Digital Fundamentals

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Transcription:

Digital Fundamentals Tenth Edition Floyd hapter 5 Modified by Yuttapong Jiraraksopakun Floyd, Digital Fundamentals, 10 th 2008 Pearson Education ENE, KMUTT ed 2009 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

ombinational Logic ircuits In Sum-of-Products (SOP) form, basic combinational circuits can be directly implemented with ND-OR combinations if the necessary complement terms are available. D J K D JK Product terms + D +... + JK Sum-of-products Product term 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

ND-OR Logic n example of an SOP implementation is shown. The SOP expression is an ND-OR combination of the input variables and the appropriate complements. X = + DE SOP D E DE 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

ND-OR-Invert Logic Summary When the output of a SOP form is inverted, the circuit is called an ND-OR-Invert circuit. The OI configuration lends itself to product-of-sums (POS) implementation. n example of an OI implementation is shown. The output expression can be changed to a POS expression by applying DeMorgan s theorem twice. X = + DE X = + DE OI D E DE X = ()(DE) DeMorgan X = ( + + )(D + E) POS 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Example Logic O or OI can be used for positive- or negative-logic control circuits 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Exclusive-OR Logic The truth table for an exclusive-or gate is Notice that the output is HIGH whenever and disagree. The oolean expression is X = + The circuit can be drawn as Symbols: Inputs Output X 0 0 0 0 1 1 1 0 1 1 1 0 X = 1 Distinctive shape Rectangular outline 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Even-parity generator/checker 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Exclusive-NOR Logic The truth table for an exclusive-nor gate is Notice that the output is HIGH whenever and agree. The oolean expression is X = + The circuit can be drawn as X Symbols: Inputs Output X 0 0 1 0 1 0 1 0 0 1 1 1 = 1 Distinctive shape Rectangular outline 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

For each circuit, determine if the LED should be on or off. +5.0 V +5.0 V 330 Ω LED +5.0 V +5.0 V 330 Ω LED +5.0 V +5.0 V 330 Ω LED (a) (b) (c) ircuit (a): XOR, inputs agree, output is LOW, LED is ON. ircuit (b): XNOR, inputs disagree, output is LOW, LED is ON. ircuit (c): XOR, inputs disagree, output is HIGH, LED is OFF. 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Implementing ombinational Logic Implementing a SOP expression is done by first forming the ND terms; then the terms are ORed together. Show the circuit that will implement the oolean expression X = + D + DE. (ssume that the variables and their complements are available.) Start by forming the terms using three 3-input ND gates. Then combine the three terms using a 3-input OR gate. D D E X = + D + DE 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Karnaugh Map Implementation For basic combinational logic circuits, the Karnaugh map can be read and the circuit drawn as a minimum SOP. Karnaugh map is drawn from a truth table. Read the minimum SOP expression and draw the circuit. changes across this boundary 1 1 1 1. Group the 1 s into two overlapping groups as indicated. 2. Read each group by eliminating any variable that changes across a boundary. 3. The vertical group is read. changes across this boundary 4. The horizontal group is read. The circuit is on the next slide: 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

ircuit: continued X = + The result is shown as a sum of products. 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

= + D + D + D 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

It is a simple matter to implement this form using only NND gates as shown in the text and following example. 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

NND Logic onvert the circuit in the previous example to one that uses only NND gates. Recall from oolean algebra that double inversion cancels. y adding inverting bubbles to above circuit, it is easily converted to NND gates: X = + 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Universal Gates Summary NND gates are sometimes called universal gates because they can be used to produce the other basic oolean functions. Inverter ND gate + + OR gate NOR gate 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Universal Gates Summary NOR gates are also universal gates and can form all of the basic gates. + Inverter OR gate ND gate NND gate 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

NND Logic Summary Recall from DeMorgan s theorem that = +. y using equivalent symbols, it is simpler to read the logic of SOP forms. The earlier example shows the idea: X = + The logic is easy to read if you (mentally) cancel the two connected bubbles on a line. 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

NND Logic 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

NOR Logic Summary lternatively, DeMorgan s theorem can be written as + =. y using equivalent symbols, it is simpler to read the logic of POS forms. For example, X = ( + )( + ) gain, the logic is easy to read if you cancel the two connected bubbles on a line. 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

NOR Logic 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Pulsed Waveforms Summary For combinational circuits with pulsed inputs, the output can be predicted by developing intermediate outputs and combining the result. For example, the circuit shown can be analyzed at the outputs of the OR gates: D D G 1 G 2 G 3 G 1 G 2 G 3 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Pulsed Waveforms lternatively, you can develop the truth table for the circuit and enter 0 s and 1 s on the waveforms. Then read the output from the table. D D G 3 G 1 G 2 G 3 0 1 0 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 1 1 0 1 0 Inputs D 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 Output X 0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 1 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved

Homework 8 hapter 5 (4, 12, 21, 32, 35) 2009 Pearson Education, Upper Saddle River, NJ 07458. ll Rights Reserved