Digital Electronics - Combinational Logic Analysis - Discussion
Discussion Forum : Combinational Logic Analysis - General Questions (Q.No. 23)
23.
How many 2-input NOR gates does it take to produce a 2-input NAND gate?
Discussion:
11 comments Page 1 of 2.
Munyitanana said:
3 years ago
I fail to understand how 4 comes out of two common inputs A and B.
Please anyone explain me.
Please anyone explain me.
SIVA said:
8 years ago
Demorgans law: (AB)'=A'+B'
(A+B)'=A'.B'.
SUPPOSE CONSIDER INPUTS ARE INVERTED THEN 2 NOR GATES ARE ENOUGH OTHERWISE 4 NOR GATES.
(A+B)'=A'.B'.
SUPPOSE CONSIDER INPUTS ARE INVERTED THEN 2 NOR GATES ARE ENOUGH OTHERWISE 4 NOR GATES.
Pradeep said:
9 years ago
It's saying two input NOR gate, not 1 input NOR gate.
Opara said:
9 years ago
Thanks @Samresh.
The given answer is correct, because they give only one NAND Gate.
The given answer is correct, because they give only one NAND Gate.
Opara said:
9 years ago
Answer should be B.
Remember this is a "2-input" NOR gate.
So if we apply A and B to the first NOR gate we get output (AB Bar) same for CD for output (CD Bar) making 2 NOR gates.
Then the output of both becomes the input for the AND gate.
Remember this is a "2-input" NOR gate.
So if we apply A and B to the first NOR gate we get output (AB Bar) same for CD for output (CD Bar) making 2 NOR gates.
Then the output of both becomes the input for the AND gate.
Arun kumar said:
1 decade ago
Option D:
1st inputs A, B are inverted using NOR gates separately and now both are given to 2 input NOR it gives AND Gate output, just invert using another NOR produces NAND Gate.
Tot gates= 4.
1st inputs A, B are inverted using NOR gates separately and now both are given to 2 input NOR it gives AND Gate output, just invert using another NOR produces NAND Gate.
Tot gates= 4.
SAMRESH CHAUDHARY said:
1 decade ago
D answer is correct.
Because, let A is given to first NOR GATE then output is A-BAR.
Same as for B, The output is B-bar.
When both output is apply on another NOR-gate the output is.
BAR OF (A-BAR +B-BAR) IS equals to A.B by De-morgan's law.
This output works as input of next NOR GATE and we get.
BAR of A.B Which equals to output of 2 input NAND gate.
SO ANSWER IS 4.
Because, let A is given to first NOR GATE then output is A-BAR.
Same as for B, The output is B-bar.
When both output is apply on another NOR-gate the output is.
BAR OF (A-BAR +B-BAR) IS equals to A.B by De-morgan's law.
This output works as input of next NOR GATE and we get.
BAR of A.B Which equals to output of 2 input NAND gate.
SO ANSWER IS 4.
Aman said:
1 decade ago
Thank you gogi.
Gogi said:
1 decade ago
No. Answer is 4 only.
A is Given to first NOR then its output is A-Bar also if B is given to Another NOR then its output is B-bar. It results in a.b
Again it is fed into nor gate which results in (a.b)bar.
A is Given to first NOR then its output is A-Bar also if B is given to Another NOR then its output is B-bar. It results in a.b
Again it is fed into nor gate which results in (a.b)bar.
Prakhar said:
1 decade ago
It must be option C.
Because, let A is Given to first NOR then its output is A-Bar also if B is given to Another NOR then its output is B-bar.
This two when given to an another NOR gate, then it will give (A.B) Bar,. So we require 3 NOR gates.
Because, let A is Given to first NOR then its output is A-Bar also if B is given to Another NOR then its output is B-bar.
This two when given to an another NOR gate, then it will give (A.B) Bar,. So we require 3 NOR gates.
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