BCD to binary conversion in 8051

In this problem, we will see how to convert 8-bit BCD number to its Binary (Hexadecimal)equivalent. The BCD number is stored at location 20H. After converting, the results will be stored at 30H.

So let us assume the data is D5H. The program converts the binary value ofD5H to BCD value 213D.

Program

   MOVR0,#20H; Initialize the address of the data
   MOVA,@R0;Get the data from an address, which is stored in R0
   MOVR2,A; Store the content of A into R2
   
   CLRA;Clear the content of A to 00H
   MOVR3,#00H
LOOP:   ADDA,#01H;increment A by 01H
   DAA; Decimal Adjust of the accumulator content
   INCR3; Increase R3 to keep track of the hex value
   CJNEA,02H,LOOP ;Run the loop until A = R2

MOVR0,#30H; Point the destination location
MOVA,R3; Move R3 to A
MOV@R0,A; Store A content to the memory location pointed by R0
HALT:   SJMPHALT

Here the logic is very simple. We are just taking the number from the memory. And storing that value into R2. By this value, we can compare while executing the loop.

The accumulator (A) and register R3 is set to 00H at first. So we are just increasing the value of A by 01H. We can use the INC A instruction to increment the value, but in this case, the value is incremented by using ADDA, #01H. The reason behind it is the INCinstruction does not affect the CY and AC flags. But these flags are needed to use DA A instruction for decimal adjusting. After increasing the value of A, DA A instruction is executed to convert the value to decimal. By using this decimal value, we can compare with the number stored inR2. In each iteration the value of R3 is incremented by 1, this is acting like a counter. So at the end, we are getting the output from register R3.

Output

Method 2

We can do the same thing using some other logic also. Here no additional loops are needed to do the task. We are just multiplying the digit of 10^th place of the BCD number with 0AH. Then adding the second digit with the result to get the number. 

If the number is 94, then it is multiplying 0AH with 9.

(9 * 0AH) = 5AH, thenadding 4 with 5AH. (5AH + 4) = 5EH.

Program

MOVR0,#20H; Initialize the address of the data
MOVA,@R0; Get the data from an address, which is stored in R0
MOVR2,A;Store the content of A into R2
SWAPA; Swap the nibbles of A register
ANLA,#0FH;Mask the Higher Nibble of A
MOVB,0AH;Take 0AH = 10D into B
MULAB ;Multiply A with B, and get result into A
MOVA,R2;Copy R2 content to A
ANLA,#0FH;Mask the higher nibble of A
ADDA,R2
MOVR0,#30H;Point the destination location
MOVA,R3;Move R3 to A
MOV@R0,A;Store A content to memory location pointed by R0
HALT:   SJMPHALT