3-8 decoder with 2 2-4 decoder

Thread Starter

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.
 

Thread Starter

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
24,974
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?
 

WBahn

Joined Mar 31, 2012
24,974
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?
 

WBahn

Joined Mar 31, 2012
24,974
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
24,974
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.
 
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