A NOR gate using a 2:1 MUX

As previously discussed we start with the equation for 2:1 MUX like following. MUX has two inputs A and B and select pin is S and output pin Out

Out = S * A + (S)bar * B

As we want to make a NOR gate we are looking for equation of the form

Out = (A + B)bar

Just keep in mind taht we don’t have to get exact same equation with same pins, we could use any of teh three input pins A, B and S.

Where do we start ? This is something I did not mentioned while forming a NAND gate, but this applies to both NAND and NOR gate. We have already made AND gate, OR gate and inverter using 2:1 MUX. You can use AND gate and inverter and combine them to make NAND. Similarly you can use OR gate and inverter and combine them to make NOR gate. That would be one option to come up with NOR gate using 2:1 MUX.

Here we will try to come up with NOR gate using alternative way.

Going back we have

Out = S * A + (S)bar * B and we want to convert this to a NOR equation.

We want form

Out = (A + B)bar which is same as

Out = (A)bar * (B)bar

In our 2:1 MUX equation, you can easily see that we need to have (B)bar in place of B and we need to zero out the ‘S * A’ term.

We can achieve both by tying (B)bar in place of B and tying 0 to input A.

Out = S * 0 + (S)bar * (B)bar

Out = (S)bar + (B)bar

Out = (S + B)bar [ DeMorgan’s rule]

That is our NOR gate.

Figure NOR using 2:1 MUX

By now you should have become expert at this process. We will look at XOR gate in next  example. Please provide your feedback, ideas and critique, through various communications channels, like comments, contact form etc. I look forward to hearing from you.


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