CPE 201/EE 236 Lab 5

(UNR/2005 Spring)

Decoder Design

Introduction (5 points):

A decoder is a multiple-input and multiple-output logic circuit that converts coded inputs into coded outputs, where the input code generally has fewer bits than the output code. 

Decoders can also be described simply as min-term (or max-term) detectors. In the decoder's one-to-one mapping, each input code produces a different output code. For example, a 3-to-8 decoder will indicate which one of eight possible min-terms was received as an input.  To design a decoder, the first step is to create the mapping table, i.e., the truth table for the decoder function.  Here, we need to consider more than what a common combinational circuit requires: 1) one 1-bit input , E, used as ENABLE input for controlling the decoder; and 2). the ENABLE input, E, and outputs are ACTIVE LOW : the decoder works ONLY when E = 0. Common decoders, such as 74155, 74138, and 74139, have active low enable inputs and active low output.

Objective (5 points):

      The purpose of this lab is to design a 2x4 decoder, more specifically, to implement a circuit (which works like 74139) with only NAND and NOT gates.

Procedure (80 points):

  1. Design and implement a 2-to-4 decoder that has active low outputs and an active low enable input(45 points).

    1. Design the circuit in EWB with only 3-input NAND and NOT gates. Include the truthtable (3 inputs and 4 outputs), logic expressions, and logic diagram in lab reports (15 points).
    2. Implement the circuit in the breadboard with 7410 and 7404 chips (check IC's layouts/schematics with EWB or the link) and show the result to the TA (30 points).
  2. Design and implement a 3-to-8 (inputs are A, B, C, and outputs: Y0, Y1, ... Y7, A and Y7 are MSBs) decoder that has active low outputs and an active low enable input. Include truth table and logic expressions in the lab reports (10 points).
  3. Design a comparator with two 2-bit binary numbers AB and CD (4 inputs:A, B, C, and D. AB and CD are 2-bit binary numbers) and 3 outputs (X, Y, and Z) with EWB (25 points).
    * If AB > CD, then output X is 1.
    * If AB < CD, Y is 1.
    * If AB = CD, Z is 1.  
      Include the following in the lab report:
    1. Truth table;
    2. K-maps (You can use ONLY ONE K-map. How and Why?);
    3. Simplified logic expressions.

Conclusion (5 points):