The Practicing Electronics Technician's Handbook

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The Practicing Electronics Technician's Handbook 2021 UPDATE THE PRACTICING ELECTRONICS TECHNICIAN’S HANDBOOK Second Edition 2021 UPDATE THE PRACTICING ELECTRONICS TECHNICIAN’S HANDBOOK Second Edition www.gbctechtraining.com © 2021 George Brown College George Brown College School of Distance Education 160 Kendal Ave, Toronto, ON M5R 1M3 PREFACE Electronics and electromechanics are incredibly exciting and dynamic fields, but to be successful means having a thorough understanding of the foundational skills and concepts that make electricity so electrifying. The Practicing Electronics Technician’s Handbook was developed in 2019 to be a compact collection of fundamental content for electronics and electromechanical technicians to refer to throughout their career. Our primary goal in preparing this second edition was to expand the range of topics in the handbook and update and refresh content. The handbook is still broken down into three sections for easy reference: articles, key equations, and supplemental information, all aimed to help users understand the content and build the necessary skills to succeed in the field. We have added subcategories to the Articles section to make it easier to find topics of interest. Covering everyday skills and topics such as areas of difficulty when working with linear DC network theorems, and math skills needed to solve simultaneous equations, the articles in the handbook clearly explain complex topics. Key equations included are commonly used equations that are divided by topic, such as solving for instantaneous voltage and average voltage. Additional topics are covered in the supplemental content section, with numerous diagrams included to help visualize theories. We hope that this handbook helps you develop a fuller understanding of electronics and becomes a useful reference to you in the future. TABLE OF CONTENTS 1. The Practicing Electronics Technician’s Series ............. 1 ELECTRICITY AND CIRCUIT ANALYSIS The Significance of Conductor Resistance and How to Calculate It. 4 Everything You Need to Know About Current Divider Circuits ...................... 6 Converting Parallel RL Circuits to their “Easier To Work with” Series Equivalents ....... 8 How to Create Correct Ohm’s Law KCL Branch Equations for Nodal Analysis ........ 10 The Utility of Finding the Thevenin Equivalent Circuit ............................ 13 The Importance of Power Factor Correction and How it can be Accomplished ........ 15 How to Solve Simultaneous Equations with Multiple Unknowns ................... 17 Using the Natural Log or “ln” Function in Circuit Analysis ........................ 20 Relation of Radians and Angular Velocity to AC Circuits ......................... 22 MAGNETISM The Importance of Inductance of a Coil ...................................... 24 An Overview of Magnetic Field Lines and its Characteristics. 28 Why Do We Need an Air Gap in Magnetic Circuit & How to Calculate it? ............ 29 FILTERS AND DIGITAL ELECTRONICS An Overview of Filters in A Crossover Network ................................. 31 Common Rules for Weighted Number Systems ................................ 34 2. Key Equations .................................. 36 3. Supplemental Content ............................ 44 Notation ............................................................... 45 Electric Charges ........................................................ 47 Resistance ............................................................. 50 RC Circuits ............................................................. 54 Schematic Symbols ...................................................... 58 THE PRACTICING ELECTRONICS TECHNICIAN’S SERIES 3 THE SIGNIFICANCE OF CONDUCTOR RESISTANCE AND HOW TO CALCULATE IT In this article, we will review the calculation required to determine the resistance of a given conductor. This often overlooked parameter can be important when trying to determine the appropriate wire diameter for a given application. It is also important to consider conductor resistance when examining application efficiency. Lower resistance means lower power dissipation by the conductor. Optimizing these two aspects of conductor resistance for your specific application can lead to significant reductions to implementation and operating costs. It is important to know the resistance offered by a given conductor as well as have an understanding of the extent to which that resistance impacts on the application and its operation. For this reason, we will review some relevant aspects of conductor resistance and briefly describe and discuss them. WHAT FACTORS DETERMINE THE RESISTANCE OF A GIVEN CONDUCTOR? There are three factors that determine the amount of resistance a given conductor will have. They are illustrated here, in the relationship used to calculate conductor resistance. Resistance = Resistivity x ( Length / Area ) R = resistance in ohms = material resistivity in ohms metres R = * ( L / A ) ρ ρ L = conductor length in metres A = cross-sectional area in square metres We begin by acknowledging the obvious fact that the length of a conductor will contribute to its overall resistance. The longer the length of a given conductor, the more resistance that conductor will have. This can be clearly seen in the relationship provided above. The material resistivity of the conductor plays a significant role in the overall resistance. This is because different materials for example gold or copper, offer different amounts of resistance to continuous current flow. Conductor materials are generally selected on a cost-benefit and suitability basis. The material resistivity of some of the most common conductors used today is listed below. Material Resistivity, ohms·m Silver 1.64 x 10-8 Copper 1.72 x 10-8 Gold 2.45 x 10-8 Aluminum 2.83 x 10-8 The last significant factor that determines conductor resistance is the cross-sectional area of the given conductor. It is important to note the inverse relationship between the cross-sectional area of a conductor, and the conductor’s resistance. The smaller the cross-sectional area of the conductor, the larger the resistance value of the conductor becomes. This means that although it may be cheaper to use smaller conductor sizes, there is a trade off with resistance. THE PRACTICING ELECTRONICS TECHNICIAN’S SERIES 4 CONDUCTOR RESISITENCE IN WHAT WAYS CAN CONDUCTOR RESISTANCE BE SIGNIFICANT? This relationship provides a means of determining conductor resistance which can be used to examine power losses that contribute to the overall efficiency of an application. This relationship can also be rearranged to solve for Area. This calculated value for cross-sectional area can then be used to provide the required wire diameter to meet specific application needs. As previously stated, these two considerations have impact on implementation and operating costs and are worth examining. EXAMPLE An example of the general relationship for calculating a conductors resistance to ascertain the minimum diameter and cross-sectional area is provided in the video animation Calculating Required Wire Diameter And Cross-sectional Area. THE PRACTICING ELECTRONICS TECHNICIAN’S SERIES 5 EVERYTHING YOU NEED TO KNOW ABOUT CURRENT DIVIDER CIRCUITS In a parallel circuit, all the components have their terminals connected together sharing the same two end nodes. This results in different paths and branches for the current to flow or pass along. However, the currents can have different values through each component, but the voltage remains the same across two end nodes. RESISTIVE CURRENT DIVIDER CIRCUITS Consider a current divider circuit with two resistors connected together in parallel as shown in below diagram. We will explain how to calculate the current flowing through each resistor in a two or more resistor parallel circuits. Parallel resistors R and R splits the supply or source current between them into two separate currents 1 2 I and I before joining together again and returning back to the source. R1 R2 I - - - R1 R2 + + + As the total current equals the sum of the individual branch currents, then the total current, IT flowing in the circuit is given by Kirchhoff’s current law (KCL) as: Equation 1: IT = IR1 + IR2 As the two resistors are connected in parallel, for Kirchhoff’s Current Law, (KCL) to hold true it must therefore follow that the current flowing through resistor R1 will be equal to I = I – I R1 T R2 Similarly, current flowing through resistor R2 will be equal to I = I – I R2 T R1 Let’s calculate voltages; V = I × R & V = I × R 1 R1 1 2 R2 2 Since the Voltage across each resistor is same I = V ⁄ R and I = V ⁄ R R1 1 R2 2 Substituting I and I in Eq (1); we now realize that I = V[ (1 ⁄ R ) + (1 ⁄ R ) ] R1 R2 T 1 2 re-arranging this we get V= I [ (R × R ) ⁄ (R + R ) ] T 1 2 1 2 THE PRACTICING ELECTRONICS TECHNICIAN’S SERIES 6 EVERYTHING YOU NEED TO KNOW ABOUT CURRENT DIVIDER CIRCUITS Solving for I gives: I = I (V ⁄ R ) = I [ R ⁄ (R + R ) ] R1 R1 T 1 T 2 1 2 Likewise, solving for I gives: I = I [ R ⁄ (R +R ) ] R2 R2 T 1 1 2 Notice that the above equations for each branch current has the opposite resistor in its numerator while solving for I we use R , and to solve for I we use R . This is because each branch current is inversely 1 2 2 1 proportional to its resistance. CURRENT DIVIDER PRACTICE EXAMPLE 1: Assuming a total current of 25mA flowing through the network. Resistor R1=6 and R2=9 is connected in parallel. Let us find out I & I using above concept. R1 R2 I - - - R1 R2 6 Ohms 9 Ohms + + + How to find I & I ? R1 R2 R I V I 2 I 9 I 9 I = T = T = T = T = 15mA
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