Friday, 26 April 2019

digital logic - Implementing a function using decoder, encoder and some gates


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



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.




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?




Full solution



schematic


simulate this circuit – Schematic created using CircuitLab


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