Power converter for a DC motor in electrical
go-kart
P5 Project
Group EE5-511
Department of Energy Technology
Aalborg University
21 December 2011
Aalborg University
Title:
Semester:
Semester theme:
Project period:
ECTS:
Supervisor:
Project group:
Power converter for a DC motor in electrical go-kart
5
Transmission and conversion of energy in electrical machines
and power system
02.09.11 to 21.12.11
17
Benoı̂t Bidoggia
EE5-511
Synopsis:
Nicolaj Nielsen
Ernir Freyr Gunnlaugsson
Eneko Unamuno Ruiz
Tamara Laencina Santolaya
Ugur Sancar
Andres Lopez-Aranguren
Copies:
8
Pages, total:
95
Appendix:
2
Supplements: CD
The purpose of this report is to design, simulate
and build a power converter for a DC motor that
has to be implemented in a go-kart. The main
goal has been to design a power converter. This
has been done by calculating the exact values of
all the elements of the converter so it can work in
specific voltage/current ranges.
There have been designed some prototypes in
order to improve the first one and reach an efficient
final version of the system.
A current controller has also been designed to
make the converter work depending on the value
of the measured current and the speed decided by
the user. All the circuits have been built in PCB
footprints in order to try to reduce noises in the
system and make it work better. A four phase
system has been designed but it was not possible
to build it, but one of the four phases works as
expected in tests and simulations.
Next step is to build all the four phases and
connect them to the DC motor.
By signing this document, each member of the group confirms that all participated in the project work and thereby all members are collectively liable for
the content of the report.
iii
Preface
This report is done by group EE5-511 on 5th semester at Department of Energy
Technology at Aalborg University. The theme for the semester is transmission and
conversion of energy in electrical machines and power system. The purpose of this report
is to design a power converter for a go-kart. The group would like to send thanks to
our supervisor Benoı̂t Bidoggia and Walter Neumayr who helped the group a lot when
working in the laboratory.
Reading guide
Through the report there will be references to sources, the references are placed in the
Bibliography. The used method for referring to sources is the Harvard method where, it
refers with [Name, Year]. If the reference is included in a sentence before the dot, the
reference covers the sentence. If the reference is after a dot, it covers the paragraph or
section. If there are more than one source with the same name and year the source gets
a letter after the year. A reference leads to the Bibliography where the source is given by
the author, title, edition, publisher, hyperlink and date. Figures, tables, equations and
calculations have numbers that indicates which chapter they belongs to and a number.
For an example the first figure in Chapter 3 has the number 3.1 and the next figure has
3.2. There is a caption to each figure and table. Appendixes are indicated with a letter.
A CD is attached to the report, which includes all data from experiments, program files,
the report in PDF and PDF-copies of all websites. All the files on the CD have a number.
A reference to a file on the CD appears by referring to the Appendix and the number on
the CD, e.g. Appendix A 3.
Publication of the entire or parts of this report is allowed only with reference and by
appointment with the authors.
v
Table of Contents
1 Introduction
1
2 Problem statement
3
3 DC-motors
3.1 Different types of DC-motors . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 The electrical system of the go-kart . . . . . . . . . . . . . . . . . . . . . . .
5
5
8
4 Power converter
4.1 Buck converter . . . . . . . .
4.2 Buck-Boost Converter . . . .
4.3 Cuk Converter . . . . . . . .
4.4 Full Bridge Converter . . . .
4.5 Choice of the power converter
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11
11
12
13
14
15
5 Design of the power converter
5.1 Topology of the converter . .
5.2 Design of the Buck converter
5.3 Multiphase . . . . . . . . . .
5.4 Design of the inductor . . . .
5.5 The converter controller . . .
5.6 Choice of the components . .
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17
17
19
24
29
32
42
6 Simulations of the converter
6.1 Simulations . . . . . . . . .
6.2 Simulation 1: Verification of
6.3 Simulation 2: Verification of
6.4 Simulation 3: Verification of
6.5 Simulation 4: Verification of
6.6 Simulation 5: Verification of
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45
45
46
48
49
55
56
7 Test of the converter
7.1 Test 1: Measurement of the currents and voltages . . . . . . . . . . . . .
7.2 Test 2: Verification of the output voltage as a function of the duty cycle
7.3 Test 3: Verification of the power losses in the converter . . . . . . . . .
7.4 Test 4: Verification of the efficiency of the converter . . . . . . . . . . .
7.5 Test 5: Verification of the border between CCM and DCM . . . . . . . .
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59
61
65
66
69
70
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the voltages and currents . . . . . .
the output voltage . . . . . . . . . .
the power losses in the converter . .
the efficiency of the converter . . . .
the border between CCM and DCM
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8 Conclusion
73
9 Improvement proposals
75
vii
Group EE5-511
Bibliography
A The
A.1
A.2
A.3
77
Buck-Boost converter
79
Topology of the converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
Design of the converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
Conclusion of the Buck-Boost converter . . . . . . . . . . . . . . . . . . . . 85
B Contents on the CD
viii
TABLE OF CONTENTS
I
Introduction
1
In recent years, electric go-karts have become more and more popular, as a competition
vehicle and also as a hobby automobile. As a consequence of this, this field has been
deeply researched and improved. Therefore the motivation of this report is to try to find
a good solution to implement into an electric go-kart in order to be able to build it and
make it work.
The purpose of this project is to build an electric go-kart up with an electric motor and
a power converter that can be controlled manually by the driver with a throttle pedal.
The focus in the report will be, how to design a DC/DC converter with an eye to build
one and install in the go-kart. The go-kart that the group will work with is a property of
Aalborg University but it is only a skeleton because there is no working machine in it.
In the report, different types of motors will be discussed and which type will be used for
the go-kart. After deciding the motor type, some types of converters will be discussed and
their function, advantages and disadvantages, will be described. When having calculated
all the values of parameters in the electric circuit and the components are chosen, the
converter will be built. The converter will be simulated to see how it should work in the
real case and afterwards the real converter will be tested. When the tests are finished and
data processing is done, the results will be compared with the simulations to see if the
real converter works as the simulations.
The motivation for making a project about a power converter is because nowadays
converters are used in all kind of electric apparatus, e.g. mobile phones or laptops where
the application relies on a battery source and also to control DC machines, which is the
point of this report.
In recent decades the market for power electronics has expanded significantly because of
the huge rise of the development of electronic apparatus. The purpose of a DC/DC
converter is to convert one DC signal to another DC signal with a higher or lower
magnitude. As seen in the block diagram in Figure 1.1 the power converter gets an
input signal from a source. The switching speed of the converter is controlled by a switch
that takes a measurement from the output signal and a reference that can be a throttle
pedal. The input signal is then handled by the converter and gives an output signal to
the load.
1
Group EE5-511
1. Introduction
Figure 1.1. A block diagram of a power converter.
[Mohan, 2003]
The purpose of a DC/DC converter is to increase or decrease the input voltage coming
from a battery or another source, in order to supply a system which needs a different
voltage than the power supply. Being specific about this project, the voltage supplied by
the batteries is between 42V and 52.4V, and the DC motor works in a range from 0.1V to
48V, therefore a DC/DC converter is needed in order to change the signal to the motor.
The converter needs a PWM signal to control the converter, which is necessary to be
created by another circuit apart from the converter. It is also necessary to regulate the
current that flows in the converter by the PWM signal. This is done to avoid that the
current in the components exceeds the value that the components can handle. Therefore,
it will be necessary to create an implemented circuit to do this.
2
Problem statement
2
The problem
The main focus of this project is to examine how a DC/DC power converter is designed.
This examination should lead to a real product that can handle some specifications that
will be decided before designing the converter. This real product will be compared with
both simulations and theory.
Procedure
To be able to design the converter for a motor in an electrical go-kart an study of how a
DC motor works has to be done and look into which types of motors exist. After that the
specifications in the DC motor, that will be used, will be described. There will also be
specified which batteries will be used.
When the motor has been studied it is necessary to look into which types of converters
exist and decide which one would fit for the given application. The converter has to be able
to work in a given electrical specifications, where a driver can control the speed manually
with a simple throttle pedal, that is a simple potentiometer. The specifications will be
described in Chapter 5. To design the converter a mathematical model of the converter
will be made and this will be designed with an eye to building it.
When the converter and the switch controller is built it will be necessary to test and
compared it with simulations, that will be made in order to check that it is working as
expected.
The simulations will show the behavior of the system. The programs used for simulations
are Multisim 11, Proteus 7.7 and PSIM 6.0. Otherwise it has been decided to design the
printed board in stead of using premade circuit boards in order to improve the result.
Therefore Orcad Layout 16.3 will be used to make the PCB Footprint.
3
DC-motors
3
In this chapter a general description of DC-motors will be made, including all the physical
characteristics of the DC-motors and how they work. Furthermore, all the specifications
of the chosen motor will be described. In the end of the section a short description of the
batteries will be done.
3.1
Different types of DC-motors
In today’s industry there are two types of motors mainly used. These motors will be
described in short terms in this section. The two types are:
• Brushed DC motor
• Brushless DC motor
Brushed DC motor
Brushed DC motors (BDC) are widely used today. This is mainly because of the low cost
and because they are easy to control. All BDC motors are constructed in the same way
and the same components. The components are: a stator, a rotor, carbon brushes and a
commutator.
The stator is where the magnetic field that surrounds the rotor is generated. The stator
is either made from electromagnetic windings or by permanent magnets. In Figure 3.1
the stator is the red part surrounding the rotor (armature).
The rotor is where electricity is produced by the magnetic field from the stator. The
rotor is made of one or more windings. When the rotor is influenced by the stator, a
magnetic field is produced around the rotor. The magnetic poles on the rotor attract
poles from the stator of the opposite type, the north pole on the rotor attracts the south
pole on the stator and reverse. This attraction causes the rotor to turn. As the rotor
turns it gets influenced with different sequences which does that the magnetic poles on
the rotor do not overrun the poles on the stator. In Figure 3.1 the rotor is the same as
the armature.
5
Group EE5-511
3. DC-motors
The BCD motor does not need a control to switch the direction of the current because
this is done inside the motor. This is done with the brushes and the commutator. The
commutator is a segmented copper ring that is placed on the axle. When the rotor
turns this copper ring turns too and the carbon brushes slide over different segments
on the copper ring and connect different windings on the rotor. This means that when
applying a voltage source to the carbon brushes a dynamic magnetic field inside the motor
is generated. The commutator and the carbon brushes can be seen in Figure 3.1.
It must be noted that the commutator and the carbon brushes are the two things that
tend to wear out the most in the BDC motor because these things slide against each other
when the motor is turning.
Figure 3.1 shows the construction of a BDC motor.
Figure 3.1. A sketch of a brushed DC motor.
[www.tutorvista.com, 2010]
There are four types of BDC motors: permanent magnet motor(PMDC), Shunt-wound
motor(SHWDC), Series-wound motor(SWDC) and Compound Wound motor(CWDC).
The PMDC motor is the most common motor in the world. Instead of electromagnet
windings in the stator it has permanent magnets. It is generally used where fractional
horsepower is involved because it is most cost effective. The disadvantage with the PMDC
is that the permanent magnets lose their magnetic power over time.
In the SHWDC motor the field coil is in parallel with the armature. This means that
the current over the coil is not the same as over the armature. As a result of this the
motor has a good speed control. These motors are generally used when more than five
horsepowers are required. A loss of magnetic power is not an issue in SHWDC so they
are more robust than PMDC.
In the SWDC motor the field coil is in series with the armature. These motors are
used where high torque is required because the current over the coil is the same as over
the armature. The disadvantage with the SWDC motor is that the speed of it is not as
accurate as for PMDC and SHWDC.
The last BDC motor type is the CWDC motor. It is a combination of SHWDC motor
and SWDC motor because the field coil is both in series and in parallel with the armature.
6
3.1. Different types of DC-motors
Aalborg University
The performance is similar to SHWDC and SWDC but has higher torque than SHWDC
and better speed control than SWDC.
[Condit, 2004]
Brushless DC motor
Brushless DC motors (BLDC) are getting more and more popular in todays industry. As
the name implies the BLDC motor does not have brushes like BDC motor but instead
uses electrical commutation. The BLDC motor has many advantages over BDC motors.
Some of these advantages are: bigger speed range, noiseless operation, higher efficiency,
higher dynamic response, better speed vs. torque characteristics and longer lifetime.
A BLDC motor is normally a synchronous machine because the magnetic field produced
by the stator is in phase with the magnetic field produced by the rotor. The BLDC motor
has three main components, the stator, the rotor and hall sensors.
Different from the BDC motor, the stator in a BLDC motor is made from stack of steel
plates and windings. The windings in the stator are placed in some slots in the stacked
plates. Each winding is constructed with numerous coils which are connected to form a
winding. These windings are distributed over the stator to form an even number of poles.
There are two main types of windings which depend on how the coils are connected. The
types are trapezoidal and sinusoidal windings. This means how the electromotive force
signal from the motor will be.
In the BLDC motor the rotor is constructed with permanent magnets and may vary
from two to eight pole pairs. The magnets material depends on the required magnetic
field density.
The last components in the BLDC motor are the hall sensors. The main difference
between BDC motor and BLDC motor is the commutation. The BLDC motor has no
commutator but instead it is electrically controlled. To rotate the rotor the stator windings
have to be energized in a sequence. To know which stator winding should be energized
the hall sensors sense the position of the rotor. They are placed in the non rotating end
of the motor at the end of the rotor as seen in Figure 3.2.
[Yedamale, 2003]
Figure 3.2. A sketch of a brushless DC motor.
[Yedamale, 2003]
7
Group EE5-511
3.2
3. DC-motors
The electrical system of the go-kart
In this section the details about the DC-motor and the batteries, inside the go-kart, will
be described in a short way.
The DC-motor
The DC-motor used in this project is a brushed DC-motor of the type Permanent Magnet
Direct Current motor (PMDC-motor). In Figure 3.3 some characteristics of the DC-motor
are shown together with a picture of the DC-motor. The full datasheet for the motor can
be found in Appendix B 1.
Mars 0708 PMDC
Voltage - no load (V )
Continuous current (ia )
Peak Current
Power
Max. power
No load speed
Loaded speed
Torque (τ )
Peak torque
Motor constant (Km )
Self-inductance (La )
Ohmic basic resistance (Ra )
motor
[0.1 ; 48]V
100A
300A
4800W
15000W
3700RPM
3200RPM
18.08Nm
38.00Nm
0.13V·s or
60µH
0.01Ω
Nm
A
Figure 3.3. Some of the characteristics of the DC-motor are shown in the table on the left
[Xixing, 2007] and [Motenergy, 2011]. The picture to the right shows the DC-motor
in its real form.
Dynamic model of the PMDC-motor
The dynamic model of this motor can be split into two parts; an electrical way and a
mechanical way. Figure 3.4 shows the electrical part of the motor and the mechanical
part.
Ra
La
Vc
τL
τJ
V
Ea
τb τc
τ
ω
J
ia
Figure 3.4. Shows the electrical and the mechanical part of the PMDC-motor.
By using Kirchoff’s loop equation on the electrical model, it is possible to write an equation
8
3.2. The electrical system of the go-kart
Aalborg University
for the voltage (V ) in the circuit. Equation (3.1) shows this connection.
V = VR + VL + Vc + Ea
m
(3.1)
V = Ra · ia + La ·
dia
+ Vc + Ea
dt
[V]
Where Ra is the resistance in the armature, La is the self inductance in the motor windings,
Vc is the voltage loss in the carbon brushes, Ea is the generated voltage and ia is the current
in the armature.
The generated voltage (Ea ) is equal to Equation (3.2).
Ea = ωm · Km
(3.2)
[V]
Where ωm is the angular velocity on the motor axis and Km is the motor constant. With
connecting Equation (3.1) and (3.2) a new expression for the voltage equation is
(3.3)
V = Ra · ia + La ·
dia
+ Vc + ωm · Km
dt
[V]
The mechanical part, in the dynamic model, is the rotating part of figure 3.4 and can be
written as the total torque in the system (τ ). This is shown in Equation (3.4).
τ = τJ + τb + τc + τL
m
(3.4)
τ =J·
dωm
+ b · ωm + τc + τL
dt
[Nm]
Where J is the moment of inertia on all the rotating parts on the go-kart, b is the viscous
friction constant, τc is the dry friction in the system and τL is the mechanical load from
the external forces.
As known the total torque (τ ) is proportional to the current in the armature (ia ), with
Km as proportional constant. Thus, Equation (3.4) is equal to Equation (3.5).
(3.5)
τ =J·
dωm
+ b · ωm + τc + τL = Km · ia
dt
[Nm]
[Andersen and Pedersen, 2011]
The batteries
The go-kart is provided by four OPTIMA YellowTop U 4.2 batteries. The batteries are
connected in series to get 48 volts as shown on the left in Figure 3.5, on the right in
Figure 3.5, one of the batteries used in the project, is shown.
The full datasheet of the batteries can be found in Appendix B 2.
9
Group EE5-511
+
48V
-
DC
3. DC-motors
Motor
DC
- +
- +
- +
- +
12V
12V
12V
12V
Figure 3.5. On the left, the circuit of the four batteries in series is shown. On the right there
is a picture of one of the battery used in the project.
The data about the batteries are shown in Table 3.1.
Performance Data
1 battery
Open circuit voltage (fully charged): 13.1V
Close circuit voltage (fully charged): 12V
Internal Resistance (fully charged): 0.0028Ω
Capacity:
55Ah
Total (4 batteries)
52.4V
48V
0.0112Ω
55Ah
Table 3.1. Optima YellowTop YT U 4.2 batteries performance data.
[SPIRALCELL technology, 2005] and [Batteries, 2005]
10
Power converter
4
The power converter is the element that has to adjust the voltage of the batteries to
the range of voltage that the DC motor works. There are many different types of
DC/DC converters, but in this chapter, a brief explanation of the ones that could suit
this application will be made. The advantageses and disadvantages of each converter will
be described and in the end, one will be chosen to be designed.
4.1
Buck converter
The Buck converter (step-down converter) delivers a lower DC voltage than the supplied
voltage. Its main application is to regulate DC power supplies and DC motor speed
control. The buck converter is known by many names: voltage step-down converter,
current step-up converter, chopper, direct converter, etc.
The main characteristic of the Buck converter is that the polarity of the output voltage
is the same as at the input voltage. The noise generated at the output is low due to a LC
circuit configuration which constitutes a low pass filter. The input tension receives pulses
from the transistor (when it conducts) and the Buck converter generates a loud noise at
the input power.
Figure 4.1. The Buck converter circuit diagram. Input voltage source (Vi ), controlled switch
(S), inductor (L), diode (D), capacitor (C) and a load resistance (Ro .
Advantages and disadvantages
The advantages of a Buck converter are many. One of the most important one is that it
is highly efficient. Besides that, it is simple to design because this kind of converter does
11
Group EE5-511
4. Power converter
not have a transformer, it is only necessary to put a minimal stress on the switch and a
relatively small output filter for low output ripple.
Some of the disadvantages of using a Buck converter are that there is no input to output
isolation, there is also a possibility of having over voltage at the output if the switch
shorts. This requires a high/low side switch driver, and has relatively high input ripple
current.
[Illan Glasner, 1996] and [Mohan, 2003]
4.2
Buck-Boost Converter
The main application of a Buck-Boost converter is in the field of regulated DC power
supplies. The value of output voltage for this converter may either be less than or greater
than the input voltage, depending on the value of duty cycle (D). Unlike the Buck
converter, the Buck-Boost converter produces an output voltage with opposite polarity to
the input voltage (Vi ). A non-isolated Buck-Boost converter circuit is shown in Figure 4.2
Figure 4.2. A circuit diagram of a Buck-Boost converter. Input voltage source (Vi ), controlled
switch (S), inductor (L), diode (D), capacitor (C) and a load resistance (Ro .
The condition of a zero volt-second product for the inductor in steady state field is shown
in Equation (4.1).
(4.1)
Vi · D · Ts = −Vo · (1 − D) · Ts
Where Ts is the period time. The output voltage (Vo ) is negative with respect to the input.
The DC voltage transfer function of the Buck-Boost converter is shown in Equation (4.2).
(4.2)
Mv =
Vo
D
=
Vi
1−D
Where Mv is the conversion ratio. The magnitude of the output voltage (Vo ) can either
be grater or smaller (if D = 0.5 , Vo = Vi ) than the input voltage as the name of the
converter implies.
[Mohan, 2003] and [Rashid, 2001]
Advantages and disadvantages
One of the advantages of the Buck-Boost converter is that the output voltage can be less
or greater than the input voltage. This is the most important feature which the others do
12
4.3. Cuk Converter
Aalborg University
not have. Another advantage is that the output stage rectifier diode is used as the reverse
blocking diode.
The disadvantages of using a Buck-Boost converter are that the output voltage is always
reversed in polarity with respect to the input. Apart from that, another disadvantage is
if the MOSFET ever shorts, there is no way to limit the current into the battery.
[Karpin, 2006]
4.3
Cuk Converter
The Cuk converter is a non-isolated type of converter. It is very similar to the Buck-Boost
converter because it can increase or decrease the output voltage depending on the duty
cycle, but the main difference is that it is designed with different elements and inverts
the polarity of the output voltage. The Cuk converter is composed of two inductors, two
capacitors, a diode and a switch to generate the duty cycle. Another difference is that it
uses a capacitor as the main element instead of an inductor like most other types of DC
converters.
Figure 4.3. Cuk converter circuit diagram. Input voltage source (Vi ), controlled switch (S),
inductor (L), diode (D), capacitor (C) and a load resistance (Ro ).
Advantages and disadvantages
The advantages of this kind of converter are that it can increase or decrease the input
voltage just by changing the value of the duty cycle. In addition, the design of this
converter allows correction of the ripple of the current because of the constant current
flow. This is very useful in order to minimize losses, because by a careful adjustment of
the values of the elements the ripple of the input and output currents can disappear.
However, Cuk converter has a lot of reactive elements and the values of them are normally
very high. Apart from that, this kind of converter is very difficult to design correctly in
order to eliminate the ripple of the input and output currents.
[Bo-Tao Lin, 1997], [Electronics, 2011] and [Xi, 2007].
13
Group EE5-511
4.4
4. Power converter
Full Bridge Converter
The Full Bridge converters an electronic circuit that enables a voltage to be applied across
a load in either direction. These circuits are often used to allow DC motors to run forward
and backward. The full bridge configuration is typically used in switching power supplies
at power levels of approximately 750W or greater.
Figure 4.4. Diagram of the Full Bridge converter. Input voltage source (Vi ), controlled switch
(S), diode (D), capacitor (C) and a load resistance (Ro ).
[Erickson, 2011]
Advantages and disadvantages
The other converters like Buck, Buck-Boost and Cuk, without any modification, can only
operate in one quadrant. This means that they can increase or decrease the output voltage
depending on the kind of converter, but they are not able to switch the sign of the current
or the voltage.
Figure 4.5. Comparison to other type of DC converters.
With the Full Bridge converter, the output voltage is Vo , which can be controlled in
amplitude as well as in polarity. The same thing happens with the current Io. Therefore,
a Full-Bridge converter can operate in all four quadrants of the Io − Vo plane, and the
power flows through the converter can be either direction.
14
4.5. Choice of the power converter
Aalborg University
Figure 4.6. Quadrants where the full bridge converter can work.
It is necessary to use this type of converter with a load that also can generate energy.
These are some of the applications where the Full Bridge converter are used:
• DC motor drives
• DC to AC (sine-wave) conversion in single-phase uninterruptible AC power supplies.
• DC to AC (high intermediate frequency) conversion in switch-mode transformerisolated AC power supplies.
It has to be mentioned that this is not a step down or a step up converter. This has to
be added to one of the other converter in order to step-up or -down. This one is only
used to change the direction of the current. Therefore this converter could be used for
regenerating energy in this project or to change the driving direction.
[Maksimovic, 1956], [Mohan, 2003], [D.A.Bradley, 1995] and [Andersson, 2011]
4.5
Choice of the power converter
After doing a brief explanation of each converter the next step is to do a comparison and
choose one depending on the necessities of the application. First of all, a table with the
main advantageses and disadvantages of each converter is shown.
Buck
Buck-Boost
Comparison of the power
Advantages
- Very efficient
- Easy to design (No transformer)
- Switch with small stress
- Small output filter
- Decrease/increase output voltage
- Easy to design (No transformer)
converters
Disadvantages
- Non isolated (Input/Output)
- Input current high ripple
- Only decreases output voltage
- Non isolated (Input/Output)
- Output voltage reversed
Table 4.1. Comparison of the power converters.
As shown in Table 4.1, only two types of power converters have been compared. Because
those two are more easier to design comparing to the Cuk converter. That is mainly
because, as mentioned in Section 4.3, the Cuk has a lot of reactive elements and the value
15
Group EE5-511
4. Power converter
of them is normally very high. That is why this type of converter has been discarded and
not used in the comparison.
After comparing the two remaining converters, and looking at the characteristics of the
batteries and the DC motor, as the input range of the converter has to be from 42V to
52.4V, and the output range from 0V to 48V. It has been chosen that the power converter
in this project has to be able to decrease or increase the output voltage, depending on
the voltage of the batteries and the position of the throttle pedal. If the batteries are
fully charged, the voltage in the output has to be decreased, but if the batteries are not
fully charged and the maximum speed is required in the go kart, this voltage has to be
increased.
This is why the Buck-Boost converter has been chosen to be designed in this project.
16
Design of the power
converter
5
In this chapter the power converter chosen in Section 4.5 will be designed for the purpose
of building a power converter for a go-kart.
The design of the Buck-Boost converter has been made and it has been concluded that this
converter is not the best for this application, primary because of the size of the components
that have to be used. Another cause is that, if the Buck-Boost converter is used for the
go-kart the Boost function of the Buck-Boost will only be used if the batteries are not
fully charged. The design of the Buck-Boost is included in Appendix A.
After discussing which converter should be better to use instead of the Buck-Boost, a
Buck converter has been chosen because the value of the capacitance will be lower and the
currents in the circuit will also be lower. The only disadvantage of using a Buck instead
of Buck-Boost is that the maximum input voltage for the motor will only be about 42V
in the case when the batteries are not fully charged.
Because of this, a Buck converter will be designed in this chapter. The chapter will
start with an analysis of the electrical circuit of the Buck converter (Figure 4.1) and the
equations of the parameters for the components will be shown and then calculated. A
inductor will be designed and a switch controller will be described and chosen. In the last
part of the chapter, all the components in the converter will be chosen.
5.1
Topology of the converter
A Buck converter works in two different states. The first state is when the switch is closed
and the diode is conducting. This is shown in Figure 5.1 (A). The other state is when
the switch is open and the diode is non-conducting, as shown in Figure 5.1 (B). These
figures are modified versions of Figure 4.1.
17
Group EE5-511
5. Design of the power converter
Figure 5.1. The two states, a Buck converter works in.
All the voltages and the currents in the converter are shown in diagrams in Figure 5.2.
Figure 5.2. All the voltages and the currents in all the components in a Buck converter. (A) is
the first state and (B) is the second state.
18
5.2. Design of the Buck converter
Aalborg University
[Mohan, 2003] and [Maksimovic, 1956]
Continuous conduction mode(CCM) and Discontinuous conduction
mode(DCM)
The condition that was described previously is called CCM. It happens when the output
current Io is higher than the half of the inductor current ripple ( ∆i2 L ). Figure 5.3 shows
the details about this condition.
Figure 5.3. A system which is working in CCM. The output current (Io ) is higher than the half
of the inductor current ripple ( ∆i2 L ).
If the load resistance (Ro ) is increased the output current will start to decrease. If the
load resistance keeps on increasing, the output current will at a time be smaller than ∆i2 L .
When this happens the system will work in DCM. Figure 5.4 shows the details about this
conditions.
Figure 5.4. A system which is working in DCM. The output current (Io ) is smaller than
∆iL
2 .
The converter will be designed in a way that it will not work in DCM.
[Mohan, 2003] and [Maksimovic, 1956]
5.2
Design of the Buck converter
In this section the Buck converter will be designed by calculating the parameters of the
converter and in the end, all the components for the converter will be chosen.
19
Group EE5-511
5. Design of the power converter
Specifications
There are few specifications for the converter and some values which have been chosen.
These specifications are shown in Table 5.1.
Specifications
Input voltage
Output voltage
Input and output power
Equivalent resistance of the load
Frequency
Switching time
Voltage ripple
Current ripple
Symbol
Vi
Vo
Pi = Po
2
Ro = VPoo
fs
Ts = f1s
∆VC
∆iL
Value
[42 ; 52.4]V
[0.1 ; 48]V
4800W
0.48Ω
30kHz
33.333µs
1% of V̂o
10% of îL
Table 5.1. Specifications for the converter.
In the calculations there will be calculated with Ro = 0.45Ω because the resistance of the
motor is variable when driving, and it has been considered necessary to leave a margin of
error because this is the worst case. The input voltage range is known from the range the
batteries can work in. The output voltage range is the input voltage range of the motor.
This requires that the converter is ideal. The power is the given power of the motor. The
frequency has been chosen to be 30kHz because this can help with decreasing the value
of the inductor and capacitor. The value of the output voltage ripple (∆VC ) is chosen to
be 1% of V̂o and becomes 0.48V, this is done to reduce the value of the capacitor. The
value of the output current ripple (∆iL ) is chosen to be fixed around 10% of the maximum
current in the inductor. The reason for choosing that is also to reduce the value of the
inductor.
Mathematical model of the converter
Knowing the values in Table 5.1 the values of the parameters of the converter can be
calculated using the following equations.
The duty cycle is a control variable that controls how the switch opens and closes. The
duty cycle is a function of the input and the output voltages of the converter. The
expression for the duty cycle is
(5.1)
D=
Vo
Vi
[−]
where D is the duty cycle, Vo is the output voltage and Vi is the input voltage.
The expression for the output current is
(5.2)
Io =
Vo
Ro
[A]
where Ro is the equivalent resistance of the load.
The average current in the diode is a function of the output current and the duty cycle
as:
(5.3)
20
< iD >= Io · (1 − D)
[A]
5.2. Design of the Buck converter
Aalborg University
The average current in the capacitor is equal to zero in one whole period
(5.4)
< iC >= 0
[A]
which means that the average current in the inductor is the same as the output current
which is shown in Equation (5.5).
(5.5)
< iL >= Io =
Vo
Ro
[A]
The average current of the switch is a function of the output current and the duty cycle
as seen in Equation (5.6).
(5.6)
< iS >= Io · D
[A]
The ripple of the output current is chosen to be fixed around 10% of the maximum current
in the inductor. This value is chosen because it is not desired to allow this value to become
too large, because in that case the peak currents of the inductor and of the switch will
be to high. This would also increase their size and price. But if the current ripple is too
small the value of the inductance will increase. The value of the inductor can be derived
from a equation for the ripple of the output current as shown in Equation (5.7).
∆iL =
Vi · Ts
· (1 − D) · D
L
[A]
Vi · Ts
· (1 − D) · D
∆iL
[H]
m
(5.7)
L=
It is necessary to choose the worst case scenario, so the value of inductor used in the
converter should be the maximum value of L. As seen in Equation (5.7), all the values
are constants except the duty cycle and Vi .
Analyzing the diagram of the D · (1 − D) in Figure 5.5, changing the duty cycle the
maximum value of D · (1 − D) is 0.25, which will be used in further the calculation.
Figure 5.5. The relationship between the duty cycle and D(1 − D).
Equation (5.8) shows how to calculate the chosen inductance.
(5.8)
Lchosen =
V̂i · Ts
· 0.25
∆iL
[H]
The value of the capacitor can be derived from an equation for the ripple of the output
voltage and the chosen value of the inductor as shown in Equation (5.9). The value of the
21
Group EE5-511
5. Design of the power converter
voltage ripple is fixed to 1% of the maximum output voltage as decribed in Table 5.1.
∆Vc =
Vi · Ts2
· (1 − D) · D
8·L·C
[V]
Vi · Ts2
· (1 − D) · D
8 · L · ∆Vc
[F]
m
C=
m
V̂i · Ts2
· 0.25
[F]
8 · Lchosen · ∆Vc
From Figure 5.2, the relations of how to calculate the maximum values of the currents
and voltages in the converter, can be seen in Equations (5.10) to (5.14).
(5.9)
Cchosen =
(5.10)
∆iL
IˆL = IˆS = IˆD =< iL > +
2
[A]
(5.11)
IˆC = IˆL − Io
[A]
(5.12)
V̂L = V̂o [V]
(5.13)
V̂C = V̂o
[V]
(5.14)
V̂S = V̂D = V̂i [V]
To be able to choose the capacitor for the converter, the Root Mean Square (RMS)
value of the current in the capacitor has to be found and this relation can be seen in
Equation (5.15).
(5.15)
IC,RMS =
∆iL
√
2· 3
[A]
The RMS value of the current in the inductor is
s
2
1
∆iL
(5.16)
·
IL,RMS =< iL > · 1 +
12
< iL >
[A]
To calculate some of the power losses in the converter it is necessary to know some more
RMS values of the current. In Equation (5.17) and (5.18) the RMS values for the current
in the switch and the diode are shown.
s
2
√
1
∆iL
(5.17)
·
[A]
IS,RMS =< iL > · D · 1 +
12
< iL >
(5.18)
ID,RMS
√
=< iL > · 1 − D ·
[Mohan, 2003] and [Maksimovic, 1956]
22
s
1
1+
·
12
∆iL
< iL >
2
[A]
5.2. Design of the Buck converter
Aalborg University
The border between CCM and DCM
It is possible to see if the designed Buck converter is working in CCM, as desired, or if it
will work in DCM. When the inductor has been chosen, the output current limit (Io lim )
can be calculated as seen in Equation (5.19). The value of Io lim has to be smaller than the
output current (Io ) of the converter to work in CCM, otherwise the converter will work
in DCM. It is normal to design a converter to always work in CCM. It works in DCM if
the current in the inductor goes to zero at the end of one period.
(5.19)
Io lim =
Vi · Ts
· (1 − D) · D
2 · Lchosen
[A]
The maximum value of Io lim is when the duty cycle is 0.5 and the input voltage is
maximum. This is shown in Equation (5.20).
(5.20)
Io lim max =
V̂i · Ts
· 2.5
2 · Lchosen
[A]
[Mohan, 2003]
Results of calculations for the converter
The calculated values from Equation (5.1) to (5.19) are shown in Table 5.2.
Parameter
Duty cycle
Output current
Avg. current in the diode
Avg. current in the inductor
Avg. current in the switch
Avg. current in the capacitor
Max. current in the inductor
Max. current in the capacitor
Max. voltage in the inductor
Max. voltage in the capacitor
Max. voltage in the switch
Voltage ripple in the capacitor
Current ripple in the inductor
Inductance
Capacitance
RMS current in the capacitor
RMS current in the inductor
RMS current in the switch
RMS current in the diode
Current limit between CCM and DCM
Symbol
D
Io
< iD >
< iL >
< iS >
< iC >
IˆL = IˆS = IˆD
IˆC
V̂L
V̂C
V̂S = V̂D
∆VC
∆iL
Lchosen
Cchosen
IC,RMS
IL,RMS
IS,RMS
ID,RMS
Io − Io lim
Value
[0.0019 ; 1][-]
[0.22 ; 106.67]A
[0 ; 8.957]A
[0.22 ; 106.67]A
[0.0004 ; 106.67]A
0A
112A
5.33A
48V
48V
52.4V
0.48V
11.2A
39µH
97.22µF
[0.0064 ; 3.08]A
[0.22 ; 106.71]A
[0.011 ; 102.13]A
[0.22 ; 30.92]A
>0
Reference
(5.1)
(5.2)
(5.3)
(5.5)
(5.6)
(5.4)
(5.10)
(5.11)
(5.12)
(5.13)
(5.14)
Tab 5.1
Tab 5.1
(5.8)
(5.9)
(5.15)
(5.16)
(5.17)
(5.18)
(5.19)
Table 5.2. Calculated values of the Buck converter.
Table 5.2 shows all the values of the Buck converter. Because the output current limit
(Io lim ) is smaller than the output current (Io ) in the whole range, the converter, with
a load resistance of 0.45Ω, will completely work in CCM. This means that the chosen
inductance is right.
23
Group EE5-511
5. Design of the power converter
The expected power losses in the converter
To calculate all the components in the converter it is necessary to have an idea of how big
the losses in all the different components will be. That is why there have been made some
calculations about how big the losses are expected to be in the converter. The maximum
total power loss in the converter is shown in Equation (5.21).
(5.21)
Plosses = Pswitch + Pinductor + Pdiode + Pcapacitor
[W]
Because of the low RMS current in the capacitor, as seen in Table 5.2, the power loss in
the capacitor can be neglected.
The efficiency of the converter is sat to 80%. That is why the total power loss in the
converter must be
η=
(5.22)
Pout
Pin − Plosses
=
Pin
Pin
[−]
The power losses in the switch and the diode can be calculated by using Equation (5.23)
and (5.24). The switching power losses are neglected in these calculations.
(5.23)
2
Pswitch = RDS · IS,RMS
[W]
(5.24)
Pdiode = VFD · < iD >
[W]
where RDS is the resistance in the switch and VFD is the voltage drop inside the diode.
These values are chosen to be some typical values of some components. From the previous
equations it is shown that the only unknown in Equation 5.21 is the power loss in the
inductor (Pinductor ), because the inductor has to be built. Table 5.3 shows the expected
power losses.
Parameter
Plosses
Pswitch
Pinductor
Pdiode
Pcapacitor
Constant
RDS = 0.036Ω
VFD = 0.93V
-
Power loss
960W
375.52W
576.15W
8.33W
≈0
Table 5.3. The expected maximum power losses in the converter.
[Maksimovic, 1956]
5.3
Multiphase
As seen in Table 5.2 the output current (Io ) value is very high, which makes it difficult to
find the components that can cope whit these values.
For this reason it has been chosen to divide the converter into different phases which
make the components values more feasible, because the current is divided by the number
of phases.
A research was made about how the different values were changing as the number of phases
were increased, in order to decide how many phases the converter should be split into.
24
5.3. Multiphase
Aalborg University
There are some points that have to be taken in account when deciding the number of
phases to use.
First of all is that the value of the inductor will increase as it is multiplied with the number
of phases. Therefore it is important to find which number of phases reduces the current
enough without increasing the inductor value to much. This is because a high value of
the inductor makes it impossible to buy or to build it.
Another point is that when building more phases the price of the converter will increase
and it takes a longer time to build more phases for the converter.
Finally, considering these points, it has been decided to split the converter into four phases,
which makes the value of the current and the inductor proper size to build it and the value
of the capacitor reasonable to buy it. Table 5.4 shows the value of the parameters in one
phase when using four phases.
Parameter
Duty cycle
Output current
Avg. current in the diode
Avg. current in the inductor
Avg. current in the switch
Avg. current in the capacitor
Max. current in the inductor
Max. current in the capacitor
Max. voltage in the inductor
Max. voltage in the capacitor
Max. voltage in the switch
Voltage ripple in the capacitor
Current ripple in the inductor
Inductance
Capacitance
RMS current in the capacitor
RMS current in the inductor
RMS current in the switch
RMS current in the diode
Current limit between CCM and DCM
Symbol
D
Io
< iD >
< iL >
< iS >
< iC >
IˆL = IˆS = IˆD
IˆC
V̂L
V̂C
V̂S = V̂D
∆VC
∆iL
Lchosen
Cchosen
IC,RMS
IL,RMS
IS,RMS
ID,RMS
Io − Io lim
Value
[0.0019 ; 1][-]
[0.055 ; 26.67]A
[0 ; 2.828]A
[0.055 ; 26.67]A
[0.0001 ; 26.67]A
0A
28A
1.33A
48V
48V
52.4V
0.48V
2.8A
156µH
24.31µF
[0.0016 ; 0.77]A
[0.055 ; 26.68]A
[0.0028 ; 25.53]A
[0.055 ; 7.73]A
>0
Table 5.4. Calculated values of the Buck converter in one phase when the converter has been
split into four phases.
When having more than one phase, the current in the different phases has to be
synchronized. The main reason for synchronizing the four phases is to reduce the current
ripple which is obtained as a consequence of this division.
The problem is that when the four phases are working at the same time, the output takes
the sum of all the current ripples, which makes the output a waveform with really high
peaks that equals to the sum of the peaks corresponding to the different phases.
Therefore it is necessary to synchronize the four phases of the converter in the most optimal
way in order to avoid this kind of complications. Figure 5.6 shows the not delayed current
in the four phases and the corresponding result.
25
Group EE5-511
5. Design of the power converter
Figure 5.6. The four phases not delayed and its corresponding result.
The solution which has been found for this problem is to delay the period in each phase,
so the output waveform will not be the sum of all the phases, on the contrary it will be
smaller and the ripple very little. Figure 5.7 shows the current in the four phases and the
corresponding result.
Figure 5.7. The four phases delayed and its corresponding result.
If all the phases are not working with the same frequency, the current ripple in each phase
could, in some cases, be very high and in others could be small. This will result in a output
waveform with a really variable ripple, something that is really undesirable. Figure 5.8
shows the current in the four phases with different frequencies and the corresponding
result.
26
5.3. Multiphase
Aalborg University
Figure 5.8. The four phases with different frequencies and its corresponding result.
It has been decided to implement a synchronization circuit with 120kHz, so each phase
works with 30kHz.
To generate the main frequency the NE555 integrated circuit will be used. This IC has
been connected as astable circuit which generates a square signal of 120kHz.
The values of the resistors and the capacitor have been calculated to obtain a higher
frequency because of the tolerance of the components which makes the values of the
calculations not as real as the values that are obtained making tests in the laboratory.
If R1 =2.15kΩ, R2 =3.16kΩ and the value of the capacitor is fixed at C1 =1nF, the frequency
can be calculated as
(5.25)
Tm = 0.7 · (R1 + R2 ) · C1 = 0.7 · (2.15 · 103 + 3.16 · 103 ) · 10−9 = 3.72 · 10−6 s = 3.72µs
(5.26)
(5.27)
(5.28)
T = Tm + Ts = 3.72 · 10−6 + 3.72 · 10−6 = 7.43 · 10−6 s = 7.43µs
fT =
1
1
=
= 134.52 · 103 Hz = 134.52kHz
T
7.43 · 10−6
fSYNC =
fT
134.52kHz
=
= 33.63kHz
4
4
Where T is the period, Tm is the mark time (output high) and Ts is the space time (output
low). R1 and R2 are the resistances and C1 is the capacitance.
Figure 5.9 shows the screen shot from the oscilloscope which represents the square signal
of 120kHz that is obtained from pin 3 which is the output pin of this IC.
27
Group EE5-511
5. Design of the power converter
Figure 5.9. The 120kHz square signal of the NE555 IC to activate the switch in all four phases.
This signal goes directly to the high clock input CP0 of a counter (HEF4017BP). This
counter generates different delayed output signals. In this project, four outputs will be
used to generate the four delayed signals that will activate respectively the switch in the
four phases of the converter. In this case, four outputs of this component have been used.
The next output pin is used to restart the counter by connecting it to the reset pin in
order to generate only four delayed signals. This can be seen in Figure 5.11.
However, these signals are not used directly from the counter. A hex inverting Schmitt
trigger (HEF40106BP) has been used for enhancing noise immunity. The synchronization
signals (Vsync1, Vsync2, Vsync3 and Vsync4) will activate the different four phases of the
converter that are obtained from the output pins of this last element. Figure 5.10 shows
the screen shot from the oscilloscope which represents the square signal of 30kHz that will
activate the switch in each phase of the converter.
Figure 5.10. The 30kHz square signal to activate the switch in one phase.
Finally, in Figure 5.11, the entire synchronization circuit can be seen with all the
components that have been used.
28
5.4. Design of the inductor
Aalborg University
Figure 5.11. The synchronization Circuit.
5.4
Design of the inductor
Because of the high value of the current in the inductor, which is needed, it is not possible
to find a inductor that has this specification. Therefore, in this section, a description of
how to design an inductor will be shown.
To design the inductor for the Buck converter some values from Table 5.2, 5.3 and 5.4
have to be used. The values are listed in Table 5.5. In Table 5.3 the expected power loss
in the inductor was calculated and in Section 5.3, it was chosen to use four phases of Buck
converters in parallel.
Parameter
L
IˆL
IL RMS
Pinductor
Value
39 µH
112A
106.71A
576.15W
Value in one of the four phases
156µH
28A
26.68A
144.04W
Table 5.5. Needed values for design of the inductor.
The first thing that has to be chosen is the material of the core that will be used. In
this case a material called N97 has been chosen. The reason for that choice is that this
material gives the core a better quality because the saturation of the flux density (Bsat ) is
very high. When looking at the datasheet in Appendix B 7 of the material, the saturation
of the flux density (Bsat ) is equal to 410mT at 100◦ C. The maximum value of the flux
density (Bmax ) is set to 350mT.
Furthermore the electrical resistivity of the copper conductor at 100◦ C (ρ100◦ C ) is 23nΩ·m
and the fill factor (ku ) has been chosen to be 0.5 because this is a typical value for a
low voltage inductor. The fill factor is an indicator of how well the wires can fill out the
winding area. If there is a big air space between the wires the fill factor will decrease.
To calculate the maximum resistance in the inductor from the chosen power loss (Pinductor ),
Equation (5.29) is used.
(5.29)
R̂L =
Pinductor
= 0.202 Ω
IL2 RMS
29
Group EE5-511
5. Design of the power converter
When the resistance in the inductor is calculated, a minimum core geometrical coefficient
(Kg min ) for the core can be computed. This coefficient tells how big the minimum size of
the core has to be. The equation is
(5.30)
Kg min =
ρ100◦ C · L2 · IˆL2
2
Bmax
· R̂L · ku
= 35 · 10−12
When the calculation of Kg min is complete the real core geometrical coefficient Kg of the
cores can be calculated. To calculate that, some parameters have to be found from the
datasheets in Appendix B 7.
The parameters from the datasheets are; the winding area (Wa ), the average length of turn
(lT ) and the effective cross section area (Ae ). Kg has to be bigger than Kg min . Otherwise
the core will not be big enough to make the inductor. To calculate Kg , Equation (5.31)
is used.
Kg =
(5.31)
A2e · Wa
lT
Values from the data sheet for some chosen cores and their calculated Kg values are listed
in Table 5.6.
Parameters
Wa
lT
Ae
Kg
Kg > Kg min
ETD 44
214mm2
77mm
173mm2
83·10−12
true
ETD 54
316mm2
96mm
280mm2
258·10−12
true
ETD 59
365.6mm2
106.1mm
368mm2
467·10−12
true
Table 5.6. The values used in calculations for different cores.
Table 5.6 shows that all the three cores can be used for the needed application. But the
chosen core is ETD 59 because this is the biggest one and therefore has the biggest safety
limit. Furthermore, the ETD 59 core is the only one that is made of the material N97.
The chosen core has no air gap inside the core (AL ), that is why it is necessary to calculate
the desired air gap length (lg ) between the two half cores. This is done in Equation (5.32).
(5.32)
lg =
µ0 · L · IˆL2
= 1.70mm
2
Ae · Bmax
·2
H
Where the vacuum permeability (µ0 ) is equal to 4 · π · 10−7 m
.
After the core is chosen and the air gap length is found, it is possible to calculate the
number of turns (N ) in the inductor. It is done in Equation (5.33).
(5.33)
N=
L · IˆL
= 33.91 ≈ 34 turns
Ae · Bmax
When the inductance of the inductor and the number of turns is known, it is possible to
calculate the flux density (Bbuilt ) in the built inductor. It is shown in Equation (5.34).
(5.34)
30
Bbuilt =
L · IˆL · Ae
= 349.1 · 10−3 T
N
5.4. Design of the inductor
Aalborg University
To see if the choice of the inductor is good enough, the maximum flux density in the
material (Bmax ) must be higher than the flux density in the built inductor (Bbuilt ). In
this case Bbuilt < Bmax is fulfilled. It is now possible to find the cross section area (Aw )
of the needed copper conductor. This can be calculated by Equation (5.35).
(5.35)
Aw =
Ku · Wa
= 5.38mm2
N
In the conductor there is a skin effect, which means that the current does not have an equal
distribution in the conductor. This skin effect does that the resistance in the conductor
will increase when the frequency is increasede, because the area where the current flows
decreases when the frequency is increased. That is why it is necessary to calculate the
skin depth diameter (Dskin ) for the conductor. It is shown in Equation (5.36).
r
ρ100◦ C
(5.36)
= 0.441mm
Dskin =
π · fs · µ0
To keep the resistance low, it is important to choose a conductor with a diameter (Dw ),
which is very close or smaller than the skin depth diameter. That is why a copper wire
that with a diameter (Dcu ) of 0.355 mm has been chosen.
Now it is possible to calculate the cross section area of the chosen conductor (Acu ) with
Equation (5.37).
(5.37)
Acu =
2
π · Dcu
= 0.0990mm2
4
After that, it is possible to find the needed number of wires (wire) that have to be in
parallel. This can be calculated by Equation (5.38).
(5.38)
wire =
Aw
= 54.32 ≈ 54 times
Acu
The final DC resistance in the inductor is as shown in Equation (5.39).
(5.39)
RL actual =
ρ100◦ C · Nnew · lT
= 0.0155Ω
wire · Acu
When the final DC resistance is known the final power loss in the core and the conductor
for DC can be computed. This is shown in Equation (5.40) to (5.42).
(5.40)
Pcu = RL actual · IL2 RMS = 11.05W
This value will be the power loss in DC, but this converter will work in AC with a frequency
of 30kHz. Because of the skin effect, the resistance inside the inductor will probably be
much higher and also the power loss, as Equation (5.40) shows.
To calculate the power loss for the core, it is necessary to know some values from the
datasheet of the core and the material. The values are the effective core volume (Ve ) and
the relative core losses per cubic meter (PV ). These values are shown in Table 5.7.
Parameters
Effective core volume
Relative core losses per cubic meter
Symbol
Ve
Pv
ETD 59
51.2 · 10−6 m3
150 · 103 mw3
Table 5.7. Values of the ETD 59 core and material N97.
31
Group EE5-511
5. Design of the power converter
Finally the power loss in the core and the total power loss can be found for DC.
(5.41)
Pcore = 2 · Ve · Pv = 12.29W
(5.42)
Pinductor = Pcu + Pcore = 23.34W
[Mohan, 2003] and [Maksimovic, 1956]
Experience by building the inductor
After building the inductor with 34 turns and 54 wires, the values of the inductor are
measured. This is shown in Table 5.8.
Parameters
RL measured-DC
RL measured-AC
Lmeasured
Built inductor (ETD 59)
0.0148Ω
0.820Ω
236.01µH
Relative deviation in %
4.8
51.3
Table 5.8. The measured values of the built inductor.
As expected, the resistance is much bigger when the converter works in AC. That is why
the power loss in the inductor will also be much higher. Equation (5.43) shows the new
power loss in the inductor for all the four phases.
(5.43)
Pinductor-built = 4 · RL measured-AC · IL2 RMS = 2334.8W
Comparing this with the expected power loss (Pinductor ) in the inductor, as shown in
Table 5.3, the real power loss (Pinductor-built ) has a relative deviation of 305%.
The reason for this big deviation could be because of the high frequency. When the
frequency is so high, there will be big losses in the wires because of the skin effect. There
will also be some losses in the connection where all the wires are connected together.
To optimize the inductor, a better choice will be to chose a wire with a smaller diameter.
This will probably decrease the resistance in the inductor and also the power loss.
As the inductance is bigger than expected, the capacitor must also change a bit. From
Equation (5.9) the new value of the capacitor can be calculated with the new value of the
inductor
(5.44)
Cmeasured = 14.71µF
[Maksimovic, 1956]
5.5
The converter controller
As one of the goals in this project is to have a variable voltage and current in the output
in order to change the speed and torque of the DC motor, it is necessary to have a
useful switch that commutates with different duty cycles in the DC converter. In order
to do that, it is needed to make a converter controller to, as the name says, control the
commutation frequency and duty cycle of the switch. In the next subsections different
types of controllers will be explained, and after choosing one, a more accurate analysis of
it will be made with a view to build it.
32
5.5. The converter controller
5.5.1
Aalborg University
Types of converter controllers
The two main families are: Voltage controllers, which measure the voltage of the motor
and actuate depending on it, and current controllers, which are almost the same but
measuring and controlling the current. Now some of the most important controllers of
each family will be explained.
Voltage controllers
The first idea was to use a voltage controller whose function is to maintain the output
voltage to a default value, independently of variations in load current, in input voltage
regulator and in temperature.
The possibility of using a Pulse Width Modulation (PWM) has been studied, where the
pulse width is controlled by voltage. The PWM output switches a transistor between off
and on with duty cycle which is controlled by the difference between the feedback voltage
and reference voltage.
The circuit that has been designed to implement that PWM oscillator is showed in
Figure 5.12 where it can be seen how the change in the value of the potentiometer compared
with the triangular signal can vary the duty cycle.
Figure 5.12. The PWM circuit example.
Another option is to use an integrated circuit known as NE555 or LM555. It has been
chosen to design an astable circuit that produces a square wave. This digital waveform
varies between the low and the high value which means that the output is changing all the
time. In Figure 5.13 it can be seen how the pedal is represented as a potentiometer which
makes the function of the resistance but with a variable value, that allows the variation
of the duty cycle of the output signal without changing the frequency.
33
Group EE5-511
5. Design of the power converter
Figure 5.13. The LM555 integrated circuit example.
Current controllers
There are different kinds of ways to design a current controller, but almost all of them
have similar structure.
To begin with, a measuring element that will measure the value of the current that is
wanted to be controlled is needed. This element can be for an example, a LEM or a shunt
resistor.
After that, the most important part of the controller, is the integrated circuit (IC). As
seen in Figure 5.14 the measured current value and the reference value established by the
user will be taken by the current controller. Depending on the difference between them
it will generate a certain signal with a concrete duty cycle to switch on/off the switch.
There are different types of integrated circuits for this application, and one of the most
common ones is in the UCC family.
Figure 5.14. The general schematic of the IC.
Finally, there are the driver and the switch. The driver adjusts the output signal of the
integrated circuit in order to get a voltage that is able to switch on/off the switch. This
last element will be the one that opens and closes the circuit in order to have the correct
current in the inductor to have a proper signal in the output. The most typical elements
that are used as a switch are the transistors or the MOSFETS.
One example of a current controller can be a PI (Proportional Integral) controller. This
controller works depending on the difference between the reference and the measured
34
5.5. The converter controller
Aalborg University
value. As seen in Figure 5.15, this controller has a closed loop to work properly.
Figure 5.15. The PI current controller example circuit.
Another example for controlling the current of the converter by generating a PWM signal
is using a IC specially made for this application. This element can be for example the
UCC3803, and in Figure 5.16 the DC converter can be seen with the implemented current
controller.
Figure 5.16. An example of the Buck converter with the current controller on it.
Choice of the converter controller
After analysing the different families of converter controllers, it is necessary to choose the
one that suits the best and is the easiest to design and build for the application. To do
that some points have to be taken into account.
First of all, one of the most important features that the controller needs to have is that the
current cannot exceed some limits because the elements of the converter can be damaged.
That is the main difference between the voltage and the current controllers. The voltage
controllers are not able to limit the current, which means that the current can get very
high. However, in the current controllers there is a maximum value for the current and it
cannot be higher than that.
It is also important that the current of the converter is controlled because this means that
the torque of the motor is controlled, as it is proportional to the current of the motor as
seen in Equation 3.5.
35
Group EE5-511
5. Design of the power converter
Another reason for choosing the correct controller can be the complexity of the circuits
of it. In the case of the PWM controller for example, the main problem is that the
triangular signal generator is quite difficult to design. The case of the 555 element and
the PI controller are similar, because they need more researching and tests in order to
calculate the values for the elements used to design them.
Finally, looking at the advantages and disadvantages of each family of controllers, it has
been decided that the best one for this application is the current controller using an IC
specially made for this purpose.
Design of the converter controller
To begin with, it has been decided to use the IC named UCC3803 for the controller,
because it has been designed for applications similar to this one. The general schematic
for the converter controller can be seen in Figure 5.17.
Figure 5.17. The converter controller general schematic.
UCC3803 Configuration Circuit
The main part of the converter is the one that contains the IC and all the elements to
adjust and configure it.
First of all it is necessary to make the UCC3803 work with the same frequency as the
converter. In order to do that a resistor has to be connected between pin 8 (VREF ) and pin
4 (RT /CT ) of the IC, and a capacitor has to be connected also from pin 4 to the ground
as seen in Figure 5.18. In this IC, VREF pin is an internal constant reference of 4V and
RT /CT pin is an input that receives the signal to generate a constant frequency of the
output signal.
36
5.5. The converter controller
Aalborg University
Figure 5.18. The adjustment circuit for the controller frequency.
[Andreycak, 1999]
The expression for calculating the values of RT and CT that appear in the schematic is
(5.45)
fPWM =
1
RT · CT
[Hz]
From that formula, choosing a frequency fPWM = 30kHz (it has to be the same as the
converter) and a capacitor CT = 3.3nF because this was available, the resistor RT can be
calculated
(5.46)
RT =
1
= 10100 ≈ 10kΩ
fPWM · CT
After adjusting the frequency is is time to add all the necessary elements to make the IC
work properly. In Figure 5.19 the entire circuit can be seen and Table 5.9 shows the values
for each component.
Figure 5.19. The complete circuit for the UCC3803.
37
Group EE5-511
5. Design of the power converter
Element
Value
R1
R2
R3
R4
R5
RT
C1
C2
C3
CT
10kΩ
50kΩ
20kΩ
10kΩ
50kΩ
10kΩ
100nF
100nF
47µF
3.3nF
Table 5.9. Values of the elements for the UCC3803 circuit.
As seen in the figure, there is a resistor and a capacitor between pin 1 (COMP) and 2
(FB). Those are connected in that way in order to make the reference voltage of the pedal
stable, so that it can be compared to the measured one afterwards. The voltage in pin 2
has to have a range of [0 ; 0.9]V to work properly in a duty cycle range of [0 ; 0.9].
Apart from that, the resistor R2 is the element that has to keep the supply voltage of the
IC at 5V, so if the supply voltage increases the resistor dissipates the rest of the voltage.
The resistor R5 has been added to make the synchronization of the phase, which will be
explained later.
In addition, there has been added a transistor T1 in series with the resistor R4 to make a
slope compensation for the voltage coming from the current measurement element.
Finally, some capacitors have been added near the pins of the IC to reduce the noises
produced by the input signals.
Current Measurement
In this application it is necessary that the current of the converter is measured, to limit
it and to generate the correct signal for the switch depending on its value.
It has been decided that the element for making this measurement will be a LEM, which
is a commercial component especially made for this purpose. There are different ranges
of currents when choosing a LEM. In this case it is necessary to have an element that
is able to handle at least 30A of peak current, so the LEM that has been chosen is the
HAIS-100TP because it can handle up to 100A of current.
Looking at the datasheet in Appendix B 5, it can be seen that it has some power pins
from where the current crosses, and other smaller pins that give a voltage depending on
the value of the current. As seen in the datasheet there is a pin which is be the power
supply of 5V (pin 1) and another two pins that are the GND (pins 2, 5). Apart from that
it has two more pins that are the internal voltage reference of 2.5V (pin 4) and the output
voltage (pin 3). Figure 5.20 shows the pins of the LEM.
38
5.5. The converter controller
Aalborg University
Figure 5.20. The pin schematic of LEM.
[Lem, 2011]
It is also necessary to know the exact voltage output that the LEM gives depending on
the value of the current flowing through it. In Figure 5.21, the relation between both
parameters can be seen.
Figure 5.21. The LEM input current vs. output voltage.
LEM Signal Value Adjustment
As seen in Figure 5.21 the output voltage of the LEM is not the same asthe UCC3803
needs in the input, so there has to be made a circuit that will adjust that value. From
this figure the values when measuring the current of the converter can be calculated. The
range of the current is [0 ; 30]A, so the voltages in the output of the LEM will be in the
range of [2.5 ; 2.6875]V respectively. This can be calculated with Equation (5.47). In
Figure 5.22 the limited range of voltage of this application can be seen.
(5.47)
Vout = Vref ± 0.625 ·
Ip
Ipn
[V]
Where Ip is the measured current and Ipn is the nominal current that the LEM can
measure.
39
Group EE5-511
5. Design of the power converter
Figure 5.22. The LEM Vout range.
In order to get the values that the UCC needs ([0V ; 0.9V]) it has been decided to use
an instrumentation amplifier. For designing that, a LM324 operational amplifier will be
used and the schematic can be seen in Figure 5.23.
Figure 5.23. The instrumentation amplifier schematic.
The main advantage of this circuit is that it can subtract V1 from V2 and it can also apply
a gain depending on the needed maximum output voltage, i.e. it will increase or decrease
the value of the output depending on the requirements of the system. The general formula
for calculating all the values for the elements in the circuit is:
Vout
2 · R1
R3
(5.48)
Gain =
= 1+
·
[−]
V2 − V1
Rgain
R2
In order to simplify the calculations it has been decided to equal the resistors:
(5.49)
R1 = R2 = R3 = R = 100kΩ
Apart from that it has also been decided to use the reference voltage of the LEM to set the
minimum output voltage to 0V. This value is constant and connected to input V1 of the
amplifier so it will be subtracted all the time. Knowing all that and taking the maximum
values in the instrumentation amplifier input voltage (2.688V) and output voltage (0.9V),
the gain can be calculated as
(5.50)
40
Gain =
Vout
0.9V
=
= 4.8
Vlem out − Vlem ref
2.6875V − 2.5V
5.5. The converter controller
Aalborg University
Finally, the value of the gain can be used to calculate the last resistor of the circuit, Rgain .
2·R
2 · 100000
=
= 52631.58Ω ≈ 51.1kΩ
Gain − 1
4.8 − 1
A 51.1kΩ resistance has been chosen because it is the nearest to the calculated value of
the available resistors.
(5.51)
Rgain =
Driver for the MOSFET
The reason why a driver is required is because from the output pin of the UCC3803, a
square signal with an amplitude of +5V is obtained. However, to activate the MOSFET
more voltage is required. Therefore, a driver is needed to raise the value of this voltage.
In this case, the MOSFET which is used can be activated with ±20V, so the used driver
will generate a square signal of 18V. In this point the MOSFET will work properly.
Figure 5.24 shows the screen shot captured from the oscilloscope which shows the 18V
square signal which the driver generates.
Figure 5.24. The square signal of 18V that is generated by the driver.
Figure 5.25 shows how the driver (TC4420) is connected with only three capacitors. These
capacitors are used in order to reduce the noises which could affect the driver. One of
them is a 1µF electrolytic capacitor and the other two are 100nF ceramic capacitors.
Figure 5.25. The circuit of the driver for the MOSFET.
[Microchip, 2001]
41
Group EE5-511
5. Design of the power converter
Power Supply
In this project the batteries feed all the circuits that will make the converter work. The
four batteries fully charged provide 52.4V. However, this power supply has to be adjusted
to the exact values of voltage that each circuit requires. For that reason, voltage regulators
have been used to avoid this problem. Specifically, the positive voltage regulators that
have been used are, L7818 and L7805 which provide +18V and +5V respectively.
Those elements can only handle a maximum voltage of 35V. Therefore, it has been decided
the voltage supply for them will be from two batteries instead of four. The voltage supply
for these component will be 24V.
These elements have to be connected in a specific way. Therefore it is required to connect a
diode and two capacitors. The values of the capacitors are 0.33µF and 0.1µF. Figures 5.26
and 5.27 show the way of connecting these components.
Figure 5.26. The L7805 circuit: Generation of +5V voltage supply.
Figure 5.27. The LT818 circuit: Generation of +18V voltage supply.
5.6
Choice of the components
In this Section all the general characteristics of each component used in the converter will
be specified. When choosing the suitable components the maximum voltage and currents
have to be taken into consideration but also the price of them.
42
5.6. Choice of the components
Aalborg University
Diode
Parameter
Average rectified forward current
Maximum repetitive reverse voltage
Maximum non-repetitive Peak current
Maximum forward voltage
Symbol
<I>
VRPM
Iˆ
VF,max
Value
30A
60V
200A
0.75V
Table 5.10. Characteristics of the chosen Schottky diode (MBR3060PT), see datasheet in
Appendix B 3.
Switch (MOSFET)
Parameter
Drain-to-Source Breakdown Voltage
Gate-to-Source Voltage - Continuous
Drain current (continuous) at T=25C
Drain current (continuous) at T=100C
Static drain-source on resistance
Symbol
VDSS
VGS
ID
ID
RDS
Value
100V
±20A
58A
41A
18.2mΩ
Table 5.11. Characteristics of the chosen MOSFET (NTP6412AN), see datasheet in Appendix B 4.
Current transducer
Parameter
Supply voltage
Primary nominal r.m.s current
Primary current measuring range
Operating Frequency Range
Response Time
(LEM)
Symbol
Vc
IPN
IPM
f
tr
Value
5V
±100A
±300A
50kHz
<5µs
Table 5.12. Characteristics of the chosen Current transducer(LEM HAIS 100-TP), see datasheet
in Appendix B 5.
As seen in Equation (5.44) the capacitor has to be 14.71µF. As this is for a voltage ripple
of 1% of the maximum output voltage, it has been decided to decrease the output ripple
because it has to be taken into account that there will probably be some noises on the
output. Therefore, a capacitor of ≈1mF has been chosen because this gives a lower output
voltage ripple. As it is difficult to find one big capacitor of this size it has been decided to
have 3 capacitors of 330µF in parallel. This gives a capacitance of 990µF. Characteristics
of the chosen capacitor can be seen in Table 5.13.
Capacitor
Parameter
Capacitance
Rated Working Voltage Range
Symbol
C
Vr
Value
3x330µF
50V
Table 5.13. Characteristics of the chosen capacitor, See datasheet in Appendix B 6.
43
Simulations of the
converter
6
In this chapter one phase of the four phase converter, designed in Chapter 5, will be
simulated. The simulations have been made in the program Proteus Isis 7.7.
There have been made five types of simulations, where:
1.
2.
3.
4.
All the currents and the voltages in the converter will be shown in graphs.
The output voltage will be simulated as a function of the duty cycle.
The power losses and the efficiency will be calculated in a fixed condition.
The efficiency of the converter will be simulated as a function of the duty cycle in
the real range of the specifications.
5. The border between CCM and DCM will be found by changing the load.
All these simulations will be compared with tests on the real converter where the same
procedure will be used.
6.1
Simulations
The real components will be simulated with real data from the producer and put into the
circuit shown in Figure 6.1.
Figure 6.1. The circuit of the Buck converter used in the simulations. The capacitor in the
input is just to take all the noise from the voltage source.
45
Group EE5-511
6. Simulations of the converter
The capacitor, which is in parallel with the input voltage, is placed with the purpose to
minimize the noise in the input voltage. The capacitor is not included in the simulations.
It will be handled as ideal and the power loss will be neglected.
6.2
Simulation 1: Verification of the voltages and currents
In this simulation it will be verified if the voltages and the currents in the simulated circuit
are the same as in the theory described in Section 5.1. A constant load of 2Ω will be used
because this will decrease the output current. The input voltage will be held constant
at 20V because this should give the same shape of signal in each component as with the
voltage, the converter is designed for. The only thing that changes is the magnitude of the
signal. The duty cycle will be held constant at 0.6 because this shows a small difference
between the two stages in one period.
Figures 6.2 to 6.5 show the simulated signal of the voltages and the currents in the
components in the converter. When comparing these diagrams with the diagrams in
Figure 5.2 it can be seen that the simulated voltages and currents resemble the diagrams
from the theory.
Input voltage (vi)
Voltage [V]
25
20
15
10
5
0
14
14.01
14.02
14.03
14.04
14.05
Time [ms]
Output voltage (vO)
14.06
14.07
14.08
14.02
14.03
14.04
14.05
14.06
Time [ms]
Output and capacitor voltage with ripple (vC and vO)
14.07
14.08
14.07
14.08
Voltage [V]
25
20
15
10
5
0
14
14.01
Voltage [V]
11.188
11.186
11.184
11.182
11.18
14
14.01
14.02
14.03
14.04
Time [ms]
14.05
14.06
Figure 6.2. Simulated input voltage, output voltage and capacitor voltage with a ripple. The
conditions are: an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω. The
output voltage ripple (∆VC ) is 2.7mV.
Figure 6.2 shows the simulated voltages in the input, the output and the capacitor. The
diagram for the capacitor is shown in a zoom to show how the output ripple looks like.
These voltages are quite similar to the diagrams in Figure 5.2. The output ripple is much
smaller than 1% of the output voltage. It is probably because of the high value of the
capacitor.
46
6.2. Simulation 1: Verification of the voltages and currents
Aalborg University
Diode voltage (vD)
Voltage [V]
20
10
0
-10
-20
14
14.01
14.02
14.03
14.04
14.05
Time [ms]
Switch voltage (vS)
14.06
14.07
14.08
14.01
14.02
14.03
14.06
14.07
14.08
14.01
14.02
14.03
14.06
14.07
14.08
25
Voltage [V]
20
15
10
5
0
14
14.04
14.05
Time [ms]
Inductor voltage (vL)
Voltage [V]
20
10
0
-10
-20
14
14.04
Time [ms]
14.05
Figure 6.3. Simulated diode voltage, switch voltage and inductor voltage. The conditions are:
an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω.
Input current (ii)
Current [A]
8
6
4
2
0
14
14.01
14.02
14.03
14.01
14.02
14.03
14.01
14.02
14.03
14.04
14.05
Time [ms]
Output current (iO)
14.06
14.07
14.08
14.09
14.04
14.05
14.06
Time [ms]
Capacitor current (iC)
14.07
14.08
14.09
14.07
14.08
14.09
Current [A]
6
5.5
5
14
Current [A]
0.5
0
-0.5
14
14.04
14.05
Time [ms]
14.06
Figure 6.4. Simulated input current, output current and capacitor current. The conditions are:
an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω. The current ripple in
the capacitor (∆iC ) is 0.7068A.
Figure 6.3 shows the voltages in the diode, the switch and the inductor. It can be noticed
47
Group EE5-511
6. Simulations of the converter
that there is a difference in time between the two stages of one period, as described in
Section 5.1, because the duty cycle is held constant at 0.6. The longer part is D · Ts and
the other is (1 − D) · Ts .
Figure 6.4 shows the currents in the input, the output and the current in the capacitor.
As before, these diagrams look like the ones from the theory.
Diode current (iD)
Current [A]
8
6
4
2
0
14
14.01
14.02
14.03
14.01
14.02
14.03
14.01
14.02
14.03
14.04
14.05
Time [ms]
Switch current (iS)
14.06
14.07
14.08
14.09
14.04
14.05
14.06
Time [ms]
Inductor current (iL)
14.07
14.08
14.09
14.07
14.08
14.09
Current [A]
8
6
4
2
0
14
Current [A]
8
6
4
2
0
14
14.04
14.05
Time [ms]
14.06
Figure 6.5. Simulated diode current, switch current and inductor current. The conditions are:
an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω. The current ripple in
the inductor (∆iL ) is 0.7073A.
Figure 6.5 shows the currents in the diode, the switch and the inductor. It can be noticed
that if the currents in the diode and the switch are put together, the result would look
like the current in the inductor. These graphs also resemble the diagrams from the theory.
The current ripple in the capacitor (∆iC ) as seen in Figure 6.4 is almost the same as the
current ripple in the inductor (∆iL ) as seen in Figure 6.5. Thats why the output current
is almost constant as expected.
The conclusion for the first simulation is that the circuit used in the simulation is correct
because the diagrams of the currents and the voltages have the same shape as the ones
from the theory.
6.3
Simulation 2: Verification of the output voltage
In this simulation, the output voltage of the converter will be simulated as a function of
the duty cycle. The input voltage will be held constant at 20V and the duty cycle will
vary from 0.1 to 1. The load will still be fixed to 2Ω. As seen in Equation (5.1), the graph
should be linear because the output voltage is proportional with the input voltage, in this
48
6.4. Simulation 3: Verification of the power losses in the converter
Aalborg University
case where the converter works in CCM. Figure 6.6 shows the result of the simulation of
this relationship.
Input and output voltage as a function of the duty cycle
22
20
18
16
Voltage [V]
14
12
10
8
6
Input voltage (Vi)
4
Output voltage (Vo)
Ideal output voltage
2
0
0.1
0.2
0.3
0.4
0.5
0.6
Duty cycle [-]
0.7
0.8
0.9
1
Figure 6.6. The curve of all the voltages. The output voltages are plotted for both the ideal
case and for the simulation, as a function of the duty cycle. The conditions are: an
input voltage at 20V and a load of 2Ω.
As seen in the Figure 6.6, the output voltage increases linear as expected when the duty
cycle is changed and the input voltage is held constant. It can also be noticed from the
diagram that the simulated output voltage does not get all the way up to the input voltage
because there is a voltage drop primary in the switch and the inductor.
6.4
Simulation 3: Verification of the power losses in the
converter
In this simulation the power losses in the inductor, diode, capacitor and the switch will be
calculated with values of the voltages and currents from simulation 1. When the power
losses in the converter are found, the efficiency of the converter can be calculated as seen
in Equation (5.22). In this simulation the input voltage is held constant at 20V and the
duty cycle at 0.6. The load is also 2Ω in this simulation.
49
Group EE5-511
6. Simulations of the converter
Switch
In the switch there are power losses both when it is conducting and when the MOSFET
is switching. This is shown in Equation (6.1).
(6.1)
Pswitch = Pon + Psw, on + Psw, off
[W]
where Psw, on and Psw, off is the power loss in the time when the MOSFET is switching
from off to on and from on to off.
To find the power loss in the switch when it is conducting, it is necessary to measure the
current in the switch and find the average value (< iS >) of it. It is also important to find
the voltage over the switch (von ) when the switch is conducting. This is multiplied with
the duty cycle as shown in Equation (6.2).
(6.2)
Pon = von · < iS > ·D
[W]
When the MOSFET is switching on and off, there will be some losses in the switching
period. It is because the switch is not ideal, but has a period where it is switching. The
fundamentals of this are shown in Figure 6.7.
Figure 6.7. The switching in the MOSFET. Because the switch is not ideal, the switching
period will not happen immediately. Thats why there are some switching losses.
The period in the on- and off switching period are simplified to be linear.
[Mohan, 2003]
To find the losses when the inductor is switching on and off, it is necessary to find the
switching on- and off time (tc on ) and (tc off ), as shown in Figure 6.7. These times are
found in the graph from the simulation. The graphs are shown in Figure 6.8 and 6.9.
50
6.4. Simulation 3: Verification of the power losses in the converter
Aalborg University
Figure 6.8. The voltage and the current in the switch from the simulation. The green boxes
indicate the periods where switching on- and off time is found and where Von and
< iS > are found. The conditions are: an input voltage of 20V, a duty cycle of 0.6
and a load of 2Ω.
Figure 6.9. The switching time when switching on- and off periods from the simulations.
Furthermore, the values of the voltage vS and the current iS are shown. The
conditions are: an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω.
The switching power losses can be calculated by Equation (6.3) and (6.4).
(6.3)
Psw, on =
1
· vS on · iS on · fs · tc on
2
[W]
(6.4)
Psw, off =
1
· vS off · iS off · fs · tc off
2
[W]
where vS is the voltage in the switch, iS is the current in the switch, fs is the frequency.
These calculations are an approximation, as shown in Figure 6.7. The real losses will in
this case be higher.
Table 6.1 shows all the values from the switch.
51
Group EE5-511
6. Simulations of the converter
Parameter
Von
< iS >
vS on /iS on
vS off /iS off
tC on
tC off
Pon
Psw, on
Psw, off
Pswitch
Value
0.0878V
5.45A
5.23V/A
5.93V/A
17.79ns
19.5ns
287.10mW
7.30mW
10.30mW
0.305W
References
Fig.6.8
Fig.6.8
Fig.6.9
Fig.6.9
Fig.6.9
Fig.6.9
Eq.(6.2)
Eq.(6.3)
Eq.(6.4)
Eq.(6.1)
Table 6.1. The power losses in the switch, with an input voltage of 20V, a duty cycle of 0.6 and
a load of 2Ω.
The power loss in the switch is now known and it is not as high as expected, with an input
voltage of 20V, a duty cycle of 0.6 and a resistance of 2Ω.
Diode
To find the power loss in the diode it is necessary to measure the current and calculate
the average value of the current. The power loss in the diode can be found using
2
Pdiode = ID,RMS
· RDS + VF · < iD >
[W]
⇓
(6.5)
Pdiode = VF · < iD > ·(1 − D)
[W]
where ID,RMS is the RMS current, RDS is the resistance, VF is the forward voltage and
< iD > is the average current. The first part of the equation can be neglected because
the resistance is almost equal to zero. The forward voltage and the average current in the
diode are found in the same way as in the switch shown in Figure 6.8. That is why (1 − D)
has to be multiplied with the average current. Table 6.2 shows all the value of the diode.
Parameter
VF
< iD >
Pdiode
Value
0.86V
4.46A
1.53W
References
Eq.(6.5)
Table 6.2. The power loss in the diode, with an input voltage of 20V, a duty cycle of 0.6 and a
load of 2Ω.
The power loss in the diode is now known and it is very similar as expected with an input
voltage of 20V and a duty cycle of 0.6.
Inductor
The power loss in the inductor depends on the ohmic resistance in the wires (RL ) and
the RMS current in the inductor. To find the RMS current, the current in the inductor
has to be measured and the average current (< iL >) and the current ripple (∆iL ) has
52
6.4. Simulation 3: Verification of the power losses in the converter
Aalborg University
to be calculated. Equation (5.16) is used to find the RMS value. The value of RL is
found in Table 5.8 for AC signal and the average current and the current ripple is found
in Figure 6.10.
Figure 6.10. The current in the inductor. Furthermore the average current and the current
ripple are shown. The conditions are: an input voltage of 20V, a duty cycle of 0.6
and a load of 2Ω.
The equation for calculating the power loss in the inductor is
(6.6)
2
Pinductor = IL,RMS
· RL
[W]
The values of the inductor are listed in Table 6.3.
Parameter
RL
< iL >
∆iL
IL,RMS
Pinductor
Value
0.820Ω
5.76A
0.707A
5.76A
27.20W
References
Tab.5.8
Fig.6.10
Fig.6.10
Eq.(5.16)
Eq.(6.6)
Table 6.3. The power loss in the inductor, with an input voltage of 20V, a duty cycle of 0.6 and
a load of 2Ω.
The power loss in the inductor is bigger than the other components in the converter.
Capacitor
The power loss in the capacitor is dependent on the RMS current in the capacitor and
the resistance in the capacitor. It is shown in Equation 6.7.
(6.7)
2
Pcapacitor = IC,RMS
· RC
[W]
The value of the resistance in the capacitor can be found using Equation 6.8.
RC = |XC | · tan(ϕ)
[Ω]
m
(6.8)
−1
· tan(ϕ)
RC = 2 · π · fs · C [Ω]
53
Group EE5-511
6. Simulations of the converter
where fs is the frequency, C is the capacitor and ϕ is the dissipation angle in the worst
case, found in the datasheet in Appendix B 6.
The RMS value of the current in the capacitor is found from Equation 5.15, where the
current ripple in the inductor (∆iL ) has to be known. Table 6.4 shows the value of the
capacitor.
Parameter
fs
C
tan(ϕ)
RC
∆iL
IC,RMS
Pcapacitor
Value
30kHz
990µF
0.19
0.001Ω
0.707A
0.204A
≈0
References
Tab.5.1
Tab.5.13
App.B 6
Eq.(6.8)
Fig.6.10
Eq.(5.15)
Eq.(6.7)
Table 6.4. The power loss in the capacitor, with an input voltage of 20V, a duty cycle of 0.6
and a load of 2Ω.
As expected, the power loss in the capacitor is very small and can be neglected from the
total power loss.
Total power loss
The total power loss in the converter is the sum of power losses in all the components.
This is was defined in Equation (5.21). Table 6.5 shows the values of all the power losses
in the converter.
Parameter
Pswitch
Pdiode
Pinductor
Pcapacitor
Plosses
Value
0.305W
1.53W
27.20W
≈0
28.94W
References
Tab.6.1
Tab.6.2
Tab.6.3
Tab.6.4
Eq.(5.21)
Table 6.5. The total power loss in the converter, with an input voltage of 20V, a duty cycle of
0.6 and a load of 2Ω.
All the relative power losses are shown in Figure 6.11.
As expected, the biggest power loss is in the inductor. The other power losses are small
compared to the inductor loss. The losses in the switch are smaller than expected.
54
6.5. Simulation 4: Verification of the efficiency of the converter
Aalborg University
Figure 6.11. All the relative power losses in the converter from the simulation.
Efficiency
The efficiency of the converter can be calculated by Equation (5.22). To find the input
power it is necessary to know the input current. It is equal to the average current of the
switch. The input power is defined in Equation (6.9).
(6.9)
Pi = Vi · < iI > ·D = Vi · < iS > ·D
[W]
Table 6.6 shows the values to calculate the efficiency in the converter and the efficiency
from the simulation.
Parameter
< iS >
Vi
Pi
Plosses
η
Value
5.654V
20V
67.848W
28.94W
0.573
References
Fig.6.8
Eq.(6.9)
Eq.(5.21)
Eq.(5.22)
Table 6.6. The total efficiency in the converter, with an input voltage of 20V, a duty cycle of
0.6 and a load of 2Ω.
The total efficiency in the converter in this simulation is 57.3% with an input voltage of
20V, a duty cycle of 0.6 and a load of 2Ω. This is quite low comparing to the expected
efficiency. This is because of the high resistance in the inductor.
[Mohan, 2003] and [Maksimovic, 1956]
6.5
Simulation 4: Verification of the efficiency of the
converter
The purpose of this test is to test the efficiency of the converter in the real conditions, with
the maximum input voltage of 52.4V and a load of 0.45Ω. In the test it was not possible
55
Group EE5-511
6. Simulations of the converter
to test the converter, with this conditions because of heating of the components. That is
why the value of the input voltage is decreased to only 45V and the load is increased to
1.5Ω. This simulation has been done by varying the duty cycle and measuring the input
and output voltage and the input and output current. This is done to be able to calculate
the input and output power. Figure 6.12 shows the efficiency of the simulated converter
as a function of the duty cycle.
Efficiency of the converter as a function of the duty cycle
1
0.9
0.8
Efficiency, η [-]
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
Duty cycle [-]
0.7
0.8
0.9
1
Figure 6.12. Efficiency of the simulated converter as a function of the duty cycle, with a constant
input voltage of 45V and a load of 1.5Ω.
As seen in the figure, the efficiency increases when the duty cycle is increased. This means
that the converter works better if the duty cycle is high.
6.6
Simulation 5: Verification of the border between CCM
and DCM
In this simulation, the border between CCM and DCM is found. It is done at three
different duty cycles (0.3, 0.5 and 0.6) and with a constant input voltage of 20V. In
the simulation the load is varied from 4Ω to 500Ω. The output voltage and current are
measured in the simulation.
Figure 6.13 shows the graph over the development in the relationship between the duty
cycle ( VVoi ) and the fraction of the output current and the maximum value of the current
Io
limit ( Io lim
). The maximum value of the current limit is found with Equation (5.20),
max
56
6.6. Simulation 5: Verification of the border between CCM and DCMAalborg University
with the real value of the inductor as seen in Table 5.8. In this case, the output voltage
is sat to be constant.
Figure 6.13. The characteristic of the converter between CCM and DCM in the simulation. In
CCM the output voltage stays constant and only the current changes. In DCM
the output voltage does not stay constant any more. The conditions are: an input
voltage of 20V and a duty cycle of 0.3, 0.5 and 0.6.
The value of io lim max is 0.353A. As seen in Figure 6.13, the output voltage is almost
constant in CCM and the only thing that changes is the current. When the curve passes
the border between CCM and DCM, the output voltage starts increasing and the current
decreases.
It can be concluded that the calculated value of the maximum value of the current limit
(Io lim max ) is very good connected to the values from the simulation. It is because the
value of the inductor is chosen correctly.
57
Test of the converter
7
After making all the simulations and theoretical tests of the converter, in this chapter the
real test results will be shown in order to compare them with the expected ones.
The system is supposed to have four phases, as shown in Figure 7.1, but only one has
been built, which means that all the tests in the converter will be from one phase instead
of all the system.
Figure 7.1. Schematic of the four phases of the converter and drivers for each phase.
Each phase of the converter has the same circuit for a controller and power converter.
Figure 7.2 shows a diagram of the circuit of the converter that has been designed, before
doing the printed circuit board in the lab. For designing the circuit and the printed
board, the program Orcad PCB Editor 16.3 has been used. It has decided to introduce
59
Group EE5-511
7. Test of the converter
some capacitors for the input signal, in order to stabilize the input voltage and minimizing
the input noises. It has also been decided to use a rectifier diode, in the output, to avoid
undesirable currents in the opposite direction from the motor. As shown in the figure
a LEM (current transducer) has been used to measure the current that flows through
the converter, as explained in Section 5.5. On this converter board there are 3 inputs
and 2 Outputs connected, besides the connection of the power input and output for the
converter:
•
•
•
•
•
+5V: Voltage supply for the LEM.
VMOS: Square signal to switch the MOSFET.
GND: General ground of the system.
VLEM: Image of the current, measured by the LEM.
VREF: Reference voltage which is given by the LEM.
Figure 7.2. Power Converter Schematic.
To test the converter, the same procedure as in the simulations has been used. This
procedure is shown below:
1. Currents and the voltages in the converter are measured and plotted to see if the
converter works as in the theory.
2. The output voltage is tested and plotted as a function of the duty cycle and a
constant input voltage.
3. Power losses in the real converter will be tested and checked if they are the same as
expected.
4. The efficiency will be tested in real conditions and compared with the simulations.
5. The border between CCM and DCM will be tested and compared with the
simulation.
Table 7.1 shows the instruments that were used to make the tests.
60
7.1. Test 1: Measurement of the currents and voltages
Instrument
Power supply
Power supply
Function generator
Oscilloscope
Electronic load
Current probe
Current probe
High voltage differential probe
High voltage differential probe
Multimeter
Name
Toellner TOE8872
GW-Instek GPS-4303
GW-Instek GFG-8216A
Tektronix DPO 2014
RBL 488
Tektronix TCP0150
Tektronix TCP0030
Tektronix P5200
Tektronix P5200
Fluke 179
Aalborg University
AAU nr/LB nr.
38626
87770
62782
79003
67152
90531
79028
87793
79010
7 25 03 A10
Table 7.1. A list of the used instruments in the tests.
Because of safety reasons, only the small power supply(GW-Instek GPS-4303) was used
in the beginning. After that, the real tests were made with the Toellner power supply
(Toellner TOE8872). An electronic load (RBL 488) was used to simulate the behavior of
the motor. To measure the output voltage, two high voltage differential probe have been
used.
7.1
Test 1: Measurement of the currents and voltages
In this test, as decided in Section 6.2, the input voltage will be held constant at 20V and
the duty cycle will be at 0.6. The load in this test will be fixed to 2Ω.
Figure 7.3 shows the voltages in the input, output and the capacitor. The input and
output voltage are almost constant as expected. The third graph is a zoom of the output
and the capacitor voltage because there is a ripple on the signal but it can also be noticed
that there is a lot of noise on the output. This is probably because the components
influence each other which results in a noise on the signal. It can also be because of the
measurement equipment or because the zoom of the signal has been made in the computer
instead of in the measurement equipment.
The output voltage ripple is bigger than in the simulation and it is 1.85% of the output
voltage. A reason for that is probaly because of all the noise in converter. But it is still
less than the value the converter is designed to handle in Table 5.4. That is why a choice
of a bigger capacitor in Section 5.6 was a good idea.
61
Group EE5-511
7. Test of the converter
Input voltage (vi)
Voltage [V]
20
15
10
5
0
0
0.01
0.02
0.03
0.04
0.05
0.06
Time [ms]
Output voltage (vO)
0
0.01
0.02
0.03
0.04
0
0.01
0.02
0.03
0.04
0.07
0.08
0.09
0.1
0.05
0.06
0.07
0.08
Time [ms]
Output and capacitor voltage with ripple (vC and vO)
0.09
0.1
0.09
0.1
Voltage [V]
20
15
10
5
0
Voltage [V]
11.5
11
10.5
0.05
0.06
Time [ms]
0.07
0.08
Figure 7.3. Test of the input voltage, output voltage and the voltage in the capacitor. The
conditions are: an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω. The
output voltage ripple (∆VC ) is 0.20V.
In Figure 7.4 the results of the tested voltage in the diode, switch and the inductor are
shown. These signals can easily be compared with the graphs of the theory because they
are very similar. It can be noticed that there is a small peak on the voltage signal in the
diode when it is stopping to conduct, which is called reverse recovery. The reason for that
is because when the diode switches off, there will be a current, that flows the other way
in the circuit. This will result in a negative voltage. Normally this negative polarity has
no effect in the converter, so this will be neglected. The same thing is the reason for the
peak in the switch.
It can be concluded from these graphs that the voltages in the components have the normal
shape for a Buck converter.
Figure 7.5 shows the tested input current and the tested output current in the converter.
It has to be noted that it has not been possible to measure the current in the capacitor, so
the current in the capacitor is calculated from the current in the inductor and the output
current (iL − io ).
62
7.1. Test 1: Measurement of the currents and voltages
Aalborg University
Diode voltage (vD)
Voltage [V]
20
10
0
-10
-20
-30
0
0.01
0.02
0.03
0.04
0.05
0.06
Time [ms]
Switch voltage (vS)
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.07
0.08
0.09
0.1
Voltage [V]
20
10
0
0.05
0.06
Time [ms]
Inductor voltage (vL)
Voltage [V]
20
10
0
-10
-20
0.05
0.06
Time [ms]
Figure 7.4. Tested voltage in the diode, switch and the inductor. The conditions are: an input
voltage of 20V, a duty cycle of 0.6 and a load of 2Ω.
Input current (ii)
Current [A]
5
4
3
2
1
0
0
0.01
0.02
0.03
0.04
0.05
0.06
Time [ms]
Output current (iO)
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.05
0.06
Time [ms]
Capacitor current (iC)
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.07
0.08
0.09
0.1
Current [A]
10
5
0
Current [A]
5
0
-5
0.05
0.06
Time [ms]
Figure 7.5. The tested currents in the input, the output and the capacitor. The conditions are:
an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω.
In Figure 7.5 it can be noticed that there is a lot of noise on the input current. This is
63
Group EE5-511
7. Test of the converter
probably because of the noise in the power supply and the instruments around the setup.
It can be noticed that the output current has also a lot of noise. Thats why it is not
possible to say anything about the ripple in the capacitor, because the noise makes a
ripple that is almost 6 times bigger than the ripple in the inductor current.
Figure 7.6 shows the tested current in the switch and the inductor. It has to be noted
that it has not been possible to measure the current in the diode, so it is calculated from
the current in the inductor and the current in the switch (iL − iS ).
Diode current (iD)
Current [A]
10
5
0
0
0.01
0.02
0.03
0.04
0.05
0.06
Time [ms]
Switch current (iS)
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.07
0.08
0.09
0.1
0
0.01
0.02
0.03
0.04
0.07
0.08
0.09
0.1
Current [A]
10
5
0
-5
0.05
0.06
Time [ms]
Inductor current (iL)
Current [A]
10
5
0
0.05
0.06
Time [ms]
Figure 7.6. The tested current in the switch and inductor. The conditions are: an input voltage
of 20V, a duty cycle of 0.6 and a load of 2Ω. The current ripple in the inductor
(∆iL ) is 0.80A.
Figure 7.6 shows that the switch current has a little peak when the switch begins and
stops to conduct. This is, as said for the voltage, because this component needs a little
power to start and stop conducting. In the switch the current has a lot of noise when
conducting, which probably is because of all the noise in the input current. The current
ripple for the inductor (∆iL ) is almost the same as in the simulation. The graph of the
inductor current seems to be almost the same as in the theory.
Figure 7.7 shows a screen shot of the screen of the oscilloscope where the output voltage,
the inductor current and the PWM signal is shown. It can be noticed in the figure that
there is a small ripple on the output voltage. This figure shows a good overview of how
the converter works.
64
7.2. Test 2: Verification of the output voltage as a function of the duty
Aalborg
cycle University
Figure 7.7. A screen shot of the oscilloscope where the output voltage, inductor current and the
PWM signal is shown. The conditions are: an input voltage of 20V, a duty cycle of
0.6 and a load of 2Ω.
7.2
Test 2: Verification of the output voltage as a function
of the duty cycle
This test is done by measuring the output voltage when changing the duty cycle and
having the input voltage constant. The input voltage is held constant at 20V and the
duty cycle is changed from 0.15 to 0.9. This range is chosen to protect the components in
the converter. Figure 7.8 shows the results of the input and output voltage as a function
of the duty cycle.
Input and output voltage as a function of the duty cycle
22
20
18
16
Voltage [V]
14
12
10
8
6
Input voltage (Vi)
Output voltage from the test (Vo)
4
Output voltage from the simulation (Vo)
Ideal output voltage
2
0
0.2
0.3
0.4
0.5
0.6
Duty cycle [-]
0.7
0.8
0.9
Figure 7.8. The tested input and output voltage as a function of the duty cycle. Furthermore,
the output voltage form the simulation and the ideal output voltage are included.
The conditions are: an input voltage of 20V and a load of 2Ω.
65
Group EE5-511
7. Test of the converter
As seen in Figure 7.8 the tested output voltage is almost linear as expected and this
confirms Equation (5.1) because the relationship between the input voltage and the duty
cycle is linear. The tested output voltage is a little bit higher than in the simulation, if the
duty cycle is small. The opposite happens when the duty cycle is high. It can probably
be because the value of the input voltage will not stay completely constant when the duty
cycle changes.
7.3
Test 3: Verification of the power losses in the converter
In this test the power losses in the components in the real converter will be calculated
from the measured data in test 1. Having found the power losses, the efficiency can be
calculated. As in the simulations the input voltage and the duty cycle will be held constant
of 20V and 0.6 respectively.
Switch
As written in the simulation, there are two types of losses in the switch; losses when the
switch is conducting and losses when switching. The same procedure has been used to
find the values that have to be found as in the simulation.
The total power loss in the switch is a sum of all the losses in the switch. Equation (6.1)
shows how this is calculated. To calculate the power loss when the switch is conducting,
the voltage and the current in the switch have to be used. Equation (6.2) will be used for
calculating that power loss.
To calculate the losses when the MOSFET is switching, Equation (6.3) and (6.4) will be
used. Figure 7.9 shows the power losses when it is switching. To find these losses the
switching time for both switching on and off has to be found. These are found from
reading the graphs in Figure 7.9. Besides that, the on and off voltage and current time
(tc on and tc off ) has to be found as seen in Figure 6.8.
Figure 7.9. The two power losses areas when the switch is switching on and off. The conditions
are: an input voltage of 20V, a duty cycle of 0.6 and a load of 2Ω.
Because of all the noise it is very difficult to find the right power loss area, so the yellow
66
7.3. Test 3: Verification of the power losses in the converter
Aalborg University
area in the figure is an approximation. Table 7.2 shows the values used for calculations
and the calculated power losses in the switch.
Parameter
Von
< iS >
vS on /iS on
vS off /iS off
tC on
tC off
Pon
Psw, on
Psw, off
Pswitch
Value
0.189V
5.654A
5.6V/A
3V/A
77ns
136ns
644.217mW
36.22mW
18.36mW
0.699W
References
Fig.6.8
Fig.6.8
Fig:.7.9
Fig.7.9
Fig.7.9
Fig.7.9
Eq.(6.2)
Eq.(6.3)
Eq.(6.4)
Eq.(6.1)
Table 7.2. The power losses in the switch, with an input voltage of 20V, a duty cycle of 0.6 and
a load of 2Ω.
Now the power loss in the switch is known at 20V input voltage and 0.6 duty cycle. The
power losses of the switch in the test are more than two times bigger than the power
losses of it in the simulation. It has to be mentioned that the switching times (tC ) and
the voltage/current (vS /iS ) have been very difficult to find precisely. That is why there is
a big uncertainty in this power loss.
Diode
The power loss in the diode can be calculated from Equation (6.5). The current and the
voltage in the diode have to be used to calculate this. As said before, it was not possible
to measure the current in the diode, so it has been calculated from iL − iS . The same
procedure has been used to find VF and < iD > as in the simulations. Table 7.3 shows the
calculated values of the forward voltage, the average current and the value of the power
loss in the diode.
Parameter
VF
< iD >
Pdiode
Value
0.335V
5.534A
0.742W
References
Fig.7.4
Fig.7.6
Eq.(6.5)
Table 7.3. The power loss in the diode with data from test 1. The input voltage is 20V, the
duty cycle is 0.6 and the load is 2Ω.
This proves that the power loss in the diode is not as big as in the simulation because of
the much lower forward voltage.
Inductor
To calculate the power loss in the inductor is it necessary to use the average current in
the inductor (< iL >). Furthermore, the average value of the voltage in the inductor,
from Figure 7.4, has to be known for the positive polarity (< VL + >) and the negative
67
Group EE5-511
7. Test of the converter
polarity (< VL - >). The power loss in the inductor can be calculated with Equation (7.1).
(7.1)
Pinductor = (< iL > · < VL + > ·D) + (< iL > · < VL - > ·(1 − D))
Parameter
< iL >
< VL + >
< VL - >
Pinductor
Value
5.508A
7.50V
-11.10V
0.33W
[W]
References
Fig.6.10
Fig.7.4
Fig.7.4
Eq.(7.1)
Table 7.4. The power loss in the inductor with data from test 1. The input voltage is 20V, the
duty cycle is 0.6 and the load is 2Ω.
The power loss in the inductor is much smaller than in simulation 3. It is because the
resistance in the inductor is much smaller than it was measured to.
Capacitor
The power loss in the capacitor has been neglected, because it is known from Table 6.4
that it does not have any influence on the total power loss.
Total power loss
Like in the simulation, the power loss in the converter is the sum of the power losses in
all the components. Table 7.5 shows the power losses in the converter.
Parameter
Pswitch
Pdiode
Pinductor
Pcapacitor
Plosses
Value
0.699W
0.742W
0.33W
≈0
1.63W
References
Tab.7.2
Fig.6.2
Fig.6.3
Eq.(5.21)
Table 7.5. The total power loss in the real converter, with an input voltage of 20V, a duty cycle
of 0.6 and a load of 2Ω.
All the relative power losses are shown Figure 7.10.
It can be concluded that the power loss in the inductor is not as big as expected. This
can possibly be because of an error in the measurement equipment when the resistance of
the inductor was measured. The diode and the switch got very hot when testing but the
inductor did not heat as much. This confirms the diagram in Figure 7.10.
68
7.4. Test 4: Verification of the efficiency of the converter
Aalborg University
Figure 7.10. All the relative power losses in the converter from the tests.
Efficiency
To calculate the efficiency, Equation (5.22) has been used. To do this, the input power has
to be calculated the same way as in the simulation with Equation (6.9). Instead of using
the average current in the switch (< iS >) the average current in the input (< ii >) has
been used. This is because the input current is not the same as the current in the switch,
in this case. The average current in the input has been calculated with Equation (7.2).
(7.2)
Pi = Vi · < ii >
[W]
Table 7.6 shows the values used for calculation of the power losses and the calculated
efficiency.
Parameter
< ii >
vi
Pi
Plosses
η
Value
3.437A
19.89V
68.362W
1.63W
0.976
References
Fig.7.5
Fig.7.3
Eq.(7.2)
Eq.(5.21)
Eq.(5.22)
Table 7.6. The total efficiency in the real converter, with an input voltage of 20V, a duty cycle
of 0.6 and a load of 2Ω.
The total efficiency in the converter, in this tests, is 97.6% with an input voltage of 20V,
a duty cycle of 0.6 and a load of 2Ω.
7.4
Test 4: Verification of the efficiency of the converter
In this test, the efficiency of the converter has been tested. This has been done as in
the simulation with a constant input voltage of 45V and a load of 1.5Ω. The duty cycle
has been varied and the input and output voltage and the input and output current have
69
Group EE5-511
7. Test of the converter
been measured. This is done to calculate the input and output power and thereafter to
calculate the efficiency. Figure 7.11 shows the efficiency in the converter as a function of
the duty cycle.
Efficiency of the converter as a function of the duty cycle
1
0.9
0.8
Efficiency, η [-]
0.7
0.6
0.5
0.4
0.3
Tested efficiency
Simulated efficiency
0.2
0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
Duty cycle [-]
0.7
0.8
0.9
1
Figure 7.11. The tested and simulated efficiency as a function of the duty cycle with a constant
input voltage of 45V and load of 2Ω.
As seen in the figure the efficiency increases as the duty cycle is increased but there is a
little jump when the duty cycle is around 0.4. This can be caused by some errors in the
measurement. It can be noticed in the figure, that the efficiency of the test is only shown
in the range [0.15 ; 0.6] and this is because there have been some problems with the switch
controller that made it impossible to increase the duty cycle more.
When comparing the tested efficiency and the simulated efficiency it is clear that the
tested efficiency is higher than the simulated one. This might be because the power loss
in the real converter is quite lower than the simulated one. This result corresponds to the
result of the calculated efficiency from simulation 3 and test 3.
7.5
Test 5: Verification of the border between CCM and
DCM
In this test the output voltage and the output current from the converter have been
measured with some different values of the resistance. The input voltage has been held
constant at 20V. The range of the resistance has been the same as in simulation 5 (4Ω to
500Ω). The duty cycle has not been completely the same as in the simulation because it is
70
7.5. Test 5: Verification of the border between CCM and DCM
Aalborg University
very difficult to tune the duty cycle on the converter (0.15, 0.25, 0.35, 0.45, 0.55, 0.65 and
0.85). Figure 7.12 shows the results from the test. The value of io lim max is still 0.353A.
Figure 7.12. The relationship between the duty cycle and the fraction of the input current and
the maximum value of the current limit. The output voltage stays almost constant
in CCM, but in DCM the output voltage increases. The conditions are: an input
voltage of 20V and a duty cycle of 0.15, 0.25, 0.35, 0.45, 0.55, 0.65 and 0.85.
As shown in Figure 7.12, the results are very similar to the simulation in Figure 6.13. The
output voltage is almost constant in CCM and increases in DCM, as expected. It can be
concluded that the calculation about the border between CCM and DCM are correct and
the chosen inductance is correct.
71
Conclusion
8
The purpose of this project has been to design, implement and test a system to control
and supply an electric go-kart. This goal has been kept in mind during the development
of the project.
Some research and analysis about different kinds of DC motors were made before
describing the one used for this application.
After researching some converters that could be used in this application, a Buck converter
was chosen because it was considered the best one for this system. Calculations were
made before building the converter, and as the value of the current in the converter was
to big it was decided to split the converter into four phases. When the new calculations
of one phase had been made, an inductor had to be built because it was not possible to
order a commercial one. After building the first prototype of the converter, it was tested
in order to check if it was working in the right way compared to some simulations. These
simulations were made with real values of the chosen components and this means the
power losses were included. There have been made two prototypes of the converter before
printing the final power converter.
At the beginning, the design of the power converter was independent from the switch
controller. It was not possible to simulate this part of the project, because there were some
commercial components on it, so it was decided to test the switch controller separately in
the lab. Therefore the first prototype was a premade circuit board until it was working
as expected. After that it was decided to print the final version of the switch controller
board.
The synchronization board was made after deciding to split the system up into four phases
and after thinking on how to make them work together.
When all the system was printed and soldered, it was tested and checked to see if it was
working as expected. First of all, each printed board was tested on its own and after that
all the system was tested together.
One phase of the converter is working as expected, with an efficiency between 83.7% and
94.4% in real conditions. The switch controller is also working as expected. When the
potentiometer, which emulates the throttle pedal, is increased the duty cycle also increases
and when decreased the duty cycle decreases. When the current, which flows through the
inductor increases, the duty cycle is decreased by the current controller in order to control
the current, but it still needs some improvements.
73
Group EE5-511
8. Conclusion
Therefore it can be concluded that one phase is working as expected.
The synchronization board is working for one phase so it is expected that it will work
properly with the four phases. But because it has not been possible to build the other
three phases, the system has not been connected to the DC motor.
Even though it has not been possible to connect the system to the motor, it has been
kept in mind. It has been decided to print the power converter board with a diode in
series with the output in order to avoid currents coming in opposite direction from the
DC motor.
In order to implement the system into the go-kart, it would just be necessary to implement
the other three boards, assemble them with the synchronization board and make all the
system work, connected to the DC motor.
Apart from that the system has not been finished and installed on the go-kart, because of
all the difficulties during the tests, it can be concluded that the results are satisfactory.
74
Improvement proposals
9
After finishing the design, making the DC converter and making tests with it, there are
some aspects of it that could be improved in order to have better results.
To begin with, the efficiency of the converter can be increased by changing some
parameters of it. The first element that can be changed is the inductor. This can be
rebuilt to try to reduce the internal resistance which will reduce the power loss in the
inductor. The other elements of the converter such as the MOSFET or the diode can
also be changed for others with better quality and less power losses, but this will mean
that the converter will be more expensive. To reduce the power loss in the diode or the
MOSFET, a bigger heat sinks can be used.
Another point to improve could be the design of the current controller. As explained in
the report, there are many different ways of controlling the current. The best controller for
the converter should be a current/voltage controller. This allows to have a more accurate
control of the system making it more efficient.
In addition, the PCB layout of the converter can be redesigned in order to add the current
controller on it and put the power elements closer. This means that the only cables between
boards will be between the board of the power supply and the synchronization, because
the cables between the power board and the control board disappear. This will also help
to reduce the size of the system what is important for these kind of applications.
Apart from that, an additional circuit can be added to the converter to cancel all the noises
generated by the switching of the MOSFET. The problem of this is that all the circuits
are affected by those noises, which changes the normal behavior of the components.
There is another important point that can be done for making a better DC converter. The
one made in this project has been designed only to make the motor work in one direction,
but a converter that can make the motor rotate in both directions can be designed.
Finally, another possibility for improving this project can be to make the converter work
in both directions, i.e. it can give power to the motor or it can receive power from it. This
is useful so as to save energy by charging the batteries when breaking the motor or going
down a hill for an example.
75
Bibliography
Andersen, T. O. and Pedersen, H. C. (2011). Electrical DC servo motor - Laboratory
Exercise. AAU.
Andersson, C. (2011). Design of a 2.5kw dc/dc fullbridge converter.
Andreycak, B. (1999). UCC 3800/1/2/3/4/5 Bicmos current mode control ICs
datasheet. Unitrode.
Batteries, O. (2005). Technical Specs.
Bo-Tao Lin, Y.-S. L. (1997). Power-factor correction using cuk converters in
discontinuous-capacitor-voltage mode operation. IEEE TRANSACTIONS ON
INDUSTRIAL ELECTRONICS, 44:6.
Condit, R. (2004). Brushed dc motor fundamentals. Microchip Technology Inc.
D.A.Bradley (1995). Power Electronics.
Electronics, J. (2011). Dc-dc converters: A primer. page 5.
Erickson, R. W. (2011). Dc - dc power converters.
Farnell.com (2011). Farnell. http://dk.farnell.com.
Illan Glasner, J. A. (1996). Advantage of boost vs. buck topology for maximum power
pointtracker in photovoltaic systems.
Karpin, O. (2006). Multi-cell li-ion/li-pol battery charger with cell-balancing and fuel
gauge function support. CYPRESS PERFORM.
Lem (2011). Current transducer hais 50..400-p and hais 50..100-tp. page 3.
Maksimovic, R. W. E. D. (1956). Fundamentalas of Power Electronics.
Microchip (2001). 6a high-speed mosfet drivers. page 8.
Mohan, Unerland, R. (2003). Power Electronics, converters, applications and design.
Wiley.
Motenergy (2011). ME0708 Product Description.
http://www.motenergy.com/me0708.html.
Rashid, M. H. (2001). Power Electronics Handbook. ACADEMIC PRESS.
SPIRALCELL technology (2005). OptimaBatteries2005. OPTIMA Batteries.
77
Group EE5-511
BIBLIOGRAPHY
www.tutorvista.com (2010). Electric motor.
http://www.tutorvista.com/content/science/science-ii/magnetic-effects-electriccurrent/electric-motor.php.
Xi, F. (2007). Design of dc-dc converters. IEEE SSCS Dallas Chapter, page 41.
Xixing (2007). Motor Dimension. Mars Electric LLC.
Yedamale, P. (2003). Brushless dc (bldc) motor fundamentals. Microchip Technology
Inc.
78
The Buck-Boost converter
A
In this appendix the Buck-Boost converter will be designed by analyzing the electric circuit
and calculating the values of the components. When the calculations are completed the
components will be chosen to be able to build the converter. In the last part of the chapter
there will be concluded if it is wise to build the converter.
A.1
Topology of the converter
The Buck-Boost converter works in two states. The first state is when the switch is closed
and the diode is not conducting (A). The second state is when the diode is conducting
and the switch is open (B). Figure A.1 shows the two states. These figures are modified
versions of Figure 4.2.
Figure A.1. The two conditions the Buck-Boost converter works in.
All the voltages and the currents in all the components in the converter are shown in
diagrams in Figure A.2.
79
Group EE5-511
A. The Buck-Boost converter
Figure A.2. All the voltages and the currents in all the components in the Buck-Boost converter.
(A) is the first condition and (B) is the second condition.
A.2
Design of the converter
In this section the Buck-Boost converter will be designed by calculating the parameters
of the converter and in the end, the components will be chosen to be able to build the
converter.
Specifications
There are few specifications which the converter has to be able to handle.
specifications are shown in Table A.1.
80
The
A.2. Design of the converter
Aalborg University
Parameter
Input voltage
Output voltage
Input power
Equivalent resistance of the load
Frequency
Switching time
Voltage ripple
Current ripple
Symbol
Vi
Vo
Pi = Po
2
Ro = VPoo
fs
Ts = f1s
∆VC
∆iL
Value
[42 ; 52.4]V
[0.1 ; 48]V
4800W
0.48Ω
50kHz
0.00002s
0.1% of Vo
20% of < iL >
Table A.1. Already known or chosen values for designing the converter.
The input voltage range is known from the range of the batteries. The output voltage
range is the input voltage range of the motor. This requires that the converter is ideal.
The output power is the power of the motor. The frequency has been chosen to be 50kHz
because this can help with decreasing the size of the inductor and the capacitor. The value
of ∆VC and ∆iL have been chosen to be that small to decrease the ripple on the output
voltage (Vo ) and the output current (Io ). The equivalent resistance of the motor can vary
but the minimum resistance is Ro = 0.48 Ω, so for the calculations, another value will be
used (0.45Ω).
When these values have been decided the parameters of the converter can be calculated
by using the following equations.
Mathematical model of the Buck-Boost converter
The duty cycle is the time that the system is active as a function of the total time (TS ).
The expression for D is
(A.1)
D=
Vo
Vi + Vo
[−]
The output current (Io ) can be calculated by Equation (A.2).
(A.2)
Io =
Vo
Ro
[A]
The average current of capacitor (< iC >) is equal to zero during the whole period.
(A.3)
< iC >= 0
[A]
This is why the average current of the diode (< iD >) is directly equal to the output
current. This is shown in Equation (A.4).
(A.4)
< iD >= Io
[A]
In swich-off time, the average current of the inductor (< iL >) is equal to the current of
the output (Io ). < iL > can be expressed as shown in Equation (A.5).
(A.5)
< iL >=
Io
1−D
[A]
81
Group EE5-511
A. The Buck-Boost converter
In swich-on time, the average current of the switch (< iS >) is equal to the average current
of the inductor (< iL >). The average current of the switch (< iS >) is calculated as seen
in Equation (A.6).
(A.6)
< iS >=
Io
·D
1−D
[A]
The value of the capacitor can be deduced from the equation for the voltage ripple (∆VC )
of the output as shown in Equation (A.7).
∆VC =
D
Vi · Ts
·
Ro · C 1 − D
[V]
−D · Ts · Vi
(D − 1) · ∆Vc · Ro
[F]
m
(A.7)
C=
The ripple of the current is chosen to be 5% of average current of the inductor as shown
in Equation (A.8).
(A.8)
∆iL =< iL > · 0.05
[A]
The size of the inductor L can be calculated as shown in Equation (A.9).
∆iL =
Vi · D · Ts
L
[A]
Vi · D · Ts
∆iL
[H]
m
(A.9)
L=
Maximum values of currents and voltages can be calculated according to the diagrams in
Figure A.2. Relations of maximum values are shown in Equation (A.10) to (A.14).
(A.10)
IˆL = IˆS = IˆD =
Io
Vi · Ts
+
·D
1−D
L·2
[A]
(A.11)
IˆC = Iˆo [A]
(A.12)
V̂L = V̂i [V]
(A.13)
V̂C = V̂o
[V]
(A.14)
V̂S = V̂D = V̂o + V̂i
[V]
The Root Mean Square (RMS) value of the current in the capacitor can be calculated by
using Equation (A.15).
(A.15)
82
IC,RMS =
∆iL
√
2· 3
[A]
A.2. Design of the converter
Aalborg University
Continuous Conduction Mode (CCM) and Discontinuous Conduction
Mode (DCM)
When the inductor has been chosen the output current limit (Io lim ) can be calculated
as seen in Equation (A.16). This value has to be smaller than the output current (Io ).
Otherwise the converter will work in DCM.
(A.16)
Io lim =
Vi · Ts
· (1 − D) · D
2·L
[A]
Results of calculations for the converter
The calculated values from Equation (A.1) to (A.16) are shown in Table A.2.
Parameter
Duty cycle
Output current
Avg. current in the diode
Avg. current in the inductor
Avg. current in the switch
Avg. current in the capacitor
Capacitance
Current ripple in the inductor
Inductance
Max. current in the inductor
Max. current in the capacitor
Max. voltage in the inductor
Max. voltage in the capacitor
Max. voltage in the switch
RMS current in the capacitor
Current limit between CCM and DCM
Symbol
D
Io
< iD >
< iL >
< iS >
< iC >
C
∆iL
L
IˆL = IˆS = IˆD
IˆC
V̂L
V̂C
V̂S = V̂D
IC,RMS
Io lim
Value
[0.0019 ; 0.53][-]
[0.22 ; 106.67]A
[0.22 ; 106.67]A
[0.23 ; 228.57]A
[0.0004 ; 121.91]A
0A
[0.85·10− 5 ; 1.14]F
[0.045 ; 45.71]A
[9.8 - 45]µH
251.43A
106.67A
52.4V
48V
100.4V
[0.22 ; 228.95]A
[0.0095 ; 1.25]A
Reference
(A.1)
(A.2)
(A.4)
(A.5)
(A.6)
(A.3)
(A.7)
(A.8)
(A.9)
(A.10)
(A.11)
(A.12)
(A.13)
(A.14)
(A.15)
(A.16)
Table A.2. Calculated values of the Buck-Boost converter.
Chosen components
To be able to chose the components there are some important criterias for each component
that have to be fulfilled.
The diode
The
voltage (Vr ) has to be larger than the maximum voltage through the diode
reverse
V̂D with some margin. The current through the diode (ID ) has to be lower than the
maximum current through the diode (IˆD ) with some margin but it can also be lower than
the average current through the diode. The forward voltage in the diode (VF ) has to
be as low as possible because this is related to power losses in the diode. The recovery
time of the diode has to be much less than the minimum time it takes the diode to close
(trr << Toff,min = (1 − D) · Ts ).
83
Group EE5-511
A. The Buck-Boost converter
As seen in Table A.2, V̂D = 100.4V and IˆD = 251.429A, the chosen diode will have to
fulfill these values.
When looking on Farnell.com [2011] for a diode that could fulfill the criterias, few diodes
were able to be chosen but one was chosen and that diode’s name is DSS 2 X 101-015A
(1080091 on the web page). In Table A.3 the values of the diode, that fulfill the criterias,
can be seen.
Vr
ID
VF
Price
150V
100A
0.77V
257.04DKK
Table A.3. The values of the DSS 2 X 101-015A (1080091 on the web page) diode that fulfill
the criterias.
The switch
The drain source voltage of the MOSFET (VDS ) has to be larger than the maximum voltage
in the switch (V̂s ) with some margin. The drain source current in the MOSFET (IDS ) has
to be lower than the maximum current through the switch (IˆS ) with some margin. The
on resistance of the MOSFET has to be as low as possible because it is related to power
loss in the MOSFET.
As seen in Table A.2, V̂s = 100.4V and IˆS = 251.429A, the MOSFET will have to fulfill
these values.
When looking on Farnell.com [2011] for a switch that could fulfill the criterias, few
MOSFETS were able to be chosen but one was chosen and that MOSFET’s name is
IRFP4568PbF (1684525 on the web page). In Table A.4 the values of the MOSFET, that
fulfill the criterias, can be seen.
VDS
IDS
RDS,on
Price
150V
171A
4.8-5.9mΩ
67.06DKK
Table A.4. The values of the IRFP4568PbF (1684525 on the web page) MOSFET that fulfill
the criterias.
The inductor
From Table A.2 a value for the inductor has to be at least 45µH. This is to keep ∆iL
low. Besides that, the inductor has to be able to cope with 251.429A. From a search on
Farnell.com [2011] there was no premade inductor that could handle that.
The capacitor
From Table A.2 a value for the capacitor has to be at least 1.1377F to keep ∆VC low.
Besides that, the capacitor has to be able to cope with 106.667A and 48V. From a search
on Farnell.com [2011] there was no capacitor that was that could handle that.
84
A.3. Conclusion of the Buck-Boost converter
A.3
Aalborg University
Conclusion of the Buck-Boost converter
From the calculations of the Buck-Boost converter is it possible to conclude, that it is not
desirable to choose a Buck-Boost converter for the go-kart. The reason for that is, that
it’s very difficult to find all the components for the converter, because there will be high
currents in the converter. Furthermore the components will be very expensive and have
a big physical size.
After seeing all the disadvantages of the Buck-Boost converter, a Buck converter will be
chosen as a converter to the go-kart. The reason for this choice is, that the value of the
capacitance in a buck converter will be smaller than the Buck-Boost converter and the
currents in the circuit will be lower. The disadvantages of this choice is that a Buck
converter is not able to boost the voltage if the output voltage of the batteries is lower
than the needed input voltage to the motor. But this case is not very typical.
85
Contents on the CD
B
B 1: Datasheets of the PMDC motor.
B 2: Datasheet of the battery, Optima YellowTop U 4,2
B 3: Datasheet of the MBR30H60CTG diode.
B 4: Datasheet of the NTP6412AN MOSFET.
B 5: Datasheet of the LEM HAIS-100-TP Current Transducer.
B 6: Datasheet of the capacitor in the converter.
B 7: Datasheets of the inductor cores and the material.
B 8: All the data from the simulations and the tests
B 9: All web pages and reports used as a reference.
B 10: P5 Report.
I