dc grids for wind farms - webfiles its...

DC Grids for Wind Farms

Olof Martander

Department of Electric Power EngineeringCHALMERS UNIVERSITY OF TECHNOLOGY

Goteborg, Sweden 2002

THESIS FOR THE DEGREE OF LICENTIATE OF ENGINEERING


Olof Martander

Department of Electric Power EngineeringCHALMERS UNIVERSITY OF TECHNOLOGY

Goteborg, Sweden 2002


Olof Martander

c© Olof Martander, 2002.

Technical Report No. 443LDepartment of Electric Power EngineeringCHALMERS UNIVERSITY OF TECHNOLOGYSE-412 96 GoteborgSwedenTelephone + 46 (0)31 772 1000

Chalmers Bibliotek, ReproserviceGoteborg, Sweden 2002

Abstract

Wind turbine technology, which is one of the oldest still in use, has undergone arevolution during the last century. The most important difference between the earliestwind mills and the new generation of wind turbines is that the latter produce electricalenergy. In order to reduce costs, the turbines are placed in wind farms and the newtrend is offshore wind farms. If the distance to the main grid is considerable, aninteresting alternative to AC transmission can be to connect the wind farms to themainland using high voltage direct current connections (HVDC).

This thesis focuses on DC/DC converters for high voltage and high power and inves-tigates different DC/DC converter topologies with respect to a wind farm applicationand presents different design concepts to determine the potential for using DC gridsin a wind farm. DC/DC converters have been examined by many authors before butmostly for low power and low voltage applications. For a high voltage and high powerapplication, like a wind park transmission system, the focus has to be on the utilizationfactor of the used components.

The chosen electrical system for a wind farm uses a boost converter as a voltage adjusterand a full-bridge converter as a DC transformer. Simulations are made with differentfault locations for a simplified system. Depending on the type of fault, different parts ofthe transmission system have to be shut down. The DC-transformer can clear all faultsby simply turning off the switches while in contrast a short circuit on the secondaryside of the boost converter cannot be cleared and leads to a permanent fault.

Finally, a complete wind farm with 25 wind turbines was simulated and the wind and,therefore, the produced power was increased up to a rated power of 50 MW. The systemshows good system performance when the different bus voltages change with differentwind speeds and, thus, different power flows.

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Acknowledgement

The financial support given by the Swedish National Energy Administration is grate-fully acknowledged.

Joachim Lindstrom introduced me to the secrets of the life of a Ph.D. student which Iappreciate. Thanks also to my roommates Andreas Petersson and Marcus Helmer forall the laughs and the sometimes not so fruitful discussions and activities. Many thanksto all my colleagues and the staff at the Department of Electric Power Engineering andespecially the ”lunch” group consisting of Tomas Petru, Rolf Ottersten, Dr. TorbjornThiringer, Mattias Jonsson and Stefan Lundberg.

Many thanks to Robert Karlsson for his assistance, answers to various questions andgood ideas.

Thanks to Dr. Jan Svensson and Dr. Ola Carlsson for their supervision, discussionsand proof-reading during the work with this thesis.

I would also like to thank my examiner Prof. Essam Hamdi.

Finally, I owe my deepest gratitude to my family and especially to Lisa for her under-standing, encouragement and love throughout this project.

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Contents

Abstract iii

Acknowledgement v

1 Introduction 1

1.1 Background and Motivation . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Literature Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Wind Energy and Wind Turbines 5

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Wind Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2.1 Wind Energy in Sweden . . . . . . . . . . . . . . . . . . . . . . 6

2.2.2 Wind Energy in Europe . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Wind Turbines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3.1 Power Control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3.2 Fixed-speed Turbine . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3.3 Variable Speed Turbine . . . . . . . . . . . . . . . . . . . . . . . 12

2.4 Grid Interaction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Wind Farm Configurations 15

3.1 Wind Farms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 Offshore Wind Farms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.3 Connecting Offshore Wind Farms . . . . . . . . . . . . . . . . . . . . . 16

3.4 System Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.5 Wind Farm Layouts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.5.1 AC Wind Farm . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.5.2 AC/DC Wind Farm . . . . . . . . . . . . . . . . . . . . . . . . 22

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3.5.3 DC1 Wind Farm . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.5.4 DC2 Wind Farm . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.6 Cost and Loss Estimation . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.7 Future Development . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.8 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

4 Hard Switching DC/DC Converter Topologies 33

4.1 Converter Environment . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.2 DC/DC Converter Topologies . . . . . . . . . . . . . . . . . . . . . . . 35

4.2.1 Boost Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

4.2.2 Cuk Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.2.3 Sepic - Single-Ended Primary Inductance Converter . . . . . . 42

4.2.4 Zeta Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.2.5 Luo Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.2.6 Flyback Converter . . . . . . . . . . . . . . . . . . . . . . . . . 49

4.2.7 Forward Converter . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.2.8 Two Two-Transistor Forward Converter . . . . . . . . . . . . . 55

4.2.9 Push Pull Converter . . . . . . . . . . . . . . . . . . . . . . . . 58

4.2.10 Half Bridge Converter . . . . . . . . . . . . . . . . . . . . . . . 61

4.2.11 Full Bridge Converter . . . . . . . . . . . . . . . . . . . . . . . . 63

4.2.12 Half Bridge Converter With Voltage Doubler . . . . . . . . . . . 66

4.3 Comparison of Switch Utilization for Different Converters . . . . . . . . 69

4.3.1 Voltage Adjustment Converter . . . . . . . . . . . . . . . . . . . 69

4.3.2 DC/DC Transformer . . . . . . . . . . . . . . . . . . . . . . . . 71

4.4 Losses for Different Converter Layouts . . . . . . . . . . . . . . . . . . 73

4.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5 Dynamic Analysis of Hard-switched DC/DC converters 75

5.1 Boost Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.1.1 Working Principle . . . . . . . . . . . . . . . . . . . . . . . . . . 75

5.1.2 Transmitted Power . . . . . . . . . . . . . . . . . . . . . . . . . 77

5.1.3 Transient Behavior . . . . . . . . . . . . . . . . . . . . . . . . . 78

5.1.4 Varying Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

5.1.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

5.2 Half Bridge Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

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5.2.3 Steady-state Behavior . . . . . . . . . . . . . . . . . . . . . . . 85


5.2.5 Half Bridge Converter With Voltage Doubler . . . . . . . . . . . 87

5.2.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

5.3 Full Bridge Converter . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90





5.3.5 Full Bridge Converter with Voltage Doubler . . . . . . . . . . . 94

5.3.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

5.4 Dual Active Bridge Converter (DAB) . . . . . . . . . . . . . . . . . . . 97


5.4.2 Comparison with General AC Theory . . . . . . . . . . . . . . . 101




5.4.6 Varying Voltage . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

5.4.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

6 Wind Farm Simulation 111

6.1 Transmission System Layout for Wind Farms . . . . . . . . . . . . . . 111

6.2 Faults . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.2.1 Test Setup . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.2.2 Line to Ground Faults . . . . . . . . . . . . . . . . . . . . . . . 114

6.2.3 Line to Line Faults . . . . . . . . . . . . . . . . . . . . . . . . . 117

6.2.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

6.3 Simulation of Complete Wind Farm . . . . . . . . . . . . . . . . . . . . 120

6.4 Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

7 Conclusions and Future Research 123

7.1 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

7.2 Future Research . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

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References 125

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Chapter 1

Introduction

1.1 Background and Motivation

Wind turbine technology, which is one of the oldest still in use, has undergone arevolution during the last century. A renewed interest in wind energy arose during theoil crisis in the mid 1970s. This attention has continued to grow as the demands onreducing polluting emissions have increased. The most important difference betweenthe earliest wind mills and the new generation of wind turbines is that the latterproduce electrical energy. Nowadays, it is not necessary to build the wind turbineclose to the user, on the contrary, people do not want the wind turbines within sight.In order to reduce cost, the turbines are placed in wind farms and the new trend isoffshore wind farms. At sea, periods of complete calm are generally extremely rare,and quite short-lived. If the distance to the main grid is considerable, an interestingalternative to AC transmission can be to connect the park to the mainland by usinga high voltage direct current connection (HVDC) and by using a local DC-grid in thewind farm. New semiconductor technologies make this possible but suitable convertertopologies have not been established yet.

A DC/DC converter can be described as the DC equivalent of an AC transformer. Itchanges the ratio between the input and output voltages and currents by introduc-ing power electronics that, with the help of passive components, transmit the powerthrough the converter. The advantages of using DC/DC converters are many: To reg-ulate the output voltage, to build subsystems supplied by the same bus and to reducetransmission losses. Presently, most applications are in the low power region, oftenbelow 200 W, and the output voltage should be controlled. Microprocessors and elec-tronics use low voltage around one to two volts and the challenge is to adjust the supplyvoltage for high efficiency. Some of the techniques used are soft switching [1], resonanceconverters [2] and synchronous rectifiers [3]. A lot of DC/DC converters are also usedtogether with diode bridges for AC/DC-conversion, for example, in a Boost Power Fac-tor Correction (PFC) [4] circuit to reduce the influence on the grid by reducing theharmonics and increasing the power factor. Today, low power DC/DC converters arecommon and stand for a major part of the turn-over in the power electronics marketbut high power and high voltage DC/DC converters in the MW range are not readilyavailable on the market.

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This thesis focuses on DC/DC converters for high voltage and high power. The objec-tive of the first part of this project was to determine the potential for using DC grids ina wind farm. Another objective was to investigate different configurations of electricalsystems for offshore wind farms.

The objective of the second part of the project was to investigate different DC/DCconverter topologies with respect to a wind farm application and present design con-cepts with different high voltage and high power converter topologies.

1.2 Literature Review

DC/DC converters have been studied by many authors before but mostly in low powerand low voltage applications.

The most common converters are covered by textbooks in power electronics for ex-ample Mohan et. al. [5], Ericson and Maksimovic [6], Hart [7], Hnatek [8] and Kas-sakian et. al. [9]. They all give a short introduction to converters and present some ofthe design issues.

A general overview of power electronic converter technology is given by Steigerwald [10]where he performs a conceptual design study and compares different high frequency,high power converters. The same author et. al. has also presented a comparison of high-power DC-DC soft-switched converter topologies in [11] and states that, in the highpower areas, the trend has been to mitigate away from more complicated topologiesthat require snubbers or auxiliary circuits and use more rugged, higher performanceswitches in a hard switched mode in order to decrease complexity and save cost.

Fuentes and Hey [12] describes a family of soft switching converters for high powerapplications and verifies the feasibility of the proposed converters by analyzing a Zero-Current-Switching (ZCS) Boost converter.

More special converters are found in articles by the inventors, such as the Luo conver-ter [13] and by many others. Luo converters are described as a series of new step-upconverters using voltage lift technique.

Special studies of the Boost Converter have been reported by Javanovic and Jang [14]and He and Jacobs [15]. The first paper describes a novel active snubber and thesecond paper describes how to connect IGBTs in parallel for the converter.

Half and Full Bridge Converters are further investigated in a paper by Kherson-sky et. al. [16] where high frequency and tail current losses are specially addressed.Cho et. al. [17] presents a Zero-Voltage-Switching (ZVS) DC/DC converter for highpower applications where ZVS is achieved in the entire line and load range withoutincreasing device voltage and current stresses. A steady state analysis with completecharacterization of the converter is given by Sabate et. al. [18].

A more recently introduced converter is the Dual Active Bridge (DAB) Converter,which has been studied by De Doncker et. al. [19], in which a three-phase DAB conv-erter is presented. The three-phase DAB has lower turn-off peak currents in the powerdevices and lower RMS current ratings in the filter capacitors. A significant increase inpower density is also attainable by using a three-phase symmetrical transformer. The

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performance of a single-phase DAB converter is described by Kheraluwala et. al. [20]and various control schemes are outlined. Zhang et. al. [21] describes a novel controlscheme for the DAB converter, which eliminates some of the problems in previous con-verters, like the large circulating energy and large current ripple for the capacitors,and achieves ZVS and ZCS for all switches. Another use of the DAB converter is forDC/AC conversion and Vangen et. al. [22, 23] describes design equations and parame-ters that influence efficiency. A Dual Bridge Converter with soft switching features forapplying a new modulation technique while eliminating the auxiliary circuits used inprevious versions, is presented by Torrico-Bascope and Barbi [24].

Papers about high power components have been presented by Satoh and Yamamoto [25]and Zeller [26] both of which describe the background to the different high-powercomponents used, and present current alternatives and future trends. Series connectionof IGBTs is reported by Busatto et. al., in [27]. The paper demonstrates that seriesconnection of IGBTs can be used to extend the trade-off between conduction andswitching losses even to high voltage modules.

Different aspects of the electrical system and components in offshore wind farms arereported by several authors and comparisons are made between AC and DC solutions.Bauer et. al. has made an inventory of electrical systems for offshore wind farms [28]and has also modeled some of the wind farm components [29]. Interest lies primarilyin identifying wind farm configurations with a low cost profile and high efficiency.Janosi et. al. [30] presents a simulation of a 12 MW traditional wind farm with ACtransmission and describes the behavior of this system from wind to electrical power tothe grid. Christensen et. al. [31] compares AC and HVDC solutions for grid integrationof offshore wind farms and states that HVDC solves most of the integration problemsand, at the same time, also provides the possibility for staged development. However,the potential should be verified through research and real scale demonstration projectsand the development time might become a critical limiting factor. A model of aHVDC transmission system for offshore wind farms has been made by Rasmussenand Pedersen [32] and some of the experiences of the model are given in the article.DC/DC conversion for offshore wind farms is reported by Macken et. al. in [33, 34] andMorren et. al. in [35, 36]. Both authors have designed an electronic DC transformerfor high power and also describe some of the issues concerning DC bus systems.

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1.3 Outline

The outline of the thesis is the following:

• Chapter 2 introduces the basics of wind energy and wind turbines;

• Chapter 3 presents wind farm configurations;

• Chapter 4 gives an overview of hard-switched DC/DC converters;

• Chapter 5 presents simulations of hard-switched DC/DC converters;

• Chapter 6 presents continued simulations with simulations of faults in a windfarm and finally a complete wind farm with a dc-grid;

• Chapter 7 includes the conclusion and future research.

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Chapter 2

Wind Energy and Wind Turbines

2.1 Introduction

In recent years, the number of wind turbines has increased greatly. In Denmark, windpower contributes 13-15% of the total electricity production. In Germany, wind powerproduces almost 3.5% of the total electricity [37]. In this chapter, the energy in thewind and different wind turbines will be presented. Moreover, the advantages anddisadvantages of using variable-speed wind turbines will be covered.

2.2 Wind Energy

Wind energy is a renewable energy source, i.e. a clean source, which does not polluteor increase green house gases. Moreover, wind resources are plentiful and wind will notrun out.

Wind turbines generate no CO2, NOx or SOx during operation, and very little energy isrequired for manufacturing, maintaining and finally scrapping the plant. In fact, withmoderate wind onshore sites, a wind turbine will recover all the energy spent in itsmanufacture, installation and maintenance in less than three months [38]. With a 20year lifetime this gives a thermal efficiency of 8000%, i.e. the wind turbine recovers theenergy about 80 times in its lifetime (comparable to a conventional coal power plant’s45% [38]). For offshore turbines, the results may be better due to the longer expectedlifetime of the turbines. Less turbulence and thus lower fatigue loads will increase thelifetime of offshore turbines to 25-30 years.

The major drawback of wind power is the variability of the wind. In large electricalgrids, however, consumers’ demand varies, and power utilities have to keep spare ca-pacity running idle in case a major generating unit breaks down. If a power utility canhandle varying consumer demand, it can technically also handle the ”negative electric-ity consumption” from wind turbines. The more wind turbines on the grid, the moreshort-term fluctuations from each turbine will be cancelled out and the total powerproduction will be smoothed out.

At most wind turbine sites around the globe, in fact, the wind varies substantially, with

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high winds occurring rather infrequently, and low winds occurring most of the time.In addition, in e.g. Europe and a number of other locations around the globe, windspeeds happen to be positively correlated with peak electricity use. More wind duringthe day than at night, more wind in winter than in summer, raising the value of thewind to the grid by 40 to 60%, compared to a completely random wind pattern [39].

2.2.1 Wind Energy in Sweden

The installed wind power capacity in Sweden was 122 MW at the beginning of year1998, coming from 334 wind turbines and has increased to 290 MW by the end of year2001 (565 turbines) [40].

2.2.2 Wind Energy in Europe

The European Wind Energy Association (EWEA) has set a target of 60000 MW ofinstalled capacity for the year 2010. This goal was set in the year 2000 when theprevious goals, set in year 1991 and 1997, 25000 MW and 40000 MW respectively, wasdiscovered to be too pessimistic. The total installed capacity as of the end of year2001 was 17000 MW and was excepted to produce some 40 TWh thus preventing theemission of 24 million tons of CO2 annually. Table 2.1 shows installed capacity andwind energy targets for different countries [37].

Table 2.1: Installed and targeted wind power capacity.

Country Year 2000 Target YearGermany 6100 MW 22000 MW 2010Spain 2400 MW 9000 MW 2010Denmark 2300 MW 50% (>6000 MW) 2030The Netherlands 450 MW 1500 MW 2010Great Britain 400 MW 2000 MW 2005Sweden 250 MW 8-10 TWh (4-5000 MW) 2010Ireland 86 MW 500 MW 2005France 80 MW 5000 MW 2010Finland 38 MW 500 MW 2010Norway 13 MW 3 TWh (1000 MW) 2010

2.3 Wind Turbines

Wind turbines are used to capture wind power. The standard wind power turbine oftoday consists of a turbine, which has three blades and the wind upwards, so the tubularsteel or concrete tower is behind the turbine (also known as the Danish concept).Figure 2.1 shows an example of the design elements of a wind turbine generator (WTG).The rotor captures the wind energy at the low-speed shaft. Since most generators aredesigned for high speed, a gear box is used and the energy is then transferred to the

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high-speed shaft and then to the generator. The generator is connected to a transformerto increase the voltage level to a suitable transmission level.

turbine

blades

nacelle

generator

tower

gearbox

low speed shaft

high speed shaft

hub

Figure 2.1: Design elements of a Wind Turbine Generator (WTG).

The power in the wind of the area A, perpendicular to the wind direction, is given bythe formula:

P =1

2ρAV 3 (2.1)

where P is the power, ρ is the air density and V is the wind speed. The fraction of theenergy captured by a wind turbine is given by a factor Cp, called the power coefficientand is defined as [41]:

Cp(λ) =16

27ηturbine(λ) (2.2)

where λ is the tip speed ratio of the blade, i.e. the tip speed divided by the wind speedand ηturbine is the efficiency of the turbine. Betz’ law [42] states that less than 16/27(or 59%) of the kinetic energy in the wind can be converted to mechanical energy usinga wind turbine. The power coefficient indicates how efficiently a turbine converts theenergy in wind to electricity. Very simply, the electrical power output is divided by thewind energy input to measure how technically efficient a wind turbine is. The powercurve divided by the area of the rotor gives the power output per square meter of rotorarea. Figure 2.2 shows a power coefficient curve for a typical Danish wind turbine. Theaverage mechanical efficiency is somewhat above 20%, but efficiency varies very muchwith wind speed and is the greatest (in this case 44%) at a wind speed around 9 m/s.This is deliberate since at low wind speeds efficiency is not very important becausethere is not much energy to capture. At high wind speeds, the turbine must waste anyexcess energy above the rated power of the generator and of course this determines the

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5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

Wind speed [m/s]

Cp

Figure 2.2: Power coefficient Cp as a function of wind speed.

mechanical loads on the structure. Therefore, efficiency matters most in the region ofwind speeds where most of the energy is to be found [41].

A wind turbine with high technical efficiency is not an aim in itself. What matters,really, is the cost of extracting energy from the wind during the lifetime of the windturbine (often 20 years). Since the fuel is free, there is no need to save it. On theone hand, the optimal turbine is not necessarily the turbine with the highest energyoutput per year. On the other hand, each square meter of rotor area costs money, so itis necessary to capture as much energy possible - as long as the costs per kilowatt-hourare kept down.

All fixed speed wind turbines have an overshoot in the rated power of the machineat the beginning of a gust of wind. Depending on the design of the power control,this overshoot increases the ratings of some components by a factor up to 50%. Thisovershoot, for example, causes excessive wear on the gearbox. With variable speed,however, this overshoot can be avoided.

There are economies of scale in wind turbines, i.e. large machines are usually able todeliver electricity at a lower cost than small machines. The reason for this is that thecosts for foundations, road building, electrical grid connection, plus a number of com-ponents in the turbine (the electronic control system etc.), are somewhat independentof the size of the machine.

Large machines are particularly well suited for offshore locations. The cost for foun-dations does not rise in proportion to the size of the machine, and maintenance costsare almost independent of the size of the machine [43].

In areas where it is difficult to find sites for more than a single turbine, a large turbinewith a tall tower use the existing wind resource more efficiently because the wind speedincreases with the height of the tower.

2.3.1 Power Control

Wind turbines are designed to produce electrical energy as inexpensively as possible.Wind turbines are therefore generally designed so that they yield maximum output at

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wind speeds around 12-15 meters per second. It is inefficient to design turbines thatmaximize their produced output power at stronger winds, because such strong windsare rare. In case of strong winds, consequently,it is necessary to waste part of the excessenergy of the wind in order to avoid damaging the wind turbine. All wind turbinesare therefore designed with some sort of power control. There are two common waysof doing this safely on modern wind turbines, either by pitch or by stall control.

Pitch Controlled Wind Turbines

On pitch controlled wind turbines, the rotor blades have to be able to turn around theirlongitudinal axis (to pitch) and thus, change the angle of attack as shown in Figure 2.3.When the power output becomes too high, the blade pitch mechanism pitches (turns)the rotor blades slightly out of the wind. Conversely, the blades are turned back intothe wind whenever the wind drops again. During normal operation the blades willpitch a fraction of a degree at a time, thus, changing the angle of attack. The pitchmechanism is usually operated by using hydraulics although some manufacturers useelectric motors.

torquewind component

from rotation

wind speed

attack angle

chord

tower

nacelle

Figure 2.3: Pitch control.

Stall-Controlled Wind Turbines

(Passive) stall-controlled wind turbines have rotor blades bolted onto the hub at a fixedangle. The geometry of the rotor blade profile, however, is aerodynamically designedto ensure that at the moment wind speed becomes too high, turbulence will be createdon the side of the rotor blade which is not facing the wind, as shown in Figure 2.4.This turbulence prevents the lifting force of the rotor blade from acting on the rotorand this phenomenon is called stall, i.e. the blades are stalled.

The blade is twisted slightly along its longitudinal axis. This is done partly in orderto ensure that the rotor blade stalls gradually rather than abruptly when the windspeed reaches its critical value. On the one hand, the basic advantage of stall-control

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torque

wind component

from rotation

high wind speeds

attack angle

chord

Figure 2.4: Stall regulation.

is that there are no moving parts in the rotor itself. On the other hand, stall-controlrepresents a very complex aerodynamic design problem, and related design challengesin the structural dynamics of the whole wind turbine, e.g. to avoid stall-inducedvibrations.

Active Stall Controlled Wind Turbines

An increasing number of large wind turbines (1 MW and up) are being developedwith an active stall power control mechanism. Technically, the active stall machinesresemble pitch controlled machines, since they have pitchable blades. In order to obtaina reasonably large torque (turning force) at low wind speeds, the machines are usuallyprogrammed to pitch their blades much like a pitch-controlled machine at low windspeeds. (Often they use only a few fixed steps depending upon the wind speed). Whenthe machine reaches its rated power, however, there is an important difference fromthe pitch-controlled machines: If the generator is about to be overloaded, the machinewill pitch its blades in the opposite direction from that of a pitch controlled machine.In other words, it increases the angle of attack of the rotor blades in order to make theblades go into a deeper stall, thus wasting the excess energy in the wind. One of theadvantages of active stall is that the output power can be controlled more accuratelycompared to passive stall. Another advantage is that the machine can run almostexactly at rated power at all high wind speeds. A normal passive stall-controlled windturbine will usually have a drop in the electrical power output for higher wind speeds,as the rotor blades go into deeper stall. The pitch mechanism is usually operated usinghydraulics or electric motors. As with pitch-control it is largely an economic questionwhether or not it is worthwhile to pay for the added complexity of the machine whenthe blade pitch mechanism is added.

10

Other Power Control Methods

Some older wind turbines use ailerons (flaps) to control the power of the rotor, justlike aircraft use flaps to alter the geometry of the wings to provide extra lift at take-off.Another possibility is to yaw the rotor partly out of the wind to decrease power. Thetechnique of yaw control is, in practice, used only for tiny wind turbines (1 kW or less),there are, nontheless, examples of wind turbines of 1.5 MW using this technique.

2.3.2 Fixed-speed Turbine

The fixed-speed turbine typically uses a squirrel cage induction generator directly con-nected to the grid, as shown in Figure 2.5.

G

Turbine

Point of common

coupling

Capacitor battery

Generator

Figure 2.5: Electrical system of fixed-speed wind turbine using induction generator andcapacitor batteries to improve power factor system.

One reason for choosing an induction generator is that it is reliable, and tends to becomparatively inexpensive. One drawback is that the induction generator consumesreactive power and that the need for reactive power increases with the produced activepower. Capacitor batteries are used to compensate for the reactive power consumptionof the induction generator and thus receive a no load power factor near one at thepoint of common coupling (PCC) of the grid. The generator also has some mechanicalproperties that are useful for wind turbines, for example, the slip of the generator makesthe grid connection softer, i.e. the speed changes with the mechanical load torque. Therobustness also provides a certain overload capability.

Some manufacturers fit their turbines with two generators, a small one for periods oflow winds, and a large one for periods of high winds. Another design is pole changinggenerators, i.e. generators which (depending on how their stator windings are con-nected) may run with a different number of poles, and thus at different rotationalspeeds. This design is still considered as fixed-speed regarding mechanical loads andgrid interaction. Whether it is worth using double generators or a higher number ofpoles for low winds depends on the local wind speed distribution, and the extra cost ofthe pole changing generator seen in comparison with the revenues the turbine ownerearns on the electricity. A good reason for having a dual generator system, however,is that the turbine can run at a lower rotational speed at low wind speeds. This ismore aerodynamically efficient, another advantage is that the noise level from the rotor

11

blades decreases with lower rotational speed (the noise is usually only a problem atlow wind speeds due to low background noise).

2.3.3 Variable Speed Turbine

A variable-speed turbine uses power electronics apparatus to vary turbine speed andstill connect the generator to the fixed frequency of the grid. The primary advantage ofthis is that wind gusts can be allowed to make the rotor turn faster, thus storing partof the excess energy as rotational energy until the gust is over. Obviously, this requiresan intelligent control strategy, since the system has to be able to separate gusts fromhigh wind speed in general. Thus, it is possible to reduce the peak torque (reducingstress on the gearbox and generator), and also reduce the fatigue loads on the towerand rotor blades. The secondary advantage is that power electronics can control thereactive power in order to improve the power quality in the electrical grid. This maybe useful, particularly if a turbine is running on a weak electrical grid. Theoretically,variable speed also gives a slight advantage in terms of annual production, since it ispossible to run the machine at an optimal rotational speed, depending on the windspeed. From an economic point of view this advantage is minor, because the apparatusneeded to obtain variable speed have losses and cost more compared with a fixed-speedsystem.

One good reason to run a turbine partially at variable speed is the fact that pitch-control is a mechanical process. This means that the reaction time for the pitch mech-anism becomes a critical factor in turbine design. However, if the variable slip generatoris used, the slip is used as a control parameter. When a wind gust occurs, the controlmechanism signals to increase generator slip to allow the rotor to run a bit faster whilethe pitch mechanism begins to cope with the situation by pitching the blades more outof the wind. Once the pitch mechanism has done its work, the slip is decreased again.In case the wind suddenly drops, the process is applied in reverse. Thus, the mechan-ical pitch system controls turbine speed and the electrical system controls torque, i.e.the electrical output power.

Running a generator at high slip releases more heat from the generator, which conse-quently runs less efficiently. This is not a problem in itself, however, since the onlyalternative is to waste the excess wind energy by pitching the rotor blades out of thewind. One of the real benefits of using the control strategy mentioned here is betterpower quality. Fluctuations in power output are decreased by varying generator slipand storing or releasing part of the energy as rotational energy in the wind turbinerotor. Slip in an induction machine is usually very small for reasons of efficiency, so therotational speed varies by 1-2% between idle and full load. Slip, however, is a functionof the (DC) resistance (measured in ohms) in the rotor windings of the generator. Thehigher the resistance, the higher the slip.

One way of varying slip and, consequently, speed is to vary the resistance in the rotor,as shown in Figure 2.6. In this way, generator slip can be increased by, e.g. 10 %,whichgives a very limited speed range. This is usually done by having a wound rotor, i.e.a rotor with copper wire windings, which are connected in a star, and connected withexternal variable resistors, plus an electronic control system to operate the resistors.The connection is usually done with brushes and slip rings, which is a clear drawback

12

compared with the elegantly simple technical design of a cage wound rotor machine.The connection also introduces parts that wear down in the generator, thus requringmore maintenance on the generator. One way to avoid the problem of introducing sliprings, brushes, and maintenance altogether is by mounting the external resistors andthe electronic control system on the rotor.

G

Turbine

Point of common

coupling

Rotor resistors

Generator

Figure 2.6: Variable rotor resistance variable speed system.

Another way of creating variable speed is the rotor cascade technique, as shown inFigure 2.7. This technique uses a wound rotor, brushes and slip rings, but the rotorwindings are connected to an ac converter with variable frequency. The rotationalspeed is proportional to the frequency difference between the stator (grid) frequencyand the rotor (converter) frequency. The speed range for a rotor cascade turbine isproportional to the converter size. If the converter size is 25% of rated power for thewind turbine generator, then the speed range is ±25%, i.e. between 50-100% of nominalspeed.

=~~ =

~~

G

Turbine

Point of common

coupling

Rotor inverter

Generator

Rotor rectifier

Figure 2.7: Rotor cascade variable speed system.

By connecting the generator to the grid via a rectifier in series with an inverter, asshown in Figure 2.8, the rotational speed of the turbine can be controlled independentof the grid frequency. Another advantage of a full size converter, also called a back-to-back converter, is that power electronics can control the reactive power and use activefiltering techniques to improve the power quality in the electrical grid. The speed rangeis 0-100% of nominal speed, since the converter can handle rated power.

13

G =~~ =

~~

Turbine

Point of common

coupling

Inverter

Generator

Rectifier

Figure 2.8: Full size converter variable speed system.

2.4 Grid Interaction

Wind turbines interact with the grid. Due to the quickly increased rated power of windturbines, the turbines affect the grid and the consumers, who are connected to the grid[44].

Flicker, i.e. short lived voltage variations in the electrical grid may cause light bulbs toflicker. This phenomenon may be relevant if a wind turbine is connected to a weak grid,since short-lived wind variations will cause variations in power output and, therefore,voltage variations. There are various ways of dealing with this issue in the design ofthe turbine, mechanically, electrically, and by using power electronics [45].

Power electronics may introduce harmonic distortion into the alternating current inthe electrical grid, thus, reducing power quality. The problem of harmonic distortionarises because the filtering process is not perfect, and it may leave some tones, whichare multiples of the grid frequency, in the output current.

If a turbine is connected to a weak electrical grid, (i.e. if it is very far away in a remotecorner of the electrical grid with a low power-carrying ability), there may be problemsof the sort mentioned above. In such cases, it may be necessary to reinforce the grid,in order to carry the fluctuating current from the wind turbine. The problems are,in fact, the same when connecting a large electricity consumer, (e.g. a factory withlarge electrical motors) to the grid. Some of the problems with a weak grid can besolved with power electronics, which can control the reactive power consumption and,consequently, the voltage in the connection point.

14

Chapter 3

Wind Farm Configurations

In this chapter, wind farms are introduced for both onshore and offshore locations. Fourexamples of complete systems together with the transmission are included. Moreover,the cost and losses of each system are estimated.

3.1 Wind Farms

To increase the electricity produced and to keep the used land area at a minimum,wind turbines should be put together in a group, a so-called wind farm or wind park.

A wind turbine always casts a wind shade in the downwind direction. In fact, there isa wake behind the turbine, i.e. a long trail of wind that is quite turbulent and sloweddown, in comparison with the wind arriving in front of the turbine. As a rule of thumb,turbines in wind parks are usually spaced somewhere between 5 and 9 rotor diametersapart in the prevailing wind direction, and between 3 and 5 diameters apart in thedirection perpendicular to the prevailing winds [46]. Typically, the energy loss due towind turbines shading one another is somewhere around 5% [46].

3.2 Offshore Wind Farms

At sea, periods of complete calm are generally extremely rare, and quite short-lived.Thus, the effective use of wind turbine generating capacity is higher at sea than onland. Wind resources above shallow waters (5 to 15 m depth) in the seas around Europecould theoretically supply all of Europe’s electricity several times over [47].

One of the primary reasons for moving wind farm development offshore is the lack ofsuitable wind turbine sites on land. Equally important, however, is the fact that windspeeds are often significantly higher offshore than onshore. An increase of some 20%at some distance from the shore is not uncommon. Given the fact that the energycontent of wind increases with the cube of wind speed, the energy yield may be some73% higher than on land. Economically optimized turbines, subsequently, will probablyyield some 50% more energy at sea than at nearby land locations.

Another argument in favor of offshore wind power is the generally smooth surface of

15

water. Thus, wind speeds do not increase as much with the height above sea level asthey do on land. This implies that it may be economic to use lower (and thus cheaper)towers for wind turbines located offshore.

The temperature difference between the sea surface and the air above is far less thanthe corresponding difference on land, particularly during the daytime. Thus, the windis less turbulent at sea than on land. This, in turn, results in lower mechanical fatigueloads and, thus, a longer lifetime for turbines located at sea.

Inside the large 120-150 MW offshore wind parks being planned in Denmark, as shownin Figure 3.1, it is likely that 30-33 kV AC grid will be used. At the center of eachpark there will probably be a platform with a 30 to 150 kV transformer station, andpossibly a number of service facilities. The connection to land will consist of 150 kV ACcables [48].

Local wind

farm grid

33 kV AC

Platform with

transformer and

service facilities

Cable transmission

to shore

150 kV AC

G

Wind Turbine

G

G

Figure 3.1: Traditional AC system for offshore wind farms planned in Denmark.

Undersea cables have to be buried in order to reduce the risk of damage due to fishingequipment, anchors, etc. If bottom conditions permit, it is most economic to washcables into the seabed (using high-pressure water jets) rather than digging or ploughingcables into the bottom of the sea. Cables have a high electrical capacitance, whichmay cause problems depending on the precise grid configuration and the distance tothe shore. If the distance to the transmission grid on land is substantial, i.e. greaterthan aproximately 100 km [49], an interesting alternative can be to connect the farmsto the land using high voltage direct current connections (HVDC).

3.3 Connecting Offshore Wind Farms

The pilot offshore wind farms that are being built today use an AC connection toshore and the distance to shore is quite moderate, about 5-50 km [50, 51]. This sectiondescribes three different solutions for a DC connection to shore and compares the costand losses of the DC systems with a traditional AC system as a reference.

16

However, the proper types of electrical systems for offshore wind farms have not beenestablished yet. A DC cable is the preferred choice when connecting large offshorewind farms to the grid for several reasons:

• The capacitive charging current for a long, high voltage AC cable is substantialand, thus, longer distances cannot be covered by AC cables;

• Losses in a DC system are less sensitive to variations in the transmission dis-tance and the cables can, therefore, be more effectively connected directly to thetransmission grid on land;

• Sanctions for new over-head lines are difficult to achieve when making the neces-sary reinforcements for connecting offshore wind farms with the transmission gridon land. A special transmission system design is, therefore, needed to compensatefor the weak grid (off course the same for AC cables);

• The cost of a DC system is presently higher than for traditional AC systems.However, a major part of the cost of a DC system is power converters and theircost pertains mainly to the semiconductors which they contain;

• Due to the extensive development of semiconductors, power density is increasingrapidly and cost is decreasing, similar to the cost/performance of microprocessors;

• Moreover, losses in the wind farm are higher due to the converter stations butare more or less compensated by lower transmission losses.

A theoretical analysis is made of different configurations for offshore wind farms withrespect to the system cost, the total losses in the system and the impact on the grid it isconnected to. Three different DC systems, from the wind turbines to the transmissiongrid, are compared with a traditional AC system, which is used as a reference system.

3.4 System Description

The system adopted consists of wind turbines, the local grid and the transmission andconnection to the transmission grid on the mainland. The proposed and investigatedwind farm consists of 100 wind turbines each with a rated output of 1.5 MW. These areconnected together electrically to a local AC or DC grid. This grid is then connectedto the transmission cables via a converter or a transformer. The AC/DC converter is avoltage source converter and the DC/DC converter is a half bridge converter, and bothuse IGBT transistors as switches. The transmission length is the total length of theoffshore and onshore distances to the transmission grid. A transformer or a convertermakes the final connection to the transmission grid onshore, as shown in Figure 3.2.

The type of turbine or the type of generator that is used is not of interest here. Variablespeed can of course be used in all layouts, but is not specifically analyzed here. However,when using individual AC to DC conversion at each wind turbine, variable speed can beobtained and each wind turbine can reach optimal production and reduce mechanicalstresses and, thus, maximize the lifetime of the turbine.

17

Offshore Distance Onshore Distance

Distribution Line

3 km

Sea cable Land cable

10

km

Transmission

Line

Transmission cable

Figure 3.2: Layout of offshore wind farm and connection to transmission grid on land.

The four rows of wind turbines are separated by nine times the turbine diameter inthe prevailing wind direction. The turbines in each row are separated by five timesthe turbine diameter transverse to the prevailing wind direction. The total size of thewind farm is ten times three kilometers. The total transmission distance is set at 20km and consists of both sea and land cables.

The communities around the proposed wind farm sites in Denmark desire longer off-shore distances and after approximately 20 - 25 km the turbines become nearly invisi-ble [48]. The wind farms, which are being planned, have in some cases longer offshoretransmission distances, as shown in Table 3.1. Moreover, the onshore transmissiondistance might also be substantial, because the connection to the onshore transmissionmust be at a point with high short-circuit capacity. One example is the planned windfarm at Horns rev in Denmark where the onshore distance is 34 km [48].

18

Table 3.1: Offshore windfarm distances from shore [50, 51].

Wind Farm Size [MW] Offshore Distance [km]

Mecklenburg-Vorpommern - D 50 5SKY 2000 - D 100 15Egmond aan Zee - NL 100 8-20Horns rev - DK 150 20Læsø - DK 300 40Borkum West - D 60 45Rødsand - DK 150 25

3.5 Wind Farm Layouts

Four different layouts are assumed and compared. The first is a traditional AC system,the second is an AC system with a DC transmission (AC/DC) and the last two havedifferent systems with a DC grid and a DC transmission (DC1 and DC2). The ACsystem with a DC transmission can reduce the frequency of the offshore grid and,thereby, has the capacity for variable speed and can produce more energy from thewind. Of course, this assumption holds true if the wind speed is the same for thewhole wind farm. The mechanical stresses of the wind turbines will, however, bethe same since the turbines are connected together. The two DC examples will haveindividual variable speed and, therefore, will be able to reduce the mechanical stresseson the turbine, as well as gain more energy. These advantages can of course be achievedin the two first examples by having a rotor cascade with a back-to-back voltage sourceconverter system or any other variable speed system, but this is not included in theestimations.

19

3.5.1 AC Wind Farm

Each wind turbine in the AC wind farm has an output voltage of 33 kV AC and eachwind turbine is connected with an AC-grid to the main transformer, 33/150 kV AC,which is located in the center of the wind farm, as shown schematically in Figure 3.3.Figure 3.4 shows the cable layout of the wind farm. The transmission from the windfarm to the connection point on the transmission line onshore is made by three 20 km150 kV AC cables. The grid inside the wind farm consists of 49 km of 33 kV ACcables. Each wind turbine has a 33 kV transformer and there is one 33/150 kV ACtransformer. There is also an inductor onshore at the connection point for reactivepower compensation, so that the power factor becomes one.

Local wind

farm grid

33 kV AC

Platform with

transformer and

service facilities

Cable transmission

150 kV AC

Wind Turbine Shore

Connection point

on land

Reactive power

compensation

G

G

G

Figure 3.3: Layout of AC wind farm consisting of wind turbines, local wind farm ACgrid, platform with transformer, AC transmission cables and reactive powercompensation at connection point on land.

20

Prevailing Wind Direction

33/150 kV Transformer

33 kV AC Cable

150 kV AC Cable

Wind turbine

Figure 3.4: Cable layout of AC wind farm.

21

3.5.2 AC/DC Wind Farm

On the AC/DC wind farm, each wind turbine has an output voltage of 33 kV AC andeach wind turbine is connected with an AC-grid to the main converter, 33/150 kV AC/DC,which is located in the center of the wind farm, as shown in Figure 3.5. Figure 3.6shows the cable layout of the wind farm. The transmission from the wind farm to theconnection point on the transmission line onshore is made by two 20 km ±75 kV DCcable. The grid inside the wind farm consists of 49 km of 33 kV AC cables. Each windturbine has a 33 kV transformer and there is one DC/AC converter onshore at theconnection point.

Local wind

farm grid

33 kV AC

Platform with

AC/DC converter

and service facilities

Cable transmission

2 x 75 kV DC

Wind Turbine Shore

Connection point

on land

DC/AC converter

on land

G

G

G

=

~~

=

~~

Figure 3.5: Layout of AC/DC wind farm consisting of wind turbines, local wind farmAC grid, platform with AC/DC converter, DC transmission cables andDC/AC converter at connection point on land.

22

150 kV HVDC Converter

33 kV AC Cable

150 kV DC Cable

Wind turbine


Figure 3.6: Cable layout of AC/DC wind farm.

23

3.5.3 DC1 Wind Farm

Each wind turbine on the DC1 wind farm has an output voltage of 15 kV DC and fiveturbines are connected with a 15 kV DC sub grid to a 15/150 kV DC/DC converter.All 15/150 kV DC/DC converters are connected in a main grid, as shown in Figure 3.7.The transmission from the wind farm to the connection point on the transmission lineonshore is done with two 20 km ±75 kV DC cables. Figure 3.8 displays the cable layoutof the wind farm. The sub grid consists of 39 km of 2x7.5 kV DC cables and the maingrid consists of 19 km of 2x75 kV DC cables. Each wind turbine has a 15 kV AC/DCconverter and there is a total of 20 15/150 kV DC/DC converters. There is also aDC/AC converter onshore at the connection point.

Local wind

farm sub grid

15 kV DC

15/150 kV DC/DC

converter and

service facilities

Cable transmission

2 x 75 kV DC

Wind Turbine Shore

Connection point

on land

DC/AC converter

on land

=

~~

==G =

~~

G =~~

G =~~

Local wind

farm main grid

150 kV DC

Figure 3.7: Layout of DC1 wind farm consisting of wind turbines, local wind farm DCsub grid, DC/DC converter, DC transmission cables and DC/AC converterat connection point on land.

24

15/150 kV DC/DC Converter

15 kV DC Cable

150 kV DC Cable

Wind turbine


Figure 3.8: Cable layout of DC1 wind farm.

25

3.5.4 DC2 Wind Farm

Each wind turbine on the DC2 wind farm has an output voltage of 6 kV DC andthey are connected with a 6 kV DC sub grid to a 6/30 kV DC/DC converter. All6/30 kV DC/DC converters are connected with another sub grid to a 30/150 kV DC/DCconverter. All 30/150 kV DC/DC converters are connected in a main grid and thetransmission from the wind farm to the connection point on the transmission grid isdone with two 20 km ±75 kV DC cables. Figure 3.10 shows the cable layout of thewind farm. The first sub grid consists of 39 km of 2x3 kV DC cables, the second subgrid consists of 13 km of 2x15 kV DC cables and the main grid consists of 8 km of2x75 kV DC cables. Each wind turbine has a 6 kV AC/DC converter and there area total of 20 6/30 kV DC/DC converters and four 30/150 kV DC/DC converters, asshown in Figure 3.9. There is also a DC/AC converter onshore at the connection point.

Local wind

farm sub grid

6 kV DC

30/150 kV DC/DC

converter

Cable transmission

2 x 75 kV DC

Wind Turbine Shore

Connection point

on land

DC/AC converter

on land

=

~~

==G =

~~

G =~~

G =~~

Local wind

farm sub grid

30 kV DC

==

6/30 kV DC/DC

converter

Local wind farm

main grid

150 kV DC

Figure 3.9: Layout of DC2 wind farm consisting of wind turbines, local wind farmDC sub grids, DC/DC converters, DC transmission cables and DC/ACconverter at connection point on land.

26


6 kV DC Cable

150 kV DC Cable

Wind turbine

30 kV DC Cable



Figure 3.10: Cable layout of DC2 wind farm.

27

3.6 Cost and Loss Estimation

The following cost analysis only includes the cost of the main components like thecables, converters, transformers, and reactive power compensation for the four differentcases shown in the figures in the previous section. The costs for erection, foundations,platforms and the cost for ploughing down the cables, etc. are not included. TheDC/AC and the AC/DC converters are traditional voltage source converters with threephase legs. The DC/DC converter has a half bridge converter topology that uses onephase leg. The data for the estimation are shown in Table 3.2. The transistor costand losses given below are for two series connected switches, which become one phaseleg. A doubling of the current for the transistors is assumed to give a 20% higher costsince the cost depends on both the silicone cost and the cost of the series connectionrequired to reach a suitable voltage level.

The power semiconductors are, next to the wind turbines, the main cost of the systemsusing DC transmission. The connection cost of a traditional AC system is approxi-mately five percent of the total cost, while the connection cost of the other three typesis approximately 25%, as shown in Table 3.3 (The total cost differs depending on con-figuration). The cost of the semiconductors is nearly 90% of the total connection costfor the AC/DC alternative and approximately 40% of the costs for the two alterna-tives with a DC grid. Consequently, the total cost of the transmission system mainlydepends on the semiconductors in the converters.

The system losses are higher in the DC alternatives and, again, it is the semiconductorsthat cause the main losses. The losses are approximately 60 % higher in the AC/DCsolution and more than double in the DC solutions compared with the AC wind farm.The semiconductors represent approximately 50 % of the total losses. The differencebetween the two DC solutions is that the second (DC2) has two conversion steps.

Findings are not in favor of DC solutions. However, this example uses a very moderatetransmission distance of 20 km and, compared with grid reinforcement onshore, DCtransmission might be the most cost-effective solution.

28

Table 3.2: Data for cost and loss estimation.

Part Parameter Value comment

Wind Size 1.5 MWturbine Price 7.5 Mkr

Diameter D=80 mGenerator voltage 690 V

Wind Number of turbines 100park Total Size 150 MW

Spacing ⊥ 9xD to the prevailing‖ 5xD wind direction

Distance - onshore 10 km- offshore 10 km

Transmission voltage 150 kV (HVAC)±75 kV (HVDC)

Converter Transformer 60 kr/kVA (50 Hz for AC)components 6 kr/kVA (1000 Hz for DC/DC)

IGBTs 260 kr/kVA High voltage130 kr/kVA Low voltage

Diodes 130 kr/kVA High voltage65 kr/kVA Low voltage

Capacitors 1500 kr/kJSwitching frequency 1000 HzMaximum ripple 5 % Input or output voltage ripple

Reactive Inductor 220 kr/kVA Compensating for thepower reactive power incompensation the AC-cable at no load

Cables AC-cable Price=(A· area + B · voltage + C) kr/mA=0.149 kr/mm2

B=1.48 kr/kVC=-19 kr

DC-cable Price=(A· area + B · voltage + C) kr/mA=0.201kr/mm2

B=0.60kr/kVC=20kr

Losses Transformer 1% (of Rated power)(estimatesd Transistors 0.7% (of Rated power)values) Diodes 0.7% (of Rated power)

AC-cable 30 W/m Determines the cable areaDC-cable 30 W/m Determines the cable areaCable resistance 2.65·10−8 Ω ·m Aluminium conductor

29

Table 3.3: Costs and loss estimation.

Today AC AC/DC DC1 DC2

Connection costTotal cost

[%] 5.3 26.4 24.7 27.9

Cost of semiconductorsConnection cost [%] 0.0 87.1 42.9 48.4

Losses at Rated PowerRated Power

[%] 5.2 8.7 11.1 14.2

Semiconductor lossesTotal losses

[%] 0.0 45.8 53.9 46.9

3.7 Future Development

Power semiconductors have undergone rapid development in recent years and it is likelythat this development will continue in the future [52]. Table 3.4 shows the expectedcosts and losses for the next ten years assuming that the price is constant and thatthe performance regarding power capability is doubled and losses are halved everythree years. Under these circumstances, the connection cost of the wind farm wouldbe approximately the same for all alternatives and the losses would also be the sameexcept for the DC2 alternative with two converter steps. Table 3.5 shows when theother alternatives are expected to perform as well as the reference AC system, i.e.the price and the losses for the semiconductors are lowered until they match the ACsystem. In 11 years, the cost for the DC1 system will be the same as for the referencesystem. For the AC/DC system, the cost will be the same after 15 years. The losses,however, will be comparable after 10 years for the AC/DC system but not until aftermore than twenty years for the DC1 system. Again, this largely depends on the shorttransmission distance of 20 km.

Table 3.4: Estimation of costs and losses year 2010.

Year 2010 AC AC/DC DC1 DC2

Connection costTotal cost

[%] 5.3 7.2 5.7 7.2

Cost of semiconductorsConnection cost [%] 0.0 40.1 23.1 23.8

Losses at Rated PowerRated Power

[%] 5.2 5.1 5.7 7.0

Semiconductor lossesTotal losses

[%] 0.0 7.7 10.4 9.5

Table 3.5: Estimation when in time DC solutions give the same performance as AC.

Years from now AC AC/DC DC1 DC2

Total cost [Year] 0 15 11 14Losses [Year] 0 10 23 - -

30

3.8 Conclusions

The pilot offshore wind farms that are being built today use an AC connection to theshore and the distances to the shore are also quite moderate. This is a well-knowntechnology and a natural step for the first offshore wind farms. However, in ten years,when the experiences from the first offshore wind farms have been evaluated and thenext expansion starts, alternatives using a DC grid will start to become cost-effectivesolutions. Future wind farms will become larger and will be situated at a fartherdistance from shore. The connection to the onshore transmission grid must be at a pointwith high short circuit capacity and the transmission distance is, therefore, expectedto be greater in the future than the 20 km used in the example. The controllabilityoffered with the three systems using power electronics is also a great advantage fortransient stability when connecting the wind farm to a weak grid.

The cost of power electronics will be significantly lower in the future and losses willcontinue to decrease, so the advantages will outweigh the disadvantages.

31

Chapter 4

Hard Switching DC/DC ConverterTopologies

This chapter, describes the working principles for hard switching DC/DC convertertopologies, particularly as regards transformer and semiconductor utilization. Theanalysis is theoretical and the overview is made partially by using material from refer-ences [7, 8, 9, 5]. The application for the DC/DC converters is in a wind farm. In orderto reduce the transmission losses from the wind farm to the shore, the transmissionvoltage has to be higher than the voltage out from the generator [53].

Voltage and current stresses on the semiconductors for the different converter topologiesare described and compared in steady state and from a wind power point of view.Series connected IGBT transistors are used as switches to achieve the necessary voltageratings. No resonance is used to decrease the switching losses in the valves. Theconverter topologies can be divided into different classes depending on their powerhandling capability, from low to high power applications or their conversion ratios,from moderate to high. Typically, moderate voltage changes are in the range of one tofive. For high voltage changes, where the converters use a transformer in the conversion,the ratio is ten or more. Two other aspects are the size of passive components and theneed for bi-directional power flow.

The characteristics of the wind turbine generator can be seen in Figure 4.1. The cut-inwind speed is 3 m/s and the output power is proportional to the third power of thewind speed up to full power at 10 m/s. Above 10 m/s excess wind is wasted, forexample, by pitching the blades. The generator has a minimum speed of 0.5 pu fromthe cut-in wind speed up to 5 m/s. The generator voltage is controlled for a constantflux in the generator from the wind speed of 5 m/s and upwards.

33

0 5 10 15 20 25

0

0.2

0.4

0.6

0.8

1

Wind [m/s]

Pow

er [

p.u.

]

Output power

0 5 10 15 20 25

0

0.2

0.4

0.6

0.8

1

Wind [m/s]

N [

p.u.

]

Generator speed

0 5 10 15 20 25

0

0.2

0.4

0.6

0.8

1

Wind [m/s]

Uin

[p.

u.]

Generator voltage

0 5 10 15 20 25

0

0.2

0.4

0.6

0.8

1

Wind [m/s]

Iin

[p.u

.]

Generator current

Figure 4.1: Output power, generator speed, generator voltage and generator currentfrom wind turbine as functions of wind speed.

34

4.1 Converter Environment

DC/DC converters are placed in a generation system where the voltage out from thegenerator is low. The converters between the generator and the transmission lineincrease the voltage to reduce transmission losses. The output voltages of the convertersare held constant by the next step in the transmission system by regulating the currentdrawn from the previous converters. A schematic layout of the wind farm is shown inFigure 4.2.

AC/DC

Converter Cable to

shore

G

DC/DC

Converter

Wind Turbine

with Generator

Low Voltage

AC from

Generator

DC/DC

Converter

Medium Voltage

DC Bus

Low Voltage

DC Bus

High Voltage

DC Bus

=~~

==

==

Direction of Power Flow

Figure 4.2: Layout of transmission system where generation is included.

The first converter in the chain is exposed to the varying input voltage of the gener-ator. A diode bridge is assumed for rectification and, therefore, a gearbox should beconnected between the turbine and the generator. Direct-drive generators normallyhave a leakage inductance that is too high to be suitable for a diode rectifier [54].

The difference between the DC transmission voltage and the voltage of the wind turbineis generally too high, maybe 100 times, to use a single DC/DC converter. Therefore, anumber of DC/DC converters are needed to convert the voltage in a number of steps.The resulting layout is very similar to the usual radial structure of AC distributionpower systems, with the DC/DC converters replacing the transformers. Normally,690 V is used as the generation voltage and, therefore, two or three steps are needed.The use of high voltage generators, with a generation voltage of ten kilovolts or more,is one possible method to reduce the number of converters. However, one problem forsome of the converters topologies is voltage reduction at low wind speeds.

4.2 DC/DC Converter Topologies

The different DC/DC converter topologies, which will be analysed, can be divided intotwo different groups: non-galvanic isolated topologies and galvanic isolated topolo-gies [5]. The difference between them is the isolating transformer that is used in thegalvanic isolated topologies. The different topologies are listed in Table 4.1. Each ofthe listed DC/DC-converters will be analysed below.

35

Table 4.1: Different investigated topologies of DC/DC-converters.

Non galvanic Galvanic isolatedisolated topologies topologies

Boost FlybackCuk ForwardSepic Two transistor forwardZeta Push pullLuo Half bridge

Full bridgeHalf bridge with voltage doubler

4.2.1 Boost Converter

The boost or the step-up converter is common when no galvanic insulation is neededand for moderate input/output voltage ratios. The converter is very simple and hasfew components, as shown in Figure 4.3. However, it can only be used when the outputvoltage Uout is higher than the input voltage Uin [5].

Iin

Iout

+

- -

+

Uin

Uout

+

-

Usw

ID

- +U

D

IS

W

+ -UL

ILL D

Sw Cout

Figure 4.3: Schematic layout of boost converter.

The boost converter works cyclically by storing energy in the inductor L when theswitch Sw is on and dumps the stored energy together with energy from the input intothe load when the switch Sw is off. The output voltage Uout is controlled and regulatedby varying the amount of energy stored and dumped each cycle. When the switch ison, the supply voltage is applied across the inductor L, and the current through theinductor increases linearly. During the on state, the capacitor C supplies the loadwith energy and, thus, the voltage across the capacitor is reduced. When the switch isturned off, the current continues through L, supplying the load via the diode D. Thus,the current decreases linearly. If the current through the inductor reaches zero beforethe next switching cycle starts, the converter works in the discontinuous conductionmode (DCM). Moreover, if the current through the inductor does not reach zero be-fore the next switching cycle starts, the converter works in a continuous conductionmode (CCM as shown in Figure 4.4). The control system has to use separate controlschemes for DCM and CCM.

In CCM, the relation between the input voltage Uin and the output voltage Uout canbe expressed as

36

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Inductor voltage and current

Time [ms]

Un1

(so

lid),

In1

(do

tted)

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Switch voltage and current

Time [ms]

Usw

(so

lid),

Isw

(do

tted)

Figure 4.4: Idealised current and voltage waveforms for inductor and switch stresses inCCM for boost converter (duty ratio D=0.8), Left: Inductor voltage UL

(line) and inductor current IL (dotted), Right: Switch voltage USW (line)and switch current ISW (dotted).

Uout

Uin

=1

1−D(4.1)

where D is the duty ratio and is defined as

D =tonTs

(4.2)

where ton is the conduction time for the switch during one switching cycle Ts.

In DCM, the relation between the input and output voltage depends on the currentand the duty ratio. The ratio between the output voltage and the input voltage canbe expressed with the duty ratio D and the relative time ∆1 as expressed in Eq. (4.3).During the relative time ∆1 , which is defined as ∆t

Ts, the switch is off and the current

in the inductor is non-zero.

Uout

Uin

=∆1 +D

∆1(4.3)

Semiconductor Stresses in CCM

In this section, the stresses on the semiconductors in CCM are analyzed. The maximumvoltage over the switch is written as

uSw = Uout (4.4)

and the maximum current through the switch becomes

37

iSw =Iout

1−D+

1

2∆IL (4.5)

where ∆IL is the peak-to-peak ripple current in the inductor L. The RMS currentthrough the switch is

ISw,RMS =

√

D((Iout

1−D)2 +

1

3(1

2∆IL)2) (4.6)

The maximum voltage over the diode is

uD = Uout (4.7)

and the maximum current through the diode can be obtained as

iSw =Iout

1−D+

1

2∆IL (4.8)

The RMS current through the diode is

ID,RMS =

√

(1−D)((Iout

1−D)2 +

1

3(1

2∆IL)2) (4.9)

Main Advantages and Drawbacks with Boost Converter

The main advantages of the boost converter are:

+ The converter is very simple with only two active components, the switch Swand the diode D, and two passive components, the inductor L and the capacitorCout;

+ Low input ripple in CCM i.e. Iout ≥ UoutD2

2Lfswwhere fsw is the switching frequency

of the converter.

The main drawbacks of the boost converter are:

- Different parasitic resistances in the switches and the inductor limit the dutycycle to approximately 0.8-0.9 and this limits the maximum boost ratio to ap-proximately 10;

- A large switch current when the duty cycle is high;

38

- Poor transient response due to the fact that stored energy in the inductor has tobe transferred to the output even if the load is removed;

- Regulator loop hard to stabilize since the gain is load dependent;

- No current limiting of output short circuits;

- The power can only flow in one direction due to the diode D.

Prototypes have been built up to approximately 100 kW with efficiency up to approx-imately 95 %.

4.2.2 Cuk Converter

The cuk converter, named after its inventor, is used when a small current ripple isneeded and for moderate voltage changes. The converter is very simple and has fewcomponents, as shown in Figure 4.5. The output is inverted and the converter canbe used when the input voltage is lower or higher than the output voltage. Galvanicinsulation can be achieved by splitting capacitor C1 into two capacitors and applyinga transformer between the two capacitors, as shown in Figure 4.6 [5].

Uin

Iin

Iout

+

-

-

+

Uout

+

-

Usw

ID

+

-

UD

IS

W

IC1

+ -

UC1

+-U

L2

+ -UL1

IL2

IL1

DSw Cout

C1

L1

L2

Figure 4.5: Schematic layout of cuk converter.

Iin

Iout

+

- -

+

Uin

Uout

+

-

Usw

ID

+

-

UD

IS

W

IC1a

+ -

UC1a

+ -U

L2

+ -UL1

IL2

IL1

IC1b

+ -

UC1b

n1

n2Sw D C

out

C1a

L1

L2C

1b

Figure 4.6: Schematic layout of cuk converter with galvanic insulation.

The cuk converter displayed in Figure 4.5, works cyclically by shifting energy betweenthe inductors L1 and L2, the capacitor C1 and the load. Unlike the previous converter,where the energy transfer is associated with the inductor, the energy transfer for theCuk converter depends on the capacitor C1.

39

When the switch Sw is on, the inductor L1 is charged from the supply and the inductorL2 is charged from the capacitor C1. In the next mode, when the switch Sw is off,the inductor L1 charges the capacitor C1 and the inductor L2 supplies the load. Theoutput is inverted in comparison with the input.

When the switch is on, the supply voltage is applied across the inductor L1 and thecurrent in L1 increases linearly. At the same time, the capacitor C1 increases thecurrent in L2 through the switch and supplies the load. When the switch is turnedoff, the current in L1 charges C1 through the diode and, thus, decreases. The currentin L2 continues to supply the load through the diode and decreases. The converternormally operates in CCM, as shown in Figure 4.7, to achieve low input and outputripple currents.

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Inductor voltage and current

Time [ms]

Ul1

(so

lid),

Il1

(do

tted)

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1


Time [ms]

Usw

(so

lid),

Isw

(do

tted)

Figure 4.7: Idealised voltage and current waveforms for inductor L1 and switch stressesin CCM for cuk converter (duty ratio D=0.8), Left: Inductor voltage UL1

(line) and inductor current IL1 (dotted), Right: Switch voltage USW (line)and switch current ISW (dotted).


Uout

Uin

= − D

1−D(4.10)

where D is the duty ratio.



usw =Uout

D(4.11)

and the maximum current through the switch becomes

40

isw =Iout

1−D+

1

2∆IL1

+1

2∆IL2

(4.12)

where ∆IL1and ∆IL2

is the peak-to-peak ripple current in the inductor L1 and L2,respectively. The RMS current through the switch is

Isw,RMS =

√

D((Iout

1−D)2 +

1

3(1

2∆IL1

+1

2∆IL2

)2) (4.13)


uD =Uout

D(4.14)

and the maximum current through the diode is obtained by

iD =Iout

1−D+

1

2∆IL1

+1

2∆IL2

(4.15)


ID,RMS =

√

(1−D)((Iout

1−D)2 +

1

3(1

2∆IL1

+1

2∆IL2

)2) (4.16)

Main Advantages and Drawbacks of Cuk Converter

The main advantages of the Cuk converter are:

+ The input and output ripples are small when operating in CCM, i.e., Iout ≥ Uout·D2

2L·fsw

where fsw is the switching frequency of the converter;

+ Galvanic isolation can be added to the basic converter by using a transformer.The capacitor C1 is divided into two parts and the transformer is inserted inbetween them. The capacitors in series with each winding remove any DC voltageand allow full utilization of the transformer;

+ Only two active components are used in the converter, the switch Sw and thediode D.

The main drawbacks of the Cuk converter are:

- The ripple current in the capacitor C1 is equal to the sum of the input Iin andoutput Iout currents;

41

- The switch must handle the sum of the input and the output currents;

- The energy can only flow in one direction.

Efficiency is up to approximately 94% and the converter is mainly used for low powerapplications.

4.2.3 Sepic - Single-Ended Primary Inductance Converter

The sepic converter works similar to the cuk converter but has positive output, asshown in Figure 4.8. The input current ripple is small and the converter is used formoderate voltage changes when the input voltage is lower or higher than the outputvoltage [55].

Iin

Iout

+

- -

+

Uin

Uout

+

-

Usw

IL2

+

-

UL2

IS

W

IC1

+ -U

C1

+ -

UD

+ -UL1

ID

IL1

D

Sw Cout

C1

L1

L2

Figure 4.8: Schematic layout of sepic converter.

The sepic converter works cyclically by shifting energy between the inductors, thecapacitor and the load. When the switch Sw is on, the inductor L1 is charged fromthe supply, the inductor L2 is charged from the capacitor C1 and Cout supplies theload. When the switch Sw is turned off, the inductor L1 charges the capacitor C1 andsupplies the load and the capacitor Cout together with inductor L2.

When the switch is on, the supply voltage is applied across the inductor L1 and thecurrent in L1 increases linearly. At the same time, the capacitor C1 increases thecurrent in L2 through the switch and the capacitor Cout supplies the load. When theswitch is turned off, the current in L1 and L2 supply the load and charge Cout. Thecurrent in L1 also charges C1.


Uout

Uin

=D

1−D(4.17)


42



uSw =Uout

D(4.18)

and the maximum current through the switch is equal to

iSw =Iout

1−D+

1

2∆IL1

+1

2∆IL2

(4.19)


is the peak-to-peak ripple current in the inductor L1 and L2,respectively. The RMS current through the switch is

Isw,RMS =

√

D((Iout

1−D)2 +

1

3(1

2∆IL1

+1

2∆IL2

)2) (4.20)

The maximum voltage over the diode is obtained as

uD1=

Uout

D(4.21)

and the maximum current through the diode becomes

iD1=

Iout1−D

+1

2∆IL1

+1

2∆IL2

(4.22)

The RMS current through the diode is equal to

ID1,RMS =

√

(1−D)((Iout

1−D)2 +

1

3(1

2∆IL1

+1

2∆IL2

)2) (4.23)

Main Advantages and Drawbacks of Sepic Converter

The main advantages of the Sepic converter are:

+ Only two active components are used in the converter, the Switch Sw and thediode D;

+ Low input ripple when operating in CCM, i.e., Iout ≥ Uout·D2

2L·fswwhere fsw is the

switching frequency of the converter.

43

The main drawbacks of the Sepic converter are:

- The switch and the diode maximum voltage and current is the sum of the inputand output voltage and current, respectively;

- Large output voltage ripple;

- The power can only flow in one direction.

Efficiency at different loads is up to 94% and the converter has, so far, been used forlow power applications.

4.2.4 Zeta Converter

The Zeta converter, sometimes called the inverted Sepic converter, works similar tothe cuk converter but has positive output, as shown in Figure 4.9. The output currentripple is small and the converter is used for moderate voltage changes when the inputvoltage is lower or higher than the output voltage [56].

Iin

Iout

+

- -

+

Uin

UOut

+

-

UL1

ID

+

-

UD

IL1

IC1

+ -U

C1

+ -UL2

+ -U

SW

IL2

ISW

DSw

Cout

C1

L1

L2

Figure 4.9: Schematic layout of Zeta converter.

The Zeta converter works cyclically by shifting energy between the inductors, thecapacitor and the load. First, the inductor L1 is charged from the input voltage, theinductor L2 is charged from the capacitors C1 and C1 together with L2, which alsosupplies the load. Second, the inductor L1 charges the capacitor C1 and the inductorL2 supplies the load.

When the switch Sw is on, the input voltage Uin is applied across the inductor L1 andthe current in the inductor L1 increases linearly. At the same time, the input voltageand the capacitor C1 increase the current through the inductor L2 and supply theload. Meanwhile, the diode D blocks. When the switch is turned off, the inductor L1charges the capacitor C1 through the diode D and the inductor L2 supplies the load,also through the diode D.


44

Uout

Uin

=D

1−D(4.24)



This section presents an analysis of the stresses on the semiconductors in CCM . Themaximum voltage over the switch is written as

uSw =Uout

D(4.25)


isw =Iout

1−D+

1

2∆IL1

+1

2∆IL2

(4.26)


is the peak-to-peak ripple current in the inductors L1 and L2,respectively. The RMS current through the switch is

Isw,RMS =

√

D((Iout

1−D)2 +

1

3(1

2∆IL1

+1

2∆IL2

)2) (4.27)


uD =Uout

D(4.28)

and the maximum current through the diode can be obtained as

iD =Iout

1−D+

1

2∆IL1

+1

2∆IL2

(4.29)


ID,RMS =

√

(1−D)((Iout

1−D)2 +

1

3(1

2∆IL1

+1

2∆IL2

)2) (4.30)

45

Main Advantages and Drawbacks of Zeta Converter

The main advantages of the boost converter are:

+ The inductor L2 and the capacitor C2 form a filter, which leads to low outputripple;

+ Only two active components are used in the converter, the switch Sw and thediode D.

The main drawbacks of the boost converter are:

- The ripple current in the capacitor C1 is equal to the sum of the input and outputcurrents;

- High semiconductor stresses;

- The switch is connected directly to the input and this causes a large input ripple;


4.2.5 Luo Converter

The Luo converter, named after its inventor [13], is derived from the Zeta converterand has positive output, as shown in Figure 4.10. The output current ripple is smalland the converter is used for moderate voltage changes when the input voltage is lowerthan the output voltage [13].

Iin

Iout

+

- -

+

Uin

Uout

+

-

UL1

ID

1

+

-

UD1

IL1

IC1

+ -U

C1

+ -UL2

+ -U

SW

IL2

ISW

IC

2

+

-

UC2

ID2

+ -U

D2

Sw

L2

L1

D2

D1

C1

C2

Cout

Figure 4.10: Schematic layout of Luo converter.

The Luo converter works cyclically by shifting energy between the inductors, the ca-pacitors and the load. First, the inductor L1 is charged from the supply, the inductorL2 and the capacitor C2 are charged from the capacitor C1. The capacitor C1 to-gether with the inductor L2 supplies the load. Second, the inductor L1 charges thecapacitor C1 and the inductor L2 together with the capacitor C2 supply the load.

When the switch Sw is on, the input voltage is applied across the inductor L1 and thecurrent in the inductor increases linearly. At the same time, the capacitor C1 increases

46

the current in the inductor L2, charges the capacitor C2 and supplies the load via thediode D2. When the switch is turned off, the inductor L1 charges the capacitor C1 viathe diode D1. The inductor L2 and the capacitor C2 supply the load.


Uout

Uin

=1

1−D(4.31)



This section presents an analysis of the stresses on the semiconductors in CCM . Themaximum voltage over the switch Sw is written as

uSw = Uout (4.32)

and the maximum current through the switch Sw is

iSw =Iout

1−D+

1

2∆IL1

+1

2∆IL2

+ δ(t) (4.33)


are the peak-to-peak ripple current in the inductors L1 and L2,respectively, and δ(t) is the inrush current for the capacitor C1 and the capacitor C2.The RMS current through the switch Sw is written as

ISw,RMS =

√

D((Iout

1−D)2 +

1

3(1

2∆IL1

+1

2∆IL2

)2) + ∆δ(D) (4.34)

where ∆δ(D) is the contribution to the RMS current from the inrush current δ(t). Themaximum voltage over the diodes becomes

uD1= uD2

= Uout (4.35)

and the maximum current through the diodes is obtained as

iD1≈√

C1L1

DUout +1

2∆IL1

(4.36)

47

when assuming that

∆IL1

IL1

=∆UC1

UC1

(4.37)

respective

iD2≈ Iout +

1

2∆IL2

+ δ(t) (4.38)

The RMS current through the diode D1 is

ID1,RMS =

√

(1−D)(C1L1

√

UoutD +1

3(1

2∆IL1

)2) (4.39)

and the RMS current through the diode D2 is equal to

ID2,RMS =

√

D(I2out +1

3(1

2∆IL2

)2) + ∆δ(D) (4.40)

Main Advantages and Drawbacks of Luo Converter

The main advantages of the Luo converter are:

+ The switch voltage is low but the switch current is high;

+ The inductor L2 and the capacitors C2 and Cout form a filter, which results inlow output ripple.

The main drawbacks of the Luo converter are:

- The ripple current in capacitor C1 is equal to the sum of the input and outputcurrents;

- The switch is connected directly to the input and this causes a large currentripple on the input;


Efficiency at different loads is up to 95 % and the converter is used for low powerapplications.

48

Iin

Iout

+

-

-

+

Uin

Uout

+

-

UswI

SW

+ -U

D2

ID2

+

-

UD1

ID

1

n1

n3

n2

Sw

D2

D1

Cout

Figure 4.11: Schematic layout of Flyback converter.

4.2.6 Flyback Converter

The Flyback converter is common when galvanic insulation is needed and for low powerapplications typically below 100 W. The converter is simple and has few components,as shown in Figure 4.11. The most complicated part of the converter is the transformer,which has a high magnetizing inductance [5].

The Flyback converter works cyclically by storing energy in the transformer and thendumping this stored energy into the load. The output voltage is controlled and regu-lated by varying the amount of energy stored and dumped each cycle.

When the switch Sw is on, the input voltage Uin is applied across the winding n1 andthe current in the transformer leakage inductance increaes linearly. The diodes D1 andD2 block and the capacitor supplies the load. When the switch Sw is turned off, theenergy in the transformer is discharged, via the winding n3 and the diode D2, to theload and the output capacitor Cout. The diode D1 and the demagnetising winding n2are only used under no-load conditions when the energy stored in the transformer istransferred back to the supply. The converter can be operated in CCM, as shown inFigure 4.12, or in DCM.


Uout

Uin

=n3n1

D

1−D(4.41)



This section presents an analysis of the stresses on the semiconductors in CCM . Themaximum voltage over the switch is written as

usw =n1n3

Uout

D(4.42)

49

0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

Time [ms]

Un1

(so

lid),

In1

(do

tted)

Transformer voltage and current

0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

Time [ms]

Usw

(so

lid),

Isw

(do

tted)


Figure 4.12: Idealised current and voltage waveforms for transformer and switchstresses in CCM for Flyback converter (duty ratio D=0.4), Left: Trans-former voltage Un1

(line) and transformer current In1(dotted), Right:

Switch voltage USW (line) and switch current ISW (dotted).


isw =n3n1

Iout1−D

+1

2∆ILm (4.43)

where ∆ILm is the peak-to-peak ripple current in the transformer leakage inductor Lm.The RMS current through the switch is

Isw,RMS =

√

D((n3n1

Iout1−D

)2 +1

12∆I2Lm) (4.44)

The maximum voltage over the diode D1 is equal to

uD1=

n1 + n2n3

1−D

DUout (4.45)

and the maximum voltage over the diode D2 becomes

uD2=

Uout

D(4.46)

The current through the diode D1 is normally zero but if the load is disconnected themaximum current can in the worse case become

iD1=

n3n2

1

1−DIout +

1

2

n1n2

∆ILm (4.47)

50

The maximum current through the diode D2 is equal to

iD2=

1

1−DIout +

1

2

n1n3

∆ILm (4.48)

The RMS current through the diode D1 is normally zero. For the diode D2, the RMScurrent becomes

ID2,RMS =

√

(1−D)((Iout

1−D)2 +

1

3(1

2

n1n3

∆ILm)2) (4.49)

Main Advantages and Drawbacks of Flyback Converter

The main advantages of the Flyback converter are:

+ The Flyback converter is, from a circuit perspective, the simplest of the low-powerisolated converters and uses only three components besides the transformer (ifthe no-load capability is omitted;

+ The switch current can be reduced by the turn’s ratio of the transformer but hasthe disadvantage of increasing the switch voltage and the diode current.

The main drawbacks of the Flyback converter are:

- The transformer design is critical and the high energy, which must be stored in thetransformer windings in the form of a DC current, requires a high inductance, ahigh current primary, which in turn requires larger cores than would be necessarywith pure AC in the windings inductance;

- Poor transformer utilization since the core is only magnetised in one direction;

- The energy is transferred when the switch is turned off and the current waveformis triangular causing a high output ripple;


4.2.7 Forward Converter

The Forward converter is very common when galvanic insulation is needed and forapplications above the power range of the Flyback converter. The converter is simpleand has few components, as shown in Figure 4.13. The most complicated part of theconverter is the transformer with three windings [5].

The Forward converter delivers the energy to the load when the transistor is on anduses the transformer in the active or forward mode. The additional winding n2 andthe diode D1 are required to reduce the voltage over the switch Sw during the turn-offtime of the switch when the core of the transformer is being demagnitized.

51

Iin

Iout

+

-

-

+

Uin

Uout

+

-

Usw

+ -U

D2

ID2

+

-

UD1

ID

1

n1

n3

n2

+

-

UD3

ID

3

+ -UL

IL

Sw

L

D1

D2

D3

IS

W

Figure 4.13: Schematic layout of Forward converter.

When the switch is on, the supply voltage is applied across the winding n1 and energyis transferred via the transformer and the diode D2, to the inductor L and the load.The diodes D1 and D3 blocks. When the switch is turned off, the inductor L suppliesthe load via the diode D3 and the energy in the leakage inductance of the transformeris transferred back to the supply, via the demagnetizing winding n2 and the diode D1.The converter can only run in discontinuous conduction mode, as shown in Figure 4.14,since the winding has to be demagnetized every cycle.

0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

Time [ms]

Un1

(so

lid),

In1

(do

tted)


0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

Time [ms]

Usw

(so

lid),

Isw

(do

tted)


Figure 4.14: Idealized current and voltage waveforms for the transformer and switchstresses in Forward Converter (duty ratio D=0.4), Left: Transformer volt-age Un1

(line) and transformer current In1(dotted), Right: Switch voltage

USW (line) and switch current ISW (dotted).

The relation between the input voltage Uin and the output voltage Uout can be expressedas

Uout

Uin

=n3n1

D (4.50)

52

Semiconductor Stresses

This section presents an analysis of the stresses on the semiconductors. The maximumvoltage over the switch is written as

usw =n1n3

Uout

D(1 +

n1n3

) (4.51)


isw =n3n1

(Iout +1

2∆IL) + iMag (4.52)

where ∆IL is the peak-to-peak ripple current in the inductor L and iMag is the peakmagnetizing current in the transformer. The RMS current through the switch can beexpressed as

Isw,RMS =

√

D((n3n1

Iout)2 +1

3(1

2

n3n1

∆IL + iMag)2) (4.53)

The maximum voltage over the diode D1 is written as

uD1=

n1 + n2n3

Uout

D, (4.54)

and the maximum voltage over the diode D2 is equal to

uD2=

n1n2

Uout

D(4.55)

and the diode D3 has the maximum voltage written as

uD3=

Uout

D(4.56)

The maximum current through the diode D1 becomes

iD1= iMag (4.57)

and the diode D2 has the maximum current as

53

iD2=

Iout1−D

+1

2∆IL (4.58)

and, finally, the diode D3 has the maximum current, written as

iD3=

IoutD

+1

2∆IL (4.59)

The RMS current through the diode D1 becomes

ID1,RMS =1√3iMag (4.60)

and the diode D2 has an RMS current equal to

ID2,RMS =

√

(1−D)((Iout

1−D)2 +

1

3(1

2∆IL)2) (4.61)

and the diode D3 has an RMS current equal to

ID3,RMS =

√

D((IoutD

)2 +1

3(1

2∆IL)2) (4.62)

Main Advantages and Drawbacks of Forward Converter

The main advantages of the forward converter are:

+ The Forward converter is simple but compared with the Flyback converter hasan extra winding on the transformer, two more diodes and an additional outputfilter inductor;

+ The output ripple is low because of the inductor L;

+ The switch current is reduced by the turn ratio of the transformer but this in-creases the duty cycle and decreases the dynamic range, i.e. the potential tochange the duty ratio when the load increases.

The main drawbacks of the Forward converter are:

- Poor transformer utilization since the core is only magnetized in one direction;

- The transformer design is critical to minimize the magnetizing current and, thus,the losses in the diode D1;

- The converter has a high input ripple due to a low duty cycle;


Converters have been built up to 15 kW at 100 kHz and efficiency is up to 97 %.

54

4.2.8 Two Two-Transistor Forward Converter

The Two-Transistor Forward converter can replace the Forward converter when theswitch voltage or the transformer design is critical. The converter has two transistors,as shown in Figure 4.15. The converter has more complicated drive circuits but has theadvantage that the switch voltage is clamped to the input voltage and no demagnetizingwinding is needed on the transformer [5].

Iin

Iout

+

-

-

+

Uin

Uout

+

-

Usw2

IS

W2

+ -U

D3

ID3

-U

D2

ID

2

n1

n2

+

-

UD4

ID

4

+ -UL

IL

+

-

Usw1

+

-

UD1

ID

1

+

IS

W1

D1

D2

Sw1

Sw2

D3

D4

L

Figure 4.15: Schematic layout of Two-Transistor Forward converter.

As with the Forward converter this converter delivers the energy to the load when thetransistor is on and uses the transformer in the active or forward mode. It uses twoswitches and two diodes but this allows the use of only two windings.

When the switches are on, energy is transferred via the transformer and the diode D2,to the inductor L and the load. The diodes D1, D2 and D4 block. When the switchesare turned off the inductor L supplies the load via the diode D4 and the energy in thetransformer leakage inductor is transferred back to the supply, via the diodes D1 andD2. The converter can only run in the discontinuous conduction mode, as shown inFigure 4.16, since the winding has to be demagnetized every cycle.


Uout

Uin

=n2n1

D (4.63)


55

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Time [ms]

Un1

(so

lid),

In1

(do

tted)


0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Time [ms]

Usw

(so

lid),

Isw

(do

tted)


Figure 4.16: Idealised current and voltage waveforms for transformer and switchstresses in Two-Transistor Forward converter (duty ratio D=0.4), Left:Transformer voltage Un1

(line) and transformer current In1(dotted),

Right: Switch voltage USW1(line) and switch current ISW1

(dotted).


This section presents an analysis of the stresses on the semiconductors. The maximumvoltage over the switches Sw1 and Sw2 is written as

uSw =n1n2

Uout

D= Uin (4.64)

and the maximum current through the switches Sw1 and Sw2 is equal to

iSw =n2n1

(Iout +1

2∆IL) + iMag (4.65)

where ∆IL is the peak-to-peak ripple current in the inductor L and iMag is the peakmagnetizing current in the transformer. The RMS current through the switches Sw1and Sw2 becomes

ISw,RMS =

√

D((n2n1

Iout)2 +1

3(1

2

n2n1

∆IL + iMag)2) (4.66)

The maximum voltage over the diodes D1 and D2 is written as

uD1= uD2

=n1n2

Uout

D= Uin (4.67)

and the diodes D3 and D4 have the maximum voltage

56

uD3= uD4

=Uout

D(4.68)

The maximum current through the diodes D1 and D2 becomes

iD1= iD2

= iMag, (4.69)

and the maximum current through the diode D3 is equal to

iD3=

Iout1−D

+1

2∆IL (4.70)

and the diode D4 has a maximum current of

iD4=

IoutD

+1

2∆IL (4.71)

The RMS current through the diodes D1 and D2 is equal to

ID1,RMS = ID2,RMS =1√3iMag, (4.72)

and the RMS current through the diodes D3 becomes

ID3,RMS =

√

(1−D)((Iout

1−D)2 +

1

3(1

2∆IL)2) (4.73)

and, finally, the diode D4 has the RMS current

ID4,RMS =

√

D((IoutD

)2 +1

3(1

2∆IL)2) (4.74)

Main Advantages and Drawbacks of Two-Transistor Forward Converter

The main advantages of the Two-Transistor Forward Converter are:

+ The switch voltage is clamped by the diodes to the input voltage;

+ Only two windings are needed in the transformer.

57

The main drawbacks of the Two-Transistor Forward Converter are:

- Requires auxiliary power supply for one switch;

- Poor transformer utilization since the core is only magnetized in one direction;

- The transformer design is critical to minimize the magnetizing current and, thus,the losses in the diodes;


4.2.9 Push Pull Converter

The Push Pull Converter also has two transistors, as shown in Figure 4.17, but simplerdrive circuits than the Two-Transistor Forward Converter. Consequently, the converterneeds a transformer with two primary windings and the switch voltage is twice the inputvoltage [5].

Iin

+

-

Uin

+ -Usw2

ISW2

-U

D2 ID2

n1

n3

Usw1

+

-U

D1

ID1

+

ISW1

+ -

Iout

-

+

Uout

+

-

UD3

ID

3

+

-

UD4

ID

4

+

-

UD5

ID

5

+

-

UD6

ID

6

n2

Cout

Sw1

Sw2

D1

D2

D3

D4

D5

D6

Figure 4.17: Schematic layout of push-pull converter.

The switches are on alternatively, thus, creating an alternating current, and energy istransferred via the transformer and the diode bridge to the load. The diodes D1 andD2 blocks and are only used under transientand no-load conditions.

When the switch Sw1 is on, the supply voltage is applied across the winding n1 anda positive flux is induced in the winding n3. The second switch Sw1 is turned off,the switch Sw2 is turned on and a negative flux is induced in the winding n3. It isimportant to have a voltage-second balance to prevent the transformer from becomingsaturated. Idealised waveforms are shown in Figure 4.18.

The relation between the input voltage Uin and the output voltage Uout when thewinding n1 is equal to the winding n2 can be expressed as

58

0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

Time [ms]

Un1

(so

lid),

In1

(do

tted)


0 0.5 1 1.5 2−1

−0.5

0

0.5

1

1.5

2

Time [ms]

Usw

(so

lid),

Isw

(do

tted)


Figure 4.18: Idealised current and voltage waveforms for transformer and switchstresses in DCM for Push Pull Converter (duty ratio D=0.4), Left: Trans-former voltage Un1

(line) and transformer current In1(dotted), Right:

Switch voltage USW1(line) and switch current ISW1

(dotted).

Uout

Uin

= 2n3n1

D (4.75)

where the maximum value of the duty ratio D is 0.5.


This section presents an analysis of the stresses on the semiconductors in the push pullconverter. The maximum voltage over the switches Sw1 and Sw2 is written as

usw =n1n3

Uout

D= 2Uin (4.76)


isw =n3n1

Iout + iMag (4.77)

where iMag is the peak magnetizing current in the transformer. The RMS currentthrough the switches Sw1 and Sw2 is

Isw,RMS =

√

D((n3n1

Iout)2 +1

3i2Mag) (4.78)

The maximum voltage over the diodes D1 and D2 is written as

59

uD1= uD2

=n1n3

Uout

D= 2Uin (4.79)

For the diodes D3 to D6, the maximum voltage is the same and becomes

uD3= uD4

= uD5= uD6

= Uout (4.80)

Under stationary conditions, the current through the diodes D1 and D2 is zero. Themaximum current through the diodes D3 to D6 becomes

iD3= iD4

= iD5= iD6

=Iout2D

(4.81)

The RMS current through the diodes D3 to D6 is the same and is written as

ID3,RMS = ID4,RMS = ID5,RMS = ID6,RMS =Iout

2√D

(4.82)

Main Advantages and Drawbacks of Push Pull Converter

The main advantages of the Push Pull Converter are:

+ Double frequency in the output reduces the requirements for the output filter;

+ Good utilization of the transformer while working with both positive and negativeflux;

+ Both switches are driven with respect to ground.

The main drawbacks of the Push Pull Converter are:

- The transformer needs two primary windings;

- The primary of the transformer must have a voltage-second balance to preventthe transformer core from saturation;

- A blanking time is necessary to ensure that no overlap between the switchesoccurs;


60

Iin

+

-

Uin

+

-

Usw2

IS

W2

-U

C2

IC

2

n1

n2

+

-

Usw1

+

-U

C1

IC

1

+

IS

W1

C1

C2

Sw1

Sw2

Iout

-

+

Uout

+

-

UD1

ID

1

+

-

UD2

ID

2

+

-

UD3

ID

3

+

-

UD4

ID

4

Cout

D1

D2

D3

D4

Figure 4.19: Schematic layout of Half Bridge Converter.

4.2.10 Half Bridge Converter

The Half Bridge Converter has two transistors working together with two capacitors,as shown in Figure 4.19. Again, the transformer is simpler in comparison with theprevious converter types and the switch voltage is the same as the input voltage, butthe drive circuit is more complicated [5].

The switches Sw1 and Sw2 are on alternatively, thus, creating an alternating voltageand the energy is transferred via the transformer and the diode bridge to the load.

When the switch Sw1 is on, half of the input voltage Uin is applied across the windingn1 and a positive flux is induced in the winding n2. When the switch Sw1 is turned off,switch Sw2 is turned on and a negative flux is induced in the winding. The primarymust have a voltage-second balance to prevent the transformer core from saturation.This can be achieved by controlling the voltage over the capacitors C1 and C2. Idealizedwaveforms are shown in Figure 4.20.


Uout

Uin

=n2n1

D (4.83)



This section presents an analysis of the stresses on the semiconductors for the halfbridge converter. The maximum voltage over the switches Sw1 and Sw2 is written as

61

0 0.5 1 1.5 2

−0.5

0

0.5

1

1.5

2

Time [ms]

Un1

(so

lid),

In1

(do

tted)


0 0.5 1 1.5 2

−0.5

0

0.5

1

1.5

2

Time [ms]

Usw

(so

lid),

Isw

(do

tted)


Figure 4.20: Idealized current and voltage waveforms for transformer and switchstresses in Half Bridge Converter (duty ratio D=0.4), Left: Transformervoltage Un1

(line) and transformer current In1(dotted), Right: Switch

voltage USW1(line) and switch current ISW1

(dotted).

usw1= usw2

=n1n2

Uout

D= Uin (4.84)


isw1= isw2

=n2n1

Iout2D

+ iMag (4.85)

where iMag is the peak magnetizing current in the transformer. The RMS currentthrough the switches Sw1 and Sw2 is equal and becomes

Isw1,RMS = Isw2,RMS =

√

D((n2n1

Iout2D

)2 +1

3i2Mag) (4.86)

The diodes D1 to D4 have the same maximum voltage which is written as

uD1= uD2

= uD3= uD4

= Uout (4.87)

The maximum current through the diodes D1 to D4 is equal and becomes

iD1= iD2

= iD3= iD4

=Iout2D

(4.88)

The RMS current through the diodes D1 to D4 is also the same and is written as


2√D

(4.89)

62

Main Advantages and Drawbacks of Half Bridge Converter

The main advantages of the Half Bridge Converter are:

+ Good utilization of the transformer while working with both positive and negativecurrents;

+ The switches can be rated to the input voltage and the anti-parallel diodes clampdestructive switching transients;

+ Double frequency ripple in the output reduces the requirements for the outputfilter.

The main drawbacks of the Half Bridge Converter are:

- Doubled switch current for the same output power in comparison with the FullBridge Converter;

- The control circuits must provide isolated drive signals to the switches;

- The primary must have a voltage-second balance to prevent the core from satu-ration;

- A blanking time is necessary to ensure that no overlap between the switchesoccurs.

4.2.11 Full Bridge Converter

The Full Bridge Converter has four transistors working together, as shown in Figure 4.21.Again the transformer is simple and the switch voltage is the same as the input voltage,but the drive circuitry is more complicated in comparison with the previous converters,since there are four switches and the drive circuits have to be separated from ground [5].

The switches Sw1 to Sw4 are on alternatively, thus, creating alternating voltage, andthe energy is transferred via the transformer and the diode bridge to the load.

First, the switches Sw1 and Sw4 are on, the supply voltage is applied across the windingn1 and a positive flux is induced in the winding n2; Then, the switches Sw1 and Sw4 areturned off and the switches Sw2 and Sw3 are turned on and a negative flux is inducedin the winding n2. The primary must have a voltage-second balance to prevent thetransformer core from saturation. Idealized waveforms are shown in Figure 4.22.


63

Iin

Iout+

-

-

+

Uin

Uout

+

-

Usw4I

SW

4

n1

n2

+

-

UD1

ID

1

+

-

Usw3

IS

W3

+

-

UD2

ID

2

+

-

Usw1

IS

W1

+

-

Usw2I

SW

2

+

-

UD3

ID

3

+

-

UD4

ID

4

Cout

Sw1

Sw2

Sw3

Sw4

D1

D2

D3

D4

Figure 4.21: Schematic layout of full bridge converter.

0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Time [ms]

Un1

(so

lid),

In1

(do

tted)


0 0.5 1 1.5 2

−1

−0.5

0

0.5

1

Time [ms]

Usw

(so

lid),

Isw

(do

tted)


Figure 4.22: Idealised current and voltage waveforms for transformer and switchstresses for Full Bridge Converter (duty ratio D=0.4), Left: Transformervoltage Un1

(line) and transformer current In1(dotted), Right: Switch

voltage USW1(line) and switch current ISW1

(dotted).

64

Uout

Uin

= 2n2n1

D (4.90)



This section presents and analysis of the stresses on the semiconductors. The maximumvoltage over the switches Sw1 to Sw4 has the same value and becomes

uSw1= uSw2

= uSw3= uSw4

=n1n2

Uout

2D= Uin (4.91)

and the maximum current through the switches Sw1 to Sw4 is equal to

iSw1= iSw2

= iSw3= iSw4

=n2n1

Iout2D

+ iMag (4.92)

where iMag is the peak magnetizing current in the transformer. The RMS currentthrough the switches Sw1 to Sw4 has the same value and becomes

ISw1,RMS = ISw2,RMS = ISw3,RMS = ISw4,RMS =

√

D((n2n1

Iout2D

)2 +1

3i2Mag) (4.93)

The maximum voltage over the diodes D1 to D4 is written as

uD1= uD2

= uD3= uD4

= Uout (4.94)

and the maximum current through the diodes D1 to D4 is equal to

iD1= iD2

= iD3= iD4

=Iout2D

(4.95)

The RMS current through the diodes D1 to D4 is the same and is written as


2√D

(4.96)

65

Main Advantages and Drawbacks of Full Bridge Converter

The main advantages of the Full Bridge Converter are:

+ Good utilization of the transformer by working with both positive and negativefluxes;

+ The switches can be rated to the input voltage and anti-parallel diodes clampdestructive switching transients;

+ Double frequency in the output reduces the requirements for the output filter;

+ The output power is doubled in comparison with the Half Bridge Converter, sincethe input voltage, instead of half of the input voltage, is applied to the primarywinding;

+ By replacing the diodes D1 to D4, the power can flow in both directions. Thiscauses extra complexity to the control circuit and drive signals have to be pro-vided for the double amount of switches.

The main drawbacks of the Full Bridge Converter are:

- The control circuits must provide isolated drive signals to the switches;

- The primary must have a voltage-second balance to prevent the transformer corefrom saturation;


Efficiency at different loads is up to 97 %.

4.2.12 Half Bridge Converter With Voltage Doubler

The Half Bridge Converter with voltage doubler has the same primary configuration asthe Half Bridge Converter but uses a voltage doubler on the secondary side, as shownin Figure 4.23 [5].

The switches are on alternatively, thus creating an alternating voltage, and the energyis transferred via the transformer and the voltage doubler to the load. First, the switchSw1 is on, half of the input voltage Uin is applied across winding n1 and a positive fluxis induced in the winding n2; Then, switch Sw1 is turned off and the switch Sw2 isturned on and a negative flux is induced in the winding. The diodes and the capacitorsat the output work as a voltage doubler, which compensates for the voltage applied tothe transformer being reduced to half of the input voltage. This, however, doubles thecurrent for the same output power. The primary must have a voltage-second balanceto prevent the transformer core from saturation but this can be achieved by controllingthe voltage over the capacitors C1 and C2.

66

Iin

Iout+

-

-

+

Uin

Uout

+

-

Usw2

IS

W2

-U

C2

IC

2

n1

n2

+

-

UD1

ID

1

+

-

Usw1

+

-U

C1

IC

1

+

IS

W1

+

-

UD2

ID

2

-

UC4I

C4

+

-

UC3

IC

3

+

D1

D2

C3

C4

C1

C2

Sw1

Sw2

Figure 4.23: Schematic layout of Half Bridge Converter with voltage doubler.


Uout

Uin

= 2n2n1

D (4.97)



In this section the stresses on the semiconductors for the Half Bridge Converter witha voltage doubler are analyzed. The maximum voltage over the switches Sw1 and Sw2is written as

uSw1= uSw2

=n1n2

Uout

2D= Uin (4.98)

and the maximum current through the switches Sw1 and Sw2 becomes

isw1= isw2

=n2n1

IoutD

+ iMag (4.99)

where iMag is the peak magnetizing current in the transformer. The RMS currentthrough the switches Sw1 and Sw2 is written as

Isw1,RMS = Isw2,RMS =

√

D((n2n1

IoutD

)2 +1

3i2Mag) (4.100)

67

The maximum voltage over the diodes D1 to D2 becomes

uD1= uD2

= Uout (4.101)

and the maximum current through the diodes D1 to D2 is equal to

iD1= iD2

=IoutD

(4.102)

The RMS current through the diodes D1 to D2 is written as

ID1,RMS = ID2,RMS =Iout√D

(4.103)

Main Advantages and Drawbacks of Half Bridge Converter with VoltageDoubler

The main advantages of the Half Bridge Converter with a voltage doubler are:

+ Good utilization of the transformer while working with both positive and negativecurrents;

+ The switches can be rated to the input voltage and anti-parallel diodes clampdestructive switching transients;

+ The double frequency ripple in the output reduces the requirements for the outputfilter;

+ By replacing the diodes D1 to D2 with switches the power can flow in bothdirections. This adds extra complexity to the control circuit and drive signalshave to be provided for the double amount of switches.

The main drawbacks of the Half Bridge Converter with a voltage doubler are:

- Doubled switch and diode currents for the same output power in comparison withthe Full Bridge Converter.- The control circuit must provide isolated drive signals to the switches;

- The primary must have a voltage-second balance to prevent the core from satu-ration;


68

4.3 Comparison of Switch Utilization for Different

Converters

For high voltage and high power applications, like a wind park transmission system, thefocus has to be on the utilization factor of the used components. Thus, the switchesshould not be exposed to voltages higher than the input voltage or currents higherthan the rated input current. Moreover, the diodes should not be exposed to voltageshigher than the rated output voltage or currents higher than the rated output current.A high voltage rating is more costly than a high current rating since a higher voltagerating means that more switches have to be series connected. A transformer wouldhave to work with bi-directional flux and for simplicity it should have as few windingsas possible.

The converters are compared on per unit basis where the output voltage and theoutput current are assumed to be the base values. The magnetizing current for thetransformers is not included in the comparison in order to simplify the analysis. Datafor such transformers is not readily available and the simplification is not believed toaffect the results. Two different applications are compared. First, the converters areused for voltage adjustment where the ratio between the input and the output voltageis between 2.5 and 5. Then, a DC transformer, where the ratio between the inputand the output voltage is 10 with a small tolerance, is used. The Voltage AdjustmentConverter is used at the point closest to the generator, as shown in Figure 4.2, toadjust the varying generating voltages at different wind speeds, as shown in Figure 4.1.The DC transformer is used after the Voltage Adjustment Converter to reach a hightransmission voltage in order to minimize transmission losses.

4.3.1 Voltage Adjustment Converter

The different theoretical ratings for the Voltage Adjustment Converter are shown inTable 4.2 and Table 4.3 below. The output voltage and current are equal to 1 pu andthe input voltage is equal to 0.4 pu. The Boost Converter, shown in Figure 4.3, is thepreferable choice with low semiconductor stresses and a simple structure. The non-galvanic isolated converters, except for the Luo Converter, have a higher switch peakvoltage while the Luo Converter has a higher number of components. The galvanicisolated converters, with the exception of the Flyback Converter, have a higher switchpeak current while the Flyback Converter has a complicated transformer.

The diode stresses in the converters are similar, with the exception of the Cuk, Sepic,Zeta and both Forward Converters that have a higher diode peak voltage. Two draw-backs of the Boost Converter are that it has no galvanic isolation and no protectionagainst short circuits on the output. The switch stresses are presented at rated power,according to Figure 4.1, and the converters are modelled for the specific input voltagevariations from the generator. This results in a high turns ratio and a low duty-ratiofor the bridge converters and, therefore, a high peak current in comparison with theBoost Converter.

69

Table 4.2: Switch stresses, duty ratio and turns ratio for converters used for voltageadjustment.

Duty Ratio Turns Ratio Switch stresses [pu]

Converter D N u i IRMS

Boost 0.60 - 1.00 2.50 1.94Cuk 0.29 - 3.50 1.40 0.75Sepic 0.29 - 3.50 1.40 0.75Zeta 0.29 - 3.50 1.40 0.75Luo 0.29 - 1.00 2.50 0.75Flyback 0.71 1 1.40 3.50 2.96Forward 0.25 10 0.80 10.0 5.00Two transistor forward 0.25 10 0.40 10.0 5.00Push Pull 0.25 5 0.80 5.00 2.50Half Bridge 0.25 10 0.40 20.0 10.0Full Bridge 0.25 5 0.40 10.0 5.00Half Bridge + VD 0.25 5 0.40 20.0 10.0

Table 4.3: Diode stresses in converters used for voltage adjustment.Diode A Diode B Diode C

Converter u i IRMS u i IRMS u i IRMS

Boost 1.0 2.5 1.6Cuk 3.5 1.4 1.2Sepic 3.5 1.4 1.2Zeta 3.5 1.4 1.2Luo 1.0 * * 1.0 1.0 0.5Flyback 0.8 * * 1.4 3.5 1.9Forward 0.8 * * 4.0 1.3 1.1 4.0 4.0 2.0Two-Transistor Forward 0.4 * * 0.4 1.3 1.1 4.0 4.0 2.0Push Pull 0.8 * * 1.0 2.0 1.0Half Bridge 1.0 2.0 1.0Full Bridge 1.0 2.0 1.0 * See text for eachHalf Bridge + VD 1.0 4.0 2.0 separate converter

70

Figure 4.24 shows the switch stresses for the switch depending on the wind speed forthe Boost Converter when it is used as a Voltage Adjustment Converter. The voltageis independent of the load and is the same as the output voltage. However, the currentvaries with wind speed. First, between the cut-in wind speed and 5 m/s, the currentchanges cubically and then, between 5 m/s and the rated wind speed, the currentchanges quadratically.

2 3 4 5 6 7 8 9 10

0

0.5

1

1.5

2

2.5

Wind speed [m/s]

Vol

tage

[pu

], C

urre

nt [

pu]

Figure 4.24: Stresses on switch depending on wind speed for Boost Converter whenused as a Voltage Adjustment Converter. Solid: Switch peak voltage uSw[pu] and dashed: Switch peak current iSw [pu].

4.3.2 DC/DC Transformer

The different theoretical ratings for the DC transformer are shown in Table 4.4 andTable 4.5 below. The output voltage and current are equal to 1 pu and the inputvoltage is equal to 0.1 pu. A Full Bridge Converter is the preferable choice, with lowsemiconductor stresses and galvanic isolated output. Non-galvanic isolated convertershave a very high switch peak voltage, especially the Cuk, Sepic and Zeta Converters.The galvanic isolated converters, with the exception of the Push Pull Converter, havea higher switch peak current while the Push Pull Converter has a higher switch peakvoltage. The Cuk, Sepic, Zeta and both forward converters have a higher diode peakvoltage and the Boost Converter has a high diode peak current. The other convertershave similar diode stresses. However, one drawback is the number of semiconductors,but the switches are fully utilized and the cost of the drive circuitry is manageable

71

with respect to the high power and high voltage ratings. If the current is manageable,the Half Bridge Converter is the best alternative although the rated current for theswitches are doubled compared with the Full Bridge Converter. Moreover, the HalfBridge Converter has a somewhat poorer fault tolerance since the transformer cannotbe completely disconnected by using the switches.

Switch stresses are presented at rated power, according to Figure 4.1 and the convertersare modelled as a DC transformer. As shown, the Full Bridge Converter has similarpeak current, but only one tenth of the peak voltage for the Boost Converter.

Table 4.4: Switch stresses in converters used as DC transformers.Duty Ratio Turns Ratio Switch stresses

Converter D N u i IRMS

Boost 0.90 - 1.00 9.80 9.29Cuk 0.09 - 10.8 1.10 0.34Sepic 0.09 - 10.8 1.10 0.34Zeta 0.09 - 10.8 1.10 0.34Luo 0.09 - 1.00 9.80 0.34Flyback 0.49 10 0.20 19.8 13.9Forward 0.49 20 0.20 20.0 14.0Two-Transistor Forward 0.49 20 0.10 20.0 14.0Push Pull 0.49 10 0.20 10.0 7.00Half Bridge 0.49 20 0.10 20.4 14.3Full Bridge 0.49 10 0.1 10.2 7.14Half Bridge + VD 0.49 10 0.10 20.4 14.3

Table 4.5: Diode stresses in converters used as DC transformers.

Diode A Diode B Diode C

Converter u i IRMS u i IRMS u i IRMS

Boost 1.00 9.80 3.13Cuk 10.8 1.10 1.05Sepic 10.8 1.10 1.05Zeta 10.8 1.10 1.05Luo 1.00 * * 1.00 1.00 0.30Flyback 0.20 * * 2.02 1.98 1.41Forward 0.20 * * 0.10 1.96 1.40 2.04 2.04 1.43Two-Transistor Forward 0.10 * * 0.10 1.96 1.40 2.04 2.04 1.43Push Pull 0.20 * * 1.00 1.02 0.71Half Bridge 1.00 1.02 0.71Full Bridge 1.00 1.02 0.71 * See text for eachHalf Bridge + VD 1.00 2.04 1.43 separate converter

72

Figure 4.25 shows the switch stresses on the switch, depending on wind speed, for theFull Bridge Converter when it is used as a DC transformer. The voltage is clampedto the input voltage and is independent of the load and the current varies with windspeed. First, between the cut-in wind speed and 5 m/s, the current changes cubicallyand then, between 5 m/s and the rated wind speed, the current changes quadratically.

2 3 4 5 6 7 8 9 10

0

0.2

0.4

0.6

0.8

1

Wind speed [m/s]

Vol

tage

[pu

], C

urre

nt [

pu]

Figure 4.25: Stresses on switch depending on wind speed for Full Bridge Converterwhen used as a DC transformer. Solid: Switch peak voltage uSw [pu] anddashed: Switch peak current iSw [pu].

4.4 Losses for Different Converter Layouts

Assuming that the Boost Converter is used as a voltage adjuster and a Half Bridge ora Full Bridge Converter is used as a DC transformer, the losses for two different windspeeds will be as presented in Table 4.6. Due to wind speed distribution, the convertersare often operated at partial load. Thus, the losses are calculated for the wind speeds5 m/s and 10 m/s, which in this example, correspond to 12.5%, respectively, 100% ofrated power. The input voltage for the voltage adjuster is 0.5 and 1.0 kV, respectively,and the output voltage is 2.5 kV. The input voltage for the DC transformer is 1.0 kVand the output voltage is 10.0 kV. The switching frequency is set to 1000 Hz. Onlythe semiconductors are taken into account and the conducting losses are calculated asthe RMS and average values for the currents and with data for typical semiconduc-tors [57, 58]. The valve type IGBT has been used in the Boost Converter and the

73

bridge converters, with the ratings 3300V/1200A and 1400V/600A, respectively, andhigh voltage diodes with the rating 3300V/600A have been used in all converters. Thecorrect voltage and current are achieved by parallel and series coupling of the compo-nents presented above. The switching losses are assumed to be of the same magnitudeas the conducting losses because of the chosen switching frequency. The results shouldonly be used as a guideline as the model is only basic. The Boost Converter has thelowest losses of the three converters, especially since it is used as a voltage adjuster.The losses are lower for the Half Bridge Converter than the Full Bridge Converter.However, the Half Bridge has a lower current handling capability.

Table 4.6: Losses for converters.

Converter 5 m/s ⇔ Pn/8 10 m/s ⇔ Pn

Boost 1.2 % 0.74 %Half Bridge 1.3 % 1.5 %Full Bridge 1.3 % 1.8 %

4.5 Summary

For a high voltage and high power application, like a wind park transmission system, thefocus has to be on the utilization factor of the used components. Designing a DC/DCconverter for an input voltage down to half the rated voltage or with a high inputto output voltage ratio might decrease the performance at rated power and voltage.Therefore, two different applications have been analyzed. One simple converter canbe used for a first adjustment of the voltage and then a second converter can be usedto raise the voltage to a suitable transmission level. The semiconductor componentstresses have been determined by a theoretical comparison of the different convertertopologies. A Boost Converter is suitable as a Voltage Adjustment Converter and aBridge Converter can be used as the DC transformer. The Half Bridge Converter isused when the current is low and the Full Bridge Converter is used for higher currents.Losses in the Boost, Half Bridge and Full Bridge Converters have also been estimatedand the converters have moderate losses.

74

Chapter 5

Dynamic Analysis of Hard-switchedDC/DC converters

In this chapter, Boost, Full Bridge, Half Bridge and Dual Active Bridge Convert-ers are analyzed and their behavior and interaction with the DC-grid and the otherDC/DC converters are investigated by using the circuit-orientated simulation packageSaber r©. The Boost and the Dual Active Bridge Converters are controlled to transmita certain amount of power while the Full and Half Bridge Converters are controlledto be DC transformers. Prior to performing transient studies, for which an analyticalsolution is readily available, the steady-state performance of the converter discussed inthe previous chapter was evaluated numerically. This of course serves to provide anindication to the degree of accuracy to be expected in all results.

5.1 Boost Converter

In this section, the Boost Converter is investigated and simulation results are pre-sented. The schematic layout of the converter with the names of the parameters andthe variables is shown in Figure 5.1. The converter input is connected to a voltagesource Udc1 in series with a source impedance of Zsource1 and the converter output isconnected to a voltage source Udc2 in series with a source impedance of Zsource2 . Thesource impedances are used to imitate the connection lines to the converter.

5.1.1 Working Principle

The switch Sw is controlled to charge the inductor L during the first period of the cycleTs when the switch is on and to discharge the inductor energy during the second periodof the cycle when the switch is turned off. Figure 5.2 shows the semiconductor stressesfor the converter during full load operation, according to the parameters presented inTable 5.1.

The converter runs in CCM, i.e., the inductor current iL never drops to zero. Thevoltage over the switch should be constant when the switch is off and the currentthrough the switch should increase linearly when the switch is on, but due to the finite

75

Iin

Iout

+

- -

+

Uin

Uout

+

-

Usw

ID

- +U

D

IS

W

+ -UL

ILL D

Sw Cout

ZSource1

Udc1

+

-

+

-U

dc2

ZSource2

Cin

Figure 5.1: Schematic layout of Boost Converter.

(A

)

0.0

1000.0

2000.0

t(s)

0.497 0.4975 0.498 0.4985 0.499 0.4995 0.5

(V

)

0.0

1250.0

2500.0

(A

)

0.0

1000.0

2000.0

(V

)

0.0

1250.0

2500.0

(A) : t(s)

i_d

(V) : t(s)

u_d

(A) : t(s)

i_sw

(V) : t(s)

u_sw

Figure 5.2: Semiconductor stresses in a Boost Converter at rated power. Top: Switchvoltage uSw [V], upper middle: Switch current iSw [A], lower middle: Diodevoltage uD [V] and bottom: Diode current iD [A].

76

size of the capacitors, Cin and Cout and the source impedance, the voltage fluctuatesand the current, therefore, does not increase linearly, as shown in Figure 5.2. Theconverter uses traditional pulse width modulation (PWM) to control the voltage and/orthe current, i.e., the reference value is compared with a sawtooth wave and the switchconducts as long as the reference value is higher than the sawtooth.

Table 5.1: System parameters of Boost Converter.

Parameter Value Component Value

Rated power 2 MW Turn-on time Tsw,on 10 µsRated input voltage 0.5-1 kV Turn-off time Tsw,off 10 µsRated output voltage 2.5 kV Inductor L 5 mHSwitching frequency 1000 Hz Inductor resistance RL 1 mΩIGBT voltage drop UT0 2.8 V Input capacitor Cin 1 mF

series resistance ron 0.45 mΩ Output capacitor Cout 2 mFDiode voltage drop UT0 1.4 V Source impedance 0.5 µH + 1 mΩ

series resistance ron 0.9 mΩ (Zsource1 = Zsource2)Blanking time Tb 5 µs

5.1.2 Transmitted Power

The transferred power of the Boost Converter is adjusted by changing the duty ratio,D. The stationary value of D, however, remains the same according to:

Uout = Uin

1

1−D(5.1)

where Uin is the average value of the voltage on the primary side and Uout is the meanvalue of the voltage on the secondary side. The losses of the converter are neglected inthe equation. If the input voltage is 1000 V and the output voltage is 2500 V, then theduty ratio, D becomes 0.6 under stationary conditions. If D >0.6, the current increasesand the converter, thus, transmits more power. Eventually, the inductor will saturateand the current will have to be reduced. If D <0.6, the inductor current decreasesand the converter, thus, transmits less power and if the current decreases so that thecurrent goes down to zero the current will go into DCM.

77

5.1.3 Transient Behavior

Step changes in the power around the operating point have been made to verify thetransient behavior of the system. The power reference is changed, +10% from anoperating point of 2 MW. The results are shown in Figure 5.3. The sampling causesa time delay of Ts ( the controller starts to react after 1 sample). The output power,used in the controller, is calculated as the average of the output voltage Uout times theoutput current Iout for each switching period and, therefore, creates another time delayof Ts. This is, however, negligible compared with the time constant for the converter,which is in the order of 50 ms.

−

0.5

0.55

0.6

0.65

0.7

t(s)

0.4 0.5 0.6 0.7 0.8

(W)

1.8meg

2meg

2.2meg

2.4meg

(W)

1.8meg

2meg

2.2meg

2.4meg

− : t(s)

D

(W) : t(s)

Power

(W) : t(s)

Pref

Figure 5.3: Transient behavior during step changes. From top: Power reference P ∗

[W], power P [W] and duty ratio D.

When the power reference increases, the duty ratio changes in order to increase thecurrent through the inductor L and, therefore, increase the transmitted power. Thiscauses the transmitted power to decline since a larger part of the cycle is used to in-crease the inductor current and less time is used to transmit power. When the desiredtransmission is reached, the duty ratio is changed back to the stationary value, accord-ing to Eq. (5.1). The power controller is assumed to use a traditional PI-controller asshown in Figure 5.4 with Kp = 50 · 10−6, Ti=50 ms and Ts=1 ms.

78

PSfrag replacements

D∑

Kp

(

1 + TsTi

11−Z−1

)+

−

P ∗

P

Powerreference

Measuredpower

PI-Controller

Dutyratio

Duty ratioexecutor

Sw

Valveswitchingstate

Figure 5.4: Power control of Boost Converter using discrete PI-controller.

5.1.4 Varying Voltage

The input voltage of the converter was changed to verify good behavior when appliedto a varying input voltage. The power reference P ∗ and the output voltage Udc2 werekept constant while the duty ratio D was adjusted, according to Eq. (5.1). The resultis shown in Figure 5.5.

The input voltage changes slowly from 500 V to 1000 V and the resulting transmittedpower is slightly above 1 MW, since the controller lags behind the power referencewhen the input voltage is increased. As the input voltage is reduced, the duty ratiohas to be increased, according to Eq. (5.1) in order to transmit the same amount ofpower and, thus, the peak current for the switches is changed according to Eq. (5.2)(ripple is not included)

isw ≈P

Uin

(5.2)

The ripple current in the inductor, is according to Eq. (5.3) and with the data givenin Table 5.1, the maximum ripple current becomes 120 A when the input voltage is1 kV. During dynamic changes, however, the maximum theoretical ripple is 200 A,i.e., the current in the inductor can be changed at a maximum of 200 A each cycle(corresponding to D = 1). A smaller inductor gives a higher ripple but a fasterconverter response since a change in the power reference, i.e., the input current, canbe achieved faster.

iripplep−p =UinTsD

L(5.3)

Figure 5.6 shows the semiconductor stresses at 750 V input voltage. The switch stressesat different input voltage levels at the power of 1 MW and a constant output voltageare shown in Table 5.2. The peak current is the same for 500V input voltage as for2 MW and 1000 V input voltage, despite that the power is halved to 1 MW.

79

(V

)

500.0

750.0

1000.0

(−

)

0.6

0.8

(W

)

900000.0

1meg

1.1meg

t(s)

0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0

(V) : t(s)

Uin

(−) : t(s)

D

(W) : t(s)

Power

Figure 5.5: Simulation of Boost Converter using varying input voltage together withconstant power reference. From top: Power P [W], duty ratio D and inputvoltage Uin [V].

Table 5.2: Duty ratio and switch peak current for different input voltages and constantpower of 1 MW.

Udc1 D isw iripplep−p Iin500 V 0.8 2040 A 80 A 2000 A750 V 0.7 1386 A 105 A 1333 A1000 V 0.6 1060 A 120 A 1000 A

80

(A

)

0.0

750.0

1500.0

t(s)

0.497 0.4975 0.498 0.4985 0.499 0.4995 0.5

(V

)

0.0

1250.0

2500.0

(A

)

0.0

750.0

1500.0

(V

)

0.0

1250.0

2500.0

(A) : t(s)

i_d

(V) : t(s)

u_d

(A) : t(s)

i_sw

(V) : t(s)

u_sw

Figure 5.6: Semiconductor stresses in a Boost Converter with 750 V input voltage andconstant power of 1 MW. Top: Switch voltage uSw [V], upper middle:Switch current iSw [A], lower middle: Diode voltage uD [V] and bottom:Diode current iD [A].

81

5.1.5 Summary

The boost converter can handle both varying input voltage and dynamic changes inthe transmitted power. The design of the inductor size is a compromise between thesize of the input ripple and the speed of the dynamic response of the converter. Asmaller inductor gives a higher ripple but a faster converter response since a changein the power reference, i.e., the input current, can be achieved faster. It has also beenshown that the input ripple is influenced by the input voltage. The input voltage hasbeen reduced to half of the rated voltage and still the converter is able to transmit thesame power by changing the duty ratio.

82

5.2 Half Bridge Converter

In this section, the Half Bridge (HB) Converter is investigated and simulation resultsare presented. The schematic layout of the converter is shown in Figure 5.7. Theconverter input is connected to a voltage source Udc1 in series with a source impedanceZsource1 and the converter output is connected to a voltage source Udc2 in series with asource impedance Zsource2 .

Iin

Iout

+

-

-

+

Uin

Uout

n1

n2

+

-

USW1

IS

W1

+

-

USW2

IS

W2

Cout

Sw1

Sw2

L1l

L2l

+

-

UD4

ID

4

+

-

UD3

ID

3+

-

UD1

ID

1

+

-

UD2

ID

2

D1

D2

D3

D4

ITfo1

ITfo2

+

-

UTfo2

+

-

UTfo1

+

-

+

-

ZSource1

Udc1

Udc2

ZSource2

C1

C2

Figure 5.7: Schematic layout of Half Bridge Converter.


The switches are controlled to create a square wave on the primary side of the trans-former. The switch Sw1 is on during the first half period and the switch Sw2 is onduring the second half of the period, thus, creating a voltage of UTFO1

between themidpoint of the capacitor leg and the dc-bus. The duty ratio D is, thus, a half period,i.e., D = 0.5. The converter is often referred to as a DC transformer, when controlledas described above, but can also be controlled with a traditional PWM control to con-trol the duty ratio D or any other control for regulating voltage and/or current. Ablanking time Tb, i.e., a time between the two half periods when no switch is on, is usedto ensure that the switches do not short-circuit the dc-bus. When the voltage UTFO2

produced on the secondary side of the transformer is larger than the secondary busvoltage Uout, the diodeswill conduct and, thus, create a power transfer. The transferredpower is determined by the difference between the two voltages Lλ and Uout and theleakage inductance in the transformer Lλ. Figure 5.8 shows the semiconductor stressesin the converter during full load operation, according to the parameters presented inTable 5.3.

The voltages over the switches should be constant when the switches are off and thecurrents through the switches should increase linearly when the switches are on, butdue to the finite size of the capacitors, C1, C2 and Cout together with the sourceimpedance, the voltage fluctuates and the current in the switches, therefore, does notincrease linearly.

83

(A

)

0.0

100.0

200.0

t(s)

0.097 0.0975 0.098 0.0985 0.099 0.0995 0.1

(V

)

0.0

5000.0

10000.0

(A

)

0.0

2000.0

4000.0

(V

)

0.0

1000.0

(A) : t(s)

i_d1

(V) : t(s)

u_d1

(A) : t(s)

i_sw1

(V) : t(s)

u_sw1

Figure 5.8: Semiconductor stresses in a Half Bridge Converter according to values inTable 5.3. Top: voltage switch one uSw1

[V], upper middle: current switchone iSw1

[A], lower middle: voltage diode one uD1[V] and bottom: current

diode one iD1[A].

Table 5.3: System parameters of Half Bridge Converter.


Rated power 1 MW Turn-on time Tsw,on 1 µsRated input voltage 1 kV Turn-off time Tsw,off 1 µs

Rated output voltage 10 kV Transformer turns ratio N = n2

n1=21

Switching frequency 1000 Hz Leakage inductance 8 µHIGBT voltage drop UT0 2.8 V (Lλ = L1λ + L′

2λ)

series resistance ron 0.225 mΩ Input capacitor Cin = C1 + C2 20 mFDiode voltage drop UT0 5.6 V Output capacitor Cout 2 mF

series resistance ron 14 mΩ Source impedance 0.5 µH + 1 mΩBlanking time Tb 5 µs (Zsource1 = Zsource2)

84


Assuming that the mean value of the primary dc bus voltage Uin and the mean valueof the secondary dc bus voltage equivalent U ′

out on the primary side is equal, i.e.,Uin = U ′

out, then the peak-to-peak current ∆i will be

∆i =12Uin − U ′

out

Lλ

Ts2

(5.4)

where Lλ is the transformer leakage impedance. Moreover, Ts is the switching time,i.e., Ts =

1fs. The average current Iin becomes

Iin =1

2∆i =

12Uin − U ′

out

Lλ

Ts4

(5.5)

The transferred power function for the FB converter can be approximated with (sinceUin is constant)

P =1

2UinIin =

12Uin(

12Uin − U ′

out)Ts

4Lλ

(5.6)

The losses for the converter are neglected in the derived function of transferred power.

5.2.3 Steady-state Behavior

The output power as a function of input voltage is shown in Figure 5.9. The deviationbetween the theoretical expression and the result of the simulated system is due toshortcomings in the Eq. (5.6) and not only to losses in the system. For example, theswitches and the diodes have voltage drops, the blanking time, turn-on and turn-offtimes of the switches are neglected. The negative value of the theoretical power due tothe equation does not take diode blocking into account. Thus, the power transmissionfor the simulated curve starts approximately 20 V above the theoretical curve and thepower is, therefore, 100 kW lower.

85

t(s)

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6

(V)

900.0

1000.0

1100.0

(W)

0.0

500000.0

1meg

(W)

0.0

1meg

(V) : t(s)

Uin

(W) : t(s)

Power

(W) : t(s)

Power (Theoretical)

Figure 5.9: Transmitted power as a function of input voltage using Half Bridge Conv-erter. From top: Power P from simulation (line) and analytical calculatedpower (dotted) [W] and input voltage Uin [V].

86


Step changes in the input voltage were made to verify the transient behavior of thesystem. The input voltage was changed +100 V from the operating point 1000 V andthere was no control of the converter, i.e., it was used as a DC-transformer. As seenin Figure 5.10, the converter manages the step without oscillations.

(V

)

1000.0

1050.0

1100.0

t(s)

0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07

(W

)

0.0

500000.0

1meg

(V) : t(s)

Uin

(W) : t(s)

Power

Figure 5.10: Transient behavior during step changes using Half Bridge Converter. Fromtop: Power P [W], input voltage Uin [V].

5.2.5 Half Bridge Converter With Voltage Doubler

A variant of the Half Bridge Converter is the Half Bridge Converter with voltage dou-bler, where the output rectifier bridge is changed to a voltage doubler. The schematiclayout of the converter is shown in Figure 5.11.

The transformer turns ratio is halved to 10.5 and the currents on the secondary side ofthe transformer are, therefore, doubled. Figure 5.12 shows the semiconductor stressesin the converter. The voltage doubler increases the voltage ripple on the secondaryside, due to the doubled currents on the secondary side of the transformer, but reducesthe number of active components by removing the diodes D3 and D4 (Figure 5.7) andby replacing the output capacitor Cout with two capacitors, C1 and C2.

87

Iin

+

-

Uin

n1

n2

+

-

USW1

IS

W1

+

-

USW2

IS

W2

Sw1

Sw2

L1l

L2l

+

-

UD1

ID

1

+

-

UD2

ID

2

D1

D2

ITfo1

ITfo2

+

-

UTfo2

+

-

UTfo1

+

-

ZSource1

Udc1

C1

C2

Iout

-

+

Uout

+

-U

dc2

ZSource2

C1

C2

Figure 5.11: Schematic layout of Half Bridge Converter with voltage doubler.

(A)

0.0

250.0

500.0

750.0

t(s)

0.097 0.098 0.099 0.1

(V)

0.0

5000.0

10000.0

(A)

0.0

2000.0

4000.0

(V)

0.0

1000.0

(A) : t(s)

i_d1

(V) : t(s)

U_d1

(A) : t(s)

i_sw1

(V) : t(s)

u_sw1

Figure 5.12: Semiconductor stresses in a Half Bridge Converter with voltage doublerat rated power. Top: voltage switch one uSw1

[V], upper middle: currentswitch one iSw1

[A], lower middle: voltage diode one uD1[V] and bottom:

current diode one iD1[A].

88

5.2.6 Summary

The half bridge DC transformer can handle dynamic changes in the transmitted powerbut cannot control the transferred power. The leakage inductance Lλ and the turnsratio for the transformer are the most important parameters of the converter design.The leakage inductance influences the current ripple, the maximum transferable powerand the sensitivity to input voltage changes. The turns ratio determines the voltagelevels for the converter.

89

5.3 Full Bridge Converter

In this section, the Full Bridge (FB) Converter is investigated and simulation resultsare presented. The schematic layout of the converter is shown in Figure 5.13. Theconverter input was connected to a voltage source Udc1 in series with a source impedanceof Zsource1 and the converter output was connected to a voltage source Udc2 in serieswith a source impedance of Zsource2 .

Iin

Iout

+

-

-

+

Uin

Uout

+

-

USW4

IS

W4

n1

n2

+

-U

SW3

IS

W3

+

-U

SW1

IS

W1

+

-

USW2

IS

W2

Cout

Sw1

Sw2

Sw3

Sw4

L1l

L2l

+

-

UD4

ID

4

+

-

UD3

ID

3+

-

UD1

ID

1

+

-

UD2

ID

2

D1

D2

D3

D4

ITfo1

ITfo2

+

-

UTfo2

+

-

UTfo1C

in

+

-

+

-

ZSource1

Udc1

Udc2

ZSource2

Figure 5.13: Schematic layout of Full Bridge Converter.


The switches are controlled in order to create a square wave on the primary side ofthe transformer. Switches Sw1 and Sw4 are on simultaneously during the first halfperiod and switches Sw2 and Sw3 are on during the second half period. The dutyratio D is, thus, a half period, i.e., D = 0.5. The converter is often referred to as aDC transformer, when controlled as described above, but can also be controlled with atraditional PWM control in order to control the duty ratio D or any other control forregulating the voltage and/or the current. A blanking time Tb is used to ensure that theswitches do not short-circuit the dc-bus. When the voltage produced on the secondaryside of the transformer UTFO2

is larger than the secondary bus voltage Uout, the diodesconduct and, thus, create a power transfer. The transferred power is determined bythe difference between the two voltages Uin and Uout and the leakage inductance in thetransformer Lλ. Figure 5.14 shows the semiconductor stresses in the converter duringfull load operation, according to the parameters presented in Table 5.4.

The voltages over the switches should be constant when the switches are off and thecurrent through the switches should increase linearly when the switches are on, butdue to the finite size of the capacitors, Cin and Cout and the source impedance, thevoltage fluctuates and the currents, therefore, do not increase linearly.

90

(A)

0.0

200.0

400.0

t(s)

0.097 0.098 0.099 0.1

(V)

0.0

5000.0

10000.0

(A)

0.0

2000.0

4000.0

(V)

0.0

1000.0

(A) : t(s)

i_d1

(V) : t(s)

u_d1

(A) : t(s)

i_sw1

(V) : t(s)

u_sw1

Figure 5.14: Semiconductor stresses in Full Bridge Converter according to values inTable 5.4. Top: voltage switch one uSw1

[V], upper middle: current switchone iSw1

[A], lower middle: voltage diode one uD1[V] and bottom: current

diode one iD1[A].

Table 5.4: System parameters of Full Bridge Converter.

Parameter Value Component ValueRated power 2 MW Turn-on time Tsw,on 1 µs cRated input voltage 1 kV Turn-off time Tsw,off 1 µs

Rated output voltage 2.5 kV Transformer turns ratio N = n2

n1=11

Switching frequency 1000 Hz Leakage inductance 18 µHIGBT voltage drop UT0 2.8 V (Lλ = L1λ + L′

2λ)

series resistance ron 0.45 mΩ Input capacitor Cin 50 mFDiode voltage drop UT0 5.6 V Output capacitor Cout 2 mF

series resistance ron 14 mΩ Source impedance 0.5 µH + 1 mΩBlanking time Tb 5 µs (Zsource1 = Zsource2)

91


Assuming that the mean value of the primary dc bus voltage Uin and the mean valueof the secondary dc bus voltage equivalent U ′

out on the primary side are equal, i.e.,Uin = U ′

out. Then the peak-to-peak current ∆i will be

∆i =Uin − U ′

out

Lλ

Ts2

(5.7)

where Lλ is the transformer leakage impedance. Ts is the switching time, i.e., Ts =1fs.

And the average current Iin becomes

Iin =1

2∆i =

Uin − U ′out

Lλ

Ts4

(5.8)

The transferred power function for the FB converter can be approximated with (sinceUin is constant)

P = UinIin =Uin(Uin − U ′

out)Ts4Lλ

(5.9)

The losses for the converter are omitted in the Eq. (5.9).


The output power as a function of input voltage is shown in Figure 5.15. As in theprevious case, the deviation between the theoretical expression and the result of thesimulated system is due to shortcomings in Eq. (5.9) and not only losses in the system.For example, the switches and the diodes have voltage drops, the blanking, turn-onand turn-off times of the switches are neglected. The negative value of the theoreticalpower is due to the equation does not take diode blocking into account. Thus, thepower transmission for the simulated curve starts at approximately 20 V above thetheoretical curve and the power is, therefore, 200 kW lower.

92

(V)

900.0

1000.0

1100.0

t(s)

0.0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65

(W)

0.0

1meg

2meg

3meg

(V)

0.0

1meg

2meg

3meg

(V) : t(s)

Uin

(W) : t(s)

Power

(V) : t(s)

Power (Theoretical)

Figure 5.15: Transmitted power as a function of input voltage using Full Bridge Conv-erter. From top: Simulated power P (line) and Theoretical power (dotted)[W] and input voltage Uin [V].

93


Step changes in the input voltage were made to verify the transient behavior of thesystem. The input voltage was changed +100 V from the operating point 1000 V andthere was no control of the converter, i.e., it was used as a DC-transformer. As seenin Figure 5.16, the converter manages the step without oscillations.

(V

)

1000.0

1050.0

1100.0

(W

)

1meg

1.5meg

2meg

t(s)

0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06 0.065 0.07

(V) : t(s)

Uin

(W) : t(s)

Power

Figure 5.16: Transient behavior during step changes using Full Bridge Converter. Fromtop: Power P [W], input voltage Uin [V].

5.3.5 Full Bridge Converter with Voltage Doubler

A variant of the Full Bridge Converter is the Full Bridge Converter with a voltage dou-bler, where the output rectifier bridge is changed to a voltage doubler. The schematiclayout of the converter is shown in Figure 5.17.

The transformer turns ratio is halved to 5.5 and the currents on the secondary side ofthe transformer are, therefore, doubled. Figure 5.18 shows the semiconductor stressesfor the converter. The voltage doubler increases the voltage ripple on the secondaryside, due to doubled currents on the secondary side of the transformer, but reduces thenumber of active components by removing the diodes D3 and D4 (Figure 5.13) and byreplacing the output capacitor Cout with two capacitors C1 and C2.

94

Iin

Iout

+

-

-

+

Uin

Uout

+

-

USW4

IS

W4

n1

n2

+

-U

SW3

IS

W3+

-U

SW1

IS

W1

+

-

USW2

IS

W2

Sw1

Sw2

Sw3

Sw4

L1l

L2l

+

-

UD1

ID

1

+

-

UD2

ID

2

D1

D2

ITfo1

ITfo2

+

-

UTfo2

+

-

UTfo1C

in

+

-

+

-

ZSource1

Udc1

Udc2

ZSource2

C1

C2

Figure 5.17: Schematic layout of Full Bridge Converter with voltage doubler.

(A)

0.0

250.0

500.0

750.0

t(s)

0.097 0.098 0.099 0.1

(V)

0.0

5000.0

10000.0

(A)

0.0

2000.0

4000.0

(V)

0.0

1000.0

(A) : t(s)

i_d1

(V) : t(s)

U_d1

(A) : t(s)

i_sw1

(V) : t(s)

u_sw1

Figure 5.18: Semiconductor stresses in a Full Bridge Converter with voltage doublerat rated power. Top: voltage switch one uSw1

[V], upper middle: currentswitch one iSw1

[A], lower middle: voltage diode one uD1[V] and bottom:

current diode one iD1[A].

95

5.3.6 Summary

The Full Bridge DC transformer can handle dynamic changes in the transmitted powerbut cannot control the transferred power. The leakage inductance Lλ and the turnsratio for the transformer are the most important parameters of the converter. Theleakage inductance influences the current ripple, the maximum transferable power andthe sensitivity to input voltage changes. The turns ratio determines the voltage levelsfor the converter.

96

5.4 Dual Active Bridge Converter (DAB)

In this section, the Dual Active Bridge Converter (DAB) is investigated and simulationresults are shown in order to give further insight to its operation and design. Theschematic layout of the converter is shown in Figure 5.19. The DAB consists of twoH-bridges and a transformer and both bridges can be controlled individually for a bi-directional power flow [19, 21]. The converter input is connected to a voltage sourceUdc1 in series with a source impedance of Zsource1 and the converter output is connectedto a voltage source Udc2 in series with a source impedance of Zsource2 .

Iin

Iout

+

- -

+

Uin

Uout

+

-

USW4

IS

W4

n1

n2

+

-U

SW3

IS

W3

+

-U

SW1

IS

W1

+

-

USW2

IS

W2

Cout

Sw1

Sw2

Sw3

Sw4

L1l

L2l

+

-

USW8

IS

W8

+

-U

SW7

IS

W7

+

-U

SW5

IS

W5

+

-

USW6

IS

W6

Sw5

Sw6

Sw7

Sw8

ITfo1

ITfo2

+

-

UTfo2

+

-

UTfo1C

in

+

-

+

-

ZSource1

Udc1

Udc2

ZSource2

Figure 5.19: Schematic layout of DAB converter.


Two sets of full bridges, one on each side of the transformer, were controlled to transferenergy to or from the low-voltage side. The switches were controlled to create a squarewave or a quasi-square wave on each side of the transformer. The switch pattern isaccording to Figure 5.20. The switch pattern to the left in Figure 5.20 is a pure squarewave with blanking time. Switches Sw1 and Sw4 conduct simultaneously for a halfperiod, i.e. ,α ≈ 180 and, after the blanking time, are replaced with switches Sw2 andSw3 during the second half period. The switch pattern to the right in Figure 5.20 isa quasi- square wave, where the output voltage is controlled by changing the controlangle β (0 < β < 180). The switches Sw1 and Sw3 conduct and, thus, create a shortcircuit in the winding. After a time delay, determined by the control angle, switch Sw3is turned off and switch Sw4 is turned on in order to create a positive voltage of Utfo1

over the winding. After the controlled on-time, switch Sw1 is turned off and switchSw4 is turned on again, thus, creating a short circuit in the winding. Thereafter, thesecond half period starts and a negative voltage is applied to the winding. All switchevents are separated by a blanking time.

The amount of transferred energy is controlled by a phase shift between the two waves,the transformer ratio and the shape and size of the wave. In the following simulations,only square waves are used.

Figures 5.21 and 5.22 show the semiconductor and the transformer stresses in theconverter with the two square waves, UTFO1

and UTFO2, phase shifted 45 (ϕ =45),

according to the parameters presented in Table 5.5. The switching period is 1 ms andcan be divided into four different intervals:

97

PSfrag replacements

α β

50/50 Controlled

ωt ωt

uu

Switches inturn-on position14 23 13 1314 24 23

Figure 5.20: Output voltage and switch patterns for Bridge Converter. Left: Squarewave α = 180 and right: Quasi square wave 0 < β < 180.

1. From the time 97 ms and 45 i.e. 125 µs forward. Switches Sw1 and Sw4 conducton the primary side and switches Sw6 and Sw7 conduct on the secondary sideof the transformer, thus, creating a positive voltage on the primary side of thetransformer UTFO1

and a negative voltage on the secondary side of the transformerUTFO2

. The applied voltage over the leakage inductance increases the currentuntil switches Sw6 and Sw7 are turned off.

2. From the time 97.125 ms to 97.5 ms, i.e., the rest of the first half period. SwitchesSw1 and Sw4 conduct on the primary side and switches Sw5 and Sw8 conducton the secondary side of the transformer, thus, creating a positive voltage onboth sides of the transformer. The current through the leakage inductance, con-sequently, does not change until switches Sw1 and Sw4 are turned off.

3. From the time 97.5 ms and 45 forward. Switches Sw2 and Sw3 conduct onthe primary side and switches Sw5 and Sw8 conduct on the secondary side ofthe transformer, thus, creating a negative voltage on the primary side and apositive voltage on the secondary side. The resulting voltage decreases the currentthrough the leakage inductance until switches Sw5 and Sw8 are turned off.

4. From the time 97.625 ms to 98 ms, i.e., the rest of the second half period. SwitchesSw2 and Sw3 conduct on the primary side and switches Sw6 and Sw7 conducton the secondary side of the transformer, thus, creating a negative voltage onboth sides of the transformer. The current through the leakage inductance, con-sequently, does not change until switches Sw2 and Sw3 are turned off and thenext switching period starts.

Ideally, the voltages over and the current through the transformer windings and theswitch should be constant during intervals two and four, but the amplitudes are de-creased due to the finite size of the capacitors, Cin and Cout and the source impedance.Component values are presented in Table 5.5.

98

(A

)

−250.0

0.0

250.0

t(s)

0.097 0.0975 0.098 0.0985 0.099 0.0995 0.1

(V

)

0.0

10000.0

(A

)

−250.0

0.0

250.0

(V

)

0.0

10000.0

(A

)

−2500.0

0.0

2500.0

(V

)

0.0

1000.0

(A

)

−2500.0

0.0

2500.0

(V

)

0.0

1000.0

(A) : t(s)

i_sw6

(V) : t(s)

u_sw6

(A) : t(s)

i_sw5

(V) : t(s)

u_sw5

(A) : t(s)

i_sw2

(V) : t(s)

u_sw2

(A) : t(s)

i_sw1

(V) : t(s)

u_sw1

Figure 5.21: Semiconductor stresses in DAB converter at ϕ =45. From top: voltageover uSw1

[V], current through iSw1[A], voltage over uSw2

[V], currentthrough iSw2

[A], voltage over uSw5[V], current through iSw5

[A], voltageover uSw6

[V] and current through iSw6[A].

99

(A

)

−250.0

0.0

250.0

t(s)

0.097 0.0975 0.098 0.0985 0.099 0.0995 0.1

(V

)

−10000.0

0.0

10000.0

(A

)

−2500.0

0.0

2500.0

(V

)

−1000.0

0.0

1000.0

(A) : t(s)

i_tfo2

(V) : t(s)

u_tfo2

(A) : t(s)

i_tfo1

(V) : t(s)

u_tfo1

Figure 5.22: Transformer stresses in a DAB converter at ϕ =45. From top: voltageover uTFO1

[V], current through 1TFO1[A], voltage over uTFO2

[V], currentthrough iTFO2

[A].

Table 5.5: System parameters of Dual Active Bridge Converter.


Rated power 2 MW Blanking time Tb 5 µs

Rated input voltage 0.5-1 kV Transformer turns ratio N = n2

n1=10

Rated output voltage 10 kV Leakage inductance 25 µHSwitching frequency 1000 Hz (Lλ = L1λ + L′

2λ)

IGBT voltage drop UT0 2.8 V Input capacitor Cin 50 mFseries resistance ron 0.45 mΩ Output capacitor Cout 2 mF

Turn-on time Tsw,on 1 µs Source impedance 0.5 µH + 1 mΩTurn-off time Tsw,off 1 µs (Zsource1 = Zsource2)

100

5.4.2 Comparison with General AC Theory

The function of the transferred power of the DAB converter is similar to two generatorsconnected together with an impedance as shown in Figure 5.23. The transferred poweris [59]:

P =U1U2Z

sinϕ (5.10)

where U1 and U2 are the RMS values of the two voltages applied on the impedance. Zis the intermediate impedance and ϕ is the phase shift between the two voltages. Theresistance of the impedance is neglected and, therefore, the input and output power isthe same, according to Eq. (5.10). The reactive power transfer is calculated as [59],where Q1 is reactive power produced by generator 1 and Q2 is consumed reactive powerin generator 2.

Q1 =U21Z− U1U2

Zcosϕ (5.11)

Q2 = −U22

Z+

U1U2Z

cosϕ (5.12)

PSfrag replacements

∼ ∼Z

Generator 1 Generator 2

−→P ,−→Q1

−→P ,−→Q2

u1(t) =√2U1 sin(ωt) u2(t) =

√2U2 sin(ωt− ϕ)

u1(t) u2(t)

Figure 5.23: Two generators with an intermediate impedance(Generator 1, Impedance Z and Generator 2).

101


The DAB converter does not have sinusoidal voltages and currents, consequently, thetransferred power is calculated differently.

PSfrag replacements

ϕ

Ts

i(t)

t

∆i

Figure 5.24: Idealized input current Iin for DAB converter

Assuming that the mean value of the primary dc bus voltage Uin and the mean valueof the secondary dc bus voltage U ′

out equivalent on the primary side are equal, i.e.,Uin = U ′

out. Then the peak-to-peak current ∆i, according to Figure 5.24 will be

∆i =ϕTs(Uin + U ′

out)

2πLλ

=UinTsπLλ

ϕ (5.13)

where Lλ is the transformer leakage impedance and ϕ the phase shift between the twosquare waves. Ts is the switching time, i.e., Ts =

1fs. The average current Iin becomes

Iin =1

2∆i

Ts − ϕ

2πTs

Ts=

UinTs2πLλ

(1− ϕ

π)ϕ (5.14)

The average power Pin is equal to (since Uin is constant)

P = UinIin (5.15)

Then the transferred power is (if |ϕ| is used, Eq. (5.16) is valid for both positive andnegative values of ϕ)

102

P =U2inTs2πLλ

(1− |ϕ|π

)ϕ (5.16)

The semiconductor stresses are dependent on the applied voltages and currents. As-suming that Uin = U ′

out, the peak current for the switches is

isw =UinTs2πLλ

ϕ (5.17)

The peak current is higher for larger Lλ at constant power P . However, the derivativeof the current becomes lower. For example, if the power is 1 MW and the leakageinductance Lλ is 50 µH, the control angle becomes approximately 20 and the peakcurrent becomes approximately 1100 A. If, however, the leakage inductance is doubled(Lλ = 100 µH), the control angle increases to approximately 50 and the peak currentincreases to approximately 1400 A. Subsequently, for the same power, the peak currentwill be approximately 25% higher with a leakage inductance that is twice as large.


If the voltages at both ends, Uin and Uout, are kept constant and, moreover, the fre-quency and the impedance are constant, the transferred power will only be dependenton the phase shift, i.e., P = f(ϕ). Simulations with two different leakage inductancesLλ and control angles ϕ that change linearly are shown in Figure 5.25. Moreover,the theoretical expression in Eq. (5.16) is displayed in Figure 5.25 for the two leakageinductances. The upper curve is Lλ = 50 µH and the lower curve is Lλ = 100 µH. Thesimulated and theoretical curves deviate due to the blanking time, the turn-on andturn-off of the switches and losses in the circuit. The irregularity around zero dependson the blanking time.

If possible, the control angle ϕ should be kept small in order to avoid high peak currentsin the switches. This can, however, be difficult since the voltages Uin and Uout are oftenfixed by the surrounding system. The frequency is determined by the switches and thetransformer leakage inductance Lλ is dependent on the transformer design. The leakageinductance Lλ also controls the current derivative for the converter and, therefore, isa critical design parameter.

103

(Deg

)

−100.0

−80.0

−60.0

−40.0

−20.0

0.0

20.0

40.0

60.0

80.0

100.0

(W)

−3meg

−2meg

−1meg

0.0

1meg

2meg

3meg

t(s)

0.2 0.4 0.6 0.8 1.0 1.2

(W)

−3meg

−2meg

−1meg

0.0

1meg

2meg

3meg

(Deg) : t(s)

fi

Power L=50uH (Theoretical)

Power L=100uH (Theoretical)

(W) : t(s)

Power L=50uH

(W) : t(s)

Power L=100uH

Figure 5.25: Transmitted power as a function of control angle P = f(ϕ). Lλ = 50 µHTheoretical (dotted) and simulated (dashed), Lλ = 100 µH Theoretical(dotted) and simulated (solid).

104


Step changes from the power operating point 2 MW were made to verify the transientbehavior of the system. The reference value was changed ±10% of the operating pointfor two different values (50 µH, 25 µH) of the leakage inductances (Lλ = 100µH usedin Figure 5.25 does not allow such a high power transfer). The results are shown inFigures 5.26 to 5.28. The sampling causes a time delay of Ts. The output poweris calculated as the average power for each switching period and, therefore, createsanother time delay of Ts.

(Deg

)

50.0

55.0

60.0

(W)

1.8meg

2meg

2.2meg

(W)

1.8meg

2meg

2.2meg

t(s)

0.095 0.1 0.105 0.11 0.115 0.12 0.125 0.13 0.135 0.14 0.145

(Deg) : t(s)

fi

(W) : t(s)

Power

(W) : t(s)

Pref

Figure 5.26: Transient behavior during step changes for DAB converter using PI-controlwith feed-forward Lλ = 50 µH. From top: Power reference P ∗ [W], powerP [W] and phase angle ϕ [].

The controllers in the first two cases have the same parameters and use a feed-forwardloop, as shown in Figure 5.29 [60]. The feed-forward controller uses Eq. (5.18) to createa faster system and the PI-controller is used to correct errors since the feed-forward isnot perfect.

ϕ =π

2(Uin −

√

T 2s U2in − 8LλTsP

Ts) (5.18)

The third case, Figure 5.28, has a PI-controller without a feed-forward term. Asshown in the figures, the feed-forward controller results in a faster response than thetraditional PI-controller. More oscillatory behavior is also observed when no feed-forward is used, because the proportional constant is larger. The simulation with the

105

(Deg

)

15.0

20.0

25.0

t(s)

0.095 0.1 0.105 0.11 0.115 0.12 0.125 0.13 0.135 0.14 0.145

(W)

1.8meg

2meg

2.2meg

(W)

1.8meg

2meg

2.2meg

(Deg) : t(s)

fi

(W) : t(s)

Power

(W) : t(s)

Pref

Figure 5.27: Transient behavior during step changes for DAB converter using PI-controlwith feed-forward Lλ = 25 µH. From top: Power reference P ∗ [W], powerP [W] and phase angle ϕ [].

(Deg

)

50.0

55.0

60.0

t(s)

0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16

(W)

1.8meg

2meg

2.2meg

(W)

1.8meg

2meg

2.2meg

(Deg) : t(s)

fi

(W) : t(s)

Power

(W) : t(s)

Pref

Figure 5.28: Transient behavior during step changes for DAB converter using PI-controlLλ = 50 µH. From top: Power reference P ∗ [W], power P [W] and phaseangle ϕ [].

106

leakage inductance Lλ, equal to 25 µH, shows more oscillations than the simulationwith the leakage inductance Lλ of 50 µH due to the stronger connection betweenthe two sides, according to Eq. (5.16). Some of the problems, for example, that thepositive step and the negative step are not equal, as seen in the figures, are an effectof applying a linear controller to a non-linear system. The data of the controllers areshown in Table 5.6.

PSfrag replacementsϕ = f(P,Uin, Uout, Lλ)

∑ PIController

∑

ϕP ∗

P

+++

−

Figure 5.29: PI-controller with feed-forward to control power using dual active bridgeconverter.

Table 5.6: Controller parameters.

Parameter PI-controller with PI-controllerfeed-forward

Kp 7.5·10−6 20·10−6Ti 3 ms 3 msTs 1 ms 1 ms

5.4.6 Varying Voltage

A varying input voltage Udc1 was applied to the converter and the power P and theoutput voltage Udc2 were kept constant while the phase angle ϕ between the two bridgeswas adjusted, according to Eq. (5.19). The simulation result is shown in Figure 5.30.The controller changes the phase angle ϕ, and thus, maintains a constant transferredpower when the input voltage is changed.

P =UinU

′outTs

2πLλ

(1− ϕ

π)ϕ (5.19)

The power is constant at 1 MW while the input voltage changes from 500 V to 1000 V.The reduced input voltage increases the control angle ϕ and, thus, the peak currentfor the switches is changed, according to Eq. (5.20) (derived from Eq. 5.17).

107

(V)

500.0

750.0

1000.0

(Deg

)

0.0

20.0

40.0

60.0

80.0

(W)

900000.0

1meg

1.1meg

t(s)

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2

(V) : t(s)

Uin

(Deg) : t(s)

fi

(W) : t(s)

Power

Figure 5.30: Simulations with varying input voltage. From the top: Power P [W],phase angle ϕ [] and input voltage Uin [V].

isw =(Uin(2ϕ− π) + πU ′

out)Ts4πLλ

(5.20)

Figure 5.31 shows transformer stresses at 750 V input voltage. The peak current ofthe transformer is the same as for 2 MW and 1000 V input voltage despite the powerbeing halved to 1 MW. The switch stresses at different voltages at 1 MW and constantoutput voltage are shown in Table 5.7. If the input voltage is halved to 500 V, the peakcurrent increases by more than three times. The maximum current for the switches isreached earlier and, therefore, the power capability of the converter is reduced whenthe input voltage is reduced, according to Eq. (5.19).

108

t(s)

0.197 0.198 0.199 0.2

(A)

−200.0

0.0

200.0

(V)

−10000.0

0.0

10000.0

(A)

−2000.0

0.0

2000.0

(V)

−750.0

0.0

750.0

(A) : t(s)

i_tfo2

(V) : t(s)

u_tfo2

(A) : t(s)

i_tfo1

(V) : t(s)

u_tfo1

Figure 5.31: Transformer stresses in a DAB converter with 750 V input voltage. Fromtop: voltage winding one uTFO1

[V], current winding one 1TFO1[A], voltage

winding two uTFO2[V], current winding two iTFO2

[A].

Table 5.7: Phase angle and switch stresses at different input voltages.

Udc1 ϕ isw1= itfo1

500V 49.7 3880A750V 28.5 2440A1000V 20.3 1130A

109

5.4.7 Summary

The converter operation can be described as two generators connected together via animpedance where the transferred power is determined by the phase angel between thevoltages, with the exception that the DAB converter does not have sinusoidal voltages.If possible, the phase angle should be kept small to avoid high switch currents forthe switches. As with the Half and Full Bridge Converters the leakage inductanceLλ and the turns ratio for the transformer are the most important parameters of theconverter. Leakage inductance influences current ripple and the maximum transferablepower. The turns ratio determines the voltage levels for the converter. Feed-forwardcontrol can be used and the converter can handle dynamic changes in the transmittedpower. The controller keeps the transferred power constant even if the input voltage isdecreased to half the rated voltage. However, a reduced input voltage increases switchcurrents.

110

Chapter 6

Wind Farm Simulation

In this chapter, wind farm simulations are performed. The wind turbines, cables andDC/DC converters are put together to form the electrical system of a wind farm. Inaddition, the proposed transmission system layout is presented. Different faults andfault locations are simulated for a simplified model of the farm and finally simulationswith the complete wind farm are conducted.

6.1 Transmission System Layout for Wind Farms

An example of a transmission system layout for a wind farm is shown in Figure 6.1.The system uses the Boost Converter as a voltage adjuster and a Full-Bridge converteras a DC transformer. For simplification, the feeder can be modelled as a currentsource, dependent on the output power to the wind turbine generator. In addition,the connection to the transmission cable can be modelled as a constant voltage source.With reference to Figure 6.1 the low voltage is the rectified generator output, which isnormally 0.5 to 1 kV and the medium voltage is about 2.5 kV. The high transmissionvoltage depends on the output power of the wind farm and the transmission distance,but 150 kV is realistic for 100-150 MW and, therefore, in comparison with Figure 6.1,an extra converter stage is needed in this case.

Boost

Converter

Controller

Variable low

voltage

Medium

voltage

Current

source

U*

I1

I2

U1

-

+ ==

Full Bridge

Converter

Controller

High

voltage

Voltage

source

U*

I3

==

Voltage

adjustment

converter

DC/DC

transformer

U2

-

+U

3-

+

=

Figure 6.1: Example layout of transmission system with converters and control system.

111

6.2 Faults

In this section, different fault types and fault locations are investigated and simulationresults are shown in order to analyze the behavior of the wind farm when faults occur.The faults are injected into the cables and two different fault types and four differentfault locations are simulated.

6.2.1 Test Setup

The schematic layout of the simplified transmission system is shown in Figure 6.3. ABoost Converter and a Full Bridge Converter are connected together with cables andsix different faults are applied. The Boost Converter is controlled for a rated powerof 2 MW with an input voltage U1 of 1 kV and an output voltage U2 of 2.5 kV. TheFull Bridge Converter is controlled as a DC transformer with a rated power of 5 MWand, thus, is not fully utilized, i.e., the converter is operated at part load. The inputvoltage U3 for the Full Bridge Converter is 2.5 kV and the output voltage U4 is 25 kV(compared with 150 kV for the proposed system).

The cables are represented by a π-equivalent model, according to Figure 6.2 [59]. Cabledata are presented in Table 6.1. Faults are injected between the four cables, one at atime, at ground or between two of the cables from the same bus. The fault locationis in the middle of the cable, as indicated in the figure, and fault impedance is set to10 mΩ [61].

PSfrag replacements

LL RR

C2

C2

C2

C2

Fault

Figure 6.2: Equivalent π-scheme with fault location.

Table 6.1: π-equivalent cable data.

Cable Length Conductor R L C2

1.25kV/800A DC 1 km 800 mm2 37.5 mΩ 0.43 mH 0.375 µF

12.5kV/400A DC 5 km 240 mm2 625 mΩ 2.4 mH 1.1 µF

112

I2

+

-

Usw

ID

- +U

D

IS

W

+ -UL

ILL D

Sw Cout

Cin

+

-

U1

ZSource1

Udc1

+

-

Iout

+

-

Uout

ZSource2

Udc2

+

-

1.25 kV

1.25 kV 12.5 kV

12.5 kV

+

-

U2

+

-

U3

+

-

U4

Iin

2

1

4

3

5 6

I3

I4

+

-

USW4

IS

W4

n1

n2

+

-U

SW3

IS

W3

+

-U

SW1

IS

W1

+

-

USW2

IS

W2

Sw1

Sw2

Sw3

Sw4

L1l

L2l

+

-

UD4

ID

4

+

-

UD3

ID

3+

-

UD1

ID

1

+

-

UD2

ID

2

D1

D2

D3

D4

ITfo1

ITfo2

+

-

UTfo2

+

-

UTfo1

C1

C2

C3

C4

Figu

re6.3:

Schem

aticlayou

tof

simplifi

edtran

smission

system

with

faultlocation

s.

113

6.2.2 Line to Ground Faults

Fault type 1 is a line to ground fault located on the cable with negative polarity be-tween the two converters, as indicated in Figure 6.3. The results from the simulationsare shown in Figure 6.4. The fault creates a voltage step at Boost Converter sec-ondary voltages U2+ and U2− . Since the Boost Converter has no galvanic isolationthe voltage disturbance is transferred through the converter and affects the input ofthe converter, i.e., both U1+ and U1− jump 1250 V. The converters are shut down forover/undervoltages (rated voltage ±20%) after approximately 2 ms and the boost in-ductor current iL is discharged. Since the input voltage to the DC-transformer dropsto half the power, transfer will stop and the secondary side of the DC-transformer,therefore, will not be affected by the short circuit.

t(s)

0.49 0.4925 0.495 0.4975 0.5 0.5025 0.505 0.5075 0.51 0.5125 0.515 0.5175 0.52 0.5225 0.525

(V

)

−1250.0

0.0

1250.0

(V

)

0.0

1250.0

2500.0

(V

)

−2500.0

−1250.0

0.0

(V

)

0.0

1250.0

2500.0

(A

)

0.0

1000.0

2000.0

(A

)

0.0

1000.0

2000.0

M

Fault

(V) : t(s)

FB : U1−

(V) : t(s)

FB : U1+

(V) : t(s)

Boost : U2−

(V) : t(s)

Boost : U2+

(A) : t(s)

Boost : i_L

(A) : t(s)

Boost : i_sw

Figure 6.4: Simulated voltages and currents during fault type 1. From top: Faultinjection, boost switch current isw [A], boost inductor current iL [A], boostsecondary side positive bus voltage U2+ [V], boost secondary side negativebus voltage U2− [V], full bridge primary side positive bus voltage U3+ [V]and full bridge primary side negative bus voltage U3− [V].

114

Fault type 2 is a line to ground fault located on the cable with positive polarity be-tween the two converters, as indicated in Figure 6.3. The results from the simulationsare shown in Figure 6.5. The fault creates a voltage drop at Boost Converter sec-ondary voltages U2+ and U2− . Again the voltage disturbance is transferred throughthe converter and both U1+ and U1− will drop 1250 V. Consequently, U1− will dropto approximately -2500 V. The converters are shut down for over/undervoltages af-ter approximately 2 ms and the boost inductor current iL is discharged. Since theinput voltage to the DC-transformer drops to half the power, transfer will stop andthe secondary side of the DC-transformer, therefore, will not be affected by the shortcircuit.

t(s)

0.49 0.4925 0.495 0.4975 0.5 0.5025 0.505 0.5075 0.51 0.5125 0.515 0.5175 0.52 0.5225 0.525

(V

)

−2500.0

−1250.0

0.0

(V

)

−1250.0

0.0

1250.0

(V

)

−2500.0

−1250.0

0.0

(V

)

0.0

1250.0

2500.0

(A

)

0.0

1000.0

2000.0

(A

)

0.0

1000.0

2000.0

M

Fault

(V) : t(s)

FB : U1−

(V) : t(s)

FB : U1+

(V) : t(s)

Boost : U2−

(V) : t(s)

Boost : U2+

(A) : t(s)

Boost : i_L

(A) : t(s)

Boost : i_sw

Figure 6.5: Simulated voltages and currents during fault type 2. From top: Faultinjection, boost switch current isw [A], boost inductor current iL [A], boostsecondary side positive bus voltage U2+ [V], boost secondary side negativebus voltage U2− [V], full bridge primary side positive bus voltage U3+ [V]and full bridge primary side negative bus voltage U3− [V].

115

Fault type 3 is a line to ground fault located on the cable with negative polarity be-tween the DC-transformer and source two, as indicated in Figure 6.3. The results fromthe simulations are shown in Figure 6.6. The fault creates a current surge in the conv-erter and the converter is, therefore, shut down for over currents after approximately1 ms. Since the DC-transformer has galvanic isolation, the voltage disturbance is nottransferred through the converter and the rest of the system , therefore, will not beaffected. The converters, however, are shut down for under voltage and the boost in-ductor current iL has to be discharged. This causes the secondary voltage U2 of theboost converter to rise to approximately 3000 V.

t(s)

0.4975 0.5 0.5025 0.505 0.5075 0.51 0.5125 0.515 0.5175 0.52

(A

)

0.0

1000.0

2000.0

(V

)

2500.0

3000.0

3500.0

(A

)

0.0

2000.0

4000.0

(A

)

0.0

2000.0

4000.0

(V

)

−12500.0

0.0

(V

)

12250.0

12500.0

12750.0

M

Fault

(A) : t(s)

Boost : i_L

(V) : t(s)

Boost : U2

(A) : t(s)

FB : i_sw2

(A) : t(s)

FB : i_sw1

(V) : t(s)

FB : U2−

(V) : t(s)

FB : U2+

Figure 6.6: Simulated voltages and currents during fault type 3. From top: Faultinjection, full bridge secondary side positive bus voltage U4+ [V], full bridgesecondary side negative bus voltage U4− [V], full bridge switch one currentisw1

[A], full bridge switch two current isw2[A], boost secondary side bus

voltage U2 [V] and boost inductor current iL [A].

Fault type 4 is a line to ground fault located on the cable with positive polarity betweenthe DC-transformer and source two, as indicated in Figure 6.3. The system behavioris the same as in fault case 3 with the only difference that the secondary voltages U4+and U4− of the DC-transformer are affected in opposite directions, i.e., U4+ becomezero, and U4− is more or less constant.

116

6.2.3 Line to Line Faults

Fault type 5 is a line to line fault located on the cables between the two converters, asindicated in Figure 6.3. The results from the simulations are shown in Figure 6.7. Thefault causes the voltage to drop to the critical level where the output diode in the BoostConverter will start to conduct and, thus, short circuit source one. Consequently, thefault can only be removed by shutting down the source Udc1 . As before, the fault doesnot affect the secondary side of the full bridge.

(V

)

−12700.0

−12600.0

−12500.0

−12400.0

(V

)

12400.0

12500.0

12600.0

12700.0

(V

)

−1250.0

0.0

1250.0

(V

)

−1250.0

0.0

1250.0

(A

)

0.0

2000.0

4000.0

6000.0

(A

)

0.0

1000.0

2000.0

t(s)

0.49 0.4925 0.495 0.4975 0.5 0.5025 0.505 0.5075 0.51 0.5125 0.515 0.5175 0.52 0.5225 0.525

M

Fault

(V) : t(s)

FB : U2−

(V) : t(s)

FB : U2+

(V) : t(s)

Boost : U2−

(V) : t(s)

Boost : U2+

(A) : t(s)

Boost : i_L

(A) : t(s)

Boost : i_sw

Figure 6.7: Simulated voltages and currents during fault type 5. From top: Faultinjection, boost switch current isw [A], boost inductor current iL [A], boostsecondary side positive bus voltage U2+ [V], boost secondary side negativebus voltage U2− [V], full bridge secondary side positive bus voltage U4+ [V]and full bridge secondary side negative bus voltage U4− [V].

117

Fault type 6 is a line to line fault located on the cables between the DC-transformerand the source Udc2 , as indicated in Figure 6.3. The results from the simulations areshown in Figure 6.8. The fault behaves like the line to ground fault even if the currentderivatives increase and both secondary voltages U4+ and U4− drop to zero. Thus, thefault creates a current surge in the converter and the converter, therefore, is shut downfor over currents after approximately 1 ms. Since the DC-transformer has galvanicisolation, the voltage disturbance is not transferred through the converter and the restof the system, therefore, is not affected. The converters, however, are shut down andthe boost inductor current iL has to be discharged. This causes the secondary voltageU2 of the boost converter to rise to approximately 3000 V.

(A

)

0.0

1000.0

2000.0

t(s)

0.49 0.4925 0.495 0.4975 0.5 0.5025 0.505 0.5075 0.51 0.5125 0.515 0.5175 0.52 0.5225 0.525

(V

)

2500.0

3000.0

3500.0

(A

)

0.0

2000.0

4000.0

(A

)

0.0

2000.0

4000.0

(V

)

−12500.0

0.0

(V

)

0.0

12500.0

M

Fault

(A) : t(s)

Boost : i_L

(V) : t(s)

Boost : U2

(A) : t(s)

FB : i_sw2

(A) : t(s)

FB : i_sw1

(V) : t(s)

FB : U2−

(V) : t(s)

FB : U2+

Figure 6.8: Simulated voltages and currents during fault type 6. From top: Faultinjection, full bridge secondary side positive bus voltage U4+ [V], full bridgesecondary side negative bus voltage U4− [V], full bridge switch one currentisw1

[A], full bridge switch two current isw2[A], boost secondary side bus

voltage U2 [V] and boost inductor current iL [A].

118

6.2.4 Summary

Depending on the type of fault and its location on the system, different parts of thelocal wind farm transmission system have to be shut down. If the fault occurs betweenthe two converters, both converters have to be shut down in order to isolate the fault.Faults between the DC-transformer and the source Udc2 can only be isolated by shuttingdown the complete system, including the source Udc2 . Faults between the source Udc1

and the boost converter have not been simulated here, but in such a case, the fault canbe isolated by shutting down the source Udc1 and the Boost Converter.

The most severe fault that can occur in this example, consequently, is a short circuit onthe secondary side of the boost converter cannot be shut off and leads to a permanentfault, which has to be shut off by shutting down the feeding source Udc1 . the bigdifference between the two converters. The DC-transformer can isolate the system forall faults by just turning off the switches.

119

6.3 Simulation of Complete Wind Farm

The principal layout of a 50 MW wind farm is shown in Figure 6.9. It consists of25 wind turbines each with a rated power of 2 MW.

=

=

==

==

==

==

==

Wind turbine with rectifier

and boost converter

0.5-1 / 2.5 kV DC

Two half bridge converters

25 / 150 kV DC

Two full bridge converters

2.5 / 25 kV DC

Transmission Cable

150 kV DC

Figure 6.9: Example of wind farm layout.

As shown in Figure 6.10, each generator is connected to a rectifier and a Boost Conv-erter and each group of five generators are connected to two parallel Full Bridge Con-verters. The Full Bridge Converters are then connected to two Half Bridge Converters.The converters in parallel are only used if the input power is above 50 % of ratedpower and, thus, the no-load losses are reduced. The configuration also gives a smallredundancy.

The wind is increased from the cut-in wind speed 3 m/s up to the rated wind speed10 m/s and the generated power, therefore, is increased from 0 to 50 MW. Resultsfrom the simulation are shown in Figure 6.11. Bus voltages change with differentwind speed and, thus, different power flow. At 8 m/s (i.e., after 0.55 s) the parallelconverters are started and this causes a voltage drop on the 2.5 kV bus, the 25 kV busand the 150 kV bus, since the two converters in parallel have a lower voltage drop. Theconnection of the parallel converters also causes some oscillations between the 25 kVand the 150 kV buses. The generator starts at 500 V and increases to 1000 V at ratedpower. The 2.5 kV bus starts at 2.2 kV and increases to approximately 2.4 kV. The25 kV bus changes between 24 and 25 kV and,finally, the 150 kV bus increases from150 kV to 152 kV.

120

Local wind

farm sub grid

2.5 kV DC

25/150 kV DC/DC

converter

Cable transmission

2 x 75 kV DC

Wind Turbine Shore

Connection point

on land

DC/AC converter

on land

=

~~

G =~~

G =~~

G =~~

Local wind

farm grid

25 kV DC

2.5/25 kV DC/DC

converter

==

==

==

==

Figure 6.10: Schematic layout of complete wind farm.

(V

)

500.0

750.0

1000.0

t(s)

0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85

(V

)

500.0

750.0

1000.0

(V

)

2200.0

2300.0

2400.0

(V

)

24000.0

25000.0

(V

)

150000.0

151000.0

152000.0

(M

W)

0.0

1.0

2.0

(m

/s)

3.0

6.0

9.0

(V) : t(s)

Voltref

(V) : t(s)

Ugen

(V) : t(s)

2.5 kV_Bus

(V) : t(s)

25 kV_Bus

(V) : t(s)

150 kV_Bus

(MW) : t(s)

Power

(m/s) : t(s)

Wind

Figure 6.11: Steady state behavior of a 50 MW wind farm from cut-in wind speed torated wind speed. From top: Wind speed [m/s], turbine power [MW],voltage at 150 kV bus [V], voltage at 25 kV bus [V], voltage at 2.5 kV bus[V], generator voltage [V] and generator reference voltage [V].

121

6.4 Results

The system investigated here uses Boost Converters as voltage adjusters and Full BridgeConverters as DC transformers. The bus voltages change with different wind speedsand thus, different power flow when the input power is changed from zero to ratedpower. The control system keeps the generator voltage at the reference voltage.

Depending on the type of fault, different parts of the transmission system have to beshut down. The DC transformer can stop the system and limit all faults by just turningoff the switches, while a short circuit on the secondary side of the Boost Convertercannot be isolated and leads to a permanent fault.

The conversion losses are reduced by parallel converters, which also provides a smallredundancy. The connection of the parallel converters causes some oscillations but theoscillations can be reduced by changing the control system.

122

Chapter 7

Conclusions and Future Research

7.1 Conclusions

DC/DC converters have been examined by many authors before although primarilyin low power and low voltage applications. This thesis focuses on DC/DC convertersfor high voltage and high power. When the first part of the project started up, theobjective was to determine the potential for using DC grids in a wind farm. Anotherobjective was to investigate different configurations of electrical systems for offshorewind farms. The objective of the second part of the project was to investigate differentDC/DC converter topologies with respect to a wind farm application and present designconcepts with different high voltage and high power converter topologies.

For a high voltage and high power application, like a wind park transmission system, thefocus has to be on the utilization factor of the used components. Designing a DC/DCconverter for an input voltage down to half the rated voltage or with a high input tooutput voltage ratio, decreases performance at rated power and voltage. Therefore,converters have been analyzed for two different applications. One simple converter canbe used for a first adjustment of the voltage and then a second converter can be usedto raise the voltage to a suitable transmission level. The semiconductor componentstresses have been determined by a theoretical comparison of the different converters.A Boost Converter is suitable as a voltage adjustment converter and a Bridge Convertercan be used as a DC transformer. Losses in the Boost and Full Bridge Convertershave also been estimated. Simple analytical expressions have been derived in order todetermine the transferred power. However, the losses and the transition times in theconverters yield significant errors in the transmitted power.

Boost and Dual Active Bridge (DAB) Converters can handle both varying input voltageand dynamic changes in the transmitted power. For the Boost Converter, the design ofthe inductor size is a compromise between the size of the input ripple and the speed ofthe dynamic response for the converter. For the DAB converter, the leakage inductanceLλ for the transformer is the most important parameter of the converter.

Half and Full Bridge DC transformers can handle dynamic changes in the transmittedpower but are not able to control the transferred power. The leakage inductance Lλ

and the turns ratio for the transformer are the most important parameters for the

123

converters.

The proposed system uses a Boost Converter as a voltage adjuster and a Full BridgeConverter as a DC transformer. System fault analysis with various faults and faultlocations have been conducted. Depending on the type of fault and its location, differ-ent parts of the local wind farm transmission system have to be shut down. If a faultcan not be isolated by the DC/DC converters, the system has to ashore that such faultbecomes isolated by other means. The DC transformer can isolate the system for allfaults simply by turning off the switches. However, a short circuit on the secondaryside of the Boost Converter cannot be shut off and leads to a permanent fault, whichhas to be isolated by shutting down the feeding source.

Finally, a complete wind farm with 25 wind turbines was simulated. The wind and,therefore, the produced power was increased up to a rated power of 50 MW. Thesystem shows good system performance when the different bus voltages change withdifferent wind speeds and, thus, different power flows. Wind farms with DC-grid isfeasible in the future according to a rough cost and loss estimation. It is also a logicalcontinuation when the experiences of todays wind farms with AC transmission havebeen fully digested and the transmission distances are greater than today.

7.2 Future Research

To reduce the losses in high power and high voltage DC/DC converters, soft switchingand resonance converter topologies can be used. However, with variations in the input-output voltage ratio and with varying load, the topology design is a great challenge.Different control strategies for operation of parallel converters, for different types offault protection and to optimize the energy production are essential to maintain reliableoperation and thus support the network on shore with energy. The control strategiesis also effected by the need of communication between the converters for protectionpurposes.

An area of research itself, is high frequency, high power and high voltage transformers,which are an essential component in DC/DC converters for wind farms. Mechanismsfor the losses, stray capacitances in the windings and the leakage inductance of thetransformer have to be determined. Another issue is the insulation problems with highvoltage derivatives from the switching actions in the converters.

It is important to validate the obtained simulated results by building an experimentalsystem in the laboratory. The system should involve the DC network of the wind farmand promising DC/DC converter topologies.

124

References

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[2] A. Isurin and A. Cook, “A novel resonant converter topology and its application”,in IEEE 32nd Annual Power Electronics Specialists Conference, PESC 2001, Van-couver, Canada, 2001, vol. 2, pp. 1039 – 1044.

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Abbreviations

Abbreviation Description

CCM Continuous conduction modeDAB Dual Active BridgeDCM Discontinuous conduction modeHB Half Bridge ConverterHBVD Half Bridge Converter with voltage doublerFB Full Bridge ConverterFBVD Full Bridge Converter with voltage doublerFF Feed-forward controllerPCC Point of common couplingPFC Power factor correctionPI Proportional, Integral controllerPWM Pulse width modulationRMS Route mean square valueSEPIC Single Ended Primary Inductance ConverterWTG Wind turbine generatorZVS Zero voltage switchingZCS Zero current switchingZVSCS Zero voltage zero current switching

131

Symbols

Symbol Description Unit

α Primary control angle

β Secondary control angle

∆IL Inductor peak-to-peak ripple current Aϕ Phase shift

Cin Input capacitance FCout Output capacitance FCx Capacitor number x FD Duty ratioDx Diode number x

iDxDiode x peak current A

IDxDiode x current A

IDx,RMS Diode x RMS current AIin Primary DC bus current AIout Secondary DC bus current A

iSwx Switch x peak current AISwx Switch x current AISwx,RMS Switch x RMS current AItfo1 Transformer primary side current VItfo2 Transformer secondary side current VKp Proportional gainLλ Total transformer leakage inductance HLλ1

Primary side leakage inductance HLλ2

Secondary side leakage inductance HLx Inductor number x Hn1 Primary winding turnsn2 Secondary winding turnsP Active power WP ∗ Active power reference WQ Reactive power WSwx Switch number xTi Integral time constantTs Switch period s

133

Symbols cont.

Symbol Description Unit

Tsw,on Switch turn on time sTsw,off Switch turn off time suDx

Diode x peak voltage VUDx

Diode x voltage VUdc1 Primary source voltage VUdc2 Secondary source voltage VUin Primary DC bus voltage VUout Secondary DC bus voltage VuSwx Switch x peak voltage VUSwx Switch x voltage VUtfo1 Transformer primary side voltage VUtfo2 Transformer secondary side voltage VZsource1 Primary source impedance ΩZsource2 secondary source impedance Ω

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