chapter 2 combinational logic circuits - lab of...

CHAPTER 2 COMBINATIONAL LOGIC CIRCUITS

What to study?- Binary Logic (Boolean Algebra) : - Gates: - How to design cost-effective circuits?

Logic Circuits:- Combinational Logic Circuits ( )- Sequential Logic Circuits ( )

2-1 Binary Logic and Gates 2-6 NAND and NOR Gates2-2 Boolean Algebra 2-7 Exclusive-OR Gates2-3 Standard Forms 2-8 Integrated Circuits2-4 Map Simplification 2-9 Chapter Summary2-5 Map Manipulation

2-1 BINARY LOGIC & GATES

1. Binary Logic (2) - Binary Logic Boolean AlgebraSwitching algebra, Two-valued Boolean Algebra 2

{, }n {, }{HIGH, LOW}n 2 {HIGH, LOW}{ON, OFF}n {ON, OFF}

: and, or, notSwitch : , , NC/NO switch

{??, ??}n {??, ??}

????, ????


2. Three Basic Logical Operations2 , () ?

2 2 1-bit 2 (Binary Logic) (Binary Digital System) (1-bit Binary Arithmetic)

, 0 1 and 0 10 0 0 0 0 01 0 1 1 0 1

,+ 0 1 or + 0 10 0 1 0 0 11 1 1 1 1 10

X X not0 1 1 0


3. (Basic) Logic Gates = 2

X Y Z X Y ZL L L 0 0 0L H L 0 1 0H L L 1 0 0H H H 1 1 1

X Y Z X Y ZL L L 0 0 0L H H 0 1 1H L H 1 0 1H H H 1 1 1

X Z X ZL H 0 1H L 1 0


4. Timing Diagram ()


5. (Multiple-Input) Gate 2 AND, 2 OR, NOT gate . (more at 2-6, 2-7)AND, OR gate AND, OR .F = ABC = ((AB)C)F = A+B+C+D+E+F = (((((A+B)+C)+D)+E)+F)

: IC

SOIC with"gull-wing" leads

PLCC withJ-type leads

LCCC withno leads

Flat package withstraight leads

Dual-in-line package(DIP)

Small-outline IC (SOIC)

: IC Marking

Manufacturer, Logic Family, Function, Package

: Pin Numbering & Diagram

Logic Families ()

. ( , , )

IC .

2-2 BOOLEAN ALGEBRA ()

Key Points: 1. Boolean Function

F(X,Y,Z) = Boolean Expression= X + YZ

* (Truth Table)* , (Logic Diagram)(Figure 2-3)

* algebraic expression logic diagram* * , () .* Informal Spec. Formal Spec.

X Y Z F0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 11 1 0 11 1 1 1

2-2 BOOLEAN ALGEBRA ( )

(Axiomatic Definition of Boolean Algebra)

(algebraic system)

V: (a set of elements or values)

O: (a set of operations)

A: (axioms or postulates)

S ( Power Set, S ) P < V = P, O = { , }> .

2-2 BOOLEAN ALGEBRA ( )

Huntington : < V={}, O={+, .} >

1. (a) V is closed with respect to the operation p1.(b) V is closed with respect to the operation p2.

2. (a) V has an identity element w.r.t the operation p1.(b) V has an identity element w.r.t the operation p2.

3. (a) V is commutative w.r.t the operation p1.(b) V is commutative w.r.t the operation p2.

4. (a) p1 is distributive over p2.(b) p2 is distributive over p1.

5. For each x in V, there exists an element y in V such that(a) x + y = 1(b) x . y = 0

6. |V| 2


2. Basic Identities of Boolean Algebra

1) X + 0 = X 2) X1 = X (identity) ()3) X + 1 = 1 4) X0 = 0 Identity Theorem 5) X + X = X 6) XX = X Idempotence Th. 7) X + X = 1 8) XX = 0 (Complement )9) (X) = X Involution Th. 10) X + Y = Y + X 11) XY = YX Commutative Law12) X+(Y+Z) = (X+Y)+Z 13) X(YZ) = (XY)Z Associative Law14) X(Y + Z) = XY+ XZ 15) X + YZ = (X+Y)(X+Z) Distributive Law16) (X + Y) = XY 17) (XY) = X + Y De Morgans Th.

* ? (inverse)?* (Principles of Duality): ANDOR, 01


3. Algebraic Manipulation ( , )- F = XYZ + XYZ + XZ

= XY(Z+Z) + XZ= XY + XZ


4. Other Useful Identities(1) (Absorption Theorem)

X + XY = X1 + XY = X(1 + Y) = XX(X + Y) = (X + 0)(X + Y) = X + 0Y = X

(2)XY + XY = X(Y + Y) = X1 = X(X + Y)(X + Y) = X + YY = X + 0 = X

(3) (Simplification Theorem)X + XY = (X + X)(X + Y) = 1(X + Y) = X + YX(X + Y) = XX + XY = 0 + XY = XY

(4) (Consensus Theorem)XY + XZ + YZ = XY + XZ(X + Y)(X + Z)(Y + Z) = (X + Y)(X + Z)

* : ,


5. Complement of a Function ()(1) (2) How to get the expression?

- DeMorgans TheoremF = XYZ + XYZ F = X(YZ + YZ)F = (XYZ + XYZ) F = {X(YZ + YZ)}

= (X+Y+Z)(X+Y+Z) = X + (Y+Z)(Y+Z)

- Using dual expression(i) take the dual expression(ii) complement each literal

F = XYZ + XYZ F = X(YZ + YZ) FD = (X+Y+Z)(X+Y+Z) FD = X + (Y+Z)(Y+Z)F = (X+Y+Z)(X+Y+Z) F = X + (Y+Z)(Y+Z)

2-3 STANDARD FORM ()

Key Points: , ,

1. - (Standard Form):

(Sum-of-Products, SOP, or Disjunctive Form)F(X,Y,Z) = XY + XZ + YZ

(Product-of-Sums, POS, or Conjunctive Form)F(X,Y,Z) = (X + Y)(X + Z)(Y + Z)

- (Non-standard, Factored, or Parenthesized Form)F(X,Y,Z) = X + (Y + Z)(Y + Z)

- (Canonical Form): (Canonical SOP, or Canonical Disjunctive Form)

F(X,Y,Z) = XYZ + XYZ + XYZ [sum-of-minterms form] (Canonical POS, or Canonical Conjunctive Form)

F(X,Y,Z) = (X+Y+Z)(X+Y+Z) [product-of-maxterms form]* .


* (product term) = ANDing of literals* (sum term) = ORing of literals* = 2 (Two-Level Form)* = (Multi-Level Form)

2. Minterms and Maxterms- minterm(, , standard product) = a product

term in which all the variables appear exactly once, either complemented or uncomplemented.

- maxterm(, , standard sum) = a sum term in which all the variables appear exactly once, either complemented or uncomplemented.

* The order of the variables should be assumed for the symbolic representation of minterms/maxterms.


Table 2-6 Minterms for Three Variables

X Y Z productterms symbolm0 m1 m2 m3 m4 m5 m6 m7

0 0 0 XYZ m0 1 0 0 0 0 0 0 00 0 1 XYZ m1 0 1 0 0 0 0 0 00 1 0 XYZ m2 0 0 1 0 0 0 0 00 1 1 XYZ m3 0 0 0 1 0 0 0 01 0 0 XYZ m4 0 0 0 0 1 0 0 01 0 1 XYZ m5 0 0 0 0 0 1 0 01 1 0 XYZ m6 0 0 0 0 0 0 1 01 1 1 XYZ m7 0 0 0 0 0 0 0 1


Table 2-7 Maxterms for Three Variables

X Y Z sumterms symbolM0 M1 M2 M3 M4 M5 M6 M7

0 0 0 X+Y+Z M0 0 1 1 1 1 1 1 10 0 1 X+Y+Z M1 1 0 1 1 1 1 1 10 1 0 X+Y+Z M2 1 1 0 1 1 1 1 10 1 1 X+Y+Z M3 1 1 1 0 1 1 1 11 0 0 X+Y+Z M4 1 1 1 1 0 1 1 11 0 1 X+Y+Z M5 1 1 1 1 1 0 1 11 1 0 X+Y+Z M6 1 1 1 1 1 1 0 11 1 1 X+Y+Z M7 1 1 1 1 1 1 1 0


3. Canonical Expressions- sum-of-minterms of F and FF(X,Y,Z) = [] + [] + [] +

= XYZ + XYZ + XYZ + XYZ= m0 + m2 + m5 + m7= m(0, 2, 5, 7), or (0, 2, 5, 7)

F(X,Y,Z) = [] + [] + [] + = XYZ + XYZ + XYZ + XYZ= m1 + m3 + m4 + m6= m(1, 3, 4, 6), or (1, 3, 4, 6)

X Y Z F F0 0 0 1 00 0 1 0 10 1 0 1 00 1 1 0 11 0 0 0 11 0 1 1 01 1 0 0 11 1 1 1 0


- product-of-maxterms of F and FF(X,Y,Z) = ()()()

= (X+Y+Z)(X+Y+Z)(X+Y+Z)(X+Y+Z)= M1M3M4M6= M(1, 3, 4, 6), or (1, 3, 4, 6)

Similarly,F(X,Y,Z) = M0M2M5M7

= (0, 2, 5, 7)Another method:F(X,Y,Z) = [F(X,Y,Z)]

= (m0 + m2 + m5 + m7)= (m0)(m2)(m5)(m7)= M0M2M5M7= (0, 2, 5, 7)

X Y Z F F0 0 0 1 00 0 1 0 10 1 0 1 00 1 1 0 11 0 0 0 11 0 1 1 01 1 0 0 11 1 1 1 0

F() = (0, 2, 5, 7)= (1, 3, 4, 6)

F() = (1, 3, 4, 6)= (0, 2, 5, 7)


4. Non-Canonical to Canonical

E = Y + XZ= (X+X)Y(Z+Z) + X(Y+Y)Z= (XY+XY)(Z+Z) + XYZ + XYZ= XYZ + XYZ + XYZ + XYZ + XYZ + XYZ= XYZ + XYZ + XYZ + XYZ + XYZ= (0, 1, 2, 4, 5)

E = Y + XZ= (Y + X)(Y + Z)= (XX+Y+Z)(X+Y+ZZ) = (X+Y+Z)(X+Y+Z)(X+Y+Z)(X+Y+Z) = (3, 6, 7)

X Y Z E0 0 0 10 0 1 10 1 0 10 1 1 01 0 0 11 0 1 11 1 0 01 1 1 0


5. Sum of Products- sum-of-minterms

. obtained directly from a truth table

. the most complex sum-of-products form

. but, a starting point to a simplified s-o-p form- logic diagram of a s-o-p form?

. two-level, AND-OR, implementation

. (complements of input variables are assumed to be available)- Non-standard form to SOP form?

. by means of the distributive law- SOP to simplified SOP?

. many ways

. Simplification Th., Consensus Th., Identity Th., .- Figure 2-5, Figure 2-6


- 2 (Figure 2-6)

6. Product of Sums- Two-level, OR-AND Implementation- Figure 2-7

2-4 MAP SIMPLIFICATION

What to Study: Map method for logic simplification () /

1. Simplification Criterion: / ?F = () + () + () + + ()(i) with a minimum number of terms(ii) with the fewest possible number of literals* There may be many, equally good expressions!

2. (Algebraic Simplification) f(a, b, c) = abc + abc + abc + abc + abc

= ab + ab + abc= a + abc= a + bc


f(a, b, c) = abc + abc + abc + abc + abc= ab + abc + bc= a(b + bc) + bc= a(b + c) + bc= ab + ac + bc= a(bc) + bc= a + bc

f(a, b, c) = abc + abc + abc + abc + abc= abc + abc + abc + abc + abc + abc= ab + ab + bc= a + bc= ???

* With proper duplication of product terms, only the distributive law would do the job.


3. Algebraic Simplification: Use any boolean axioms and theorems, or(1) two adjacent product terms into one bigger product term(2) a term may be used more than once during combination(3) a term may be partitioned into smaller ones

But, still we have some difficulties:(1) no specific rules to predict each succeeding step(2) difficult to determine whether the simplest expression has been

achieved

- Simplify the following expression:f(A,B,C,D) = ABCD + ABCD + ABC + ABCD + ABCD

= ABD + ABC + ABCD + ABCD= ??????= ABC + ABC + BD


4. Karnaugh Map, (K-map, Veitch Diagram)

(1) 2 ( )(2) (3) (not for computer-aided design)(4) (not apply directly to

simplification in non-standard forms)(5) up to 4 variables? ==> depends on YOU!

* possible to find two or more simplified expressions.

Glue logic


5. 2-, 3-, 4-Variable K-Maps * Gray Code

XYXYXY

XY1

0

0 1

X

Y

m1m3m2

m01

0

0 1

X

Y

1

32

0

1

0

0 1

X

Y

X Y Z0 0 00 1 01 0 01 1 1

X Y Z0 0 00 1 11 0 11 1 1

0

X

Y0

1

0

1

0

0 1

0

X

Y0

1

0

1

1

1 1

Note the adjacency of minterms

YX YXYXYX Z

+=++=


[EXAMPLE 2-3] F(X,Y,Z) = (2,3,4,5)* rectangles of 1, 2, 4, 8, cells or squares

0

X

Y

Z

00

1

11 10

0

011 3

5

2

4 7 6

m0

X

Y

Z

00

1

11 10

0

01

m6m7m5m4

m2m3m1

X

Y

Z

00

1

11 10

0

01

ZYX

ZXYXYZZYXZYX

YZXZYXZYX

0

X

Y

Z

00

1

11 10

0

01

1 01 0

110


[Examples] F(X,Y,Z) = (0,2,4,6) F(X,Y,Z) = (0,1,2,3,6,7)

1

X

Y

Z

00

1

11 10

0

01

1 1

1 1

X

Y

Z

00

1

11 10

0

01

1 1

111

[EXAMPLE 2-4] F(X,Y,Z) = (3,4,6,7) F(X,Y,Z) = (0,2,4,5,6)

X

Y

Z

00

1

11 10

0

01

1 1

1 1

X

Y

Z

00

1

11 10

0

01

1 1 1

1


[Example] F(X,Y,Z) = (1,3,4,5,6)

X

Y

Z

00

1

11 10

0

01

1 1 1

11

[Example] F(X,Y,Z) =

X

Y

Z

00

1

11 10

0

01

YXZ YZZYXYXZX +=>+++


[EXAMPLE 2-5] Four-Variable MapF(W,X,Y,Z) = (0,1,2,4,5,6,8,9,12,13,14)

0

W

X

Y

Z

00

01

11 10

00

01

11

10

1 3

5

2

4 7 6

12 13 14

8 9 11 10

15

W

X

Y

Z

00

01

11 10

00

01

11

10


[EXAMPLE 2-6]

A

B

C

D

00

01

11 10

00

01

11

10

DBCACBADCBCBAF +++=


What to Study: Map method for logic simplification () /

1. Simplification Criterion: / ?F = () + () + () + + ()(i) with a minimum number of terms(ii) with the fewest possible number of literals* There may be many, equally good expressions!

2. (Algebraic Simplification) f(a, b, c) = abc + abc + abc + abc + abc

= ab + ab + abc= a + abc= a + bc