I was wondering how it would be without using the double dabble algorithm in binary bcd conversion. I tried two methods. This circuit was more complicated than the other. Because I did it by creating a table and deriving formulas. However, I did not use the Karnaugh map method. Maybe it can be simplified by using the Karnaugh method. The formulas I created are as follows. If we call the inputs ABCDE and the outputs Q0, Q1…..Q5
Q0= A.(B+C)
Q1= (B.(C+D)+A).(q0+(B.C.D))
Q2=(a.q1.B)+(((A.D.Q1)+(A.B.C)).(Q0XORQ1))
Q3=((a.q1.C)+(q0.((B.C.D)+A))+(q1.(A.B))).q2
Q4=D XOR Q1u
Q5=E
In the other method, I used almost the same number of Logic elements. I couldn't simplify it any further. I will share that too.
