Digital Electronics - Combinational Logic Circuits - Discussion

I guess the answer lies in what "enable" really means. If it means that the output does not resemble any possible value that can logically be expected from the circuit then the answer is not accurate. To make my idea clear, consider a 2-4 decoder.

The output values that can be expected are (1000, 0100, 0010, and 0001). It is obvious that if you want to make the circuit inactive, you would use the output (0000), which does not resemble any possible output. If you wish to use the decoder as a part of a bigger circuit, it is easy to make a logic to detect the output bits are all low and thus the device is inactive.

This cannot be the case for a MUX. This is because a low output is a valid output for the circuit. In other words, if this MUX is used as a part of a bigger circuit, it would be impossible to say whether the circuit is off or in a valid logical state.

This is why tristate logic is the best solution in this case and in cases where multiple MUXs are used on a bus.

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.

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.

@All.

Multiplexer only works when (enable is low), as Here enable is high that means multiplexer is not working, so the output is low.

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.

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.

I think the answer is C.

The answer would be D. As the IC is in off mode.

Answer is A.

Because when enable is low, the circuit will be disabled, so output will be low.

Because the D input out do the S input so the answer can be A, though the input is not enabled.