Unit 7Unit 7Multi-Level Gate Circuits / Multi-Level Gate Circuits /

NAND and NOR GatesNAND and NOR Gates

Ku-Yaw ChangKu-Yaw Changcanseco@mail.dyu.edu.twcanseco@mail.dyu.edu.tw

Assistant Professor, Department of Assistant Professor, Department of Computer Science and Information EngineeringComputer Science and Information Engineering

Da-Yeh UniversityDa-Yeh University

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ObjectivesObjectives

Design a minimal two-level or multi-level Design a minimal two-level or multi-level circuit of AND and OR gates to realize a circuit of AND and OR gates to realize a given function.given function.

Design or analyze a two-level gate circuit Design or analyze a two-level gate circuit using any one of the eight basic forms.using any one of the eight basic forms.

Design or analyze a multi-level NAND-Design or analyze a multi-level NAND-gate or NOR-gate circuit.gate or NOR-gate circuit.

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ObjectivesObjectives

Convert circuits of AND and OR gates to Convert circuits of AND and OR gates to circuits of NAND gates or NOR gates, and circuits of NAND gates or NOR gates, and conversely, by adding or deleting inversion conversely, by adding or deleting inversion bubbles.bubbles.

Design a minimal two-level, multiple-Design a minimal two-level, multiple-output AND-OR, OR-AND, NAND-NAND, output AND-OR, OR-AND, NAND-NAND, or NAND-NOR circuit using Karnaugh or NAND-NOR circuit using Karnaugh maps.maps.

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OutlineOutline

7.1 Multi-Level Gate Circuits7.1 Multi-Level Gate Circuits7.27.2 NAND and NOR GatesNAND and NOR Gates7.37.3 Design of Two-Level Circuits Using NANDDesign of Two-Level Circuits Using NAND and NOR Gates and NOR Gates7.47.4 Design of Multi-Level NAND and NOR GateDesign of Multi-Level NAND and NOR Gate Circuits Circuits7.57.5 Circuit Conversion Using Alternative GateCircuit Conversion Using Alternative Gate Symbols Symbols7.67.6 Design of Two-Level, Multiple-OutputDesign of Two-Level, Multiple-Output Circuits Circuits7.77.7 Multiple-Output NAND and NOR CircuitsMultiple-Output NAND and NOR Circuits

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7.1 Multi-Level Gate Circuits7.1 Multi-Level Gate Circuits

LevelLevel Maximum number of gates cascaded in series Maximum number of gates cascaded in series

between a circuit input and the outputbetween a circuit input and the output

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Four-Level Realization of ZFour-Level Realization of Z

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TerminologyTerminology

AND-OR circuitAND-OR circuit A two-level circuit composed of a level of AND A two-level circuit composed of a level of AND

gates followed by an OR gate at the outputgates followed by an OR gate at the output

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TerminologyTerminology

OR-AND circuitOR-AND circuit A two-level circuit composed of a level of OR gates A two-level circuit composed of a level of OR gates

followed by an AND gate at the outputfollowed by an AND gate at the output

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TerminologyTerminology

OR-AND-OR circuitOR-AND-OR circuit A three-level circuit composed of a level of OR A three-level circuit composed of a level of OR

gates followed by a level of AND gates followed by gates followed by a level of AND gates followed by an OR gate at the outputan OR gate at the output

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TerminologyTerminology

Circuit of AND and OR gatesCircuit of AND and OR gates No particular ordering of the gatesNo particular ordering of the gates The output gate may be either AND or OR.The output gate may be either AND or OR.

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Trade-offTrade-off

Number ofNumber of GatesGates Gate inputsGate inputs LevelsLevels

Gate delayGate delay

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Example 7-1Example 7-1

Z = (AB+C) (D+E+FG) + HZ = (AB+C) (D+E+FG) + H

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Example 7-1Example 7-1

Partially multiplying outPartially multiplying out

Z = (AB+C) (D+E+FG) + HZ = (AB+C) (D+E+FG) + H

= (AB+C) [ = (AB+C) [(D+E)(D+E)+FG] + H+FG] + H

= = AB(D+E) + C(D+E) + ABFG + CFGAB(D+E) + C(D+E) + ABFG + CFG + H + H

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Example 7-1Example 7-1

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Example 7-1Example 7-1

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ProblemProblem

Find a circuit of AND and OR gates to Find a circuit of AND and OR gates to realizerealize

f(a, b, c, d) = f(a, b, c, d) = ∑ m(1, 5, 6, 10,13, 14)∑ m(1, 5, 6, 10,13, 14)

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SolutionSolution

First, simplify f by using a Karnaugh mapFirst, simplify f by using a Karnaugh map

f(a, b, c, d) =f(a, b, c, d) =∑ m(1, 5, 6, 10,13, 14)∑ m(1, 5, 6, 10,13, 14)

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SolutionSolution

This leads directly to a two-level AND-OR gate This leads directly to a two-level AND-OR gate circuit f = a’c’d + bc’d + bcd’ + acd’circuit f = a’c’d + bc’d + bcd’ + acd’

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SolutionSolution

FactoringFactoring

f = a’c’d + bc’d + bcd’ + acd’ f = a’c’d + bc’d + bcd’ + acd’

yieldsyields

f = c’d (a’ + b) + cd’(a + b) f = c’d (a’ + b) + cd’(a + b)

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SolutionSolution

This leads directly to a three-level OR-AND-OR This leads directly to a three-level OR-AND-OR gate circuitgate circuit

f = c’d (a’ + b) + cd’(a + b)f = c’d (a’ + b) + cd’(a + b)

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SolutionSolution

Both solutions have an OR gate at the Both solutions have an OR gate at the outputoutput Sum of productsSum of products Using 1s on the Karnaugh mapUsing 1s on the Karnaugh map

How about having an AND gate at the How about having an AND gate at the output?output? Product of sumsProduct of sums Using 0s on the Karnaugh mapUsing 0s on the Karnaugh map

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SolutionSolution

Secondly, simplify f by using a Karnaugh mapSecondly, simplify f by using a Karnaugh map(considering 0s)(considering 0s)

f’ = c’d’ + ab’c’ +f’ = c’d’ + ab’c’ + cd + a’b’c cd + a’b’c

f = (c+d) (a’+b+c)f = (c+d) (a’+b+c) (c’+d’) (a+b+c’) (c’+d’) (a+b+c’)

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SolutionSolution

This leads directly to a two-level OR-AND circuitThis leads directly to a two-level OR-AND circuit

f = (c+d)(a’+b+c)(c’+d’)(a+b+c’)f = (c+d)(a’+b+c)(c’+d’)(a+b+c’)

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SolutionSolution

Partially multiplying outPartially multiplying out

f = (c+d)(a’+b+c)(c’+d’)(a+b+c’) f = (c+d)(a’+b+c)(c’+d’)(a+b+c’)

using (X+Y)(X+Z) = X+YZusing (X+Y)(X+Z) = X+YZ

f = [c + d(a’+b)] [c’ + d’(a+b)] f = [c + d(a’+b)] [c’ + d’(a+b)]

= (c+a’d+bd)(c’+ad’+bd’) = (c+a’d+bd)(c’+ad’+bd’)

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SolutionSolution

This leads directly to a three-level AND-OR-This leads directly to a three-level AND-OR-AND circuitAND circuit

f = (c+a’d+bd)(c’+ad’+bd’)f = (c+a’d+bd)(c’+ad’+bd’)

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ComparisonComparison

f =f = LevelLevel GatesGates GateGateInputsInputs

a’c’d + bc’d + bcd’ + acd’a’c’d + bc’d + bcd’ + acd’ 22 55 1616

c’d(a’+b) + cd’(a+b)c’d(a’+b) + cd’(a+b) 33 55 1212

(c+d)(a’+b+c)(c’+d’)(a+b+c’)(c+d)(a’+b+c)(c’+d’)(a+b+c’) 22 55 1414

(c+a’d+bd)(c’+ad’+bd’)(c+a’d+bd)(c’+ad’+bd’) 33 77 1616