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