Assemblylanguageprogrammingof8085 100523023329-phpapp02

Assembly Language
Programming of 8085
Unit-2

Topics
1. Introduction
2. Programming model of 8085
3. Instruction set of 8085
4. Example Programs
5. Addressing modes of 8085
6. Instruction & Data Formats of 8085

1. Introduction
• A microprocessor executes instructions given by
the user
• Instructions should be in a language known to
the microprocessor
• Microprocessor understands the language of 0’s
and 1’s only
• This language is called Machine Language

• For e.g.
01001111
– Is a valid machine language instruction of
8085
– It copies the contents of one of the internal
registers of 8085 to another

A Machine language program to
add two numbers

00111110 ;Copy value 2H in register A
00000010
00000110 ;Copy value 4H in register B
00000100
10000000 ;A = A + B

Assembly Language of 8085
• It uses English like words to convey the
action/meaning
• 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, 2H ;Copy value 2H in register A
MVI B, 4H ;Copy value 4H in register B
ADD B ;A = A + B

Note:
• Assembly language is specific to a given
processor
• For e.g. assembly language of 8085 is different
than that of Motorola 6800 microprocessor

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 Machine
Assembler
Language Language
Program
Program 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

2. Programming model of 8085

Accumulator 16-bit
Address Bus
Register Array
ALU

Memory Pointer
Flags Registers 8-bit Data
Bus
Instruction
Decoder

Control Bus
Timing and Control Unit

Accumulator (8-bit) Flag Register (8-bit)
S Z AC P CY

B (8-bit) C (8-bit)
D (8-bit) E (8-bit)
H (8-bit) L (8-bit)
Stack Pointer (SP) (16-bit)
Program Counter (PC) (16-bit)

8- Lines 16- Lines
Bidirectional Unidirectional

Overview: 8085 Programming model

1. Six general-purpose Registers
2. Accumulator Register
3. Flag Register
4. Program Counter Register
5. Stack Pointer Register

1. Six general-purpose registers
– B, C, D, E, H, L
– Can be combined as register pairs to
perform 16-bit operations (BC, DE, HL)
2. Accumulator – identified by name A
– This register is a part of ALU
– 8-bit data storage
– Performs arithmetic and logical operations
– Result of an operation is stored in
accumulator

3. Flag Register
– This is also a part of ALU
– 8085 has five flags named
• Zero flag (Z)
• Carry flag (CY)
• Sign flag (S)
• Parity flag (P)
• Auxiliary Carry flag (AC)

• These flags are five flip-flops in flag register
• Execution of an arithmetic/logic operation can
set or reset these flags
• Condition of flags (set or reset) can be tested
through software instructions
• 8085 uses these flags in decision-making
process

4. Program Counter (PC)
– A 16-bit memory pointer register
– Used to sequence execution of program
instructions
– Stores address of a memory location
• where next instruction byte is to be fetched
by the 8085
– when 8085 gets busy to fetch current
instruction from memory
• PC is incremented by one
• PC is now pointing to the address of next
instruction

5. Stack Pointer Register
– a 16-bit memory pointer register
– Points to a location in Stack memory
– Beginning of the stack is defined by loading
a 16-bit address in stack pointer register

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
1. Data Transfer (Copy)
2. Arithmetic
3. Logical and Bit manipulation
4. Branch
5. Machine Control

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

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

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

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

3. Logical & Bit Manipulation
Operations
1. AND two 8-bit numbers
2. OR two 8-bit numbers
3. Exclusive-OR two 8-bit numbers
4. Compare two 8-bit numbers
5. Complement
6. Rotate Left/Right Accumulator bits

Example Logical & Bit Manipulation
1. Logically AND Register H ANA H
with Accumulator
2. Logically OR Register L with ORA L
Accumulator
3. Logically XOR Register B XRA B
with Accumulator
4. Compare contents of CMP C
Register C with Accumulator
5. Complement Accumulator CMA
6. Rotate Accumulator Left RAL

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

Example Branching
1. Jump to a 16-bit Address JC 2080H
2080H if Carry flag is SET
2. Unconditional Jump JMP 2050H
3. Call a subroutine with its 16-bit CALL 3050H
Address
4. Return back from the Call RET
5. Call a subroutine with its 16-bit CNC 3050H
Address if Carry flag is RESET
6. Return if Zero flag is SET RZ

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

4. 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

Program 8085 in Assembly language to add two 8-
bit numbers and store 8-bit result in register C.

1. Analyze the problem
– Addition of two 8-bit numbers to be done
2. Program Logic
– Add two numbers
– Store result in register C
– Example
10011001 (99H) A
+00111001 (39H) D
11010010 (D2H) C

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

4. Stop • Stop processing

4. Make a Flowchart
Start
• Load 1st no. in register D
Load Registers D, E • Load 2nd no. in register E

Copy D to A • Copy register D to A
Add A and E

Copy A to C • Copy A to register C

• Stop processing
Stop

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

Program 8085 in Assembly language to add two 8-
bit numbers. Result can be more than 8-bits.

1. 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 99H B

1
0 00110010
10011001 32H
99H A
CY

• Storing result in Register memory
CY A
1 10011001 32H

Register B Register C

Step-1 Copy A to C
Step-2
a) Clear register B
b) Increment B by 1

2. Program Logic

1. Add two numbers
2. Copy 8-bit result in A to C
3. If CARRY is generated
– Handle it
4. Result is in register pair BC

Translation to 8085
3. Algorithm
operations

1. Load two numbers in • Load registers D, E
registers D, E
• Copy register D to A
2. Add them
• Copy A to register C
3. Store 8 bit result in C
4. Check CARRY flag • Use Conditional Jump
5. If CARRY flag is SET instructions
• Store CARRY in
• Clear register B
register B
• Increment B
4. Stop
• Stop processing

4. Make a Flowchart
Start

Load Registers D, E If False
CARRY Clear B
NOT SET
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
• Copy A to register C ADD E
MOV C, A
• Use Conditional Jump
JNC END
instructions
• Clear register B MVI B, 0H
• Increment B INR B
• Stop processing END: HLT

4. Addressing Modes of 8085
• Format of a typical Assembly language instruction
is given below-
[Label:] Mnemonic [Operands] [;comments]
HLT
MVI A, 20H
MOV M, A ;Copy A to memory location whose
address is stored in register pair HL
LOAD: LDA 2050H ;Load A with contents of memory
location with address 2050H
READ: IN 07H ;Read data from Input port with
address 07H

• The various formats of specifying
operands are called addressing modes
• Addressing modes of 8085
1. Register Addressing
2. Immediate Addressing
3. Memory Addressing
4. Input/Output Addressing

1. Register Addressing
• Operands are one of the internal registers
of 8085
• Examples-
MOV A, B
ADD C

2. Immediate Addressing
• Value of the operand is given in the
instruction itself
• Example-
MVI A, 20H
LXI H, 2050H
ADI 30H
SUI 10H

3. Memory Addressing
• One of the operands is a memory location
• Depending on how address of memory
location is specified, memory addressing
is of two types
– Direct addressing
– Indirect addressing

3(a) Direct Addressing
• 16-bit Address of the memory location is
specified in the instruction directly
• Examples-
LDA 2050H ;load A with contents of memory
STA 3050H ;store A with contents of memory

3(b) Indirect Addressing
• 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

4. Input/Output Addressing
• 8-bit address of the port is directly
specified in the instruction
• Examples-
IN 07H
OUT 21H

5. Instruction & Data Formats
8085 Instruction set can be classified
according to size (in bytes) as
1. 1-byte Instructions

1. 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

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

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

Assemblylanguageprogrammingof8085 100523023329-phpapp02

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