Introduction to 8085 & it's description(includes basic lab experiments)

INTRODUCTION TO 8085

INTEL 8085 is one of the most popular 8-bit microprocessor capable of addressing 64 KB of memory and its architecture is simple. The device has 40 pins, requires +5 V power supply and can operate with 3MHz single phase clock.

8085 MICROPROCESSOR FUNCTIONAL BLOCK DIAGRAM

Architecture of 8085 Microprocessor

a) General purpose registerIt 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) AccumulatorIt 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 unitIt 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 registerIt is an 8 bit register associated with ALU hold data, entering an operation, used by

the microprocessor and not accessible to programs.

g) FlagsFlag 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 arei) Carry flag (C) ii) Parity flag (P) iii) Zero flag (Z) iv) Auxiliary carry flag (AC) v) Sign flag (S)

Carry flag:If an arithmetic operation results in a carry, the carry flag is set. The carry flag also

serves as a borrow flag for subtraction.

Parity flag:After arithmetic – logic operation, if the result has an even number of 1’s the flag is

set. If it has odd number of 1’s it is reset.

Zero flag:The zero flag is set if the ALU operation results in zero. This flag is modified by the

result in the accumulator as well as in other registers.

Auxiliary carry flag:In an arithmetic operation when a carry is generated by digit D3 and passed on to D4,

the auxiliary flag is set.

Sign flag:After the execution of the arithmetic - logic operation if the bit D7 of the result is 1,

the sign flag is set. This flag is used with signed numbers. If it is 1, it is a negative number and if it is 0, it is a positive number.

h) Timing and control unitSynchronous all microprocessor, operation with the clock and generator and control

signal from it necessary to communicate between controller and peripherals.

i) Instruction register and decoderInstruction is fetched from line memory and stored in line instruction register decoder

the stored information.

j) Register ArrayThese are used to store 8 bit data during execution of some instruction.

PIN Description

Address Bus1. The pins Ao – A15 denote the address bus. 2. They are used for most significant bit

Address / Data Bus1. 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 / M1. 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.

RD1. This is an active low signal 2. This signal is used to control READ operation of the microprocessor.

WR1. WR is also an active low signal 2. Controls the write operation of the microprocessor.

HOLD1. This indicates if any other device is requesting the use of address and data

bus. 2. It indicates whether the hold signal is received or not.

HLDA1. HLDA is the acknowledgement signal for HOLD 2. It indicates whether the hold signal is received or not.

INTR1. INTE is an interrupt request signal 2. IT can be enabled or disabled by using software

INTA1. Whenever the microprocessor receives interrupt signal 2. It has to be acknowledged.

RST 5.5, 6.5, 7.51. These are nothing but the restart interrupts 2. They insert an internal restart junction automatically.

8 BIT ADDITION

AIM:

To write a assembly language program for Addition two (8) bit numbers by

using-8085 micro processor kit.

APPARATUS REQUIRED:

1. 8085 microprocessor kit2. keyboard

ALGORITHM (ADDITION)

Step1 : Start the microprocessor

Step2 : Initialize the carry as ‘Zero’

Step3 : Load the first 8 bit data into the accumulator

Step4 : Copy the contents of accumulator into the register ‘B’

Step5 : Load the second 8 bit data into the accumulator.

Step6 : Add the 2 - 8 bit datas and check for carry.

Step7 : Jump on if no carry

Step8 : Increment carry if there is

Step9 : Store the added request in accumulator

Step10 : More the carry value to accumulator

Step11 : Store the carry value in accumulator

Step12 : Stop the program execution.

Flow chart

NO

YES

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

Increment carry by one

Store the added value in accumulator

Store the value of carry in accumulator

Move the contents of carry into accumulator

END

Check for carry

START

PROGRAM:

ADDRESS

LABEL MNEMONICS OPCODE COMMENTS

4100 MVI C,00 0E,00 Clear the content of C Register.

4102 LDA 4300 3A,00,43Load the content of the accumulator in 4300

4105 MOV B, A 47Move the content from accumulator to B register


4109 ADD B 80Add the content from accumulator to B register

410A JNC 410E ( LOOP 1) D2,0E,41Jump on no carry to the label (LOOP 1)

410D INR C 0C Increment the C register content

410E LOOP 1 STA 4302 32,02,43Store the sum at accumulator at in location 4300

4111 MOV A,C 79Move the carry content from C Register to accumulator

4112 STA 4303 32,03,43Store carry content in accumulator at the location 4303

4115 HLT 76 End the program

InputWithout carry

Input Address Value4300 044301 02

OutputOutput Address Value

4302 064303 00 (carry)

With carryInput Address Value

4300 FF4301 FF

Output Address Value Output 4302 FE

4303 01 (carry)

Calculation 1111 1111 0000 01001111 1111 0000 0010

---------------(1) 1111 1110 0000 0110

=========

F E

8 BIT SUBTRACTION

Aim:To write a assembly language program for subtracting two (8) bit numbers by using-

8085 micro processor kit.

Apparatus required:1. 8085 microprocessor kit2. Keyboard

ALGORITHM


Step2 : Intialize the carry as ‘Zero’

Step3 : Load the first 8 bit data into the accumulator

Step4 : Copy the contents of accumulator into the register ‘B’

Step5 : Load the second 8 bit data into the accumulator.

Step6 : Add the 2 - 8 bit datas and check for carry.


Step8 : Increment carry if there is

Step9 : Store the added request in accumulator

Step10 : More the carry value to accumulator


Step12 : Stop the program execution.

Flow chart

NO

YES

Intialise the carry as zero

Load the 1st 8 bit number

Transfer the 1st number to register ‘B’

Load the 2nd 8 bit number

Subtract the two values

Increment carry by one

1’s compliment of 2nd value

Add 1 to 1’s compliment for 2’s compliment

END

Check for carry

START

Store the value of result in accumulator

Move the carry into the accumulator

Store the value of carry in accumulator

PROGRAM:

ADDRESS LABEL MNEMONICS OPCODE COMMENTS4100 MVI C,00 0E,00 Clear the content of C Register.


4105 MOV B, A 47Move the content from accumulator to B register


4109 SUB B 90Sub the content from accumulator to B register

410A JNC 410E ( LOOP 1) D2,0E,41 Jump on no carry to the label (LOOP 1)410D INR C 0C Increment the C register content

410E LOOP 1 STA 4302 32,02,43Store the content of accumulator at in location 4302

4111 MOV A,C 79Move the carry content from C Register to accumulator

4112 STA 4303 32,03,43Store carry content in accumulator at the location 4303


InputWithout borrow

Input Address Value4300 054301 07


4302 024303 00 (borrow)

With carry borrowInput Address Value

4300 074301 05

Output Address Value4302 024303 01 (borrow)

Calculation 05 – 0707 – 011110000001------1001

05 - 0101------1110 (-2)

ADDITION OF TWO 16 – BIT NUMBERS

AIM:

To write an assembly language program for adding two 16 bit numbers using 8085 micro processor kit.

APPARATUS REQUIRED:

1. 8085 microprocessor kit2. Key board

ALGORITHM (ADDITION):

Step1 : Get the 1st 8 bit in ‘C’ register (LSB) and 2nd 8 bit in ‘H’

Register (MSB) of 16 bit number.

Step2 : Save the 1st 16 bit in ‘DE’ register pair

Step3 : Similarly get the 2nd 16 bit number and store it in ‘HL’ register

Pair.

Step3 : Get the lower byte of 1st number into ‘L’ register

Step4 : Add it with lower byte of 2nd number

Step5 : tore the result in ‘L’ register

Step6 : Get the higher byte of 1st number into accumulator

Step7 : Add it with higher byte of 2nd number and carry of the lower bit

Addition.

Step8 : Store the result in ‘H’ register

Step9 : Store 16 bit addition value in ‘HL’ register pair

Step10 : Stop program execution

Flow chart

NO

YES

Address Label Mnemonics Hex Code Comments4500 MVI C,00 0E C = 00H

C = 00H

Load ‘HL’ with 1st Data

Transfer HL - DE

Load ‘HL’ with 2nd Data

DE + HL = HL

C = C + 01

Store ‘HL’ in memory (SUM)

Transfer C - A

END

Check for carry

START

Store ‘A’ in memory (Cy)

4501 004502 LHLD 4800 2A HL – 1st No.4503 004504 484505 XCHG EB HL – DE4506 LHLD 4802 2A HL – 2nd No.4507 024508 484509 DAD D 19 Double addition DE +

HL450A JNC Ahead D2 If Cy = 0, G0 to 450E

450E450B 0E450C 45450D INR C 0C C = C + 01450E AHEAD SHLD 4804 22 HL – 4804 (sum)450F 044510 484511 MOV C,A 79 Cy – A4512 STA 4806 32 Cy – 48064513 064514 484515 HLT 76 Stop execution

Input without carry

Input Address Value4800 01 (addend)4801 044802 02 (augend)4803 03 (augend)


4804 03 (sum)4805 07 (sum)4806 00 (carry)

Calculation 0000 0100 0000 00010000 0011 0000 0010---------------------------------0000 0111 0000 00110 7 0 3

With carry

Input Address Value

4800 FF (addend)

4801 DE (addend)

4802 96 (augend)

4803 DF (augend)

Output Address Value

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

SUBTRACTION OF TWO 16 – BIT NUMBERS

Aim:To write an assembly language program for subtracting two 16 bit numbers using

8085 microprocessor kit.

Apparatus required:1. 8085 microprocessor kit2. Keyboard

ALGORITHM


Step2 : Get the 1st 16 bit in ‘HL’ register pair

Step3 : Save the 1st 16 bit in ‘DE’ register pair

Step4 : Get the 2nd 16 bit number in ‘HL’ register pair

Step5 : Get the lower byte of 1st number

Step6 : Get the subtracted value of 2nd number of lower byte by

subtracting it with lower byte of 1st number

Step7 : Store the result in ‘L’ register

Step8 : Get the higher byte of 2nd number

Step9 : Subtract the higher byte of 1st number from 2nd number with

borrow

Step10 : Store the result in ‘HL’ register

Step11 : Stop the program execution

Flow chart

C = 00H

Load ‘HL’ with 1st Data

Transfer HL - DE

Load ‘HL’ with 2nd Data

Transfer E – A (LSB)

END

START

A = A – L (LSB)

Store ‘A’ in memory (LSB)

Transfer D – A (MSB)

A – A – H – Borrow (MSB)

Store ‘A’ in memory (MSB)

PROGRAM:

Address Label Mnemonics Hex Code Comments4500 MVI C,00 0E C = 00H4501 004502 LHLD 4800 2A L – 1st No.4503 004504 484505 XLHG EB HL – DE4506 LHLD 4802 2A HL – 2nd No.4507 024508 484509 MOV A,E 7B LSB of ‘1’ to ‘A’450A SUB L 95 A – A – L450B STA 4804 32 A – memory450C 04450D 48450E MOV A,D 7A MSB of 1 to A450F SBB H 9C A- A – H4510 STA 4805 32 A – memory4511 054512 484513 HLT 76 Stop execution

Input

Without borrow

Input Address Value4800 074801 084802 054803 06


4804 024805 02

4807 00

Output

With borrow

Input Address Value4800 074801 084802 054803 06


4804 024805 024807 00

Calculation

05 06 - 07 08

05 06 0101 0110 07 08 0111 10001010 1001 1000 01110000 0001 0000 0001--------------- --------------1010 1010 1000 1000

05 06 + 07 081010 10101000 1000---------------

(1) 0010 001002 02

8 BIT MULTIPLICATION & DIVISION

AIM:

To perform 8085 assembly language program for 8 bit multiplication .

APPARATUS REQUIRED:


ALGORITHM (MULTIPLICATION):

Algorithm:


Step2 : Get the 1st 8 bit numbers

Step3 : Move the 1st 8it number to register ‘B’

Step4 : Get the 2nd 8 bit number

Step5 : Move the 2nd 8 bit number to register ‘C’

Step6 : Intialise the accumulator as zero

Step7 : Intialise the carry as zero

Step8 : Add both register ‘B’ value as accumulator


Step10 : Increment carry by 1 if there is

Step11 : Decrement the 2nd value and repeat from step 8, till the 2nd

value becomes zero.

Step12 : Store the multiplied value in accumulator

Step13 : Move the carry value to accumulator


Flowchart

NO

YES

No

YES

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

Increment carry

Decrement 2nd number

Store the value f carry in accumulator

END

Check for carry

start

2nd Number

PROGRAM:

Address Label Mnemonics Hex Code Comments4100 LDA 4500 3A, 00, 45 Load the first 8 bit number4103 MOV B,A 47 Move the 1st 8 bit data to

register ‘B’4104 LDA 4501 3A, 01, 45 Load the 2nd 16 it number4107 MOV C,A 4F Move the 2nd 8 bit data to

register ‘C’4108 MVI A, 00 3E, 00 Intialise the accumulator as

zero410A MVI D, 00 16, 00 Intialise the carry as zero410C ADD B 80 Add the contents of ‘B’ and

accumulator410D INC D2 11, 41 Jump if no carry4110 INR D 14 Increment carry if there is4111 DCR C OD Decrement the value ‘C’4112 JNZ C2 0C, 41 Jump if number zero4115 STA 4502 32 02, 45 Store the result in

accumulator4118 MOV A,D 7A Move the carry into

accumulator4119 STA 4503 32,03,45 Store the result in

accumulator411C HLT 76 Stop the program execution

InputInput Address Value

4500 044501 02


4502 084503 00

DIVISION OF TWO 8 – BIT NUMBERS

Aim:To write an assembly language program for dividing two 8 bit numbers using

microprocessor kit.

Apparatus required:8085 microprocessor kit Key board

Algorithm:Step1 : Start the microprocessor

Step2 : Intialise the Quotient as zero

Step3 : Load the 1st 8 bit data

Step4 : Copy the contents of accumulator into register ‘B’

Step5 : Load the 2nd 8 bit data

Step6 : Compare both the values

Step7 : Jump if divisor is greater than dividend

Step8 : Subtract the dividend value by divisor value

Step9 : Increment Quotient

Step10 : Jump to step 7, till the dividend becomes zero

Step11 : Store the result (Quotient) value in accumulator

Step12 : Move the remainder value to accumulator

Step13 : Store the result in accumulator

Step14 : Stop the program execution

Flow chart

NO

YES

YES

END

Store the Quotient in accumulator

Move the remainder to accumulator

Dividend

Decrement 2nd number

Increment carry

Check for carry

Add the accumulator with 1st number

Compare the dividend & divisor

Get the divisor

Intialise the Quotient as zero

Get the divided

START

Store the remainder in accumulator

PROGRAM:

ADDRESS LABEL MNEMONICS OPCODE COMMENTS4100 MVI B,00 0E,00 Clear the content of B Register.

4102 LXIH 4200 21,00,42 The content of data 4200 transfer to HL pair.

4105 MOV A,M 7E Move the content of memory location to A Register.

4106 INX H 23 Increment memory location.

4107 LOOP 1 SUB M 96 Subtract the content of memory location with accumulator.

4108 INR B 04 Increment the B register.4109 JNC 4107 (LOOP 1) D2,07,41 If carry = 0, jump to address 4107.410C DCR B 05 Decrement the B register.

410D ADD M 86 Add the content of memory location with accumulator.

411E STA 4301 32,01,43 Store the remainder in address 4301.

4111 MOV A,B 78 Move the B Register content to the accumulator.

4112 STA 4302 32,02,43 Store the quotient in address 4302.4115 HLT 76 End the program

OBSERVATION:

INPUT OUTPUTADDRESS DATA ADDRESS DATA

4200 43004201 4301

RESULT:

Thus the ALP for 8 bit multiplication and division were performed and output was verified

EXPT .NO:

BLOCK TRANSFER OPERATION

AIM:

To perform a forward block transfer and reverse block transfer operation using 8085 microprocessor.

APPARATUS REQUIRED:


ALGORITHM (Forward block transfer):

1. Initialize the source address in HL register pair.2. Initialize the destination address in DE register pair3. Move the memory content pointed by HL pair to accumulator.4. Move accumulator content to memory location pointed by storing DE register.5. Increment the HL pair6. Increment the DE registers pair7. Decrement C register8. If there is no zero jump to step 49. Stop the Program.

PROGRAM:

ADDRESS LABEL MNEMONICS OPCODE COMMENTS

4100 MVI C,05 0E,05 Clear the content of C registers.(Count 5)

4102 LXIH 4200 21,00,42 The content of data 4200 transfer to HL pair

4105 LXID 4300 21,00,43 The content of data 4300 transfer to DE pair

4108 LOOP 1 MOV A,M 7E Move the content of memory to accumulator

4109 STAX D 12 Store the accumulator content at D register.

410A INX H 23 Increment the H register410B INX D 13 Increment the D register410C DCR C 0D Decrement the C register410D JNZ 4108 (LOOP 1) C2,08,41 If carry = 0, jump to address 4108.4110 HLT 76 End the program

Flow chart

No

OBSERVATION:


4200 4300

Clear the content of C registers. (Count value)

Load the content to H reg

Load the content to D reg

Store the Accumulator content D register

INR H Reg

END

Check for carry

START

INX D Reg

Move the content of memory to accumulator

4201 43014202 43024203 43034204 4304

ALGORITHM (Reverse transfer):

1. Initialize the source address in HL register pair.2. Initialize the destination address in DE register pair3. Move the memory content pointed by HL pair to C register4. Move the content of C register to E register5. Increment the HL pair6. Move the content of memory to accumulator7. Store the accumulator content at D register8. Decrement the D register9. Decrement the C register10. If carry = 0, jump to address 410711. End the program

PROGRAM:

ADDRESS LABEL MNEMONICS OPCODE COMMENTS


4103 LXID 4300 21,00,43 The content of data 4200 transfer to DE pair

4106 MOV C,M 4E Move the content of memory to C register

4107 LOOP 1 MOV E,C 59 Move the content of C register to E register.

4108 INX H 23 Increment the H register

4109 MOV A,M 7E Move the content of memory to accumulator

410A STAX D 12 Store the accumulator content at D register.

410B DCX D 1B Decrement the D register410C DCR C 0D Decrement the C register410D JNZ 4107 (LOOP 1) C2,07,41 If carry = 0, jump to address 4107.4110 HLT 76 End the program

OBSERVATION:


4200 (Count 4) 43004201 43014202 43024203 43034204

Flow Chart

No

RESULT:

Load the content to HL Pair (Count value)

Transfer content HL to DE Pair

Transfer the content of C register to E register.

INR H Reg

END

Check for carry

START

DCR C Reg

Transfer the content of memory to C Reg

Transfer the content of memory to accumulator

Store the accumulator content at D register.

DCX D Reg

Thus the forward block transfer and reverse block transfer operation were performed and output was verified.

EXPT .NO:

ARRAY OPERATION

AIM:

To perform a smallest and largest of numbers using 8085 microprocessor.

APPARATUS REQUIRED:

1. 8085 microprocessor kit2. Keyboard

ALGORITHM (LARGEST NUMBER):

1. Place all the elements of an array in the consecutive memory locations.2. Fetch the first element from the memory location and load it in the accumulator.3. Initialize a counter (register) with the total number of elements in an array.4. Decrement the counter by 1.5. Increment the memory pointer to point to the next element.6. Compare the accumulator content with the memory content (next Element).7. If the accumulator content is smaller, then move the memory content (Largest element) to the accumulator. Else continue.8. Decrement the counter by 1.9. Repeat steps 5 to 8 until the counter reaches zero10. Store the result (accumulator content) in the specified memory location

PROGRAM:

ADDRESS LABEL MNEMONICS OPCODE COMMENTS4100 MVI C,00 0E,00 Clear the content of C Register





4108 DCR C 0D Decrement the C register4109 LOOP 1 INX H 23 Increment the C register

410A CMP M BE Compare the content of memory with accumulator

410B JNC 410F (Loop 2) D2,0F,41 If carry = 0, jump to address 410F410E MOV A,M 7E Move the content of memory to

accumulator410F Loop 2 DCR C 0D Decrement the C register4110 JNZ 4109 (LOOP 1) C2,09,41 If carry = 0, jump to address 4109

4113 STA 4300 32,00,43 Store the largest number in the memory location

4116 HLT 76 End the program.OBSERVATION:


4200 03 (Count) 4300420142024203

ALGORITHM (SMALLEST NUMBER):

1. Place all the elements of an array in the consecutive memory locations.2. Fetch the first element from the memory location and load it in the accumulator.3. Initialize a counter (register) with the total number of elements in an array.4. Decrement the counter by 1.5. Increment the memory pointer to point to the next element.6. Compare the accumulator content with the memory content (next Element).7. If the accumulator content is smaller, then move the memory content (Largest element) to the accumulator. Else continue.8. The counter by 1.9. Repeat steps 5 to 8 until the counter reaches zero10. Store the result (accumulator content) in the specified memory location

PROGRAM:






4108 DCR C 0D Decrement the C register4109 LOOP 1 INX H 23 Increment the C register

410A CMP M BE Compare the content of memory with accumulator

410B JC 410F (Loop 2) DA,0F,41 If carry = 1, jump to address 410F

410E MOV A,M 7E Move the content of memory to accumulator

410F LOOP 2 DCR C 0D Decrement the C register4110 JNZ 4109 (LOOP 1) C2,09,41 If carry = 0, jump to address 4109

4113 STA 4300 32,00,43 Store the Smallest number in the memory location


OBSERVATION:


4200 03(Count) 4300420142024203

RESULT:

Thus the largest and smallest number operation were performed and output was verified.

EXPT .NO:

SORTING

AIM:

To perform a ascending and descending order of using 8085 microprocessor.

APPARATUS REQUIRED:


ALGORITHM (ascending order):1. Get the numbers to be sorted from the memory locations.2. Compare the first two numbers and if the first number is larger than second then Interchange the number.3. If the first number is smaller, go to step 44. Repeat steps 2 and 3 until the numbers are in required order

PROGRAM:ADDRESS LABEL MNEMONICS OPCODE COMMENTS


4103 MOV B,M 46 Move the content of memory to B register

4104 DCR B 05 Decrement the B register

4105 LOOP 1 MOV C,B 48 Move the content of B register to C register.



410A INX H 23 Increment the H register

410B MOV D,M 56 Move the content of memory to D register

410C CMP D BA Compare the content of D register with accumulator

410D JC 4114 (LOOP 3) DA,14,41 If carry = 1, jump to address 4114

4110 MOV M,A 77 Move the content of memory to accumulator

4111 DCX H 2B Decrement the H register

4112 MOV M,D 72 Move the content of D register to memory

4113 INX H 23 Increment the H register4114 LOOP 3 DCR C 0D Decrement the C register4115 JNZ 4109 (LOOP 2) C2,09,41 If zero flag = 0, jump to address 41094118 DCR B 05 Decrement the B register

4119 JNZ 4105 (LOOP 1) C2,05,41 If zero flag = 0, jump to address 4105411C HLT 76 End the program

OBSERVATION:


4200 (Count 05) 42014201 42024202 42034203 42044204 42054205

ALGORITHM (Descending order):1. Get the numbers to be sorted from the memory locations.2. Compare the first two numbers and if the first number is smaller than second then I Interchange the number.3. If the first number is larger, go to step 44. Repeat steps 2 and 3 until the numbers are in required order

PROGRAM:ADDRESS LABEL MNEMONICS OPCODE COMMENTS


4103 MOV B,M 46 Move the content of memory to B register

4104 DCR B 05 Decrement the B register

4105 LOOP 1 MOV C,B 48 Move the content of B register to C register.



410A INX H 23 Increment the H register

410B MOV D,M 56 Move the content of memory to D register

410C CMP D BA Compare the content of D register with accumulator

410D JNC 4114 (LOOP 3) DA,14,41 If carry = 0, jump to address 4114

4110 MOV M,A 77 Move the content of memory to accumulator

4111 DCX H 2B Decrement the H register

4112 MOV M,D 72 Move the content of D register to memory

4113 INX H 23 Increment the H register4114 LOOP 3 DCR C 0D Decrement the C register4115 JNZ 4109 (LOOP 2) C2,09,41 If zero flag = 0, jump to address 41094118 DCR B 05 Decrement the B register

4119 JNZ 4105 (LOOP 1) C2,05,41 If zero flag = 0, jump to address 4105411C HLT 76 End the program

OBSERVATION:


4200 (Count 05) 42014201 42024202 42034203 42044204 42054205

RESULT:

Thus the Ascending and Descending operation were performed and output was verified.

EXPT .NO:

CODECONVERSION

AIM:

To perform a code conversion for Following

1.HEXA DECIMAL TO ASCII2.ASCII TO HEXA DECIMAL3APPARATUS REQUIRED:


PROGRAM:


4102 LHLD 4200 2A,00,42 Load the augend in DE pair through HL pair

4105 XCHG EB Transfer the content of HL pair to DE pair

4106 LHLD 4202 2A,02,42 Load the addend in HL pair

4109 MOV A,E 7B Move the content of E register to A register

410A SUB L 95 Subtract the content of L register with accumulator

410B STA 4204 32,04,42 Store the LSB in address location 4204

410E MOV A,D 7A Move the Content of D register to register A

410F SBB H 9C Subtract the content of H register and carry flag with accumulator

4110 STA 4205 32,05,42 Store the MSB in address location 42054113 JNC 4117 ( LOOP 1) D2,17,41 If there is no carry go to label (LOOP 1)

4116 INR C 0C Increment the C register

4117 LOOP 1 MOV A,C 79 Move the content of C Register to A register

4118 STA 4206 32,06,42 Store the carry in 4206 through accumulator ( MSB of sum )

411B HLT 76 End the program

Introduction to 8085 & it's description(includes basic lab experiments)

Engineering

carry flag c

auxiliary flag

flag z

carry value

g flags flag register

hold signal

address data bus

carry step8