3-8 decoder with 2 2-4 decoder

Leite33

Joined Nov 28, 2015
57
Dear friends.
I looked a lot at google but i cant find a solution. How i can make one 3-8 decoder with (2) 2-4 decoders with out use enable input and without inverse outputs.

bertus

Joined Apr 5, 2008
22,233
Hello,

You will NEED the select or enable inputs for making a 3->8 using 2 X 2->4 decoders.

Bertus

Leite33

Joined Nov 28, 2015
57
But my exercise says "Design a circuit that will produce a digit of even parity for a number of 3 binary digit (xyz) using:
a) decoder 2-4 (without using enable input and inverse outputs) and gates OR

WBahn

Joined Mar 31, 2012
29,519
But my exercise says "Design a circuit that will produce a digit of even parity for a number of 3 binary digit (xyz) using:
a) decoder 2-4 (without using enable input and inverse outputs) and gates OR
In other words, the problem is NOT asking you to create a 3-to-8 decoder at all!

How would you make an even parity generator for a three-bit input using XOR or XNOR gates?

How would you make an XOR of XNOR gate using 2-4 decoders and OR gates?

Leite33

Joined Nov 28, 2015
57
So you say to me that exercise is wrong?

WBahn

Joined Mar 31, 2012
29,519
There is nothing wrong with the exercise as you presented it in Post #3:

Design a circuit that will produce a digit of even parity for a number of 3 binary digit (xyz) using:
a) decoder 2-4 (without using enable input and inverse outputs) and gates OR
But this is NOT the same problem that you described in your original post:

How i can make one 3-8 decoder with (2) 2-4 decoders with out use enable input and without inverse outputs.
These are two essentially unrelated problems. The only thing they happen to share is that each involves 2-4 decoders.

Which problem are you trying to solve?

Leite33

Joined Nov 28, 2015
57
Sorry the first one.

WBahn

Joined Mar 31, 2012
29,519
Which brings us back to my previous question, but let's rephrase it a bit.

Produce a circuit that produces an even-parity output for a three-bit input using nothing but XOR and XNOR gates.

Do that much first, and then we can worry about using 2-4 decoders and OR gates.

WBahn

Joined Mar 31, 2012
29,519
Hello,

You will NEED the select or enable inputs for making a 3->8 using 2 X 2->4 decoders.

Bertus
Actually, you don't. You can use a 2->4 decoder (without enable) to produce an AND gate and to produce a NOT gate. That means that it is a universal gate and that you can implement ANY Boolean function using just a bunch of 2-> decoders. Now, it may not be pretty, but it can be done.