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.
Input : 72H (0111 0010)2 Output : 48H (in hexadecimal) (0011 0000)2 ((4x16)+(8x1))=72
- Load the BCD number in the accumulator
- Unpack the 2 digit BCD number into two separate digits. Let the left digit be BCD1 and the right one BCD2
- 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
- Load the BCD number from the memory location (201FH, arbitrary choice) into the accumulator
- Temporarily store the accumulator’s value in B
- Obtain BCD2 by ANDing the accumulator with 0FH and store it in C
- Restore the original value of the accumulator by moving the value in B to A. AND the accumulator with F0H
- If the value in the accumulator equals 0, then BCD2 is the final answer and store it in the memory location, 2020H (arbitrary)
- Else, shift the accumulator to right 4 times to obtain BCD1. Next step is to multiply BCD1 by 0AH
- 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
- Finally, add C to the accumulator and store the result in 2020H
2020H contains the result.
|2003H||MOV B, A|
|2006H||MOV C, A|
|2007H||MOV A, B|
|2011H||MOV D, A|
|2013H||MVI E, 0AH|
Store the BCD number in 201FH. 2020H contains its binary equivalent.
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