IS 151 Digital Circuitry 1

Integrated Circuit (IC) Logic Families

• Integrated – composed of various components• 3 digital IC families

– TTL – Transistor-Transistor Logic• The use of bipolar junction transistors in the circuit

technology used to construct the gates at the chip level.– CMOS – Complementary Metal Oxide Semiconductor

• Uses field effect transistors; logic functions are the same (whether the device is implemented with TTL or CMOS), the difference comes in performance characteristics.

– ECL – Emitter Coupled Logic• Bipolar circuit technology; has the fastest switching speed

but it’s power consumption is much higher

IS 151 Digital Circuitry 2

Boolean Operations and Expressions

• Boolean algebra – the mathematics of digital systems

• Variable – a symbol used to represent a logical quantity.

• Any single variable can have a 0 or a 1 value. E.g. A, B, C

• Complement – the inverse of a variable– E.g. A’, B’, C’– If A = 0, A’ = 1 and vice versa

• Literal – a variable or the complement of a variable

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Boolean Addition

• Equivalent to the OR operation• Basic rules

• 0 + 0 = 0• 0 + 1 = 1• 1 + 0 = 1• 1 + 1 = 1

• e.g. determine the values of A, B, C and D which makes the sum term A + B’ + C + D’ equal to 0.

• Solution: for the sum to be 0, each of the literals on the term must be 0. Therefore A = 0, B = 1 (so that B’ = 0), C = 1, D = 1

• Exercise: determine the values of A and B which makes the sum term A’ + B = 0 (A = 1, B = 0)

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Boolean Multiplication

• Equivalent to the AND operation• Basic rules

• 0.0 = 0• 0.1 = 0• 1.0 = 0• 1.1 = 1

• e.g. determine the values of A, B, C and D which make the product A.B’.C.D’ equal to 1

• Solution: for the product term to be 1, each one of the literals in the term must be 1. Therefore A = 1, B = 0, C = 1, D = 0

• Exercise: determine the values of A and B which make the product A’B’ equal to 1 (A = 0, B = 0)

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Laws and Rules of Boolean Algebra

• Laws– Commutative Laws

• A + B = B + A; AB = BA

– Associative Laws• A + (B + C) = (A + B) + C; A(BC) = (AB)C

– Distributive Laws• A(B +C) = AB + AC

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Laws and Rules of Boolean Algebra

• Rules1. A + 0 = A2. A + 1 = 13. A.0 = 04. A.1 = A5. A + A = A6. A + A’ = 17. A.A = A8. A.A’ = 09. A’’ = A10. A + AB = A11. A + A’B = A + B (same as A + A’B’ = A + B’)

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De Morgan’s Theorems

• The complement of a product of variables is equal to the sum of the complements of the variables: (XY)’ = X’ + Y’

• The complement of a sum of variables is equal to the product of the complements of the variables: (X +Y)’ = X’.Y’

• Example: Apply De Morgan’s theorems to the expressions– (XYV)’ = X’ + Y’ + Z’– (X + Y + Z)’ = X’.Y’.Z’

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De Morgan’s Theorems - Exercises

• Apply De Morgan’s theorems to the expressions– (X’ + Y’ +Z’)’ = X’’.Y’’.Z’’ = X.Y.Z– ((A + B + C)D)’ = (A + B + C)’ + D’ – = A’.B’.C’ + D’– (ABC + DEF)’ = (ABC)’.(DEF)’ – = A’ + B’ C’ . D’ + E’ + F’– (AB’ + C’D + EF)’ = (AB’)’.(C’D)’.(EF)’ – = A’ + B’’.C’’ + D’.E’ + F’ – = A’ + B.C + D’.E’ + F’

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De Morgan’s Theorems - Exercises

• The Boolean expression for an ex-OR gate is AB’ + A’B. Develop an expression for the ex-NOR gate– Ex-OR = AB’ + A’B;– Ex-NOR = (AB’ + A’B)’

= (AB’)’.(A’B)’ = (A’ + B’’) . (A’’ + B’) = (A’ + B) . (A + B’)

= A’A + A’B’ + BA + BB’ = 0 + A’B’ + AB + 0

= A’B’ + AB

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Simplification using Boolean Algebra

• The aim is to reduce the number of gates used to implement a circuit

• Examples– Simplify the following expressions using Laws

and Rules of Boolean Algebra, and De Morgan’s Theorems where necessary

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Simplification using Boolean Algebra - Examples

1. AB + A(B + C) + B(B + C)

AB + AB + AC + BB + BC

AB + AB + AC + B + BC

AB + AC + B + BC

AB + AC + B

B + AC

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Simplification using Boolean Algebra - Examples

2. [AB’(C + BD) + A’B’]C(AB’C + ABB’D + A’B’)C(AB’C + AD0 + A’B’)C(AB’C + 0 + A’B’)C(AB’C + A’B’)CAB’CC + A’B’CAB’C + A’B’CB’C(A + A’)B’C(1)B’C

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Simplification using Boolean Algebra - Examples

3. A’BC + AB’C’ + A’B’C’ + AB’C + ABCBC(A’ + A) + AB’(C’ + C) + A’B’C’BC(1) + AB’(1) + A’B’C’BC + AB’ + A’B’C’BC + B’(A + A’C’)BC + B’(A + C’) (rule 11)BC + B’A + B’C’BC + AB’ + B’C’

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Simplification using Boolean Algebra - Exercises

• Simplify the following1. AB’ + A(B + C)’ + B(B + C)’

2. [AB(C + B’D) + (AB)’]D

3. ABC’ + A’B’C + A’BC + A’B’C’

4. (AB + AC)’ + A’B’C

5. (AB)’ + (AB)’ + A’B’C

• End of lecture

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