1 votes 1 votes Minimum number of 2 input NAND gate required to implement 4 input NAND gate is __________________ A_i_$_h asked Jul 19, 2017 A_i_$_h 978 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments joshi_nitish commented Jul 19, 2017 reply Follow Share ~(ab) and ~(cd) given to nand gate will not give 'abcd',it will give... ~(~(ab).~(cd))= ab + cd..... 0 votes 0 votes A_i_$_h commented Jul 19, 2017 reply Follow Share got it !!! thank you :) 0 votes 0 votes $ruthi commented Jul 20, 2017 reply Follow Share how can it be 4 nand gate? first nand gate produces (ab)' and next nand gate produces (cd)' third nand gate produces ((ab)'.(cd)')' = ((ab)')'+((cd)')' = ab+cd 4th nand gate produces (ab+cd)' = (a'+b')(c'+d') 0 votes 0 votes Please log in or register to add a comment.
Best answer 2 votes 2 votes 5 AND gates are required, Please correct me if i am wrong, AnilGoudar answered Jul 20, 2017 selected Jul 22, 2017 by A_i_$_h AnilGoudar comment Share Follow See all 0 reply Please log in or register to add a comment.