how many two input NAND gates are required to perform the action of a two input OR gate?
Answers
Answered by
3
We can use 3 NAND gates to perform the action of an OR gate.
We first use 2 NAND gates, and their outputs are connected to the third NAND gate, which then produces the same result as an OR gate.
If X is the initial input in the first NAND gate, we get
(X.X)`
= X`
If Y is the initial input in the second NAND gate, we get
(Y.Y)`
= Y`
When we connect these two outputs to the input of the third NAND gate, we get
(X` . Y`)`
= (X`)` + (Y`)`
= X + Y
which is an OR gate.
We first use 2 NAND gates, and their outputs are connected to the third NAND gate, which then produces the same result as an OR gate.
If X is the initial input in the first NAND gate, we get
(X.X)`
= X`
If Y is the initial input in the second NAND gate, we get
(Y.Y)`
= Y`
When we connect these two outputs to the input of the third NAND gate, we get
(X` . Y`)`
= (X`)` + (Y`)`
= X + Y
which is an OR gate.
Answered by
5
correct Answer is : Three NAND gates
Explanation :
⇒We can use three NAND gates to perform the action of an two input OR gate.
⇒NAND gate is called as Universal gate.
⇒See the attachment for the figure:
⇒First two NAND gates are used and the ouput is given to Third NAND gate which results in OR gate .
____
_ _ = =
⇒The Boolean Expression is : Y= A.B = A +B = A+B read as Y= A or B
Explanation :
⇒We can use three NAND gates to perform the action of an two input OR gate.
⇒NAND gate is called as Universal gate.
⇒See the attachment for the figure:
⇒First two NAND gates are used and the ouput is given to Third NAND gate which results in OR gate .
____
_ _ = =
⇒The Boolean Expression is : Y= A.B = A +B = A+B read as Y= A or B
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