Digital Electronics - Combinational Logic Circuits - Discussion

Discussion Forum : Combinational Logic Circuits - General Questions (Q.No. 4)
4.

For the device shown here, let all D inputs be LOW, both S inputs be HIGH, and the input be LOW. What is the status of the Y output?

LOW
HIGH
Don't Care
Cannot be determined
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
19 comments Page 2 of 2.

Shardul kandari said:   1 decade ago
here EN denotes that it is enable & if we keep the enable low then how high the other outputs are the circuit remains low

Ravi kumar sinha said:   1 decade ago
Since s1 and s2 are 1 so 11 is the state which means d3 will be selected and since d3 is low so output will be low.

MVP said:   8 years ago
Here active low enable used so whatever the input is there that will appear at the output when enable is low.

Vatsal Salla said:   1 decade ago
Here the circuit works like a mux. S1 S2 as select lines.

So D3 is selected, and will be D3.S1.S2 = LOW.

Maulik said:   1 decade ago
s1=1&s2=1
&d3 is selected

EN=0
d3=0

So, y=0

Dattika said:   9 years ago
Here, whatever value store in D3 that would pass to O/p. So that point how we can say o/p is low?

Kokila said:   1 decade ago
S1 and S2 are high. Then input of D3 will be the output. Hence answer is low since D3 is low.

RAMESH ROY said:   10 years ago
Here inputs that is D low & S be high.

EN=0, then output (Y) is low.

Praveen said:   8 years ago
How it is low if EN is low?


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