74×151 8-to-1 multiplexer — To select data input I3, what binary levels must be applied to the select inputs (S2, S1, S0)?

Difficulty: Easy

Correct Answer: S2 = 0, S1 = 1, S0 = 1

Explanation:


Introduction / Context:
The 74151 is an 8-to-1 multiplexer with three select inputs choosing one of eight data inputs I0–I7. Understanding how the select lines encode the chosen channel is fundamental to using multiplexers in datapaths and control logic.

Given Data / Assumptions:

  • Three select inputs labeled S2 (MSB), S1, S0 (LSB).
  • Input lines I0–I7 are selected by the binary value on S2 S1 S0.
  • Active-low enable must be asserted for the device to function; assume it is enabled.


Concept / Approach:
With binary encoding, the select code corresponds to the index of the input: code 000 selects I0, 001 selects I1, …, 011 selects I3, up to 111 selecting I7. Therefore, to route I3 to the output, the select inputs must represent binary 3, namely 0 1 1 on S2 S1 S0 respectively.

Step-by-Step Solution:

Map desired input: I3 → binary index 3.Write code in S2 S1 S0 order: 0 1 1.Verify alternate choices correspond to I4 (100), I1 (001), I6 (110), not I3.


Verification / Alternative check:

Consult truth tables: S2 S1 S0 = 011 is documented to select I3.


Why Other Options Are Wrong:

S2 = 1, S1 = 0, S0 = 0 → selects I4.S2 = 0, S1 = 0, S0 = 1 → selects I1.S2 = 1, S1 = 1, S0 = 0 → selects I6.


Common Pitfalls:

Reversing bit order and treating S0 as MSB.Forgetting to enable the device, which tri-states or forces the outputs to known inactive states.


Final Answer:

S2 = 0, S1 = 1, S0 = 1

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