Test by Bikram | Mock GATE | Test 2 | Question: 54

Question

The number of $NAND$ gates are required to realize the following function $f$ given below is ________.
 $f = AB + BC + CD + ..... + YZ$

Assume that it is Multi input $NAND$ Gate.

Answer 1

There are 25 minterms in the SOP.  Number of NAND gates = Number of minterms + 1 = 25+1=26.

Comments

yes for sure, multi input nand gate is used.

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hope it helps!!!

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correct!

Answer 2

AB+BC+AC=((AB)'(BC)'(AC)')'

(A+B)(A+C)(B+C)=((A+B)'+(A+C)'+(B+C)')'

 

Number of NAND gates = Number of minterms + 1 = 25+1=26.

Comments

ab+bc+cd should be there na?? why ab+bc+ac?

plz elaborate the soln

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I took a random example .. not the exact terms present in the qs..

Answer 3

AND-OR realization is equal to NAND-NAND realization. So total AND and OR is (25 + 1) = 26 NAND gate needed.

Answer 4

• (SOP) AND-OR realisation $\equiv$ NAND-NAND realisation
   
• (POS) OR-AND realisation $\equiv$ NOR-NOR realisation

See this: https://www.electrical4u.com/electrical-mcq.php?subject=digital-electronics&page=18

http://www.ece.ualberta.ca/~lkurgan/EE280/SN-15.pdf

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AB, BC, CD...YZ require $25$ 2-input AND gates.

Then, we can OR all of them with a $single$ 25-input OR gate.

 

This is equivalent to using NAND gates of the exact same pattern. We will need $25$ 2-input NAND gates, and $1$ 25-input NAND gate.

So, 26 NAND gates.

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See the image provided in Akash Dinkar's comment for clarity.