Page 137 - DCAP210_INTRODUCTION__TO_MICROPROCESSORS

Unit 9: Introduction to 8085 Instructions

For example: Notes

Task Op code Operand Binary Code Hex Code
Copy the
contents of the
accumulator in
the register C. MOV C,A 01001111 4FH

Add the contents
of register B to
the contents of
the accumulator ADD B 1000 0000 80H
Invert (compliment)
each bit in the
accumulator. CMA 00101111 2FH

These instructions are 1-byte instructions performing three different tasks. In the first instruction,
both operand registers are specified. In the second instruction, the operand B is specified and the
accumulator is assumed. Similarly, in the third instruction, the accumulator is assumed to be the
implicit operand. These instructions are stored in 8- bit binary format in memory; each requires
one memory location.
MOV rd, rs
rd <— rs copies contents of rs into rd.

Coded as 01 ddd sss where ddd is a code for one of the 7 general registers which is the destination
of the data, sss is the code of the source register.

Example: MOV A, B

Coded as 01111000 = 78H = 170 octal (octal was used extensively in instruction design of such
processors).
ADD r
A <— A + r
Two-Byte Instructions

In a two-byte instruction, the first byte specifies the operation code and the second byte specifies
the operand. Source operand is a data byte immediately following the opcode. For example:

Task Op code Operand Binary Code Hex Code
Load an 8-bit Data 3E First Byte
data byte in
the accumulator. MVI A, 0011 1110 Data Second Byte

Assume that the data byte is 32H. The assembly language instruction is written as

Mnemonics Hex code
MVI A, 32H 3E 32H

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