Programming with 8085

ASSEMBLY LANGUAGE
PROGRAMMING OF 8085
Presentation By: Shehrevar Davierwala
Contact : shehrevar@live.com
Lecturer at GES’S Katgara Polytechnic Institute

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 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, 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
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

2. PROGRAMMING MODEL OF 8085
Accumulator
ALU
Flags
Instruction
Decoder
Register Array
Memory Pointer
Registers
Timing and Control Unit
16-bit
Address Bus
8-bit Data
Bus
Control Bus

Accumulator (8-bit) Flag Register (8-bit)
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)
S Z AC P CY
16- Lines
Unidirectional
8- Lines
Bidirectional

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)
1. 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

(COPY)
OPERATIONS /
INSTRUCTIONS
1. Load a 8-bit number 4F in
register B
2. Copy from Register B to
Register A
3. Load a 16-bit number 2050
in Register pair HL
4. Copy from Register B to
Memory Address 2050
5. Copy between Input/Output
Port and Accumulator
MVI B, 4FH
MOV A,B
LXI H, 2050H
MOV M,B
OUT 01H
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
Accumulator
2. Add contents of Register B to
Accumulator
3. Subtract a 8-bit number 32H
from Accumulator
4. Subtract contents of Register C
from Accumulator
5. Increment the contents of
Register D by 1
6. Decrement the contents of
Register E by 1
ADI 32H
ADD B
SUI 32H
SUB C
INR D
DCR E

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

MANIPULATION
OPERATIONS /
INSTRUCTIONS
1. Logically AND Register H
with Accumulator
2. Logically OR Register L
with Accumulator
3. Logically XOR Register B
with Accumulator
4. Compare contents of
Register C with
Accumulator
5. Complement Accumulator
6. Rotate Accumulator Left
ANA H
ORA L
XRA B
CMP C
CMA
RAL

4. BRANCHING OPERATIONS
These operations are used to control the flow of
program execution
1.Jumps
 Conditional jumps
 Unconditional jumps
1.Call & Return
 Conditional Call & Return
 Unconditional Call & Return

EXAMPLE BRANCHING
OPERATIONS /
INSTRUCTIONS
1. Jump to a 16-bit Address 2080H
if Carry flag is SET
2. Unconditional Jump
3. Call a subroutine with its 16-bit
Address
4. Return back from the Call
5. Call a subroutine with its 16-bit
Address if Carry flag is RESET
6. Return if Zero flag is SET
JC 2080H
JMP 2050H
CALL 3050H
RET
CNC 3050H
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
1. Program Logic
 Add two numbers
 Store result in register C
 Example
10011001 (99H) A
+00111001 (39H) D
11010010 (D2H) C

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

4. MAKE A FLOWCHART
Start
Load Registers D, E
Copy D to A
Add A and E
Copy A to C
Stop
• Load 1st
no. in register D
• Load 2nd
no. in register E
• Stop processing

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

0
 How 8085 does it?
 Adds register A and B
 Stores 8-bit result in A
 SETS carry flag (CY) to indicate carry bit
10011001
10011001
A
B
+
99H
99H
10011001 A1
CY
00110010 99H32H

 Storing result in Register memory
10011001
A
32H1
CY
Register CRegister B
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
1. Result is in register pair BC

3. ALGORITHM
1. Load two numbers
in registers D, E
2. Add them
3. Store 8 bit result in
C
4. Check CARRY flag
5. If CARRY flag is
SET
• Store CARRY in
register B
4. Stop
 Load registers D, E
Translation to 8085
operations
• Stop processing
• Use Conditional
Jump instructions
• Clear register B
• Increment B

4. MAKE A FLOWCHART
Start
Load Registers D, E
Copy D to A
Add A and E
Copy A to C
Stop
If
CARRY
NOT SET
Clear B
Increment B
False
True

5. ASSEMBLY LANGUAGE PROGRAM
MVI D, 2H
MVI E, 3H
MOV A, D
ADD E
MOV C, A
HLT
• Load registers D, E
• Stop processing
• Use Conditional
Jump instructions
• Clear register B
• Increment B
JNC END
MVI B, 0H
INR B
END:

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 location with address
2050H
STA 3050H ;store A with contents of
memory location with address
3050H

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
30HA 20H
H
50H
L
30H2050H

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
0011 0010
3EH
32H
MVI B, F2H 0000 0110
1111 0010
06H
F2H

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

HEXADECIMAL NUMBERS AND
STORE IT IN TWO DIFFERENT
LOCATIONS
 LDA 2200H ; Get the packed BCD number
 ANI F0H ; Mask lower nibble
0100 0101 45
1111 0000 F0
---------------
0100 0000 40
 RRC
 RRC
 RRC ; Adjust higher digit as a lower
digit.
 RRC 0000 0100 after 4 rotations

CONTD.
 STA 2300H ; Store the partial result
 LDA 2200H ; Get the original BCD no.
 ANI 0FH; Mask higher nibble
0100 0100 45
0000 1111 0F
---------------
0000 0100 05
 STA 2301H ; Store the result
 HLT ; Terminate program
execution

BLOCK DATA TRANSFER
 MVI C, 0AH ; Initialize counter i.e no. of
bytes
Store the count in Register C, ie
ten
 LXI H, 2200H ; Initialize source memory pointer
Data Starts from 2200 location
 LXI D, 2300H ; Initialize destination memory
pointer
BK: MOV A, M ; Get byte from source memory
block
i.e 2200 to accumulator.
 STAX D ; Store byte in the destination
memory block i.e 2300 as stored in
D-E pair

CONTD.
 INX H ; Increment source memory
pointer
 INX D ; Increment destination
memory pointer
 DCR C ; Decrement counter
to keep track of bytes moved
 JNZ BK ; If counter 0 repeat steps
 HLT ; Terminate program

Programming with 8085

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