8085 Instruction Memory Occupancy

The question asks to identify the 8085 microprocessor instruction that occupies 3 bytes in memory. We need to determine the size of each instruction based on its opcode and operands.

Analyzing Instruction Sizes

The size of an 8085 instruction in bytes is determined by its opcode and the type of addressing mode used. Instructions using direct addressing or immediate data operands are generally larger than those using register addressing.

• $MVI A, 45H$: This instruction loads the immediate hexadecimal value $45H$ into the Accumulator (register A). It requires one byte for the opcode ($MVI A$) and one byte for the immediate data ($45H$). Total size = 2 bytes.
• $LDA 2000H$: This instruction loads the Accumulator with data from a specific memory location. It requires one byte for the opcode ($LDA$) and two bytes for the 16-bit memory address ($2000H$). Total size = 3 bytes. This instruction meets the criteria.
• $ADD C$: This instruction adds the contents of register C to the Accumulator. It uses register addressing, which requires only one byte for the opcode ($ADD C$). Total size = 1 byte.
• $INR B$: This instruction increments the value in register B. It also uses register addressing and requires only one byte for the opcode ($INR B$). Total size = 1 byte.

Determining the 3-Byte Instruction

By examining the byte allocation for each instruction:

• $MVI A, 45H$ is 2 bytes.
• $LDA 2000H$ is 3 bytes.
• $ADD C$ is 1 byte.
• $INR B$ is 1 byte.

Therefore, the $LDA 2000H$ instruction is the one that occupies 3 bytes in memory.