8085 program to convert a BCD number to binary

Problem – Write an assembly language program for converting a 2 digit BCD number to its binary equivalent using 8085 microprocessor.

Examples:

Input : 72H (0111 0010)2
Output : 48H (in hexadecimal) (0011 0000)2
((4x16)+(8x1))=72

Algorithm:

1. Load the BCD number in the accumulator
2. Unpack the 2 digit BCD number into two separate digits. Let the left digit be BCD₁ and the right one BCD₂
3. Multiply BCD₁ by 10 and add BCD₂ to it

If the 2 digit BCD number is 72, then its binary equivalent will be
7 x OAH + 2 = 46H + 2 = 48H

Steps:

1. Load the BCD number from the memory location (201FH, arbitrary choice) into the accumulator
2. Temporarily store the accumulator’s value in B
3. Obtain BCD₂ by ANDing the accumulator with 0FH and store it in C
4. Restore the original value of the accumulator by moving the value in B to A. AND the accumulator with F0H
5. If the value in the accumulator equals 0, then BCD₂ is the final answer and store it in the memory location, 2020H (arbitrary)
6. Else, shift the accumulator to right 4 times to obtain BCD₁. Next step is to multiply BCD₁ by 0AH
7. Multiplication: Move BCD₁ to D and initialise E with 0AH as the counter. Clear the accumulator to 0 and add D to it E number of times
8. Finally, add C to the accumulator and store the result in 2020H

2020H contains the result.

Store the BCD number in 201FH. 2020H contains its binary equivalent.

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