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I have no idea where to start. I know how to create a MUX but how can I create it using three state buffers ?

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  • \$\begingroup\$ Have you looked at existing or similar designs yet? \$\endgroup\$ – Ignacio Vazquez-Abrams Aug 7 '13 at 1:24
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Look at it this way. A digital signal MUX allows selection of one of several inputs gated to a single output. Such logic normally has to include an OR stage in the logic path to bring the selected input out to the output path. Now note that tri-state buffers can be used to make a wired-OR type bus (i.e. one on at a time).

Take that information and see what you can do with it to produce the MUX required by what sounds like homework.

I do not want to take away from your learning opportunity by showing a direct solution. Hopefully others here will take that same consideration!!

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  • \$\begingroup\$ Considering this is quite old now, would the input to the MUX just be used to enable the tri-state buffers that would receive the inputs? That is, the only logic required would be to select one of the 8 gates with the 3 select line bits? \$\endgroup\$ – sherrellbc Jul 25 '14 at 12:53
  • \$\begingroup\$ It is a little hard to understand your question so let me describe the logic. You use eight separate three state buffers that have all of the outputs connected together. The eight inputs of course are the MUX inputs. Next you have to design a way to select a single one of the eight tri-state buffer controls. A common way to do that is to start with a 3-bit select code on three wires and then decode the eight combinations of these wires to produce the eight individual enables for the buffers. You could build this out of a batch of 3-input AND gates and three INVERTERs or (continued) \$\endgroup\$ – Michael Karas Jul 25 '14 at 13:03
  • \$\begingroup\$ (continued from above) use a ready made 3-8 decoder chip. \$\endgroup\$ – Michael Karas Jul 25 '14 at 13:05
  • \$\begingroup\$ That is exactly what I was (trying) to say. thanks! \$\endgroup\$ – sherrellbc Jul 25 '14 at 18:53

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