# Turning MUX's into gates?

Discussion in 'General Electronics Chat' started by Fear58, Jan 28, 2010.

1. ### Fear58 Thread Starter New Member

Jan 11, 2010
9
0
Hey all, I'm a little confused on how to turn mux's into gates...
I have an exam tommorow, and my professor expects to give us a problem where we are only given devices like mux's and the such, and we have to use these to complete a boolean algebra statement.

An example he gave us was:

F=AB+BC

with table:

A|B|C |F
0|0|0 |0
0|0|1 |0
0|1|0 |0
0|1|1 |1
1|0|0 |0
1|0|1 |0
1|1|0 |1
1|1|1 |1

So I can only use MUX's! Anyone have a clue on how to do this?

2. ### Papabravo Expert

Feb 24, 2006
11,927
2,557
Run A,B, & C to the select lines on the MUX
Copy the F outputs from the set {0,1} to the mux inputs according to the table and you're DONE.

Each value of {A,B,C} selects one of 8 inputs which happens to be the correct value of the boolean function.

Pretty SLICK eh?

3. ### Fear58 Thread Starter New Member

Jan 11, 2010
9
0
Papabravo, thanks but i'm still a little lost.
What do you mean by from the set {0,1}?
Also I forgot to mention, in this problem, each Mux only has ONE select. For example one 2-1 mux can only have a B select, etc...

4. ### Fear58 Thread Starter New Member

Jan 11, 2010
9
0
for example, something like this...

I'm really confused on how it was made....
Thanks for any help

5. ### Papabravo Expert

Feb 24, 2006
11,927
2,557
It means you get to choose either a 0 or a 1 for each input. There are no other "elements" in Boolean Algebra.

Fine, so you can only use 2:1 muxes.

Take each pair of successive lines in the table, use the C input to select either a 0 or a 1 for each pair of lines. That is a total of four 2:1 muxes.

Now use the B line to select pairs of outputs from the C multiplexers. That requires two more 2:1 multiplexers.

Finally use the A line to select one of the two outputs from the B multiplexers. This requires only a single multiplexer.

No inverters required.

Depending on the actual function you might observe that some multiplexers may be eliminated because there is no difference in the output if both inputs are tied to Vcc or GND

QED