Digital Electronics - Combinational Logic Analysis - Discussion


Implementing the expression AB + CDE using NAND logic, we get:

[A]. (A)
[B]. (B)
[C]. (C)
[D]. (D)

Answer: Option A


No answer description available for this question.

Amit Kumar said: (Mar 3, 2015)  
Answer is C because only OR gate can perform + operation.

Priya said: (May 21, 2015)  
Any one can explain me?

Jai said: (Sep 26, 2015)  
But there is answer A? Is right or wrong?

Md Nazim said: (Oct 5, 2015)  
Bubble at the end of and gate and bubble at the starting of or gate will cancel each other so A is correct answer. Because 1st and gate will give ab and 2nd and gate will give cde and or gate taking input 1st input=ab, 2nd input=cde will give the result ab+cde.

Gaurav Chavda said: (Sep 4, 2016)  
A is the correct answer. Continuous 2 bubble cancels each other. So circuit in option A is simple to implement AB + CDE.

Vineet said: (Jan 22, 2018)  
First of all, "read" the question. They asks using NAND gate. So option c having OR gate is canceled out, option D having AND gate also cancel out, now there are A and B. We are to find first AB we see option A has AB. So, accurately answer is A.

Rohit said: (May 17, 2018)  
I am not understanding this. Please explain.

Asharaf Ansari said: (Dec 7, 2018)  
No, we have to use only NAND gates.

Paul said: (Jun 17, 2019)  
Yes, A is correct.

(A+B)' + (CDE)' = from the output of NAN Gates.
(A+B)+(CDE) = to the input of OR gate.

Raji Oruganti said: (Jun 4, 2020)  
A is correct answer as per rule A"=A.
(AB)"+(CDE)" =(AB)+(CDE).

Kranti said: (Jul 6, 2020)  
The bubbled OR gate is equivalent to NAND gate, so(ab* (.)cde*)*= ab + cde.

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