10 votes 10 votes The number of $NAND$ gates are required to realize the following function $f$ given below is ________. $f = AB + BC + CD + ..... + YZ$ Assume that it is Multi input $NAND$ Gate. GATE tbb-mockgate-2 numerical-answers digital-logic digital-circuits circuit-output + – Bikram asked Jan 24, 2017 retagged Sep 12, 2020 by ajaysoni1924 Bikram 1.3k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 7 votes 7 votes There are 25 minterms in the SOP. Number of NAND gates = Number of minterms + 1 = 25+1=26. Bikram answered Jan 24, 2017 selected Jan 24, 2017 by Tendua Bikram comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments sushmita commented Feb 2, 2017 reply Follow Share yes for sure, multi input nand gate is used. 1 votes 1 votes akash.dinkar12 commented Sep 16, 2017 reply Follow Share hope it helps!!! 11 votes 11 votes Kabi commented Dec 26, 2017 reply Follow Share correct! 1 votes 1 votes Please log in or register to add a comment.
5 votes 5 votes AB+BC+AC=((AB)'(BC)'(AC)')' (A+B)(A+C)(B+C)=((A+B)'+(A+C)'+(B+C)')' Number of NAND gates = Number of minterms + 1 = 25+1=26. Shreya Roy answered Jan 27, 2017 Shreya Roy comment Share Follow See all 2 Comments See all 2 2 Comments reply sushmita commented Feb 2, 2017 reply Follow Share ab+bc+cd should be there na?? why ab+bc+ac? plz elaborate the soln 0 votes 0 votes Shreya Roy commented Feb 2, 2017 reply Follow Share I took a random example .. not the exact terms present in the qs.. 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes AND-OR realization is equal to NAND-NAND realization. So total AND and OR is (25 + 1) = 26 NAND gate needed. Tanmoy Mondal answered Feb 4, 2017 Tanmoy Mondal comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes (SOP) AND-OR realisation $\equiv$ NAND-NAND realisation (POS) OR-AND realisation $\equiv$ NOR-NOR realisation See this: https://www.electrical4u.com/electrical-mcq.php?subject=digital-electronics&page=18 http://www.ece.ualberta.ca/~lkurgan/EE280/SN-15.pdf AB, BC, CD...YZ require $25$ 2-input AND gates. Then, we can OR all of them with a $single$ 25-input OR gate. This is equivalent to using NAND gates of the exact same pattern. We will need $25$ 2-input NAND gates, and $1$ 25-input NAND gate. So, 26 NAND gates. See the image provided in Akash Dinkar's comment for clarity. JashanArora answered Jan 13, 2020 JashanArora comment Share Follow See all 0 reply Please log in or register to add a comment.