Unit 4 year 12 electrical-electronics

UNIT 4
Principals of Electrical &
Electronic Engineering
12/05/2017 1

Unit Aims
Electrical systems and electronic devices are present in almost every aspect
of modern life – and it is electrical and electronic engineers who design, test
and produce these systems and devices.
This unit will develop learners’ knowledge and understanding of the
fundamental principles that underpin electrical and electronic engineering.
By completing this unit learners will develop an understanding of:
fundamental electrical principles, alternating voltage and current, electric
motors and generators, power supplies and power system protection,
analogue electronics, digital electronics.
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Learning Objective 1
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1.1 application of the defining equations for:
• resistance
• power
• energy
• resistors connected in series
• resistors connected in parallel
1.2 measurement of voltage, current and resistance
in a circuit using a:
• voltmeter
• ammeter
• ohmmeter
• multimeter
1.3 Circuit theory, i.e.
• calculation of the total resistance and total current for a circuit
• Kirchhoff’s first law and its application
• Kirchhoff’s second law and its application
• the maximum power transfer theorem

LO1.1 Defining Equations
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1.1 application of the defining equations for:
1. resistance
2. power
3. energy
4. resistors connected in series
5. resistors connected in parallel

Aspire :
Challenge:
Resistance, Voltage, Current
Ohm's law defines the relationship
between the voltage, current,
and resistance in an electric circuit: I
= V/R. The current is directly
proportional to the voltage and
inversely proportional to
the resistance.
R = V/I
V = I x R
I = V/R
To understand the fundamental nature of these relationships
To calculate using the defining equations

Aspire :
Challenge:
The Hydraulic Analogy

Aspire :
Challenge:
Resistance

Aspire :
Challenge:
Resistance
For the following sets of values
Calculate the total resistance in
series and then in parallel.
10Ω 10Ω 10Ω 10Ω
10Ω 20Ω 30Ω 40Ω
For a 6 volt power source how to
each effect the current flow?
And for 12 volts ?

Aspire :
Challenge:
Power
Energy In
Power Out
Electric power, is the rate of doing work measured
in watts and represented by the letter P. The
term wattage is used colloquially to mean "electric
power in watts." The electric power
in watts produced by an electric current I consisting
of
a charge of Q coulombs every t seconds (current)
passing through an electric potential(voltage)
difference of V is

Aspire :
Challenge:
Power
When electric charges move through from the positive (+) terminal to the negative (−)
terminal through a ‘Load’ , work is done by the charges on the device.
The potential energy of the charges due to the voltage between the terminals is
converted to a different form of energy by the device.
These devices are called passive components or loads; they 'consume' electric power
from the circuit, converting it to other forms of energy such as heat, light, etc.
Examples are electrical appliances, such as light bulbs, electric motors, and electric
heaters.

Aspire :
Challenge:
Energy
A 2 kW electrical fire is switched on for 3 hours. It uses 2 × 3 = 6 kWh of
electrical energy.
The amount of electrical energy transferred to an appliance depends on its
power and the length of time it is switched on. The amount of mains
electrical energy transferred is measured in kilowatt-hours, kWh.
1 watt = 1 Joule/second
Joule, unit of work or energy in the International System of Units (SI); it is
equal to the work done by a force of one Newton acting through one metre.

Aspire :
Challenge:
OUT FOX THE CLASS
Based on this lesson – design a question
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Aspire :
Challenge:
Capacitors & Timing
A Capacitor is an electronic temporary
storage device, it can be thought of as a
mini battery.
It consists of two metal plates separated
by a non conductor.
The plates can become positively or
negatively charged and it is this ‘charge’
that is held by the capacitor and then
released.
To understand capacitance and it’s role in creating a time constant.
To calculate the time constant of a capacitor resistor combination.

Aspire :
Challenge:
Capacitors - Equation
The amount of charge stored is a
product of the capacity available and the
voltage supplied.

Aspire :
Challenge:
UNITS
Capacitors’ capacity is measured in Farads and has the symbol C. 1 Farad is
quite a large charge and so we often use sub-devsions of Farads instead.

Aspire :
Challenge:
Parallel and Series

Aspire :
Challenge:
1 ÷ 0.1 x 10-6 = 100,000,000
1 ÷ 0.2 x 10-6 = 50,000,000
1 ÷ 0.2 x 10-6 = 30,000,000*
Total Charge = 1/180,000,000 = 5.5 x 10-9
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0.1 x 10-6 = 0.0000001
0.2 x 10-6 = 0.0000002
0.2 x 10-6 = 0.0000003
Total Charge = 0.0000006 = 0.6 x 10-6

Aspire :
Challenge:
Charging
The time it takes a capacitor to
charge up fully is dependant on
the voltage, a resistor in the
circuit will effectively reduce the
voltage available to the
capacitor and therefore the
capacitor will take longer to
charge up.

Aspire :
Challenge:
Charging
The voltage and current of a charging capacitor can be found using
Where ‘e’ is……well let’s let some one else try and explain -
https://www.youtube.com/watch?v=yTfHn9Aj7UM
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Vc = VS (1 - e-t/RC)
i = Vse-t/CR

Aspire :
Challenge:
Discharging
The time it takes a capacitor to
discharge fully is dependant on
the flow of voltage away from
the capacitor

Aspire :
Challenge:
Discharging
The voltage across a discharging capacitor is
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VC=Vse-t/CR
Note not 1- as when charging !!!

Aspire :
Challenge:
Kirchhoff's First (Current) Law
Kirchhoff's first law when officially
stated sounds more complicated than it
actually is. Generally speaking, it says,
the total current entering a junction
must equal the total current leaving the
junction. After all, no charges can
simply disappear or get created, so
current can't disappear or be created
either. A junction is any place in a
circuit where more than two paths
come together.
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Aspire :
Challenge:
Kirchhoff's Second (Voltage) Law
Kirchhoff's second law when officially
stated also sounds more complicated
than it actually is. Generally speaking, it
says, around any loop in a circuit, the
voltage rises must equal the voltage
drops. Another way of thinking about
this is to consider that whatever energy
a charge starts with in a circuit loop, it
must end up losing all that energy by
the time it gets to the end. Or we could
say that by the time a charge makes it
to the end of a circuit, it must have
given all its energy to do work.
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Aspire :
Challenge:
Integrated Circuits
http://www.radio-electronics.com/info/circuits/opamp_basics/operational-
amplifier-basics-tutorial.php
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Aspire :
Challenge:
Operational Amplifiers
The operational amplifier (op-amp) is a high gain amplifier commonly used in
analogue circuit design.
Video Link
Vocab
Impedance is the measure of the opposition that a circuit presents to
a current when a voltage is applied – In DC circuits it is like resistance.
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To calculate the gain of for non-inverting and inverting op-amps.
To recognise defining characteristics of non-inverting/inverting op-amps.

Aspire :
Challenge:
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Aspire :
Challenge:
Non-Inverting/Inverting
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Aspire :
Challenge:
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Aspire :
Challenge:
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Aspire :
Challenge:
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Aspire :
Challenge:
Summing op-amp
The summing op-amp is the final configuration for learners to investigate. The
following web-based resource includes a complete explanation of the
summing op-amp along with worked calculations –
http://www.electronics-tutorials.ws/opamp/opamp_4.html
https://www.youtube.com/watch?v=y0Q0ERSP24A
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Aspire :
Challenge:
Analogue to Digital
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To convert analogue to digital or digital to analogue signals.
To understand the fundamental difference in analogue and digital.

Aspire :
Challenge:
Analogue to Digital
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Aspire :
Challenge:
Digital to Analogue
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Aspire :
Challenge:
Digital to Analogue
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Aspire :
Challenge:
Combinational Logic
Logic gates are often used in combination in order to produce a desired logic
function (ie a desired set of output conditions for a given set of input
conditions).
Operation of combinational logic combinations (ie input to output truth
tables) could be determined manually, or by simulation.
The following webbased resource is an online logic simulation too
http://www.neuroproductions.be/logic-lab/
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Aspire :
Challenge:
Combinational Logic Exercise
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Aspire :
Challenge:
Boolean
The starting point for many logic designs is with a truth table describing how
the circuit should operate. Another way of representing and manipulating
combinational logic functions is using Boolean expressions. Teachers might
used web-based resources to explain how combinational logic functions can
be represented by Boolean expressions – such as
http://www.allaboutcircuits.com/vol_4/chpt_7/9.html
Learners could undertake simple logic problems in order to develop and
recognise simple Boolean expressions.
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Aspire :
Challenge:
555 - Revision
Here the 555 timer is connected as a
basic Astable Multivibrator. Pins 2 and 6
are connected together so that it will re-
trigger itself on each timing cycle.
Capacitor, C1 charges up through resistor,
R1 and resistor, R2 but discharges only
through resistor, R2 as the other side of
R2 is connected to the discharge
terminal, pin 7.
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Then the timing period of t1 and t2 is given as:
t1 = 0.693 (R1 + R2) C1
t2 = 0.693 (R2) C1
T = t1 + t2 = 0.693 (R1 + 2R2) C1

Aspire :
Challenge:
Sequential Logic
A multivibrator circuit oscillates between
a “HIGH” state and a “LOW” state
producing a continuous output.
Astable multivibrators generally have an
even 50% duty cycle, that is that 50% of
the cycle time the output is “HIGH” and
the remaining 50% of the cycle time the
output is “OFF”.
Sequential logic circuits that use the
clock signal for synchronization.
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Lesson 1&2
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MOTORS & GENERATORS

Aspire :
Challenge:
Pre-Starter
Apply defining equations from last lesson.
Using this circuit, Calculate the;
Current.
Power.
Energy
Where the circuit is on for 2 minutes.
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To apply defining equations for DC Motors and Generators.
To understand the difference between Motors and Generators.

Aspire :
Challenge:
Starter
With your shoulder partner try to develop a definition of a motor and a
generator which describes how they are different..
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Aspire :
Challenge:
Defining Equation for DC Motors
The defining equation for the DC motor is
V = Eb + Ia Ra
Where V is the supply voltage, E is the back emf produced by the motor, Ia Ra
are the armature current and armature resistance.
Note the plus sign for motors
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Aspire :
Challenge:
Shunt Wound (Parallel) DC Motor
In a shunt wound Motor the field
windings are in parallel to the
armature.
Speed stable irrespective of load.
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Aspire :
Challenge:
Series Wound DC Motor
In a series wound Motor the field
windings are in series with the
armature.
Speed varies with load
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Aspire :
Challenge:
Generators
The defining equation for the DC Generator is
V = Eb - Ia Ra
Where V is the output voltage, E is the back emf produced by the motor, Ia Ra
are the armature current and armature resistance.
Note the minus sign for generators.
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Aspire :
Challenge:
Shunt Wound Generator
In shunt wound DC generators the field
windings are connected in parallel with
armature conductors as shown in figure
below. In these type of generators the
armature current
Ia divides in two parts. One part is the
shunt field current Ish flows through
shunt field winding and the other part is
the load current IL goes through the
external load..
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Ia = Ish + Il

Aspire :
Challenge:
Series Wound Generator
Conversely to shunt wound generators
the current isn’t shared, therefore in
these types of generators the field
windings, armature windings and
external load circuit all are connected in
series and have the same current.
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Ia = Ish = Il

Aspire :
Challenge:
Exercises
1. Explain the difference between a motor and a generator (1)
2. Show with a diagram how the field winding is connected to the armature of a series
wound self excited motor. (2)
3. Show with a diagram how the field winding is connected to the armature of a shunt
wound self excited motor. (2)
4. A d.c. shunt wound generator has armature resistance of 0.2 ohms and shunt field
resistance of 40 ohms. The generator is delivering 50 amperes at 240 volts.
Calculate:
a) The field current. (3)
b) The armature current. (3)
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Resources
http://www.docircuits.com/#home
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Delivery Guide
For starting a DC motor (ie due to its high starting current) this high starting
current is overcome using a motor starter. Web-based resources might prove
useful in explaining this – with the following web pages explaining 3 and 4
point starters that include a ‘no volt coil’ and overload protection -
http://www.electrical4u.com/starting-methodsto-limit-starting-current-
torque-of-dc-motor/. Learners could practice explaining the reasons for using
a DC motor starter and how a typical starter operates.
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Starter
1. Show with a diagram how the field winding is connected to the armature of
a series wound self excited motor. (2)
2. Show with a diagram how the field winding is connected to the armature of
a shunt wound self excited motor. (2)
12/05/2017 53/30

Aspire :
Challenge:
Starting a DC-Motor (DANGER)
The starting of DC motor is somewhat different from the starting of all other
types of electrical motors. This difference is credited to the fact that a DC
motor unlike other types of motor has a very high starting current that has
the potential of damaging the internal circuit of the armature winding of DC
motor if not restricted to some limited value.
http://www.electrical4u.com/starting-methods-to-limit-starting-current-
torque-of-dc-motor/
12/05/2017 54/30

Aspire :
Challenge:
LO-2-Understanding AC
research how alternating current (AC) is generated and introduce learners to
the mathematical theory of alternating waveforms, including the terms
amplitude, frequency and periodic time
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To apply mathematical theory for AC waveforms.
To understand how Alternating current is generated.

Generating an Alternating Current
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Aspire :
Challenge:
You’re already familiar with
the rotation generators use to
produce Alternating Current.
As the coil rotates through the
magnetic field a current is
induced in the coil.
As the coil completes a
revolution current is induced
in the opposite direction
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Flemming’s Left Hand Rule

Aspire :
Challenge:
This alternating current
when measured produces a
particular pattern in
voltage and current, the
pattern is a Sine wave and
the pattern is known as:-
Sinusoidal
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Aspire :
Challenge:
The AC Cycle
The stronger the magnetic
field and or the more coils
on the armeture the more
voltage and current is
induced.
The faster the armeture
rotates the more times in a
second the current changes
direction.
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Aspire :
Challenge:
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Identify
Amplitude
Peak Value
Peak to Peak
RMS (Max voltage = 6)

Aspire :
Challenge:
The voltage at any given
intermediate time can
be calculated based on;
Maximum Voltage
The angle of the
armeture at that time.
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Vi = Vmax SINθ

Aspire :
Challenge:
Q1 – Calculate the intermediate voltage of a generator where the maximum
voltage is 330V and the coil is at;
15° , 30°, 45°, 60° , respectively.
Q2 – Another generator is measured and the intermediate voltage is 240V
when the armature is at 60° what is the peak voltage of the generator?
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Vi = Vmax SINθ

Aspire :
Challenge:
Sinusoidal Wave Form Construction
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Vi = Vmax SINθ

Aspire :
Challenge:
Angular Velocity (rate of turning)
The Angular velocity basically describes
how fast the armature is rotating and
therefore what the frequency of the
Alternating Current is.
Frequency in Hertz = cycles per second.
So for 50 Hertz the angular velocity is 314.2
radians every second.
There are 2 times pi radians in one rotation
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Ѡ = 2πf (radians/second)
UK = 50 Hertz so;
Ѡ = 2xπx50 = 314.2
(radians/second)

Aspire :
Challenge:
AC - So Far
So we have
•generator, rotating in a magnetic field of flux density (B),
•with a number of coils of length (l)
•at velocity (v) in metres per second m/s
•With an angular velocity (w) measured radians/second
•giving a frequency in Hertz (Hz)
•producing an induced electromotive force measured in volts (E)
•with a current of (I)
•through a transformer (Vin : Vout)
Can you use your notes to create a simplified explanation taking you through this.
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To describe how Alternating current is behaves in a component.

Aspire :
Challenge:
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Aspire :
Challenge:
Inductive Reactance
When the current in an Inductor changes,
a back emf is created that opposes the
change in current, and the faster the
initial change in current the greater the
back emf. As we know the frequency is
dependent on the angular velocity.
Like resistance, reactance it is measured
in ohms.
XL = 2πfL
(Where L = Heneries)
This is a phasor diagram
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Aspire :
Challenge:
Capacitive Reactance
When an alternating voltage is applied,
current flows first in one direction, and
then the other.
The capacitor is first charging, and then
discharging but the capacitor never
reaches its fully charged state and
current continues to flow all the time.
The amount of current flowing will
depend on the angular velocity of the
applied voltage, and on
the capacitance of the capacitor.
Again we can use a Phasor Diagram
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Aspire :
Challenge:
Impedance Z with scale phasor
Impede – to delay, or prevent (impede progress, to
slow it down).
A scale phasor diagram can provide a solution to
calculating impedance in an AC circuit.
Step 1 - inductive reactance XL=2πfL
XL = 2 x 3.142 x 50 Hz x 0.035 mH*
XL=11
Step 2 – Pythagoras a2 + b2 = c2
121 + 100 = 221 √221= 14.86
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Aspire :
Challenge:
Impedance Z without scale phasor
Without a scale phasor (e.g. in an exam) you can still
work it through.
Step 1 - inductive reactance XL=2πfL
XL = 2 x 3.142 x 50 Hz x 0.035 mH*
XL=11
Step 2 – SOHCAHTOA, θ = tan-1 Opposite/Adjacent
tan-1 11/10 = 47.7°
Step 3 – SOHCAHTOA = Cosθ = Adjacent/Hypotenuse
Hypotenuse = Adjacent / Cosθ
Hypotenuse = 10 / Cos47.7° = 14.858Ω
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Aspire :
Challenge:
Exercises
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In all of these cases you
may also be given a
voltage V and asked to
calculate the Current I.
Remember I = V/Z
e.g. this circuit has a
voltage V of 100V and
an impedance Z of 22Ω
What is the current?
100V/22Ω = 6.7A

Aspire :
Challenge:
Phase Shift- In Phase
Here the Voltage
and current are
said to be ‘in
phase’ with each
other.
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To apply mathematical theory for Phase Shift.
To understand Phasor Diagrams.

Aspire :
Challenge:
Phase Shift – Out of Phase
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Here the current
is leading the
voltage or the
voltage is lagging
behind the
current.
Leading and
lagging are simply
relevant to what
you are focusing
on.
Current is leading by 30°

Aspire :
Challenge:
90° out of Phase = cosine wave form
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Aspire :
Challenge:
We’ve just seen that sinusoidal waveforms of the same frequency can have a
Phase Difference between themselves which represents the angular
difference of the two sinusoidal waveforms. Also the terms “lead” and “lag”
as well as “in-phase” and “out-of-phase” were used to indicate the
relationship of one waveform to the other.
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Aspire :
Challenge:
12/05/2017 76/30

Aspire :
Challenge:
Phasor diagram
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Aspire :
Challenge:
Phasor addition (using a force parallelogram)
Sometimes it is necessary
when studying sinusoids to
add together two
alternating waveforms,
Consider two AC voltages, .
The total voltage, VT of the
two voltages can be found
constructing a scale
parallelogram in which two
of the sides are the
voltages, V1 and V2 are as
shown.
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Aspire :
Challenge:
Phasor Addition Exercises
Construct suitable scale force parallelograms for the following;
1. 240V is leading 330V by 30° what is the total voltage?
2. 40V is lagging 30V by 30° what is the total voltage?
3. 240V and 40V are in phase and are leading the in phase 330V and 30V
what is the total voltage?
4. Can the first two new vectors drawn to prove and answer to 3?
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Aspire :
Challenge:
Phasor subtraction
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If this is addition
This must be?
Subtraction

Aspire :
Challenge:
Phasor Addition Exercises
Construct suitable scale force parallelograms for the following;
1. 240V is leading -330V by 30° what is the total voltage?
2. -40V is lagging 30V by 30° what is the total voltage?
3. Can you take the angles from the previous two questions to work out the
overall combination?
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Aspire :
Challenge:
Three Phase Phasor Diagram
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Aspire :
Challenge:
Finally
• Vectors, Phasors and Phasor Diagrams ONLY apply to sinusoidal AC waveforms.
• A Phasor Diagram can be used to represent two or more stationary sinusoidal
quantities at any instant in time.
• Phasor diagrams can be drawn to represent more than two sinusoids. They can be
either voltage, current or some other alternating quantity but the frequency of all
of them must be the same.
• Generally, the length of a phasor represents the R.M.S. value of the sinusoidal
quantity rather than its maximum value.
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Aspire :
Challenge:
Power supplies
Single Phase
Three Phase
Diode Rectification AC - DC
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Aspire :
Challenge:
Transformers Recap
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A very useful property of transformers
is the ability to transform voltage and
current levels according to a simple
ratio, determined by the ratio of input
and output coil turns.
If the energized coil of a transformer is
energized by an AC voltage, the
amount of AC voltage induced in the
empowered coil will be equal to the
input voltage multiplied by the ratio of
output to input wire turns in the coils.

Aspire :
Challenge:
Diode Rectification (Half Wave)
Rectification. Simply defined, rectification is the conversion of alternating current (AC)
to direct current (DC).
This involves a device that only allows one-way flow of electrons. As we have seen,
this is exactly what a semiconductor diode does.
The simplest kind of rectifier circuit is the half-wave rectifier. It only allows one half of
an AC waveform to pass through to the load.
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Aspire :
Challenge:
Diode Rectification (Full Wave)
If we need to rectify AC power to obtain the full use of both half-cycles of the sine
wave, a different rectifier circuit configuration must be used.
Such a circuit is called a full-wave rectifier. One kind of full-wave rectifier, called the
center-tap design, uses a transformer with a center-tapped secondary winding and
two diodes.
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Aspire :
Challenge:
Diode Rectification (Full Wave Bridge)
Another, more popular full-wave rectifier design exists, and it is built around a four-
diode bridge configuration, this design is called a full-wave bridge.
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Aspire :
Challenge:
Polyphase Rectification
When polyphase AC is rectified, the phase-shifted pulses overlap each other to
produce a DC output that is much “smoother”
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Aspire :
Challenge:
http://www.electronics-tutorials.ws/accircuits/sinusoidal-waveform.html
http://www.learnabout-electronics.org/index.php
http://www.schoolphysics.co.uk/age16-19/
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Aspire :
Challenge:
12/05/2017 91/30

Unit 4 year 12 electrical-electronics

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Unit 4 year 12 electrical-electronics