Search for notes by fellow students, in your own course and all over the country.
Browse our notes for titles which look like what you need, you can preview any of the notes via a sample of the contents. After you're happy these are the notes you're after simply pop them into your shopping cart.
Title: Transformers and Induction Motors
Description: Utilisation of Electrical Energy - Assignment 1. Operation and construction of power transformers. Operation and construction of a poly-phase induction motors. Grade received: DISTINCTION Tutor: Mark Nailis Institute: Middlesbrough College Programme: Higher National Certificate (HNC) in Electrical and Electronic Engineering
Description: Utilisation of Electrical Energy - Assignment 1. Operation and construction of power transformers. Operation and construction of a poly-phase induction motors. Grade received: DISTINCTION Tutor: Mark Nailis Institute: Middlesbrough College Programme: Higher National Certificate (HNC) in Electrical and Electronic Engineering
Document Preview
Extracts from the notes are below, to see the PDF you'll receive please use the links above
HNC Electrical and Electronic Engineering
Year One - 2013/14
Module: Module Digital & Analogue Devices
Power
Supplies
Keith A
...
Hudson
2
Digital / Analogue
03 December, 2013
Contents
1
Linear Power Supply (Q1a)
...
1
Transformer
...
2
Rectifier
...
3
Smoothing
...
4
Regulator
...
5
The Power Supply
...
7
3
Switched Mode vs
...
8
4
Construction of a Linear Power Supply (Q3a)
...
1
Power Source
...
2
Transformer (TR1)
...
3
Rectifier (BR1)
...
4
Smoothing (C1)
...
5
Regulator (U1)
...
6
Load (RV1)
...
7
Complete circuit - simulation
...
8
Complete circuit - wired
...
13
6
Power Supply Analysis and Improvements (Q3c)
...
17
HNC Electrical and Electronic Engineering
Year One: 2013/14
Keith A
...
4
Figure 2: The changes in the wave form of the voltage / current
...
4
Figure 4: Bridge using separate diodes
...
5
Figure 6: Regulator TO-220 package
...
6
Figure 8: Block diagram of a switched mode power supply
...
10
Figure 10: The power supply under full load
...
11
Figure 12: Almost 5V output
...
7mA
...
7mA
...
12
Tables
Table 1: Types of transformer
...
7
Table 3: Advantages / disadvantages - Switched mode and Linear power supply
...
13
Table 5: Line Regulation
...
13
Table 7: Power loss
...
15
Table 9: Efficiency of the Rectifier (calculated at full load)
...
16
Table 11: Efficiency of the Rectifier (calculated at full load)
...
Hudson
Digital / Analogue
03 December, 2013
4
1 Linear Power Supply
(Q1a)
The purpose of a linear power supply is to provide a suitable voltage and current to a specific load
...
The available high voltage (mains), alternating current must be
converted into a low voltage, regulated, direct current
...
C
...
C
...
The values of the components
are dependent upon the input voltage, current and the requirements of the load
...
A
...
Voltage
Input
Transformer
Rectifier
Smoothing
Regulator
D
...
Voltage
Output
Figure 1: The conversion steps
...
The components required to convert A
...
mains voltage to a suitable D
...
low voltage are shown in Figure 1
...
1
...
(electrical4u
...
d
...
This causes a
magnetic flux that is concentrated by the soft iron core
...
This induces an EMF (voltage) in the
secondary winding
...
For further
explanation see Faradayβs Law of Electro-magnetic Induction
...
Hudson
Digital / Analogue
03 December, 2013
5
Table 1: Types of transformer
Number of
Primary
Windings
Number of
Secondary
Windings
Transformer
Type
N
>N
Step-up
N
N
1:1
(Isolation)
N
Step-down
Result
O/P Voltage
>
I/P Voltage
O/P Voltage
=
I/P Voltage
O/P Voltage
<
I/P Voltage
For our power supply a step-down transformer will be used to reduce the high voltage (see Figure 2
...
2), high current power
...
1
...
2 from the transformer into direct current
...
3 shows the result of full-wave rectification
...
(There may be circumstances were only negative voltages are required
...
) Half wave rectification is achieved by only allowing
positive (or negative) current to flow
...
A half-wave rectifier combined with a centre-tapped transformer, also results in a full wave rectification
...
Figure 4: Bridge using separate diodes
Figure 5: Bridge diode package
1
...
3)
...
4)
...
They are usually electrolytic type capacitors with a value over
470ΞΌF
...
Hudson
6
Digital / Analogue
03 December, 2013
1
...
To produce a constant output voltage (see Figure 2
...
The most popular form for these solid state devices is the TO-220
package (see Figure 6)
...
The βxxβ indicates the output voltage,
common are 5 and 12V
...
5 The Power Supply
Figure 7 shows a complete power supply circuit
...
These are not part of the power supply
...
Next are the smoothing capacitor and the regulator
...
Hudson
Digital / Analogue
03 December, 2013
7
2 Load Regulation
(Q1b)
Load regulation is a measure of how the output voltage of a power supply is affected by changes to the output
load (current)
...
Table 2: Load regulation of a 12VDC, 0-5A power supply
π πππππππ_ππππ = 12
...
00 π (Implied in the question
...
95 π
πΏπππ π πππ’πππ‘πππ =
(π πππ _ππππ β π πππ₯ _ππππ )
β 100
π πππππππ_ππππ
HNC Electrical and Electronic Engineering
(12
...
95)
β 100
12
...
05
πΏπππ π πππ’πππ‘πππ =
β 100
12
...
00
πΏπππ π πππ’πππ‘πππ =
12
...
417 %
πΏπππ π πππ’πππ‘πππ =
Year One: 2013/14
Keith A
...
Linear
(Q2)
In a linear power supply, the supply voltage is reduced by a transformer and then it is rectified
...
Figure 8 shows the
steps involved in a switched mode power supply (Wellforces Ltd, 2012)
...
M
...
The βInrush current controlβ protects the power supply at start up
...
C
...
C
...
The βConvertorβ is a
high frequency transformer
...
C
...
C
...
Feedback is another difference
between the switch mode supply and a linear one
...
Large/heavy transformer and heat sink
...
Cost
Cheap for low power output
...
Reliability
High (decreases as temperature rises)
...
Input voltage
Efficiency significantly reduced if a wide
input range is required
...
C
...
001 β 0
...
Small/light (typically 75 β 90% less than linear
equivalent)
...
Cheap for low power output
...
(Getting cheaper
...
More components but fit on a
single PCB
...
M
...
Can be reduced using filters
...
Can accept D
...
input
...
1 β 3
...
5 β 10ms)
Year One: 2013/14
Keith A
...
1 Power Source
Amplitude of 339V and a frequency of 50Hz were selected for the generator
...
e
...
4
...
This results in an approximate step-down of 12:1
...
This actually drops the input voltage down to about 20V
...
3 Rectifier (BR1)
BRIDGE
A bridge rectifier was chosen to convert the alternating current from the transformer to direct current
...
4
...
The default value of 1mF was used
...
0V (no load) down to 20
...
4
...
The minimum input voltage for this component is
19
...
0V (Farnell UK Limited, 2013)
...
4
...
HNC Electrical and Electronic Engineering
Year One: 2013/14
Keith A
...
7 Complete circuit - simulation
Figure 9 and Figure 10 show the complete power supply circuit
...
C
...
C
...
These values could then be used to calculate various
performance figures for the power supply
...
Hudson
Digital / Analogue
03 December, 2013
11
4
...
(It is show without wires for clarity
...
A number of
meters readings were taken at various points in the circuit (See Figure 12, Figure 13 and Figure 14)
...
Figure 11: Wiring the circuit on a Digiac 3000 β Power supply module
Figure 12: Almost 5V output
Figure 13: Full load - 51
...
7mA
Year One: 2013/14
Keith A
...
The output is βfuzzyβ due to interference from electrical devices
in the classroom
...
Hudson
13
Digital / Analogue
03 December, 2013
5 Calculations
(Q3b)
Using the values in Figure 9 and Figure 10 various performance values can be calculated for the power supply
...
00 π
π πππ _ππππ = 12
...
12 π΄
π πππ₯ _ππππ = 11
...
79 π΄
(π πππ _ππππ β π πππ₯ _ππππ )
%πΏπππ π πππ’πππ‘πππ =
β 100
π πππππππ_ππππ
(12
...
90)
β 100
12
...
10
πΏπππ π πππ’πππ‘πππ =
β 100
12
...
00
πΏπππ π πππ’πππ‘πππ =
12
...
833 %
πΏπππ π πππ’πππ‘πππ =
Table 5: Line Regulation
The following are input voltages:
π πππ = 220 π (π΄πππππ‘π’ππ = 310 π)
π πππ₯ = 250 π (π΄πππππ‘π’ππ = 352 π)
Ξπππ = π πππ₯ β π πππ
Ξπππ = 250 β 220
Ξπππ = 30 π
%πΏπππ π πππ’πππ‘πππ =
Ξπ ππ’π‘
β 100
Ξπππ
0
β 100
30
πΏπππ π πππ’πππ‘πππ = 0
...
e
...
0 π
π πππ₯ = 12
...
0 β 12
...
The
above tests were carried out under nominal load
...
Table 6: %Efficiency of the power supply (calculated at full load)
π΄πΆ πΌπππ’π‘ πππ€ππ = ππΌπ β πΌ πΌπ
π΄πΆ πΌπππ’π‘ πππ€ππ = 239 β 0
...
85 π
π·πΆ ππ’π‘ππ’π‘ πππ€ππ = π πππ β πΌ πππ
π·πΆ ππ’π‘ππ’π‘ πππ€ππ = 11
...
79
π·πΆ ππ’π‘ππ’π‘ πππ€ππ = 9
...
401
%πΈπππππππππ¦ =
β 100
35
...
22% (2ππ)
%πΈπππππππππ¦ =
As Table 6 show, this linear power supply is not efficient
...
Hudson
14
Digital / Analogue
03 December, 2013
Table 7: Power loss
πππ€ππ πΌπ = πΌπππ’π‘ π£πππ‘πππ β πΌπππ’π‘ ππ’πππππ‘
πππ = 239 β 0
...
85 π
πππ€ππ ππ’π‘ = ππ’π‘ππ’π‘ π£πππ‘πππ β ππ’π‘ππ’π‘ ππ’πππππ‘
π ππ’π‘ = 11
...
79 = 9
...
85 β 9
...
45 π (2ππ)
Table 7 shows that 35
...
4 W leaves as electrical power
...
45W leaves the power supply as other forms of energy
...
A small amount will be lost through the transformer, some down to internal resistance of
the windings, some due flux leakage and some as sound
...
Hudson
15
Digital / Analogue
03 December, 2013
6 Power Supply Analysis and Improvements
(Q3c)
By comparing the power going into individual components against the power out we will identify the inefficient
component
...
9 β 1
...
422 π
π΄πΆ ππ’π‘ππ’π‘ πππ€ππ
β 100
π΄πΆ πΌπππ’π‘ πππ€ππ
35
...
85
%πΈπππππππππ¦ = 98
...
Table 9: Efficiency of the Rectifier (calculated at full load)
π πππ‘πππππ πΌπππ’π‘ πππ€ππ = ππΌπ β πΌ πΌπ
π πππ‘πππππ πΌπππ’π‘ πππ€ππ = 25
...
8
π πππ‘πππππ πΌπππ’π‘ πππ€ππ = 20
...
9 β 0
...
401 π
π πππ‘πππππ ππ’π‘ππ’π‘ πππ€ππ
β 100
π πππ‘πππππ πΌπππ’π‘ πππ€ππ
9
...
72
%πΈπππππππππ¦ = 45
...
Here is where most of the power is
βlostβ
...
9V to 12V without altering the current significantly
...
By reducing the output value of the transformer we can reduce the voltage supplied to the regulator, and hence
reduce the amount of energy converted to heat
...
Linear regulators will only produce the rated output voltage if the input voltage is at least a specific number of volts
above the rated output voltage
...
π·πππππ’π‘ ππππ‘πππ (π ππ ) = ππππππ’π πΌπππ’π‘ ππππ‘πππ β π ππ‘ππ ππ’π‘ππ’π‘ ππππ‘πππ
The 7812 Regulator in the power supply has a Dropout Voltage, π ππ = 2π, although the minimum supply voltage is
specified as 14
...
HNC Electrical and Electronic Engineering
Year One: 2013/14
Keith A
...
5V
...
1
22
...
4
20
...
2
18
...
0
17
...
1
15
...
1
14
...
5V which equals the
minimum input voltage for the Regulator
...
7 β 0
...
025 π
π πππ‘πππππ ππ’π‘ππ’π‘ πππ€ππ = π πππ β πΌ πππ
π πππ‘πππππ ππ’π‘ππ’π‘ πππ€ππ = 11
...
74
π πππ‘πππππ ππ’π‘ππ’π‘ πππ€ππ = 8
...
806
%πΈπππππππππ¦ =
β 100
11
...
87% (2ππ)
%πΈπππππππππ¦ =
Table 11 shows a massive increase in efficiency, up from about 45% to almost 80%
...
79A down to 0
...
This equates to an output power of 8
...
5W)
...
The value of the smoothing
capacitor may be the cause
...
33ΞΌF capacitor
on the input side of the regulator and a 0
...
A heat sink on the regulator would also be an improvement to the power supply
...
HNC Electrical and Electronic Engineering
Year One: 2013/14
Keith A
...
All About Circuits : Free Electric Circuits Textbooks
...
allaboutcircuits
...
electrical4u
...
d
...
[Online]
Available at: http://www
...
com/what-is-transformer-definition-working-principle-of-transformer/
[Accessed 24 11 2013]
...
L7812CP - STMICROELECTRONICS - IC, V REG +12V, 7812, TO-220FP | Farnell UK - Op
...
farnell
...
LJ Create, 2013
...
[Online]
Available at: http://www
...
com/products/product
...
Marian, P
...
Step Down Transformer
...
electroschematics
...
Wellforces Ltd, 2012
...
[Online]
Available at: http://nz
...
com/resources/introduction-linear-switching-power-supplydifferences/197/#
...
Wikipedia, 2013
...
[Online]
Available at: http://en
...
org/wiki/Diode_bridge
[Accessed 25 11 2013]
...
Load regulation - Wikipedia, the free encyclopedia
...
wikipedia
...
HNC Electrical and Electronic Engineering
Year One: 2013/14
Title: Transformers and Induction Motors
Description: Utilisation of Electrical Energy - Assignment 1. Operation and construction of power transformers. Operation and construction of a poly-phase induction motors. Grade received: DISTINCTION Tutor: Mark Nailis Institute: Middlesbrough College Programme: Higher National Certificate (HNC) in Electrical and Electronic Engineering
Description: Utilisation of Electrical Energy - Assignment 1. Operation and construction of power transformers. Operation and construction of a poly-phase induction motors. Grade received: DISTINCTION Tutor: Mark Nailis Institute: Middlesbrough College Programme: Higher National Certificate (HNC) in Electrical and Electronic Engineering