Digital Electronics - Logic Gates - Discussion

@Ethan.

Check the answer using two inputs it also gives the same answer that I explain using three inputs.

Considering the truth table of NAND gate.
A B OUTPUT
0 0 1
0 1 1
1 0 1
1 1 0 = here we get the output HIGH when the two inputs are LOW.

Considering the truth table for OR gate.
A B OUTPUT
0 0 0
0 1 1
1 0 1
1 1 1 = here the output is HIGH when the two inputs are HIGH.

BUT question said first input must be LOW when the output is HIGH. So we can't get OR gate as the answer.

@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........

@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).

If a signal passing through a gate is inhibited by sending a LOW into one of the inputs, and the output is HIGH, the gate is a(n):

Here already signal is passing through the gate, it is inhibited by applying a low into one of its inputs, that means now the two inputs are low, and the output is high so it may be either NAND or NOR.

Drawing table:

A - B AND NAND OR NOR
0 - 0 --->0 1 0 1
0 - 1 --->0 1 1 0
1 - 0 --->0 1 1 0
1 - 1 --->1 0 1 0

Now let us take attention here in this table which will clear all the confusion.

In this situation consider 3 inputs NAND gate.
for instance,

Where X column is the result and A B C are inputs,

A B C X
0 0 0 1
0 0 1 1
1 0 0 1
0 1 0 1
1 1 0 1
0 1 1 1
1 1 1 0

For every low input, there will be a high output.
Therefore it satisfies the given condition.

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.

It combination of "AND" gate followed by "NOT" gate it is reverse operation of "AND" gate.

If both inputs are high output is low, if any one input is low output is high.

Expression:- y=A. B.

0 0=1 (high).

0 1=1 (high).

1 0=1 (high).

1 1=0 (low).

@Mehak.

IF one the I/P is low,then OR-GATE OUT COMPLETELY DEPENDS ON SECOND I/P only.

1.If second gate I/P is low, then OR-GATE O/P is LOW.

2.If Second gate I/P is HIGH, then OR-GATE O/P is HIGH.

Then our condition won't satisfy.

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.