We can use NAND gate only to get ( XOR gate ) of 2 input ( A and B ) :(adsbygoogle = window.adsbygoogle || []).push({});

By using (4) NAND gate :

The output of 1'st NAND : (AB)'

The output of 2'nd NAND : ((AB)'.A)' = (A'B)

The output of 3'rd NAND : ((AB)'.B)' = (AB')

The output of the hole circuit will be: ((A'B).(AB'))' = AB' + A'B ( Which is an XOR gate )

The truth table of ( XOR gate ) is:

00 0

01 1 A'B

10 1 AB'

11 0

What about in case of 3 input ( A, B, and C ) ?!!

I Know the output should be by using (8) NAND gates :

000

001 (A'B'C)

010 (A'BC')

011

100 (AB'C')

101

110

111 (A'B'C')

Y= (A'B'C)+(A'BC')+(AB'C')+ (A'B'C')

But how to drive it ? I mean the output at each NAND ?

Any HELP !!!

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# Homework Help: 3 input XOR gate using NAND gate only ( Logic )

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