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Physics – Further Electronics “the essentials” lectures, 2014 author(s) These materials represent the collective effort of many teachers across the state. The principal author of this booklet is: Dr. Greg Wilmoth, B. Sc (Hons)., Ph.D., Dip. Ed., Grad. Dip. Computing (Senior VCE Teacher – Haileybury College). lecturer(s) To ensure that students are afforded every possible advantage in their examinations, our lectures are prepared and delivered by qualified, currently practising VCE teachers and official VCAA exam markers who possess the knowledge and experience to demonstrate the means by which students can achieve the higher ATAR scores. Further details regarding our teachers (including qualifications and experience) may be obtained at http://www.tsfx.com.au/what-is-tsfx/ourteachers/. important notes Our policy at TSFX is to provide students with the most detailed and comprehensive set of notes that will maximise student performance and reduce study time. These materials, therefore, include a wide range of questions and applications, all of which cannot be addressed within the available lecture time i.e. Due to time constraints; it is possible that some of the materials included in this booklet will not be addressed during the course of these lectures. Where applicable, fully worked solutions to the questions in this booklet will be handed to students on the last day of each subject lecture. Although great care is taken to ensure that these materials are mistake free, an error may appear from time to time. If you believe that there is an error in these notes, please let us know asap ([email protected]). Errors, as well as clarifications and important updates, will be posted at www.tsfx.com.au/vce-updates The views and opinions expressed in this booklet and corresponding lecture are those of the authors/lecturers and do not necessarily reflect the official policy or position of TSFX. TSFX - voted number one for excellence and quality in VCE programs. copyright notice These materials are the copyright property of The School For Excellence and have been produced for the exclusive use of students attending this program. Reproduction of the whole or part of this document constitutes an infringement in copyright and can result in legal action. No part of this publication can be reproduced, copied, scanned, stored in a retrieval system, communicated, transmitted or disseminated, in any form or by any means, without the prior written consent of The School For Excellence (TSFX). The use of recording devices is STRICTLY PROHIBITED. Recording devices interfere with the microphones and send loud, high-pitched sounds throughout the theatre. Furthermore, recording without the lecturer’s permission is ILLEGAL. Students caught recording will be asked to leave the theatre, and will have all lecture materials confiscated. it is illegal to use any kind of recording device during this lecture REGULATED DC POWER SUPPLY Electricity comes to our home as 240 volt 50 Hz alternating current (AC). AC has many advantages over DC: • More efficiently transmitted. • Easily transformed. • AC generators are simpler and cheaper. • More efficient safety devices work on AC. (The generation and distribution of AC electricity will be covered in Unit 4.) Most of our electronic devices require a low voltage direct current (DC). This unit of work looks at the design of a regulated DC power supply that is able to convert an AC supply to DC. Generally there are four main stages involved in the establishment of a regulated DC supply from an AC input. Transformer: Converts the AC supply to a suitable value for conversion. Rectification: Converts alternating current into current flowing in a single direction. Filtering: Smoothing of the pulsating voltage. Regulation: Maintaining a constant voltage over a range of input and load conditions. Block diagram for a regulated DC power supply The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 1 WHY IS A VOLTAGE DIVIDER AN UNREGULATED POWER SUPPLY? 1. Effect of varying load on a voltage divider. Consider the following voltage divider circuit. What is the voltage output when the voltage divider is unloaded (no load resistance placed across VOUT)? Enter your answer into the table below. Now consider a single 100 Ω load placed across VOUT. This load resistance is effectively arranged in parallel to R2. Determine the combined resistance and apply voltage divider principles to find VOUT? Enter your answer in the table below. Add another, then another 100 Ω load across VOUT. Again determine combined resistance with R2, as well as the output voltage. Number of 100 ohm loads Combined R2 / load Resistance (Ω) 0 100 Output voltage 1 2 3 What happens to VOUT when the voltage divider is loaded? 2. Effect of varying input voltage. Consider what would happen if the input voltage reduces to 10 V from the original 12 V. Voltage divider calculations will reveal that VOUT decreases to 6.7 V. Dropping the input even further to 9 V gives an output voltage of only 6.0 V. Clearly there is a direct relationship between VIN and VOUT. In an unregulated power supply, as the load current increases, the voltage across the load decreases. This is unsuitable for many applications. Voltage regulation involves the supply of a constant DC voltage over a range of input voltages and varying load conditions. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 2 ALTERNATING CURRENT The current in an AC circuit changes direction many times each second. The free electrons in the conducting wires move first in one direction and then the other, creating an oscillating voltage ‘wave’. In an AC circuit the electrons simply oscillate back and forth. At 50 Hz, the electrons change direction 100 times per second. In a DC circuit, such as a battery, the electrons move in only one direction. Alternating supply voltage The peak value of the domestic AC voltage supply oscillates between +340 V and –340 V. Sometimes the supply is described by its peak to peak variation, in this case 680 V. i.e. V peak-to-peak or V p-p = 680 V Both the Vp and the V p-p values are of limited use because they are not representative of the effective voltage delivered. These values are only maintained for an instant and a lesser amount is supplied at all other times. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 3 An arithmetic average would yield zero voltage and current because of the alternating nature. The most useful measurement of an AC voltage is what is called its root mean square (RMS) value. This term refers to the equivalent DC voltage that would be needed to produce the same power. Unless otherwise specified, it is usual to assume RMS values as this is the most effective and useful value. RMS voltage V rms = V peak 2 RMS current I rms = I peak 2 Note that the relationship for powerRMS is somewhat different: PRMS = VRMS × I RMS = VPEAK I PEAK V ×I P × = PEAK PEAK = PEAK 2 2 2 2 The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 4 QUESTION 1 A cathode ray oscilloscope was used to measure the maximum voltage of an AC power supply at 18 V. What is the RMS voltage of the power supply? A 9.0 V B 12.7 V C 25.5 V D 3.0 V QUESTION 2 What is the peak to peak voltage of a 25 V RMS AC supply? A 17.7 V B 35.4 V C 50.0 V D 70.7 V The following information refers to Questions 3 to 5. An AC current with an RMS voltage of 12 V is applied to a 50 ohm load. QUESTION 3 What is the RMS current? A 0.24 A B 0.03 A C 0.01 A D 0.48 A QUESTION 4 What is the peak current? A 0.04 A B 0.34 A C 0.10 A D 0.68 A The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 5 QUESTION 5 What is the peak power dissipated in the load? A 4.08 W B 2.89 W C 5.77 W D 8.15 W QUESTION 6 What is the RMS equivalent of 0.50 A peak current? A 0.250 A B 0.353 A C 0.707 A D 0.354 A QUESTION 7 If the voltage at point A is 200 V, what is the RMS voltage for this supply? A 141 V B 283 V C 71 V D 566 V QUESTION 8 The frequency for this supply is 50 Hz. What is the time represented by point D on the graph? Give your answer in milliseconds (ms). A 20 ms B 60 ms C 30 ms D 40 ms The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 6 TRANSFORMER PRINCIPLES A transformer is a device that can change the size of AC voltage. A transformer is a very simple device that consists of two sets of insulated coils that are wound onto a soft iron core. The first coil is called the primary or input, and the other is the secondary coil or output. Load Primary Winding Secondary Winding AC Power Supply Iron Core The operating principles of a transformer will be explored more thoroughly in Unit 4. In this unit it is only a requirement to know what the transformer does, not how it does it. For the point of this exercise, it is worth noting that the number of coils in the primary coil compared to the secondary coil will affect the ratio of voltages and currents according to the following relationship: Transformer equation: N P VP I S = = N S VS IP You may also come across the term ‘an ideal transformer’. This simply means that it is a transformer that is 100% efficient, i.e. power in = power out. QUESTION 9 A transformer has 1000 primary coils and an input of 240 V rms. How many secondary coils would be needed in order to have an output of 6 volts? A 18 coils B 25 coils C 17 coils D 40 coils The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 7 QUESTION 10 The diagram below shows a simple transformer, which may be regarded as 100% efficient. The input terminals P and Q are connected to a 50 V AC power supply. The output terminals R and S are connected to an AC voltmeter. Based on this diagram, what is the approximate voltmeter reading? A 13 V B 1.0 V C 10 V D 14 V QUESTION 11 The power input into an ideal transformer is 50 W. If the output voltage is 10 V, what current would be flowing in the secondary circuit? A 0.5 A B 2.0 A C 0.2 A D 5.0 A The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 8 MEASURING DEVICES VOLTMETER Voltmeters are placed in parallel across the component being measured. Their purpose is to measure the difference in electrical potential (energy level) between the two points of connection. It is important that any measuring device being used has minimal effect on the actual circuit being used. Voltmeters achieve this by having very high resistance. This way they draw very little current and place negligible additional load on the circuit being tested. However, under circumstances of high circuit resistance, inclusion of a voltmeter can affect the amount of current drawn. AMMETER Ammeters have very low resistance and are designed to be placed in series within the circuit being tested. The low resistance of the ammeter minimises any effect it may have on the resistive load of the circuit. If the circuit has low resistance however, some effect may be noted. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 9 OHMMETER Ohmmeters are designed to measure the resistance of components or sections of circuits. Ohmmeters can be used with individual components or on open circuits such as the one shown below. The ohmmeter applies a potential difference (hence it needs its own power source) to the component and measures the current that flows. The amount of current is related to the resistance of the circuit. An ohmmeter can be used to measure continuity, i.e. whether a circuit is continuous or open. An example of this is the testing of a faulty lead. If the lead has an internal break, the ohmmeter will record this as infinite resistance and go off scale. If the lead is continuous, the ohmmeter will record a near zero resistance. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 10 MULTIMETER (From Jacaranda Physics 2 – 2nd Ed) A multimeter is a device that can perform a range of functions. Most can measure DC and AC voltages, current and resistance; however, others can also test diodes and transistors, as well as measure temperature and frequency. Multimeters also have a range of scales. If the approximate value being measured is unknown, start with the maximum range, and then progressively switch to an appropriate narrower range. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 11 CATHODE RAY OSCILLOSCOPE An oscilloscope is a laboratory instrument commonly used to display and analyse the waveform of electronic signals. In effect, the device draws a graph of the instantaneous signal voltage as a function of time. The CRO produces a beam of electrons that makes a spot on the screen. The beam can be made to sweep across the screen using the time base control. The vertical deflection results from the applied input. (Heinemann Physics 12 – 2nd Ed) Since a CRO measures voltage, it is connected in parallel with the component across which the voltage variation is to be measured. If connected to an AC source it would produce an image such as this. The grid allows voltage and time to be measured and calculated in conjunction with the horizontal and vertical scale settings. A cathode ray oscilloscope is a particularly useful measuring device in this unit because it not only reveals the AC nature of the input voltage, but it visualises rectification, the extent of filtering, and the effectiveness of the final regulated output. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 12 An oscilloscope is connected to an AC signal generator. The vertical scale is set on 5 V/cm, and the horizontal scale on 20 ms/cm. QUESTION 12 What is the peak-to-peak voltage of the signal generator? A 35 V B 17.5 V C 7.5 V D 15 V QUESTION 13 What is the voltage RMS of the signal generator? A 5.30 V B 24.7 V C 10.6 V D 12.4 V QUESTION 14 What is the frequency of the signal generator? A 20 Hz B 12.5 Hz C 50 Hz D 80 Hz The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 13 RECTIFICATION Power diode – has a very high breakdown voltage to prevent reverse flow of current. Note the different scales for reverse and forward voltages, and for forward and reverse current. When the voltage across the diode is positive, the diode is said to be forward biased. Before a silicon diode starts to conduct, or behave like a low resistance conductor, it must have a voltage of about 0.7 V across it. When a negative voltage is placed across the diode, it is said to be reverse biased; a small leakage current of only a few microamperes will flow at best. If the negative voltage is large enough, the semiconductor will break down and a large current will flow. Note that when a diode is reverse biased, as shown below, very little current flows through the diode and the resistor. The voltage drop across the resistor will be negligible and all of the voltage drop will be across the diode. In forward bias the diode conducts well and the Vdiode is around 0.7 V. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 14 A rectifier is a circuit element that is connected to the output of a transformer and converts the AC voltage into a DC voltage. Rectification can be half-wave or full-wave. Half-wave rectification simply blocks the negative part of the AC cycle, full-wave rectification converts the negative cycle into a positive one. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 15 HALF-WAVE RECTIFICATION A half-wave rectifier uses a single power diode to pass only one direction of the AC voltage from the transformer onto another part of the device that is sometimes called the load. The remaining part of the voltage signal is across the diode, since the diode has a very large resistance when reverse biased. A half-wave rectifier circuit is shown below. Since the load has an effective resistance, it is represented as a resistor. If the input voltage from the transformer has a peak voltage of +9.0 V, the maximum voltage drop across the load resistance will be +8.3 V, because silicon-based diodes will have a voltage drop of 0.7 V when they are forward biased. When the input voltage is negative, the diode effectively prevents current from flowing through the load resistance and the voltage drop across the load will be zero. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 16 FULL-WAVE RECTIFICATION A bridge rectifier uses four diodes to produce full-wave rectification. Trace the current through the ‘four diode bridge’ for both halves of the AC cycle to determine the direction of current flow across the load. There will be an approximate voltage drop of 0.7 V across each diode that the current passes through. Each half of the AC cycle passes through two diodes in a bridge rectifier. This means that the peak voltage from a bridge rectifier will be approximately 1.4 V less than the peak voltage of the AC signal. If the frequency of the AC supply is 50 Hz, what is the frequency of the rectified output? The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 17 A centre-tapped transformer uses a different method to achieve full-wave rectification. The centre-tap full-wave rectifier uses only two diodes. The diodes are connected to either end of the secondary coil of a transformer. The centre of the coil is connected (tapped) to the earth, hence it is maintained at 0 V. In this case there will also be an approximate voltage drop of 0.7 V across each diode that the current passes through. Here each half of the AC cycle passes through only one diode in the centre-tap rectifier. This means that the peak voltage from a centre-tap rectifier will be approximately 0.7 V less than the peak voltage of the AC signal. The output of a centre-tap rectifier can also be smoothed using a capacitor, as shown in the following section. QUESTION 15 Which one or more of the above circuits will produce a full wave DC output? The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 18 QUESTION 16 The bridge rectifier consists of four diodes mounted side by side on a heat sink, and connected by wires. Which one of the following circuits shows the diodes correctly connected? The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 19 EXAMPLE 1 What is the peak voltage across the load? Don’t forget to allow for the voltage drop across the rectifying diodes when assigning a value to VPEAK beyond the rectifier. Solution This problem requires three steps to be completed. Step 1: Determine the RMS voltage output of the transformer. N P VP I S = = N S VS I P 20 249 = 1 VS Vs = 12 V RMS Step 2: Determine the peak voltage output of the transformer. VPEAK = VRMS × 2 = 12 × 2 = 17.0V Step 3: Allow for voltage drop across the two diodes. At any one time current passes through two diodes. Therefore, two potential decreases of 0.7 V occur. The peak voltage across the load will be 17.0 – 2 x 0.7 = 15.6 V. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 20 CAPACITORS The process of converting the varying DC voltage from the rectifying diodes into a steady DC voltage is called filtering or smoothing. They can be achieved by placing a capacitor into the circuit. In general, circuits containing a resistor and a capacitor are known as RC circuits. This is an example of a simple RC circuit. When the switch is closed, current flows through the circuit. Electrons are pulled off one plate, flow through the battery, and are deposited onto the other plate of the capacitor. This is called charging the capacitor and continues until the voltage across the plates is equal to the emf of the cell. The closer the voltage across the capacitor gets to the supply voltage, the smaller the current becomes. When the voltages of the cell and the capacitor are equal, the current ceases. When the battery is disconnected the capacitor retains its charge. If the capacitor is now placed into a circuit, because it has potential difference across its plates, it can release this stored up charge. This is known as discharging the capacitor. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 21 The magnitude of the current is affected by the resistance of the resistor. The bigger the resistance, the smaller the current will be and the slower the capacitor will charge (and discharge). If this resistance is decreased, the larger current will allow faster charging. The time constant, τ, (the Greek letter tau) for an RC circuit is the time in seconds it takes for the capacitor to reach 63% (or approximately two-thirds) of its final voltage, E, when charging. The time constant can be calculated by finding the product of the resistance in ohms and capacitance in farads for the RC circuit. τ = RC In a discharging cycle, the time constant is the time it takes for 63% of the charge to discharge from the plates. As a rule of thumb, 5 time constants is the time to either fully charge or discharge the capacitor, but note however, that different time constants may apply to the charging and discharging cycles as these may involve different resistances. The charge (Q) stored on a capacitor is proportional to the voltage drop (V) across the plates. Q=CxV The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 22 The relationship between the potential difference across the capacitor and the current flowing to or from the capacitor is interesting: The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 23 Capacitors in series behave like one capacitor with a plate separation greater than that of the individual capacitors. Because capacitance is inversely related to plate separation, the capacitance of capacitors in series is less than that of the individual capacitor. 1 C SERIES = 1 1 + C1 C 2 Capacitors in parallel behave like one capacitor with a greater plate area than the individual capacitors. The capacitance of parallel capacitors is the arithmetic sum of the individual capacitors i.e. CPARALLEL = C1 + C2 In this unit the most likely arrangement is in parallel. For example, if there is insufficient capacitance in a filtering circuit it could be handled by placing an additional capacitor in parallel (or replacing with a larger capacitor). The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 24 The following information refers to Questions 17 to 19. When switch S1 is closed, capacitor C charges as depicted in the graph below. QUESTION 17 What is the value of capacitor C? Give your answer in mF. A 10 mF B 25 mF C 15 mF D 20 mF QUESTION 18 What is the current passing through the 200 ohm resistor 5.0 seconds after switch S1 closes? A 0.082 A B 0.041 A C 0.091 A D 0.009 A The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 25 QUESTION 19 After one minute of charging switch S1 is opened and switch S2 closes. What is the voltage across the capacitor 30 seconds after the closing of S2? A 10 V B 3.7 V C 2.5 V D 9.0 V The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 26 FILTERING (OR SMOOTHING) The varying voltage from the rectifier is unsuitable for most electronic circuitry. It is important to be able to smooth out the signal and make it a constant DC value. This is where the capacitor is used. This is a smoothing circuit for a half-wave rectified load. While the voltage is reaching its peak the capacitor is charging as shown with the half-wave rectified signal shown below. As the voltage from the input starts to drop from its peak value the potential difference of the capacitor is now larger than from the source, so the capacitor starts to release its charge, maintaining a current flow through the resistor and potential difference across the load. The size of the time constant determines how quickly this charge is released. The larger the time constant of the capacitor and resistor combination the smoother the current will be. Full-wave rectifier output signals can be smoothed in the same way as half-wave rectifier signals. The main difference is that for a given RC combination, the capacitor has less time to discharge before it is recharged by the output signal. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 27 Using a capacitor with a time constant five times longer than the period of the supply gives a ripple voltage about 20% of Vpeak. This is an acceptable value for many devices. QUESTION 20 How long will it take the capacitor to fully charge when switch S1 is closed? A 0.01 s B 0.21 s C 0.05 s D 0.50 s QUESTION 21 Switch S1 is opened when the capacitor was fully charged, then switch S2 closes. How long will it take for the capacitor to discharge? A 1.0 s B 0.2 s C 0.5 s D 1.4 s The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 28 The following circuit was built to provide high voltage DC to a cathode-ray tube. The DC voltage has to be approximately 10 kV and the effective resistance of the cathode-ray tube in the circuit is 5 M Ω (5 x 106 Ω). QUESTION 22 The primary windings of the transformer have 50 turns. Which of the following best gives the number of windings required in the secondary windings? A 5 B 30 C 1500 D 7200 The bridge rectifier consists of four diodes connected appropriately, and the smoothing capacitor has a value of 10 µF (10 x 10-6 F). QUESTION 23 Which of the following best gives the time constant for the smoothing circuit of the cathoderay tube? A 5 x 10-6s B 0.5s C 5s D 50s The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 29 QUESTION 24 Which of the following graphs best shows the signal as observed across XY? QUESTION 25 Which of the following graphs best shows the signal as observed across CD? The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 30 QUESTION 26 Initially the circuit operates correctly providing the required smoothed output, however, a fault then develops and the output across EF is shown below: Which of the following is the most likely cause of this fault? A failure of: A B C D The transformer. The capacitor used for smoothing the signal. All the diodes in the bridge rectifier are blown. Only one of the diodes in the bridge rectifier is blown. QUESTION 27 After fixing this fault another develops, causing the following signal across EF: Which of the following is the most likely cause of this fault? A failure of: A B C D The transformer. The capacitor used for smoothing the signal. All the diodes in the bridge rectifier are blowing. Only one of the diodes in the bridge rectifier blowing. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 31 QUESTION 28 What would be the lowest power transformer that would be able to provide the cathode-ray tube with the required power? A B C D 1W 5W 25 W 100 W QUESTION 29 If the cathode-ray tube and its power supply, including the transformer, was connected to the 110 VRMS 60 Hz mains supply in the USA, some changes would need to be made? The best reason(s) for this change is: A B C D A different transformer is needed to allow for the different power required. A different transformer is needed to allow for the different input voltage only. A different transformer is needed for the different frequency only. A different transformer is needed to allow for both the different input voltage and different frequency. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 32 RIPPLE VOLTAGE Ripple voltage is the periodic variation in a DC voltage that results from the rectification/smoothing of an AC voltage. The peak-to-peak value of the ripple voltage is: Vripple = Vmax - Vmin The average voltage will be approximately the mid value between the maximum and minimum voltage. The peak-to-peak value of the ripple voltage depends on the following: • The maximum value of the voltage (Vmax) supplied by the diode(s) of the rectifier. • The period (T) of the unsmoothed voltage supplied by the diode(s) of the rectifier. • The capacitance of the capacitor. • Load resistance. And can be expressed as: Since: V = I LOAD R Then: Vr ( p − p ) ∝ I LOAD Vr ( p − p ) = V max × T R×C = I LOAD × T C Note that with a 50 Hz supply the period is: Half-wave rectified: Full-wave rectified: T = 20 ms T = 10 ms Note: The study design clearly specifies that only a qualitative understanding of ripple voltage is required. Therefore you won’t be asked to perform calculations such as the one above. The formula has deliberately been included here because it provides a convenient means for addressing some of the more likely qualitative questions. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 33 QUESTION 30 The following circuits have identical AC input, loads, capacitance and use similar diodes. CIRCUIT A CIRCUIT B How would the ripple voltages across each compare? Justify your choice. QUESTION 31 Indicate whether the following changes to CIRCUIT A from the previous question will alter the magnitude of the ripple voltage measured across the LOAD. (a) (b) Place a second capacitor in parallel to the original capacitor. A Increase B Decrease C No change Place a second LOAD in parallel to the original LOAD. A Increase B Decrease C No change The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 34 (c) (d) Place a second LOAD in series with the original LOAD. A Increase B. Decrease C No change Double voltage across the LOAD and double the capacitance. A B C Increase Decrease No change QUESTION 32 A rectifier supplies power to a 1 kΩ resistor. Which of the following capacitors connected across the output would give the smoothest DC supply? A B C D 200 μF 2000 μF 20000 μF 200000 μF The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 35 VOLTAGE REGULATION ZENER DIODE SHUNT Zener diodes have a relatively low reverse breakdown voltage, meaning that they can conduct in reverse bias when a specified voltage is applied. The breakdown voltage of a zener diode is significantly lower than a power diode. In this example, the lamp placed in parallel to the zener has a potential difference equivalent to the zener breakdown voltage. There are limitations which will restrict the use of this circuit. The current flowing through the dropping resistor, as well as through the zener diode, will result in power lost as wasted heat. If too much power is lost this way either the dropping resistor or the zener diode may burn out. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 36 EXAMPLE 2 A zener with a breakdown voltage of 6 V is placed in series with a voltage dropping resistor connected to a 9 V DC supply. This circuit delivers a constant 120 mA current. The load is an ohmic conductor with a resistance of 60 ohms. (a) What value voltage dropping resistor should be used? (b) How much power is dissipated by the load? (c) How much power is dissipated across the zener diode? (d) How much energy is wasted as heat across the dropping resistor and the zener diode during operation of this circuit in 60 seconds? Solution (a) The voltage across the resistor should be 3.0 V and the current 0.12 A. Applying Ohm’s law gives: R = (b) V 3 = = 25 Ω I 0.12 How much power is dissipated by the load? I = V 6 = = 0 .1 A R 60 PLOAD = VLOAD x ILOAD = 6 x 0.1 = 0.6 W (c) (or PLOAD = V 2 62 = = 0.6 W ) R 60 The current through the zener will equal the total current less what flows through the load i.e. 0.02 A. PZENER = VZENER x IZENER = 6 x 0.02 = 0.12 W (d) The total power consumption of the circuit is: Ptotal = VI = 9 x 0.12 = 1.08 W With 0.6 W dissipated by the load, the balance of the electrical power, which is 0.48 W, is dissipated as heat through the rest of the circuit (i.e. within the zener and the dropping resistor). Energy wasted = power x time = 0.48 x 60 = 28.8 J The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 37 The following information refers to Questions 33 to 35. QUESTION 33 What is the voltage observed across R1? A 3.2 V B 6.0 V C 8.0 V D 0. 7 V QUESTION 34 What is the voltage observed across R2? A 7.3 V B 1.0 V C 4.8 V D 2.0 V QUESTION 35 Which one of the following best gives the current through the Zener diode? A B C D 1 mA 3 mA 8.5 mA 10 mA The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 38 The following information refers to Questions 36 to 38. QUESTION 36 A 12 V battery is used to run a 9 V mobile phone by using a voltage dropping resistor in series with a zener diode. The zener has a breakdown voltage of 9.0 V. The telephone uses 180 mA of current and 20 mA flows through the zener diode. What should be the resistance of the voltage dropping resistor? A 0.15 Ω B 2.40 Ω C 0.54 Ω D 15.0 Ω QUESTION 37 How much power is the mobile phone using? A 0.18 W B 2.40 W C 180 W D 1.62 W QUESTION 38 How much energy is being wasted in the circuit each second? A 2.40 J B 0.78 J C 1.62 J D 2.22 J The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 39 The following information refers to Questions 39 to 43. An oscilloscope is placed across the 500 ohm load and the output is shown in the graph below: QUESTION 39 Which one of the following is the best estimate of the average power dissipated in the 500 Ω load resistor? A B C D 0.02 W 0.2 W 0.4 W 5000 W QUESTION 40 Which one of the following is the best estimate of the peak-to-peak ripple voltage? A B C D 0.5 V 5.0 V 8.5 V 13.5 V The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 40 Voltage regulation is attempted by adding R1 and a zener diode as shown below. QUESTION 41 Which one of the following best shows the output she will observe? QUESTION 42 What is the function of resistor R1? QUESTION 43 Which one of the following changes would best reduce the ripple voltage? A B C D Replace R1 with a 50 ohm resistor. Replace the 6 V Zener diode with a 9 V one. Replace the capacitor with a 500 uF capacitor. Replace the transformer with one with a 9.0 VRMS output. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 41 IC REGULATION A voltage regulator is an integrated circuit (IC) device that delivers a steady terminal voltage despite variations in the input voltage or variations in the current drawn by the load. The three terminals of an IC regulator are: • The input from the power supply. • The common or ground (0 V). • The output, the smoothed and regulated voltage ready for use. There are two types of commonly used IC voltage regulators. A fixed regulator gives a set output voltage, and an adjustable regulator can give a range of output voltages. It is important that the input into the IC regulator is maintained a few volts above the regulated output voltage. The minimum amount of difference required by the IC regulator is called the drop-out voltage. The ripple must not go below the output voltage of the IC regulator. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 42 Because the input to the regulator must be maintained higher than the regulated output, it goes that there must be a potential drop across the regulator. The power dissipated in the IC regulator equals: PLOSS = Vdrop x ILOAD But how can you determine Vdrop if the input is not constant? VIN ≈ Vmax + Vmin i.e. the input voltage is approximately the mid-ripple voltage 2 and Vdrop = V IN − VOUT . EXAMPLE 3 A 9 volt regulator is connected to a full-wave rectified power supply with a 500 μF smoothing capacitor. The load has resistance of 100 Ω and the diodes have a forward threshold of 0.7 V. 12V rms LOAD – 100 ohms (a) What is the maximum voltage supplied to the regulator? (b) With a ripple voltage of 3.1 V, what is the power loss in the voltage regulator? (c) What is the power loss in the load? Solution (a) V peak = Vrm × 2 = 16.97 less 2 x Vdiode = 15.6 V (b) The minimum voltage is the maximum voltage less the ripple voltage. i.e. 15.6 – 3.1 = 12.5 V The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 43 To find the power loss we need to determine the average potential drop across the regulator, as well as the load current. Vav = Vmax + Vmin 15.6 + 12.5 = = 14.05 V (a close approximation) 2 2 Vdrop = 14.05 – 9.0 = 5.05 V I LOAD = V LOAD 9 = = 0.09 A R LOAD 100 PLOSS = VDROP x ILOAD = 5.05 x 0.09 = 0.45 W Note that the value ‘V’ used in Ohm’s Law and in power formulae does not represent voltage, rather potential difference. It is the drop in electrical potential across the element considered. (c) PLOAD = VLOAD x ILOAD = 9.0 x 0.09 = 0.81 W The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 44 The following information applies to Questions 44 to 51. The RMS output of the transformer is 12 V AC 50 Hz. The diodes are typical silicon power diodes with a forward threshold of 0.7 V. The load has 200 Ω, and C = 1000 μF. QUESTION 44 What is the peak voltage at point X? A 10.6 V B 16.3 V C 24.0 V D 15.6 V QUESTION 45 With S1 closed and S2 open, what is the value of Vripple at X? A 12.0 V B 0.0 V C 15.6 V D 10.6 V QUESTION 46 With S1 open and S2 closed, what is Vripple? A 15.6 V B 12.0 V C 0.80 V D 0.0 V The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 45 With both S1 and S2 closed there is a ripple voltage of about 0.8 V. QUESTION 47 If a second 1000 μF capacitor is now arranged in parallel to the original capacitor, what is the closest value of the ripple voltage? A 1.6 V B 0.4 V C 0.6 V D 0.9 V QUESTION 48 A second identical load is added in parallel to the original. What happens to the ripple voltage? A 1.6 V B 0.4 V C 0.6 V D 0.9 V QUESTION 49 Explain why is this circuit classified as unregulated? It is decided to add a 7812 voltage regulator (12 V) to the circuit. The ripple voltage received at the input pin to the regulator is 0.8 V and the regulator has a drop out voltage of 2.4 volts. QUESTION 50 With a single 200 Ω load across the voltage regulator, what power loss is dissipated within the regulator? A 0.72 W B 0.14 W C 0.89 W D 0.19 W The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 46 QUESTION 51 With another load applied to the circuit it was found that a 2.5 V ripple was observed at the input to the regulator. What does minimum voltage value at the input pin drop down to? A 15.5 V B 13.1 V C 14.4 V D 12.0 V The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 47 HEAT DISSIPATION Under certain circumstances, voltage regulators can get so hot that they can lead to circuit failure. Voltage regulators are often mounted on heat sinks that help to disperse the heat produced. If for instance the supply averaged 3 V above the output for a 1 amp current, that would equate to 3 W of heat being dissipated within the regulator (P = V x I). This is a considerable amount that would cause it to heat up considerably. Heat can move from one place to another by three processes: Conduction, radiation and convection. Features of heat sinks that facilitate heat dissipation include: • They are usually made from aluminium because this material is an excellent conductor of heat. Copper is sometimes used. • The component that needs to have thermal energy removed from it must make good contact with the heat sink. Thermal grease smeared between the surfaces facilitates heat transfer. • Heat sinks may be coloured black because this enables them to better radiate heat energy. • Heat sinks are designed so that they have a large surface area in contact with the air. This allows the thermal energy to be better conducted from the sink to the air. The hot air expands, becomes less dense and rises, carrying the energy away with it. This process is known as convection. Electronic circuits that use heat sinks usually need to have ventilation holes in the appliance casing so that cool air can enter the casing, come into contact with the heat sink, heat up and carry heat energy away through the ventilation holes. Cooling fans can also be added in cases of extreme heat accumulation. If voltage regulators do overheat, they are designed to automatically shutdown if their operating temperature exceeds a certain value. As they cool down they switch back on again. A drop in their output voltage may be experienced prior to thermal shutdown, as well as a reduction in the amount of current that they can conduct. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 48 SOLUTIONS FOR ERRORS AND UPDATES, PLEASE VISIT WWW.TSFX.COM.AU/VCE-UPDATES QUESTION 1 Answer is B 12.7 V QUESTION 2 Answer is D 70.7 V QUESTION 3 Answer is A 0.24 A QUESTION 4 Answer is B 0.34 A QUESTION 5 Answer is C 5.76 W QUESTION 6 Answer is D 0.354 A QUESTION 7 Answer is A 141 V QUESTION 8 Answer is C 30 ms QUESTION 9 Answer is B 25 coils QUESTION 10 Answer is D 14 V The reading on the voltmeter will be less than 50 V because the number of turns of wire is less. An approximate calculation is: Vs = Vp × Ns 3 = 50 × = 14 V (approximately). Np 11 QUESTION 11 Answer is D 5.0 A QUESTION 12 Answer is A 35 V QUESTION 13 Answer is D 12.4 V QUESTION 14 Answer is C 50 Hz QUESTION 15 Answer is A and D QUESTION 16 Answer is B QUESTION 17 Answer is C 15mF One time constant is 3.0 s = RC; C = 0.015 F = 15 mF QUESTION 18 Answer is D 0.009 A At 5.0 s VCAP = 8.2 V, VRESISTOR = 1.8 V; Current = 1.8 / 200 = 0.009 A QUESTION 19 Answer is B 3.7 A 30 seconds = one discharge time constant. Therefore it has discharged 63%, giving remaining voltage at 3.7 V. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 1 QUESTION 20 Answer is C 0.05 s or 50 ms QUESTION 21 Answer is A One second. QUESTION 22 Answer is C QUESTION 23 Answer is D QUESTION 24 Answer is D QUESTION 25 Answer is A QUESTION 26 Answer is B QUESTION 27 Answer is D QUESTION 28 Answer is C QUESTION 29 Answer is B QUESTION 30 Circuit A has a frequency of 50 Hz, whereas Circuit B has a frequency of 100 Hz. Therefore, the period of Circuit A (20 ms) is double that of Circuit B (10 ms). Since VRIPPLE ∝ Period , Circuit A has approximately double the ripple of Circuit B. QUESTION 31 (a) Answer is B Decrease (b) Answer is A Increase (c) Answer is B Decrease (d) Answer is C No change QUESTION 32 Answer is C 2000 µF QUESTION 33 Answer is B 6.0 V QUESTION 34 Answer is D 2.0 V QUESTION 35 Answer is C 8.5 mA QUESTION 36 Answer is D 15.0 Ω Answer is D 1.62 W R= V 3 = = 15 Ω I 0.2 QUESTION 37 P = V × I = 9 × 0.18 = 1.62 W The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 2 QUESTION 38 Answer is B 2.22 J Ptotal = VI = 12 x 0.2 = 2.4 W Power wasted is 2.4 – 1.62 = 0.78 W. Therefore 0.78 J is wasted each second. QUESTION 39 Answer is B 0.2 W QUESTION 40 Answer is C 8.5 V QUESTION 41 Answer is A QUESTION 42 R1 is a dropping resistor which takes the potential difference between the zener diode and the rectifier output. QUESTION 43 Answer is C Replace the capacitor with a 500 µF capacitor QUESTION 44 Answer is D 15.6 V 15.6 V (convert Vrms to Vpeak then subtract 2 x Vdiode) QUESTION 45 Answer is B 0.0 V Zero ripple because there is no load current, therefore the capacitor cannot discharge. QUESTION 46 Answer is A 15.6 V (as there is no smoothing.) QUESTION 47 Answer is B 0.4 V With an inverse relationship between ripple voltage and capacitance, the ripple will approximately halve. QUESTION 48 Answer is A 1.6 V The second load doubles the current (or halves the total resistance depending on how you want to look at it). This doubles the ripple voltage to about 1.6 V. QUESTION 49 It is classed as unregulated because the average voltage supplied to the load will decrease as the current through the load increases. With a high resistive load: Vaverage ≈ V peak = 15.6 V . While the peak voltage will be maintained, the minimum voltage decreases as the current increases due to increasing ripple. Consequently the mean voltage falls as the current increases. Also, if the primary supply voltage changes, so too will the peak voltage, and hence the mean voltage to the load. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 3 QUESTION 50 V AVERAGE = Answer is D 0.19 W VMAX + V MIN 15.6 + 14.8 = = 15.2 2 2 V regulator = 15.2 − 12.0 = 3.2 V I LOAD = VLOAD 12 = = 0.06 A RLOAD 200 PLOSS = Vregulator × I LOAD = 3.2 × 0.06 = 0.19 W QUESTION 51 Answer is B 13.1 V There will be a small voltage fluctuation from the regulator. With a dropout voltage of 2.4 V, the output of the regulator will fall if its input drops below 14.4 V. With a ripple of 2.5 V, the minimum voltage at the input pin of the regulator drops to 13.1. This is below the level needed to maintain constant output. The School For Excellence 2014 The Essentials – Physics – Further Electronics Page 4