chapter 2 – combinational logic circuitslogic circuitshoangtrang/lecture...

Computer Engineering 1 (ECE290)

Chapter 2 – Combinational Logic CircuitsLogic Circuits

Part 2 – Circuit Optimization

HOANG Trang

Reference: © 2008 Pearson Education IncReference: © 2008 Pearson Education, Inc.

Overview Part 1 – Gate Circuits and Boolean Equations

• Binary Logic and Gatesy g• Boolean Algebra• Standard Forms

Part 2 – Circuit OptimizationPart 2 Circuit Optimization• Two-Level Optimization• K-map• Map Manipulation• Map ManipulationPart 3 – Additional Gates and Circuits• Other Gate Types• Exclusive OR Operator and Gates• Exclusive-OR Operator and Gates• High-Impedance Outputs

Hoang Trang Reference: Chapter 2 - Part 2 2

Circuit Optimization

Goal: To obtain the simplest implementation for a given functionimplementation for a given function Optimization requires a cost criterion to

measure the simplicity of a circuitmeasure the simplicity of a circuit Distinct cost criteria we will use:

• Literal cost (L)• Literal cost (L)• Gate input cost (G)• Gate input cost with NOTs (GN)p ( )


Literal Cost

Literal – a variable or it complement Literal cost – the number of literal

appearances in a Boolean expression corresponding to the logic circuit diagram Examples:

• F = BD + A C + A L = 8DB C

D• F = BD + A C + A + AB L = • F = (A + B)(A + D)(B + C + )( + + D) L =

B B D CB C


( )( )( )( )• Which solution is best?

Gate Input Cost Gate input costs - the number of inputs to the gates in the

implementation corresponding exactly to the given equation or equations. (G - inverters not counted, GN - inverters counted)

For SOP and POS equations, it can be found from the equation(s) by finding the sum of:equation(s) by finding the sum of:• all literal appearances• the number of terms excluding single literal terms,(G) and• optionally, the number of distinct complemented single literals (GN).

Example: G=L+term, GN=L+term + complement!• F = BD + A C + A G = 8 GN = 11DB C• F = BD + A C + A G = 8, GN = 11• F = BD + A C + A + AB G = , GN = • F = (A + )(A + D)(B + C + )( + + D) G = , GN =

DB CB B D C

B D B C


F (A + )(A + D)(B + C + )( + + D) G , GN • Which solution is best?

B D B C

Cost Criteria (continued) Example 1:

F A + B C + B C L = 5GN = G + 2 = 9

F = A + B C +

BC

B C G = L + 2 = 7

AC

F

L (literal count) counts the AND inputs and the singleliteral OR input.

G (gate input count) adds the remaining OR gate inputs GN(gate input count with NOTs) adds the inverter inputs


GN(gate input count with NOTs) adds the inverter inputs

Cost Criteria (continued) Example 2:

F = A B C + BCABCA F = A B C +

L = 6 G = 8 GN = 11 F (A + )( + C)( + B)

BC

A

CF

C B

A

F = (A + )( + C)( + B) L = 6 G = 9 GN = 12

S f ti d

AC B

Same function and sameliteral cost

But first circuit has betterF

ABC

gate input count and bettergate input count with NOTsS l t it!

FG, GN: is proportional to the number

of transistors and wires used to implement a logic circuit


Select it!implement a logic circuit

=> G, GN: good measure to choose!

Boolean Function Optimization Minimizing the gate input (or literal) cost of a (a

set of) Boolean equation(s) reduces circuit cost.set of) Boolean equation(s) reduces circuit cost. We choose gate input costgate input cost. Boolean Algebra and graphical techniques areBoolean Algebra and graphical techniques are

tools to minimize cost criteria values. Some important questions:

• When do we stop trying to reduce the cost?• Do we know when we have a minimum cost?

T t ti ti t f ti Treat optimum or near-optimum cost functionsfor two-level (SOP and POS) circuits first.

Introduce a graphical technique using Karnaugh


Introduce a graphical technique using Karnaugh maps (K-maps, for short)

Karnaugh Maps (K-map)

A K-map is a collection of squares• Each square represents a minterm• Each square represents a minterm• The collection of squares is a graphical representation

of a Boolean function• Adjacent squares differ in the value of only one variable

(Gray Code)• Alternative algebraic expressions for the same functionAlternative algebraic expressions for the same function

are derived by recognizing patterns of squares The K-map can be viewed as

• A reorganized version of the truth table• A topologically-warped Venn diagram as used to

visualize sets in algebra of sets


visualize sets in algebra of sets

Some Uses of K-Maps Provide a means for:

Fi di i i• Finding optimum or near optimum SOP and POS standard forms, and two-level AND/OR and OR/AND circuit two-level AND/OR and OR/AND circuit

implementationsfor functions with small numbers of variables

• Visualizing concepts related to manipulating Boolean expressionsBoolean expressions

• Demonstrating concepts used by computer-aided design (CAD) programs to simplify


aided design (CAD) programs to simplify large circuits

Two Variable Maps A 2-variable Karnaugh Map:

• Note that minterm m0 and y = 0 y = 1• Note that minterm m0 andminterm m1 are “adjacent”and differ in the value of the

x = 0 m0 = m1 =yx yxand differ in the value of the

variable y• Similarly, minterm m0 and

x = 1 m2 = m3 =yx yx

yx yxSimilarly, minterm m0 andminterm m2 differ in the x variable.

• Also, m1 and m3 differ in the x variable as

y

Also, m1 and m3 differ in the x variable as well.

• Finally, m2 and m3 differ in the value of the


yvariable y

K-Map and Truth Tables

The K-Map is just a different form of the truth table. Example Two variable function: Example – Two variable function:

• We choose a,b,c and d from the set {0,1} to implement a particular function, F(x,y).p e e t a pa t cu a u ct o , ( ,y).

Function Table K-MapInput Function y = 0 y = 1Values(x,y)

ValueF(x,y)

0 0 a

y yx = 0 a b

1 d0 0 a0 1 b1 0 c

x = 1 c dThe K-map:


1 1 d a reorganized version of the truth table

K-Map Function Representation

Example: G(x,y) = x + y G = x+y y = 0 y = 1

x = 0 0 1

x = 1 1 1

For G(x y) two pairs of adjacent cells containing

x 1 1 1

For G(x,y), two pairs of adjacent cells containing 1’s can be combined using the Minimization Minimization TheoremTheorem: yyyxyyyx x x

yxyxxyyxyx)y,x(G

D li t


Duplicate xy

Three Variable Maps

A three-variable K-map:yz=00 yz=01 yz=11 yz=10y y y y

x=0 m0 m1 m3 m2

x=1 m4 m5 m7 m6

Where each minterm corresponds to the product terms:

x 1 m4 m5 m7 m6

yz 00 yz 01 yz 11 yz 10yz=00 yz=01 yz=11 yz=10

x=0 zyx zyx zyx zyx

Note that if the binary value for an index differs in only one bit position the minterms are adjacent on the K-

x=1 zyx zyx zyx zyx


one bit position, the minterms are adjacent on the K-Map

Alternative Map Labeling

useful

y

useful

y10 23x

yyy z

10 23

x0

00 01 11 10

zx 4 5 67

x

zz4 5 67

0

1xz zz z


Example Functions

By convention, we represent the minterms of F by a "1" in the map and leave the minterms of blankFin the map and leave the minterms of blank

Example: y10 23 11(2,3,4,5) z)y,F(x, m

F

Example:

x 4 5 67

111

1

z Example:

Learn the locations of the 8

zy

10 231(3,4,6,7) c)b,G(a, m

indices based on the variable order shown (x, most significantand z least significant) on the

x 4 5 671 111

z


and z, least significant) on themap boundaries

z

Simplify? => Combining Squares

By combining squares

reduce number of literals in a product term,

reducing the literal cost, thereby reducing the other two cost criteria


Example: Combining Squares

Example: Let y10 23 11

m(2,3,6,7)F

x10 2

4

3

5 671 111

Applying the Minimization Theorem three times:

z

times:

zyyz zyxzyxzyxzyx)z,y,x(F

Thus the four terms that form a 2 × 2 square y

zyyz


correspond to the term "y".

Three-Variable Maps

Example Shapes of 2-cell Rectangles:yy

0 1 3 2

5 64 7xxzz


Three-Variable Maps

Example Shapes of 4-cell Rectangles:yy

0 1 3 2

5 64 7xxzz


Three Variable Maps K-Maps can be used to simplify Boolean functions by

systematic methods. Terms are selected to cover thesystematic methods. Terms are selected to cover the“1s”in the map.

Example: Simplify )(1,2,3,5,7z)y,F(x, m

yz yx)( , , , ,)y,( ,

11

x 1 1

1

x

z

1 1


z)y,F(x, z yx

Three-Variable Map Simplification Use a K-map to find an optimum SOP

i f 7)(0 1 2 4 6Z)YF(X equation for ,7)(0,1,2,4,6 Z)Y,F(X, m


Four Variable Maps

Map and location of minterms:

0 1 3 2Y

5 64 7

1011

12 13 1415 X

W8 9 1011W


Z

Four-Variable Maps

K-map: usefulYZ

0 1 3 2

YZWX 00 01 11 10

005 64 7

00

01

8 9 1011

12 13 141511

8 9 101110


Four-Variable Maps

Example of combining squares

0 1 3 2Y

5 64 7

8 9 1011

12 13 1415 X

W8 9 1011


Z

Four-Variable Maps

Example of combining squaresYY

0 1 3 2

5 64 7

X

9

12 13 1415W 8 9 1011W


Z

Four-Variable Map Simplification )8,10,13,152,4,5,6,7,(0,Z)Y,X,F(W, m


Four-Variable Map Simplification

3,14,15 )(3,4,5,7,9,1Z)Y,X,F(W, m


Systematic Simplification:prime implicant, essentialp p ,

A Prime Implicant (PI) => a product term obtained by bi i h i ibl b f djcombining the maximum possible number of adjacent squares

(the number of squares a power of 2)

A prime implicant: is called an Essential Prime ImplicantA prime implicant: is called an Essential Prime Implicant

IF it is the only prime implicant that covers (includes) one or more mintermsmore minterms.

Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map.determined by inspection of a K Map.

A set of prime implicants "covers all minterms" if, for each minterm of the function, at least one prime implicant in the


p pset of prime implicants includes the minterm.

Example of Prime Implicants

Find ALL Prime ImplicantsCD

DBESSENTIAL Prime Implicants

CCD

DB C

1 1

1 1

1

BD BD

1 1

1 1

1

B

A1 1

B

A1 1

1 11 1

D

1 11 1

D

BA

Hoang Trang Reference: Chapter 2 - Part 2 30CB

DMinterms covered by single prime implicant

DAD

Prime Implicant Practice

Find all prime implicants for:p p13,14,15),10,11,12,(0,2,3,8,9D)C,B,F(A, m


Another Example

Find all prime implicants for:15)12 13 14(0 2 3 4 7D)CBG(A

• Hint: There are seven prime implicants!15),12,13,14,(0,2,3,4,7D)C,B,G(A, m


Five Variable or More K-Maps

5 variables => use two adjacent K-maps. It becomes harder to visualize adjacent minterms forIt becomes harder to visualize adjacent minterms for selecting PIs.

6 variables => use 4 K-Maps.p

YV = 0 V = 1

Y

X XX

W

X

W


Z Z

Don't Cares (X) in K-Maps

Sometimes a function table or map contains entries for which it is known:for which it is known:• the input values for the minterm will never occur, or• The output value for the minterm is not usedThe output value for the minterm is not used

=> the output value need not be defined Instead, the output value is defined as a “don't , p

care” (“X”): value “0” or “1”. Placing “X” in the function table or map=> the cost (L, G, GN) of the logic circuit may be

lowered.


X in K-Maps: example

Example 1: A logic function has:+ input: the binary codes for the BCD

digits.digits. Only the codes for 0 through 9 are

sed The si codes 1010 thro ghused. The six codes, 1010 through 1111 never occur,

=> output values for these codes are “X” to represent “don’t cares.”


p

Example: “X” BCD “5 or More”BCD 5 or MoreK-Map: Function F1(w,x,y,z) is defined as "5 or

more" over BCD inputsmore over BCD inputs. 1. “X” used for the 6 non-BCD combinations as “1”

F1 (w x y z) = w + x z + x y G = 7F1 (w,x,y,z) = w + x z + x y G = 72. “X” are treated as "0s."

G = 120 1 3 20 0 0 0

y

yxwyxwzxwz) y,x,F2(w, G 12=> F1 is lower in cost than F2 For this particular function, cost G for the

w

4 5 7 6

12 13 15 14

1 11

X X X X

0x

yy)y,,2( ,

POS solution for F1(w,x,y,z) is not changed by using the don't cares.

z

w8 9 11 10

11 X X

Placing “X” in the function table or map => the cost (L


Placing X in the function table or map => the cost (L, G, GN) of the logic circuit may be lowered

Optimization Algorithm

Find all prime implicants. Include all essential prime implicants in the

solutionS l t i i t t f ti l Select a minimum cost set of non-essential prime implicants to cover all minterms not yet covered:covered:• Obtaining an optimum solution: See Reading

Supplement - More on Optimization• Obtaining a good simplified solution: Use the

Selection Rule


Prime Implicant Selection Rule

Minimize the overlap among prime implicants as much as possible.

In particular, in the final solution, make sure that each prime implicant selectedsure that each prime implicant selected includes at least one minterm not included in any other prime implicantincluded in any other prime implicant selected.


Selection Rule Example

Simplify F(A, B, C, D) given on the K-E i lmap.

C CEssentialSelected

11 1

11

1 11 1

11

1

1B

A

1

1B

A

1

11

D

11

D


D DMinterms covered by essential prime implicants

Selection Rule Example with Don't Cares

Simplify F(A, B, C, D) given on the K-map. E ti lSelected

C CEssential

1

x

x x

1

1 1

x

x x

1

1x

xB

A

1 x

xB

A

1

1 x

D

1 1 x

D

1


Minterms covered by essential prime implicantsD D

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