1. Microprocessor
T. Y. B. Sc.

2. ADDRESSING MODES OF 8085 Immediate addressing Mode
Register addressing Mode
Direct addressing Mode
Indirect addressing Mode
Implied addressing Mode

3. Immediate addressing mode
Immediate data is transferred to address or register.
Example:
MVI A,20H. Transfer immediate data 20H to accumulator.
Number of bytes:
Either 2 or 3 bytes long.
1st byte is opcode.
2nd byte 8 bit data .
3rd byte higher byte data of 16 bytes.

4. Register addressing mode
Data is transferred from one register to other.
Example:
MOV A, C :Transfer data from C register to accumulator.
Number of bytes:
Only 1 byte long.
One byte is opcode.

5. Direct addressing mode
Data is transferred from direct address to other register or vice-versa.
Example:
LDA C200H .Transfer contents from C200H to Acc.
Number of bytes:
These are 3 bytes long.
1st byte is opcode.
2nd byte lower address.
3rd byte higher address.

6. Indirect addressing mode
Data is transferred from address pointed by the data in a register to other register or vice-versa. Example: MOV A, M: Move contents from address pointed by M to Acc. Number of bytes: These are 3 bytes long. 1st byte is opcode. 2nd byte lower address. 3rd byte higher address.

7. A memory pointer register is used to store the address of the memory location
Example-
MOV M, A ;copy register A to memory location whose address is stored in register pair HL H L A
30H
20H
50H
2050H
30H

8. Implied addressing mode
These doesn’t require any operand. The data is specified in Opcode itself. Example: RAL: Rotate left with carry. No.of Bytes:1 These are single byte instruction or Opcode only.

9. Instruction & Data Formats
8085 Instruction set can be classified according to size (in bytes) as
1-byte instructions
2-byte instructions
3-byte instructions

10. One-byte Instructions
Includes Opcode and Operand in the same byte
Examples-
Opcode
Operand
Binary Code
Hex Code
MOV
C, A
4FH
ADD B
80H
HLT
76H

11. Two-byte Instructions
First byte specifies Operation Code
Second byte specifies Operand
Examples-
Opcode
Operand
Binary Code
Hex Code
MVI
A, 32H
3EH
32H
B, F2H
06H
F2H

12. Three-byte Instructions
First byte specifies Operation Code
Second & Third byte specifies Operand
Examples-
Opcode
Operand
Binary Code
Hex Code
LXI
H, 2050H
21H
50H
20H
LDA
3070H
3AH
70H
30H

13. Assembly Language of 8085
It uses English like words to convey the action/meaning called as MNEMONICS
For e.g.
MOV to indicate data transfer
ADD to add two values
SUB to subtract two values

14. Assembly language program to add two numbers
MVI A, 02H ;Copy value 2H in register A
MVI B, 04H ;Copy value 4H in register B
ADD B ;A = A + B

15. Microprocessor understands Machine Language only!
Microprocessor cannot understand a program written in Assembly language
A program known as Assembler is used to convert a Assembly language program to machine language
Assembly
Language
Program
Assembler
Program
Machine
Language
Code

16. Low-level/High-level languages
Machine language and Assembly language are both
Microprocessor specific (Machine dependent)
so they are called
Low-level languages
Machine independent languages are called
High-level languages
For e.g. BASIC, PASCAL,C++,C,JAVA, etc.
A software called Compiler is required to convert a high-level language program to machine code

17. 3.Instruction Set of 8085 Consists of 74 operation codes, e.g. MOV
246 Instructions, e.g. MOV A,B
8085 instructions can be classified as
Data Transfer (Copy)
Arithmetic
Logical and Bit manipulation
Branch
Machine Control

18. Data Transfer (Copy) Operations
Load a 8-bit number in a Register
Copy from Register to Register
Copy between Register and Memory
Copy between Input/Output Port and Accumulator
Load a 16-bit number in a Register pair
Copy between Register pair and Stack memory

19. Example Data Transfer (Copy) Operations / Instructions
Load a 8-bit number 4F in register B
Copy from Register B to Register A
Load a 16-bit number 2050 in Register pair HL
Copy from Register B to Memory Address 2050
Copy between Input/Output Port and Accumulator
MVI B, 4FH
MOV A,B
LXI H, 2050H
MOV M,B
OUT 01H
IN 07H

20. Arithmetic Operations
Addition of two 8-bit numbers
Subtraction of two 8-bit numbers
Increment/ Decrement a 8-bit number

21. Example Arithmetic Operations / Instructions
Add a 8-bit number 32H to Accumulator
Add contents of Register B to Accumulator
Subtract a 8-bit number 32H from Accumulator
Subtract contents of Register C from Accumulator
Increment the contents of Register D by 1
Decrement the contents of Register E by 1
ADI 32H
ADD B
SUI 32H
SUB C
INR D
DCR E

22. Logical Operations AND two 8-bit numbers OR two 8-bit numbers
Exclusive-OR two 8-bit numbers
Compare two 8-bit numbers
Complement
Rotate Left/Right Accumulator bits

23. Example Logical & Bit Manipulation Operations / Instructions
Logically AND Register H with Accumulator
Logically OR Register L with Accumulator
Logically XOR Register B with Accumulator
Compare contents of Register C with Accumulator
Complement Accumulator
Rotate Accumulator Left
ANA H
ORA L
XRA B
CMP C
CMA
RAL

24. Branching Operations
These operations are used to control the flow of program execution
Jumps
Conditional jumps
Unconditional jumps
Call & Return
Conditional Call & Return
Unconditional Call & Return

25. Example Branching Operations / Instructions
Jump to a 16-bit Address 2080H if Carry flag is SET
Unconditional Jump
Call a subroutine with its 16-bit Address
Return back from the Call
Call a subroutine with its 16-bit Address if Carry flag is RESET
Return if Zero flag is SET
JC 2080H
JMP 2050H
CALL 3050H
RET
CNC 3050H RZ

26. Machine Control Instructions
These instructions affect the operation of the processor. For e.g.
HLT Stop program execution
NOP Do not perform any operation

27. Writing a Assembly Language Program
Steps to write a program
Analyze the problem
Develop program Logic
Write an Algorithm
Make a Flowchart
Write program Instructions using Assembly language of 8085

28. Analyze the problem Program Logic
Program 8085 in Assembly language to add two 8-bit numbers and store 8-bit result in register C.
Analyze the problem
Addition of two 8-bit numbers to be done
Program Logic
Add two numbers
Store result in register C
Example
(99H) A
(39H) D
(D2H) C

29. Translation to 8085 operations
3. Algorithm
Translation to 8085 operations
Get two numbers
Add them
Store result
Stop
Load 1st no. in register D
Load 2nd no. in register E
Copy register D to A
Add register E to A
Copy A to register C
Stop processing

30. 4. Make a Flowchart Load 1st no. in register D
Start
Load 1st no. in register D
Load 2nd no. in register E
Load Registers D, E
Copy D to A
Copy register D to A
Add register E to A
Add A and E
Copy A to register C
Copy A to C
Stop processing
Stop

31. 5. Assembly Language Program
Get two numbers
Add them
Store result
Stop
Load 1st no. in register D
Load 2nd no. in register E
MVI D, 2H
MVI E, 3H
Copy register D to A
Add register E to A
MOV A, D
ADD E
Copy A to register C
MOV C, A
Stop processing
HLT

32. Program 8085 in Assembly language to add two 8-bit numbers
Program 8085 in Assembly language to add two 8-bit numbers. Result can be more than 8-bits.
Analyze the problem
Result of addition of two 8-bit numbers can be 9-bit
Example
(99H) A
(99H) B
(132H)
The 9th bit in the result is called CARRY bit.

33. How 8085 does it? Adds register A and B Stores 8-bit result in A
SETS carry flag (CY) to indicate carry bit
99H A + B
99H 1
32H
99H A CY

34. Storing result in Register memoryA CY 1
32H
Register B
Register C
Step-1 Copy A to C
Step-2
Clear register B
Increment B by 1

35. 2. Program Logic Add two numbers Copy 8-bit result in A to C
If CARRY is generated
Handle it
Result is in register pair BC

36. Translation to 8085 operations
3. Algorithm
Translation to 8085 operations
Load two numbers in registers D, E
Add them
Store 8 bit result in C
Check CARRY flag
If CARRY flag is SET
Store CARRY in register B
Stop
Load registers D, E
Copy register D to A
Add register E to A
Copy A to register C
Copy A to register C
Use Conditional Jump instructions
Clear register B
Increment B
Stop processing

37. 4. Make a Flowchart Start Stop False Load Registers D, E Clear BIf
CARRY
NOT SET
Load Registers D, E
False
Clear B
Copy D to A
Increment B
True
Add A and E
Copy A to C
Stop

38. 5. Assembly Language Program
Load registers D, E
MVI D, 2H
MVI E, 3H
Copy register D to A
Add register E to A
MOV A, D
ADD E
Copy A to register C
Copy A to register C
MOV C, A
Use Conditional Jump instructions
JNC END
Clear register B
Increment B
MVI B, 0H
INR B
Stop processing
HLT
END:

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