Microprocessor T. Y. B. Sc.

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

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.

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.

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.

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.

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

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.

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

One-byte Instructions Includes Opcode and Operand in the same byte Examples- Opcode Operand Binary Code Hex Code MOV C, A 0100 1111 4FH ADD B 1000 0000 80H HLT 0111 0110 76H

Two-byte Instructions First byte specifies Operation Code Second byte specifies Operand Examples- Opcode Operand Binary Code Hex Code MVI A, 32H 0011 1110 0011 0010 3EH 32H B, F2H 0000 0110 1111 0010 06H F2H

Three-byte Instructions First byte specifies Operation Code Second & Third byte specifies Operand Examples- Opcode Operand Binary Code Hex Code LXI H, 2050H 0010 0001 0101 0000 0010 0000 21H 50H 20H LDA 3070H 0011 1010 0111 0000 0011 0000 3AH 70H 30H

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 10011001 (99H) A +00111001 (39H) D 11010010 (D2H) C

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

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

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

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 10011001 (99H) A +10011001 (99H) B 100110010 (132H) The 9th bit in the result is called CARRY bit.

How 8085 does it? Adds register A and B Stores 8-bit result in A SETS carry flag (CY) to indicate carry bit 10011001 99H A + 10011001 B 99H 1 00110010 10011001 32H 99H A CY

Storing result in Register memory A CY 1 10011001 32H Register B Register C Step-1 Copy A to C Step-2 Clear register B Increment B by 1

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

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

4. Make a Flowchart Start Stop False Load Registers D, E Clear B If 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

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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