19 votes 19 votes Design a logic circuit to convert a single digit BCD number to the number modulo six as follows (Do not detect illegal input): Write the truth table for all bits. Label the input bits $I_1, I_2, \ldots$ with $I_1$ as the least significant bit. Label the output bits $R_1, R_2\ldots$ with $R_1$ as the least significant bit. Use $1$ to signify truth. Draw one circuit for each output bit using, altogether, two two-input AND gates, one two-input OR gate and two NOT gates. Digital Logic gatecse-2000 digital-logic min-no-gates descriptive + – Kathleen asked Sep 14, 2014 edited May 16, 2018 by Milicevic3306 Kathleen 2.9k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 17 votes 17 votes $${\begin{array}{|cccc|c|ccc|}\hline \bf{I_4}& \bf{I_3}& \bf{I_2}&\bf{ I_1}& &\bf{R_3}& \bf{R_2} & \bf{R_1}\\\hline 0&0&0&0&\bf{0} &0&0&0\\\hline 0&0&0&1&\bf{1}& 0&0&1 \\ \hline 0&0&1&0&\bf{2}& 0&1&0 \\ \hline 0&0&1&1&\bf{3}& 0&1&1 \\ \hline 0&1&0&0&\bf{4}& 1&0&0 \\ \hline 0&1&0&1&\bf{5} &1&0&1 \\ \hline 0&1&1&0&\bf{6}& 0&0&0 \\ \hline0&1&1&1&\bf{7}& 0&0&1\\ \hline 1&0&0&0&\bf{8}& 0&1&0 \\ \hline 1&0&0&1&\bf{9}& 0&1&1 \\ \hline \end{array}}$$ $R_1 = I_1$ $R_2 = I_2\overline{ I_3} + I_4$ $R_3 = I_3\overline{I_2}$ This requires $2$ NOT gates, $2$ two-input AND gates and $1$ two-input OR gate. sabir answered Oct 15, 2016 edited Apr 16, 2019 by ajaysoni1924 sabir comment Share Follow See all 8 Comments See all 8 8 Comments reply srestha commented Sep 24, 2017 reply Follow Share Simply Y0=D 2 votes 2 votes arch commented May 17, 2018 reply Follow Share why other inputs not consider here that is after 9. i think input should go from 0 to 15. 0 votes 0 votes krishn.jh commented May 28, 2018 i edited by krishn.jh Jul 29, 2018 reply Follow Share No. of Logic Gates given in question would be insufficient, so how you are going to implement your expression in circuit diagram??? and for Expression why you did not use don't care for (10,11,12,13,14,15) ? 1 votes 1 votes Nikhil Patil commented Jul 29, 2018 reply Follow Share why you did not use don't care for (10,11,12,13,14,15) ? 1 votes 1 votes mehul vaidya commented Oct 9, 2018 reply Follow Share wrong answer selected as best answer 0 votes 0 votes Arjun commented Oct 9, 2018 reply Follow Share corrected. 0 votes 0 votes talha hashim commented Nov 17, 2018 reply Follow Share R2=(i2)'(i4) + (i2)(i3') 0 votes 0 votes nsaisirisha commented Nov 14, 2019 reply Follow Share The question says that a single digit BCD is converted. BCD values range from 0 to 9 only. So we can consider all the output values for (10,11,12,13,14,15) as don't cares. The answer is right. :) 0 votes 0 votes Please log in or register to add a comment.
12 votes 12 votes After using Don't care (10,11,12,13,14,15) and after K-Map simplification you will get R1=I1 R2=I2.I3' + I4 R3= I3.I2' R4 = 0 Here, 2 input AND Gate used=2 2 input OR Gate used=1 NOT Gate used=2 krishn.jh answered May 28, 2018 krishn.jh comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes Don't cares are not shown in the table but considered while constructing K-map for $R_1,R_2,R_3.$ jatinmittal199510 answered Apr 10, 2021 jatinmittal199510 comment Share Follow See all 0 reply Please log in or register to add a comment.