Electronics

Electronics Use your mouse to move around
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click anywhere on the screen to
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Always move the mouse before
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ã TPS 1
2002

2
Electronics
Introduction
Ohm’s Law
Power Calculations
Resistors in Series and Parallel
Capacitors
Alternating Current
Waveforms
The Potential Divider
Transistor Circuits
Questions

3
Introduction
Basic Concepts
Simple Circuits
Questions
Main Menu

4
Basic Concepts
Return previous slide
Electric current is due to the flow of charge.
In a solid conductor, the charge is carried by electrons.
In a solid conductor, an electric current is due to the flow
of electrons.

5
Basic Concepts
Conductors include:
copper
gold
silver
lead
All metals
And water (not distilled) which is why you should not use mains
appliances in the presence of water.

6
Basic Concepts
Insulators include:
Rubber
Plastic
Most solid non metals
Glass
Glass, unless it is very hot, is one of the best insulators available.

Electric current (I) is measured in ampere (A) - I is the symbol used to
indicate current.
The “amp” is a rather large unit for most electronic applications so we
use the following sub-multiples:
1 mA = 0.001A that is 1 / 1 000 th of an ampere
You already know that 1mm is 1/1000th of an metre so there is nothing
7
Basic Concepts
new here.
1 mA = 0.000 001A that is 1 / 1 000 000 th of an ampere

Common sub-multiples of the volt (less than a volt) include:
8
Basic Concepts
Voltage is measured in volt (V)
1 mV = 0.001V that is 1 / 1 000 th of a volt
1 mV = 0.000 001V that is 1 / 1 000 000 th of a volt
Common multiples of the volt (greater than a volt) include:
1 kV = 1 000 V
1 MV = 1 000 000 V
The kV and the MV are not common in electronics.

Common sub-multiples of the ohm (less than an ohm) include:
9
Basic Concepts
Resistance is measured in ohm (W)
1 m W = 0.001 W that is 1 / 1 000 th of an ohm
1 m W = 0.000 001 W that is 1 / 1 000 000 th of an ohm
This is pronounced micro ohm
Common multiples of the ohm (greater than an ohm) include:
1 k W = 1 000 W
1 M W = 1 000 000 W
The mW and the m W are not common in electronics.

Common sub-multiples of the farad (less than a farad) include:
10
Basic Concepts
Capacitance is measured in farad (F)
1 m F = 0.001 F that is 1 / 1 000 th of a farad
1 m F = 0.000 001 F that is 1 / 1 000 000 th of a farad
1 n F = 0.000 000 001 F that is 1 / 1 000 000 000 th of a farad
this is written in full as a nano farad
1 p F = 0.000 000 000 001 F that is 1 / 1 000 000 000 000 th of a farad
This is written in full as pico farad

Here is a summary of many of the available multiples and sub-multiples
Symbol Prefix Multiplication factor
T tera 1012 1 000 000 000 000
G giga 109 1 000 000 000
M mega 106 1 000 000
k kilo 103 1 000
h hecto 102 100
da deca 101 10
d deci 10-1 0.1
c centi 10-2 0.01
m milli 10-3 0.001
m micro 10-6 0.000 001
n nano 10-9 0.000 000 001
p pico 10-12 0.000 000 000 001
f femto 10-15 0.000 000 000 000 001
a atto 10-18 0.000 000 000 000 000 001
The most
frequently used
are in bold.
11
Basic Concepts

This gold band indicates that the tolerance
of the resistor is ±5% (plus or minus 5 percent).
This means that its resistance is between 3 400 W
and 3 800 W. We say it is nominally 3 600 W.
The third band is red. This means that there are 2
zeros.
The second band is blue. This means the second digit is 6.
12
Basic Concepts
Resistors are marked with a series of coloured rings to give us an idea of
how big their resistance is.
The first band is orange. This means the first digit is
3.
So the resistor is nominally 3 600 W.or 3k6

The colours used for the first three bands and their meanings are as
follows:
Colour Number Number of zeros
Black 0 none
Brown 1 0
Red 2 00
Orange 3 000
Yellow 4 0 000
Green 5 00 000
Blue 6 000 000
Violet 7 0 000 000
Grey 8 00 000 000
White 9 000 000 000
13
Basic Concepts

The colours used for the last band and their meanings are as follows:
Gold ± 5 %
Silver ± 10 %
No band ± 20 %
Resistors are manufactured in “preferred values”. That
means that you can only buy certain values.
The preferred values for resistors with a tolerance of
±20% are: 10,15,22,33,47,68 and 100. These are just
the first two significant figures. You can buy a 1500W
but not a 2000 W.
14
Basic Concepts

15
Basic Concepts
The preferred values for 10% resistors are:
10
12
15
18
22
27
33
39
47
56
58
82
100
&

16
Basic Concepts
The preferred values for 5% resistors are:
10
11
39
12
43
13
47
15
51
16
18
&
56
62
20
68
22
75
24
82
27
91
30
100
33
36

17
Basic Concepts
In summary:
Voltage is measured in
Current is measured in
Resistance is measured in
Capacitance is measured in
Volt (V)
Ampere (A)
ohm (W)
farad (F)
3MV = 3 000 000 V
2kV = 2 000 V
5mV = 0.005 A
7mA = 0.000 007 A
& 1nF = 0.000 000 001 F
Home
1pF = 0.000 000 000 001 F

This is called a block diagram.
•The processor is the decision-making part of the system.
•The input is a sensor that transforms everyday phenomena such as
temperature and heat to an electric signal that the processor can deal
with.
•The output is a device that converts an electric signal from the processor
into something that we want.
18
Simple Circuits
Simple circuits have three main blocks of components in common that
perform the same type of job. These are:
Input Processor Output
Return to menu slide

19
Simple Circuits
Return to previous slide
Examples of input devices include:
Pressure pads
LDRs
Thermistors
Reed switches

20
Simple Circuits
Examples of output devices include:
Lamps
LEDs
Motors
Solenoids

21
Simple Circuits
Examples of basic processors include:
Transistors
Operational amplifiers

22
Simple Circuits
Home
What would the block diagram look like for a system that brought on a
light when it got dark?
LDR Processor Lamp

Questions - Simple Circuits
1 Write down the units of voltage, capacitance, resistance and
current. What is the symbol for current?
2 Write out 22mA and 420 pF in full.
3 Give three examples of input devices.
4 Write out a block circuit diagram for a device that could lift up a
trap door when a beam of light was broken.
5 What is the nominal value and tolerance of this resistor?
6 What would the colours of a 47MW resistor be?
23
ANSWER
ANSWER
ANSWER
ANSWER
ANSWER
ANSWER

Solutions - Simple Circuits
1 Write down the units of voltage, capacitance, resistance and
current. What is the symbol for current?
24
Voltage volt (V)
Capacitance farad (F)
Resistance ohm (W)
Current ampere (A)
The symbol for current is I
Return

25
2 Write out 22mA and 420 pF in full.
22mA = 22 x 0.001 A = 0 . 22 A
420 pF = 420 x 0.000 000 000 001 F = 0 . 000 000 000 42 F
Return

26 Return
3 Give three examples of input devices.
Pressure pads
LDRs
Thermistors
Reed switches

4 Write out a block circuit diagram for a device that could lift up a
trap door when a beam of light was broken.
LDR Processor Motor
27
or
LDR Processor Solenoid
Note that a bulb is not an input device as it gives out light.
Return

The third band is red. This means that there are 2
zeros.
The second band is black. This means the second digit is 0.
The first band is brown. This means the first digit is 1.
28
So the resistor is nominally 1 200 W.
Return

6 What would the colours of a 47MW resistor be?
The third band is blue. This means that there are 6 zeros.
The second band is violet. This means the second digit is 7.
The first band is yellow. This means the first digit is 4.
29
The resistor is nominally 47 000 000 W.
Return

30
Ohm’s Law
Ohm’s Law states that so long as the physical conditions remain
constant, the current through a conductor is proportional to the voltage
across it.
This gives us the formula:
Voltage = current x resistance
V = I R
We can rearrange this equation to give either
R = V / I
or
I = V / R

31
Ohm’s Law
What does it mean?
“Physical conditions remaining constant” - This really means as long
as the temperature remains constant. Usually it does.
“The current through a conductor is proportional to the voltage across
it” - this means that if you double the voltage, you get twice the
current. Triple the voltage and you triple the current etc.
Voltage
Current
Low resistance
High resistance

32
Ohm’s Law
Calculations using Ohm’s law fall into three types:
What is the resistance
What is the current
if ?
if ?
What is the voltage
(Use R = V / I) if ? (Use V = I x R)
E.G.
What resistance could
you use with a 10V
supply to limit the
current to 15mA?
R = V / I = 10 / 0.015
667 W so use 680 W
(Use I = V / R)
E.G.
A 430 W resistor
protects an LED in a
5V circuit. What is
the current through
the LED?
I = V / R = 5 / 430
= 12 mA
E.G.
12mA runs through
a prorctive resistor
of resistance 820 W.
What is the voltage
across the resistor ?
V = IR = 0.012x820
= 9.84 V

The voltage across the diode is 0.7
V and the cell produces 1.5 V.
What is the current through the
resistor?
0.7 V
820 W Diode
If you can’t see how to do it straight away, write the values
given onto the diagram.
33
Ohm’s Law
1.5V
Voltage across the resistor = 1.5V (provided by the cell)
- 0.7V (lost across the diode) = 0.8V
Using I = V / R = 0.8 / 820 = 1mA

Home
1 A power supply drives a current of 500mA through a bulb with a
working resistance of 3W. What voltage is the power supply?
2 A power supply provides 12 V to a bulb passing 3 A. What is the
working resistance of the bulb?
3 A 47 kW resistor has a pd of 9 V across it. What current passes
through the resistor?
4 An 18V power supply is placed across a resistor of resistance
10kW. What current will flow through the resistor?
5 The effective resistance of a small motor is 5 W. What current
passes through it if a cell of voltage 6 V is placed across it?
Solution 1 Solution 2 Solution 3 Solution 4 Solution 5
34
Ohm’s Law - Questions

35
Ohm’s Law - Solutions
V = I x R
so V = 0.5 x 3
=1.5 volt
Return

2 A power supply provides 12 V to a bulb passing 3 A. What is the
working resistance of the bulb?
36
V = I x R
So R = V / I
= 12 / 3
= 4 A
Return

37
3 A 47 kW resistor has a pd of 9 V across it. What current passes
through the resistor?
V = I x R
So I = V / R
= 9 / 47 000
= 0.000 191 A
= 191 mA
Return

38
4 An 18V power supply is placed across a resistor of resistance
10kW. What current will flow through the resistor?
V = I x R
so I = V / R
= 18 / 10 000
= 0.001 8 A
= 1.8 mA
Return

5 The effective resistance of a small motor is 5 W. What current
passes through it if a cell of voltage 6 V is placed across it?
39
V = I x R
So I = V / R
= 6 / 5
= 1.2 A
Return

40
Power Calculations
The detail from the bottom of an
electrical appliance shown here
gives a very useful, commonly
used method of writing the power
of the appliance.
20 VA is exactly the same as 20 W (20 watts).
The more powerful an appliance is, the greater the number will be.
An electric fire might well be 2 or 3 kW (2 000 or 3 000 W).
1W is sometimes called 1VA because you can calculate the
power by multiplying “the volts by the amps”!
Power = current x voltage or P = I V
Return main menu

This device runs from a 230V mains supply. What can we learn from
this information? Well, we know that P = IV; we also know the
voltage and the power, so we can calculate the current I.
41
Power Calculations
Remember that :
20VA means a power of 20W
and that
P = I x V
I = P / V
So I = 20 / 230
= 87 mA

Can you match these typical power ratings with the device that they
describe?
42
Power Calculations
Light emitting diode
Halogen desk lamp
Power station
Electric lamp
Torch Bulb

43
Power Calculations
Light emitting diode
Halogen desk lamp
Power station
Electric lamp
Torch Bulb

Power Calculations - the formulae
44
Power = current x voltage
P = IV
so I = P / V and V = P / I
But from Ohm’s Law, V = I x R
So P = I x IR
P = I2R
And from Ohm’s Law, I = V / R
So P = (V/R) x V
P = V2/R

Power Calculations - Questions
Home
1 A diode has a voltage of 0.7V across it and a current of 100mA
flowing through it. What is the power dissipated in the diode?
2 A wire carries a current of 5 mA and the power dissipated in the
wire is 2.5 mW. What is the voltage across the wire?
3 What is the current passing through a coil that dissipates 40mW
when a voltage of 5V is applied across it?
4 A current of 10 mA passes through a 10W resistor. What is the
power dissipated in the resistor?
5 A voltage of 9 V is applied across a 10 k W resistor. What power
is dissipated in the resistor?
45

Power Calculations - Solutions
1 A diode has a voltage of 0.7V across it and a current of 100mA
flowing through it. What is the power dissipated in the diode?
46
P = IV
so P = 0.1 x 0.7
= 0.07
= 70 mW

2 A wire carries a current of 5 mA and the power dissipated in the
wire is 2.5 mW. What is the voltage across the wire?
47
P = IV
So V = P / I
= 0.000 0025 / 0.005
= 0.000 5 W
= 0.5 mW
= 500 mW

3 What is the current passing through a coil that dissipates 40mW
48
P = IV
So I = P / V
= 0.04 / 5
= 0.008 W
= 8mW

4 A current of 10 mA passes through a 10W resistor. What is the
power dissipated in the resistor?
49
P = I2R
= 0.012 x 10
= 0.001
=1mW

5 A voltage of 9 V is applied across a 10 k W resistor. What power
is dissipated in the resistor?
50
P =V2/R
= 92 / 10 000
0.0081
= 8.1 mW

Resistors are said to be connected in series when the same current has to
pass through each resistor i.e. the current does not have to split.
These three resistors are connected in series.
51
And so are these 5 resistors
Return to main menu

Resistors are said to be connected in parallel when the current has to
split to pass through each resistor i.e. the current through each resistor
might not be the same.
These three resistors are connected in
52
parallel.
And so are these two.

R1 R2 R3
47 MW 47 MW 47 MW
These three resistors connected in series, could be replaced by
one resistor of resistance 141MW.
53
That is 47 MW + 47 MW + 47 MW = 141 MW
(This is not available so we might use a 150 MW)
As a general formula we could write:
R total= R1 + R2 + R3 or R = R1 + R2 + R3

54
R1
R2
R3
The resistance of each of the resistors
in this parallel network is 47 MW.
The effective resistance of three
resistors is 15.7 MW (use 16 MW).
You would expect the resistance to be less than any of the individual
resistances in the network as there are three possible routes for the
electricity to take.
The formula used to add the resistances is:
1 = 1 + 1 +
1
R R R R total
1 2 3
or
R R R
1 2 3
RTotal = + +
R R R R R R
1 2 2 3 3 1

Resistors in Series and Parallel - Questions
1 What is the combined resistance of a 39k W and a 47k W resistor
connected in series?
connected in parallel?
3 What is the combined resistance of a 10k W, 20k W and 47k W
resistor connected in series?
resistor connected in parallel?
5 Suppose you need a 30k W resistor but you only have a 27k W
and a handful of assorted resistors. What other resistor would you search
for and how would you connect them together?
55
Home

Resistors in Series and Parallel - Answers
56
So R = 39 000 + 47 000 = 86 000
= 86kW

1 1 1
R R R Total
1 2
57
So 1/R = (1/39 000) +(1/47 000)
= 0.00002564 + 0.00002128 = 0.00004692
So R = 1/0.00004692 = 21 314W
= 21.3kW
= 22kW
= +

3 What is the combined resistance of a 10 kW, 20 kW and 47 kW
resistor connected in series?
58
So R = 10 000 + 20 000 + 47 000
= 77 000 W
= 77 k W

4 What is the combined resistance of a 10kW, 20kW and 47kW
1 = 1 + 1 +
1
R R R R Total
1 2 3
The numbers here could get very big so let us omit three zeros and give
the answer in kW as all the resistances are in kW anyway.
59
R = (1/10) +(1/20) +(1/47)
= 0.1000 + 0.0500 + 0.0213 = 0.1713
=5.84 kW
Use 5.8 kW

5 Suppose you need a 30 kW resistor but you only have a 27 kW
You are looking to increase the resistance by adding another resistor.
This can only be done by adding a resistor in series.
60
We know that Rtotal is 30 kW and R1 is 27 kW
So 30 = 27 + R2
R2 = 30 - 27
= 3 kW

Capacitors
Capacitors store charge. The greater the
voltage that you apply to them, the greater the
charge that they store. In fact the ratio of the
charge stored to the voltage applied is called
the capacitance.
Capacitance = Charge / Voltage
61
or C = Q / V
Capacitance is measured in farad (F) but the farad is a large unit of
capacitance so you usually see microfarad mF (millionth of a farad 10-6
or 0. 000 001 F), nanofarad nF (10-9) or picofarad pF (10-12).
The capacitor in the picture is a 470 mF (0.000 47 F) polar (you must
connect it the correct way around in the circuit) capacitor rated at 40V
(the working voltage should not exceed 40V).
Return to main menu

62
Capacitors - Symbols
This is a non-polar capacitor -
it does not matter which way
around you place it in the
circuit.
This is a polar capacitor. It
is essential that the
capacitor is connected into
the circuit the correct way
around.

63
Capacitors
Capacitor Characteristics
Time Constant Calculations
Capacitors in Series
Capacitors in Parallel
Questions

64
Closing the switch allows the capacitor
to charge. As this happens, the voltage
across the capacitor will rise in line
with the fall of current through it as it
becomes fully charged.
V
A
Voltage
Current
Time
Time

65
The circuit has now been adapted so
that closing the switch allows the
capacitor to discharge through the
resistor. Note now that the current will
fall as the voltage falls.
V
A
Voltage
Current
Time
Time

66
The circuit on the left allows us to
investigate the charging and discharging
of a capacitor simply.
Connecting the flying lead S to point X
will charge the capacitor from the cell
through the resistor R.
Connecting the flying lead S to Y will
then discharge the capacitor through the
resistor.
S
It has been found that increasing either the capacitance or the
resistance will increase the time taken for the capacitor to charge.
A
V
X Y
C
R

67
When the flying lead S is connected to X,
the capacitor will charge up through the
resistor.
At first there will be little or no charge in the
capacitor so the current flows into the
capacitor (via the resistor), quite rapidly.
The current through the resistor develops a
voltage over the resistor. The voltage across
the capacitor will be proportional to the
charge in it. Since the charging has only just
begun, it will be small but growing.
S
The capacitor begins to charge: It gets harder for more charge to flow
into the capacitor so the current decreases. As the charge on the
capacitor is increasing, the voltage across it increases too.
A
V
X Y
C
R

Home
The capacitor is said to be fully charged..
68
A
V
X Y
C
S
R
Eventually the capacitor will “fill”. This
really means that it approaches the condition
such that the voltage across it is equal to the
supply voltage.
There will no longer be any current flowing.
The time taken to achieve this increases with
increased capacitance and /or resistance.
Increasing the supply voltage makes no difference to the time taken
for the voltage across the capacitor to approach the voltage across the
supply.

69
The time constant is the time taken for:
the current or voltage to have fallen to 37% of its original value
or
the voltage to have risen to 63% of its original value
We can calculate the time constant for a circuit by multiplying the
capacitance of the capacitor by the resistance of the resistor:
T = C x R
The units of the time constant are seconds if the resistance is in ohms
and the capacitance in farads.
Return to menu

E.G. One
A 10M0 resistor is connected in series with a 470 pF capacitor. How
long will it take to discharge the capacitor to 37V from an initial voltage
of 100V?
Note that the voltage is falling to 37% of its initial value, so we are
looking at one time constant.
70
Using T = C R
T = 0.000 000 000 470 x 10 000 000
= 0.004 7 s

E.G. Two
A 10M0 resistor is connected in series with a capacitor. If the time
constant is 0.001s, what is the capacitance of the capacitor?
71
Using T = C R
0.001 = C x 10 000 000
C = 0.001 / 10 000 000
= 0.000 000 000 1 F
= 100 pF
Home

72
C1 C2 C3
These three capacitors are
connected in series. Their combined
capacitance is given by:
1 = 1 + 1 +
1
C C C C total
1 2 3
or
C C C C Total + +
1 2 3
C C C
1 2 3
=
Return to menu

Home
4 What is the combined capacitance of a 10 mF, 20 mF and 47 mF
capacitor connected in series?
C C C C Total + +
1 2 3
C C C
The numbers here could get very small so let us omit 5 zeros and give
the answer in mF as all the capacitances are in mF anyway.
73
C = (10 x 20 x 47) / (10 + 20 + 47)
= 9400 / 77
=122 mF
1 2 3
=
C1 C2 C3

74
C1
C2
C3
These three capacitors are connected in
parallel with each other.
Note that because they are in parallel,
they must have the same voltage
across each other.
The combined capacitance of the network
of capacitances is given by:
Ctotal = C1 + C2 + C3
Return to menu

Home
Suppose that the capacitances are 10 mF, 20
mF and 47 mF.
75
C1
C2
C3
So C = 10 + 20 + 47
= 77 mF

76
Capacitors - Questions
1 What is the combined capacitance of a 10pF and a 20pF capacitor
3 What is the time constant for a 1000 pF capacitor connected to a
100kW resistance?
4 What capacitance would you need to combine with a 200 mF
capacitor to give the combinations a capacitance of 100 mF? How would
you connect them together to achieve this?
5 What resistor could you connect with a 47 mF capacitor to give a
time constant of 1.83 s ?

C C C C Total + +
C C C
77
Capacitors - Answers
1 2 3
the answer in pF as all the capacitances are in pF anyway.
C = (10 x 20) / (10 + 20)
= 200 / 30
=6.7 pF
1 2 3
=
C1 C2 C3
Return to menu

78
C1
C2
C3
So C = 10 + 20
= 30 pF
Return to menu

79
3 What is the time constant for a 1000 pF capacitor connected to a
100kW resistance?
T = C x R
= (1000 x 0.000 000 000 001) x 100 000
= 0.000 1 s
Return to menu

C C C C Total + +
C C C
80
4 What capacitance would you need to combine with a 200 mF
capacitor to give the combinations a capacitance of 100 mF? How would
you connect them together to achieve this?
In order to get a smaller capacitance, you need to connect them in series.
1 2 3
1 2 3
=
C1 C2 C3
So 100 = 200 x C2 / (200 + C2)
100 (200 + C2) = 200C2
20 000 + 100C2 = 200C2
20 000 = 200C2 - 100C2
100 C2 = 20 000
C2 = 200 mF Return to menu

81
5 What resistor could you connect with a 47 mF capacitor to give a
time constant of 1.83 s ?
T = C x R
1.83 = 0.000 047 x R
R = 1.83 / 0.000 047
= 38 936 W
Use 39 kW
Return to menu

82
Alternating Current
Direct current (DC) is the current that comes from a cell or battery.
It is unidirectional. That is to say that the net drift of electrons is in one
direction. This one direction will always be from positive to negative for
electrons but negative to positive for conventional flow.
It is easier to convert voltages from one value to another if the direction
of the current is rapidly changing.
This is called an alternating current (AC).
Return to main menu

83
Alternating Current
Alternating current has some strange properties:
•it can appear to pass through a capacitor
•it produces the discharges that you see in a plasma ball
•it can be stepped up (to higher voltages and lower current)
•it can be stepped down (to lower voltages and higher current)
Mains voltage is always due to an alternating current. It is used because
it can be stepped up or down easily.

84
Alternating Current
Throughout Europe, mains voltage is supplied at a frequency of 50 Hz.
You will remember that Hz is the abbreviation for hertz - the unit of
frequency.
This means that the electricity goes through one complete cycle 50 times
every second.
This means that the voltage will:
•start from zero and build up in one direction until it reaches a
maximum value (about 325 V).
•Fall back to zero
•Change direction and start to build up to a maximum value (about -
325 V)
•Fall back to zero
•50 times per second
Alternating voltages (and currents) can have extremely high frequencies.
The current that produces radio waves can be many MHz (millions of
hertz). Return to previous slide

85
Alternating Current - Questions
Home
1 What is the frequency of the mains voltage in the UK and
Europe?
2 What is the frequency of the mains voltage in the USA?
3 How long is a complete cycle of mains ac at 50 Hz?
4 How many times does the electricity in 50 Hz ac mains, change
direction?
5 Write out 250 MHz in full.

Alternating Current - Answers
1 What is the frequency of the mains voltage in the UK and
Europe? 50Hz
2 What is the frequency of the mains voltage in the USA? 60Hz
3 How long is a complete cycle of mains ac at 50 Hz?
86
1/50th of a second = 0.02s
Home
4 How many times does the electricity in 50 Hz ac mains, change
direction?
Twice each cycle so 100 times
5 Write out 250 MHz in full
250 000 000 hertz
Return to menu

87
Waveforms
As alternating currents and voltages vary with time, it is useful to have a
graphical representation of them.
The electronic device that does this for us is called an oscilloscope.
The essence of an oscilloscope is its ability to plot a graph for us,
showing the variation of voltage with time. It is particularly useful
because it works extremely quickly and it barely interferes with the
circuit from which you are taking the measurements.
When you first look at an oscilloscope, the number of knobs, levers and
buttons can be bewildering but you will soon get used to it. In fact, you
rarely need to use most of them.
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88
Waveforms
Screen showing the trace On / off switch
x input
The trace shows a graph with voltage on the y axis (vertical) and time on
the x axis (horizontal).

89
Waveforms
The waveform produced by mains voltage looks like this:
Voltage
time
This shape of waveform is called a sinusoidal wave.

90
Waveforms
Voltage
time
The arrows indicate a complete cycle. A cycle is the time that it takes to
reach the next adjacent identical position on the waveform.

Waveforms
91
Voltage
time
This arrow indicates the peak voltage. For mains it is about 325V in the
UK. The value of 230V that is quoted is the peak divided by the square
root of 2. (It is the dc equivalent voltage that would produce the same
heating effect as an ac with a peak voltage of 325V. Dividing by root 2
only works for sinusoidal waveforms). The peak voltage is the maximum
voltage.

Waveforms
92
Voltage
time
This arrow indicates the peak to peak voltage. Notice that the peak to
peak voltage is twice the peak voltage. For mains it is about 650V in the
UK and Europe.

93
Waveforms
Another common waveform called the square wave looks like this:
Voltage
time

94
Waveforms
Voltage
time
This arrow indicates the peak voltage.

95
Waveforms
Voltage
time
This arrow indicates the peak to peak voltage. It will be twice the peak
voltage

96
Waveforms
Another common waveform called the saw-tooth wave looks like this:
Voltage
time

97
Waveforms
Voltage
time
This arrow indicates the peak voltage.

98
Waveforms
Voltage
time
This arrow indicates the peak to peak voltage.

99
Waveforms - Questions
Home
1 What is the name of the
trace shown on the oscilloscope?
2 How many complete cycles
can you see?
3 If the scale on the y-axis is 2V per division, estimate the peak
voltage. What is the peak to peak voltage?
4 If the scale on the x-axis is 2ms per division, estimate the
length of a cycle.
5 Using your answer to question 4, calculate the frequency of
the wave.

100
Waveforms - Answers
1 What is the name of the
trace shown on the oscilloscope?
Return to menu

101
Waveforms - Answers
2 How many complete cycles
can you see?
ONE TWO THREE
Return to menu

3 If the scale on the y-axis is
2V per division, estimate the peak
voltage. What is the peak to peak
voltage?
102
Waveforms - Answers
5 divisions
so 5 divisions x 2 V per division
= 10V
Return to menu

103
Waveforms - Answers
4 If the scale on the x-axis is
2ms per division, estimate the length
of a cycle.
Count as many complete cycles as you can to get as accurate an
answer as possible.
3 cycles is about 35 divisions
1 cycle is about 11.7 divisions
11.7 divisions is 11.7 x 2 ms
= 23.4 ms
Return to menu

104
Waveforms - Answers
5 Using your answer to question 4, calculate the frequency of the
wave.
From question 4 one cycle is about = 23.4 ms
23.4ms = 23.4 x 0.001s
= 0.0234s
How many of these can we get into 1 second?
= 1 / 0.0234
= 42.7 Hz
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105
Voltage is sometimes referred to as
potential difference. The potential divider
simply divides up a potential or voltage.
In its simplest form it is two resistors
placed across a power supply. The voltage
across of each resistor is less than the
supply voltage. Adding the voltage across
each resistor will give the supply voltage.
It is probably easiest to understand if you
look at the diagram.
+9V
0 V
Here the power supply is 9V. Note that both
the resistances are the same.
The voltage from the supply will be split (divided) equally as 4.5V. Of
course 4.5V + 4.5V = 9V.
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106
+9V
0 V
In this potential divider circuit, the
resistances are not the same.
2 000 W 1 000 W
The bigger resistance here
means that there will be a
bigger voltage here.
The smaller resistance here
means that there will be a
smaller voltage here.
It is 2/3rds
of the
resistance.
It is 2/3rds
of the
resistance.
It is 1/3rd
of the
resistance.
It is 1/3rd
of the
resistance.
It develops
2/3rds of
the
voltage.
It develops
2/3rds of
the
voltage.
It develops
1/3rd of
the
voltage.
It develops
1/3rd of
the
voltage.
2/3 x 9V = 6V
1/3 x 9V = 3V
If you would like to work through this again, step back through
the sequence using the left arrow key.

107
+V
0 V
R1
R2
V1
V2
You can calculate the voltage across each
resistor using a formula too.
VV11 VV 11 == VV xx RR11 // ((RR11 ++ RR22))
V1 = 9 x 2 000 / (1 000 + 2000) = 9 x 2 / 3 = 6V
VV 22 == VV xx RR22 // ((RR11 ++ RR22))
V1 = 9 x 1 000 / (1 000 + 2000) = 9 x 1 / 3 = 3V

108
The potential divider does not have to be made up of two fixed resistors.
One of them could be variable, or even both.
R1
R2
V1
V2
As R1 increases so does V1 but V2
will fall.
As R1 decreases so does V1 but V2
will rise.
It is just as if there is only one cake to
go around (the voltage). If R1
increases then V1gets more cake so
there is less left for V2!

109
Now the variable resistor has been moved to the lower position in the
network..
R1
R2
V1
V2
As R2 increases so does V2 but V1
will fall.
As R2 decreases so does V2 but V1
will rise.
It can be handy to change the position
of the variable resistor. Later you will
see that it can change the action of a
transistor circuit so make sure that
you follow it.

110
+V
0V
With this potential divider, the “tap” in the
middle is a “slider”. It probably moves
along a track of carbon.
R1 and V1 will be the resistance and
voltage “above” the slider.
R1 V1
R2 and V2 will be the resistance and
voltage “below” the slider.
R2 V2
As R1 increases, R2 decreases. This will
result in V1 increasing and V2 decreasing.
As R1 decreases, R2 increases. This will
result in V1 decreasing and V2 increasing.

Here is a component that could be used as a potential divider.
The black ring is
the carbon track.
You adjust it by
putting a
screwdriver in
here and turning
the outer metal
ring.
111

You adjust it by
putting a
screwdriver in
here and turning
the outer metal
ring.
112
Here is another:

It is possible to use many different components that vary their resistance
in a potential divider circuit. Here are a few that you might find and the
physical conditions that change their resistance.
Light Dependent Resistor (LDR) - decreases resistance with increased
illumination.
Thermistor - decreases its resistance with increased temperature
(negative temperature coefficient).
Microphone - changes resistance with sound.
Strain gauge - changes its resistance when stressed.
Photodiode - decreases resistance with increased illumination.
113

The Potential Divider - Questions
1 A 2k and a 3k resistor are used in potential divider using a 10V
supply. Sketch a possible set up and label the resistors and the voltages
across them.
2 Suggest two possible fixed resistors that could be used to obtain
3V from a 15V supply.
3 A 27k and a 62k resistor are used in potential divider using a
12V supply. Sketch a possible set up and label the resistors and the
voltages across them.
4 A 15V supply is attached across a potential divider. If one of the
resistors is a 390k and there is a voltage of 9V across the other, what is
the second resistance?
5 A potential divider is created from a fixed resistor and an LDR.
Explain how the network produces different voltages.
114
Home
Answer
Answer
Answer
Answer
Answer

The Potential Divider - Answers
1 A 2k and a 3k resistor are used in potential divider using a 10V
supply. Sketch a possible set up and label the resistors and the voltages
across them.
115
+10V
0 V R 1 = 2 000 W
V1= 4V
V2 = 6V
VV11 VV 11 == VV xx RR11 // ((RR11 ++ RR22))
V1 = 10 x 2 000 / (3 000 + 2000) = 10 x 2 / 5 = 4V
VV 22 == VV xx RR22 // ((RR11 ++ RR22))
V1 = 10 x 3 000 / (3 000 + 2000) = 10 x 3 / 5 = 6V
R 2 = 3 000 W
Return to menu

2 Suggest two possible fixed resistors that could be used to obtain
3V from a 15V supply.
VV11 VV 11 == VV xx RR11 // ((RR11 ++ RR22)) So 3 = 15 x R1 / (R1 + R2)
We could use more or less any combination. However, they should be
quite high so that they do not drain a lot of current.
116
Let us choose R1 as being 100k.
3 = 15 x 100 / (100 +R2)
3(100 + R2) = 1 500
300 + 3R2 = 1 500
3R2 = 1 500 - 300 = 1 200
R2 = 1 200 / 3 = 400k.
Return to menu

3 A 27k and a 62k resistor are used in potential divider using a
12V supply. Sketch a possible set up and label the resistors and the
voltages across them.
VV11 VV 11 == VV xx RR11 // ((RR11 ++ RR22))
V1 = 12 x 27 / (27 + 62) = 12 x 27 / 89 = 3.64V
VV 22 == VV xx RR22 // ((RR11 ++ RR22))
117
+12V
0 V R 1 = 27kW
V1= 3.64V
V2 = 8.36V
V1 = 12 x 62 / (27 + 62) = 12 x 62 / 89 = 8.36V
R 2 = 62kW
Return to menu

4 A 15V supply is attached across a potential divider. If one of the
resistors is a 390k and there is a voltage of 9V across the other, what is
the second resistance?
118
So V1 = 15 - 9 = 6V
VV11 VV 11 == VV xx RR11 // ((RR11 ++ RR22))
Now 6 = 15 x R1 / (R1 +390)
6 (R1 + 390) = 15 R1
6 R1 + 2 340 = 15 R1
9R1 = 2 340
R1 = 2 340 / 9 = 260k so use 270k
Return to menu

5 A potential divider is created from a fixed resistor and an LDR.
Explain how the network produces different voltages.
In bright illumination the LDR’s resistance falls. The voltage across the
LDR will consequently fall while the voltage across the fixed resistance
will rise.
As it goes dark, the LDR’s resistance will increase, increasing the
voltage across the LDR. As this happens, the voltage across the resistor
will fall.
119
Return to menu

The transistor is a three connection component that can be used either as
an amplifier or a switch.
+V
0V
120
Transistor Circuits
+0.7V
base
collector
emitter
Essentially the circuit is set up so as to try to force electrons through the
emitter and out of the collector. This might be to light a bulb.
However, under normal circumstances, there is a very high resistance
between the emitter and the collector so the bulb will not light.
If we make the base go positive, the collector / emitter junction conducts
and the bulb will light. Return to menu

121
Transistor Circuits
The clever part now is to control the base bias voltage that turns the
transistor on.
+V
0V
The base bias voltage is the voltage between
the base and the emitter. If it is anything
much less that 0.7V, the transistor will be off.
The transistor switches
on when it is 0.7V.
You should never allow the base bias voltage
to get too high as this will overheat the base
and burn out the transistor. For this reason
you will frequently find a resistor connected
to the base.
c
e
b
Rb
Return

Correct selection of the
two resistors Rand R1 2
will take the base to
0.7V and turn the
transistor on.
Suppose Rwas much
2 higher than R. The
1voltage across Rwould
2 be high so the transistor
would switch on. 122
Return
Transistor Circuits
transistor on.
+V
0V
c
e
b
This can be achieved
using a potential divider.
Rb
R1
R2

123
Transistor Circuits
transistor on.
+V
0V
c
e
b
Rb
R1
R2
Now suppose R2 was an
LDR.
In the bright light, its
resistance would be low
so the voltage across it
would be low, the
transistor switched off
and the lamp off.
But suppose that it
now goes dark!
Return

It has just gone dark!
The resistance of the
LDR rises.
The voltage across the
LDR rises.
The base bias voltage
reaches 0.7V
The transistor
switches on. The bulb lights. 124
Return
Transistor Circuits
transistor on.
+V
0V
c
e
b
Rb
R1
R2

125
Transistor Circuits
transistor on.
+V
0V
c
e
b
Rb
R1
R2
Suppose that we now
swap the positions of
the resistor and the
LDR.
The bulb will now come
on in daylight! It might
be useful as a warning
light circuit in certain
circumstances.
Return

126
Transistor Circuits
transistor on.
+V
0V
c
e
b
Rb
R1
R2
Now let us consider:
•Ib the base current
that flows into the
transistor
•Ie the emitter
current that flows
out of the transistor
Ib
Ie
•IC the collector
current that flows
into the transistor
Ie = Ib + Ic
Ic
Return

127
Transistor Circuits
transistor on.
+V
0V
c
e
b
Rb
R1
R2
Ib
Ie
Ie = Ib + Ic
Ic The base current will
be very small as it has
passed through R1 and
Rb so it is almost true
that Ie = Ic.
The ratio of Ic : Ib is
important. It shows
that the transistor is
amplifying. It is often
around about 100. Return

Ie = Ib + Ic
That is to say that the Ic
collector current is a
always a constant
amount bigger than
the base current.
Feed a small current
to the base and you
get a big current in
the collector.
128
Transistor Circuits
transistor on.
+V
0V
c
e
b
Rb
R1
R2
Ib
Ie
Return

129
Transistor Circuits
transistor on.
+V
0V
c
e
b
Rb
R1
R2
Ib
Ie
Ie = Ib + Ic
Ic The ratio is called hfe.
hfe = Ic / Ib
Return

130
Click on a
component to find
out what it does.
Transistor Circuits
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

131
Transistor Circuits
Click on a
component to find
out what it does.
Capacitor
This stores charge. It
acts as a time delay
to any switching. If
the transistor is on
and tries to go off, it
will act as a reservoir
and keep the
transistor on for a
while longer.
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

132
Click on a
component to find
out what it does.
LDR Light
Dependent Resistor
Its resistance
decreases with
increased
illumination. In the
dark, the resistance
goes up turning the
transistor on.
Transistor Circuits
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

133
Click on a
component to find
out what it does.
Base Bias Resistor
This fixed resistor
protects the base
from too much
current.
Transistor Circuits
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

134
Click on a
component to find
out what it does.
Potential Divider
The LDR and R2 are
a potential divider.
Transistor Circuits
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

135
Click on a
component to find
out what it does.
Transistor
A small voltage at
the base will allow
current to flow
through the emitter
from the collector.
Transistor Circuits
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

136
Click on a
component to find
out what it does.
Diode - this only
allows current to
flow in the direction
of the arrow head.
Rapid changes in the
magnetic field of the
relay can cause high
voltage that would
damage the
transistor.
The diode
diverts the
currents
formed by
this
process.
Transistor Circuits
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

137
Click on a
component to find
out what it does.
Relay - current
through the relay
produces a magnetic
field that throws a
switch in another
external circuit. The
external circuit can
be a much higher
powered circuit.
Transistor Circuits
+V
0V
c
e
b
Rb
R1
R2
C1
Relay
CONTINUE
Return

Transistor Circuits - Questions
1 The collector current in a circuit is 120mA when the base current
is 3mA. What is hfe and the emitter current?
2 Why do we connect a resistor directly to the base of the
transistor?
3 Sketch a circuit that will throw a relay in the dark that in turn will
turn on a switch.
4 Why should a diode be connected across a relay in the collector
circuit of a network?
5 Explain what happens when the resistance of the base bias
resistor falls in a transistor circuit controlling a motor.
138
Answers:
1 2 3 4 5

Transistor Circuits - Answers
1 The collector current in a circuit is 120mA when the base current
is 3mA. What is hfe and the emitter current?
139
hfe = Ic / Ib
= 120 / 3
= 40
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2 Why do we connect a resistor directly to the base of the
transistor?
The resistor limits the current entering the base. This stops the base from
overheating due to excessive currents which would burn the transistor
140
out.
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3 Sketch a circuit that will throw a relay in the dark that in turn will
turn on a switch.
141
+V
0V
c
e
b
R1
C1
Relay
LDR
Return to menu

4 Why should a diode be connected across a relay in the collector
circuit of a network?
The relay can create large voltages due to rapid changes in magnetic
fields as they switch off. The diode provides a short circuit for the
current due to this voltage. As diodes only carry current in one direction,
the diode connected to the relay will have no effect in normal operation
142
of the relay.
Return to menu

5 Explain what happens when the resistance of the base bias
resistor falls in a transistor circuit controlling a motor.
As the resistance falls, so will the voltage across it.
The voltage across the base will consequently fall.
143
This will turn off the transistor.
The collector current will fall to zero (or extremely close to zero).
The motor will switch off.
Return to menu

144
Electronics Questions
3 What is the current passing through a heater that dissipates 40W
Return to menu

5 Suppose you need a 50k W resistor but you only have a 47k W
8 What is the time constant for a 1000 mF capacitor connected to a
10MW resistance?
145

9 If the scale on the y-axis is 4mV per division, estimate the
peak voltage. What is the peak to peak voltage?
10 If the scale on the x-axis is 50 ms per division, estimate the
length of a cycle. Using your answer, calculate the frequency of the
wave.
146

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The third band is black. This means that there are no zeros.
The second band is green. This means the second digit is 5.
The first band is violet. This means the first digit is 7.
148
So the resistor is nominally 75 W.
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Electronics Answers

149
V = I x R
so V = 0.15 x 10
=1.5 V
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Electronics Answers

3 What is the current passing through a coil that dissipates 40W
150
P = IV
So I = P / V
= 40 / 10
= 4 W
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Electronics Answers

4 What is the combined resistance of a 30kW, 20kW and 47kW
R R R R Total + +
1 2 3
R R R
1 2 3
151
=
The numbers here could get very big so let us omit three zeros and give
the answer in kW as all the resistances are in kW anyway.
R = (30 x 20 x 47) / (30 + 20 + 47)
= 28 200 / 97
=291 kW
Use 300 kW
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Electronics Answers

5 Suppose you need a 50 kW resistor but you only have a 47 kW
You are looking to increase the resistance by adding another resistor.
This can only be done by adding a resistor in series.
152
We know that Rtotal is 50 kW and R1 is 47 kW
So 50 = 47 + R2
R2 = 50 - 47
= 3 kW
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Electronics Answers

C C C C Total + +
1 2 3
C C C
the answer in pF as all the capacitances are in pF anyway.
153
C = (47 x 20) / (47 + 20)
= 940 / 67
=14 pF
1 2 3
=
C1 C2 C3
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Electronics Answers

154
C1
C2
C3
So C = 47 + 20
= 67 pF
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Electronics Answers

8 What is the time constant for a 1000 mF capacitor connected to a
10MW resistance?
155
T = C x R
= (1000 x 0.000 001) x 10 000 000
= 10 000 s
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Electronics Answers

9 If the scale on the y-axis is
4mV per division, estimate the
peak voltage. What is the peak to
peak voltage?
156
5 divisions
so 5 divisions x 4mV per division
= 20mV
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Electronics Answers

Electronics Answers
10 If the scale on the x-axis is 50
ms per division, estimate the length of
a cycle.
Count as many complete cycles as you can to get as accurate an answer
as possible.
157
3 cycles is about 35 divisions
1 cycle is about 11.7 divisions
11.7 divisions is 11.7 x 2 ms
= 23.4 ms

10 continued Using your previous answer, calculate the frequency of
the wave.
From question 4 one cycle is about = 23.4 ms
158
23.4 ms = 23.4 x 0.000 001s
= 0.000 023 4s
How many of these can we get into 1 second?
= 1 / 0.000 023 4s
= 42.7 kHz
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Electronics Answers

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Introduction
Ohm’s Law
Power Calculations
Capacitors
Alternating Current
Waveforms
Transistor Circuits
Components
Questions

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