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IS 151 Lecture 4 - UDSM 2013

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- 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 1
- 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 IS 151 Digital Circuitry 2
- 3. 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) IS 151 Digital Circuitry 3
- 4. 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) IS 151 Digital Circuitry 4
- 5. 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 IS 151 Digital Circuitry 5
- 6. Laws and Rules of Boolean Algebra • Rules 1. A + 0 = A 2. A + 1 = 1 3. A.0 = 0 4. A.1 = A 5. A + A = A 6. A + A’ = 1 7. A.A = A 8. A.A’ = 0 9. A’’ = A 10. A + AB = A 11. A + A’B = A + B (same as A + A’B’ = A + B’) IS 151 Digital Circuitry 6
- 7. 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’ IS 151 Digital Circuitry 7
- 8. 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’ IS 151 Digital Circuitry 8
- 9. De Morgan’s Theorems Exercises • The Boolean expression for an ex-OR gate is AB’ + A’B. Develop an expression for the exNOR 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 IS 151 Digital Circuitry 9
- 10. 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 IS 151 Digital Circuitry 10
- 11. 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 IS 151 Digital Circuitry 11
- 12. 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’)C AB’CC + A’B’C AB’C + A’B’C B’C(A + A’) B’C(1) B’C IS 151 Digital Circuitry 12
- 13. Simplification using Boolean Algebra - Examples 3. A’BC + AB’C’ + A’B’C’ + AB’C + ABC BC(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’ IS 151 Digital Circuitry 13
- 14. Simplification using Boolean Algebra - Exercises • Simplify the following 1. 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 IS 151 Digital Circuitry 14
- 15. • End of lecture IS 151 Digital Circuitry 15

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