1. Logic Gates
Digital Electronics

2. Interest
Sensor Technology in digital cameras use CCD and CMOS technology to convert light into electrons. CMOS has many TTL logic gates that you can read a schematic and determine simplified output.
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3. Using Logic Circuits . Boolean Algebra
. Developed by George Boolean, a 19th century mathematician.
. His theories were used to develop an assembly of gears and pulleys to be used to drive a grain elevator.
. A Boolean expression is nothing more than a description of the input conditions necessary to get a desired output.
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4. CENT-113 Digital Electronics
Tracing Logic Gates A A
A + B + C B B
(A + B + C)(C) C C C C
CENT-113 Digital Electronics

5. Tracing Logic Gates A AB A (AB)(B + C) B B B (AB)(B + C) C B + C C
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6. Questions Q. Draw the logic circuit for AB + CD + ABC = Y A. A B C Y D
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7. Simplified Boolean Expression
Karnaugh Maps
In 1953, Maurice Karnaugh published a paper about his system of mapping & simplifying Boolean expressions.
Example:
OR Truth Table
Karnaugh Map
Boolean Expression
Inputs
Output B A Y 1 B B
AB + AB + AB = Y x A
Simplified Boolean Expression A B B
A + B = Y x A
2 Terms:
1 Variable = 2 Blocks
2 Variables = 1 Block A
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8. Simplified Boolean Expression
Karnaugh Maps
Example using 3 variables:
OR Truth Table
Karnaugh Map
Boolean Expression
Inputs
Output C B A Y 1 B B
ABC + ABC + ABC +
ABC = Y x A A C C C
Simplified Boolean Expression B B x
A + B + C = Y A
3 Terms:
1 Variable = 4 Blocks
2 Variables = 2 Blocks
3 Variables = 1 Block A C C C
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9. Questions Q. Simplify AB + CBA + BAC + CB + AC. A. A + C x B A C
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10. Questions Q. Simplify ABC + CB + AC + BCA + CA. A. A + BC + BC x X B A
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11. Simplified Boolean Expression
Karnaugh Maps
Example using 4 variables:
OR Truth Table
Karnaugh Map
Boolean Expression
Inputs
Output D C B A Y 1 B B
ABCD + ABCD + ABCD +
ABCD + ABCD + ABCD
= Y x D A D A D C C C B B
Simplified Boolean Expression x D A
A + B + C + D = Y D
4 Terms:
1 Variable = 8 Blocks
2 Variables = 4 Blocks
3 Variables = 2 Blocks
4 Variables = 1 Block A D C C C
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4 corner rule

12. Questions Q. Simplify ABCD + DC + ABD + DAC. A. AB + AD + DC x B D A C
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13. Questions Q. Simplify ABCD + DC + ABC + BC + ABD.x
A. BC + AD + AC + CD (be aware of 4 corners)
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14. Questions
Q. Simplify ABCD + ABCD + ACB + CB AB + ACD + ABCD + ABCD ABC + ACBD. A B C D x
A. CD + BC + AB + AC + BD
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15. Karnaugh Maps Example using 5 variables: BD = Y ABCDE + ABCDE + ABDE +
Boolean Expression A B C D
ABCDE + ABCDE + ABDE +
ABD + ABCDE + ABCDE = Y X X X E X A B C D X X
Simplified Boolean Expression X E X
BD = Y
5 Terms:
1 Variable = 16 Blocks
2 Variables = 8 Blocks
3 Variables = 4 Blocks
4 Variables = 2 Blocks
5 Variables = 1 Block
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4 corner rule

16. Questions
Q. Simplify ABCDE + ABCDE + ABCDE + ABCDE + ABCD + ABCE + ABCDE = Y. A B C D E X
A. ABD + ABC
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17. Questions
Q. Simplify ABCDE + ABCDE + ABD + ABC ABCDE + ABCD + ABCD + CD = Y. A B C D E X
A. AC + CD + BD = Y
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18. Questions
Q. Simplify ABCDE + ABCD + ACD + ABCDE + ABD + ABD + ABCDE = Y. A B C D E X
A. AD + ABCE
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19. Using NAND Logic A NAND TTL chip is the Quad 2 input 7400.
NAND gates are used in many logic circuits because they solve many logic problems with fewer chips.
A two input inverted OR gate is the same as a NAND gate.
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20. Data Selectors (multiplexers)
A single pole 8 position rotary switch works similar to a data selector chip.
Logic Symbol for a 1 of 8 data selector.
Data Selector 1 2 3 4 5 6 7
Rotary Switch 1
Data
Inputs 1 1 2 W
Output 3 1 1 4
Output 5 6 7 C B A
Data
Selector
Inputs 1 1
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21. Data Selector IC Logic Problems
Inputs
Output D C B A Y 1
Data Inputs 1 2 3 4 5 6 7 W
Output 8 9 10 11 12 13 14 15 16 C B A D
Data Select Inputs
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22. Data Selector Folding Technique
Demonstrate using page 76.
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23. More Complex Logic Problems
Programmable array of logic gates that within a single IC with multiple outputs:
PAL: Programmable Array Logic: 16 inputs, 8 outputs.
ROMs: Read-only memories or gate arrays.
PROMs: Programmable ROMs: User programmable gate arrays.
EPROMs: Erasable PROMs: User programmable gate arrays.
Programmable gate arrays
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24. De Morgan’s Theorems = = First Theorem Second Theorem A A Y Y B B
AB = Y
A + B = Y
Second Theorem A A Y Y B B =
AB = Y
A + B = Y
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25. De Morgan’s Theorems Maxterm to Minterm. Given: (AB) (B + C) = Y
Step 1: Change all ORs to ANDs and all ANDs to ORs.
(A + B) + (BC) = Y
Step 2: Complement each individual variable (short overbar).
Step 3: Complement the entire function (long overbar).
Step 4: Eliminate all groups of double overbars.
Solution: (A + B) + (BC) = Y
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26. Conclusion Q. What is the minterm of A + B? A. AB
Q. What is the chip number for a data selector chip? A
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