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:
- 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
Steps:
- 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.
ADDRESS | LABEL | MNEMONIC |
---|---|---|
2000H | LDA 201FH | |
2001H | ||
2002H | ||
2003H | MOV B, A | |
2004H | ANI 0FH | |
2005H | ||
2006H | MOV C, A | |
2007H | MOV A, B | |
2008H | ANI F0H | |
2009H | ||
200AH | JZ SKIPMULTIPLY | |
200BH | ||
200CH | ||
200DH | RRC | |
200EH | RRC | |
200FH | RRC | |
2010H | RRC | |
2011H | MOV D, A | |
2012H | XRA A | |
2013H | MVI E, 0AH | |
2014H | ||
2015H | SUM | ADD D |
2016H | DCR E | |
2017H | JNZ SUM | |
2018H | ||
2019H | ||
201AH | SKIPMULTIPLY | ADD C |
201BH | STA 2020H | |
201CH | ||
201DH | ||
201EH | HLT |
Store the BCD number in 201FH. 2020H contains its binary equivalent.
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