Digital Electronics - Logic Gates - Discussion
Discussion Forum : Logic Gates - General Questions (Q.No. 3)
3.
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):
Discussion:
29 comments Page 2 of 3.
Shobi said:
1 decade ago
First fix the any one i/p = 0.
Then whatever i/p given to the second pin always lead to high , so
pin1 pin2 o/p(nand)
0 0 1
0 1 1
So even 1 i/p is low o/p will be high in NAND.
Then whatever i/p given to the second pin always lead to high , so
pin1 pin2 o/p(nand)
0 0 1
0 1 1
So even 1 i/p is low o/p will be high in NAND.
Krishna said:
1 decade ago
Simply inhibited low signal as a input means one input is fixed that is 0 what ever may be the 2nd input out is constant for that combination i.e. NAND gate check it by drawing NAND gate diagram then it is more clear.
Arun said:
1 decade ago
TAKE THIS IS IN CONSIDERATION.
Here one input is low and another would be high or low. In both cases i.e.either in (0, 1) or in (0, 0) the output should be high. And this condition satisfy by NAND gate.
Here one input is low and another would be high or low. In both cases i.e.either in (0, 1) or in (0, 0) the output should be high. And this condition satisfy by NAND gate.
Atul said:
7 years ago
It cannot be nor, as it both the input need to low or high as the same time, not nor will not be an option, only left with Nand as one input should be low at the time which is mentioned in the here.
Rajashekar said:
1 decade ago
If signal passing through a gate is inhibited by sending a LOW into one of the those inputs then output is HIGH.
Because of at least one of the input is low then output is high of the nand gate.
Because of at least one of the input is low then output is high of the nand gate.
Ankit said:
1 decade ago
NAND - if any one of input is low than output is high.
But.
In OR if both input is low than output is low.
So here only one input is specified which is low, and output is high so NAND gate.
But.
In OR if both input is low than output is low.
So here only one input is specified which is low, and output is high so NAND gate.
P.Narendra kumar said:
1 decade ago
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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Suhani thakur said:
10 years ago
The simple answer would be to combat the confusion between NAND and OR that whenever there is an i/p signal at 0 the o/p must be 1, whatever be the 2nd i/p.
Bayaw said:
1 decade ago
Take this for instance. (i.e. Input A-1 Input B-0)
OR 1 + 0 = 1 (NOT) = 0 (logic low) ----> NOR.
AND 1 x 0 = 0 (NOT) = 1 (logic high)----> NAND.
OR 1 + 0 = 1 (NOT) = 0 (logic low) ----> NOR.
AND 1 x 0 = 0 (NOT) = 1 (logic high)----> NAND.
(1)
Mehak said:
1 decade ago
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?
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