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Instruction Word Size in Microprocessor

  • Last Updated : 21 Aug, 2019

The 8085 instruction set is classified into 3 categories by considering the length of the instructions. In 8085, the length is measured in terms of “byte” rather then “word” because 8085 microprocessor has 8-bit data bus. Three types of instruction are: 1-byte instruction, 2-byte instruction, and 3-byte instruction.

1. One-byte instructions –
In 1-byte instruction, the opcode and the operand of an instruction are represented in one byte.

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  • Example-1:
    Task- Copy the contents of accumulator in register B.



    Mnemonic- MOV B, A
    Opcode- MOV
    Operand- B, A
    Hex Code- 47H
    Binary code- 0100 0111 
  • Example-2:
    Task- Add the contents of accumulator to the contents of register B.
     Mnemonic- ADD B
    Opcode- ADD
    Operand- B
    Hex Code- 80H
    Binary code- 1000 0000 
  • Example-3:
    Task- Invert (complement) each bit in the accumulator.
    Mnemonic- CMA
    Opcode- CMA
    Operand-  NA
    Hex Code- 2FH
    Binary code- 0010 1111 

Note – The length of these instructions is 8-bit; each requires one memory location. The mnemonic is always followed by a letter (or two letters) representing the registers (such as A, B, C, D, E, H, L and SP).

2. Two-byte instructions –
Two-byte instruction is the type of instruction in which the first 8 bits indicates the opcode and the next 8 bits indicates the operand.

  • Example-1:
    Task- Load the hexadecimal data 32H in the accumulator.
    Mnemonic- MVI A, 32H
    Opcode- MVI
    Operand- A, 32H
    Hex Code- 3E
    32
    Binary code- 0011 1110
    0011 0010 
  • Example-2:
    Task- Load the hexadecimal data F2H in the register B.
    Mnemonic- MVI B, F2H
    Opcode- MVI
    Operand- B, F2H
    Hex Code- 06
    F2
    Binary code- 0000 0110
    1111 0010 

Note – This type of instructions need two bytes to store the binary codes. The mnemonic is always followed by 8-bit (byte) data.

3. Three-byte instructions –
Three-byte instruction is the type of instruction in which the first 8 bits indicates the opcode and the next two bytes specify the 16-bit address. The low-order address is represented in second byte and the high-order address is represented in the third byte.

  • Example-1:
    Task- Load contents of memory 2050H in the accumulator.
    Mnemonic- LDA 2050H
    Opcode- LDA
    Operand- 2050H
    Hex Code- 3A
    50
    20
    Binary code- 0011 1010
    0101 0000
    0010 0000 
  • Example-2:
    Task- Transfer the program sequence to the memory location 2050H.
    Mnemonic- JMP 2085H
    Opcode- JMP
    Operand- 2085H
    Hex Code- C3
    85
    20
    Binary code- 1100 0011
    1000 0101
    0010 0000 

Note – These instructions would require three memory locations to store the binary codes. The mnemonic is always followed by 16-bit (or adr).

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