3 Area of Study 2 m pl e pa ge s Electronics and photonics Sa Unit outcome On completion of this area of study, you should be able to investigate, describe, compare and explain the operation of electronic and photonic devices, and analyse their use in domestic and industrial systems. chapter 4 Electronics E Sa m pl e pa ge s lectronics is one of the main drivers of modern technology. In a very real sense, many of the technological advances of the 20th century were made possible because of rapid advancements in electronics. Continual miniaturisation of electronic components (especially complex transistor circuits) has meant that the cost of many electronic devices has decreased while their functionality has dramatically increased. Computers are a perfect example—they are very complex electronic devices. New computers get more and more powerful and include more and more features, while staying the same price (in relative terms) or becoming cheaper. Continual miniaturisation of electronic components has made this possible. Many of our common household appliances, such as fridges, washing machines and air conditioners, are controlled by microcontrollers—small but powerful single-chip computers. We shall revise some key concepts in electronics that you will probably be familiar with, and introduce some new concepts and devices. We shall then use these concepts and devices to analyse and interpret the operation of some simple electronic circuits. by the end of this chapter you will have covered material from the study of electronics, including: • concepts of current, resistance, potential difference (voltage drop) and power as applied to the operation of simple DC electronic circuits • the operation of diodes, resistors and thermistors • calculate the effective resistance of circuits • Ohm’s law, V = IR • power, P = IV. 4.1 Analysing electronic circuits Electronics review Interactive Physics file An ideal battery maintains a constant EMF regardless of how much current is drawn from it. In practice, the potential difference across the terminals of a real battery decreases as the current drawn from the battery increases. This becomes significant when relatively large currents are drawn from the battery. The phenomenon is attributed to the contact resistance, ionic resistance of the chemical cell and the limitation of the rate at which charge can be supplied by the chemical reaction inside the battery. In electronic circuits, we often combine all these limitations (of a real battery) and model them as one internal resistance in series with an ideal battery, as shown in Figure 4.2. pa ge s In Year 11 you studied how the chemical reactions inside a battery separate positive and negative electric charges, converting energy released by chemical reactions to electrical potential energy between the battery terminals. This potential energy can be converted to other useful forms of energy (heat, light, sound etc.) when an electronic circuit is connected across the terminals. The electronic circuit provides a conducting path between the oppositely charged battery terminals, and mobile charge can therefore move from one terminal to the other under the influence of the electric force (giving up its electrical potential energy in the process). It is important to realise that the electrical potential energy is the only thing that decreases as the charge carriers move around the circuit—the number of carriers and the charge on each carrier always remain constant (i.e. charge carriers are not used up). Remember that instead of referring to the total electrical potential energy between two points in a circuit, in electronics it is more convenient to talk about the electric potential (potential energy per unit charge). This is often simply called potential difference or voltage. For a battery, the EMF is the energy supplied per unit charge by the chemical reactions responsible for the charge separation. The SI unit for both electric potential and EMF is the volt (V). It is convenient to choose a zero electric potential reference point when analysing electronic circuits. This reference point is often called the circuit ground or earth point. Any voltage that appears to refer to only one specific point in a circuit (rather than the difference between two points) can be assumed to be measured with respect to circuit ground. pl e – – + EMF of battery +– +– – – direction of conventional current – +– chemical reaction separates charge +– negative terminal +– – – – – I R resistance of circuit potential difference (V) ideal battery between the terminals producing an of the real battery EMF of E negative terminal of real battery Figure 4.1 A battery connected to a light light globe – – – current (I ) drawn from the real battery Figure 4.2 A battery with internal resistance. – I'm full of energy and ready to share it with the world. – electrical potential energy converted to light and heat here V = IR voltage drop across resistance of globe V E – battery – – – – – – – – – – – – – + Rint – Sa E – I + + + + + + + + + + + +– – m positive terminal positive terminal of real battery internal resistance of real battery My energy is exhausted. I'm heading off home. – flow of electrons The charge carriers that transport the electrical energy throughout the circuit are usually electrons, although in certain circumstances they can be positive ions or a combination of both. For this reason, when physically globe. The large arrow shows the direction of ‘conventional’ current; the movement of electrons is in the opposite direction. Electrons have a lot of potential energy as they leave the negative battery terminal but little as they return to the positive terminal. Energy is converted to light and heat in the light globe. The current leaving the battery equals the current returning to the battery. Prac 17 Chapter 4 Electronics 117 or Wires crossed at a junction Variable resistor or Fixed resistor or Unspecified circuit element Diode or Earth or ground Battery or DC supply or Capacitor or Time-varying or AC supply or Ammeter A Voltmeter V Ohmmeter Ω + v( t) or pl e Transistor (npn) Lamp/bulb (closed) or Figure 4.3 Summary of circuit symbols used in Chapter 4. Current (I) Potential difference (V) Figure 4.4 Current versus potential difference for an ohmic resistor. 118 i Sa (open) m Thermistor or temperaturedependent resistor Switch explaining how electronic devices work, it is often convenient to talk about the flow of the actual positive or negative charge around a conducting path. In electronic circuits, we talk about conventional electric current (I), which is defined in terms of the transfer of positive charge with respect to time. Hence: Q I= t The direction of conventional current around a conducting path is therefore defined to be the same as the direction of positive charge flow around the circuit. If the charge carriers happen to be positive, then the directions of mobile charge flow and conventional current are the same. If the charge carriers are negative (i.e. electrons), then the direction of electron flow is opposite to that of conventional current. In simple electronic circuits, the direction of conventional current goes from the positive to negative battery terminals, although the actual flow of electrons is in the opposite direction (see Figure 4.1). The circuit symbols for the electronic devices introduced in this chapter are summarised in Figure 4.3. The energy stored or dissipated in any electronic device is equal to the potential difference across it multiplied by the charge flowing through it (E = VQ). Hence, it follows that the power dissipated in any device (power is the rate of change of energy with respect to time) is equal to the potential difference multiplied by the current: Q E P = t = V( t ) = VI The conductors and resistors you studied in Year 11 were mainly ohmic resistors; that is, they behaved according to Ohm’s law. The graph of current versus voltage for any ohmic device is shown in Figure 4.4. The resistance of a circuit element determines the amount of current able to flow when a given potential difference is applied. An ohmic device has a constant resistance, regardless of the applied voltage. Most resistors and metal conductors at a constant temperature are ohmic devices. pa ge s Wires crossed not joined Electronics and photonics Ohm’s law states that the current flowing in a conductor is directly proportional to the potential difference across it, i.e. I ∝ V. The constant of proportionality is called the resistance: V = IR Using Ohm’s law together with the expression of power, we can show that: P= VI = I 2R V2 = R In Year 11 you also learned that the equivalent resistance of any number of resistors in series (see Figure 4.5a) is given by: Req = R1 + R2 + R3 + ... + Rn and in a series circuit: E or Vsupply = V1 + V2 + V3 + ... and Itotal = I1 = I2 = I3 = ... (a) Also recall that the equivalent resistance of any number of resistors in parallel (see Figure 4.5b) is given by: 1 1 1 1 1 Req = R1 + R2 + R3 + … + Rn and in a parallel circuit: E(a)or Vsupply = V1 = V2 = V3 = ... and Itotal = I1 + I2 + I3 + ... I I R1 (b) same current through all resistors connected in series R2 R3 ............. Rn (b) same current through all resistors connected in series Consider the circuit shown in the diagram. aIf E = 12 V, R 1 = 10 kΩ, R 2 = 5 kΩ and R1 R R2. 2 R3 = 8 kΩ, calculate the voltage across bIf E = 10 V, V1 = 3 V, R2 = 10 kΩ and R3 = 4 kΩ, calculate the resistance of R1. R1 R3 ............. Rn R1 V1 R2 V2 R3 V3 R2 R3 ........... R n V I E Solution aThe total resistance of the series circuit is: same voltage across all resistors connected in parallel Figure 4.5 (a) Resistors in series; (b) resistors in parallel. Sa m pl e Rt= R1 + R2 + R3 = 10 + 5 + 8 = 23 kΩ and so the current flowing around the conduction path is: I = E Rt = 12 3 23 × 10 = 5.2 × 10–4 = 0.52 mA and hence the voltage across R2 is: V2= IR2 = (0.52 × 10–3)(5 × 103) = 2.6 V bThe total resistance of R2 in series with R3 is: Rt= R2 + R3 = 10 + 4 = 14 kΩ The total voltage across the two resistors is: Vt= E – V1 = 10 – 3 = 7 V Hence the current flowing around the conduction path is: V I= t Rt 7 = 14 × 103 = 5.0 × 10–4 = 0.5 mA and the resistance of R1 is: V R1= 1 I 3 = 0.5 × 10–3 = 6 kΩ pa ge s Worked example 4.1A (a) Physics file The total current flowing from the battery terminals and into the circuit is called the line current. The current flowing in any particular branch of the circuit is called a branch current. Chapter 4 Electronics 119 Worked example 4.1B Consider the circuit shown in the diagram. I I1 E R1 V1 I2 R2 V2 I3 R3 V3 I aIf E = 15 V, R1 = 10 kΩ, R2 = 5 kΩ and R3 = 8 kΩ, calculate the total current flowing through the circuit. bIf E = 10 V, I = 10 mA, R2 = 10 kΩ and R3 = 4 kΩ, calculate the resistance of R1. pa ge s Solution aThe total resistance of the parallel circuit is: 1 = 1 + 1 + 1 Rt R 1 R 2 R 3 1 = + 1 + 1 10 × 103 5 × 103 8 × 103 and therefore Rt = 2.4 kΩ Hence the total current flowing around the conduction path is: I= E Rt 15 2.4 × 103 = 6.3 × 10–4 = 0.63 mA bThe total current through R2 and R3 is: I2 + I3= E + E R2 R3 m pl e = 10 + 10 10 × 103 4 × 103 = 1 × 10–3 + 2.5 × 10–3 = 3.5 × 10–3 = 3.5 mA Thus the current flowing through R 1 is: I1= I – (I2 + I3) = 10 – 3.5 = 6.5 mA Hence the resistance of R1 is: R 1= E I1 10 = 6.5 × 10–3 = 1.5 kΩ Sa = Worked example 4.1C Simplify the following resistor network (between points A and B) to one single equivalent resistor (Req), by using the idea of series and parallel circuit simplification. (Hint: Remember that there is no conduction path (and hence no current flow) between the points X and Y. This is often called an open circuit (o/c). As the current is zero, the resistance between the points X and Y is effectively infinite, i.e. RXY = ∞ Ω.) 120 Electronics and photonics 10 Ω A 20 Ω 20 Ω 10 Ω X A o/c B Req Y B Solution series 10 Ω A 20 Ω 20 Ω A ∞Ω 10 Ω B 10 Ω ∞Ω 10 Ω 20 Ω B parallel 10 Ω pa ge s A 10 Ω 20 Ω B series A 20 Ω 20 Ω pl e B parallel A eq m 10 Ω = B Sa Voltage dividers When we have a number of resistors in series, we can work out the potential difference across a particular individual resistor by using the voltage division principle. For example, look at the two-resistor series circuit shown in Figure 4.6. Clearly the potential difference (Vin) between points A and B exists across the two resistors R1 and R2. In this circuit, the same current (I) flows through both resistors. Hence, using Ohm’s law, we see that the potential difference across each resistor is directly proportional to its resistance, and that the potential difference across both resistors is proportional to the total resistance. If one of the resistors has a larger resistance than the other, there will be a greater fraction of Vin across the larger resistor. In mathematical terms, using Ohm’s law, we have: Vin = I(R1 + R2) and Vout = IR2 Eliminating I gives: Vout R2 Vin = R1 + R2 A R1 V in I R2 V out B Figure 4.6 Simple two-resistor voltage divider. Physics file The ratio of voltage output to voltage input of a circuit is called its voltage gain. The relationship between output and input voltages is studied again when we look at amplifiers. Chapter 4 Electronics 121 As we are often interested in the output voltage of a voltage-dividing circuit, this relationship is usually written as: i Vout = R2 ×V R1 + R2 in where Vout = output potential (V) Vin = input potential (V) R2 = resistance across which the output potential is measured (Ω) Note: This rule must be adjusted if Vout is taken from R1 rather than R2. The voltage division principle can be used for series circuits with more than two resistors, and in general we have: R VX = R + R +X… + R × Vin 1 2 n where VX is the voltage across resistor RX. pa ge s Worked example 4.1D Find the voltage across R2 in the three-resistor circuit shown. Assume R 1 = 2 kΩ, R 2 = 3 kΩ, R 3 = 5 kΩ and Vin = 20 V. R1 V in Solution R2 V out Vout= R2 V R1 + R2 + R3 in 3 × 103 × 20 (2 + 3 + 5) × 103 =6V = R3 pl e m Note that R2 has 30% of the total resistance of the series circuit and therefore quite logically uses 30% of the supply voltage! Using this approach can allow you to calculate Vout values quite quickly in voltage dividing circuits. Mathematically, this idea can be expressed as: V2 R2 Vin = Rt for series circuits. Sa Prac 18 Determine the effective resistance for the circuit shown. Then find the total current drawn by the circuit. X 30 Ω 18 Ω 3Ω Va 12 V 12 Ω Vb Ib Y 122 Solution Z Ia Worked example 4.1E Electronics and photonics 12 Ω Any simple resistor network can be replaced with one single resistor that has the same overall characteristics as the original circuit, i.e. one resistor that draws exactly the same current from the battery as the original network. Redraw the circuit in a linear form, so that the voltage across it drops from top to bottom. This can help visualise what is going on. Now use the series and parallel resistor rules to simplify the network. Eventually you are left with one single resistor that is equivalent to the whole circuit. 30 Ω 30 Ω 3Ω 18 Ω 30 Ω 3Ω 18 Ω series 12 V 18 Ω 12 Ω 12 Ω 9Ω 6Ω parallel para llel series 30 Ω pa ge s 36 Ω 6Ω So the entire network is equivalent to one 36 Ω resistor (i.e. this single equivalent resistor behaves exactly like the original network with respect to the current drawn from the power supply). Using Ohm’s law, we see that the current flowing from the battery and into the V network (or its equivalent resistance) is It = supply = 12 = 0.33 A. Rtotal 36 I = 0.33 A 36 Ω 18 Ω I = 0.33 A 12 V Sa m 12 V Physics file X pl e 30 Ω 12 Ω 3Ω Z 12 Ω Y Thermistors and voltage dividing Voltage dividing circuits allow us to manipulate the size of a voltage in response to a given input. The principles of voltage dividing will be very important later in this chapter when we examine amplifier circuits and in the next chapter when we look at electrical-optical circuits. Voltage dividing circuits can allow us to detect and control a wide range of physical phenomena. Can you think of any electronic devices in your home that respond to light levels, air temperature, pressure or movement? It is very likely that they use the principle of voltage division to create an output voltage in response to some type of input. A simple circuit for an electronic temperature sensor is shown in Figure 4.7. It contains a thermistor, which is a device whose resistance decreases sharply as its temperature rises. Its symbol is shown in Figure 4.3, page 118. The thermistor is placed in series with a selected resistor and a power supply. The potential difference of the supply is shared and the output for the circuit (Vout) is recorded across R2. When the temperature is low, the thermistor R1 will have a high resistance and therefore use a larger share of the supply Superconducting materials have negligible resistance to the flow of electrons. First discovered in 1911, they have been in commercial use since 1995; but since they only possess this property at temperatures close to absolute zero, their use is limited. Currently, the highest temperature at which a material can be superconductive is 55 kelvin— this is –218°C! R1 Vin R2 Vout Figure 4.7 This circuit contains a thermistor. Chapter 4 Electronics 123 voltage. This means that Vout will be low. Alternatively, if the temperature rises, the thermistor will have lower resistance and therefore use a smaller share of the supply voltage. Thus, Vout increases. Worked example 4.1F Resistance (kΩ) A potential divider circuit includes a thermistor and is used as a temperature sensor. The resistance–temperature characteristics of the thermistor are shown. Using this graph and the information included on the circuit diagram, determine the: a resistance of the thermistor at 30°C b current in the circuit c output potential difference, Vout d output potential if the temperature fails to 20°C. Vin 5V pa ge s R1 20 15 5 kΩ R2 Vout 10 pl e 0 20 30 40 Temperature (°C) m Solution aThe resistance of the thermistor can be found directly from the resistance–temperature Sa graph. At 30°C the thermistor would have a resistance of 15 kΩ. V b I = R +inR = 15 × 103 +5 15 × 103 = 20 ×5 103 = 2.5 × 10–4 A 1 2 R2 c Vout= R + R × Vin 1 2 5 kΩ × 5 = 1.25 V 15 kΩ + 5 kΩ d At 20°C the thermistor’s resistance will be 20 kΩ and 5 kΩ Vout = ×5=1V 20 kΩ + 5 kΩ = Using a multimeter for accurate measurements As well as calculating the various voltages and currents in a circuit, we also need to be able to measure these values accurately. In electronics we do this with measuring instruments such as voltmeters, ammeters and ohmmeters. These devices give a numeric readout of a particular voltage, current or resistance in a DC circuit. The numeric readout can also represent RMS values of sinusoidal voltages and currents in AC circuits. These various functions can be combined into one instrument called a multimeter. The act of measuring a voltage or current in an electronic circuit will inevitably cause some change in that circuit, since some electrical energy must be diverted from the circuit into the measuring instrument. Multimeters are 124 Electronics and photonics (a) (b) Physics file The potential divider relationship assumes equal current in both resistors, i.e. no current is drawn by a voltmeter measuring the output potential. In fact, voltmeters always draw a small current, making it difficult to draw accurate predictions of resistance values. This ‘loading’ effect can amount to significant errors and should be considered during any practical investigations. pa ge s Figure 4.8 A multimeter combines the function of a voltmeter and an ammeter. Both analogue (a) and digital (b) forms are available. Although multimeters have been replaced by computer-based systems in some applications, their ease of use and portability still make them useful for making quick measurements of instantaneous circuit conditions. Measuring voltage V (b) Sa m Figure 4.9 shows how to use a multimeter to measure the potential difference between two points in a circuit. • The multimeter should be switched to the appropriate range for the measurement. If this is unknown, always start on the least sensitive (largest) range. Note that some multimeters are auto-ranging. • The multimeter leads must be connected in parallel across the two points whose potential difference we wish to measure. In voltmeter mode, the multimeter has a very large resistance between its two terminals, so that it draws very little current from the circuit and hence minimises its effect on the voltage being measured. Measuring current Figure 4.10 shows how to use a multimeter to measure the current flowing past a particular point in a circuit. • If the multimeter is not auto-ranging, it should be switched to the appropriate range for the measurement. If this is unknown, always start on the least sensitive (largest) range. • The multimeter leads must be connected in series with the electronic component through which we wish to measure the current. In ammeter mode, the multimeter has a very small resistance between its two terminals so that it has very little effect on the current being measured. V1 1 pl e very cleverly designed so as to minimise this energy diversion, but we still need to understand how multimeters work so that we can use them correctly. Most multimeters draw a very small current from the circuit. This current is used to sense the quantity being measured and is then converted to some form of visual indicator, usually either a moving pointer (analogue display) or a digital screen (digital display) as shown in Figure 4.8. (a) Circuit elements +V 2 V2 V COM 3 V3 (c) 2.11 V A Ω V A V COM Figure 4.9 (a) Circuit symbol for a voltmeter. (b) Using a voltmeter to measure voltage between two points in a circuit. (c) A multimeter in voltmeter mode. Chapter 4 Electronics 125 (a) (b) (c) 1 1.30 mA +A A A A Ω COM A V I 3 V A 2 COM V B Figure 4.10 (a) Circuit symbol for an ammeter. (b) Using an ammeter to measure the current flowing through a point in a circuit. (c) A multimeter in ammeter mode. Measuring resistance Figure 4.11 shows how to use a multimeter to measure the resistance of a particular component in a circuit. • If the multimeter is not auto-ranging, it should be switched to the appropriate range for the measurement. If this is unknown, always start on the least sensitive (largest) range. • One end of the electronic component must be disconnected (and therefore isolated) from the rest of the circuit. The multimeter leads are then connected in parallel across this component. The meter has a battery that passes a very small current through the unknown resistive component. Ohm’s law tells us that, since the battery has a constant voltage, the current flowing through the resistive element will be inversely proportional to the unknown resistance. The meter senses the magnitude of this current and displays the corresponding resistance. pl e Many multimeters provide a choice of scales so they can be used to measure a large range of voltage, current or resistance. For example, in voltmeter mode, a multimeter might have ranges of 0–3, 0–10 and 0–30 V etc. Since multimeters are usually used to read unknown voltage, current or resistance, it is a useful precaution to always set meters on the highest available range first. Read the value and then check if it could be measured on a lower range. Using a lower range will enable the value to be measured with greater accuracy. pa ge s Physics file m (a) (c) 1 4.13 Ω A A 2 V Ω Ω 3 B V Figure 4.11 (a) Circuit symbol for an ohmmeter. (b) Using an ohmmeter to measure the resistance between two points that are disconnected from the circuit. (c) A multimeter in ohmmeter mode. 4.1 summary Analysing electronic circuits • The direction of conventional current is defined as the same direction that positive charge would flow in an electrical circuit. • The operation of electronic circuits can be analysed using the following relationships: Q I= t V = IR V2 P = VI = I2R = R 126 Electronics and photonics V A Sa Ω (b) • For a series circuit: Req = R1 + R2 + R3 + ... + Rn Vsupply = V1 + V2 + V3 + ... It = I1 = I2 = I3 = ... COM Analysing electronic circuits (continued) • The voltage division principle states: R VX = R + R +X… + R × Vin 1 2 n • Thermistors are devices whose resistance decreases sharply with increases in temperature. • For a parallel circuit: 1 1 1 1 1 R = R + R + R +…+ R eq 1 2 3 n Vsupply = V1 = V2 = V3 = ... It = I1 + I2 + I3 + ... 4.1 questions Analysing electronic circuits 1 A circuit consists of three identical lamps connected to a battery with a switch, as shown in the diagram. When the switch is closed, what happens to the: a intensity of lamp A? b intensity of lamp C? c current in the circuit? d potential difference across lamp B? e potential difference across lamp C? f total power dissipated in the circuit? I A B switch C R2 E m E V X 10 V 20 kΩ V Y A Voltmeter X displays the higher voltage drop. B Voltmeter Y displays the higher voltage drop. C Both voltmeters display the same voltage. 4 A voltmeter is used to measure the voltage (relative to ground) at various points around a circuit. The voltages at points X and Y are as shown in the diagrams. For circuits (a) and (b) determine the magnitude and direction (P or Q) of the current flowing through the 5 kΩ resistor and the 2 kΩ resistor. (b) +5 V X (a) +5 V X X or Y R3 P Q R5 a Which resistor(s) has the highest current flowing through it? b Which resistor(s) has the lowest current flowing through it? c Which resistor(s) has the highest power dissipated through it? 3 a For the following circuits, which of the following statements is correct? Assume that the ammeters are identical. A X 10 V 10 V 10 kΩ 5 kΩ V R4 Sa R1 10 kΩ pl e 2In the following three circuits the batteries and resistors are all identical. Assume the batteries are ideal (i.e. no internal resistance). E 10 V pa ge s E b For the following circuits, which of the following statements is correct? Assume that the voltmeters and batteries are ideal and identical. 10 kΩ A Y A Ammeter X displays the highest current. B Ammeter Y displays the highest current. C Both ammeters display the same current. –3 V 5 kΩ P Q 2 kΩ Y 2 kΩ –3 V Y 5The following circuit is a simple voltage divider con sisting of two variable resistors of resistance R1 and R2. Copy and complete the table. R1 (Ω) R1 1000 20 V R2 R2 (Ω) Vout 10 1000 400 5.0 100 900 2.0 Vout (V) 2.0 3.0 6 Three variable resistors, R1, R2 and R3, are connected to a switch S. Determine the values of Vout in the table. Continued on next page Chapter 4 Electronics 127 Analysing electronic circuits (continued) R1 10 If the two globes are connected in series across the 12.0 V supply, which: a has the greater potential difference across it? b has the greater current through it? c glows more brightly? R2 20 V R3 S Vout 11 5Ω Vout (V) R1 (Ω) R2 (Ω) R3 (Ω) 200 200 100 open 300 100 25 open 50 50 75 open 100 100 1000 open 200 100 100 closed S 150 V 25 Ω a Which of the following combinations of resistors in the circuit shown are wired in series with each other? A30 Ω, 5 Ω B20 Ω, 5 Ω C20 Ω, 25 Ω D5 Ω, 10 Ω, 25 Ω E10 Ω, 15 Ω F5 Ω, 10 Ω, 25 Ω, 20 Ω Vout pl e m 800 600 400 200 Sa 1000 Resistance (Ω) pa ge s 7 A portable refrigerator is controlled by the thermistor circuit shown. The control circuit is supplied by the 12 V battery of a car and it is required to keep the refrigerator temperature below 10°C. An output voltage of 4.0 V is required to turn the cooling system on. Using the characteristic curve of the thermistor shown, determine the required maximum value of the variable resistor. 12 V 10 20 30 Temperature (°C) 10 Ω 15 Ω 20 Ω 30 Ω 40 The following information applies to questions 8–10. A manufacturer makes two different globes for a 12.0 V torch: one is rated at 0.50 W and the other at 1.00 W. 8What is the resistance of each globe when it is operating correctly? 9 If the two globes are connected in parallel across a 12.0 V supply, which: a has the greater potential difference across it? b has the greater current through it? c glows more brightly? b Which of the following combinations of resistors in the circuit are wired in parallel with each other? A30 Ω with 5 Ω B20 Ω with 15 Ω C30 Ω with 10 Ω D15 Ω with 10 Ω E(5 Ω, 10 Ω) with (15 Ω, 25 Ω) F(5 Ω, 10 Ω, 25 Ω) with 15 Ω c Which of the four resistors in the circuit shown has the least power dissipated through it? A20 Ω B15 Ω C25 Ω D5 Ω dWhich of the four resistors in the circuit shown has the most power dissipated through it? A20 Ω B30 Ω C25 Ω D5 Ω 1 2 Consider the circuit shown. 40 kΩ 10 V 10 kΩ 30 kΩ 30 kΩ 5 kΩ a Determine the total resistance connected across the battery terminals. b Determine the voltage across the 40 kΩ resistor. c Determine the current through the 5 kΩ resistor. d Determine the voltage across the 10 kΩ resistor. Worked Solutions 128 Electronics and photonics Chapter review Electronics 1In the following two circuits the batteries and resistors are identical. Assume the batteries are ideal (i.e. no internal resistance). E1 R1 a b c d power that can safely be dissipated in any one resistor is 25 W. 100 Ω E2 R4 R2 R5 R3 100 Ω A B Z Which resistor(s) has the highest current flowing through it? Which resistor(s) has the lowest current flowing through it? Which resistor(s) has the highest power dissipated through it? Which battery is supplying the largest current? The following information applies to questions 3 and 4. I V1 RA= R V2 RB= 3R 7The circuit below combines variable resistors R1 and R2 with fixed resistors to make a complex voltage divider. Complete the table by determining the output voltage for each row. pl e V a What is the maximum potential difference that can be applied between points A and B? b What is the maximum power that can be dissipated in this circuit? pa ge s 16 kΩ? 10 kΩ? 1 kΩ? 5.33 kΩ? The circuit shows two resistors connected in series across a battery with a terminal voltage of V. The resistance of RB is three times that of RA. Y 100 Ω X R6 2 You have four 4 kΩ resistors. How would you arrange the four resistors to give a total effective resistance of: a b c d 6Three 100 Ω resistors are connected as shown. The maximum 3 Show mathematically that one-quarter of the battery’s voltage is dropped across RA. R1 (Ω) R2 (Ω) Switch 1000 2000 open 2000 4000 open 4000 2000 open 8000 5000 closed 90 V Vout (V) R1 1000 Ω 4If V = 12 V and I = 200 mA, determine the: m a resistance values of RA and RB b power dissipated in RB. Sa 5 A thermistor is a semiconductor device whose resistance depends on the temperature. The graph shows the resistance versus temperature characteristic for a particular thermistor. Determine the temperature of the thermistor in the circuit if Vout is 1 V. T thermistor 6V R 1 kΩ Vout = 1 V 2000 Ω R2 Vout 8 Explain the meaning of the following statements about amplifier circuits. a At saturation the amplifier gain is not linear. bThe amplifier circuit must be correctly biased before receiving an AC input signal. 9 For the circuit shown, find the: a current in the 20 Ω resistor b potential difference between points X and Y. Resistance (kΩ) 20 15 10 X 5 5Ω 0 144 10 Ω 10 20 30 40 50 Temperature (°C) Electronics and photonics 10 Ω 5Ω 20 Ω 25 V Y It may be useful to draw the circuit in a linear form, so that the voltage across it drops from top to bottom, as shown below. 25 V 12 For the circuits shown in diagrams a and b, assume the diode has a switch-on voltage Vs = 0.7 V and a thermal leakage current It = 1 nA. Determine the current flowing through the: i100 Ω resistor ii500 Ω resistor iiidiode. 25 V a 100 Ω 3V 0V 500 Ω circuit drawn in linear form b 10Two electronic components, X and Y, are operating in the circuit shown. The voltage–current characteristics of each device are shown in the accompanying graph. The reading on the voltmeter is 60 V. pa ge s I (mA) 50 X 30 20 100 10 50 75 X Y X m V 100 V (V) pl e 25 A 60 40 20 0 0.2 0.4 0.6 0.8 1.0 Potential difference (V) Which device is non-ohmic? What is the resistance of X? What is the reading on the ammeter? Determine the EMF of the battery. Calculate the total power consumption in the circuit. A E Ω 11 A silicon diode has a switch-on voltage of ~0.7 V and a280 thermal leakage current of ~100 nA. Determine the6 Vvoltage across the diode (Vd) and the current (Id) through the diode for: Id 1.2 V Sa E Current (mA) 80 0 a b c d e 500 Ω 13 All the diodes in the circuit below are identical and have the voltage–current characteristics shown in the graph. Y 40 100 Ω 3V Vd The ammeter reading is 52 mA. a What is the reading on the voltmeter? b What is the EMF of the battery? a circuit 1 b circuit 2, where the polarity of the battery Circuit has been 1 reversed. 280 Ω 6V Id Vd Circuit 1 Id Circuit 2 280 Ω 6V 280 Ω 6V Vd Continued on next page Chapter 4 Electronics 145 Electronics (continued) The following information applies to questions 14–16. Vout (V) The amplifier circuit shown below has a voltage gain of –100. Assume it is correctly biased so that an input of 0 V results in an output of 0 V. The small varying input signal shown below is then fed into the amplifier. 5 4 C D 3 2 supply 1 –500 –300 –100 D C Vout Vin amplifier A B 6 100 –1 300 500 Vin (mV) –2 –3 –4 B A Vout(max) = 2 V –5 –6 a If the input voltage is as shown in the diagram, sketch the output voltage for amplifiers B and D. Vout(min) = –2 V Vin (mV) 200 Vin (mV) 5 100 pa ge s 10 –100 1 2 3 4 1 2 3 t (ms) 4 5 6 7 8 –200 Time (ms) –5 –10 b If the input voltage is as shown in the diagram, sketch the output voltage for amplifiers A and C. 400 14 Quote the peak-to-peak voltage of the input signal shown. Vin (mV) 15 Draw a graph of the resulting output voltage pl e m Vout (V) 0 1 –1.0 0 –10 1.0 3 4 5 6 7 8 t (ms) A cathode ray oscilloscope is used to monitor the output of a linear amplifier. An AC signal generator is used to produce a sinusoidal input voltage with a peak-to-peak voltage of 50.0 mV. oscilloscope 10 –2.0 2 The following information applies to questions 19 and 20. Sa 20 200 100 16The size of the input signal is now tripled but its frequency is unaltered. Draw the resulting output voltage graph. 17The input–output characteristics of an amplifier are shown in the graph. 300 1.0 V 2.0 Vin (V) –20 a What type of amplifier has these characteristics? b Calculate the voltage gain for that amplifier. c What is the maximum peak-to-peak voltage that this amplifier can amplify without distortion? 18The graph shows the characteristics of four different voltage amplifiers (A–D). The input voltage to these amplifiers is limited to between ±600 mV. AC signal generator voltage amplifier 0.5 ms 19a Is the amplifier operating within its limits? How can you determine this? b Analyse the gain of the amplifier. 20 a What is the frequency of the input signal? b What is the frequency of the output signal? Worked Solutions 146 Electronics and photonics Chapter Quiz

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