ELEXBO Electrical - Experimentation Box 1
Table of contents 2 Introduction...3 Basics...3 The current......4 The voltage...6 The resistance....9 Measuring resistance...10 Summary of the electrical values...11 From the photo to the diagram...11 The Ohm's Law...12 The serial connection...15 Replacing de picture of the multimeter by a symbol...16 The voltage divider......17 The parallel connection 18 Calculation and measurement in the parallel connection 19 Combination circuit...20 The electric power...21 Electric energy...22 The potentiometer......23 The NTC resistor...24 The PTC resistor...25 The LDR resistor...26 The capacitor...27 The switch...30 The push button switch... 30 The light bulb...31 The relay...32 Types of relays...33
3 Introduction Hello, I am pleased to accompany you through the fascinating world of electrotechnics. Only two tips for you...: 1. Be pleased if a circuit works 2. Take your time I tried to not burden you with too much theory, but to let you experience the electricity. Basiscs Let's get started: Construct the first circuit as shown in the photo. Flip switch Operate the switch and be pleased when the lamp lights up. You have created your first electrical circuit. Remove now a cable at any place of the circuit. What's happening? No matter where you interrupt the circuit, the lamp always extinguishes. Thus, there must be something happening inside de cables? But what?
4 Now pull out both plugs from the switch and hold both silver ends together. Now the lamp lights up! Now hold both plugs together at the red end. Now the lamp doesn't light up. Why? There must be something moving through the cable, that the lamp lights up only by the metal. Now we need to talk about theory. The current If you look closely into a cable, you can see that it consists of an insulation made of plastic and the inside part is of a metal wire (copper wire). If you look now quite exactly into the copper metal, you realize that it is composed of atoms. Atoms are the smallest components. Now, if we have a look at such a copper atom, we see that it is composed of protons, neutrons and electrons. Electron
5 The electrons revolve around the atom's nucleus, which is composed of protons and neutrons. All materials are composed of different atoms. Metals have a special property. They have a single electron on the outermost shell, which can be easily moved away. So, when electrons move through the wire, we call that electric current or electricity. And we can measure it. Construct the circuit as shown in the picture (red cable to A, and black cable to COM ). Therefore we let flow the electrons through the measuring device (multimeter). Turn the rotary switch on A - Close the switch, the lamp lights up and the multimeter indicates 0.08 A. When we switch off the lamp, the display indicates 0.00 A. We can say: It flows a current of 0.08 amperes. The electric current is measured in units called amperes, abbreviated with a majuscule A. Why, the electrons should want to flow through the wire? Who "presses" the electrons through the wire?
6 The voltage Now, again just a bit of theory... When we look at the battery which is already installed on the labour board... It consists of a housing with two metallic connections The upper connexion is marked by a plus (+) sign and is therefore called Positive terminal The lower connexion is marked by a minus (-) sign and is therefore called Negative terminal The battery is additionally marked with the following symbol. The wide upper bar indicates that this is the positive pole. You can also see the plus sign. Now, what is inside the battery? The battery consists on two separate containers, which are filled with metall. In the electrotechnology we are only interested in freely moving electrons in the atom and therefore only these electrons are shown in the drawing of the battery. In our case, in both containers there are the same number of electrons. The battery is not charged.
7 Now, we connect the empty battery to a long piece of coiled wire (technical term = the coil) Yet nothing is happening... n Now, when we move past the coil a permanent magnet (as shown in the picture), we can move the electrons in the interior of the wire. Now the electrons get "sucked out" of the negative terminal. The moving permanent magnet "sucks" the electrons out of the negative terminal and pump them into the positive terminal. In the positive pole, the space gets narrower and the "crush" is increasing. In the negative pole there is always more space The battery is being charged.
8 Now, when all electrons are in the positive pole, the battery is charged. There is a big crowd in the positive pole. The battery is charged. w In the positive pole of the battery there is a big crowd. The excess electrons want to return to the negative pole. The amount of pressure with which the electrons are pushing out of the positive pole represents the measure of the voltage in volts. Name Unit Unit abbrev. Voltage Volt V And we can measure this voltage. Switch on your multimeter and select the unit V---. with the turn knob switch. Connect the red wire of the battery positive (+) terminal to the terminal V of the multimeter. Connect the black wire of the battery negative (-) terminal to the terminal Com of the multimeter. Depending on the measurement device (multimeter) the voltmeter should now indicate a voltage of 4.2 to 4.8 V. 4.8 V indicates, that the battery is fully charged. The voltage of a battery indicates the amount of pressure that pushes the electrons out of the battery.
9 Now, construct this circuit. Now when you operate the switch and the lamp lights up there is a voltage of about 3.3 to 4.6 V across the lamp. When you open the switch and the lamp extinguishes, there is no voltage across the lamp and the voltmeter indicates 0 V. Observe the brightness of the lamp and try to keep in mind the brightness. Now complete the circuit with the 47 Ω component. Now observe the brightness and measure the voltage. The lamp shines less brightly and the voltage across the lamp is only about 2.2 V. The electrons can no longer flow "unbraked" through the light bulb. The 47 Ω resistor "slows down" the electrons. And this brings us to the third value!!! The resistance I we would connect the positive pole directly with a cable to the negative pole, the electrons could flow back again "unbraked" to the negative pole and heat thereby the cable so strongly, that the cable melts. This would be a short circuit and destroy the components
10 The electrons must be "decelerated". This is made by the resistor. Our picture shows the resistor-component. It has coloured rings around its circumference. This indicates the value of the resistor. In addition, the component is described as a symbol with the associated value. The number 47 indicates how strongly the electrons are "decelerated". The larger the number the more the electrons will be "decelerated". The larger the number the more the electrons are "decelerated". Ω means ohm and is the unit for the electric resistance We say: The resistor has a value of 47 ohms Measuring resistance Switch on vour multimeter and select the unit Ω. Connect the red cable of battery + to the red terminal of the multimeter Connect the black cable of battery - to the black terminal Com (common ground) of the multimeter. Depending on the multimeter, the ohmmeter should indicate a value of 46 Ω to 48 Ω. The resistance value has a tolerance and therefore this value fluctuation is possible. Attention: when measuring resistance, all connexions should be disconnected from the electrical circuit. Here, you can see several types of resistors. On top, with indication of the value, below, with color codes. (color code table, see component data reg. 9)
11 Summary of the electrical values The voltage in volts indicates how strongly the electrons are pushed through the wire. The current in amperes indicates how much electrons flow through the wire. The resistance in ohms indicates how strongly the electrons are "slowed down". From the photo to the diagram A photo needs a lot of memory space and we don't optimally recognize the components. That's why in the electronics, the components are displayed with symbols. Instead of the photo, only the symbol is drawn. Attention: 47R = 47Ω Each component is marked with a symbol. Now look at all components and the corresponding symbol. Measure the resistor (components) with your ohmmeter. You will find all components with the symbols in register 6, details about components. Now construct this circuit, which is only shown with symbols. It is the same circuit as on the picture on top of page 9.
12 The Ohm's Law Construct this circuit: Now measure the voltage across the lamp. Result = about 2.1 V But the battery voltage is about 4.6 V Where is the voltage difference of about 2.5 V? Now measure the voltage across the resistor: Measure the voltage drop across the resistor of 47 Ω. Result = about 2.5 V So we see that with a resistor, the voltage across the lamp can be lowered. The electrons have to flow through the 47 Ω resistor and they are "slowed down". Thereby the voltage drops. The voltages across the resistor and across the lamp are split up. Now measure the current as shown in the circuit The ammeter indicates a current of 50 ma. We also could calculate de current: Current = Voltage : Resistance= Current = 2.5 V : 47 ohms = 0.05 A = 50mA If we know the values of any two of the three quantities (voltage, current, and resistance) in this circuit, we can use Ohm's Law to determine the third..
13 Voltage in volts = Resistance in ohms x Current in amperes V = Ω x A In the formula these abbreviations are common Name (quantity) Formula sign Unit Unit abbreviated Voltage U Volts V Resistance R Ohms Ω Current I Amperes (amp) A With the Ohm's Law triangle you can transpose a formula, without any mathematical knowledge: It's easy to work with this triangle... Let's make an example:
14 Calculation of values 1: What voltage will indicate the voltmeter, when a current of 0.05 A flows? The voltage across the resistor can be calculated: U R 47Ω = R x I = 47 Ω x 0.05 A = 2.35 V Battery voltage - resistance of the battery = voltage across de lamp U Batt - U R = U L = 4.5 V - 2.35 V = 2.15 V Calculation of values 2: Construct this circuit and measure the voltage across the resistor Result = 0.7 V How much current is flowing through the wires? I = U : R = 0.7 V : 10 Ω = 0.07 A What is the resistance of the filament of the lamp? Battery voltage - voltage across the resistor = voltage across the lamp U batt - U R = U L = 4.6 V - 0.7 V = 3.9 V R lamp = U lamp : I lamp = 3.9 V : 0.07 A = 56 Ω
15 The serial connection Construct this circuit: The ohmmeter indicates about 10 Ω Extend the circuit as follows: Now the ohmmeter indicates 57 Ω Statement: The resistors are connected in series. Law: In the series circuit, the individual resistances are added up. R total = R 47Ω + R 10Ω = 47 Ω + 10Ω = 57 Ω What is the total resistance in this circuit? R total = R 47Ω + R 100Ω = 47 Ω + 100Ω = 147 Ω R = R 1 + R 2
16 Replacing the picture of the multimeter by a symbol Also the multimeter can be represented in a simplified form Measuring device Symbol Description The voltmeter is indicated by a circle with the letter V. The setting on the multimeter must be chosen by yourself. The ohmmeter is indicated by a circle with the letter Ω. The setting on the multimeter must be chosen by yourself. The ammeter is indicated by a circle with the letter A. The setting on the multimeter must be chosen by yourself. Calculation of series circuits: Calculating the total resistance: R total = R 47Ω + R 100Ω + R 100Ω = 47 Ω + 100Ω + 100Ω =247 Ω Check the result by measuring. Calculation of the total resistance: R total = R 2.2kΩ + R 100Ω + R 100Ω = 2200 Ω + 100Ω + 100Ω =2400 Ω Check the result by measuring.
17 The voltage divider Construct this circuit, without connecting the voltmeter. Would it be possible to calculate the values which the voltmeters should indicate, on the basis of this information? 1. Calculation of the total resistance R = R 10 + R 47 = 10 Ω + 47 Ω = 57 Ω 2. Calculation of the current I = U : R = 4.5 V : 57 Ω = 0.078 A = 78 ma 3. Voltage across the 10 Ω resistor U = R x I = 10 Ω x 0.078 A = 0.78 V 4. Voltage across the 47 Ω U = U Batt U 10Ω = 4.5 V 0.78 V = 3.72 V With resistors the voltage can be split up. Now measure and evaluate the calculated values. Now, calculate and check this circuit like shown above
18 The parallel connection Construct this circuit: The ohmmeter indicates 50 Ω Now measure the total resistance! The ohmmeter indicates about 9.1 Ω The total resistance is less than the smallest of the individual resistances A circuit with two resistors can be calculated as follows: R = R 10 x R 100 = 10 Ω x 100 Ω = 9.09 Ω R 10 + R 100 10 Ω + 100 Ω R= R 1 x R 2 R 1 + R 2 Construct this circuit and measure the total resistance: Result = 8.3 Ω To calculate this circuit, a different formula must be used: R = 1 = 1 = 8.33 Ω 1 + 1 + 1 1 + 1 + 1 R 1 R 2 R 3 100 100 10 R = 1 = 1 + 1 + 1 R 1 R 2 R 3
19 Calculation and measurement in the parallel connection Construct this circuit: Observe the brightness of the lamp: The lamp lights up pratically as brightly as without any resistors. The resistors are connected in parallel. What is the total resistance of both resistors? R = R 10 x R 47 = 10 Ω x 47 Ω = 8.24 Ω R 10 + R 47 10 Ω + 47 Ω Justification: The electrons can flow through both resistors, the 10 Ω and the 47 Ω as well. The effect of the two resistors is as if a 8.2 Ω resistor would be installed. Now measure the total current = 73 ma Now, what current flows through the 10 Ω resistor? It represents a series circuit of both resistors with the lamp. The voltage across the resistors is U = R : I = 8.2 Ω x 0.073 A = 0.6 V I R10 = U R : R 10 = 0.6 V : 10 Ω = 0.06 A Now, what current flows through the 47 Ω resistor? I R47 = U R : R 47 = 0.6 V : 47 Ω = 0.012 A = 12 ma or differently calculated: I R47 = I I R10 = 73 ma 60 ma = 13 ma Now measure and check all calculated values with the multimeter
20 Combination circuit: A combination circuit is composed of a series and a parallel connection. The two 100 Ω resistors are connected in parallel and to these resistors is connected the 10 Ω resistor in series. First, the value of the parallel connection must be calculated: When two parallel connected resistors have the same value, the formula is: R = R indiv. R= 100 Ω = 50 Ω n = number of resistors n 2 Now the equivalent resistance of 50 Ω in series with the 10 Ω resistor can be calculated: R = R 50 + R 10 = 50 Ω + 10 Ω = 60 Ω Calculate the total resistance of this circuit 1.The parallel circuit of the two 100 Ω resistors = 50 Ω 2. 50 Ω resistor in series with the 10 Ω resistor = 60 Ω 3. These 60 Ω resistors are connected in parallel with the 1 kω resistor. Calculation with the parallel-formula: R= R 60 x R 1000 = 60 Ω x 1000 Ω = 56.6 Ω R 60 + R 1000 60 Ω + 1000 Ω
21 The electric power Construct this circuit Now measure the current that flows through the lamp: Result: = 78 ma or 0.078 A Now measure the voltage across the lamp: Result: = ca. 4.5V When a current of 0.078 A with a voltage of 4.5 V flows through the lamp, this is a power, because the lamp illuminates and brightens the room. The power can be calculated by multiplying the voltage by the current. P = U x I The formula symbol for power is the capital letter P (engl. Power) The formula symbol for voltage is the capital letter U The formula symbol for current is the capital letter I The power of the lamp is: P = U x I = 4.5 V x 0.078 A = 0.35 W The unit of power is Watt, abbreviated W Here, the Ohm's Law triangle can also be used: P = U x I P U x I U = P I I = P U
22 Construct this circuit: Electric energy: What's now the power consumption of the lamp? Search in the above circuits the following values: Current: = 51 ma = 0.051A Voltage across the resistor: = 2.5 V P L = U x I = (4.5 V 2.5 V) x 0.051 A = 0.1 W Now the lamp consumes a power of 0.1 W How long can the lamp light up with this battery? La lamp consumes a power of 0.35 W The battery has a capacity of 4800 mah = 4.8 Ah (according components descriptions). The capacity of 4.8 Ah means that this battery can deliver a current of 4.8 A for 1 hour. One can imagine that there are so many electrons in the positive terminal of the battery, that for one hour a current (electrons) of 4.8 A can continuously flow to the negative terminal. Through the lamp flows a current of: I = P : U = 0.35 W : 4.5 V = 0.078 A = 78 ma Time = Capacity = 4800 mah = 61.5 hours Current 78 ma With this battery, the lamp could light up for 61.5 hours. What is the energy consumption of the lamp in this case? Energy = Power x Time = 0.35 W x 61.5 h =21.5 Wh 1000 Wh electric current costs sfr. 0.25
23 The potentiometer: Construct this circuit and turn the potentiometer By turning the potentiometer the resistance changes from 0 Ω up to 220 Ω. The potentiometer is a variable resistor The hatched surface is a carbon film which has connexions on both sides. The rotatable part is a metal that ensures the electrical contact Construct this circuit and turn the potentiometer By turning the potentiometer, the brightness of the lamp can be variably changed (adjusted). (Example: Car interior lights)
24 The NTC resistor (Negative Temperature Coefficient) Construct this circuit: Construct this circuit: Findings: Use the green LED as symbolically depicted in the diagram Findings: The LED does not light up, no current flows Now, heat carefully the NTC resistor with a lighter Findings: The LED begins to light up, current flows What value shows the ohmmeter at room temperature? Result: = about 11000 Ω (written = 11 kω) This is also described in the symbol At 11000 ohms resistance, almost no current can flow Now, heat the resistor By heating, the NTC resistor changes its resistance value By heating, the resistance decreases Result: = the resistance drops below 1000 Ω This is indicated by the two arrows. Additionally there is the sign t The left arrow is pointing upwards and indicates the value of the temperature The right arrow is point downwards and indicates the value of the resistor t means, that this is a temperature dependent resistor at 20 C it has a resistance of 13kΩ
25 The PTC resistor (Positive Temperature Coefficient) Construct this circuit: Use the green LED as symbolically depicted in the diagram Findings: The LED lights up, current flows Now, heat carefully the PTC resistor with a lighter Findings: The LED begins to darken, less current flows Construct this circuit: Findings: What value shows the ohmmeter at room temperature? Result: = about 50-100 Ω This is also described in the symbol At 67 ohms resistance, current can flow Now, heat the resistor Result: = the resistance increases above 1000 Ω By heating, the PTC resistor changes its resistance value By heating, the resistance increases This is indicated by the two arrows. Additionally there is the sign t The left arrow is pointing upwards and indicates the value of the temperature The right arrow is pointing upwards and indicates the value of the resistor t means, that this is a temperature dependent resistor at 20 C it has a resistance of 67Ω
26 The LDR resistor (LDR =Light Dependent Resistor) Construct this circuit: Use the green LED as symbolically depicted in the diagram. Findings: The LED lights up, current flows Now, keep your finger on the LDR Findings: The LED begins to darken, less current flows Construct this circuit: What value shows the ohmmeter at brightness? Result: = about At 1000 ohms resistance, current can flow Now obscure the LDR Result: = the resistance increases above 60000 Ω Findings: The LDR resistor changes its resistance value when it is exposed to light In the dark the resistance is high This is indicated by the two arrows. The two arrows indicate that light acts on the surface The square symbol shows a resistor with the inscription LDR, this indicates that it is a light dependent resistor
27 The capacitor Construct this circuit: Close the switch! The voltmeter indicates about 4.5 V Open the switch and observe the voltmeter Findings: Although no battery is connected, the voltmeter indicates first 4.5 V and then the voltage drops slowly. How is that possible? Now, again just a bit of theory... The capacitor is made up of two metal plates which are only separated by a thin plastic film. Since metal atoms of the outermost shell have only one electron, only these electrons are depicted in the metal plates. Metal plate above (+plate) Plastic film Metal plate below (-plate) Charging of the capacitor When a battery is connected to a capacitor, electrons "push" into the upper positive metal plate.
28 This filling of the positive metal plate creates a magnetic force that pushes the electrons out of the negative metal plate and let them flow into the negative terminal of the battery. Now the capacitor is charged and acts like a charged battery. Now, when we connect a voltmeter, the electrons push out of the charged battery and indicate the battery voltage 47 μ F means 47 microfarad and is the unit for the size of the metal plates. The greater the number the greater the plates The second important quantity of the inscription is the voltage in V. It indicates the maximum voltage that the plastic film can (insulate) prevent electrons jumping. Now construct this circuit: To do this, use pushbutton switches and not switches First press shortly the left pushbutton switch. What is happening now? Electrons flow from the positive battery terminal into the positive plate of the capacitor. As a result, the electrons are pushed out of the negative metal plate and by this current flow through the LED and it lights up briefly and then it is dark.
29 Discharging of the capacitor Now, when the upper pushbutton switch is pressed, the capacitor discharges. The electrons flow from the positive plate to the negative plate and balance each other out (neutralize). Construct this circuit: Now, when you close the left pushbutton switch, the capacitor is being charged. Open the left pushbutton switch and press the right pushbutton switch. Result: The LED lights up until the capacitor is discharged. Construct this circuit: To understand the circuit, you need to know the function of the LED. Only this: The LED only lights up when current flows in the direction of the arrow. Now, flip the switch back and forth several times and observe the function. When the switch is positioned like in the above diagram, the capacitor is being charged. The current will flow through the left LED which flashes briefly. By flipping the toggle switch down, the capacitor is being discharged and the current flows through the right LED which flashes briefly. This is the proof that the capacitor works. You will experience applications of the capacitor in further circuits.
30 The switch Picture Schematic symbol Sign for latching The switch is a metallic connector. The contact is latching, that means that the switch remains closed by itself. This is shown in the diagram by the latching sign. This is a toggle switch Examples: light switch, On-Off Switches for electrical devices. The push button switch Picture Schematic symbol latching symbol is missing The pushbutton switch is also a metallic connector. However, it is usually triggered by a pushbutton. When released, it automatically disconnects the contact again. Therefore the latching symbol is missing. Examples: doorbell, car horn etc.
31 The light bulb Picture Symbol The light bulb consists of a filament of tungsten wire. Electrons cannot flow well through the tungsten wire and have to "fight". Thus, the wire is heated up to 3000 C and begins to glow. Thus the glowing wire does not burn and melt, he is in a glass envelope without air (vacuum) Construct this circuit and measure the resistance of the wire, when it is cold. Result: about 7.5 Ohm By calculating, at a battery voltage of 4.5 V a certain current should flow. I = U : R = 4.5 V : 7.5 Ω = 0.6 A or 600 ma However, on page 5 we have measured a current of 0.08 A or 80 ma. Why such a difference? At a cold tungsten wire, the atoms do not move and the current can flow well. Therefore, the light bulb has a small resistance of 7.5 Ω when it is cold. When the tungsten wire gets hot, the atoms vibrate strongly and slow down massively the electrons. Therefore the current will then be about 8 times smaller. The calculated resistance is now about 57 Ω. The light bulb has a PTC behaviour and changes its resistance according to the temperature. In The practice, relay the measured values will be different
32 The relay Picture Symbol It consists of a coil and a pushbutton switch When current flows through the coil, the anchor becomes magnetic and closes the contact. Construct this circuit: When you close the switch, you will hear a click in the relay and the lamp lights up. Now measure the current across the coil: Result: 37 ma With 37 ma it is now possible to switch currents up to 5A. This gives an amplification factor of = 5 A : 0.037 A = 135 The current which flows through the coil is called control current and is the small current The current which flows through de switch is called load current and is the high current Applications: Relays are used for switching high currents Relays can also be used for logic functions.
33 Types of relays When a voltage is applied to the relay coil terminals 85 and 86, the contact is attracted and closes. This is a Normally Open Relay When a voltage is applied to the relay coil terminals 85 and 86, the contact is attracted and opens. This is a Normally Closed Relay When a voltage is applied to the relay coil terminals 85 and 86, the contact is attracted and depending on the connection, the contact closes or opens. This is a Changeover Relay Construct this circuit This is a logic circuit. When the left pushbutton switch is activated, the lamp begins to light up and remains bright. By pressing the right pushbutton switch, the lamp will go out again. This is a first circuit for collecting information and thus the first computer was built up.
34 Importances: Here are two important issues that often lead to discussions. Construct this circuit, but do not close the switch yet. Will the lamp light up, if the voltmeter is connected this way? The resistance between the terminal V and the terminal COM is very high ( 20 000 000 Ω = 20MΩ). That means, no current can flow through the voltmeter and therefore the lamp cannot light up. What value the voltmeter will indicate when the switch is closed? The voltmeter indicates the voltage of the battery, about 4.6 V although de 47 Ω resistor is installed. Why? When the switch is closed, no electrons can flow; that means, they "push" to the front of the cable with 4.6 V. Now an important statement in the electrotechnology: No current, no voltage drop. Now, construct this circuit: What value the voltmeter will now indicate, when the switch is closed, since the resistor is installed in the negative line? The voltmeter indicates about 2.15 V. The voltage is divided in the same way as when the resistor is installed in the positive line. The electrons are only slowed down after the light bulb and can only flow with full speed into the negative terminal after the 47 Ω resistor. This was the electrotechnology course. In order to consolidate these many impressions, we recommend you to work through this set once again. Then you can deepen the reached knowledges in section 4 Repetition with instructions.