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Minimum number of 2 input NAND gate required to implement 4 input NAND gate is __________________
in Digital Logic by Boss (10.2k points) | 551 views
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5 ??
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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
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~(ab) and ~(cd) given to nand gate will not give 'abcd',it will give...

~(~(ab).~(cd))= ab + cd.....
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got it !!! thank you :)
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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')

1 Answer

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Best answer

5 AND gates are required,

Please correct me if i am wrong,

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