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Understanding Minterms in 3-Input Combinational Logic Circuits

This lecture explores the concept of minterms in combinational logic, specifically focusing on circuits involving three inputs. It illustrates the schematic representation of all eight minterms and discusses why not all minterms can be present in practical applications. The lecture also explains how to implement logical expressions using present minterms, detailing how outputs only occur when specific minterms are activated. Additionally, the importance of managing unused inputs is highlighted, including strategies for simplifying logic gates for better circuit efficiency.

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Understanding Minterms in 3-Input Combinational Logic Circuits

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  1. Digital Systems I EEC 180A Lecture 4 Bevan M. Baas

  2. m0 MintermExample m1 • This circuit schematic shows all 8 minterms “present” for a 3-input combinational logic function • In practice, all possible minterms would never all be present in a circuit (do you see why?) • There is one possible minterm for each row in the truth table m2 m3 Z m4 m5 m6 m7

  3. m0 MintermExample 0 0 0 1 m1 1 • By construction, one and only one minterm is active (equals 1) at any point in time m2 0 m3 0 1 Z m4 0 m5 0 m6 0 m7 0

  4. m0 MintermExample 1 0 0 1 m1 0 • By construction, one and only one minterm is active (equals 1) at any point in time m2 0 m3 0 1 Z m4 0 m5 1 m6 0 m7 0

  5. m0 MintermExample 1 0 0 1 m1 0 • Z = m0 + m1 + m7 • To implement an expression, a circuit is built with only the present minterm(s) • The output can be 1 only when one of the present minterms forces the output to 1 m2 0 Z m7 0

  6. m0 MintermExample 1 0 1 1 m1 0 • Z = m0 + m1 + m7 • To implement an expression, a circuit is built with only the present minterm(s) • The output can be 1 only when one of the present minterms forces the output to 1 m2 1 Z m7 1

  7. m0 MintermExample 1 0 1 1 m1 0 • Z = m0 + m1 + m7 • Of course gate inputs can not be left unconnected (unspecified). There are two solutions: • Tie unused inputs to a value that disables those inputs. For an OR gate, inputs would be tied to 0 (or False or Gnd) • The best solution is to simplify the gate. In this example, the 8-input OR gate is simplified to a 3-input OR gate m2 1 Z m7 1

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