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?
Discussion:
19 comments Page 2 of 2.
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.
So D3 is selected, and will be D3.S1.S2 = LOW.
RAMESH ROY said:
10 years ago
Here inputs that is D low & S be high.
EN=0, then output (Y) is low.
EN=0, then output (Y) is low.
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?
Praveen said:
8 years ago
How it is low if EN is 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.
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.
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.
Maulik said:
1 decade ago
s1=1&s2=1
&d3 is selected
EN=0
d3=0
So, y=0
&d3 is selected
EN=0
d3=0
So, y=0
Sneha jose said:
1 decade ago
EN is low means ic is enabled. When s0, s1 are high, it selects d3 to output. Since d3 is low, output is low.
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