3 . 2 0 Construct a 1 6 - 1 multiplexer using four 4 - to - 1 multiplexers, one 4 - input OR gate, and one 2 - 4 decoder as the mux selector. Assume the multiplexers have enable inputs.
Question:
3 . 2 0 Construct a 1 6 - 1 multiplexer using four 4 - to - 1 multiplexers, one 4 - input OR gate, and one 2 - 4 decoder as the mux selector. Assume the multiplexers have enable inputs.
This is AI generated Response.
Asked by: Abhishek
Created at: 2025-09-28 22:58:05
Asked by: Abhishek
Created at: 2025-09-28 22:58:05
Thought Process:
Rendered
Raw Text
**Clarifying multiplexer design**
I need to parse the problem of constructing a 16-to-1 multiplexer using four 4-to-1 multiplexers, one 4-input OR gate, and a 2-to-4 decoder as the selector. The prompt seems a bit jumbled but indicates that we should show how to implement this design while considering the enable inputs for the multiplexers. I'm pondering if the 4-input OR gate is for combining outputs; typically, you would implement a 16:1 multiplexer using four 4:1 multiplexers feeding into a final 4:1 multiplexer to select between groups.**Reviewing multiplexer functionality**
I realize that I can't use a final multiplexer in this design—only a 4-input OR gate and a 2-to-4 decoder as selectors. The enable pins are crucial since they'll gate the four 4-to-1 multiplexers, resulting in the OR gate combining their outputs only when activated. I need to ensure that only one 4-to-1 output is active at a time, otherwise, the OR gate would need to handle inactive outputs properly. From the datasheets I've seen, some define outputs as zero when disabled, which means the OR will yield just the selected output.**Detailing the wiring process**
For the wiring, I need to set up 16 data inputs labeled D0 to D15. I’ll group these into four groups, each containing four inputs: group 0 has D0 to D3, group 1 has D4 to D7, group 2 has D8 to D11, and group 3 has D12 to D15. The low-order select lines, S0 and S1, connect to each 4-to-1 multiplexer, while the higher-order lines, S2 and S3, go to a 2-to-4 decoder to generate one-hot enables Y0 to Y3 for activating the multiplexers. Each output from the 4-to-1 multiplexers feeds into a 4-input OR gate, whose output forms the final Y. Only one multiplexer should be enabled at a time, producing the corresponding output while the others yield zero.**Planning the block diagram**
I want to include a block diagram in my description, detailing the wiring steps, logic equations, and signal naming. It’s important to note that decoders may have enable inputs; while it's not always necessary, I'll mention that the 2-to-4 decoder serves as the higher-order selector using S2 and S3. If the 4-to-1 MUX has active-high enables, the decoder outputs must also be active-high; if not, I'd need to invert or use muxes with active-low enables. Since the OR gate combines four outputs, it could face issues if outputs are tri-stated unless we add pull-downs. Assumptions about mux enables should cover this. I'll also clarify that the decoder outputs are one-hot and name the inputs I0 through I15, along with signals S3, S2, S1, and S0, with S3 being the most significant bit.**Crafting a structured solution**
I’m focusing on how the 2-to-4 decoder outputs (E0 to E3) will connect to the relevant groups of inputs. I’ll map MUX0 inputs to I0 to I3, MUX1 inputs to I4 to I7, and so forth. The selection lines S1 and S0 link to each multiplexer, and their enables are driven by the decoder outputs. ORing the outputs gives Y, but I need to clarify if outputs are valid when tri-stated.
I’ll outline that the hierarchical selection allows low-order selects for group selections and provide a clear component list: four 4-to-1 MUXs (active-high enable), one 2-to-4 decoder, and one 4-input OR gate, with inputs D0 to D15 and select lines S3, S2, S1, S0. The connection steps will involve linking the select lines and ensuring outputs are managed so that only one of the sixteen inputs is active, depending on the selection. Using bullet points for the steps will keep it readable while I present an example to illustrate how specific selections lead to the correct output. Lastly, I’ll make notes on the active high/low considerations for better clarity.
Final Answer:
Rendered
Raw Text
Did you like this answer?
Goal: Build a 16-to-1 multiplexer using:
- four 4-to-1 multiplexers (each with an enable input),
- one 2-to-4 decoder (to generate one-hot enables),
- one 4-input OR gate (to combine the four intermediate outputs).
Signal names:
- Data inputs: D0 … D15
- Select lines: S3 S2 S1 S0 (S3 is MSB)
- Output: Y
How to connect:
1) Group the 16 inputs into four groups of four:
- MUX0 inputs: D0, D1, D2, D3
- MUX1 inputs: D4, D5, D6, D7
- MUX2 inputs: D8, D9, D10, D11
- MUX3 inputs: D12, D13, D14, D15
2) Low-order select to all 4:1 MUXes:
- Connect S1 and S0 to the select inputs of every 4-to-1 multiplexer (these choose one of the four inputs within each group).
3) High-order select to the 2-to-4 decoder:
- Feed S3 and S2 into the 2-to-4 decoder’s select inputs.
- The decoder produces one-hot outputs E0, E1, E2, E3.
4) Use the decoder outputs as MUX enables:
- Connect E0 to the Enable of MUX0
- E1 to Enable of MUX1
- E2 to Enable of MUX2
- E3 to Enable of MUX3
Assumption: MUX enable is active-high and, when disabled, the MUX output is forced low (not floating). If your parts use active-low enables or different disable behavior, see the notes below.
5) Combine the MUX outputs:
- Connect the outputs of MUX0, MUX1, MUX2, MUX3 to the four inputs of the 4-input OR gate.
- The OR gate output is Y.
Why this works:
- S3 S2 selects exactly one of the four 4:1 MUXes via the one-hot decoder enables.
- S1 S0 selects one of four data lines within the enabled MUX.
- The enabled MUX drives its selected data to its output, while the three disabled MUXes drive 0. The 4-input OR gate therefore passes only the active MUX’s output, giving Y = D[ (S3 S2 S1 S0 as a binary index) ].
Example:
- S3 S2 S1 S0 = 10 01
- Decoder enables MUX2 (E2 = 1).
- S1 S0 = 01 selects the second input of MUX2, which is D9.
- OR gate sees MUX2 output = D9, others = 0, so Y = D9.
Notes and alternatives:
- If your 4:1 MUX enable is active-low, either use an inverter on each decoder output or a decoder with active-low outputs, so that only the intended MUX is enabled.
- If a disabled MUX goes high-impedance rather than driving 0, the OR gate inputs must not be left floating. Ensure your MUX drives 0 when disabled (common for parts like 74HC153), or add pull-downs or replace the OR with wired-OR logic appropriate to tri-state outputs.
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :
VIPSolutions