Digital Electronics - Logic Gates - Discussion

In question he is not discussed about second input .we can take second input as a dont care if we apply the 0&dont care combination NAND gate will give the high output
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Ok, then can anyone explain how its not NOR.

@Mayphics

Consider one situation that if you apply 2 inputs to NOR gate in that one of the input is LOW(i,e 0)..

Then the OUTPUT is entirely depends on the second input which will you provide.......

means ,,,,,,,2 Conditions here

1.IF second input is LOW(0) Then OUTPUT of the NOR gate is HIGH(1)

2.IF second input is HIGH(1), then OUTPUT of NOR gate Iis LOW(0)

But in this question we req. OUTPUT SHOULD be HIGH means OUTPUT independent of INPUT.......

So ANSWER is NAND GATE........

Here second input is not given, then we can take any i/p. 0 or 1.

In nor gate if one input is 0, and other is 1, then the output will be 0. If both input are low then the o/p is 1, so we conform that nor gate is not the answer but in nand gate if one input is 0 and other is either 0 or 1, the output is high. So the answer is nanad gate.

Can any one say why it is not OR?

Why it is NAND gate not NOR gate?

Hey when we give 1 low input and get high output the OR gate also gives the same result. So why it is only NAND gate not OR gate?

In OR gate, if one input is zero and the other is 1, then O/P is 1. If both I/P's are 0, then O/P is 0. So the answer is NAND gate because in NAND gate if one I/P is 0 and the other is either 0 or 1, then O/P will be 1.

@Krishna.

NOR GATE means if any of the INPUT is HIGH OUTPUT is LOW.
But here condition is ONE of the INPUT is low and OUTPUT should be HIGH.
Its not valid in NOR GATE.
SO NAND-GATE will satisfy above condtion in all cases.
A B AND NAND OR NOR
0 0 0 1 0 1
0 1 0 1 1 0

From above table we can say answer is NAND-GATE(By-Default A=0, and B=0 or 1, o/p of NAND-GATE is HIGH only).