Digital Electronics

This topic is an extension of the topic of Boolean Algebra which includes a more thorough description of the category in terms of determining whether a circuit results in a TRUE or FALSE value using truth tables or how to simplify a circuit to as few gates as possible. Electrical engineers use the following symbols to design electrical circuits that are used inside the computer’s hardware. The following table illustrates the equivalent Boolean algebra expression and truth table for each gate.

Definitions

[math]NAME[/math]	[math]GRAPHICAL SYMBOL[/math]	[math]ALGEBRAIC EXPRESSION[/math]	[math]TRUTH TABLE[/math]
BUFFER	Symbol	X = A	truth table
NOT	Symbol	X = [math]\overline{A}[/math]	! A X	0 1	1 0
AND	Symbol	X = [math]AB[/math] or [math]A*B[/math]	A B X	0 0 0	0 1 0	1 0 0	1 1 1
NAND	Symbol	X = [math]\overline{AB}[/math] or [math]\overline{A*B}[/math]	A B X	0 0 1	0 1 1	1 0 1	1 1 0
OR	Symbol	X = [math]A+B[/math]	A B X	0 0 0	0 1 1	1 0 1	1 1 1
NOR	Symbol	X = [math]\overline{A+B}[/math]	A B X	0 0 1	0 1 0	1 0 0	1 1 0
XOR	Symbol	X = [math]A \oplus B[/math]	A B X	0 0 0	0 1 1	1 0 1	1 1 0
XNOR	Symbol	X = [math]\overline{A \oplus B}[/math]	A B X	0 0 1	0 1 0	1 0 0	1 1 1

Sample Problems

Sample Problem 1

Find all ordered 4-tuples (A, B, C, D), which make the following circuit FALSE:

Solution: This circuit diagram translates to: [math](\overline{C+D}+B) \oplus (\overline{A}B) \oplus (\overline{C+D})[/math]. The table has the following headings: 1=[math]\overline{C+D}[/math], 2=[math]1+\overline{B}[/math], 3=[math]\overline{A}B[/math], 4=[math]A \oplus B[/math], and 5=[math]4 \oplus 1[/math]. Thus, the 4-tuples (0,0,0,0), (1,0,0,0), (1,1,0,0), (1,1,0,1), (1,1,1,0), and (1,1,1,1) all make the circuit FALSE.

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Sample Problem 2

Find all ordered triplets (A, B, C) which make the following circuit FALSE:

Solution: The circuit translates to the Boolean expression [math]\overline{AB}+C[/math]. To find when this is FALSE we can equivalently find when the [math]\overline{\overline{AB}+C}[/math] is TRUE. We can simplify this by applying DeMorgan’s Law and cancelling the double not over AB to yield [math]AB\overline{C}[/math] . This is TRUE when for (1,1,0) only.

Sample Problem 3

Simplify the following digital circuit as a Boolean algebra expression using the fewest gates possible:

Solution: This translates to: [math]\overline{A*B+\overline{C}*D}[/math] . Simplify: [math]\overline{(A*B+\overline{C})*D}=\overline{AB+\overline{C}}+\overline{D}[/math] [math]=\overline{AB}*\overline{\overline{C}}+\overline{D}=(\overline{A}+B)C+\overline{D}[/math] This has 5 gates and [math]\overline{A}C+BC+\overline{D}[/math] has 6 so the above expression is best.

[math]A[/math]	[math]B[/math]	[math]C[/math]	[math]D[/math]	[math]1[/math]	[math]2[/math]	[math]3[/math]	[math]4[/math]	[math]5[/math]
0	0	0	0	1	1	0	1	0
0	0	0	1	0	1	0	1	1
0	0	1	0	0	1	0	1	1
0	0	1	1	0	1	0	1	1
0	1	0	0	1	1	1	0	1
0	1	0	1	0	0	1	1	1
0	1	1	0	0	1	1	1	1
0	1	1	1	0	0	1	1	1
1	0	0	0	1	1	0	1	0
1	0	0	1	0	1	0	1	1
1	0	1	0	0	1	0	1	1
1	0	1	1	0	1	0	1	1
1	1	0	0	1	1	0	1	0
1	1	0	1	0	0	0	0	0
1	1	1	0	0	0	0	0	0
1	1	1	1	0	0	0	0	0