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 1.0k views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply joshi_nitish commented Jul 19, 2017 reply Follow Share 5 ?? 0 votes 0 votes A_i_$_h commented Jul 19, 2017 reply Follow Share how ...?? why cant it be 4? output of one nand is ab complement output of another one cd complement these both given to a nand gate to give abcd complement cpmplement => abcd now give to another nand => abcd complement therefore 4 nand gates right 0 votes 0 votes joshi_nitish commented Jul 19, 2017 reply Follow Share ~(ab) and ~(cd) given to nand gate will not give 'abcd',it will give... ~(~(ab).~(cd))= ab + cd..... 0 votes 0 votes A_i_$_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.