# 3-8 decoder with 2 2-4 decoder

Discussion in 'Homework Help' started by Leite33, Dec 6, 2015.

1. ### Leite33 Thread Starter Member

Nov 28, 2015
57
0
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.

Apr 5, 2008
15,796
2,384
Hello,

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

Bertus

3. ### Leite33 Thread Starter Member

Nov 28, 2015
57
0
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

4. ### WBahn Moderator

Mar 31, 2012
18,080
4,917
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?

5. ### Leite33 Thread Starter Member

Nov 28, 2015
57
0
So you say to me that exercise is wrong?

6. ### WBahn Moderator

Mar 31, 2012
18,080
4,917
There is nothing wrong with the exercise as you presented it in Post #3:

But this is NOT the same problem that you described in your original post:

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?

7. ### Leite33 Thread Starter Member

Nov 28, 2015
57
0
Sorry the first one.

8. ### WBahn Moderator

Mar 31, 2012
18,080
4,917
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.

9. ### WBahn Moderator

Mar 31, 2012
18,080
4,917
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.