13 Jun 2026

Addition, Subtraction, and Multiplication Using the 8085 Microprocessor

practical pg-iv electronics 8085 microprocessor arithmetic

Aim

To write and execute 8085 programs for addition, subtraction, and multiplication of two 8-bit numbers.

Apparatus

8085 microprocessor trainer, keyboard, monitor or serial terminal, power supply, and connecting leads.

Experimental arrangement

The trainer accepts the program and data, executes the arithmetic instructions, and displays the result and status flags.

Theory

The 8085 stores every operand as an 8-bit binary word. Before an arithmetic instruction can operate, the program places one operand in the accumulator and the other in a general register or memory location. During the fetch-decode-execute cycle, the program counter supplies the address of the next opcode, the control unit decodes it, and the arithmetic-logic unit performs the specified binary operation. The result normally returns to the accumulator.

For addition, ADD r performs

\[A\leftarrow A+r,\]

whereas ADC r also includes the previous carry. If the sum exceeds $FF_{16}$, only the lower eight bits remain in $A$ and the carry flag becomes one. Subtraction is carried out internally by adding the two’s complement of the subtrahend. Thus SUB r performs $A\leftarrow A-r$; in 8085 convention, the carry flag is set when a borrow is required. SBB includes the previous borrow.

Each result updates the sign, zero, auxiliary-carry, parity, and carry flags. These flags are not decorative outputs: conditional jump instructions use them to control program flow. Since the 8085 has no hardware multiply instruction, multiplication is implemented by repeated addition. The multiplicand is added to a partial sum while the multiplier register is decremented to zero. For an 8-bit by 8-bit product, a register pair should be used because the result may require sixteen bits. A correct practical therefore verifies both the numerical result and the relevant flag or high-byte condition.

Sample verification

Operation	Operand 1	Operand 2	Expected result
addition	25H	17H	3CH
subtraction	35H	12H	23H
multiplication	06H	04H	0018H

Calculation

The hexadecimal operands are first interpreted as their decimal values for checking:

\[25_H+17_H=37+23=60=3C_H.\]

For subtraction,

\[35_H-12_H=53-18=35=23_H.\]

The 8085 has no multiplication instruction, so the program adds $06_H$ four times:

\[06_H+06_H+06_H+06_H=24_{10}=18_H.\]

Since the product is larger than one byte only when it exceeds $FF_H$, it is stored here as the two-byte result $0018_H$.

Result

The three programs execute correctly and the displayed accumulator/memory values agree with the expected arithmetic results.

Viva Questions

1. Why is multiplication done by repeated addition? The 8085 instruction set has no direct multiply instruction.
2. What does the carry flag indicate? Carry out of the most significant bit in unsigned arithmetic.
3. What is the accumulator used for? It is the main operand and result register of the ALU.

Maxima Code

Download the 8085 arithmetic check.