Digital Electronics - Combinational Logic Circuits - Discussion
Discussion Forum : Combinational Logic Circuits - General Questions (Q.No. 5)
5.
For the device shown here, let all D inputs be LOW, both S inputs be HIGH, and the 
input be HIGH. What is the status of the Y output?
Discussion:
28 comments Page 1 of 3.
Jayant indLa said:
8 years ago
In any combinational or sequential digital circuit if they have En then only the circuit gives proper output value otherwise it should stay at a previous value, another wise circuit will enter into the meta-stable state.
So, it should remain or hold the previous value until next En High.
Here previously we have s1s2 as 11 so, it should be enabled d3 = 0.
so, the answer is LOW "0"
So, it should remain or hold the previous value until next En High.
Here previously we have s1s2 as 11 so, it should be enabled d3 = 0.
so, the answer is LOW "0"
Suren said:
6 years ago
The answer is A.
Enable high = 1.
Enable high (complement) = 0.
So in question, EN specified as complement.
Everybody knows in digital electronics any letter with bar is treated as a complement of that letter. So EN bar = 0.
It means EN bar = high it is equal to active low. That is = 0.
Enable high = 1.
Enable high (complement) = 0.
So in question, EN specified as complement.
Everybody knows in digital electronics any letter with bar is treated as a complement of that letter. So EN bar = 0.
It means EN bar = high it is equal to active low. That is = 0.
Jay said:
3 years ago
Here, It is a multiplexer. Regardless of the input and current state of the output, the enable line is directly connected to the internal AND Gates.
Thus, when EN bar = HIGH, all AND Gates are LOW.
This is connected to the OR Gate seen in the sources above.
Thus, the answer is A.
Thus, when EN bar = HIGH, all AND Gates are LOW.
This is connected to the OR Gate seen in the sources above.
Thus, the answer is A.
Neelaji said:
1 decade ago
Enable input is used only for on the device. The o/p is depends on EN and also data lines. When EN is HIGH the o/p depends upon data lines where as EN low o/p is depends on EN input, what ever it may be data lines low or high.
Vikas said:
1 decade ago
An enable input makes the multiplexer operate. When EN = 0, the output is 0. When EN = 1, the multiplexer performs its operation depending on the selection line. In this case it is EN not.
Zion said:
1 decade ago
Assuming that the IC is not enable so, the output is, nothing, zero, 0 = low.
Even if all the inputs are high the output is still 0 due to the IC is not enable.
Even if all the inputs are high the output is still 0 due to the IC is not enable.
Priyanka singh said:
9 years ago
We know that when enable pin is high then MUX will be operate but when enable pin is low then MUX willn't operate so option is incorrect.
Passi said:
7 years ago
As (EN not) is HIGH therefore EN should be low & whenever the enable is low the device don't work hence resultant o/p should be low.
Passi said:
7 years ago
As (EN not) is HIGH therefore EN should be low & whenever the enable is low the device don't work hence resultant o/p should be low.
Arshid Bhat said:
7 years ago
@All.
Multiplexer only works when (enable is low), as Here enable is high that means multiplexer is not working, so the output is low.
Multiplexer only works when (enable is low), as Here enable is high that means multiplexer is not working, so the output is low.
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