1. EXPERIMENT NO. 1
Aim: Transfer Block of data from one memory location to other.
Apparatus: - Trainer Kit of Microprocessor 8085.
Assumption: - Assume that the data block of 5 bytes is present from memory location 7000H
& it is to be transfer from memory location 7500H.
FLOW CHART
START
HL
BC
D
7000H
7500H
05H
A
A
HL+1
BC+1
D–1
M (HL)
(BC)
HL
BC
D
IF ZF=0
Yes
Z=0
No
STOP
RESULT: - Thus we have transferred the data block from one memory location to other
memory.

2. EXPERIMENT NO. 1
Aim: Transfer Block of data from one memory location to other.
Apparatus: - Trainer Kit of Microprocessor 8085.
Assumption: - Assume that the data block of 5 bytes is present from memory location 7000H
& it is to be transfer from memory location 7500H.
PROGRAME
(MNEMONICS)
LXI H,7000H
LXI B,7500H
MVI D, 05H
BACK: MOV A,M
STAX B
INX H
INX B
DCR D
JNZ BACK(6008H)
HLT
HEX
CODE
21,00,70
01,00,75
16,05
7E
02
23
03
15
C2,08,60
76
MEMORY
LOCATION
6000,6001,6002
6003,6004,6005
6006,6007
6008
6009
600A
600B
600C
600D,600E,600F
6010
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of memory pointer, I.e. load BC pair with 7500H
Initialization of counter to count 5 bytes to be transfer
Move content of memory pointed by HL pair to accumulator
Store content of accumulator to memory pointed by BC pair
Increment HL pair to point next memory location
Increment BC pair to point next memory location
Decrement counter D to count NO. of bytes transferred
Jump to location specified in the instruction if result is not zero
Halt state
RESULT: - Thus we have transferred the data block from one memory location to other
memory.

3. EXPERIMENT NO. 02
Aim: Apparatus: -
Find the greatest and smallest No.
Trainer Kit of Microprocessor 8085.
Part A: -
A data block of 10 bytes is present from memory location 7000H
write a program to find greatest No. from given block of data and store the
result in register B as well as in Memory location 7500H.
FLOW CHART
START
HL
C
A
7000H
0AH
00H
A - M (HL)
NO
IF CF=1
YES
Z=0
A
M
HL
C
HL+1
C-1
IF ZF=1
Z=0
A
A
YES
7500H
B
STOP
NO

4. EXPERIMENT NO. 02
Aim: Apparatus: -
Find the greatest and smallest No.
Trainer Kit of Microprocessor 8085.
Part A: -
A data block of 10 bytes is present from memory location 7000H
write a program to find greatest No. from given block of data and store the
result in register B as well as in Memory location 7500H.
PROGRAME
(MNEMONICS)
LXI H,7000H
MVI C,0AH
MVI A,00H
BACK: CMP M
JNC 600A
MOV A,M
INX H
DCR C
JNZ BACK(6007H)
STA 7500H
MOV B,A
HLT
HEX
CODE
21,00,70
MEMORY
LOCATION
6000,6001,6002
6003,6004
6005,6006
6007
6008,6009,600A
600B
600C
600D
600E,600F,600G
600F.6010.6011
6012
6010
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of counter to count 10 bytes
Store lowest No. to compare to find largest No.
Compare data of Memory with accumulator to find largest No.
Jump to location specified in the instruction if carry is not generated
Move data of memory into a accumulator if it is greater
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero
Store accumulator content i.e. greatest No. into memory 7500H
Move data of register A i.e. greatest No. into register B
Halt state
RESULT: - Thus we have transferred the data block from one memory location to other
memory.

5. Part B: -
A data block of 10 bytes is present from memory location 7000H
write a program to find smallest No. from given block of data and store the
result in register B as well as in Memory location 7501H.
FLOW CHART
START
HL
C
A
7000H
0AH
FFH
A - M (HL)
NO
IF CF=0
YES
Z=0
A
M
HL
C
HL+1
C-1
IF ZF=1
NO
Z=0 YES
A
A
7501H
D
STOP
RESULT: - Thus we have found greatest and smallest No. from the given data block.

6. Part B: -
PROGRAME
(MNEMONICS)
LXI H,7000H
MVI C,0AH
MVI A,FFH
BACK: CMP M
JC 600A
MOV A,M
INX H
DCR C
JNZ BACK(6007H)
STA 7500H
MOV B,A
HLT
A data block of 10 bytes is present from memory location 7000H
write a program to find smallest No. from given block of data and store the
result in register B as well as in Memory location 7501H.
HEX
CODE
21,00,70
MEMORY
LOCATION
6000,6001,6002
6003,6004
6005,6006
6007
6008,6009,600A
600B
600C
600D
600E,600F,600G
600F.6010.6011
6012
6010
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of counter to count 10 bytes
Store largest No. to compare to find smallest No.
Compare data of Memory with accumulator to find largest No.
Jump to location specified in the instruction if carry is generated
Move data of memory into a accumulator
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero
Store accumulator content i.e. smallest No. into memory 7500H
Move data of register A i.e. smallest No. into register B
Halt state
RESULT: - Thus we have found greatest and smallest No. from the given data block.

7. EXPERIMENT NO. 03
Aim: Apparatus: -
Arrange the data block in the ascending & descending order.
Trainer Kit of Microprocessor 8085.
Part A: -
A data block of 10 bytes is present from memory location 7000H
write a program to arrange the Nos. in ascending order from memory location
7500H.
FLOW CHART
START
SP
HL
B
A
SMALLEST
7FFFH
7500H
0AH
FFH
PUSH H
HL
7000H
C
0AH
A-M
CALL SMALLEST
NO
(HL) M
A
(DE) M
HL
B
A
FFH
A
HL+1
B-1
IF CF=0
Z=0
A
DE
YES
M
HL
YES
IF ZF=0
Z=0
NO
STOP
HL
C
HL+1
C-1
IF ZF=1
Z=0
YES
POP H
RET
NO

8. EXPERIMENT NO. 03
Aim: -
Arrange the data block in the ascending & descending order.
Apparatus: -
Trainer Kit of Microprocessor 8085.
Part A: -
A data block of 10 bytes is present from memory location 7000H
write a program to arrange the No. in ascending order from memory location
7500H.
MAIN
PROGRAME
(MNEMONICS)
LXI SP,7FFFH
LXI H,7500H
MVI B,0AH
MVI A,FFH
BACK:CALL Smallest(6500H)
MOV M,A
MVI A,FF
STAX D
INX H
DCR B
JNZ BACK(600AH)
RST 1
SUB PROGRAME
(MNEMONICS)
SMALLEST NO.
PUSH H
LXI H,7000H
MVI C,0AH
AGAIN:CMP M
JC SKIP(650D)
MOV A,M
HEX
CODE
31.FF,7F
21,00,75
06.0A
3E,FF
CD,00.65
77
3E,FF
12
23
4E
C2, 0A,60
CF
HEX
CODE
E5
21,00,70
0E.0A
BE
DA,0D,65
7E
MEMORY
LOCATION
6000,6001,6002
6003,6004,6005
6006,6007
6008,6009
600A,600B,600C
600D,
600E,600F
6010
6011
6012
6013,6014,6015
6016
MEMORY
LOCATION
6500
6501,6502,6503
6504,6505
6506
6507,6508,6509
650A
COMMENT
Initialization of stack pointer.
Initialization of memory pointer, I.e. load HL pair with 7500H
Initialization of counter to count 10 bytes
Store largest No. to compare to find smallest No.
Call subroutine to find smallest No.
Jump to location specified in the instruction if carry isn’t generated
Store greatest No. in Accumulator.
Store greatest No. in place of smallest No. pointed by DE pair.
Increment HL pair to point next memory location
Decrement counter B
Jump to location specified in the instruction if result is not zero
Transfer the control to ISR of RST 1
COMMENT
Save the content of HL pair in to stack memory.
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of counter to count 10 bytes
Compare data of Memory with accumulator to find smallest No
Jump to location specified in the instruction if carry is generated
Move data of memory into a accumulator

9. MOV D,H
MOV E,L
SKIP:INX H
DCR C
JNZ AGAIN(6506H)
POP H
RET
54
5D
23
0D
C2,06,65
E1
C9
650B
650C
650D
650E
650F,6510,6512
6013
6014
Save higher 8 bit address of memory to register D
Save lower 8 bit address of memory to register E
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero
Load HL pair by content of stack top two locations.
Return to main program.
CONCLUSION:-Thus we have studied how data block can be arranged in ascending order using assembly
language program.
Part B: -
A data block of 10 bytes is present from memory location 7000H
write a program to arrange the No. in descending order from memory location
7400H.
FLOW CHART
GREATEST
START
SP
HL
B
A
7FFFH
7400H
0AH
00H
YES Z=0
STOP
IF CF=1
YES
Z=0
A
00H
A
HL+1
B-1
IF ZF=1
A - M (HL)
NO
CALL GREATEST
(HL) M
A
(DE) M
HL
B
PUSH H
HL
7000H
C
0AH
A
DE
M
HL
HL
C
NO
HL+1
C-1
IF ZF=1
Z=0
YES
POP H
RET
NO

10. RESULT: - Thus we have arranged the data block in ascending and descending order.
Part B: -
A data block of 10 bytes is present from memory location 7000H
write a program to arrange the No. in descending order from memory location
7500H.
MAIN
PROGRAME
(MNEMONICS)
LXI SP,7FFFH
LXI H,7500H
MVI B,0AH
MVI A,00H
J1:CALL Greatest(6650H)
MOV M,A
MVI A,00
STAX D
INX H
DCR B
JNZ J1 (600AH)
RST 1
SUB PROGRAME
(MNEMONICS)
GREATEST NO.
HEX
CODE
31.FF,7F
21,00,75
06.0A
3E,00
CD,50.66
77
3E,00
12
23
4E
C2, 0A,60
CF
HEX
CODE
MEMORY
LOCATION
6000,6001,6002
6003,6004,6005
6006,6007
6008,6009
600A,600B,600C
600D,
600E,600F
6010
6011
6012
6013,6014,6015
6016
MEMORY
LOCATION
COMMENT
Initialization of stack pointer.
Initialization of memory pointer, I.e. load HL pair with 7500H
Initialization of counter to count 10 bytes
Store smallest No. to compare to find greatest No.
Call subroutine to find smallest No.
Jump to location specified in the instruction if carry isn’t generated
Store smallest No. in Accumulator.
Store greatest No. in place of smallest No. pointed by DE pair.
Increment HL pair to point next memory location
Decrement counter B
Jump to location specified in the instruction if result is not zero
Transfer the control to ISR of RST 1
COMMENT

11. PUSH H
LXI H,7000H
MVI C,0AH
J3:CMP M
JNC J2(665DH)
MOV A,M
MOV D,H
MOV E,L
J2:INX H
DCR C
JNZ J3 (6656H)
POP H
RET
E5
21,00,70
0E.0A
BE
D2,5D,66
7E
54
5D
23
0D
C2,56,66
E1
5F
6650
6651,6652,6653
6654,6655
6656
6657,6658,6659
665A
665B
665C
665D
665E
665F,6660,6662
6663
6664
Save the content of HL pair in to stack memory.
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of counter to count 10 bytes
Compare data of Memory with accumulator to find smallest No
Jump to location specified in the instruction if carry is not generated
Move data of memory into a accumulator
Save higher 8 bit address of memory to register D
Save lower 8 bit address of memory to register E
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero
Load HL pair by content of stack top two locations.
Return to main program.
CONCLUSION:-Thus we have studied how data block can be arranged in descending order using
assembly
language program.
RESULT: - Thus we have arranged the data block in ascending and descending order.
EXPERIMENT NO. 04

12. Aim: Apparatus: -
Find Even, Odd, Zero Nos.
Trainer Kit of Microprocessor 8085.
Part A: -
A data block of 10 bytes is present from memory location 7000H
write a program to find Even Nos. and store them from memory location
7500H.
FLOW CHART
START
HL
DE
C
A
YES
7000H
7500H
0AH
M (HL)
A - 00H
IF ZF=1
Z=0
NO
RAR
IF CF=0
NO
Z=0
YES
RAL
(DE)
A
DE
DE+1
HL
C
HL+1
C-1
IF ZF=1
Z=0
YES
STOP
NO

13. EXPERIMENT NO. 04
Aim: Apparatus: -
Find Even, Odd, Zero Nos.
Trainer Kit of Microprocessor 8085.
Part A: -
A data block of 10 bytes is present from memory location 7000H
write a program to find Even Nos. and store them from memory location
7500H.
PROGRAME
(MNEMONICS)
LXI H,7000H
LXI D,7500H
MVI C,0AH
BACK: MOV A,M
CPI 00H
JZ SKIP(6015)
RAR
JC SKIP(6015)
RAL
STAX D
INX D
SKIP:INX H
DCR C
JNZ BACK(6008)
RST 1
HEX
CODE
21,00,70
11,00,75
0E,0A
7E
FE,00
CA,15,60
1F
DA,15,60
17
12
13
23
0D
C2,08,60
CF
MEMORY
LOCATION
6000,6001,6002
6003,6004,6005
6006,6007
6008
6009,600A
600B,600C,600D
600E
600F,6010,6011
6012
6013
6014
6015
6016
6017,6018,6019
601A
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of memory pointer, I.e. load DE pair with 7500H
Initialization of counter to count 10 bytes
Move content of memory into register A.
Compare data with 00H.
Jump to location specified in the instruction if result is zero
Rotate accumulator right to check first bit of data.
Jump to location specified in the instruction if carry is generated.
Rotate accumulator left to get original data.
Store accumulator content i.e. even No. into memory pointed by DE
Increment DE pair to point next memory location
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero.
Transfer control to ISR of RST 1.
Conclusion: - Thus we have studied how to find even Nos. using assembly language program.

14. Part B: -
A data block of 10 bytes is present from memory location 7000H
write a program to find odd Nos. and store them from memory location 7500H.
FLOW CHART
START
HL
DE
C
A
7000H
7500H
0AH
M (HL)
RAR
IF CF=1
Z=0
YES
RAL
(DE)
A
DE
DE+1
HL
C
NO
HL+1
C-1
IF ZF=1
YESZ=0
STOP
NO

15. Part B: -
PROGRAME
(MNEMONICS)
LXI H,7000H
LXI D,7500H
MVI C,0AH
BACK: MOV A,M
RAR
JNC SKIP(6010)
RAL
STAX D
INX D
SKIP:INX H
DCR C
JNZ BACK(6008)
RST 1
A data block of 10 bytes is present from memory location 7000H
write a program to find odd Nos. and store them from memory location 7500H.
HEX
CODE
21,00,70
11,00,75
0E,0A
7E
1F
D2,10,60
17
12
13
23
0D
C2,08,60
CF
MEMORY
LOCATION
6000,6001,6002
6003,6004,6005
6006,6007
6008
6009
600A,600B,600C
600D
600E
600F
6010
6011
6012,6013,6014
6015
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of memory pointer, I.e. load DE pair with 7500H
Initialization of counter to count 10 bytes
Move content of memory into register A.
Rotate accumulator right to check first bit of data.
Jump to location specified in the instruction if carry is generated.
Rotate accumulator left to get original data.
Store accumulator content i.e. odd No. into memory pointed by DE
Increment DE pair to point next memory location
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero.
Transfer control to ISR of RST 1.
Conclusion: - Thus we have studied how to find odd Nos. using assembly language program.

16. Part C: -
A data block of 10 bytes is present from memory location 7000H
write a program to find zeros and store them from memory location 7500H.
START
HL
DE
C
A
NO
7000H
7500H
0AH
M (HL)
A - 00H
IF ZF=1
Z=0
YES
(DE)
DE
A
DE+1
HL
C
HL+1
C-1
IF ZF=1 NO
Z=0 YES
STOP
Result:- Thus we have found Even Odd and Zero Nos.

17. Part C: -
PROGRAME
(MNEMONICS)
LXI H,7000H
LXI D,7500H
MVI C,0AH
BACK: MOV A,M
CPI 00H
JNZ SKIP(6010)
STAX D
INX D
SKIP:INX H
DCR C
JNZ BACK(6008)
RST 1
A data block of 10 bytes is present from memory location 7000H
write a program to find zeros and store them from memory location 7500H.
HEX
CODE
21,00,70
11,00,75
0E,0A
7E
FE,00
02,10,60
12
13
23
0D
C2,08,60
CF
MEMORY
LOCATION
6000,6001,6002
6003,6004,6005
6006,6007
6008
6009,600A
600B,600C,600D
600E
600F
6010
6011
6012,6013,6014
6015
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Initialization of memory pointer, I.e. load DE pair with 7500H
Initialization of counter to count 10 bytes
Move content of memory into register A.
Compare data with 00H.
Jump to location specified in the instruction if result is zero
Store accumulator content i.e. even No. into memory pointed by DE
Increment DE pair to point next memory location
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero.
Transfer control to ISR of RST 1.
Conclusion: - Thus we have studied how to find zero Nos. using assembly language program.
Result:- Thus we have found Even Odd and Zero Nos.

18. EXPERIMENT NO. 05
Aim: Addition of 10 BCD Nos.
Apparatus: - Trainer Kit of Microprocessor 8085.
Problem: Write a Program to add 10 BCD Nos. present from memory location 7000H &
store
16bit result after the data.
FLOW CHART
START
HL
A
B
C
7000H
00H
00H
0AH
A+M (HL)
DAA
NO
A
CY=1
YES
E
A
A
B
A
A+01
DAA
B
A
A
E
HL
C
HL+1
C-1
IF ZF=1
YES
Z=0
M
HL
M
A
HL+1
B
STOP
NO

19. Result: - Thus we have performed addition of 10 BCD Nos.
EXPERIMENT NO. 05
Aim: Apparatus: Problem: store
Addition of 10 BCD Nos.
Trainer Kit of Microprocessor 8085.
Write a Program to add 10 BCD Nos. present from memory location 7000H &
16bit result after the data.
PROGRAME
(MNEMONICS)
LXI H,7000H
MVI A,00H
MVI B,00H
MVI C,0AH
BACK: ADD M
DAA
JNC SKIP
MOV E,A
MOV A,B
ADI 01H
DAA
MOV B,A
MOV A,E
SKIP: INX H
DCR C
JNZ BACK(6008H)
MOV M,A
INX H
MOV M,B
RST 1
HEX
CODE
21,00,70
3E,00
46
MEMORY
LOCATION
6000,6001,6002
6003,6004
6005
23
4E
80
0D
6006
6007
6008
6009
C2,08,60
57
23
77
CF
600A,600B,600C
600D
600E
600F
6010
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Reset accumulator for successive addition
Reset register B to save carry.
Initialization of counter for 10 BCD Nos.
Add content of memory location with content of accumulator.
Decimal adjust accumulator to get legal BCD No.
Jump to location specified in the instruction if carry is not generated
Save content of A i.e. intermediate result.
Move data of register B into accumulator to save carry.
Add 01H with the data of accumulator.
Decimal adjust accumulator to get legal BCD No.
Save carry generated in register B.
Move the content of register E in A to get intermediate result in A
Increment HL pair to point next memory location
Decrement counter C
Jump to location specified in the instruction if result is not zero
Save 8bit LSB result in memory
Increment HL pair to point next memory location
Save 8bit MSB result in memory.
Transfer control to ISR of RST 1.
Conclusion: - Thus we have studied how to add BCD Nos. using assembly language program &
result found to
be……….
Result:- Thus we have performed BCD addition.

20. EXPERIMENT NO. 06
Aim: Apparatus: -
To study 8bit multiplication.
Trainer Kit of Microprocessor 8085.
Part A: -
Two 8bit data are present from memory location 7000H write a program to
multiply these data and store 8bit result in register D & after the data.
FLOW CHART
START
HL
A
B
HL
C
7000H
00H
M (HL)
HL+1
M (HL)
A+B
C
A
C-1
IF ZF=1
NO
Z=0
YES
D
HL
(HL) M
A
HL+1
A
STOP

21. EXPERIMENT NO. 06
Aim: Apparatus: -
To study 8bit multiplication.
Trainer Kit of Microprocessor 8085.
Part A: -
Two 8bit data are present from memory location 7000H write a program to
multiply these data and store 8bit result in register D & after the data.
PROGRAME
(MNEMONICS)
LXI H,7000H
MVI A,00H
MOV B,M
INX H
MOV C,M
BACK: ADD B
DCR C
JNZ BACK(6008H)
MOV D,A
INX H
MOV M,A
RST 1
HEX
CODE
21,00,70
MEMORY
LOCATION
6000,6001,6002
6003,6004
6005
6006
6007
6008
6009
600A,600B,600C
600D
600F
6010
6011
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Reset accumulator
Move second operand of multiplication into register C.
Increment HL pair to point next memory location
Move first operand of multiplication into register C.
Add data of register B and accumulator.
Decrement counter C
Jump to location specified in the instruction if result is not zero
Save the result of multiplication into register D.
Increment HL pair to point next memory location
Save the result of multiplication into memory.
Transfer control to ISR of RST 1.
Conclusion: - Thus we have studied how to multiply two 8bit data using assembly language
program and 8bit result found to be………

22. Part B: -
Two 8bit data are present from memory location 7000H write a program to
multiply these data and store 16bit result in register DE pair & after the
data.
FLOW CHART
START
HL
A
D
7000H
00H
00H
B
HL
C
M (HL)
HL+1
M (HL)
A+B
A
IF CF=1
Z=0 YES
D
C
NO
D+1
C-1
IF ZF=1
Z=0
E
HL
(HL)M
HL
M
YES
A
HL+1
A
HL+1
D
STOP
NO

23. RESULT: - Thus we have multiplied two 8bit data.
Part B: -
Two 8bit data are present from memory location 7000H write a program to
multiply these data and store 16bit result in register DE pair & after the
data.
PROGRAME
(MNEMONICS)
LXI H,7000H
MVI A,00H
MOV B,M
INX H
MOV C,M
BACK: ADD B
JNC SKIP(600D)
INR D
SKIP: DCR C
JNZ BACK(6008H)
MOV E,A
INX H
MOV M,A
INX H
MOV M,D
RST 1
HEX
CODE
21,00,70
MEMORY
LOCATION
6000,6001,6002
6003,6004
6005
6006
6007
6008
6009,600A,600B
600C
600D
600E,600F,6010
6011
6012
6013
6014
6015
6016
COMMENT
Initialization of memory pointer, I.e. load HL pair with 7000H
Reset accumulator
Move first operand of multiplication into register C.
Increment HL pair to point next memory location
Move second operand of multiplication into register C.
Add data of register B and accumulator.
Jump to location specified in the instruction if no carry is generated
Increment data of register D to save carry.
Decrement counter C
Jump to location specified in the instruction if result is not zero
Save the 8bit LSB of multiplication result into register E.
Increment HL pair to point next memory location
Save the 8bit LSB of multiplication result into memory.
Increment HL pair to point next memory location
Save the 8bit MSB of multiplication result into memory.
Transfer control to ISR of RST 1.
Conclusion: - Thus we have studied how to multiply two 8bit data using assembly language
program and 16bit result found to be…….
RESULT: - Thus we have multiplied two 8bit data.

24. EXPERIMENT NO. 07
Aim: Apparatus: Problem: -
To Interface 8255 with Microprocessor 8085.
Trainer Kit of Microprocessor 8085 & Trainer kit of 8255.
Interface 8255 PPI with µP 8085 and write a program to add content of Port A and
Port B and store the result at Port C.
Address Decoding Table:IC
ports
Hex Address
8255 Port A
PPI Port B
Port C
CWR
F0H
F1H
F2H
F3H
Binary address
A15 A14 A13 A12 A11 A10 A9 A8
A7 A6 A5 A4 A3 A2
A1 A0
1
1
1
1
0
0 0 0
1
1
1
1
0
0 0 1
1
1
1
1
0
0 1 0
1
1
1
1
0
0 1 1
To enable
Decoder
Interfacing:RESET OUT
A15 – A14
ALE
RESET
PA7-PA0
8
OE
IC 74373
Latch
AD7-AD0
µp 8085
6
A1
8255PPI
A0
D7-D0
PB7-PB0
LATCH
5V
G1A G1B G1
IC 74138
3:8
Decoder
IO/M
RD
WR
Y5 IORD
RD
PC7-PC0
Y6 IOWR
WR
A5
A6 A7
Y0
Y
A4
A3
A2
G1A G1B G1
IC 74138
3:8
Decoder
Y4
CS
CS
Input to
decoder
Address

25. EXPERIMENT NO. 07
Aim: Apparatus: Problem: -
To study 8255 PPI.
Trainer Kit of Microprocessor 8085 & Trainer kit of 8255.
Interface 8255 PPI with µP 8085 and write a program to add content of port A and
Port B and store the result at port C.
CWR Format to initialize port A and port B as input ports and port C as output port.
D7
D6
D5
D4
D3
D2
D1
D0
1
0
0
1
0
0
1
0 = 92H
PROGRAME
(MNEMONICS)
MVI A,92H
OUT F3H
IN F0H
MOV B,A
IN F1H
ADD B
OUT F2
RST 1
HEX
CODE
3E,92
D3,F3
DB,F0
47
DB,F1
80
D3,F2
CF
MEMORY
LOCATION
6000,60001
6002,6003
6004,6005
6006
6007,6008
6009
600A,600B
600C
COMMENT
Move CWR format to accumulator.
Out CWR format to CWR for initialization of Ports as input or output
Take data of port A
Save port A data
Take data of port B
Add content of register B and A
Save result to Port C
Transfer control to ISR of RST 1.
CONCLUSION: - Thus we have added data of port A and port B and result saved at port C.
RESULT: -
Thus we have studied have to transfer data on ports.

26. EXPERIMENT NO. 09
Aim: Apparatus: Problem: -
To Interface DAC with Microprocessor 8085.
Trainer Kit of Microprocessor 8085 & Trainer kit of DAC.
Interface DAC WITH μp 8085 and write a program to generate saw tooth wave. The
port A is 08H.
Address Decoding Table:IC
ports
Hex Address
8255 Port A
PPI Port B
Port C
CWR
08H
09H
0AH
0BH
Binary address
A15 A14 A13 A12 A11 A10 A9 A8
A7 A6 A5 A4 A3 A2
A1 A0
0
0 0
0
1
0 0 0
0
0 0
0
1
0 0 1
0
0 0
0 1
0
1 0
0
0 0
0 1
0 1 1
To enable
Decoder
Input to
Decoder
Address
Interfacing:RESET OUT
A15 – A14
ALE
RESET
8
OE
IC 74373
Latch
AD7-AD0
µp 8085
6
A1
A0
D7-D0
LATCH
5V
G1A G1B G1
IC 74138
3:8
Decoder
IO/M
RD
WR
A5
8255PPI
Y5 IORD
RD
Y6 IOWR
S
WR
A6 A7
Y0
Y
A4
A3
G1A G1B G1
IC 74138
3:8
Decoder
Y2
CS
CS
PA7
PA6
PA5
PA4
PA3
PA2
PA1
PA0
IOUT
DAC
0808
+VREF
- VREF

27. EXPERIMENT NO. 09
Aim: Apparatus: Problem: -
To Interface DAC with Microprocessor 8085.
Trainer Kit of Microprocessor 8085 & Trainer kit of DAC.
Interface DAC WITH μp 8085 and write a program to generate saw tooth wave. The
port A is 08H.
CWR Format to initialize port A as output port.
D7
D6
D5
D4
D3
D2
D1
1
0
0
0
0
0
0
D0
0 = 80H
Let the saw tooth wave of amplitude + 5V is to be generated.
MVI A, 80H
OUT 0B (CWR)
MVI A, 00H
BACK: OUT 08H
CALL DELAY; for some time
INR A
JMP BACK
Saw tooth wave form
Result: - Thus we have interfaced DAC with Microprocessor 8085.

28. EXPERIMENT NO. 10
Aim: Apparatus: Problem: -
To Interface ADC with μP 8085.
Trainer Kit of Microprocessor 8085 & Trainer kit of ADC.
Interface ADC with µP 8085 and write a program to take 100 samples at the rate of 40
samples/sec. and
store these samples from address 7000H. The port A address is 08H.
Address Decoding Table:IC
ports
Hex Address
Binary address
A15 A14 A13 A12 A11 A10 A9 A8
A7 A6 A5 A4 A3 A2
A1 A0
08H
0
0
0
0
1
0 0 0
PPI Port B
09H
0
0
0
8255 Port A
0
0
0
1
1
Port C
CWR
0AH
0BH
0
0
0
0
0
0
To enable
Decoder
0
0
1
1
Input to
Decoder
0
0
1
1
0
1
Address
Interfacing:RESET OUT
A15 – A14
ALE
RESET
8
Vin
AD7-AD0
OE
IC 74373
Latch
µp 8085
PC0
6
A1
A0
D7-D0
8255PPI
LATCH
IO/M
RD
WR
RD
5V
G1A G1B G1
IC 74138
3:8
Decoder
A5
A6 A7
Y5 IORD
PA7
PA6
PA5
PA4
PA3
PA2
PA1
PA0
PC7
PC0
PB3
WR
PB2
PB1
PB0
Y6 IOWR
CS
IN0
ADC
0809
OE
EOC
SOC
ALE
A
B
C
S&H

29. Y0
A4
A3
A2
G1A G1B G1
IC 74138
3:8
Decoder
Y2
Result:- Thus we have interfaced ADC with μP 8085.
EXPERIMENT NO. 10
Aim: Apparatus: Problem: -
To Interface ADC with μP 8085.
Trainer Kit of Microprocessor 8085 & Trainer kit of ADC.
Interface ADC with µP 8085 and write a program to take 100 samples at the
rate of
40samples/sec. and store these samples from address
7000H. The port A address is 08H.
CWR Format to initialize port A as input port, port B and port C as output port.
D7
D6
D5
D4
D3
D2
D1
D0
1
0
0
1
1
0
0
0 = 98H
Data Required for the Channel Selection. Let channel 0 is to be select.
PB7
PB6
PB5
PB4
PB3
PB2
PB1
PB0
x
x
x
x
1
0
0
0 = 08H
Generation of SOC Signal
PC7
PC6
PC5
x
x
x
x
x
x
x
x
x
PC4
x
x
x
PC3
x
x
x
PC2
x
x
x
PC1
x
x
x
PC0/SOC
0
= 00H
1
= 01H
0
= 00H
Data rate is given as 40samples/sec.
Therefore time required to take one sample or time delay between two samples=
1/40=25msec

30. Program:LXI SP, FFFFH
LXI H, 7000H
MVI C, 64H
MVI A, 98H
OUT 0BH (CWR)
MVI A, 08H
OUT 09 (PORT B)
MVI A , 00H
OUT 0AH (PORT C)
INR A
OUT 0AH (PORT C)
DCR A
BACK: OUT 0AH (PORT C)
IN 0AH (PORT C)
RAL
JNC BACK
IN 08
MOV M,A
CALL DELAY;25msec
INX H
DCR C
JNZ AGAIN
RST 1
Result:- Thus we have interfaced ADC with μP 8085.