Digital Electronics - Combinational Logic Circuits - Discussion
Discussion Forum : Combinational Logic Circuits - General Questions (Q.No. 24)
24.
For the device shown here, assume the D input is LOW, both S inputs are LOW, and the 
input is LOW. What is the status of the 
outputs?
All are HIGH.
All are LOW.
All but 
are LOW.
are LOW.All but 
are HIGH.
are HIGH.Discussion:
10 comments Page 1 of 1.
Lakshmi priya said:
2 years ago
It is acting as a Demux and the value for input D as 0 represents 1 for Y0 and the rest should be 0.
Ravishanker said:
5 years ago
Right @Dipta Ranaghat.
Other options are also high as it is an active-low circuit.
Other options are also high as it is an active-low circuit.
Chaynnitt agarwal said:
8 years ago
C is the correct option.
s2 s1 D y3 y2 y1 y0
0 0 0 1 1 1 1
s2 s1 D y3 y2 y1 y0
0 0 0 1 1 1 1
Shwetha said:
9 years ago
Given EN is low. Since it is being inverted, it becomes high.
Now, EN=1, and given that S0=0, S1=0 and D=0. Therefore, Y0 goes high, irrespective of other inputs.
Now, EN=1, and given that S0=0, S1=0 and D=0. Therefore, Y0 goes high, irrespective of other inputs.
(1)
Chandrapaul said:
10 years ago
Please say fluently.
Dipta ranaghat said:
10 years ago
We know S0 and S1 are selection bits;
s0 & s1 both are low so o/p y0 is activate. But i/p D is low.
So o/p y0 is low but y0 is inverted so y0 is high and other o/p is low.
s0 & s1 both are low so o/p y0 is activate. But i/p D is low.
So o/p y0 is low but y0 is inverted so y0 is high and other o/p is low.
(1)
Aligati avinash said:
1 decade ago
Here y0 must be low remaining all are high.
Jyothsna said:
1 decade ago
Actually, S0 and S1 are selection bits....
Input S0 S1 Output
D=0 0 0 Y0
D=0 0 1 Y1
D=0 1 0 Y2
D=0 1 1 Y3
The inputs are S0 and S1 are Low means 0,so output is Y0
Input S0 S1 Output
D=0 0 0 Y0
D=0 0 1 Y1
D=0 1 0 Y2
D=0 1 1 Y3
The inputs are S0 and S1 are Low means 0,so output is Y0
(1)
Madhan said:
1 decade ago
I don't think so because the output will be high for all.
Shefa said:
1 decade ago
I don't think that it is correct because the input will apear inverted at the first line output which Y'=1.
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