Download uncorrected page proofs

CHAPTER 17 How can AC electricity charge a DC
device?
Contents
Controlling voltage in circuits
Basic voltage dividers
Transducers
FS
Thermistors
Diodes
O
240 V AC to 6 V DC
PR
O
Introduction
The task
Risk assessment
G
E
AC voltage and current
Cathode ray oscilloscope
PA
How to use a cathode ray oscilloscope
Dual trace CROs
Using a multimeter
TE
D
Transformers
Converting AC to DC
Half-wave rectification
Capacitor smoothing
O
Voltage regulators
R
Ripple voltage
R
Full-wave rectification
EC
Charging and discharging capacitors
C
Internal resistance
N
Choosing a suitable voltage regulator
U
Voltage regulator using a Zener diode
Power dissipation
Photonics
Light-emitting diodes (LEDs)
Light-dependent resistors (LDRs)
Photodiodes
Phototransistors
Communications carriers
Twisted-pair copper cable
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Coaxial cable
Optical-fibre cable
Attenuation
Bandwidth
Modulation
Decoding signals
Demodulation
Chapter review
FS
Summary
PR
O
O
Questions
U
N
C
O
R
R
EC
TE
D
PA
G
E
Note to students and teachers: This PDF has been provided as an offline solution for times when you do not have
internet access or are experiencing connectivity issues. It is not intended to replace your eBook and its suite of
resources. While we have tried our best to replicate the online experience offline, this document may not meet
Jacaranda's high standards for published material. Please always refer to your eBook for the full and latest
version of this title.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
PA
A circuit diagram of a voltage regulated power supply system
G
E
PR
O
O
FS
CHAPTER 17 How can AC electricity charge a DC
device?
TE
D
REMEMBER
Before beginning this chapter, you should be able to:
apply the concepts of charge (Q), current (I), voltage (V), energy (U) and power (P) to electric circuits
■
analyse electric circuits using mathematical relationships and graphs
■
model resistance in series and parallel circuits using voltage-versus-current graphs
■
state and apply Ohm’s Law, V = IR, for ohmic devices at a constant temperature
■
find the effective resistance for resistors in series
■
find the effective resistance for resistors in parallel
R
R
O
C
N
recall and apply the principle of conservation of energy: Energy cannot be created nor destroyed, but it can be
transformed from one form to another
U
■
EC
■
■
distinguish between ohmic and non-ohmic resistors
■
calculate the effective resistance of devices connected in series and in parallel
■
calculate the voltage drop across, and current through, devices connected in series or in parallel
■
■
recall that a voltage divider consists of two or more resistors arranged in series to produce a smaller voltage
at its output
recall that a diode allows current to pass through it in only one direction
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
■
recall that a capacitor is a device that stores charge
KEY IDEAS
After completing this chapter, you should be able to:
■
apply the concepts of current, potential difference (voltage drop) and power to the operation of circuits that
use diodes and resistors
calculate the effective resistance of series and parallel circuits and unloaded voltage dividers
■
apply the rules V = IR and P = VI to electric circuits
■
specify AC voltages by peak, root mean square, and peak-to-peak values
■
describe the role of a transformer
■
describe how to use a cathode ray oscilloscope to measure voltage as a function of time
■
explain how diodes are used to convert alternating currents into direct currents
■
describe the differences between half-wave and full-wave rectifiers
■
explain what happens to the voltage and current of capacitors when they are charged and discharged
■
calculate the time constant for charging and discharging capacitors using the relationship τ = RC
■
describe how capacitors are used to smooth the outputs of DC power supplies
■
describe the characteristics and use of voltage regulators
■
apply the current-voltage characteristics of voltage regulators when designing a circuit
■
explain the effects that a multimeter has on measurements of voltage, resistance and current
■
design, construct and explain the operation of a low-voltage AC to DC regulated voltage power supply system
■
O
PR
O
G
E
PA
TE
D
EC
R
R
O
C
explain the use of diodes in half-wave and full-wave bridge rectification
N
■
analyse the role of the transformer in the power supply system, including the analysis of voltage ratio (not
including induction or its internal workings)
explain the effect of capacitors with reference to voltage drop and current change when charging and
discharging (time constant for charging and discharging, τ = RC) leading to smoothing for DC power supplies
U
■
FS
■
■
describe the use of voltage regulators, including Zener diodes and integrated circuits
■
analyse systems, including fault diagnosis, following selection and use of appropriate test equipment
■
interpret a display on an oscilloscope with reference to voltage as a function of time
■
apply the use of heat and light sensors such as thermistors and light-dependent resistors (LDRs) to trigger an
output device such as lighting or a motor
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
■
evaluate the use of circuits for particular purposes using technical specifications related to potential difference
(voltage drop), current, resistance, power, temperature and illumination
■
compare different light sources (bulbs, LEDs, lasers) for their suitability for data transfer
■
explain the use of optical fibres for short and long distance telecommunications
■
explain why thermistors are transducers and describe the relationship between temperature and resistance in
different types of thermistors
carry out calculations for circuits that respond to temperature using thermistors
■
describe how photonic transducers transform energy from electrical to optical, and from optical to electrical.
FS
■
PR
O
O
Controlling voltage in circuits
It is often necessary to control the voltage at various points in the electronic circuits that make up electronic
systems.
G
E
The electrical potential, or voltage, is defined as the amount of electrical potential energy at a point in a circuit per
unit charge. The potential difference, or voltage drop, across a device is the difference in potential on either side
of the device.
PA
The SI unit for voltage is the volt (V). A voltage of 1 volt at a point in a circuit means that every coulomb of charge
passing that point has the ability to transfer 1 joule of energy to the rest of the circuit.
TE
D
The electrical potential energy is provided to a circuit by a source of electromotive force (emf). Emf is also
measured in volts. An emf of 1 volt means that the source of emf provides the circuit with 1 joule of energy for
every coulomb of charge passing through the source. Examples of sources of emf include cells, batteries,
generators, photovoltaic cells and thermocouples.
U
N
C
O
R
R
EC
The source of voltage is sometimes represented in circuit diagrams by two horizontal lines known as supply
rails. These indicate the voltage that the source supplies to the circuit. They are shown in figure (a). Figure (b)
shows the traditional way of showing a DC source. Supply rails are used in circuit diagrams when the actual
nature of the source is not important, other than being a DC supply of a particular value.
(a) Supply rail circuit and (b) equivalent traditional circuit
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
The voltage drop, or potential difference, across a device in a circuit is a measure of the amount of energy
transformed in that device by the source of emf for every coulomb of charge that passes through it. If a resistor
has a voltage drop of 4.5 V across it, 4.5 J of energy are transformed by the source of emf for every coulomb of
charge passing through the resistor.
The current flowing in the circuit is the means by which energy is transferred from the source of emf to the
devices in the circuit.
Basic voltage dividers
TE
D
PA
G
E
PR
O
O
FS
When two resistors are placed in series with a supply, the supply voltage is shared, or divided, between the
resistors. The way that the voltage is divided depends on the resistance values of the resistors. Such an
arrangement of resistors, as shown in the figure at below, is known as a voltage divider and is used when it is
necessary to provide a useful voltage to a part of the circuit that is smaller than the supply voltage.
The basic voltage divider
V in
R1  R 2
R
I 
EC
The effective resistance of the voltage divider is R1 + R2. The current flowing through each resistor is:
O
R 2 V in
R1  R 2
C
V out 
R
Therefore, the value of Vout is the current times the resistance:
U
N
When using voltage dividers, it is assumed that the resistance of the part of the
circuit being provided with the output voltage (Vout in the figure) will be so great
that it will not affect the current flowing through R2 and the calculated value of
Vout. This is because when a resistance is placed in parallel with another resistor,
the effective resistance is reduced.
eModelling:
Voltage divider
Spreadsheets model the
circuits of voltage dividers.
doc-0036
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 17.1
Variable resistors are often used in voltage dividers to give an adjustable output voltage. Reducing the size of one
of the resistances increases the current flowing through the divider and therefore increases the voltage drop
across the other resistor.
Calculate the output voltage in the voltage divider circuit shown below left.
Solution:
R 2V in
R1  R 2
FS
V out 
3.3 k   5.0V
2.2 k   3.3 k 
 3.0 V
PR
O
O

G
E
Note that it was not necessary to convert kΩ to Ω in this problem, as the
multipliers cancel. If the resistances are given in different units, they must be
converted to the same units before calculations are attempted.
PA
SAMPLE PROBLEM 17.2
What would the resistance of the variable resistor have to be to give an output voltage of 4.0 V in the circuit
shown?
 R1 
R 2V in
R1  R 2
R 2V in
 R2
V out
2.2 k   5.0 V
 2.2 k 
4.0 V
 0.55 k 
So R1  0.55 k  or 550 
U
N
C
O
R
R

EC
V out 
TE
D
Solution:
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
REVISION QUESTION 17.1
Calculate the value of the variable resistor in sample problem 17.3 that would give an output voltage of
2.0 V.
Transducers
FS
Transducers are devices that transform energy from one form to another. They are used in electronic
communications, instrumentation and control systems. Transducers are the interface between the environment
and the electronic system. There are two types of transducer: input and output.
PR
O
O
In electronic systems, a transducer that changes non-electrical energy into electrical energy is called an input
transducer. For example, photocells or solar cells convert light energy into electricity. They are used for automatic
exposure control in cameras. Microphones transform sound energy into an electrical signal. Another example of
an input transducer is a thermocouple, which consists of two different metals such as copper and iron that are
joined together. They are used in electronic thermometers because the size of the voltage they produce is related
to the temperature. Input transducers are used in sensing devices in household and industrial systems.
G
E
Thermistors
PA
A thermistor is an input transducer made from a mixture of semiconductors.
The resistance of a thermistor varies with temperature. Thermistors are
considered to be input transducers because they convert thermal energy into a
voltage when they are used in potential dividers.
EC
TE
D
Thermistors are used in fire alarms, temperature circuits in car engines,
electronic thermometers and thermostats for controlling the temperature in
household and industrial equipment. They are also included in protection
circuits used to protect equipment from sudden current and voltage surges.
You can investigate the way in which the resistance of a thermistor varies with
temperature by doing investigation 17.1.
Digital doc
Investigation 17.1:
Thermistors
The aim of this investigation
is to construct a resistanceversus-temperature
characteristic for a thermistor.
U
N
C
O
R
R
Figure (a) below shows the circuit symbol for a thermistor. Figures (b) and (c)
are examples of graphs showing how the resistance of different thermistors
varies with temperature. Note that in figure (c) the graph shows the
characteristic curves for three separate thermistors and the vertical scale is logarithmic. Each line represents the
next whole number multiple of the previously stated power of ten. For example, on the vertical scale the lines
above that labelled 10 are 20, 30, 40 and so on. Similarly, the lines above that labelled 100 are 200, 300, 400 and
so on.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
(a) Circuit symbol for a thermistor; (b) and (c) resistance-versus-temperature characteristics for different
thermistors
U
N
C
O
R
R
EC
TE
D
Thermistors are produced in two forms. The most common is the negative temperature coefficient thermistor,
which has a resistance that decreases as the temperature increases. This type has a resistance-versustemperature graph like that shown in figures (b) and (c). Thus an increase in temperature leads to an increase in
the current in a temperature-sensing circuit.
The two common forms of thermistor
The other type of thermistor is the positive temperature coefficient thermistor, which has a resistance that
increases as the temperature increases. For positive coefficient thermistors, an increase in the temperature will
lead to a decrease in the current in a temperature-sensing circuit.
Thermistors are usually connected to variable resistors in voltage dividers.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 17.3
A thermistor has the temperature-resistance characteristic shown by the bottom curve in figure (c) on page xx. It
is placed in the voltage divider shown below.
a. Is this a positive or negative coefficient thermistor?
b. What is the resistance of the thermistor when the temperature
is 150 °C?
c. What is the value of the variable resistor if the temperature is
200 °C and Vout is 6.0 V?
FS
Solution:
b. From the graph, when the temperature is 150 °C, the
resistance is 20 Ω.
V out 
R 2V in
R1  R 2
6V 
10   9 V
R1  10 
G
E
From the graph, when the temperature is 200 °C, the
resistance is 10 Ω. Vin = 9.0 V, Vout = 6.0 V, R2 = 10 Ω, R1 = ?
PA
c.
PR
O
O
a. This is a negative coefficient thermistor bacause the
resistance decreases as the temperature increases.
6 V  R1  60 V  90 V
TE
D
6 V  R1  30 V
EC
R1  5 
R
REVISION QUESTION 17.2
O
R
The thermistor in sample problem 17.3 is replaced with one that has the temperature–resistance
characteristic shown by the top resistance-versus-temperature curve in figure (c) in the previous log
scale graph.
U
N
C
a. What is the resistance of the thermistor when the temperature is 200 °C?
b. Calculate the value of the variable resistor in the voltage divider if the temperature is 100 °C and the
output voltage is 4.5 V.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
PHYSICS IN FOCUS
Thermistors in fire alarms and thermostats
Fire alarm sensors are used in areas where smoke detectors are not suitable, for instance in smoky or dusty
environments. They use a sensing circuit that employs a thermistor that is in contact with the air and is placed on
the ceiling of the room. The circuit gives a voltage output that is related to the temperature of the air because the
resistance of the thermistor varies with temperature. When the air temperature reaches a predetermined value, a
control system is activated. This may sound alarms, turn on sprinklers and cause a light-emitting diode to flash.
FS
The thermistor is used as an input transducer that transforms thermal energy into electrical energy.
O
A thermostat is a device that controls the heating or cooling of an object in order to keep it at a constant
temperature. It consists of a temperature-sensing device connected to a switching device.
PR
O
As in fire alarm systems, thermistors are used in thermostats as transducers to convert thermal energy into
electrical energy. They are used in voltage dividers so that when a predetermined temperature is reached, the
voltage output from the divider causes a switch to be triggered. This activates a heating or cooling device.
O
R
R
EC
TE
D
PA
G
E
Thermostats are used in ovens, hot-water systems, space-heating systems and refrigerators.
U
N
C
A fire alarm sensor
Diodes
A diode is a device that allows current to flow through it in only one direction. Diodes act like a valve in a car tyre.
They effectively provide little resistance to a current flowing in one direction, but a very large resistance to a
current flowing in the other direction. The circuit symbol for a diode is shown in figure (a) below, with different
types of diode commonly used in electronics shown in figure (b). Diodes are usually made from semiconductors,
the most common being silicon. The chip of a diode is made by fusing a p-type semiconductor to an A-type
semiconductor.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
(a) Circuit symbol for a diode and (b) different types of diode
TE
D
PA
A diode conducts current in the direction indicated by the arrow in the circuit diagram. The bar indicates that
current will not flow in the opposite direction. The ends of a diode are known as the cathode and the anode.
Current flows from the anode to the cathode. Diodes are marked with a band around the cathode. This
corresponds to the bar on the circuit symbol.
U
N
C
O
R
R
EC
The characteristic curve of a device is a graph showing how a particular device behaves electrically. The
characteristic curve for a typical silicon diode is shown at below.
The current–voltage characteristic of a typical silicon diode
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
A diode connected in series with a variable voltage supply and a resistor
PR
O
O
FS
Note that the scales on the graph are different on the positive and negative sides of both axes. 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 it has low resistance, it has to have a voltage of about 0.6–0.7 V across it.
G
E
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. If the negative voltage is large enough, the semiconductor will break down and
a large current will flow.
PA
Diodes are usually placed in series with other circuit elements such as resistors, as shown at below.
TE
D
In the circuit shown, when the voltage from the supply is increased from 0 V to 0.6 V, little current flows and the
voltage drop is almost entirely across the diode because it has a very large resistance. As the voltage of the
supply is further increased, the voltage drop across the diode stays at about 0.7 V, and the rest of the voltage
drop is across the resistor. The size of the current through the circuit is then largely determined by the resistor.
U
N
C
O
R
R
EC
Note that when a diode is reverse biased, as shown in the figure at below, very little current flows through the
diode and the resistor. The voltage drop across the resistor will be negligible and all the voltage drop will be
across the diode.
The voltage drop across a reverse-biased diode
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 17.4
A diode and resistor are connected in series with a power supply. If the emf of the supply is 10 V, and the value of
the resistor is 100 Ω, estimate the current flowing through the diode.
Solution:
The diode can be assumed to have a voltage drop of about 0.7 V across it. Therefore, the voltage drop across the
resistor is 9.3 V. The current through the resistor is found using Ohm’s Law.
V
R
9.3 V

100 
 0.093 A, or 93 mA
PR
O
O
FS
I 
As the diode and resistor are in series, the current through the diode is also 93 mA.
G
E
REVISION QUESTION 17.3
R
R
EC
TE
D
PA
The current N-voltage characteristic curve of a diode is shown below.
O
The diode is placed in the circuit shown at right.
U
N
C
Calculate the current flowing in the circuit. Express your answer in mA.
240 V AC to 6 V DC
Introduction
In this option you will investigate the processes involved in transforming the alternating current delivered by the
electrical supplier into low-voltage direct current for use with small-current electrical devices. You will investigate a
variety of circuits to explore processes including transformation, rectification, smoothing and regulation.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
You will then investigate how data is transferred from one point to another using optical fibre networks. This
involves transducer circuits that:
•
•
•
transform energy, such as heat and light, into electrical signals
transform electrical signals into light signals that can be transmitted along optical fibres
transform light signals back into electrical signals.
In this area of study, you will design, build and test a low-voltage AC to DC regulated voltage power supply
system that could be used to charge a device.
To help you achieve this goal, you will study theory relating to:
FS
using diagnostic equipment such as a cathode ray oscilloscope (CRO) and a multimeter
converting an AC voltage signal to a DC voltage signal (voltage rectification) using diodes
smoothing the output of a rectifier using a capacitor
selecting and using a voltage regulator that will give a constant voltage output (The voltage rectifier will do this
even though there are changes to the voltage provided by the rectifier and there are changes to the load.)
O
•
•
•
•
PR
O
The task
Design a system that gives a constant output voltage when the load draws different currents. For example:
Design, construct and test a regulated voltage supply to provide a steady 6.0 V output voltage. Test the output
of the regulator when it is connected to a variable load resistance.
G
E
•
PA
The initial power supply should be an unregulated AC power supply of no more than 12 V RMS (RMS — root
mean square — is the effective value of an AC voltage, corresponding to the DC voltage that gives the same
heating effect.). The regulated output voltage should be a steady 6.0 V DC.
The equipment
EC
an unregulated AC power supply
4 × power diodes (for example, 1N4004)
a range of capacitors
a range of resistors
a suitable solid-state voltage regulator (for example, a 7806, which gives a steady output of 6.0 V)
a CRO
a multimeter.
O
R
Risk assessment
R
•
•
•
•
•
•
•
TE
D
The following is a list of the recommended equipment for this detailed study:
N
C
The activities in this option involve the use of a low-voltage power source. Make sure that the leads from the
mains to the power supply are in good condition. Check the output voltage using an AC voltmeter. Never use a
multimeter to measure mains voltages.
U
If using a soldering iron, take care to prevent burns to hands and clothing. Wear shoes that completely cover your
feet, and also wear safety glasses to prevent splashes of molten solder from entering your eyes. Voltage
regulators become very hot because they operate by converting electrical energy into thermal energy. Do not
touch one while it is operating and for a short time after it has been switched off.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
AC voltage and current
PA
The variation of voltage with time for a sinusoidal AC signal
An AC signal is described in terms of:
TE
D
•
VRMS 
EC
•
1

frequency, f  f  

T
amplitude. If the voltage cycle is centred on 0 V, the voltage amplitude is Vpeak. Similarly, if the current cycle is
centred on 0 A, the current amplitude is Ipeak.
VRMS. This is the value of the DC voltage that generates the same power as the AC voltage when placed
across a resistor.
V peak
R
•
period, T
2
R
•
G
E
PR
O
O
FS
An AC current is one that periodically changes direction. The figure below shows a graph of how voltage varies
with time for an AC voltage signal. Note that the voltage is sinusoidal, which means that the graph has the same
form as a sine curve.
•
I peak
N
2
U
I RMS 
C
O
Similarly, IRMS is the value of the DC current that generates the same power when it flows through a resistor.
The maximum variations in the voltage and current are called the peak-to-peak voltage, Vp-p, and the peak-topeak current, Ip-p, respectively.
The figure below shows how the current through a resistor varies with time when an AC voltage is placed across
it.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
The variation of current with time for a sinusoidal AC signal
G
E
Cathode ray oscilloscope
PA
A cathode ray oscilloscope (CRO) is an instrument that shows how voltage drop across a device varies with time.
A cathode ray consists of a beam of electrons that strike a phosphorescent screen at the front of the instrument,
creating a spot. The beam can be made to sweep across the screen at different speeds and the voltage across
the device affects the vertical position of the beam.
U
N
C
O
R
R
EC
TE
D
By adjusting the controls, a picture, called a trace, of how voltage varies with time is displayed. The trace is
identical to a graph where the y-axis gives the voltage and the x-axis gives time. The scales for the x- and y-axes
may be altered using the controls. This is explained in the next section.
A cathode ray tube
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
How to use a cathode ray oscilloscope
EC
TE
D
PA
G
E
PR
O
O
FS
Figure (a) below shows the basic controls of a single trace CRO and (b) how to connect it to a device, in this case
a signal generator that you may use when studying sound. The ground terminal of the CRO is held at 0 V. The
other terminal (marked ‘input’) samples the voltage of the signal generator output.
R
(a) The front panel of a single trace CRO and (b) connecting a signal source to a CRO
R
Before connecting the CRO to the device, turn it on and use the focus and intensity knobs to adjust the trace. The
horizontal and vertical deflection knobs are used to position the trace at the centre of the screen.
N
Time scale
C
O
The screen has a centimetre grid marked on it to assist in centring the trace and in interpreting the values that the
trace represents.
U
The timebase control is adjusted so that at least one complete waveform is displayed on the screen. This
enables the period of the signal to be read. The period is the time it takes for the signal to complete one cycle.
Measure the length of one complete cycle of the trace from the screen and multiply this value by the factor
indicated on the timebase control.
Voltage scale
The vertical scale of the trace is set by the vertical amplifier. This should be set so that the top and bottom of the
trace fall within the screen grid. The amplitude of the trace is then multiplied by the factor indicated by the vertical
amplifier control.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 17.5
What are the period, frequency and amplitude of the trace shown at right
if the following timebase and voltage settings are used?
a. Timebase 5 ms cm–1, voltage 1 V cm–1
b. Timebase 0.1 ms cm–1, voltage 5 mV cm–1
Solution:
FS
a. Each centimetre in the horizontal direction represents 5 ms. The
cycle is completed in 4 cm; therefore the period is:
O
T  4 cm × 5 ms cm 1
= 20 ms.
PR
O
Frequency is the inverse of period; therefore:
1
2.0  10 2 s
 50 Hz.
f 
G
E
Each centimetre in the vertical direction represents 1 V. The amplitude of the trace is 2 cm; therefore:
PA
A  2 cm  1 V cm 1
 2 V.
b. Each centimetre in the horizontal direction represents 0.1 ms. The cycle is completed in 4 cm; therefore
the period is:
TE
D
T  4 cm  0.1 ms cm 1
 0.4 ms.
Frequency is the inverse of period; therefore:
EC
1
4  10 4 s
 2.5  103 H z
R
f 
R
Each centimetre in the vertical direction represents 5 mV. The amplitude of the trace is 2 cm; therefore:
N
C
 10 V.
O
A  2 cm  5 V cm 1
U
REVISION QUESTION 17.4
Repeat sample problem 17.1 if the timebase is 20 ms cm–1 and the voltage setting is 10 mV cm–1.
Dual trace CROs
A dual trace CRO has two inputs, each with separate vertical deflection controls, that enable the signal at different
parts of a circuit to be compared.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Using a multimeter
R
R
Switched range multimeter
EC
TE
D
PA
G
E
PR
O
O
FS
The cathode ray oscilloscope discussed in the preceding section is a benchtop diagnostic tool. It is a good device
for measuring the period and amplitude of AC signals, or the output of voltage rectifiers. A multimeter is a handheld diagnostic device. It is useful for analysing DC circuits that you might construct. Multimeter have a digital
display of the quantities being measured. A multimeter can be used as an ammeter or a voltmeter in either DC or
AC circuits, or as an ohmmeter. The figure at below shows the front of a typical multimeter. This type is known as
a ‘switched range multimeter’ because you can select its function and range of readout by rotating the central
control dial.
N
C
O
Always switch off the circuit’s power supply before connecting a multimeter to a circuit. Choose the quality
(voltage, current or resistance) you wish to measure with the selection knob. Estimate the magnitude of the
reading you expect to find and select the appropriate range. If you have absolutely no idea what the reading might
be, start on the greatest range — for example, the 10 A range for current. This will reduce the risk of damaging
the multimeter.
U
The first of the following figures shows a simple DC circuit consisting of a power source, a switch and two
resistors. We will now consider how to use a multimeter to analyse this circuit.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
(a) The original circuit. Connecting the multimeter as (b) an ammeter, (c) a voltmeter and (d) an ohmmeter
TE
D
PA
When using a multimeter as an ammeter to measure current through a component or branch of a circuit, the
ammeter must be part of the circuit. The ammeter must therefore be connected in series with the circuit
components whose current is to be measured. It is necessary to break the circuit at an appropriate point and
connect the meter, as shown in part (b) of the figure above. In this mode, the multimeter has very little resistance
so that it has a minimal effect on the current flowing through that branch of the circuit. The resistance of the
ammeter will cause a slight reduction in the current.
If there is no current detected flowing through the circuit branch you are testing, check for breaks, loose
connections and open switches. If there is still no current detected after doing this, try using a lower range on the
multimeter.
R
EC
When using a multimeter as a voltmeter to measure the voltage drop between points in the circuit, the voltmeter is
connected across the two points (see part (c) of the figure above). If measuring the voltage drop across a
component, the multimeter is connected in parallel with the component. In this mode the multimeter has a very
high resistance, of the order of half a million ohms. A negligible current will still flow through the multimeter.
C
O
R
You must never use a multimeter to measure mains voltages. This will blow the fuse, damage other internal
circuitry and could injure or kill you. When using a multimeter to measure current or voltage in a DC circuit,
connect the common terminal of the multimeter to the part of the circuit closest to the ground or negative terminal
of the power supply.
U
N
The ohmmeter is not used when the circuit is connected to the power supply. It uses the internal battery of the
meter to send a current through the desired part of the circuit and shows the resistance by analysing the current
(see part (d) of the figure above). If the ohmmeter indicates that there is an infinite resistance in that part of the
circuit, it means that there is either a break in the circuit, a faulty connection or that a switch is open. Ohmmeters
are difficult to use with circuits that have already been constructed because it is not always apparent when other
components are connected in parallel with the component being tested.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Transformers
FS
A transformer is a device that increases or decreases the voltage from an AC supply. The circuit symbol for a
transformer is:
O
A transformer consists of two coils of wire linked by a soft iron core. The operation of a transformer is studied in
detail in unit 3.
PR
O
The two coils have different numbers of turns of wire wrapped around the core. If there are more turns of wire on
the input side than the output side of the transformer, the input voltage will be greater than the output voltage.
This type of transformer is called a step-down transformer.
G
E
The input and output voltages (V1 and V2) are related to the number of turns of wire on the input and output sides
(N1 and N2) by the relationship:
PA
N 1 V1
 .
N 2 V2
SAMPLE PROBLEM 17.6
TE
D
Transformers operate using AC voltages. The input and output voltages are alternating currents.
EC
A transformer is used to step down a 240 V voltage to 6.0 V. What is the ratio of the input turns to output turns of
wire in this transformer?
Solution:
R
V1  240 V
R
V2  6.0 V
C
240
6.0
 40
O
N 2 V1

N 1 V2
U
N

REVISION QUESTION 17.5
A transformer is used to provide an AC voltage of 6.0 V from a mains supply of 240 V AC. What is the
value of the following ratio?
Number of turns on the transformer secondry
Number of turns on the transformer primary
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Converting AC to DC
Many electronic devices such as iPads and mobile phones require a DC supply to operate. Electric power is
usually supplied to buildings in AC form. It is therefore necessary to convert or ‘rectify’ the voltage supplied to the
electronic circuits inside many appliances from AC to DC.
Many portable digital radios operate using four 1.5 V cells in series as a power supply. This means that the
circuits inside the radio are designed to run from a six-volt supply. When the player is plugged into the mains, the
voltage must be converted from 240 VRMS AC to 6 V DC. The first step in such a conversion is to use a
transformer to reduce the AC voltage to a suitable level. At this stage the current is still AC. What is needed next
is a device that allows current to pass through it in only one direction. Such a device is called a diode.
PR
O
Charging and discharging capacitors
O
FS
Diodes are discussed in later in this chapter. Another component that is important in systems that convert AC
voltages to DC voltages is the capacitors. They are used to produce a more constant DC voltage. We will now
look at what happens when capacitors charge and discharge.
TE
D
PA
G
E
Circuits containing a resistor and a capacitor are known as RC circuits. An example of a simple RC circuit is
shown at below.
EC
An RC circuit
R
R
When the switch in the figure is closed, a current flows through the circuit. As charge builds up on the capacitor,
the current decreases until the voltage across the plates is equal to the emf of the cell. Remember that a current
flows from a high voltage point to a low voltage point. When the voltages of the cell and the capacitor are equal,
the current ceases.
C
O
The magnitude of the current is affected by the resistance of the resistor. The bigger the resistance, the smaller
the current and the slower the capacitor will charge.
U
N
The amount of time it takes to fully charge the capacitor depends on the capacitance of the capacitor and the
resistance of the resistor. The greater the capacitance, the more charge can be stored on the plates per unit
voltage across the plates. The figure below shows how the voltage across the capacitor varies with time.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
How voltage varies with time when charging a capacitor
The time constant, τ or t, 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.
G
E
  RC
R
EC
TE
D
PA
The circuit shown in part (a) of the figure below will discharge the capacitor when the switch is closed. The way
voltage varies with time for discharging the capacitor is shown in part (b). In this case, the time constant is the
amount of time it takes for the voltage across the capacitor to fall to 37% of its original value.
O
R
(a) A circuit for discharging a capacitor and (b) how voltage varies with time when discharging a capacitor
U
N
C
A capacitor is considered to be fully charged or discharged after five time
constants have elapsed.
eModelling:
Charging a capacitor
This spreadsheet helps in
understanding how the
build-up of charge on a
capacitor in an RC circuit
acts to slow down the rate
of charging.
doc-0039
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
AS A MATTER OF FACT
An electronic heart pacemaker is an example of an RC circuit. Pacemakers provide regular voltage pulses that
start and control the frequency of the heartbeat. They can be worn externally or implanted beneath the skin.
Pacemakers contain a resistor and a capacitor with electrodes implanted in or near the heart. The charge on the
capacitor builds up and then discharges through the electrodes. Then it starts to build up again. The time between
pulses depends on the values of R and C.
SAMPLE PROBLEM 17.7
FS
A camera flash charging circuit consists of a 100 μF capacitor in series with a 220 kΩ resistor as shown.
G
E
PR
O
O
The emf of the cell is 1.5 V. Approximately how long will it take for the voltage across the capacitor to reach 1.0
V?
PA
Solution:
TE
D
The time constant for this circuit is 220  103   100  10 –6 F, which is 22 s. 1.0 V is 0.67 of the final voltage of the
capacitor. This will be reached after approximately one time constant; therefore, the answer is 22 s.
REVISION QUESTION 17.6
EC
Repeat sample problem 17.7 but with a capacitance of 220 μF and a resistance of 33 kΩ.
R
R
Half-wave rectification
O
A rectifier is a circuit element that is connected to the output of a transformer and converts the AC voltage into a
DC voltage. The AC voltage from the transformer gives a sine wave when graphed against time.
U
N
C
A half-wave rectifier uses a 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 in part (a)
of the figure below, and how the input voltage varies with time is shown in part (b). Since the load has an effective
resistance, it is represented as a resistor.
(a) A half-wave rectifier circuit and (b) how the input voltage varies with time
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Digital doc:
Investigation 17.1:
Constructing a half-wave
rectifier
The aim of this investigation is
to connect a voltage regulator
to a rectifier circuit to produce a
steady-value voltage supply.
TE
D
PA
G
E
PR
O
O
FS
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 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 load now has only a positive voltage drop across it
and the current flows through the load in only one direction. A DC voltage
has been achieved, but, as the figure below shows, this voltage is far from
being steady. This rectifier is called a half-wave rectifier because only one
half of the sine wave has been allowed to pass through it.
The transformer voltage and the voltage drop across the load for a half-wave rectifier
EC
Capacitor smoothing
U
N
C
O
R
R
Most electronic circuits require a steady DC voltage to operate effectively. The process of converting the DC
voltage shown in the above figure into a steady DC voltage is called smoothing. Smoothing is achieved by
placing a capacitor in parallel with the load resistance as shown in the circuit at below.
A half-wave rectifier with a smoothing capacitor
When the positive part of the transformer voltage passes through the diode, it produces a voltage drop across the
load resistance and charges the capacitor. When the diode blocks the negative part of the transformer voltage,
the capacitor discharges through the load. The voltage drop across the load is shown in the figure below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
The voltage drop across the load for a half-wave rectifier with a smoothing capacitor
Full-wave rectification
TE
D
PA
The voltage drop shown in the figure above is still not constant. The variations in voltage are called ripples. The
size of the ripples depends on the capacitance of the capacitor, the load resistance and the frequency of the
waveform from the transformer. The bigger the time constant when the capacitor is discharging through the load,
the smoother the voltage drop across the load. It is for this reason that good quality DC power supplies require
smoothing capacitors that have large capacitance.
EC
A half-wave rectifier makes use of only half the input AC current or voltage. A full-wave rectifier makes use of both
half-cycles of an AC waveform to supply a current that flows in one direction through a load.
R
Bridge rectifier
U
N
C
O
R
A bridge rectifier uses four diodes to produce full-wave rectification. The figure below shows a circuit diagram for
a bridge rectifier with input and output signals.
A bridge rectifier
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
The figure below shows the pathway taken by the current through the load of a bridge rectifier.
TE
D
PA
G
E
PR
O
O
FS
Remember that 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.
Capacitor smoothing
EC
Direction of current through the load of a bridge rectifier when the AC current (e.g. from the transformer) is (a)
positive and (b) negative
U
N
C
O
R
R
Full-wave rectifier output signals can be smoothed in the same way as half-wave rectifier signals (see the figure at
below). 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. This is shown in the figure below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
TE
D
EC
R
R
O
C
N
U
Capacitor smoothing of a bridge rectifier (a) circuit diagram and (b) comparison of signals
Centre-tap full-wave rectifier
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. The arrangement is shown in the
figure to the below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Centre-tap rectifier circuit diagram
EC
TE
D
PA
G
E
PR
O
O
FS
In this case there is also an approximate voltage drop of 0.7 V across each diode that the current passes through.
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 figure.
R
Ripple voltage
R
Capacitor smoothing of a centre-tapped rectifier (a) circuit diagram and (b) comparison of signals
C
O
Ripple voltage is the periodic variation in a DC voltage that results from the rectification of an AC voltage. The
peak-to-peak value of the ripple voltage is:
N
V r (p  p)  V max V min .
U
The value of the peak-to-peak value of the ripple voltage depends on the
following factors:
•
•
•
the maximum value of the voltage (Vmax) supplied by the diode(s) of the
Weblink:
rectifier
the period (T) of the unsmoothed voltage supplied by the diode(s) of the
Rectifiers
rectifier. This is half the period of the input AC voltage supplied to a full-wave
rectifier, or the same as the period of the input AC voltage supplied to a half-wave rectifier. In Australia, the
frequency of the AC power supply is 50 Hz. This means that the AC power supply has a period of 0.020 s.
the time constant (τ) of the capacitor and load resistance: τ = RC.
V r ( p p ) 
V maxT
RC
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Voltage regulators
Most electronic circuits operate using a steady DC voltage. This voltage generally needs to remain constant.
There might be variations in the input voltage — for example, from the AC voltage supplied to your house by the
power lines, or from a battery that is slowly ‘going flat’. There might also be variations in the effective resistance of
the load circuit. Most digital logic circuits and processors require a 5 V DC power supply.
G
E
PR
O
O
FS
The diode- and capacitor-based rectifier circuits discussed so far do not achieve provide a steady DC voltage.
Attaching a load across the output of a rectifier, as shown in the figure below, will reduce the effective resistance
of the rectifier and therefore affect the time constant of the smoothing capacitor. The solution to this problem is to
use a voltage regulator.
PA
Attaching a load resistor in parallel with the resistor of a half-wave rectifier reduces the effective resistance of the
rectifier.
TE
D
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. In contrast, the voltage drop across the
output terminals of a school power supply varies as its current changes.
EC
Many DC power supplies used in school laboratories consist of a
transformer connected to a rectifier circuit. Changes to the input AC
voltage of the transformer can alter the magnitude of the output voltage of
the supply.
Digital doc:
Investigation 17.2: Constructing
a voltage regulator
Produce a steady-value voltage
by connecting a voltage regulator
to a rectifier circuit.
U
N
C
O
R
R
Voltage regulators are usually integrated circuits that are placed between
the power supply and the load. The following figure shows a 7805 threepin voltage regulator and a circuit diagram showing how a voltage
regulator can be used in a practical situation. Note that voltage regulators provide an output voltage that is less
than their input voltage.
(a) A 7805 voltage regulator, (b) schematic circuit diagram of a voltage regulator and (c) how a voltage regulator
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
is placed in a circuit to give a constant output voltage.
Internal resistance
Students are often required to carry out experiments to verify that when resistors are connected in parallel, the
voltage across them remains constant. When they do the practical work, students usually find that adding extra
resistors in parallel results in the voltage across them falling.
O
FS
The drop in voltage occurs because power supplies are not ideal. They also offer resistance to the current drawn
from the supply. This resistance is known as ‘internal resistance’. Even though a student connects resistors in
parallel, they are also in series with the internal resistance of the power supply. When the student adds an extra
resistor to the parallel array, the total resistance of the circuit is reduced. The current through the power supply
increases and the voltage drop across the ‘internal resistance’ increases. This results in the terminal voltage of
the supply decreasing.
PR
O
PHYSICS IN FOCUS
Using voltage regulators
G
E
Voltage regulators are used in portable sound systems. The amplifying circuits need a constant voltage so that
sound is reproduced accurately. Voltage regulators are used because the output voltage of batteries and dry cells
falls as their internal resistance increases. (That is, the batteries go flat!).
TE
D
PA
Voltage regulators, such as the 7805, are available in three-pin integrated circuit (IC) packages. One pin connects
to the unregulated input voltage, one pin provides the regulated output voltage and the other pin connects to the
ground. There are two types of IC voltage regulator you could consider for your project. A fixed regulator gives a
set output voltage, and an adjustable regulator can give a range of output voltages.
Choosing a suitable voltage regulator
EC
Voltage regulators come in different types and sizes. They are able to operate with different input and output
voltages as well as current ranges. If you are going to use a voltage regulator in a circuit, you will need to choose
the best one for your requirements. You can choose a voltage regulator by studying data sheets.
U
N
C
O
R
R
Another way to choose a voltage regulator is to examine the current–voltage characteristics of the regulators.
These are graphs that show how the regulator’s voltage varies as its output current changes. The figure below
shows the characteristics for a 7805 voltage regulator.
The current–voltage characteristic curve for a 7805 voltage regulator
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
If you required a steady supply voltage of 5.0 V operating from a 12.0 V supply, you would choose a 7805 voltage
regulator because it is designed to give a steady 5.0 V DC output. A 7809 voltage regulator is designed to give a
steady 9.0 V DC output from an unregulated supply with an output voltage greater than 10 V.
Voltage regulator using a Zener diode
FS
A Zener diode is a diode that breaks down at a specific voltage when it is reverse biased. This means that a
Zener diode will stop a reverse current from flowing through it until the reverse voltage applied across it reaches a
fixed value known as the breakdown voltage. Zener diodes are designed to break down in a reliable and nondestructive way so that they can be used in reverse to maintain a fixed voltage across their terminals.
PR
O
O
Circuit diagram symbol for a Zener diode
Zener diodes are used to provide a stable output voltage from an unstable source. One application is to provide a
stable voltage from the storage battery of a solar cell array. The output of such a battery might vary depending on
the amount of charge stored. Zener diodes are also used in circuits to maintain a fixed voltage across a load.
U
N
C
O
R
R
EC
TE
D
PA
G
E
The diagram below shows the Zener diode’s voltage versus current characteristics and how it can be connected
to an unstable source. The resistor is placed in series to limit the current from the unstable source.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Zener diodes can be distinguished from ordinary diodes by their code and breakdown voltage, which are printed
on them. Zener diode codes begin BZX or BZY. Their breakdown voltage is printed with V in place of a decimal
point, so 4V7 means 4.7 V. The minimum voltage Zener diode available is 2.7 V.
If a Zener diode is placed in parallel with one of the resistors in a voltage divider
circuit (as shown at below), no current flows through the Zener diode when the
voltage across the resistor is less than the breakdown voltage. When the voltage
is equal to the breakdown voltage, some current passes through this resistor, the
rest passes through the diode. The result is that the voltage across the diode and
the resistor R2 remains constant.
FS
If the input voltage increases, the current through the voltage divider increases
and any increase in voltage happens across the other resistor, R1.
Power dissipation
PR
O
O
In electric and electronic circuits, power is the rate at which electrical energy is transformed into other forms of
energy, usually heat. When this happens, power is said to be dissipated.
The amount of power dissipated in a circuit element can be calculated using the relationship:
P VI
G
E
where V is the potential difference (or voltage drop) across the circuit element and I is the current flowing through
the element.
PA
If the circuit element has a fixed resistance, R, the formula above can be expanded to include the resistance using
Ohm’s Law:
TE
D
V  IR
So, if you only know the current I and the resistance R, you can use the relationship:
P  I 2R
EC
If you only have the potential difference V and the resistance R, you can use the relationship:
R
P  V 2 /R
O
R
SAMPLE PROBLEM 17.8
U
P  I 2R
N
Solution:
C
A 100 Ω resistor in series with a diode carries a current of 5.0 mA. Calculate the power dissipated in the resistor.
 (5.0  10 3 )2  100
 2.5  10 3 W or 2.5 mW
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Photonics
Photonics is the branch of science that studies or uses the manipulation of
photons (the basic unit of light) to manipulate, encode, transmit, decode and/or
store information. It involves:
•
Weblink:
Photonics resources for
teachers
FS
•
•
•
using electronic systems to convert information into light (or other
electromagnetic radiation) energy signals
transmitting the light — for example, through optical fibres
converting the light signal back into useful information — for example, sound
storing information using light — for example, by burning CDs (compact discs)
or DVDs (digital video discs).
O
Photonics does not just involve the use of visible light. It also involves infra-red and ultraviolet sections of the
electromagnetic spectrum.
G
E
PR
O
A television remote control transmits information to the television set using infra-red radiation. An integrated circuit
chip detects when you press a button on the remote. It then produces a coded signal that is something like morse
code. There is a different code for each button. The coded signal is amplified and sent to an infra-red LED that
sends invisible radiation into the room. The code varies the intensity of the radiation. This varying of intensity is
known as modulation. On the television set there is a photodetector that converts the infra-red radiation back
into an electric signal. The detector might be a photocell — such as a photovoltaic (solar) cell — or a photodiode
or phototransistor. The magnitude of the current produced by the photodetector is proportional to the intensity of
the infra-red radiation.
U
N
C
O
R
R
EC
TE
D
PA
The infra-red LED is an example of a transducer that transforms electrical energy to electromagnetic radiation.
The infra-red radiation is the carrier of the signal. The photodetector is an example of a transducer that transforms
electromagnetic radiation to electrical energy. This photonic system is illustrated below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
(a) Television remote control and (b) block diagram of television remote control system
Light-emitting diodes (LEDs)
PA
G
E
You have seen light-emitting diodes (LEDs) on many electronic devices. An LED is a small semiconductor
diode that emits light when a current passes through it. LEDs are used in electronic displays. They can be made
to emit any colour (red, green and yellow are the most common) by the choice of impurities added to the base
semiconductor used in their construction. They are used on amplifiers to indicate the amplitude of the sound
signal. They also appear in cash register displays and as indicators on modems
and routers.
Light-emitting diodes have the following advantages over ordinary light globes:
•
TE
D
EC
•
R
•
R
•
LEDs are very durable and can be used in situations where excessive
vibrations would damage the filaments of ordinary light globes — for example,
in aircraft cockpits, in the space shuttle and on the control panels of vibrating
machinery.
LEDs emit light using very low voltages, so they are useful indicator lights for
electronic systems.
An LED will operate for over 500 000 hours (depending on the operating
current and temperature) compared with about 1000 hours for a light globe.
LEDs are much faster to respond than light globes. A light globe takes many
milliseconds to turn on and off, while red or yellow LEDs take about 75
nanoseconds to turn on and off.
LEDs consume much less power than light globes.
O
•
Digital doc
Investigation 17.2 Lightemitting diodes
The aims of this
investigation are to
construct an I–V
characteristic for an LED
and determine how an LED
behaves when it is forward
and reverse biased.
N
C
Before considering how LEDs operate, you should revise the information on
diodes covered previously in the chapter.
U
The structure of LEDs
An LED consists of a small semiconductor chip mounted on, and in electrical contact with, a metal base. The
base is reflective to ensure much of the radiation is projected forwards through the top of the LED. The cathode
lead is connected to the metal base. A fine wire is connected to the upper surface of the chip and leads to the
anode lead of the LED. This assembly is covered with an epoxy resin that acts as a lens to intensify the light
passing through the top of the LED. This structure is shown in figure (a) below. The circuit diagram symbol for an
LED is shown in figure (b). The colour of the LED determines the voltage at which the LED reaches saturation or
the maximum voltage across the LED when it is conducting a current. The I–V characteristic curve for a red LED
is shown in figure (c).
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
(a) The structure of a typical LED, (b) the circuit diagram symbol for an LED and (c) I–V characteristic curve for a
red LED
PR
O
Commercial LEDs are designed to release visible light or infra-red radiation. The chip is based on a
semiconducting material made from a gallium–arsenic–phosphorus compound. Adjusting the ratio of phosphorus
to arsenic enables different wavelengths (and colours) of light to be emitted.
G
E
As with other diodes, the chip of an LED is made by fusing a p-type semiconductor to an n-type semiconductor.
The p-type material forms the anode and the n-type material forms the cathode. An LED will emit light when a
current passes through it; that is, when it is forward biased. It is important that the current is not too high, as the
LED might be damaged, or not too low as not enough light will be emitted. The forward bias voltage drop across
an LED is between 1.6 V and 2.8 V, depending on the colour of the LED and the current through it.
R
EC
TE
D
PA
A red LED has a forward bias voltage drop of approximately 1.6 V, whereas for a green or yellow LED, the value
is approximately 2.3 V. To ensure that the current passing through the LED is suitable, the LEDs are protected by
being placed in series with resistors (known as limiting resistors). The value of the resistor is calculated to reduce
the potential drop across the LED to the required value. This arrangement is shown below.
O
R
Limiting resistor in series with a forward-biased LED
C
SAMPLE PROBLEM 17.9
N
The LED shown has a voltage drop across it of 1.8 V and carries a current of 20 mA.
U
a. What is the voltage drop across the limiting resistor?
b. What is the value, R, of the limiting resistor?
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Solution:
a. The emf of the source equals the sum of the voltage drops around the circuit. So the voltage drop across
the resistor is 4.2 V (6.2 V – 1.8 V).
b. The current through the resistor is the same as that through the LED: 20 mA or 0.020 A.
The value of R is calculated using the formula:
V  IR

V
I
4.2 V
0.020 A
FS
R
PR
O
O
So
R  210 .
REVISION QUESTION 17.7
PA
G
E
The resistor in sample problem 17.9 is replaced with another resistor. The voltage drop across the LED
remains at 1.8 V and the current changes to 5.0 mA. Calculate the resistance of the limiting resistor.
TE
D
AS A MATTER OF FACT
U
N
C
O
R
R
EC
The seven-segment number display is used in clock radios, microwave ovens and many other appliances. The
figure here shows a unit with the number seven activated. Each segment of the display consists of an LED and a
lens. When the LED conducts a current, it emits light that passes through the top of the lens. By activating
different segments, different numbers can be formed.
The seven-segment number display
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Infra-red LEDs emit electromagnetic radiation with a longer wavelength than that
of visible light. They are used widely in optical communications, especially in
television and audio system remote controls.
Laser diodes are made by polishing the ends of suitable p–n junction crystals.
They are used in DVD players and laser pointers.
Weblink:
LEDs
AS A MATTER OF FACT
FS
How do LEDs produce light?
EC
AS A MATTER OF FACT
TE
D
where
λ = wavelength of the electromagnetic radiation
c = speed of light
f = frequency of the radiation
h = Planck’s constant.
G
E
c hc

f Eg
PA

PR
O
O
When an electron at the bottom of the conduction band of a semiconductor falls into a hole at the top of a valence
band, an amount of energy, Eg, is released, where Eg is the energy gap width. In silicon-based semiconductors
this energy is transformed into thermal energy in the vibrating lattice of the atoms. In some semiconducting
materials, the released energy can appear as electromagnetic radiation. The wavelength of this radiation can be
calculated using the following formula:
R
R
Optical-fibre telecommunication systems usually use LEDs and laser diodes that operate in the infra-red region of
the electromagnetic spectrum. Laser diodes produce a much smaller range of wavelengths than LEDs. Certain
wavelengths of infra-red radiation are used because they travel better through the optical fibres than visible light.
O
Light-dependent resistors (LDRs)
N
C
Photonic transducers that transform light energy to electrical energy can be used
in electronic circuits to produce current or voltage signals.
U
A light-dependent resistor (LDR) is a semiconductor device that has a
resistance which decreases with the amount of light falling on it. Figure (b) below
gives an example of a graph that shows the relationship between resistance and
light intensity for an LDR. Note that in this example, the scales on both axes are
logarithmic.
Digital doc:
Investigation 17.3:
Light-dependent resistor
The aim of this investigation
is to examine the way in
which the resistance of a
light-dependent resistor
varies with light intensity.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
TE
D
(a) Circuit symbols for an LDR, (b) resistance-versus-light intensity graph for an LDR and (c) an LDR
U
N
C
O
R
R
EC
LDRs are used in cameras (see the next ‘Physics in focus’, box) and in a wide
range of automatic devices controlled by light. They are often found in voltage
dividers in conjunction with variable resistors; the output voltage is used to
switch another device on or off. The variable resistor is used to control the light
intensity at which the switching device is activated.
Chapter 17 How can AC electricity charge a DC device?
Digital doc:
Investigation 17.4: Lightdependent resistor:
voltage divider method
The aim of this investigation
is to gather information to
produce a graph of how the
resistance of a lightdependent resistor varies
with light intensity.
© John Wiley & Sons Australia, Ltd
SAMPLE PROBLEM 17.10
A shop minder circuit consists of a beam of light that shines onto the voltage divider circuit, as shown below. The
LDR has the same characteristic curve as that shown in figure (b) on page xxx.
PR
O
O
FS
a. What is the resistance of the LDR when the light intensity is 1.0 lux?
b. If the variable resistor is set at 500 Ω, calculate the value of Vout when the light intensity is 1.0 lux.
c. What is the value of the variable resistor if the light intensity is 0.1 lux and Vout is 6.0 V?
G
E
Solution:
b. V in  12 V, R1  500 , R 2  1000 , V out  ?
R 2V in
R1  R 2
TE
D
V out 
1000   12 V
500   1000 
12 000  V

1500 
 8.0V
EC

R
From the graph, when the intensity is 0.1 lux, R 2  10 k .
R
c.
PA
a. Reading from the graph, when the intensity is 1.0 lux, R 2  1 k .
C
R 2V in
R1  R 2
10 000   12 V
R1  10 00 
N
V out 
O
V in  12 V, V out  6 V, R 2  10 000 , R1  ?
U
6V 
6 V  R1  60 000  V  120 000  V
6 V  R1  60 000  V
R1  10 000 
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
REVISION QUESTION 17.8
An LDR with the resistance-versus-light characteristic shown in the
previous log scale graph is placed in the voltage-divider circuit shown
below.
PR
O
O
FS
a. What is the resistance of the LDR when the light intensity is 200
lux?
b. If the variable resistor were set to 320 Ω, what would Vout be for this
intensity?
c. What is the value of the variable resistor if the light intensity were
1.0 lux and Vout were 6.0 V?
REMEMBER THIS
G
E
Ohmic devices obey Ohm’s Law (V = IR). When V is plotted against I, a straight line results. A device is
considered to be non-ohmic if, when V is plotted against I, a straight line does not result.
PA
PHYSICS IN FOCUS
LDRs in cameras
R
EC
TE
D
To get the best results from a camera it is necessary to control the
amount of light that falls on the detector. This is achieved by
changing the aperture of the lens diaphragm and the shutter speed.
High shutter speeds are necessary when photographing fast-moving
objects. If you use a high shutter speed you have to use a wide
aperture opening so that enough light falls on the detector to get
proper resolution. Similarly, in dull conditions you need a wide
aperture opening so that enough light enter the camera.
U
N
C
O
R
Many professional level cameras have a light meter so that photographers can adjust the shutter speed and
aperture opening. These meters are either connected to a photocell that converts light directly into an electric
current or they are connected to a voltage divider that has an LDR in it. In a camera that uses a voltage divider,
part of the light passing through the lens system is reflected onto the LDR by a small mirror. The output voltage of
the voltage divider is fed to the meter that converts the reading to the appropriate units for light intensity. It is then
up to the photographer to make the necessary adjustments.
Photodiodes
Photodiodes are used to detect light and to convert light energy to electrical energy. A photodiode consists of an
active p–n junction that is operated in reverse bias. When a photon of sufficient energy strikes an electron in the
junction, the photon the electron transfers its energy to the electron allowing the electron to escape the crystal
lattice. This creates a free electron and a hole that move in opposite directions under the influence of the electric
field provided by the source of electromotive force.
The intensity of light depends on the number of photons present. The current produced by a photodiode is
therefore proportional to the intensity of light incident on the photodiode. The next figure shows a circuit with a
reverse-biased photodiode and a response characteristic curve.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Photodiodes are fast to respond to light. It takes only a few nanoseconds (10–9 s) for a photodiode to produce a
current after the photons arrive. This makes them useful for converting light signals from fibre-optic
communication systems back into electrical signals. Although the response time of a photodiode seems to be
extremely rapid, it does limit the bandwidth (amount of information carried per second) of an optical-fibre
communications system. The response time determines both the minimum duration of pulses and the maximum
number of pulses that can be transmitted down the fibre each second.
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
FS
Different types of photodiode have different optimum wavelengths of light to which they respond. Silicon-based
photodiodes operate best with wavelengths between 700 and 900 nm. Germanium-based photodiodes operate
best with wavelengths between 1400 and 1500 nm.
Photodiode (a) circuit symbol, (b) reverse biased and (c) sample characteristic curves
Phototransistors
A phototransistor is similar to an ordinary transistor except that incident light on the p–n junction controls the
response of the device. Phototransistors have a built-in gain and are more sensitive than photodiodes.
A phototransistor is usually an n-p-n transistor. The base region is large and usually does not have a terminal
attached to it. The circuit symbol for a phototransistor is shown in the figure at below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
PR
O
O
FS
The collector–base junction is sensitive to light in the same way as a photodiode. In fact, the collector–base
junction is also a reverse-biased p–n junction, just like a photodiode. When light shines on it, a base current flows.
The base current is proportional to the light intensity and produces a collector current that is also proportional to
the light intensity.
Phototransistor circuit symbol
PA
G
E
Phototransistors are more sensitive than photodiodes because transistors amplify the collector–base current.
They are not as quick to respond to light as photodiodes. This reduces the bandwidth of photonic systems that
use phototransistors rather than photodiodes.
AS A MATTER OF FACT
EC
TE
D
Optical-fibre systems that have a bandwidth of 43 Tbps (terabits per second) have been trialled. This means that
the system can transmit up to 43 × 1012 bits of data per second. By comparison, coaxial cable, which is used to
carry telephone calls between cities and is also used to transmit data over shorter distances, has a bandwidth of
up to 100 Mbps.
R
Communications carriers
O
R
There are different types of communication carriers involved in carrying telephone calls and data between users.
Most houses are still connected to the telephone network by twisted pairs of wires.
C
Twisted-pair copper cable
U
N
Bundles of copper cables are commonly used throughout all telecommunications networks. Each pair of cables
carries one call.
A twisted-pair copper cable consists of two independently insulated wires twisted around one another. One wire
carries the signal while the other wire is grounded and absorbs signal interference. Twisted-pair cables are used
by older telephone networks. They are the least expensive type of local area network (LAN) cable. Most networks
still contain some twisted-pair cabling at some point along the network.
Coaxial cable
Coaxial cable consists of a centre wire surrounded by insulation and then a grounded shield of braided wire. The
shield minimises electrical and radio frequency interference. It was once the main cable used in the trunk network.
It carried calls between major cities. Each cable could carry up to 2700 calls at a time.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Coaxial cabling is widely used in the cable television industry and computer networks. It is much less susceptible
to interference than twisted-pair cable and can carry much more data. Coaxial cables are gradually being
replaced by optic fibres and microwave systems.
Optical-fibre cable
O
FS
An optical fibre is a thin filament of glass or plastic that has a central core and a cladding with a lower refractive
index, so that total internal reflection will happen.
PR
O
Light reflections in an optical fibre
U
N
C
O
R
R
EC
TE
D
PA
G
E
An optical-fibre cable consists of several optical fibres in a common package. A strength member, such as a steel
wire, supports the fibres. This protects the delicate optical fibres from stretching. A single optical fibre can carry 30
000 telephone calls at a time or hundreds of television signals. Some optical-fibre systems in development have
demonstrated a capacity of over 40 million telephone calls.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
TE
D
EC
R
R
O
C
N
U
Communications carriers (a) twisted pair (b) coaxial cable and (c) optical-fibre cable
Attenuation
Attenuation is the loss of power of a signal along a communicating channel. Attenuation is measured in decibels
(dB).
Telephone systems that use electrical signals travelling through copper cables experience attenuation. The
strength of the signal decreases with the distance travelled. These systems use amplifiers to increase the power
of the signals. One reason for the attenuation of signals in metal wire carriers is the skin effect.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
The skin effect is the uneven distribution of current carriers throughout the cross-section of a conductor when it is
carrying an alternating current. The current density becomes greater at the surface than it is at the centre. The
skin effect is caused by electromagnetic effects, and it becomes more significant as the frequency of the
alternating current increases. One consequence of the skin effect is that the effective resistance of a conductor is
greater when it is carrying an alternating current than its true or direct current resistance.
Bandwidth
Bandwidth is the amount of data that can be transmitted in a fixed amount of time. For digital devices, the
bandwidth is usually expressed in bits per second (bps) or bytes per second. For analog devices, the bandwidth is
expressed in cycles per second, or hertz (Hz).
FS
The bandwidth of an information system is essentially the highest frequency or rate at which the data can be
transmitted.
PR
O
O
System bandwidth is the range of frequencies that are carried by the transmission system. In practice, a
commercial quality speech transmission system need carry only frequencies between 300 Hz and 3400 Hz. This
is a bandwidth of 3400 Hz because the maximum frequency it can carry is 3400 Hz.
Fibre-optic cables have a much greater bandwidth than metal cables. This means that they can carry more data.
The transmission rate of optical fibres is in the order of 1 trillion bits (1 terabit) per second per fibre. The
transmission rate of telephone wire pairs is in the range of 16–100 million bits per second (16–100 Mbps).
G
E
Modulation
U
N
C
O
R
R
EC
TE
D
PA
Modulation is the changing of one wave train caused by another wave. An example of this is amplitude
modulation (AM) used in radio transmission systems. In this case, the amplitude of the carrier signal is changed
by superimposing the waveform of a sound such as music or voice. This is shown below.
Amplitude modulation of a radio signal
Optical intensity modulation in photonic communications systems involves varying the output intensity of a carrier
light source (for example, an LED or laser diode) by using the electric signal from a sound-to-electricity transducer
such as a microphone, or from a sound playback device. The figure below shows an analog modulation system
where the output from an LED is modulated using a microphone and amplifier.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
(a) Optical intensity modulation system block diagram and (b) adding the signals
AS A MATTER OF FACT
TE
D
PA
The output of the amplifier is an amplified version of the information signal. The carrier is the light from the LED or
laser. This light is very close to being monochromatic. Monochromatic light has a constant wavelength. The
output of the amplifier makes the light produced by the output transducer brighter or dimmer. The output
transducer can respond to variations in intensity very rapidly.
R
R
EC
LEDs are widely used in data transmission systems to transmit data at rates below 270 Mbps (2.7 × 108 bits per
second). This means they can be switched on and off at this rate. Laser diodes are used when faster transmission
rates are required.
C
O
Decoding signals
N
Once the modulated signal has been transmitted through the carrier system, it needs to be changed back to its
original form so that it can be used. The first step in this process is known as demodulation.
U
Demodulation
Demodulation is the process whereby the transmitted signal is converted back into an electrical signal. This can
then be amplified and used to drive a loudspeaker (for example, in an AM radio) or used by a controlling system
to control another device (for example, a television set). In photonic systems, the light signal falls on a transducer
called a detector. The most common devices used as detectors are the photodiode or phototransistor. The
detector produces an electrical current whose magnitude is proportional to the intensity of the light. This produces
an identical electrical signal to that which modulated the light carrier waveform.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
(a) Optical intensity demodulation system block diagram and (b) restoring the electrical signal
PA
Chapter review
Summary
A cathode ray oscilloscope shows how voltage varies with time.
■
A multimeter is a diagnostic tool that can be used as an ammeter, an AC or DC voltmeter, or an ohmmeter.
■
DC currents flow in one direction; AC currents periodically change direction.
■
Capacitors store charge.
EC
R
The time constant (τ) of a resistor–capacitor (RC) circuit is the time it takes the voltage across the capacitor
to reach 63% of its final value when charging, or to fall to 37% of its original value when discharging.
R
■
TE
D
■
A rectifier converts alternating currents to direct currents.
■
A half-wave rectifier uses one diode to allow half an AC signal to produce a DC signal.
■
A full-wave rectifier uses an array of diodes to make both directions of an AC signal flow in one direction.
■
Capacitors are used to smooth the outputs of rectifiers.
C
N
U
■
O
■
A voltage regulator is a device that keeps the terminal voltage of a voltage supply within required limits
despite variations in the input voltage or the load.
■
A Zener diode breaks down at a specific voltage when reverse biased. It can be used as a voltage regulator.
■
Ripple voltage is the periodic variation in a DC voltage that results from the rectification of an AC voltage.
■
■
A voltage divider consists of two or more resistors arranged in series to produce a smaller voltage at its
output.
The output of a basic voltage divider can be calculated using the equation:
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
V out 
R 2V in
R1  R 2
A transducer is a device that can be affected by, or affect, the environment.
■
An input transducer transforms non-electrical energy into electrical energy.
■
A thermistor is a semiconductor device whose resistance varies with temperature.
■
The resistance of a negative coefficient thermistor falls as the temperature increases.
■
The resistance of a positive coefficient thermistor rises as the temperature increases.
■
The relationship between resistance and temperature for a thermistor is usually shown graphically.
■
A diode is a semiconductor device that allows current to pass through it in one direction only.
■
Photonics involves the use of photons to manipulate, encode, transmit, decode and/or store information.
■
A photonic transducer transforms light energy into electrical energy, or electrical energy into light energy.
■
A light-emitting diode (LED) is a semiconductor diode that emits light when a current passes through it.
■
■
O
PR
O
G
E
The relationship between resistance and light intensity for an LDR is shown graphically.
PA
■
A light-dependent resistor (LDR) is a semiconductor device that has a resistance that decreases as the
amount of light falling on it increases.
Photodiodes produce a current when they are reverse biased and light falls on the junction. The current is
proportional to the intensity of the light.
TE
D
■
FS
■
Phototransistors generally have two terminals. A photosensitive collector–base junction provides the base
current.
Photodiodes are faster at responding to light than phototransistors, but phototransistors are more sensitive.
■
An optical fibre can carry 30 000 telephone calls at a time.
■
Attenuation is the loss of power of a signal along a communicating channel.
R
R
O
Optical intensity modulation in photonic communications systems involves varying the output intensity of a
carrier light source by using the electric signal from a transducer.
N
■
The bandwidth of an information system is essentially the highest frequency or rate at which the data can be
transmitted.
C
■
EC
■
U
Relative light
intensity IL
0.60
0.40
0.20
0.10
0.05
Current through
LDR (μA)
350
220
90
40
20
Voltage across
LDR (V)
38.7
39.0
39.1
39.0
39.0
Chapter 17 How can AC electricity charge a DC device?
Resistance
(kΩ)
Log
RL
Log
IL
© John Wiley & Sons Australia, Ltd
Questions
AC voltage and current
The variation of voltage drop with time for a 100 Ω resistor is shown below.
FS
1.
PR
O
Vpeak
Vp-p
VRMS
T
f
IRMS
G
E
a.
b.
c.
d.
e.
f.
O
What is the value of the following quantities?
The trace obtained on a CRO is shown below.
O
R
R
EC
TE
D
2.
PA
CROs and multimeters
N
the period
the frequency
the peak voltage
the RMS voltage.
U
i.
ii.
iii.
iv.
C
For the voltage and timebase settings in (a) and (b) below, write down:
a. 1 V cm–1, 1 s cm–1
b. 5 mV cm–1, 2 ms cm–1
3.
A signal generator is connected to a CRO as shown in figure (a) below. The CRO trace obtained is shown in
figure (b). The timebase setting is 5 ms cm–1, the voltage setting is 1 V cm–1.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
TE
D
EC
a. What are the values of the following quantities?
R
T
f
Vpeak
VRMS
R
i.
ii.
iii.
iv.
U
N
C
O
b. Sketch the CRO trace obtained if a 1.5 V cell is connected between the signal generator and the CRO
as shown below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
a. Describe how you would use a multimeter as a voltmeter.
b. Why must you never use a hand-held multimeter to measure the mains voltage?
PA
4.
Sketch the CRO trace obtained if the cell in part (b) were reversed.
G
E
c.
Converting AC to DC
a.
b.
c.
d.
TE
D
Convert the following capacitances to farads:
0.1 μF
220 μF
100 pF
140 nF.
EC
5.
a. Define the expression time constant for an RC circuit.
b. A capacitor is charged to 10 V. What will be the voltage drop across the capacitor after one time
constant if it is discharged through a resistor?
7.
Calculate the time constants for the following RC combinations:
R
R
6.
O
C
The variation of voltage across a capacitor with time for an RC circuit is shown above below.
U
8.
390 Ω, 100 μF
33 kΩ, 100 μF
68 kΩ, 10 μF
470 kΩ, 0.47 μF.
N
a.
b.
c.
d.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
What is the voltage drop across the capacitor when t = 1 ms?
What is the time constant for this RC circuit?
If R = 100 Ω, what is the value of C ?
If C = 0.1 μF, what is the value of R?
FS
a.
b.
c.
d.
Define rectification.
10.
Describe the output of a half-wave rectifier for a sinusoidal input voltage.
11.
a. What is a diode?
b. What is the maximum voltage drop possible across a silicon diode?
c. How many diodes are needed for half-wave rectification?
12.
a. Describe the effect of placing a capacitor in parallel with the load resistor of a half-wave rectifier.
b. Why does this effect occur?
c. What happens to the output voltage as the value of the capacitor is increased?
13.
a. Sketch the circuits of a bridge rectifier and a centre-tap rectifier.
b. If the peak voltage entering the bridge and centre-tap rectifier is 12.0 V, calculate the peak voltage of
the output of each rectifier.
14.
Calculate the peak-to-peak ripple voltage of a voltage signal if the maximum voltage is 6.3 V and the
minimum voltage is 5.8 V.
15.
A full-wave rectifier has an output with a peak value of 12.5 V and a frequency of 100 Hz. Calculate the
ripple voltage when the following combinations of load resistance and smoothing capacitor are used.
PR
O
G
E
PA
TE
D
EC
Load resistance (kΩ)
10
3.3
10
25
100
O
R
R
(a)
(b)
(c)
(d)
(e)
Smoothing capacitor (μF)
100
100
5.0
20
25
For a transformer, the ratio
C
16.
O
9.
U
N
N umber of turns on the transformer secondry
N umber of turns on the transformer primary
is
21
. If the input voltage is 120 V AC, what is the output voltage?
1200
Voltage regulators
17.
A simple RC circuit is shown below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
O
A half-wave rectifier circuit using a silicon diode is shown in the following figure (a), along with an input
voltage signal (figure (b)).
PR
O
18.
FS
a. How long does it take for the voltage drop across the capacitor to reach approximately 1.0 V after the
switch is closed?
b. How long will it take for the capacitor to be fully charged?
c. What will be the voltage drop across the capacitor when it is fully charged?
d. How much charge is stored on the capacitor when it is fully charged?
e. When the voltage drop across the capacitor is 1.0 V, what is the voltage drop across the resistor?
a. What is the peak output voltage for this circuit?
b. Sketch the output voltage for this circuit.
c. Sketch the output voltage if the following capacitors are placed in parallel with the load resistor:
G
E
10 μF
100 μF.
A half-wave rectifier circuit is shown. The capacitor can be engaged by closing the switch. The input voltage
to the rectifier is 9.0 V (RMS) at 50 Hz.
U
N
C
19.
O
R
R
EC
TE
D
PA
i.
ii.
a. What is the peak input voltage?
b. What is the period of the output voltage waveform when the switch is open?
c. Sketch the waveform of Vout when the switch is open, paying attention to the maximum voltage and the
period. (Assume that the maximum voltage drop across the diode is 0.7 V.)
d. Sketch the waveform of VD, the voltage across the diode, when the switch is open.
e. What is the time constant if the capacitor discharges through the resistor?
f. Sketch the waveform of Vout when the switch is closed.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
g. List two changes that would give a smoother output voltage.
A square waveform of period 2.0 s (figure (a)) is fed into a resistor capacitor circuit as shown in figure (b).
What is the time constant for this RC combination?
Will the capacitor fully charge before the input voltage falls to zero? Justify your answer.
Sketch the waveform for VC, the voltage drop across the capacitor.
Sketch the waveform for VR, the voltage drop across the resistor.
G
E
a.
b.
c.
d.
PR
O
O
FS
20.
PA
Voltage regulators
List three causes for variations in the output of DC power supplies.
22.
a. What is meant by voltage regulation?
b. Why are voltage regulators necessary in circuits?
c. Where are voltage regulators placed in circuits?
d. Compare the output voltage of a voltage regulator with its input voltage.
23
Why can Zener diodes be used as voltage regulators?
24.
What is the breakdown voltage of a Zener diode?
25.
Describe a circuit containing a Zener diode that can provide a constant voltage across a load from a
rectifier.
R
R
EC
TE
D
21.
U
N
C
O
Questions 26 to 28 refer to the following information and circuit diagram. A student sets up the circuit shown
in the figure below.
26.
Vin is 40 V, R1 is 50 Ω, RL is 100 Ω and the breakdown voltage of the Zener diode is 20 V.
a.
b.
c.
d.
e.
What is the voltage drop across the load resistance RL?
Calculate the current through the load resistor RL.
What is the voltage drop across R1?
Calculate the current through R1.
What is the current through the Zener diode?
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
27.
R1 is 4.0 k Ω, RL is 10 k Ω and the breakdown voltage of the Zener diode is 30 V. The input voltage ranges
between 70 V and 100 V.
a. What is the range of voltage drops across R1?
b. Calculate the maximum and minimum currents through R1?
c. What is the current in the load resistor?
d. Calculate the maximum and minimum currents through the Zener diode.
28.
The voltage drop across RL is 12 V, Vin has values between 20 V and 35 V. The minimum current through RL
is 100 mA, and the minimum current through the Zener diode is 8.0 mA.
Consider the circuit shown in figure (a). The characteristic curve of the diode curve shown in figure (b).
EC
TE
D
PA
G
E
PR
O
O
29.
What is the rating of the Zener diode?
What is the minimum voltage drop across R1?
What is the minimum current through R1?
Calculate the value of RL.
Calculate the value of R1.
FS
a.
b.
c.
d.
e.
If the current is measured to be 4.0 mA:
R
R
what is the voltage drop across the diode
what is the voltage drop across the resistor
what is the resistance of the resistor?
Calculate the power dissipated in the resistor.
O
a.
b.
c.
d.
U
N
C
The diode is now reversed, as shown below.
e. What is the voltage drop across the diode?
f. What is the current flowing in the circuit?
g. What is the voltage drop across the resistor?
30.
a. Find the output voltage for the voltage divider shown in circuit (a) below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
O
FS
b. What is the output voltage of circuit (b) if a load of resistance 4.4 kΩ is connected across the output
terminals of the voltage divider?
Find the value of R2 in the voltage divider in the following figure which would give an output voltage of
2.0 V.
32.
Find the voltage drop between A and B in each of the following voltage divider circuits.
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
31.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
TE
D
What happens to the voltage drop across a variable resistor in a two-element voltage divider when its
resistance decreases and the other resistance is unchanged?
34.
Find the value of the unknown resistor in the following voltage dividers.
U
N
C
O
R
R
EC
33.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
PA
TE
D
EC
A thermistor has the characteristic curve shown below.
R
35.
R
Transducers
C
O
a. What is the resistance of the thermistor at the following temperatures?
i. 20 °C
ii. 80 °C
U
N
b. What is the temperature when the thermistor has the following resistances?
i. 4 kΩ
ii. 1.5 kΩ
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
The thermistor from question 35 is placed in a voltage divider as shown.
PA
G
E
PR
O
O
36.
37.
TE
D
a. If the variable resistor is set at 4 kΩ and the temperature is 40 °C, what is the output voltage?
b. An output voltage of 6.0 V is required when the temperature is 80 °C. What must the value of the
variable resistor be?
A voltage divider circuit is set up as shown in the following figure (a). The thermistor has the characteristic
curve shown in figure (b).
Is this a positive or negative coefficient thermistor? Explain your answer.
What is the temperature when the resistance of the thermistor is 5 kΩ?
What is the resistance of the thermistor when the temperature is 90 °C?
What is the value of the variable resistor that gives an output voltage of 3.0 V when the temperature is
90 °C?
e. What happens to the value of the output voltage as the temperature falls and the variable resistor
remains at a fixed value? Explain your answer.
U
N
C
O
R
R
EC
a.
b.
c.
d.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
38.
A temperature sensing system in an oven consists of a thermistor connected into a voltage divider as shown
in figure (a) below. The thermistor has the characteristic curve shown in figure (b).
R
EC
TE
D
PA
G
E
PR
O
O
FS
a. What is the resistance of the thermistor when the temperature in the oven is 100 °C?
b. What is the temperature in the oven when the resistance of the thermistor is 400 Ω?
c. Calculate the resistance of the variable resistor when the temperature in the oven is 200 °C and the
output voltage Vout is 8.0 V.
O
The current–voltage characteristic curve of a diode is shown below.
U
N
C
39.
R
Diodes
The diode is placed in the circuit shown below.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
Calculate the current flowing in the circuit. Express your answer in mA.
Consider the circuit shown in the following figure (a). The characteristic curve of the diode is shown in figure
(b).
TE
D
PA
G
E
PR
O
O
FS
40.
If the current is measured to be 4.0 mA:
EC
a. What is the voltage drop across the diode?
b.
What is the voltage drop across the resistor?
c. What is the resistance of the resistor?
R
The diode is now reversed, as shown below.
U
N
C
O
R
d. What is the voltage drop across the diode?
e. What is the current flowing in the circuit?
f. What is the voltage drop across the resistor?
Photonic transducers
41.
What is a transducer?
42.
Explain the relationship in a light-dependent resistor (LDR) between the resistance and the amount of light
falling on the LDR.
43.
What is the role of LDRs in cameras?
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
44.
A LDR is used in a voltage divider circuit as shown in figure (a) below. The characteristic curve of the LDR is
given in figure (b).
45.
TE
D
PA
G
E
PR
O
O
FS
a. What is the illumination when the LDR has a resistance of 2.0 k Ω?
b. Calculate Vout for this illumination.
c. Describe what happens to Vout as the illumination increases.
A student carries out a practical activity to construct a characteristic curve of an LDR. She obtained the data
in the table below.
R
EC
a. Copy and complete the table, calculating the resistance of LDR for each light intensity, the log of RL and
the log of IL.
b. Plot a graph of log RL against log IL.
c. Determine the relationship between RL and IL from your graph.
d. Is the LDR an ohmic or non-ohmic device? Explain your answer.
a. Describe the structure of LEDs in terms of p-type and n-type semiconductors.
b. What is the meaning of the terms ‘forward biased’ and ‘reverse biased’?
c. Draw circuit diagrams to show how to:
i. forward bias an LED
ii. reverse bias an LED.
47.
Give three examples of a situation where an LED would be preferable to an ordinary light source. Analyse
each case to justify your choices.
48.
Why is it that not all diodes emit light?
49.
Why are limiting resistors placed in series with LEDs?
50.
a. Consider the circuit shown in figure (a) below. The characteristic curve of the light-emitting diode is
shown in figure (b). If the current is measured to be 20 mA:
U
N
C
O
R
46.
i. What is the voltage drop across the LED?
ii. What is the voltage drop across the resistor?
iii. What is the resistance of the resistor?
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
A student carries out a practical activity on an LED. She initially sets up the circuit shown in figure (a) below.
She is able to measure the voltage drop across the diode and the resistor as well as the emf supplied by the
variable DC supply. This enables her to construct the I–V characteristic curve for the diode, as shown in
figure (b).
EC
51.
TE
D
PA
b. The LED is now reversed, as shown in figure (c):
i. What is the voltage drop across the LED?
ii. What is the current flowing in the circuit?
iii. What is the voltage drop across the resistor?
R
When the current in the circuit is 60 mA, calculate the following quantities:
U
N
C
O
R
a. the voltage drop across the diode
b the voltage drop across the resistor
c. the emf of the supply.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
FS
O
PR
O
G
E
52.
If the student from question 51 reverses the polarity of the supply, very little current flows through the circuit.
53.
PA
a. Explain why this occurs.
b. If the emf of the supply is –2.5 V, what is the voltage drop across the diode? Justify your answer.
The LED shown below has a voltage drop across it of 1.8 V and carries a current of 40 mA.
a.
b.
c.
d.
N
C
Refer to the circuit diagram shown below. If the LED has a voltage drop of 1.7 V across it and carries a
current of 20 mA, calculate the value of the limiting resistor if the emf of the power supply is:
9.0 V
12 V
24 V
50 V.
U
54.
O
R
R
EC
TE
D
a. What is the voltage drop across the limiting resistor?
b. What is the value, R, of the limiting resistor?
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd
A student connects a 100 Ω resistor in series with an LED and a 15 V power supply.
56.
Describe the purpose of a photodetector.
57.
a.
b.
c.
d.
e.
O
a. Calculate the current through the LED if it has a voltage drop of 2.0 V across it.
b. What might happen to the LED in this situation?
G
E
PR
O
Describe the operation of a photodiode.
Describe the operation of a phototransistor.
Which device has the fastest response time?
Which device is the most sensitive?
Describe the effect of response time on bandwidth.
Modulation and demodulation
FS
55.
a. What is the bandwidth of an information system?
b. State the bandwidth of a communications system that can carry a maximum frequency of 15 000 Hz.
c. Compare the bandwidths of metal wires and optical fibres.
d. Why does the skin effect occur?
59.
a.
b.
60.
The diagram below shows part of an optical-fibre telephone system.
PA
58.
TE
D
What is optical intensity modulation?
Describe how an analog signal — for example, the electrical output signal from a microphone — can
be used to alter the intensity of light produced by a light-emitting diode.
U
N
C
O
R
R
EC
a. Explain the terms modulation and demodulation as they apply to the transmission of sound by this
system.
b. State a device that could be used as a modulator in this system.
c. State three devices that could be used to demodulate the light signal.
Chapter 17 How can AC electricity charge a DC device?
© John Wiley & Sons Australia, Ltd