Digital Electronics - Combinational Logic Analysis - Discussion

Discussion Forum : Combinational Logic Analysis - General Questions (Q.No. 6)
6.
Before an SOP implementation, the expression would require a total of how many gates?
1
2
4
5
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
10 comments Page 1 of 1.

Shruthi said:   1 decade ago
AB = 1.
C' = 1.
C'D = 1.
EF = 1.
C'D + EF=1.
AB(C'D+EF) = 5.

Totally 5 gates are used.

Nandan said:   1 decade ago
This expression require 6 gates.
AB makes AND gate.
EF makes AND gate.
c bar is also a NOT gate.
C'D makes AND gate.
(C'D+EF) makes OR gate.
Finally AB(C'D+EF) makes AND gate.

Akshay said:   1 decade ago
No you can always put AB& (C'D+EF) in a 3 input and gate.

So it will be 5 gates only not 6.

Nasaru said:   1 decade ago
I don't understand what @Akshay said.

Eliza said:   10 years ago
First go for the C'D+EF.

Up to that step you need 4 gates.

Then make an and gate of three inputs which are A, B, and (C'D+EF).

Now the no. of gates required altogether is 5.

DWISH said:   10 years ago
Yes it's 5 no need of final and gate because you can use the or gate output with ab to an and gate.

Aiysha said:   10 years ago
AB - And gate.

C'D - Nand gate.

EF - And gate.

C'D+EF - OR gate.

AB(C'D+EF) - And gate.
(2)

Gopi said:   8 years ago
Why we are not taking a single bubbled and gate for C'D.

I have small doubt. When we take single bubbled AND Gate is it considered as two logic gates?

Debraj Simlai said:   7 years ago
Why not gate is not considered?

Can anyone please explain because C bar means a bubble connected to c which is not the gate?

Ranjitha said:   5 years ago
AB(C'D+EF).
ABC'D+ABEF.
C'=1.
ABC'D = 1.
ABEF = 1.
ABC'D + ABEF = 1.
So, 4 gates are needed.

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