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
| NAME | GRAPHICAL SYMBOL | ALGEBRAIC EXPRESSION | TRUTH TABLE | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| BUFFER | 
|
X = A |
| ||||||||||||||||||
| NOT | 
|
X = [math]\overline{A}[/math] |
| ||||||||||||||||||
| AND | 
|
X = [math]AB[/math] or [math]A \cdot B[/math] |
| ||||||||||||||||||
| NAND | 
|
X = [math]\overline{AB}[/math] or [math]\overline{A\cdot B}[/math] |
| ||||||||||||||||||
| OR | 
|
X = [math]A+B[/math] |
| ||||||||||||||||||
| NOR | 
|
X = [math]\overline{A+B}[/math] |
| ||||||||||||||||||
| XOR | 
|
X = [math]A \oplus B[/math] |
| ||||||||||||||||||
| XNOR | 
|
X = [math]\overline{A \oplus B} \text{ or } A \odot B[/math] |
| ||||||||||||||||||
Sample Problems
Sample Problem 1
Find all ordered triplets (A, B, C) which make the following circuit FALSE:
Solution:
One approach to solving this problem is to reason about that inputs and outputs are necessary at each gate. For the circuit to be FALSE, both inputs to the file OR gate must be false. Thus, input C must be FALSE, and the output of the NAND gate must also be false. The NAND gate is false only when both of its inputs are TRUE; thus, inputs A and B must both be TRUE. The final answer is (TRUE, TRUE, FALSE), or (1, 1, 0).
Another approach to solving this problem is to translate the circuit into a Boolean Algebra expression and simplify the expression using the laws of Boolean Algebra. This 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. The expression becomes [math]\overline{\overline{AB}}\cdot C[/math] after applying DeMorgan’s Law. The double NOT over the AB expression cancels out, to become [math]AB\overline{C}[/math]. The AND of 3 terms is TRUE when each term is TRUE, or A=1, B=1 and C=0.
Sample Problem 2
How many ordered 4-tuples (A, B, C, D) make the following circuit TRUE?
Solution:
We'll use a truth table to solve this problem. The rows in the truth table will correspond to all possible inputs - 16 in this case, since there are 4 inputs. The output columns will be the output of each gate, other than the NOT gates.
| INPUT | OUTPUT | |||||||
| A | B | C | D | NOR | OR | AND | XOR | XOR |
|---|---|---|---|---|---|---|---|---|
| [math]p=\overline{C+D}[/math] | [math]q=p+\overline{B}[/math] | [math]r=\overline{A}B[/math] | [math]s=q \oplus r[/math] | [math]p \oplus s[/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 |
From the truth table, there are 10 rows where the final output is TRUE.
Video Resources
ACSL Advisors
The following YouTube videos show ACSL students and advisors working out some ACSL problems that have appeared in previous contests. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
| Digital Electronics -1 (Ravi Yeluru (hemsra))
Introduces the AND, OR, and NOT gates, and works through a couple of simple circuits. | |
| Digital Electronics -2 (Ravi Yeluru (hemsra))
Solves a 3-gate circuit (the AND of an OR and a NOR) given by the Boolean expression [math](\overline{A}+B)(\overline{B+C})[/math]. | |
| Digital Electronics Boolean Algebra (Tangerine Code)
Solves a 3-gate circuit (the AND of an OR and a NOR) given by the Boolean expression [math]\overline(A+B)}(B+C)[/math]. | |
| ACSL Digital Electronics Worksheet Sample (misterminich)
Solves a handful of ACSL problems. |
Other Videos
The topic of Digital Electronics is fundamental and there are many videos on YouTube that teach this subject. We found the following videos to be nice introductions to digital logic gates. Some of the videos contain ads; ACSL is not responsible for the ads and does not receive compensation in any form for those ads.
| [URL Digital Electronics Basics (Beginning Electronics)] | |
| Logic Gate Expressions (Kevin Drumm) | |
| Logic Gate Combinations (Kevin Drumm) |