Checklist
ELECTRICITY AND ELECTRONICS
The knowledge and understanding for this unit is given below.
Electric fields and resistors in circuits
1.
State that, in an electric field, an electric charge experiences a force.
2.
State that an electric field applied to a conductor causes the free electric charges in it to move.
3.
State that work W is done when a charge Q is moved in an electric field.
4.
State that the potential difference (V) between two points is a measure of the work done in moving one coulomb of charge between the two points.
5.
State that if one joule of work is done moving one coulomb of charge between two points, the potential difference between the points is one volt.
6.
State the relationship V = W/Q.
7.
Carry out calculations involving the above relationship.
8.
State that the e.m.f. of a source is the electrical potential energy supplied to each coulomb of charge which passes through the source.
9.
State that an electrical source is equivalent to a source of e.m.f. with a resistor in series, the internal resistance.
10.
Describe the principles of a method for measuring the e.m.f. and internal resistance of a source
11.
Explain why the e.m.f. of a source is equal to the open circuit p.d. across the terminals of a source
12.
Explain how the conservation of energy leads to the sum of the e.m.f.’s round a closed
circuit being equal to the sum of the p.d.’s round the circuit.
13.
Derive the expression for the total resistance of any number of resistors in series, by
consideration of the conservation of energy.
14.
Derive the expression for the total resistance of any number of resistors in parallel, by
consideration of the conservation of charge.
15.
State the relationship among the resistors in a balanced Wheatstone bridge.
16.
Carry out calculations involving the resistance in a balanced Wheatstone bridge.
17.
State that for an initially balanced Wheatstone bridge, as the value of one resistor is changed by a small amount, the out of balance p.d. is proportional to the change in resistance.
18.
Use the following terms correctly in context: terminal p.d., load resistor, bridge circuit, lost volts, short circuit current.
Alternating Current and Voltage
1.
Describe how to measure frequency using an oscilloscope.
2.
State the relationship between peak and r.m.s. values for a sinusoidally varying voltage and current.
3.
Carry out calculations involving peak and r.m.s. values of voltage and current.
4.
State the relationship between current and frequency in a resistive circuit.
Physics: Electricity and Electronics (H) 1

Checklist
Capacitance
1.
State that the charge Q on two parallel conducting plates is directly proportional to the p.d. V between the plates.
2.
Describe the principles of a method to show that the p.d. across a capacitor is directly
proportional to the charge on the plates.
3.
State that capacitance is the ratio of charge to p.d.
4.
State that the unit of capacitance is the farad and that one farad is one coulomb per volt.
5.
Carry out calculations using C = Q/V
6.
Explain why work must be done to charge a capacitor.
7.
State that the work done to charge a capacitor is given by the area under the graph of
8.
charge against p.d.
1
State that the energy stored in a capacitor is given by (charge × p.d.) and equivalent
2
9.
expressions.
1
Carry out calculations using QV or equivalent expressions.
10.
Draw qualitative graphs of current against time and of voltage against time for the charge and discharge of a capacitor in a d.c. circuit containing a resistor and capacitor in series.
11.
Carry out calculations involving voltage and current in CR circuits (calculus methods are not required).
12.
State the relationship between current and frequency in a capacitive circuit.
13.
Describe the principles of a method to show how the current varies with frequency in a capacitive circuit.
14.
Describe and explain the possible functions of a capacitor: storing energy, blocking d.c. while passing a.c.
Analogue Electronics
1.
State that an op-amp can be used to increase the voltage of a signal.
2.
State that for the ideal op-amp: a) input current is zero; i.e. it has infinite input resistance b) here is no potential difference between the inverting and non-inverting inputs; i.e. both input pins are at the same potential.
3.
Identify circuits where the op-amp is being used in the inverting mode.
4.
State that an op-amp connected in the inverting mode will invert the input signal.
5.
State the inverting mode gain expression: V
0
/ V
1
= -R f
/ R i
.
6.
Carry out calculations using the above gain expression.
7.
State that an op-amp cannot produce an output voltage greater that the positive supply voltage or less that the negative supply voltage.
8.
Identify circuits where the op-amp is being used in the differential mode.
9.
State that a differential amplifier amplifies the potential difference between its two inputs.
10.
State the differential mode gain expression - V o
= (V
2
- V
1
) R f
/ R i
.
11.
Carry out calculations using the above gain expression.
12.
Describe how to use the differential amplifier with resistive sensors connected in a
Wheatstone bridge arrangement.
13.
Describe how an op-amp can be used to control external devices via a transistor.
Physics: Electricity and Electronics (H) 2

Checklist
Uncertainties
1.
State that measurement of any physical quantity is liable to uncertainty.
2.
Distinguish between random uncertainties and recognised systematic effects.
3.
State that the scale-reading uncertainty is a measure of how well an instrument scale
can be read.
4.
Explain why repeated measurements of a physical quantity are desirable.
5.
Calculate the mean value of a number of measurements of the same physical quantity.
6.
State that this mean is the best estimate of a ‘true’ value of the quantity being measured.
7.
State that where a systematic effect is present the mean value of the measurements will be ‘offset’ from a ‘true’ value of the physical quantity being measured.
8.
Calculate the approximate random uncertainty in the mean value of a set of measurements using the relationship: approximate uncertainty in the mean = maximum value - minimum value number of measurements taken
9.
Estimate the scale reading incurred when using an analogue display and a digital display.
10.
Express uncertainties in absolute or percentage form.
11.
Identify, in an experiment where more than one physical quantity has been measured, the quantity with the largest percentage uncertainty.
12.
State that this percentage uncertainty is often a good estimate of the percentage in the final numerical result of the experiment.
13.
Express the numerical result of an experiment in the form: final value

uncertainty
Units, prefixes and scientific notation
1.
Use SI units of all physical quantities in the above checklist.
2.
Give answers to calculations to an appropriate number of significant figures.
3.
Check answers to calculations.
4.
Use prefixes (p, n, µ, m, k, M, G)
5.
Use scientific notation.
Physics: Electricity and Electronics (H) 3

Summary Notes
ELECTRIC FIELDS AND RESISTORS IN CIRCUITS
Force Fields
In Physics, a field means a region where an object experiences a force without being touched.
For example, there is a gravitational field around the Earth. This attracts masses towards the earth’s centre. Magnets cause magnetic fields and electric charges have electric fields around them.
Electric Fields
In an electric field, a charged particle will experience a force. We use lines of force to show the strength and direction of the force. The closer the field lines the stronger the force. Field lines are continuous - they start on positive and finish on negative charge. The direction is taken as the same as the force on a positive “test” charge placed in the field.
Electric Field Patterns
Positive point charge Negative point charge
+ test charge has a force
‘outwards’
+ test charge has a force
‘inwards’
These are called radial fields. The lines are like the radii of a circle. The strength of the field decreases as we move away from the charge.
Electric Field Patterns
Positive and negative point charges Parallel charged plates
The field lines are equally spaced between the parallel plates. This means the field strength is constant. This is called a uniform field.
Electric fields have certain similarities with gravitational fields.
Physics: Electricity and Electronics (H) – Student Material 1

Summary Notes
Gravitational Fields h
If a mass is lifted or dropped through a height then work is done i.e. energy is changed.
If the mass is dropped then the energy will change to kinetic energy.
If the mass is lifted then the energy will change to gravitational potential energy.
Change in gravitational potential energy = work done.
Electric Fields
Consider a negative charge moved through a distance in an electric field. If the charge moves in the direction of the electric force, the energy will appear as kinetic energy. If a positive charge is moved against the direction of the force as shown in the diagram, the energy will be stored as electric potential energy.
Change in electric potential energy = work done
If the charge moved is one coulomb, then the work done is the potential difference or voltage.
If one joule of work is done in moving one coulomb of charge between two points in an electric field, the potential difference, (p.d.) between the two points is one volt.
1 volt = 1 joule per coulomb
W = QV
In this section W will be used for the work done i.e. energy transferred.
Example:
A positive charge of 3 µC is moved, from A to B, between a potential difference of 10 V.
(a) Calculate the electric potential energy gained.
(b) If the charge is now released, state the energy change.
(c) How much kinetic energy will be gained on reaching
B the negative plate?
(a) W
= QV= 3 × 10 -6
× 10
= 3 × 10 -5
J
+
+
+
+
(b) Electric potential energy to kinetic energy
(c)
By conservation of energy the energy will be the same, i.e. 3 × 10 -5
J.
+ Q
A
-
-
-
-
Physics: Electricity and Electronics (H) – Student Material 2

Summary Notes
Moving Charges in Electric Fields
From the previous example, when the positive charge is released at plate B then the electical potential energy is converted to kinetic energy.
QV =
1
2 mv
2
Example
An electron is accelerated (from rest) through a potential difference of 200 V.
Calculate (a) the kinetic energy, E k
gained.
(b) the final speed of the electron.
(Mass of an electron = 9.1 × 10 -31
kg, charge on an electron = -1.6x10
-19
C)
(a) E k
=
1
2 mv
2
= QV = 1.6x10
-19
× 200
= 3.2 × 10 -17
J
(b)
1
2 mv
2
= 3.2 × 10
-17 v
2 
 
17  m

 
17 

31 v = 8.4
×
10
6
m s
-1
Applications of electric fields (for background interest)
A television involves the use of electron guns. The electrons gain kinetic energy by accelerating through an electric field. Deflection of the electrons is usually done by electromagnetic coils, although flat screen tubes are now dependent on electrostatic deflection.
An oscilloscope also depends on electric fields acting on electrons.
Electrostatic Spraying makes use of electric fields. Paint or powder particles are blown from a nozzle, where they acquire a charge. The object to be coated is earthed. The charged paint or powder particles follow the field lines and so reach the object, some reaching the back of the object as well as the front.
Other applications include photocopiers, ink jet and laser printers.
Physics: Electricity and Electronics (H) – Student Material 3

Summary Notes
Electric Current
When S is closed, the free electrons in the conductor experience an electric field which cause them to move in one direction.
Note: The electron current will flow from the negative terminal to the positive terminal of the battery. The energy required to drive the electron current around the circuit is provided by a chemical reaction in the battery or by the mains power supply. The electrical energy which is supplied by the source is converted to other forms of energy in the components which make up the circuit.
The lamp has resistance R. In any circuit providing the resistance of a component remains constant, if the potential difference V across the component increases the current I through the component will increase in direct proportion. This is Ohm’s Law which is summarised by the equation below.
V = IR
A component which has a constant resistance when the current through it is increased is said to be ohmic. Some components do not have a constant resistance, their resistance changes as the p.d. across the component is altered, for example the transistor.
Electric Power
For a given component the power P = I V where I is the current through that component and
V is the potential difference across the component.
Resistive Heating
The expression I2 R gives the energy transferred in one second due to resistive heating.
Apart from obvious uses in electric fires, cookers, toasters, etc. consideration has to given to heating effects in resistors, transistors and integrated circuits and care taken not to exceed the maximum ratings for such components.
Electromotive Force (e.m.f)
2
P = I R =
V
2
R
The energy given to a coulomb of charge by a source of electrical energy is called the e.m.f. of the source. This is measured in J C -1 or volts.
Note: the potential difference between two points in the external circuit is also measured in volts, but this is concerned with electrical energy being transformed outside the source.
Physics: Electricity and Electronics (H) – Student Material 4

Summary Notes
Sources of e.m.f.
E.M.F’s can be generated in a great variety of ways e.g: chemical cells, solar cell, thermocouple, electromagnetic generator. piezo - electric generator,
Resistors in Series - Conservation of Energy
Applying the conservation of energy to resistors in series for one coulomb of charge.
Energy supplied by source = energy converted by circuit components e.m.f. = IR
1
+ IR
2
+ IR
3
R s = equivalent series resistance
IR
S
= IR
1
+ IR
2
+ IR
3
R
S
= R
1
+ R
2
+ R
3
Resistors in Parallel - Conservation of Charge
The total charge per second (current) passing through R
1
, R
2
and R
3
must equal the charge per second (current) supplied from the cell i.e. passing through R
P
.
Conservation of charge gives:
I = I
1
+ I
2
+ I
3
(Since I =
Q t
for each resistor)
E
R
1
E
=
R
2
E
+
R
3
E
+
R
4
R p
= equivalent parallel resistance
Physics: Electricity and Electronics (H) – Student Material 5

Summary Notes
Internal Resistance
In choosing a suitable power supply for a circuit, you would have to ensure that it :
 gives the correct e.m.f
 is able to supply the required maximum current
When a power supply is part of a closed circuit, it must itself be a conductor. All conductors have some resistance. A power supply has internal resistance, r.
Energy will be wasted in getting the charges through the supply (the heat from the supply will be noticeable) and so the energy available at the output (the terminal potential difference ) will fall. There will be “lost volts” . The lost volts = I r
The greater the current, the more energy will be dissipated in the power supply until eventually all the available energy (the e.m.f.) is wasted and none is available outside the power supply. This maximum current is the short circuit current. This is the current which will flow when the terminals of the supply are joined with a short piece of thick wire.
Open Circuit
When no current is taken from the power supply, no energy is wasted. The terminal potential difference is therefore the maximum available and equals the e.m.f.
General Circuit r
E
R
S
V
Any power supply can be thought of as a source of constant e.m.f. E, in series with a small resistance, the resistance circuit).
. internal
With S open, the voltmeter reading gives the e.m.f. (an open
With S closed, the voltmeter reading will fall (lost volts). The voltmeter now gives the difference, t.p.d.
With S closed and using the conservation of energy e.m.f. = lost volts + t.p.d. e.m.f. = lost volts + output voltage output voltage, the terminal potential which is:
E
E
= Ir +
= Ir +
V
IR
Notice that the total resistance of the circuit is R + r, giving the equation:
E = I ( R + r)
The short circuit current is the maximum which can be supplied by a source. This occurs when there is no external component and R = 0.
Physics: Electricity and Electronics (H) – Student Material 6

Example
E r
V
R = 28 
A cell of e.m.f. 1.5 V is connected in series with a 28
 resistor.
A voltmeter measures the voltage across the cell as 1.4 V.
Calculate:
(a) the internal resistance of the cell
(b) the current if the cell terminals are short circuited
(c) the lost volts if the external resistance R is increased to 58 Ω.
(a) E = Ir + IR = Ir + V
Lost volts = Ir = E - V = 1.5 - 1.4 = 0.1 V r = lost volts
I
=
0.1
1
I =
V
R
=
1.4
28
= 0.05 A
0.1
r =
0.05
= 2

(b)
A short circuit occurs when R = 0 (no external resistance)
I
R + r
=
E r
=
1.5
2
= 0.75 A
(c) Lost volts = Ir
I
R + r
=
= 0.05 A
E
28 + 2
= 1.5
Lost volts = 0.05 × 2
= 0.1 V
Summary Notes
Physics: Electricity and Electronics (H) – Student Material 7

Summary Notes
Wheatstone Bridge Circuit
Any method of measuring resistance using an ammeter or voltmeter necessarily involves some error unless the resistances of the meters themselves are taken into account. The use of digital voltmeters largely overcomes this as they tend to have very high resistances.
Further error may be introduced if the meter is not correctly calibrated. The only situation where neither of these errors matter is if the meter reading is zero. The Wheatstone bridge circuit is one such example of using a meter as a null deflection indicator.
Bridge Circuit
If V
OA
= V
OB
there is no p.d. between A and B hence no current flows. The potentials at A and at B depend on the ratio of the resistors that make up each of the two voltage dividers.
The voltmeter forms a ‘bridge’ between the two voltage dividers to make up a bridge circuit.
Balanced Bridge
5 V
R1
R2
A




V
R3

B

R4
No potential difference will exist across AB when
R
1 
R
2
R
3
R
4
0 V O
The bridge is balanced when the voltmeter or galvanometer
(milliammeter) reads zero.
Alternative ‘diamond’
Representation with galvanometer
Unbalanced Bridge
If the bridge is initially balanced, and the resistor × is altered by a small amount  × then the out of balance p.d. (reading on G) is directly proportional to the change in resistance, provided the change is small.
+1.5 V
Hence for small  ×,
Reading on G reading on G
  ×
R1 R2
 X/ 
G
X+  X R4
Physics: Electricity and Electronics (H) – Student Material 8

Summary Notes
ALTERNATING CURRENT AND VOLTAGE
Peak and r.m.s. values
The graph of a typical alternating voltage is shown below.
The maximum voltage is called the peak value .
From the graph it is obvious that the peak value would not be a very accurate measure of the voltage available from an alternating supply.
In practice the value quoted is the root mean square (r.m.s.) voltage.
The r.m.s. value of an alternating voltage or current is defined as being equal to the value of the direct voltage or current which gives rise to the same heating effect (same power output).
Consider the following two circuits which contain identical lamps.
The variable resistors are altered until the lamps are of equal brightness. As a result the direct current has the same value as the effective alternating current (i.e. the lamps have the same power output). Both voltages are measured using an oscilloscope giving the voltage equation below. Also, since V=IR applies to the r.m.s. valves and to the peak values a similar equation for currents can be deduced.
V r.m.s.
= 1
2
Vpeak and
I r.m.s.
= 1
2
Note: a moving coil a.c. meter is calibrated to give r.m.s. values.
Ipeak
Physics: Electricity and Electronics (H) – Student Material 9

Summary Notes
Graphical method to derive relationship between peak and r.m.s. values of alternating current
The power produced by a current I in a resistor R is given by I
2
R. A graph of I
2
against t for an alternating current is shown below. A similar method can be used for voltage.
The average value of I
2
is
I
2
Peak
2
An identical heating effect (power output) for a d.c. supply = I
2 r.m.s
R
[since I (d.c.) = I r.m.s.]
Average power output for a.c. =
I
2
Peak
2
I
2 r.m.s
2
R = hence
2
R 
R
I
2 r.m.s.
=
Frequency of a.c.
 I
2
giving I r.m.s. =
I
Peak
2 
2
To describe the domestic supply voltage fully, we would have to include the frequency i.e.
230 V 50 Hz.
An oscilloscope can be used to find the frequency of an a.c. supply as shown below.
Time base = 0.005 s cm
Wavelength = 4 cm
-1
Time to produce one wave = 4 × 0.005
= 0.02 s
Frequency =
1 time to produce one wave
1
= = 50 Hz
Physics: Electricity and Electronics (H) – Student Material 10

Summary Notes
Mains supply
The mains supply is usually quoted as 230 V a.c. This is of course 230 V r.m.s.
The peak voltage rises to approximately 325 V. Insulation must be provided to withstand this peak voltage.
Example
A transformer is labelled with a primary of 230 V r.m.s. and secondary of 12 V r.m.s.
What is the peak voltage which would occur in the secondary?
V
V
V peak peak peak
=
 2 × V
= 17.0 V r.m.s.
= 1.41 × 12
Physics: Electricity and Electronics (H) – Student Material 11

Summary Notes
CAPACITANCE
The ability of a component to store charge is known as capacitance.
A device designed to store charge is called a capacitor.
A typical capacitor consists of two conducting layers separated by an insulator.
Circuit symbol

Relationship between charge and p.d.
The capacitor is charged to a chosen voltage by setting the switch to A. The charge stored can be measured directly by discharging through the coulomb meter with the switch set to B. In this way pairs of readings of voltage and charge are obtained.
Q
Charge is directly proportional to voltage.
0 V
Q
V
= constant
For any capacitor the ratio Q/V is a constant and is called the capacitance. coulombs (C) farad (F) capacitance = charge voltage volts (V)
The farad is too large a unit for practical purposes.
In practice the micro farad (µF) = 1 × 10 -6
F and the nano farad (nF) = 1 × 10
-9
F are used.
Example
A capacitor stores 4 × 10 -4 C of charge when the potential difference across it is 100 V.
What is the capacitance ?
C =
Q
V
= 4 µF

4

10
-6
100
Physics: Electricity and Electronics (H) – Student Material 12

Summary Notes
Energy Stored in a Capacitor
A charged capacitor can be used to light a bulb for a short time, therefore the capacitor must contain a store of energy. The charging of a parallel plate capacitor is considered below.
There is an initial surge of electrons from the negative terminal of the cell onto one of the plates (and electrons out of the other plate towards the +ve terminal of the cell).
Once some charge is on the plate it will repel more charge and so the current decreases.
In order to further charge the capacitor the electrons must be supplied with enough energy to overcome the potential difference across the plates i.e. work is done in charging the capacitor.
Eventually the current ceases to flow. This is when the p.d. across the plates of the capacitor is equal to the supply voltage.
For a given capacitor the p.d. across the plates is directly proportional to the charge stored
Consider a capacitor being charged to a p.d. of V and holding a charge Q.
Charge
Q
The energy stored in the capacitor is given by the area under graph
V p.d.
Area under graph= 1 Q x V
2
Energy stored = 1 Q x V
2
If the voltage across the capacitor was constant work done = Q x V, but since V is varying, the work done = area under graph.
Q = C × V and substituting for Q and V in our equation for energy gives:
Energy stored in a capacitor =
1
QV =
2
1
2
2
CV =
1
2
Q
2
C
Example
A 40  F capacitor is fully charged using a 50 V supply. How much energy is stored?
Energy =
1
2
2
CV =
1
2

40
 6
10

2500
= 5 × 10
-2
J
Physics: Electricity and Electronics (H) – Student Material 13

Summary Notes
Capacitance in a d.c. Circuit
Charging
Consider the following circuit:-
When the switch is closed the current flowing in the circuit and the voltage across the capacitor behave as shown in the graphs below. current p.d. across capacitor
Supply voltage
0 time
Consider the circuit at three different times.
0 0
0 time
0
A
As soon as the switch is closed there is no charge on the capacitor the current is limited only by the resistance in the circuit and can be found using Ohm’s law.
A
+ +
- -
As the capacitor charges a p.d. develops across the plates which opposes the p.d. of the cell as a result the supply current decreases.
A
+ +
+ +
The capacitor becomes fully charged and the p.d.
across the plates is equal and opposite to that across the cell and the charging current becomes zero.
Physics: Electricity and Electronics (H) – Student Material 14

Summary Notes
Discharging
Consider this circuit when the capacitor is fully charged, switch to position B
- - - -
+ + + +
A
A B
If the cell is taken out of the circuit and the switch is set to A, the capacitor will discharge
A
A B
While the capacitor is discharging the current flowing in the circuit and the voltage across the capacitor behaves as shown in the graphs below.
Current
0 time p.d. across capacitor
0
Supply voltage time
Although the current/time graph has the same shape as that during charging the currents in each case are flowing in opposite directions. The discharging current decreases because the p.d. across the plates decreases as charge leaves them.
Factors affecting the rate of charge/discharge of a capacitor

When a capacitor is charged to a given voltage the time taken depends on the value of the capacitor. The larger the capacitor the longer the charging time, since a larger capacitor requires more charge to raise it to the same
Current large capacitor small capacitor

When a capacitor is charged to a given voltage the time taken depends on the value of the resistance in the circuit. The larger the resistance the smaller the initial charging current, hence the longer it takes to charge the capacitor as Q = It
Current
Time small resistor large resistor
Time
(The area under this I/t graph = charge. Both curves will have the same area since Q is the same for both.)
Physics: Electricity and Electronics (H) – Student Material 15

Example
The switch in the following circuit is closed at time t = 0
10 V
Vs
1 M 
Immediately after closing the switch what is:
(a) the charge on C
(b) the p.d. across C
(c) the p.d. across R
(d) the current through R.
When the capacitor is fully charged what is:
1 µF
(e) the p.d. across the capacitor
(f) the charge stored.
(a) Initial charge on capacitor is zero.
(b) Initial p.d . is zero since charge is zero.
(c) p.d. is 10 V = V s
- V c
= 10 - 0 = 10 V
(d)
I
R
= V =
10
10
6
3
= 10 A
(e) Final p.d. across the capacitor equals the supply voltage = 10 V.
(f)
Q = CV = 2 × 10 -6
× 10 = 2 × 10
-5
C.
Summary Notes
Physics: Electricity and Electronics (H) – Student Material 16

Summary Notes
Resistors and Capacitors in a.c. Circuits
Frequency response of resistor
The following circuit is used to investigate the relationship between current and frequency in a resistive circuit.
The results show that the current flowing through a resistor is independent of the frequency of the supply.
Frequency response of capacitor
The following circuit is used to investigate the relationship between current and frequency in a capacitive circuit.
The results show that the current is directly proportional to the frequency of the supply.
To understand the relationship between the current and frequency consider the two halves of the a.c. cycle.
The electrons move back and forth around the circuit passing through the lamp and charging the capacitor one way and then the other (the electrons do not pass through the capacitor).
The higher the frequency the less time there is for charge to build up on the plates of the capacitor and oppose further charges from flowing in the circuit More charge is transferred in one second so the current is larger.
Physics: Electricity and Electronics (H) – Student Material 17

Summary Notes
Applications of Capacitors (for background interest)
Blocking capacitor
A capacitor will stop the flow of a steady d.c. current . This is made use of in the a.c./d.c. switch in an oscilloscope. In the a.c. position a series capacitor is switched in allowing passage of a.c. components of the signal, but blocking any steady d.c. signals.
Flashing indicators
A low value capacitor is charged through a resistor until it acquires sufficient voltage to
1 - 2 M  fire a neon lamp. The neon lamp lights when the p.d. reaches 100 V. The capacitor is quickly discharged and the lamp goes out when the p.d. falls below 80V.
120 V
Crossover networks in loudspeakers
In a typical crossover network in low cost loudspeaker systems, the high frequencies are routed to LS-2 by the capacitor.
LS 1 LS 2
Smoothing
The capacitor in this simple rectifier circuit is storing charge during the half cycle that the diode conducts. This charge is given up during the half cycle that the diode does not conduct.
This helps to smooth out the waveform.
Capacitor as a transducer
A parallel plate capacitor can be used to convert mechanical movements or vibration of one of its plates into changes in voltage. This idea forms the basis of many measuring systems, e.g. by allowing a force to compress the plates we have a pressure transducer.
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Summary Notes
ANALOGUE ELECTRONICS
Analogue and Digital Signals
An analogue system transmits information in the form of a continuously varying signal.
A digital system has the information broken up into a series of discrete values or steps.
A simple example showing these definitions would be analogue and digital watches.
 an analogue watch has a set of numbers in a circle and hands that point to them. The hands sweep round the face in a continuous way.
 a digital watch has a series of numbers that are displayed, the numbers changing at the end of each second, minute or hour.
The Op-Amp (Operational Amplifier)
As with any amplifier, the operational amplifier (op-amp) will change the size of an input electrical signal. A diagram of the basic set-up of an op-amp is shown below.
The op-amp has two separate inputs - an inverting input (“-” terminal) and a non-inverting input (“+” terminal). The amplifier must have an energy supply and this is provided by the supply voltages +Vs and -Vs across the amplifier. Often the power supply voltages are not shown on circuit diagrams of op-amps.
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Activities
The Inverting Amplifier
If an amplifier is set up in the configuration shown below, it is said to be in the inverting mode.
R f
R
1
V
1
-
+
V
0
0 V
The input potential, V
1
, is applied to the inverting input (-ve input). The non-inverting input (+ve input) is connected straight to “ground”, 0 V.
There is also a resistor, R f
, (feedback resistor) connected between the output and the inverting input. This feedback resistor reduces the overall gain of the amplifier and allows the gain to be stabilised controlled.
When the input voltage signal to the op-amp, d.c or a.c, is compared to the output voltage signal it is found that the sense or ‘sign’ of the output is opposite to that of the input - the voltage has been inverted , hence the reason why this circuit is called the “inverting mode”.
Some examples of this are given below:
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Activities
Inverting Mode Gain Equation
The ideal op-amp fulfils two conditions:
 no current flows into the op-amp. Its input resistance is infinite.
 there is no potential difference between the inverting and non-inverting inputs.
The following inverting mode gain equation can be verified by experiment.
Saturation
From the inverting mode gain equation, it would seem that by inserting any pair of resistors
R
1
and R f
in the inverting amplifier circuit it is possible to provide any gain required. It could therefore be possible to produce either very low or very high output voltages from any input voltage.
This is not possible. The output of an op-amp circuit is limited by the size of the power supply used (conservation of energy). In theory, the maximum output voltage possible would be equal to the supply voltage. (However, in reality it is limited to approximately
85% of the supply voltage.)
When the maximum output voltage has been reached, the amplifier is said to be saturated.
-
+15V
V
0
In theory :
[In practice :
-15V
-13V


V
V
0
0


+15 V
+13 V (13V  85% of 15 V)]
+
-15V
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Summary Notes
Example
An inverting mode operational amplifier is set up as shown below.
(a) If V
1
is set at +0.8 V, calculate the output voltage V
0
.
(b) If an a.c signal of peak voltage 1.5 V is applied to V
1
, sketch the input voltage, V
1
, and the output voltage, V
0
.
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Activities
The Differential Amplifier
If the amplifier is set up in the configuration shown below, it is said to be in the differential mode.
There are two input potentials,V
1
and V
2
, one applied to each of the input terminals of the opamp.
There is a feedback resistor, R f
, connected between the output and the inverting input. This allows control over the gain of the amplifier as it did for the inverting mode.
When the op-amp is used in this mode, it amplifies the difference between the inputs V
1
and
V
2
, with a gain set by the ratio
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Summary Notes
Example
A differential amplifier is set up as shown below.
For the following values shown, calculate the output voltage, V
0
.
(a) V
1
= +5.0 V, V
2
= +4.8 V
(b) V
1
= -2.0 V, V
2
= +4.5 V
(a) V
1
= +5.0 V, V
2
= +4.8 V, R
1
=R
2
= 10 kΩ, R f
= R
3
= 100 kΩ, V
0
= ?
V
0
=
R f
R
1
(V
2
- V
1
) V
0
=
100
10
(4.8 - 5.0)
V
0
= - 2.0 V
(b) V
1
= -2.0 V, V
2
= +4.5 V, R
1
=R
2
= 10 kΩ, R f
= R
3
= 100 kΩ, V
0
= ?
V
0
=
R f
R
1
(V
2
- V
1
) V
0
=
100
10
(4.5 - (-2.0)) V
0
= + 65 V in theory.
But supply voltage = + 15 V hence output voltage, V
0
, will saturate at + 15 V
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Summary Notes
The differential amplifier as part of a monitoring system
+Vs
10 k  t
V1
10 k 
0-10 k
V2

R1
10 k 
R2
10 k 

R3
-
+
100 k 
Rf
100 k 
V0
0 V 0 V
Wheatstone
Bridge input
Differential
Amplifier
The setting of the variable resistor can be adjusted so as to achieve an output voltage of zero for a particular temperature setting. The bridge circuit would then be balanced , that is the potential difference V
2
- V
1
= 0 V.
The potential, V
2
, will remain constant as long as the resistance of the variable resistor is not changed. Any change in temperature will change the potential, V
1
, and will therefore produce a potential difference between V
2
and V
1
.
The amplifier will then amplify the difference between V
2
and V
1
, giving an output potential,
V
0
. The output voltage will increase as the change in thermistor resistance, ∆R t
, increases.
The amplifier is, effectively, amplifying the out-of-balance potential difference from the
Wheatstone Bridge.
Practical Application
The output voltage could be calibrated by placing the thermistor in melting ice (0 o
C), then in boiling water (100 o
C), noting the output potential, V
0
, for each case. The range could then be divided into 100 equal divisions to give an electronic thermometer over the range 0 - 100 o
C.
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Activities
Control Circuits
A transistor, such as those shown below, can act as an electrical switch.
If the input voltage to these transistors, V i
, is positive , then it switches on allowing a current to flow between the collector and emitter or source and drain, otherwise it is off.
There are two types of transistor switches: bipolar n-p-n transistor n-channel enhancement MOSFET
+V
S base collector
-V
S drain gate
0V
V i emitter
V i
0V source
(i) A positive (+ve) potential (>+0.7 V) is needed for the input, V i
, to switch transistor on .
(ii) The collector arm is connected to a positive (+ve) supply rail.
(iii) If the input potential is negative
(<0.7 V), the transistor is off.
(i) A positive (+ve) potential >1.8 V is needed for the input, V i
, to switch transistor on .
(ii) The drain is connected to a positive
(+ve) supply rail.
(iii) if the input potential is negative or
<1.8 V the transistor is off.
These transistors can be used with a Wheatstone Bridge/differential amplifier circuit to switch a device on or off.
Low light-level indicator
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Activities
The variable resistor in the Wheatstone Bridge is adjusted so that, at a particular light level, the output of the op-amp is zero or less so that the transistor is off. In practice, the variable resistor would be adjusted until the warning lamp is just off. In this situation, V
1
≈ V
2
.
If the light level falls, the resistance of the LDR will increase causing the voltage V
1 to fall .
If V
1
falls, then V
1
< V
2
, therefore (V
2
- V
1
) will be positive (+ve).
This will cause the output voltage, V
0
, to be positive also, switching on the transistor and the Warning Lamp lights .
If the light level to the LDR increases, then the opposite will be true.
The resistance of the LDR will decrease causing the voltage V
1
to increase.
If V
1
increases, then V
1
>V
2
, (V
2
- V
1
) will be negative causing the output to be negative and the transistor will not switch on.
The gain of the circuit (about 1000) is deliberately chosen to be large so that a small variation from the balance point of the Wheatstone Bridge produces a large enough output voltage to switch the transistor on.
Modifications
This circuit does have its limitations however. If the output device to be used has a high power rating requiring a current larger than the transistor can safely supply, then a relay switch must be used with the transistor to switch on an external circuit.
An example of this type of switch is shown below. In this case, a relay can be energised by the current through the transistor (the collector current) and its contacts can then be used to switch on a high-power device such as a heater.
+ V s
Rf
+Vs t
0-10K
100K relay
230 V mains
V1
R1
1K
R2
1K
-
+
V2
R3
V0 Vi
0V heater
Wheatstone
Bridge input
Differential
Amplifier
Transistor Switch and Output
Notice that as the temperature falls the resistance of the thermistor increases, causing
V
1
to fall. this means V
1
< V
2
giving a positive output voltage V
0
. The transistor will be switched on, which will energise the relay switch. The heater will be turned on.
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Activities
Control Circuit Diagrams
The Wheatstone Bridge arrangement can be shown as two straight potential dividers or as a diamond arrangement. They are the same circuit , just drawn a different way.
Straight potential dividers
+Vs t
0V
Diamond arrangement bridge circuit
V1
V2 to differential amplifier
10 k 
V2
+Vs t
V1
V2
+Vs to op-amp
10 k 
0V
10 k  to op-amp t
10 k 
0V
All the above circuits are identical examples of the sensor bridge circuit that can be used with a differential amplifier. It is important to recognise each circuit for what it is and how it behaves.
V1
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