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 1 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
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.
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
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.
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.
Akhilesh maurya said:
1 decade ago
Here, S1 and S2 are enabled i.e. both are at HIGH(1 1) state then D3 must be enable & D3 will be as output but all D are LOW so output is LOW(D3)
Shama said:
1 decade ago
Since both S1 and S2 are high and EN is low given but due to bubble in front of it it behaves as high. So in such a case value of d3 is considered for the output.
As value of d3 is low so value of output Y is also low.
As value of d3 is low so value of output Y is also low.
Nagaraj Hegde said:
1 decade ago
If enable is low we are taking that only that's why and the if S1=S2=1, d3 will be selected then d3=0 when the enable is low. So the output will be low only.
Sangee said:
1 decade ago
This is a multiplexer operation.
The multiplexer operation needs many number of input and one output. That is it needs 2^n input and it produce one output.
The multiplexer operation needs many number of input and one output. That is it needs 2^n input and it produce one output.
Nitesh kumar said:
1 decade ago
d0 = d1 = d2 = d3 = 0;
s0 = s1 = 1.
For producing output en =0;
When s1 = s2 = 1.
Then d3 is selected.
Hence, y = d3s1s2 = 0;
s0 = s1 = 1.
For producing output en =0;
When s1 = s2 = 1.
Then d3 is selected.
Hence, y = d3s1s2 = 0;
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers