Instruction type MOV M, r in 8085 Microprocessor

In 8085 Instruction set, this instruction MOV M, r will copy 8-bit value from the register r to the memory location as pointed by HL register pair.This instruction uses register addressing for specifying the data.

As “r” can have any one of the seven values −

r = A, B, C, D, E, H, or L

Thus there are seven opcodes for this type of instruction. It occupies only 1-Byte in memory.

Mnemonics, Operand Opcode(in HEX) Bytes
MOV M, A 77 1
MOV M, B 70 1
MOV M, C 71 1
MOV M, D 72 1
MOV M, E 73 1
MOV M, H 74 1
MOV M, L 75 1

MOV M, E is an example instruction of this type. It is a 1-Byte instruction. Let us suppose, E is having the initial value ABH, HL register pair is pointing to the memory location 4050H, 4050H memory location’s content is CDH. Then after execution of the instruction MOV M, E, E register’s content will be CDH. The result of execution of this instruction is shown with this example is shown below −

Before After

(E)

 ABH ABH

(HL)

 (4050H) (4050H)

(4050H)

 CDH ABH

Address Hex Codes Mnemonic Comment
2005 273 MOV M, E Memory location pointer by HL register pair <- E

Here is the timing diagram of the instruction MOV M, E as below.

Mov MR

Summary − So this instruction MOV M, E requires 1-Byte, 2-Machine Cycles (Opcode Fetch, Memory Read) and 7 T-States for execution as shown in the timing diagram.

Updated on: 30-Jul-2019

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