Implementing a function using decoder, encoder and some gates

Question

Implement the function F(A,B,C,D,E) = A’B’C’DE’+ABCD’E using only the components required from the ones given below:

• One 3:8 decoder with active high outputs and an active high enable input
• One 8:3 Priority Encoder with input no. 7 at highest priority with one active high enable which if disabled forces the outputs to logic low
• One 2 input XOR gate
• One 2 input OR gate
• One 2 input AND gate

My attempts:

• I have noticed that the function has the minterms 2 and 29 - 00010 and 11101.
• I can make the decoder have four inputs using an enable pin (for a variable).
• Drawing the K-Map doesn't seem to simplify anything.
• Applying De-Morgan's law doesn't seem to simplify things.
• Tried using B,C and E in the decoder and A or D in the enable. This provides me with 8 minterms of B,C and E.

I am stuck on how to implement it using only these.

How do I approach this question(and other such design questions) further?

Answer 1

There is no set procedure for solving these kinds of problems. It requires a lot of creativity and insight.

Some insights that may prove useful:

• The two patterns are complements of each other.
• Priority encoders are particularly good at detecting combinations of zeros.
• Decoders are particularly good at detecting combinations of ones.

There is a solution that uses exactly the gates listed. (It does not require a "valid" output on the encoder, although that is a normal feature of such a chip.) I'll post it in a day or two if you're still stuck.

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The truth table for a priority encoder looks like this:

      Inputs         Outputs
E 7 6 5 4 3 2 1 0    V C B A
-----------------    -------
0 x x x x x x x x    0 0 0 0   <--
1 1 x x x x x x x    1 1 1 1
1 0 1 x x x x x x    1 1 1 0
1 0 0 1 x x x x x    1 1 0 1
1 0 0 0 1 x x x x    1 1 0 0
1 0 0 0 0 1 x x x    1 0 1 1   <--
1 0 0 0 0 0 1 x x    1 0 1 0
1 0 0 0 0 0 0 1 x    1 0 0 1
1 0 0 0 0 0 0 0 1    1 0 0 0
1 0 0 0 0 0 0 0 0    0 0 0 0

The key insight here is that both the first and sixth lines of this table are significant for this problem. Pay attention to the C output. If you wire the inputs correctly, you can make it go low for ABCDE = 000x0 or ABCDE = xxx0x. The remaining question is, how can you use the XOR gate to distinguish between these two cases?

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Full solution

^{simulate this circuit – Schematic created using CircuitLab}

Answer 2

The fact that your problem is expressed as a Sum Of two Products gives you a big pointer towards the solution, as you have two complex components and an OR gate available. So it should just be a matter of getting one component to generate one of the products, the other component to generate the other product, and then using the OR gate to combine the two to get the output. You also have two apparently randomly chosen gates to help you out.

The term ABCD’E is the easy one. You have four inputs that need to be 1 and one that needs to be 0. Combine two of the inputs that need to be 1 using the AND gate and you're down to three inputs that need to be 1 and one that needs to be 0, and that can be achieved very easily using your 3:8 decoder.

The second is harder so I'll leave you to try and figure it out, but write out the truth table and study it to see how you can combine the 8:3 encoder with the XOR gate to get the second product.