word
63. Draw the block schematic of a typical data word flow diagram and explain the
same.
Ans. The data word flows via the data bus into the accumulator. The data source can be a
memory device or an input device. Data from the accumulator is then manipulated in
ALU under control of T & C unit. The manipulated data is then put back in the
accumulator and can be sent to memory or output devices.
The block schematic of data flow is shown below.
Data Bus
Accumulator
Registers Output
ALU
bus
CPU Control
Fig. 2.22: Flow of data word
64. Why a microprocessor based system is called a sequential machine?

Ans. It can perform the jobs in a sequential manner, one after the other. That is why it is called
a sequential machine.
65. Why a microprocessor based system is called a synchronous one?
Ans. All activities pertaining to the µP takes place in synchronism with the clock. Hence it
is called a synchronous device.
66. Draw the diagram which will show the three buses separately, with the help
of peripheral ICs.
Ans. The scheme of connections is shown below. The octal bus driver 74LS244 drives the higher
order address bus A15 – A8 while 74LS373 Latch drives the lower order address bus
A7 – A0 (with the help of ALE signal). The bidirectional bus driver 74LS245 drives the
data bus D7 – D0 while the 74LS138 (a 3 to 8 decoder chip) outputs at its output pins IOW,
IOR, MEMW, MEMR, signals from the IO/ M , RD and WR signals of 8085.
8
74LS244
A15 – A8 A15 – A8
Octal bus driver
(21-28) (Higher order address bus)
�P
8 8 74LS373
0 AD7 – A 0 Latch A7 – A0
8 (12 – 19) G (Lower order address bus)
5
(30) ALE
74LS245
Bidirectional D7 – D0
bus driver (Data bus)
74LS138 IOW
(34) IO/M 3-to-8
IOR Control
(32) RD Decoder
MEMW bus
(31) WR G2
MEMR
(3) Reset
out
Fig. 2.23: Demultiplexing of address/data bus and separation of control signals

ææ ææ
67. Discuss the status of WR and RD signals of 8085 at any given instant of time.
ææ ææ
Ans. At any given instant, the status of the two signals WR and RD will be complementary
to each other.
It is known that microprocessor based system is a sequential device, so that at any
given point of time, it does the job of reading or writing—and definitely not the two jobs
ææ ææ
at the same time. Hence if WR signal is low, then RD signal must be high or vice versa.
68. Which registers of 8085 are programmable?
Ans. The registers B, C, D, E, H and L are programmable registers in that their contents can
be changed by programming.
69. Suggest the type of operations possible on data from a look of the architecture
of 8085.
Ans. The architecture of 8085 suggests that the following operations are possible.
z Can store 8-bits of data.

z Uses the temporary registers during arithmetic operations.
z Checks for the condition of resultant data (like carry, etc.) from the flag register.
70. What are the externally initiated operations supported by 8085?
Ans. 8085 supports several externally initiated operations at some of its pins. The operations
possible are:
z Resetting (via Reset pin)
z Interruptions (via Trap, RST 7.5, RST 6.5, RST 5.5 and Interrupt pin)
z Ready (via Ready pin)

z Hold (via Hold pin)
71. Name the registers not accessible to the programmer.

Ans. The following registers cannot be accessed by the programmer:
z Instruction register (IR)
z Memory address register (MAR) or supply address register (AR)
z Temporary registers.
72. Name the special purpose registers of 8085.

Ans. The special purpose registers used in 8085 microprocessor are:
z Accumulator register (A)
z Program counter register (PC)
z Status (or Flag) register

z Stack pointer register (SP)
73. What should be the size of the Instruction Register if an arbitrary microprocessor
has only 25 instructions?
Ans. The length of the Instruction Register would be 5-bits, since 25 = 32, since a 5-bit
Instruction Register can decode a maximum of 32 instructions.
74. Explain the difference between HLT and HOLD states.
Ans. HLT is a software instruction. Its execution stops the processor which enters into a HALT
state and the buses of the processor are driven into tri-state.
HOLD is an hardware input to the processor. When HOLD input = 1, the processor
goes into the HOLD state, but the buses don’t go into tri-state. The processor gives out
a high HLDA (hold acknowledge) signal which can be utilised by an external device (like
a DMA controller) to take control of the processor buses. HOLD is acknowledged by the
processor at the end of the current instruction execution.
75. Indicate the length of the Program Counter (PC) to access 1 KB and 1 MB
memory.
Ans. 1 KB = 1024 bytes
and 1 MB = 1024 K bytes
Thus the required number of bits in PC to access 1 KB are 10 (210 = 1024) and 20
20
(2 = 1024 K) respectively.
76. What determines the number of bytes to be fetched from memory to execute an
instruction?
Ans. An instruction normally consists of two fields. These are:
Opcode Operand
Thus, while the system starts executing an instruction, it first decodes the opcode
which then decides how many more bytes are to be brought from the memory—its
minimum value is zero (like RAR) while the maximum value is two (like STA 4059 H).
77. A Microprocessor’s control logic is ‘microprogrammed’. Explain.
Ans. It implies that the architecture of the control logic is much alike the architecture of a
very special purpose microprocessor.
78. Does the ALU have any storage facility?
Ans. No, it does not have any storage facility. For this reason, the need for temporary data
registers arise in ALU–it has two inputs: one provided by the accumulator and the other
from the temporary data register. The result of summation is stored in the accumulator.
3
Instruction Types and
Timing Diagrams
1. What is an instruction?
Ans. An instruction is a command given to the microcomputer to perform a specific task or
function on a given data.
2. What is meant by instruction set?
Ans. An instruction set is a collection of instructions that the microprocessor is designed to
perform.
3. In how many categories the instructions of 8085 be classified?
Ans. Functionally, the instructions can be classified into five groups:
z data transfer (copy) group
z arithmetic group
z logical group
z branch group
z stack, I/O and machine control group.
4. What are the different types of data transfer operations possible?

Ans. The different types of data transfer operations possible are cited below:
z Between two registers.
z Between a register and a memory location.
z A data byte can be transferred between a register and a memory location.
z Between an I/O device and the accumulator.
z Between a register pair and the stack.
The term ‘data transfer’ is a misnomer—actually data is not transferred, but copied
from source to destination.
5. Mention the different types of operations possible with arithmetic, logical,
branch and machine control operations.
Ans. The arithmetic operations possible are addition, subtraction, increment and decrement.
The logical operations include AND, OR, EXOR, compare, complement, while branch
operations are Jump, Call, Return and Restart instructions.
The machine control operations are Halt, Interrupt and NOP (no operation).
6. What are the different instruction word sizes in 8085?
Ans. The instruction word sizes are of the following types:
Instruction Types and Timing Diagrams 33
z 1-byte instruction
z 2-byte instruction
z 3-byte instruction.
7. What an instruction essentially consists of?
Ans. An instruction comprises of an operation code (called ‘opcode’) and the address of the data
(called ‘operand’), on which the opcode operates. This is the structure on which an
instruction is based. The opcode specifies the nature of the task to be performed by an
instruction. Symbolically, an instruction looks like
Operation code Address of data

opcode operand
8. Give one example each of 1-byte, 2-byte and 3-byte instructions.
Ans. The examples are given below:
z 1-byte instruction : ADD B
z 2-byte instruction : MVIC, 07
z 3-byte instruction : LDA 4400
In 1-byte instruction, the opcode and the operand are in the same byte i.e.,
Opcode/Operand
A 2-byte instruction looks like this:

Opcode 1st byte
Operand or data/address 2nd byte
While a 3-byte instruction looks like the following:

Opcode 1st byte
Low order byte of address 2nd byte
High order byte of address 3rd byte

9. What is meant by ‘addressing mode’? Mention the different addressing modes.
Ans. Each instruction indicates an operation to be performed on certain data. There are
various methods to specify the data for the instructions, known as ‘addressing modes’.
For 8085 microprocessor, there are five addressing modes. These are:
z Direct addressing
z Register addressing
z Register indirect addressing
z Immediate addressing
z Implicit addressing.
10. Give one example each of the five types of addressing modes.
Ans. The examples for each type of addressing mode are given below:
(a) Direct Addressing: In this mode, the operand is specified within the instruction itself.
Examples of this type are:

LDA 4000H, STA 5513H, etc.
IN/OUT instructions (like IN PORT C, OUT PORT B, etc.) also falls under this
category.
(b) Register Addressing: In this mode of addressing, the operand are in the general
purpose registers.
Examples are: MOV A, B ; ADD D, etc.
(c) Register Indirect Addressing: MOV A, M; ADD M are examples of this mode of
addressing. These instructions utilise 1-byte. In this mode, instead of specifying a
register, a register pair is specified to accommodate the 16-bit address of the operand.
(d) Immediate Addressing: MVI A, 07; ADI 0F are examples of Immediate Addressing
mode.
The operand is specified in the instruction in this mode. Here, the operand address
is not specified.
(e) Implicit Addressing: In this mode of addressing, the operand is fully absent. Examples
are RAR, RAL, CMA, etc.
11. Let at the program memory location 4080, the instruction MOV B, A (opcode 47H)
is stored while the accumulator content is FFH. Illustrate the execution of this
instruction by timing diagram.
Ans. Since the program counter sequences the execution of instructions, it is assumed that
the PC holds the address 4080H. While the system Opcode fetch
executes the instruction, the following takes place T1 T2 T3 T4
one after another.
CLK
1. The CPU places the address 4080H (residing
in PC) on the address bus—40H on the high A15
order bus A15 – A8 and 80H on the low order High-order
40H memory address
Unspeci-
fied
A8
bus AD7 – AD0. Low-Order
2. The CPU raises the ALE signal to go high— AD7 80H 47H Opcode
the H to L transition of ALE at the end of the AD0 Memory address
first T state demultiplexes the low order bus.
3. The CPU identifies the nature of the ALE
machine cycle by means of the three status IO/M
S0 Status IO/M = 0, S = 1, S = 1 Opcode Fetch
signals IO/ M , S0 and S1.
0 1
S1
IO/ = 0, S1 = 1, S0 = 1 RD
4. In T2, memory is enabled by the signal.

The content of PC i.e., 47H is placed on the
Fig. 3.1: 8085 timing for execution of the
data bus. PC is incremented to 4081H. instruction (MOV B, A)
5. In T3, CPU reads 47H and places it in the
instruction register.
6. In T 4, CPU decodes the instruction, places FF H (accumulator content) in the
temporary register and then transfers it to register B. Figure 3.1 shows the
execution of the above. It consists of 4T states.
4
8085 Interrupts
1. Mention the interrupt pins of 8085.

Ans. There are five (5) interrupt pins of 8085—from pin 6 to pin 10. They represent TRAP,
RST 7.5, RST 6.5, RST 5.5 and INTR interrupts respectively. These five interrupts are
‘hardware’ interrupts.
2. Explain maskable and non-maskable interrupts.
Ans. An interrupt which can be disabled by software means, is called a maskable interrupt.
Thus an interrupt which cannot be masked is an unmaskable interrupt.
3. Which is the non-maskable interrupt for 8085?
Ans. TRAP interrupt is the non-maskable interrupt for 8085. It means that if an interrupt
comes via TRAP, 8085 will have to recognise the interrupt.
4. Do the interrupts of 8085 have priority?
Ans. Yes, the interrupts of 8085 have their priorities fixed—TRAP interrupt has the highest
priority, followed by RST 7.5, RST 6.5, RST 5.5 and lastly INTR.
5. What is meant by priority of interrupts?
Ans. It means that if 8085 is interrupted by more than one interrupt at the same time, the
one which is having highest priority will be serviced first, followed by the one(s) which
is (are) having just next priority and so on.
For example, if 8085 is interrupted by RST 7.5, INTR and RST 5.5 at the same time,
then the sequence in which the interrupts are going to be serviced are as follows: RST
7.5, RST 5.5 and INTR respectively.
6. Mention the types of interrupts that Table 4.1: Vector addresses for software
8085 supports. interrupts
Ans. 8085 supports two types of interrupts— Instruction Corresponding Vector
hardware and software interrupts. HEX code addresses
RST 0 C7 0000H
7. What are the software interrupts of
RST 1 CF 0008H
8085? Mention the instructions, their RST 2 D7 0010H
hex codes and the corresponding RST 3 DF 0018H
vector addresses. RST 4 E7 0020H
Ans. 8085 has eight (8) software interrupts RST 5 EF 0028H
from RST 0 to RST 7. The instructions, RST 6 F7 0030H
RST 7 FF 0038H
hex codes and the vector locations are
tabulated in Table 4.1:
8. How the vector address for a software interrupt is determined?

Ans. The vector address for a software interrupt is calculated as follows:
Vector address = interrupt number × 8
For example, the vector address for RST 5 is calculated as
5 × 8 = 40)10 = 28)H
\ Vector address for RST 5 is 0028H.
9. In what way INTR is different from the other four hardware interrupts?
Ans. There are two differences, which are discussed below:
1. While INTR is not a vectored interrupt, the other four, viz., TRAP, RST 7.5, RST
6.5 and RST 5.5 are all vectored interrupts. Thus whenever an interrupt comes via
any one of these four interrupts, the internal control circuit of 8085 produces a
CALL to a predetermined vector location. At these vector locations the subroutines
are written.
On the other hand, INTR receives the address of the subroutine from the external
device itself.
2. Whenever an interrupt occurs via TRAP, RST 7.5, RST 6.5 or RST 5.5, the
corresponding returns address (existing in program counter) is auto-saved in STACK,
but this is not so in case of INTR interrupt.
10. Indicate the nature of signals that will trigger TRAP, RST 7.5, RST 6.5, RST 5.5
and INTR.
Ans. TRAP interrupt is both positive edge and level triggered, RST 7.5 is positive edge
triggered while RST 6.5, RST 5.5 and INTR are all level triggered.
11. Why the TRAP input is edge and level sensitive?
Ans. TRAP input is edge and level sensitive to avoid false triggering caused by noise and
transients.
12. Draw the TRAP interrupt circuit diagram and explain the same.
Ans.
1
D Q Call 2400H
Trap Input
CLR Q
Reset in
Trap
acknowledge
Fig. 4.1: The TRAP interrupt circuit
The positive edge of the TRAP signal sets the D F/F, so that Q becomes 1. It thus enables
the AND gate, but the AND gate will output a 1 for a sustained high level at the TRAP
input. In that case the subroutine written at vector memory location 2400H corresponding
to TRAP interrupt starts executing. 2400H is the starting address of an interrupt service
routine for TRAP.
8085 Interrupts 37
There are two ways to clear the TRAP interrupt:

1. When the microprocessor is resetted, then via the inverter, a high comes out of the
OR gate, thereby clearing the D F/F, making Q = 0.
2. When a TRAP is acknowledged by the system, an internal TRAP ACKNOWLEDGE
is generated thereby clearing the D F/F.
13. Discuss the INTR interrupt of 8085.
Ans. The following are the characteristics of INTR interrupt of 8085:
z It is a maskable interrupt
z It has lowest priority
z It is a non-vectored interrupt.
Sequentially, the following occurs when INTR signal goes high:
1. 8085 checks the status of INTR signal during execution of each instruction.
2. If INTR signal remains high till the completion
æææææ
of an instruction, then 8085 sends
out an active
æ
ææææ
low interrupt acknowledge (INTA) signal.
3. When INTA signal goes low, external logic places an instruction OPCODE on the
data bus.
4. On receiving the instruction, 8085 saves the address of next instruction (which
would have otherwise been executed) in the STACK and starts executing the ISS
(interrupt service subroutine).
14. Draw the diagram that outputs RST 3 instruction opcode on acknowledging the
interrupt.
Ans. The diagram is shown below:
+5V
1K
74LS244
DB0
DB1
DB2
DB3
DB4
DB5
DB6
DB7
INTA
Fig. 4.2: Hardware circuit diagram to implement RST 3 (Opcode DF H)

æ
ææææ
When an INTR is acknowledged by the microprocessor, it outputs a low INTA . Thus
an RST 3 is gated onto the system bus. The processor then first saves the PC in the STACK
and branches to the location 0018H. At this address, the ISS begins and ends with a RET
instruction. On encountering RET instruction in the last line of the ISS, the return address
saved in the stack is restored in the PC so that normal processing in the main program
(at the address which was left off when the program branched to ISS) begins.
15. What is to be done if a particular part of a program is not to be interrupted by
RST 7.5, RST 6.5, RST 5.5 and INTR?
Ans. Two software instructions—EI and DI are used at the beginning and end of the particular
portion of the program respectively. The scheme is shown schematically as follows:
Main program
Dl
Program portion not to be

interrupted by RST 7.5, RST 6.5, RST 5.5
and INTR
El
Fig. 4.3: Employing EI and DI in a program
16. Explain the software instructions EI and DI.

Ans. The EI instruction sets the interrupt enable flip-flop, thereby enabling RST 7.5, RST 6.5,
RST 5.5 and INTR interrupts.
The DI instruction resets the interrupt enable flip-flop, thereby disabling RST 7.5,
RST 6.5, RST 5.5 and INTR interrupts.
17. When returning back to the main program from Interrupt Service Subroutine
(ISS), the software instruction EI is inserted at the end of the ISS. Why?
Ans. When an interrupt (either via RST 7.5, RST 6.5, RST 5.5, INTR) is acknowledged by
the microprocessor, ‘any interrupt acknowledge’ signal resets the interrupt enable
F/F. It thus disables RST 7.5, RST 6.5, RST 5.5 and INTR interrupts. Thus any future
interrupt coming via RST 7.5, RST 6.5, RST 5.5 or INTR will not be acknowledged unless
the software instruction EI is inserted which thereby sets the interrupt enable F/F.
8085 Interrupts 39
18. Mention the ways in which the three interrupts RST 7.5, RST 6.5 and RST 5.5
are disabled?
Ans. The three interrupts can be disabled in the following manner:
z Software instruction DI
æææææææ
z RESET IN signal
z Any interrupt acknowledge signal.
19. Draw the SIM instruction format and discuss.
Ans. The SIM instruction format is shown below:
Serial output
control
Interrupt control logic
D7 D6 D5 D4 D3 D2 D1 D0
SOD SOE X R7.5 MSE M¢7.5 M¢6.5 M¢5.5
Masking
Serial output data bits Set up 5.5 mask
SOD enable Set up 6.5 mask
Set up 7.5 mask
Mask set enable
Reset RST 7.5
Fig. 4.4: The SIM instruction format
D7 and D6 bits are utilised for serial outputting of data from accumulator. D5 bit is a
don’t care bit, while bits D4–D0 are used for interrupt control.
D4 bit can clear the D F/F associated with RST 7.5.
D 3 bit is mask set enable (MSE) bit, while bits D 2 –D 0 are the masking bits
corresponding to RST 7.5, RST 6.5 and RST 5.5 respectively.
None of the flags are affected by SIM instruction.
By employing SIM instruction, the three interrupts RST 7.5, RST 6.5 and RST 5.5
can be masked or unmasked. For masking any one of these three interrupts, MSE (i.e.,
bit D3) bit must be 1.
For example let RST 7.5 is to be masked (disabled), while RST 6.5 and RST 5.5 are
to be unmasked (enabled), then the content of the bits of the SIM instruction will be like
0000 1100 = 0CH
For this to be effective the following two instructions are written,
MVI A, 0CH
SIM
Execution of SIM instruction allows copying of the contents of the accumulator into
the interrupt masks.
20. Show the RIM instruction format and discuss the same.
Ans. RIM stands for ‘Read interrupt mask’ and its format is as follows:
Pending
Serial input interrupts Interrupt masks
SID I 7.5 I 6.5 I 5.5 IE M�7.5 M�6.5 M�5.5
Serial input
data Interrupt masks
1 = masked
0 = unmasked
Interrupt enable flag
Pending interrupts 1 = enable
1 = pending 0 = disable
Fig. 4.5: The RIM instruction format
When RIM instruction is executed in software, the status of SID, pending interrupts
and interrupt masks are loaded into the accumulator. Thus their status can be monitored.
It may so happen that when one interrupt is being serviced, other interrupt(s) may occur.
The status of these pending interrupts can be monitored by the RIM instruction. None
of the flags are affected by RIM instruction.
21. Write a program which will call the interrupt service subroutine (at 3C00H)
corresponding to RST 7.5 if it is pending. Let the ACC content is 20 H on
executing the RIM instruction.
Ans. The program for the above will be as hereunder:
SID I7.5 I6.5 I5.5 IE M¢7.5 M¢6.5 M¢5.5
RIM fi ACC
0 0 1 0 0 0 0 0
ANI 40 H fi AND immediate

0 1 0 0 0 0 0 0
with 40 H
Isolate 17.5 bit

CNZ 3C00H fi Call interrupt service subroutine corresponding to RST 7.5, if RST 7.5
is pending.
22. Write a program which will call the subroutine (say named ‘SR’) if RST 6.5 is
masked. Let content of ACC is 20 H on executing the RIM instruction.
Ans. RST 6.5 is masked if bit M¢6.5 (D1 bit of RIM) is a 1 and also D3 bit (i.e., IE) is 1
The program for the above will be as hereunder:
SID I7.5 I6.5 I5.5 IE M¢7.5 M¢6.5 M¢5.5
RIM fi ACC
0 0 1 0 0 0 0 0
ANI 0A H fi AND immediate 0 0 0 0 1 0 1 0

with 0A H
Isolate M¢6.5 bit
JNZ SR fi Jump to subroutine “SR” if RST 6.5 is masked.
8085 Interrupts 41
23. For what purpose TRAP interrupt is normally used?

Ans. TRAP interrupt is a non-maskable one i.e., if an interrupt comes via the TRAP input,
the system will have to acknowledge that. That is why it is used for vital purposes which
require immediate attention like power failure.
If the microprocessor based system loses power, the filter capacitors hold the supply
voltage for several mili seconds.
During this time, data in the RAM can be written in a disk or E2PROM for future
usage.
24. Draw the interrupt circuit diagram for 8085 and explain.
Ans. Figure 4.6 is the interrupt circuit diagram of 8085. It shows the five hardware interrupts
TRAP, RST 7.5, RST 6.5, RST 5.5 and INTR along with the software interrupts RST n:
n = 0 to 7.
Trap is both edge and level sensitive interrupt. A short pulse should be applied at the
trap input, but the pulse width must be greater than a normal noise pulse width and also
long enough for the µP to complete its current instruction. The trap input must come
down to low level for it to be recognised for the second time by the system. It is having
highest priority.
Next highest priority interrupt is RST 7.5 which responds to the positive edge
(low to high transition) of a pulse. Like trap, it also has a D F/F whose output becomes
1 on accepting the RST 7.5 input, but final call to vector location 3C00H is reached only
if RST 7.5 remains unmasked and the program has an EI instruction inserted already.
These are evident from the circuit. If R 7.5 (bit D4 of SIM instruction) is 1, then RST 7.5
instruction will be overlooked i.e., it can override any RST 7.5 interrupt.
Like RST 7.5, final call locations 3400H and 2C00H corresponding to interrupts at
RST 6.5 and RST 5.5 are reached only if the two interrupts remain unmasked and the
software instruction EI is inserted.
The three interrupts RST 7.5, RST 6.5 and RST 5.5 are disabled once the system
accepts an interrupt input via any one of these pins—this is because of the generation
of ‘any interrupt acknowledge’ signal which disables them.
Any of the software RST instructions (RST n : n = 0 to 7) can be utilised by using
INTR instruction and hardware logic. RST instructions are utilised in breakpoint service
routine to check register(s) or memory contents at any point in the program.
25. The process of interrupt is asynchronous in nature. Why?
Ans. Interrupts may come and be acknowledged (provided masking of any interrupt is not
done) by the microprocessor without any reference to the system clock. That is why
interrupts are asynchronous in nature.
26. In how many categories can ‘interrupt requests’ be classified?
Ans. The ‘interrupt requests’ can be classified into two categories—maskable interrupt and
non-maskable interrupt.
A maskable interrupt can either be ignored or delayed as per the needs of the system
while a non-maskable interrupt has to be acknowledged.
42
Trap is both edge Short pulse but greater than noise
and level sensitive pulse and long enough that microprocessor
Understanding 8085/8086 Microprocessors and Peripheral ICs through Questions and Answers
(Trap responds to a is able to complete the current
positive edge and instruction, so that IC 7 stays enabled
a sustained high
level at its Trap input)
CLK
Reset In CLR
Trap
4
RST 7.5 responds to
positive edge of a pulse
R 7.5 comes from D4 bit I 7.5
of SIM.If D4 = 1
CLK
then RST7.5 F/F (here,
IC2) is reset.It is used to CLR
override RST7.5
without servicing it.
Acknowledge
RST 6.5 RST 5.5

respond to high
level only
Trap F/F can be cleared by

1 Reset In Pin 9 M5.5
2 Trap acknowledge RST 5.5
RST0(0000H)
Get
EI RST
from
DI ext
When any of the four interrupts hard
occur- Trap, RST 7.5, 6.5 or 5.5, ware
then return address (in PC) is auto 16
saved in stack before branching
to the specific address occurs.
Fig. 4.6: Interrupt circuit description

8085 Interrupts 43
27. When the interrupt pins of 8085 are checked by the system?
Ans. Microprocessor checks (samples) interrupt pins one cycle before the completion of an
instruction cycle, on the falling edge of the clock pulse. An interrupt occurring at least
160 ns (150 ns for 8085A-2, since it is faster in operation) before the sampling time is
treated as a valid interrupt.
28. Is there a minimum pulse width required for the INTR signal?
Ans. Microprocessor issues a low INTA signal as an acknowledgement on receiving an INTR
interrupt input signal. A CALL instruction is then issued so that the program branches
to Interrupt Service Subroutine (ISS). Now the CALL requires 18 T-states to complete.
Hence the INTR pulse must remain high for at least 17.5 T-states. If 8085 is operated
at 3 MHz clock frequency, then the INTR pulse must remain high for at least 5.8 mS.
29. Can the microprocessor be interrupted before completion of existing Interrupt
Service Subroutine (ISS)?
Ans. Yes, the microprocessor can be interrupted before the completion of the existing ISS.
Let after acknowledging the INTR, the microprocessor is in the ISS executing
instructions one by one. Now, for a given situation, if the interrupt system is enabled
(by inserting an EI instruction) just after entering the ISS, the system can be interrupted
again while it is in the first ISS.
If an interrupt service subroutine be interrupted again then this is called ‘nested
interrupt’.
30. Bring out one basic difference between SIM and DI instructions.
Ans. While by using SIM instruction any combinations or all of RST 7.5, RST 6.5 and RST 5.5
can be disabled, on the other hand DI disables RST 7.5, RST 6.5, RST 5.5 and in addition
INTR interrupt also.
31. What is RIM instruction and what does it do?
Ans. The instruction RIM stands for Read Interrupt Mask. By executing this instruction in
software, it is possible to know the status of interrupt mask, pending interrupt(s), and
serial input.
32. In an interrupt driven system, EI instruction should be incorporated at the
beginning of the program. Why?
Ans. A program, written by a programmer in the RAM location, is started first by system reset
and loading the PC with the starting address of the program.
Now, with a system reset, all maskable interrupts are disabled. Hence, an EI
instruction must be put in at the beginning of the program so that the maskable
interrupts, which should remain unmasked in a program, remain so.
33. How the system can handle multiple interrupts?
Ans. Multiple interrupts can be handled if a separate interrupt is allocated to each peripheral.
The programmable interrupt controller IC 8259 can also be used to handle multiple
interrupts when they are interfaced through INTR.
34. When an interrupt is acknowledged, maskable interrupts are automatically

disabled. Why?
Ans. This is done so that the interrupt service subroutine (ISS) to which the program has
entered on receiving the interrupt, has a chance to complete its own task.
35. What is meant by ‘nested interrupts’? What care must be taken while handling
nested interrupts?
Ans. Interrupts occurring within interrupts are called ‘nested interrupts’.
While handling nested interrupts, care must be taken to see that the stack does not
grow to such an extent as to foul the main program—in that case the system program
fails.
36. ‘A RIM instruction should be performed immediately after TRAP occurs’—Why?
Ans. This is so as to enable the pre-TRAP interrupt status to be restored with the
implementation of a SIM instruction.
37. What does the D4 bit of SIM do?
Ans. Bit D4 of SIM is R 7.5 which is connected to RST 7.5 F/F via a OR gate. If D4 of SIM is
made a 1, then it resets RST 7.5 F/F. This thus can be used to override RST 7.5 without
servicing it.
38. Comment on the TRAP input of 8085.
Ans. Trap input is both edge and level sensitive. It is a narrow pulse, but the pulse width should
be more than normal noise pulse width. This is done so that noise cannot affect the TRAP
input with a false triggering. Again the pulse width should be such that the TRAP input
which is directly connected to the gate stays high till the completion of current instruction
by the mP. In that case, only the program gets diverted to vector call location 2400 H.
TRAP cannot respond for a second time until the first TRAP goes through a high to
low transition.
TRAP interrupt, once acknowledged, goes to 2400 H vector location without any
external hardware or EI instruction, as is the case for other interrupt signals to be
acknowledged.
39. Discuss about the triggering levels of RST 7.5, RST 6.5 and RST 5.5.
Ans. RST 7.5 is positive edge sensitive and responds to a short trigger pulse. The interrupt
that comes via RST 7.5 is stored in a D F/F, internal to mP. The final vector call location
3C00 H is invoked only if RST 7.5 remains unmasked via SIM and software instruction
EI is inserted in the program.
RST 6.5 and 5.5 respond to high level (i.e., level sensitive) at their input pins—thus
these two pins must remain high until microprocessor completes the execution of the
current instruction.
40. Discuss the utility of RST software instruction.
Ans. For an RST instruction (RST n, n : 0 to 7) to become effective, external hardware is
necessary, along with INTR interrupt instruction.
When debugging is required in a program to know the register(s) or memory
contents, breakpoints are inserted via RST instruction.
8085 Interrupts 45
A breakpoint is an RST instruction inserted in a program. When an RST instruction

is recognised, the program control is transferred to the corresponding RST vector
location. From this vector location, it is again transferred to the breakpoint service
routine so that programmer can check the contents of any register or memory content
on pressing specified key(s). After testing, the routine returns to the breakpoint in the
main program.
Thus RST instructions can be inserted within a program to examine the register/
memory content as per the requirement.
41. Under what condition, an RST instruction is going to be recognised?
Ans. Any RST instruction is recognised only if the EI instruction is incorporated via software.
42. Can the ‘TRAP’ interrupt be disabled by a combination of hardware and
software?
Ans. Yes, it can be disabled by SIM instruction and hardware, as shown in Fig. 4.7.
The following two instructions are executed.
MVIA, 40 H
SIM
It ensures that a ‘0’ logic comes out via SOD pin (pin 4) of 8085. This is then ANDed
with TRAP input.
SOD (pin 4)
To pin 6 (TRAP of 8085)
TRAP signal
from device
Fig. 4.7
Thus pin 6 (TRAP) always remains at ‘0’ logic and hence TRAP input is disabled or
‘MASKED’.
43. Level wise, how the interrupts can be classified? Distinguish them.
Ans. Level wise, interrupts can be classified as
• single level interrupts
• multi level interrupts
Their distinguishing features are shown below:
Single level interrupt Multi level interrupt
1. Interrupts are fed via a single pin 1. Interrupts are fed via different pins of
of microprocessor (like INTR of 8085) microprocessor (like RST 7.5, RST 6.5 etc),
each interrupt requiring a separate pin of
microprocessor.
2. CPU polls the I/O devices to identify 2. Since each interrupt pin corresponds
the interrupting device. to a single I/O device, polling is not
necessary.
3. Slower because of sl. no. 2. 3. Faster because of sl. no. 2.

MPD

Uploaded by

Copyright:

Available Formats

MPD

Uploaded by

Document Information

Copyright

Available Formats

Share this document

Share or Embed Document

Sharing Options

Did you find this document useful?

Is this content inappropriate?

Copyright:

Available Formats

MPD

Uploaded by

Copyright:

Available Formats

MICROPROCESSOR & MICROCONTROLLER LAB MANUAL

ADDITION OF TWO 8 BIT NUMBERS

To perform addition of two 8 bit numbers using 8085.

1) Start the program by loading the first data into Accumulator.

MVI C, 00 Initialize C register to 00

Thus the program to add two 8-bit numbers was executed.

SUBTRACTION OF TWO 8 BIT NUMBERS

To perform the subtraction of two 8 bit numbers using 8085.

1. Start the program by loading the first data into Accumulator.

Thus the program to subtract two 8-bit numbers was executed.

MULTIPLICATION OF TWO 8 BIT NUMBERS

To perform the multiplication of two 8 bit numbers using 8085.

MVI D, 00 Initialize register D to 00

Thus the program to multiply two 8-bit numbers was executed.

DIVISION OF TWO 8 BIT NUMBERS

To perform the division of two 8 bit numbers using 8085.

Output: 01 (4152) ---- Remainder

Thus the program to divide two 8-bit numbers was executed.

LARGEST NUMBER IN AN ARRAY OF DATA

1) Load the address of the first element of the array in HL pair

LXI H,4200 Set pointer for array

Input: 05 (4200) ----- Array Size

SMALLEST NUMBER IN AN ARRAY OF DATA

1) Load the address of the first element of the array in HL pair

LXI H,4200 Set pointer for array

Input: 05 (4200) ----- Array Size

ARRANGE AN ARRAY OF DATA IN ASCENDING ORDER

To write a program to arrange an array of data in ascending order

1. Initialize HL pair as memory pointer

Input: 4200 05 (Array Size)

Output: 4200 05(Array Size)

Thus the given array of data was arranged in ascending order.

ARRANGE AN ARRAY OF DATA IN DESCENDING ORDER

To write a program to arrange an array of data in descending order

1. Initialize HL pair as memory pointer

Input: 4200 05 (Array Size)

Output: 4200 05(Array Size)

Thus the given array of data was arranged in descending order.

BCD TO HEX CONVERSION

1) Initialize memory pointer to 4150 H

Input: 4150 : 02 (MSD)

HEX TO BCD CONVERSION

1) Initialize memory pointer to 4150 H

LXI H,4150 Initialize memory pointer

Output: 4151 : 55 (LSB)

HEX TO ASCII CONVERSION

1. Load the given data in A- register and move to B – register

LDA 4200 Get Hexa Data

Input: 4200 E4(Hexa data)

Output: 4201 34(ASCII Code for 4)

ASCII TO HEX CONVERSION

1. Load the given data in A- register

SQUARE OF A NUMBER USING LOOK UP TABLE

To find the square of the number from 0 to 9 using a Table of Square.

1. Initialize HL pair to point Look up table

LXI H,4125 Initialsie Look up table address

Output: 4151 25 (Square)

Output: 4151: FF (Error Indication)

INTERFACING WITH 8085

INTERFACING 8251 (USART) WITH 8085 PROCESSOR