Design, Control, Simulation and Energy Evaluation of a DC ... · Design, Control, Simulation and...
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Design, Control, Simulation and Energy Evaluation of a DC Offshore Wind Park André Madeira Marques Dissertação para obtenção do Grau de Mestre em Engenharia Electrotécnica e de Computadores Júri: Presidente: Prof. Paulo José da Costa Branco Orientador: Prof. José Fernando Alves da Silva Vogal: Prof. João José Esteves Santana Setembro de 2009
Transcript of Design, Control, Simulation and Energy Evaluation of a DC ... · Design, Control, Simulation and...
Design, Control, Simulation and Energy Evaluation of a DC Offshore Wind Park
André Madeira Marques
Dissertação para obtenção do Grau de Mestre em
Engenharia Electrotécnica e de Computadores
Júri:
Presidente: Prof. Paulo José da Costa Branco Orientador: Prof. José Fernando Alves da Silva Vogal: Prof. João José Esteves Santana
Setembro de 2009
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Acknowledgments
First, I would like to thank my supervisor in Chalmers University of Technology in Göteborg, Dr.
Torbjörn Thiringer. I would also like to thank my VESTAS contacts, Dr. Lars Helle. Also to help finishing my
supervisor in Portugal, Dr. Fernando Silva.
I also would like to thank my friends that gave me support and fun moments in 4 years in Técnico and
in the last one in ERASMUS in Chalmers, it was really a pleasure being with you!
Also I want to thank my mother, father and brother for giving me support, love and advice.
André Marques
Lisbon
September, 2009
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Resumo
Esta tese investiga as capacidades de controlo e produção de energia de um parque eólico de 200MW
DC, situado a 300km da costa. O dimensionamento e o controlo de velocidade e binário da Gerador Síncrono de
Ímans Permanentes (PMSG) é feito para 3 conversores: o conversor elevador, o conversor de ponte completa e o
Rectificador Trifásico Activo (ATR). O controlo de tensão e corrente é explicado para o Conversor de Tensão
(VSC) em terra que entrega potência para a rede. Simulações são feitas usando o MATLAB/Simulink. Cálculos
de Perdas são feitos para todas as velocidades de vento operáveis.
O tempo de resposta do conversor elevador é 8ms, 5ms para o ATR e 20ms para o conversor de ponte
completa. Usando 12kV como tensão de saída, o conversor com menos perdas é o ATR seguido pelo conversor
elevador (ambos com 1%). Se a tensão de saída for 60kV, o melhor conversor é o de ponte completa com
controlo de factor de ciclo (1.62% à potência nominal).
O controlo de corrente do VSC em terra é feito desacopulando Id e Iq para controlar a tensão do cabo
submarino DC e a potência reactiva entregue à rede independentemente. O sistema responde rápido e aguenta
perturbações grandes e pequenas.
A melhor topologia para a rede DC é ligando 5 turbinas em paralelo para 12kV, depois subir 60kV, e
depois para 200kV, onde o cabo submarino transporta a energia para terra.
Palavras-chave: Parque Eólico DC offshore, conversor elevador, conversor de ponte completa,
Rectificador Trifásico Activo, Conversor de Tensão, Gerador Síncrono de Ímans Permanentes.
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Abstract
This thesis investigates the control capabilities and energy production of a 200MW DC wind park,
placed 300km offshore. The design as well as the torque and speed control of the Permanent Magnet
Synchronous Generator (PMSG) is done for three converters: the boost converter, the full bridge, and the Active
Three-phase Rectifier (ATR). The voltage and current control is explained for the Voltage Source Converter
(VSC) onshore that delivers power to the grid. Simulations are made using MATLAB/Simulink. Loss
calculations are done for all wind speeds operable.
The response time for the boost converter is 8ms, for the ATR is 5ms and 20ms for the full bridge.
Using 12kV as output voltage, the converter with less losses is the ATR followed by the boost converter (both
around 1%). Using 60kV as output voltage the best converter is the full bridge using duty cycle control (1.62%
at rated power).
The current control of the VSC onshore is done by decoupling Id and Iq to control the voltage in the DC
submarine cable and the reactive power to the grid independently. The system responds fast and able to
withstand small and large perturbations.
The best topology for the DC grid is connecting 5 turbines in parallel to 12kV, then to 60kV, then to
200kV, where a submarine cable transports the energy to shore.
Keywords: DC offshore wind park, boost converter, full bridge converter, Active three-phase rectifier, Voltage
Source converter, Permanent Magnet Synchronous Generator.
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Table of Contents Acknowledgments ..................................................................................................................2 Resumo ..................................................................................................................................3 Abstract..................................................................................................................................4 Table of Contents ...................................................................................................................5 Symbols .................................................................................................................................8 Glossary ...............................................................................................................................11 List of Figures ......................................................................................................................12 Chapter 1 – Introduction.......................................................................................................14
1.1. Problem background – Why wind power?..................................................................14 1.2. Why study offshore wind turbines? ............................................................................14 1.3. Why develop DC wind farms? ...................................................................................14 1.4. Layout of the Report ..................................................................................................15 Chapter 2 – Background Theory and park specifications...................................................16 2.1. Aerodynamic principals of Wind Turbines .................................................................16
2.1.1. Power from the Wind ..........................................................................................16 2.1.2. Mechanical Power Extracted from the Wind........................................................17 2.1.3. Blade Pitching System.........................................................................................18 2.1.4. Pitch control........................................................................................................18 2.1.5. Model of the turbine and gearbox........................................................................19 2.1.6. Model of the generator ........................................................................................20
2.2. The problem specifications ........................................................................................21 2.2.1. Wind Park Specifications ....................................................................................21 2.2.2. Characteristics of the turbine ...............................................................................21 2.2.3. Characteristics of the Generator...........................................................................21 2.2.4. Characteristics of the Gear Box ...........................................................................22 2.2.5. Characteristics of the Transformers .....................................................................22
Chapter 3 – Control Fundamentals........................................................................................23 3.1. – Method of dominant pole ........................................................................................23
3.1.1. – Setting ξ ...........................................................................................................24 3.1.2. – Setting the bandwidth of the system .................................................................24
3.2. Symmetry criterion method........................................................................................24 3.3. Method of Ziegler Nichols .........................................................................................26
Chapter 4 – Torque control solutions ....................................................................................27 4.1. Boost converter connected to PMSG..........................................................................27
4.1.1. Description..........................................................................................................28 4.1.2. Sizing..................................................................................................................28 4.1.3. Control................................................................................................................28 4.1.4. Simulation Results: .............................................................................................30
4.2. Full Bridge Converter connected to the PMSG...........................................................32
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4.2.1. Description..........................................................................................................32 4.2.2. PMSG, transformer and full bridge converter ......................................................33 4.2.3. Sizing the output inductor and input capacitor .....................................................34 4.2.4. Control and determination of the controller parameters .......................................34 4.2.5. Simulation results................................................................................................36
4.3. PMSG with a ATR.....................................................................................................39 4.3.1. Determination of the LATR and Udc.......................................................................39 4.3.2. Control and Simulation .......................................................................................41
4.4. Efficiency calculation of the three previous converters...............................................42 4.4.1. 60kV Boost converter..........................................................................................42 4.4.2. 60kV Full Bridge ................................................................................................44 4.4.3. 60kV ATR ..........................................................................................................45
4.5. Conclusion:................................................................................................................46 Chapter 5 – Speed and pitch control of the turbine................................................................48
5.1. Description: Presentation of the matlab model used ...................................................48 5.1.2. Block “Turbine Model”.......................................................................................49 5.1.3. Block “Drive Train” ............................................................................................49 5.1.4. Block “PMSG and ATR” ....................................................................................49 5.1.5. Block “Speed Control”........................................................................................49
5.2. Simulation .................................................................................................................50 5.3. Conclusion.................................................................................................................51
Chapter 6 - Control of the main inverter onshore ..................................................................52 6.1. Block diagram of the Plant .........................................................................................52 6.2. Inner Current Control.................................................................................................54
6.2.1. Determination of the voltage level on the primary side of the transformer ...........56 6.2.2. Determination of the PI parameters for the current control...................................57
6.3. Simulation of the system with current control ............................................................57 6.3.1. Test with Id and Iq coupled ..................................................................................59
6.4. Voltage control ..........................................................................................................60 6.5. Simulation of the voltage control for large perturbations ............................................61 6.6. Simulation of the voltage control for small perturbations ...........................................62 6.7. Conclusion.................................................................................................................63
Chapter 7 – Analysis of the connection of the wind park.......................................................64 7.1. Best Connection: Parallel or Series?...........................................................................64
7.1.1. Wake effect .........................................................................................................66 7.2. Option 1: 6kV PMSG with ATR to 12kV...................................................................66
7.2.2. Loss calculation ..................................................................................................66 7.3. Calculation of the cable length ...................................................................................70
7.3.1. Length and resistance for 12kV cables ................................................................70 7.3.2. . Length and resistance for 60kV cables...............................................................71 7.3.3. Length and resistance for the main cable .............................................................73
7.4. Option2: 10kV PMSG with Full Bridge Converter to 60kV .......................................73 7.5. Energy production of the park....................................................................................74 7.6. Conclusion.................................................................................................................75
Conclusion: ..........................................................................................................................76 Future Work .....................................................................................................................77
References............................................................................................................................77 Appendix..............................................................................................................................79
Sizing of the components for the boost converter..............................................................79 Sizing of the components for the full bridge converter ......................................................79
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Sizing of the components for 2MW ATR..........................................................................80 Design and current control of the 2MW 60kV boost converter ......................................81 Design and current control of the 2MW 60kV full bridge converter ..............................82 Design and current control of the 2MW 60kV ATR ......................................................83 Parameters for the VSC onshore ...................................................................................83 Efficiency of the 2MW 12kV VSC and 200MW 200kV VSC onshore ..........................84 Efficiency of the 10MW 12/60kV and 200MW 60/200kV full bridge converters ..........84 Design, Control and efficiency evaluation of the 2MW 60kV boost converter with transformer ...................................................................................................................86 Design of the transformer for the full bridge converter..................................................86 Design of the Inductor ..................................................................................................88
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Symbols Bm rotating damping coefficient [Nmsrad-1]
c scale parameter of the Weibull distribution
Ci capacitance input capacitor in the full bridge converter [F]
Conshore capacitance to be used onshore [F]
Ctrl(s) transfer function representing the controller dynamics
Cp Power coefficient
Depth_buried depth of the buried cable [m]
Depth_sea depth of the sea [m]
ed, eq voltages e1, e2, e3 in the dq frame [V]
E Energy [J]
Em peak line to ground voltage of the PMSG [V]
e1, e2, e3 RMS voltage line to ground primary transformer onshore [V]
Eproduced energy produced by the park in one year [Wh]
Err reverse recovery energy of the diodes [J]
f frequency of the PMSG [Hz]
fcom switching frequency [Hz]
ferated rated electrical speed of the PMSG [Hz]
( )vf Probability density function of the wind
F(s) function transfer between V(s) and P(s)
Ia current in phase a in the PMSG [A]
id, iq currents i1, i2, i3 in the dq frame [A]
IDC DC current that enters the VSC onshore [A]
Idiode current in the diode [A]
IIEGT current in the IEGTs [A]
Ig current from the submarine cable [A]
Iload current in the load of the converter [A]
Iin input current [A]
Iqnom rated current in the q axis [A]
Iout output current [A]
Isec current of the secondary of the transformer [A]
Igroup1,2 current of the 12kV cables of group1,2 [A]
iL current in the inductor [A]
∆iL current variation in the inductor [A]
i1, i2, i3 currents from the VSC to the grid [A]
Jeq inertia seen from the generator side [kgm2]
Jg inertia from the generator [kgm2]
Jw inertia from the turbine [kgm2]
K factor relating the time and fall times and switching losses
k shape parameter of the Weibull distribution
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Kd derivative constant of the PID controller
Ki integral constant of the PI or PID controller
Kp proportional constant of the PI or PID controller
KPcrit critical proportional constant
Kv modulator gain
L inductance of the PMSG [H]
LATR inductance in the AC side in the ATR [H]
L(opt6) length of the cable in option6 [m]
L(12kVopt1) length of the 12kV cables for option1 [m]
L(60kVopt1) length of the 60kV cables for option1 [m]
edgeL5 Length of the 12kV cables of group 1 [m]
Lboost inductance of the inductor in the boost converter [H]
Lcable cable length [m]
Ld, Lq inductance of the generator in the dq-axis frame [H]
Length_cable_buried Length of the cable buried [m]
Lload output inductor inductance in the full bridge converter [H]
LG1 cable length in group1 [m]
Ltrans inductance of the transformer [H]
m& Mass flow rate [kg/s]
Mod(s) transfer function of the modulator dynamics
n transformer ratio in the full bridge converter
ng speed ratio of the gear box
np number of pole pairs in the generator
ns number of semiconductors in series
p pressure of the air [Pa]
P Power [W]
Pavg average power of the wind park [W]
PCDiodes conduction losses in the diodes [W]
PCIEGT conduction losses in the IEGT [W]
Pfriction friction power [W]
Pmain_cable ohmic losses in the main cable [W]
Pmech mechanical power in the turbine [W]
Pnom rated power [W]
Pclusters1in
input power in the clusters 1 [W]
Pgroup1 power in group1 [W]
PlossesDRc conduction losses in the diode rectifier [W]
PlossesDRs switching losses in the diode rectifier [W]
Pohmic12kV ohmic losses in the 12kV cables [W]
Pohmic60kV ohmic losses in the 60kV cables [W]
Pout electric power of the PMSG [W]
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Ptrans power losses in the transformer [W]
P(s) dynamic perturbation
( )vP Power curve of each turbine [W]
Q Reactive Power [Var]
Pw Power from the turbine [W]
R gas constant [m3PaK-1kg-1]
Rb Radius of the rotor [m]
Ra resistance of the PMSG in the stator windings [Ω]
Rcable resistance per length of the cable [Ω/m]
Rgroup1,2 total resistance of the 12kV cables of group1,2 [Ω]
RL resistance of the inductor [Ω]
Rload output resistance [Ω]
Rtrans resistance of the transformer [Ω]
Rsec resistance of the secondary of the transformer [Ω]
S Cross section area of the cable [m2]
T switching period [s]
Tair Temperature of the air [K]
TCcrit critical period of the oscillations [s]
Td time constant of the derivative part in the controller [Nm]
Te Electromagnetic Torque [Nm]
tf fall time of the semiconductor [s]
Ti time constant of the integral part in the controller [Nm]
Tower Height of the tower [m]
tr rise time of the semiconductor [s]
Tw torque in the turbine side [Nm]
Tw_g Aerodynamic torque in the turbine from the generator side [Nm]
Tz time constant of the zero of the PI controller [s]
Uα, Uβ Alpha and Beta Voltages [V]
Ub voltage line-to-line base [V]
Uc control voltage [V]
Udiode ON voltage of the diode [V]
ud, uq voltages u1, u2, u3 in the dq frame [V]
Udc DC voltage for the VSC onshore and offshore [V]
Uin input voltage of the converter [V]
UP voltage across the switch in the boost converter [V]
Urated, Prated Rated voltage and Power of the PMSG [V, W]
Uqnom rated voltage in the q axis [V]
Uo output voltage in the converter [V]
u1, u2, u3 RMS voltage line to ground after the VSC onshore [V]
v wind speed [m/s]
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v Average wind speed [m/s]
VIEGT ON voltage of the IEGT [V]
Vref(s) reference of the variable to be controlled
V(s) variable to be controlled
Zb impedance base [Ω]
δ duty cycle
θ pitch angle [degrees]
λ Tip speed Ratio
ξ damping factor
ρcopper resistivity of the copper [Ωm]
ρair air density [kg/m3]
τ time constant of Mod(s) [s]
Ψ flux in the permanent magnets in the rotor [Wb]
ωe angular electrical speed of the PMSG [rad/s]
ωerated rated angular electrical speed of the PMSG [rad/s]
ωgrid angular frequency of the grid [rad/s]
ωm angular speed of the generator from the generator side [rad/s]
ωn frequency of the oscillations [rad/s]
ωref reference speed for the turbine [rad/s]
ωt turbine speed from the turbine side [rad/s]
ωrotor mechanical speed of the rotor from the generator side [rad/s]
Z gain of the series of Mod(s) and F(s)
Glossary
ATR Active Three Phase Rectifier
DR Diode Rectifier
FD Freewheeling Diodes
IEGT Injection Enhanced Gate Transistor
IGBT Insulated Gate Bipolar Transistor
PMSG Permanent Magnet Synchronous Generator
PWM Pulse Width Modulation
SVPWM Space Vector Pulse Width Modulation
T1 Single Phase Transformer
VSC Voltage source converter (in this thesis it is done with IEGT)
WS Wind Speed
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List of Figures Fig. 2.1 – Cp as function of λ and θ. Fig. 2.2 – Power curve that will be followed by the pitch angle control. Fig. 3.1 – Block diagram used in the method of the dominant pole. Fig. 3.2 – Block diagram simplified. Fig. 3.3 – Block diagram using the symmetry criterion. Table 3. 1 – Ziegler-Nichols and Tyreus-Luyben tunning rules. Fig 4.1 – Diagram with the first implementation using the boost converter. Fig 4.2 – Current control block used in the simulation of the boost converter. Fig 4.3. – Block diagram representing the dynamics of the system. Fig 4.4– Inductor current Response. Fig 4.5. – Voltage and Current in the diode. Fig 4.6. – Voltage and Current in the IEGT. Fig 4.7. – Currents Ia, Iq and Id in the stator in the PMSG and electromagnetic torque. Fig 4.8. – Full bridge converter used in this application. Fig 4.9 – Schematic of the inverter. Fig 4.10 – Phase shift control block. Fig 4.11 – Waveforms for the control of switch 1 and switch 3 (switches 2 and 4 are the negative). Fig 4.12 – Step response of the full bridge with different controllers. Fig 4.13 – Input voltage, control voltage Uc and input and output currents of the full bridge converter. Fig 4.14 – Stator Ia, Iq, Id and electromagnetic torque of the PMSG. Fig 4.15 – Voltage Vp and the current in the primary side of the transformer. Fig 4.16 – Current in the Rectifier1 and in the IEGTs. Fig 4.17 – Equivalent electric circuit. Fig 4.18 – Simulation model to determine the maximum percentage of voltage harmonics. Fig. 4.19. – Current control for Id and Iq in the ATR. Fig. 4.20. – PMSG measurements with the ATR control. Table 4.1 – Losses for all wind speeds for the diode rectifier plus boost converter to 60kV. Table 4.2 – Losses for the full bridge converter to 60kV using phase shift control. Table 4.3– Losses for the full bridge to 60kV using duty cycle control. Table 4.4 – Losses for the 60kV ATR. Fig. 4.21. – Losses for the three converters discussed in this chapter in percentage of transmitted power. Fig. 5.1 – Simulink model for the simulation of the one turbine. Fig 5.2 – Frequency reference and the frequency of the generator. Fig 5.3 – Electric power in generator and mechanical power in the turbine. Fig 6.1 – Wind park from the submarine cable to the grid. Fig 6.2 – Electric circuit that represents the wind park from the submarine cable to the grid. Fig 6.3 – Simplified electric circuit. Fig 6.4 – Block diagram of the Plant, relating Ud and Uq with Id and Iq. Fig 6.5 – Block diagram with the Current Control System and the Plant. Fig 6.6 – Block diagram with the Plant and the Current Control System altered in order to de-couple Id and Iq. Fig 6.7 – Block diagram with Id and Iq de-coupled. Fig 6.8 – Id and Iq using the Simulation Model. Fig 6.9 – Transient of Id zoomed in. Fig 6.10 – Transient of Iq zoomed in.
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Fig 6.11 – Id and Iq with the control system with Id and Iq coupled, Fig 6.5. Fig 6.12 – Transfer Function of the Current Control System and the Plant (Fig 6.6) in close loop. Fig 6.13 – Circuit with the currents and Voltages used in the voltage control system. Fig 6.14 – Block diagram with Voltage control in close loop. Fig 6.15 – Id response simulated in Simulink (Large perturbations). Fig 6.16 – UDC response simulated in Simulink (Large Perturbations). Fig 6.17 – UDC response simulated in Simulink for small Perturbations. Fig 6.18 – Idref response simulated in Simulink for small Perturbations. Fig. 7.1 – Proposed connections for the Wind park. Fig. 7.2 – Losses for 12kV converters in this thesis. Table 7.1 – Losses for the 12kV converter, currents and power in the 2 groups. Table 7.2 – Losses for cables and 12/60kV full bridge converter. Fig. 7.3 – Losses for the 12/60kV converter. Fig. 7.4 – Losses for the 60/200kV converter. Table 7.3 – Losses in the main cable, VSC onshore and output power of the park. Fig. 7.5 – Losses in the VSC onshore. Fig. 7.6 – Representation of one quarter of the wind park. Each circle is one turbine. Fig. 7.7 – Representation of half of the wind park. Table 7.4 – Losses for 60kV converters and currents in both groups. Table 7.5 – Losses in the 60kV cables, 60/200kV converter and in the main cable. Fig. 7.8 – Efficiency of the whole park using option1 and option2. Fig. A.1 – Dimensions of the single-phase transformer. Fig. A.2 – Total Losses and relative price for the transformer for the best iron area. Fig. A.3 – Dimensions of the Inductor.
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Chapter 1 – Introduction
1.1. Problem background – Why wind power?
Nowadays there is a huge demand for electric power in order for societies to develop. In countries with
quickly growing economics like China and India where the electricity consumption is increasing exponentially
daily, there is a necessity for reliable and cheap sources of electric power.
The usual sources have a huge problem that cannot be hidden: they cause very serious environmental
problems like green house effect on the long run. An alternative that doesn’t have this problem is wind power.
1.2. Why study offshore wind turbines?
In most of the western European countries there is already a great utilization of wind power inland.
There are so many wind turbines operating that there is a lack of space in land for wind power to continue to
increase. Also too many wind turbines inland have the problem that they can spoil the view of the landscape.
For these reasons engineers all over the world start to focus their attention to place wind turbines
offshore. Offshore turbines have a lot of advantages over inland turbines: the wind offshore is more constant and
has higher effective speed over the year, there is space in the North Sea and Atlantic Ocean more than enough to
supply the consumers, there is no spoil in the landscape, no people live nearby to be affected by the noise. More
energy can be extracted, with less environment problems compared to what inland turbines have. Studies even
say that offshore turbines are beneficial for the fauna in the sea. The great downside is that installing turbines
offshore is much more expensive than inland, but bigger turbines can be installed.
1.3. Why develop DC wind farms?
Studies show the wind speed is higher and more constant in a reasonable distance from the shore. Also
people cannot see the turbines when they are placed far from the shore.
When the distances are too large, if all wind turbines are connected in an AC connection there is a
problem of reactive power created that lowers the power factor. If the connection is DC instead, there is no
reactive power and the frequency in the generators can be independent from the onshore grid frequency. The
cable resistance of a DC cable is lower than an AC cable for the same cross section leading to lower losses. With
the development of power electronic semiconductors like Insulated Gate Bipolar Transistor (IGBT) and Injection
Enhanced Gate Transistor (IEGT) the active and reactive power delivered to the grid can be controlled like in an
AC wind farm.
The downside is that nowadays the efficiency of the transformers to raise the voltage is above 99%
while the efficiency of DC/DC converters is lower. A DC wind farm will have more losses than a conventional
AC park.
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1.4. Layout of the Report
This thesis is going to design, control the torque and speed, simulate and compute the efficiency of 3
converters for 12kV and 60kV DC. These are connected to the PMSG, and they will be compared in order to
find the best one to use in a DC offshore wind farm.
With the efficiency data computed, the losses in the whole park are known and thus the efficiency. This
will be done for two grid topologies, with two and three voltage levels. They will be compared in order to
determine the best efficiency. The yearly energy produced by the park is computed with the wind speed
distribution that is assumed to be a Rayleigh distribution with an average wind speed equal to 10 m /s.
In chapter 2 some aerodynamic principals are presented, and the modelling of the PMSG and wind
turbine is done. It is also displayed the problem specifications and the characteristics of the equipment used. In
chapter 3 the control theory necessary to explain the control strategies ahead is exhibited.
In chapter 4 the torque control, simulation and efficiency calculation are done for the 3 converters: the
boost converter, the full bridge converter and the ATR. In chapter 5 is performed the speed control for the same
previous three converters.
In chapter 6 the voltage and reactive power control is conducted for the VSC onshore. In chapter 7 the
best configuration for the park grid is investigated with the goal to minimize the losses, for all wind speeds. With
these results the yearly energy production is computed.
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Chapter 2 – Background Theory and park specifications
In this section a model for static and dynamic approach of a wind turbine is developed. For the
development of a control of the speed turbine for maximization of power extracted from the wind, the dynamic
modelling approach is chosen.
In the first section “Aerodynamic principals of Wind turbines” the equations that give the Cp, λ and
power from the wind are given.
In the second section “Model of the Wind turbine and gearbox”, the equations that give the Turbine
Torque, Turbine Power and Turbine Speed are presented.
In the third section “Model of the PMSG” the equations that relate voltage, current and electrical speeds
and the equation that gives the electromagnetic torque of the PMSG are presented.
2.1. Aerodynamic principals of Wind Turbines
2.1.1. Power from the Wind
The kinetic energy of the air with mass m and speed v is:
2
2
1mvE = . (2.1)
The power is:
2
2
1vm
dt
dEP &== . (2.2)
Where m& is the mass flow rate. When the air passes across an area A, in this case the area swept by the
rotor blades, the power of the air will be:
323
2
1
2
1vRAvP
bairairπρρ == . (2.3)
Where Rb is the radius of the blades, and airρ is the air density. It will vary with air pressure and
temperature along with:
air
airRT
p=ρ . (2.4)
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Where p is the pressure, Tair the temperature and R the gas constant. At sea level at a temperature of
Tair=288K=30ºC, it will have the value airρ =1.225kg/m3. This will be the value considered in this thesis.
2.1.2. Mechanical Power Extracted from the Wind
In [1] it is explained that no energy converter can extract all the energy of the wind into mechanical
energy. The efficiency of this energy conversion is the coefficient Cp. This coefficient depends on two
parameters: the tip-speed ratio λ and the blade pitch angle θ. The tip-speed ratio is defined as:
v
Rbt
ωλ = , (2.5)
where Rb is the radius of the blade, ωt is the angular speed in the turbine side, and v is the wind speed.
The blade pitch angle is defined as the angle between the plane of rotation and the blade cross-section
chord. Cp for a three bladed-rotor can be computed with:
−
−−=
βθ
β
5.12exp54.0
11622.0pC . (2.6)
1
035.0
08.0
11
3 +−
+
=
θθλ
β (2.7)
In [1] and in Fig.2.1 is seen that the best Cp possible for the three-bladed rotor is λ=7. For this reason,
except when the wind speed is higher than the rated speed, the turbine speed ω is going to be controlled to
ensure that λ is 7 for medium and low wind speeds. This procedure maximizes Cp and consequently maximizes
the power that can be converted into mechanical energy from the wind.
Fig. 2.1 show Cp as a function of λ and θ.
Fig. 2.1 – Cp as function of λ and θ.
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It can be seen that in order to maximize Cp the pitch angle should be as low as possible, for example 2
degrees like in Fig. 2.1 and in [2]. This will be possible unless the wind speed is higher than the rated speed as
explained in the next section.
2.1.3. Blade Pitching System
There are two ways of controlling the amount of mechanical power extracted from the wind: the stall
and pitch control. The first is used in fixed-speed turbines and the second in variable speed turbines.
In many previous works it was concluded that the pitch control has numerous advantages over stall
control, like less noise and less power fluctuation. For these reasons, in this thesis only variable speed turbines
with pitch control are studied.
2.1.4. Pitch control
This control system consists of having a motor that according to the direction and speed of the wind will
position the blades in order to get the reference mechanical power. This reference power varies as a function of
the wind speed as described in Fig. 2.2 :
Fig. 2.2 – Power curve that will be followed by the pitch angle control.
This figure was taken from [6], which is a wind turbine model with the desired rated power for the
turbines of the wind park in study: 2 MW. The wind turbine will start operating at 4m/s and stops rotating at
19
25m/s. It can be seen in the figure that for the lowest sound level the rated speed will be 15m/s, but for the
highest sound level the rated speed is 12m/s. As these turbines will be placed 300km from the shore, almost
nobody will hear them. So the rated speed will be chosen to be 12m/s, more power can be extracted at medium
wind speed.
In order for the turbine profile to follow Fig. 2.2 , the following steps must be taken:
-for wind speeds lower than 12m/s (this value for the rated wind speed was chosen in [6]), the goal is to
extract the maximum power possible from the wind, so Cp should be maximum, equal to 0.4. From Fig. 2.1 it is
obtained λ=7 and θ=2º.
-for wind speeds higher than 12m/s, Cp is should be lowered in order for the power from the turbine to
be constant and equal to its nominal value, in this example 2 MW.
The control of the maximum extraction point will follow these rules. From (2.3)(2.3.) it is known:
33232
650
225.140
222
vv
MW
vR
PC
b
mech
p=
×==
πρπ. (2.8)
Since the rated power of the turbine is 2MW, the mechanical power will have to be a bit higher than
2MW to take into account the mechanical and electrical losses. Those losses were computed in latter
simulations, and knowing the electric power (2MW) the mechanical power is known and thus the constant that
relates wind speed and Cp in (2.8). However the results won’t be altered significantly by the losses.
2.1.5. Model of the turbine and gearbox
The turbine mechanical system could be considered as a two mass lumped system; if the mechanical
stresses in the shaft are to be studied (detailed models are needed). However, here a rigid shaft is considered.
Between the wind turbine and the generator there is a gear box to raise the speed and decrease the
torque in order for the generator to be smaller (lower torque gives lower current which gives lower losses).
Newton’s law of motion for this system gives:
( )mmgwe
eq
m BTTJdt
dω
ω−−= _
1 (2.9)
Where
mω is the angular speed of the shaft in the generator side, Te is the electromechanical torque in
the generator, Tw_g is the aerodynamic torque from the turbine seen from the generator side, Bm is the rotating
damping coefficient and Jeq is the inertia seen from the generator side as well. Jeq and Tw_g can be computed as:
2g
w
geqn
JJJ += .
g
w
gwn
TT =_ . (2.10)
Where Jg is the inertia of the generator, Jw is the inertia of the turbine, ng is the speed ratio of the gear
box, and Tw is the torque in the turbine side. This torque, with Cp and λ computed, can be calculated as:
20
λρπ
p
bw
CvRT
23
2
1=
. (2.11)
And the power from the turbine can be computed as:
wtw TP ω= . tgm n ωω = . (2.12)
The reference frequency for the generator is:
pg
b
ref nnR
vλω = . (2.13)
where ng is the speed ratio of the gear box and np is the number of pole pairs in the generator.
As a conclusion, the turbine is modelled using Newton’s equation. The Cp and λ give the torque and the
reference speed of the turbine. The reference speed will give the current that will give the torque Te from the
generator. With this torque and the torque from the turbine Tw_g, with the Newton equation, the speed can now be
computed.
2.1.6. Model of the generator
In this thesis the generator could be the Synchronous Generator, either wound or with permanent
magnets. However, from now forward only the option of permanent magnets is going to be used.
In [2] it is explained the background theory. The dynamics of this generator is derived using the two-
phase synchronous frame, where two axes are considered – the d axis aligned with the rotor position and the q
axis which is 90º ahead of the d-axis.
Applying Faraday’s Law to the three windings in the stator, three equations will be obtained. Applying
Park’s Transformation to these equations the following can be obtained:
+
+−−=
++−=
q
d
q
d
eq
q
aq
d
d
q
d
q
ed
d
ad
uLL
iL
Li
L
R
dt
di
uL
iL
Li
L
R
dt
di
11
1
ψω
ω
(2.14)
where id, iq, ud, uq, Ld and Lq are the currents voltages and inductances in the d-axis and q-axis respectively, Ra is
the resistance in the stator windings and Ψ is the flux in the permanent magnets in the rotor. For many PMSG,
Ld=Lq=L [17]. Then, it will become:
21
+
+−−=
++−=
qdeq
aq
dqed
ad
uLL
iiL
R
dt
di
uL
iiL
R
dt
di
11
1
ψω
ω. (2.15)
The electromagnetic torque is computed using:
( )( ) qpdqde iniLLT ψ+−=
. (2.16)
2.2. The problem specifications
2.2.1. Wind Park Specifications
In this section the specification given for the wind park to study are displayed.
• Distance to the shore: 300km.
• Nominal power: 200MW.
• Number and disposal of turbines: 100 turbines disposed 10*10 squared, each turbine apart of
400 meters.
• Nominal power of each turbine: 2MW.
2.2.2. Characteristics of the turbine
• Cut-in wind speed: 4m/s
• Nominal wind speed (2000kW): 12m/s
• Cut-out wind speed: 25m/s
• Rotor diameter: 80 meters
2.2.3. Characteristics of the Generator
The generator used is going to be a PMSG (Permanent Magnet Synchronous Generator). It will have the
following characteristics, taken from [2].
4
25.0
008.0
=
=
=
p
a
n
puL
puR
.
Nominal Power: 2MW
Nominal Frequency: 100Hz
ωrotor=p
n
fπ2 =
4
1002π=157rad/s
22
The nominal voltage will differ from case to case. It can be a low voltage generator; in this case it will
present 690V. It can be a high voltage generator; in this case it could be 5kV. Ra is the stator resistance, L is the
stator inductance, np is the number of pole pairs.
For the 690V PMSG, Ra=2mΩ and L=0.09mH. For the 5kV PMSG, Ra=100mΩ and L=5mH.
2.2.4. Characteristics of the Gear Box
The inertia seen from the generator side, Jeq in (2.9) is Jeq=8000 kgm2, and includes the turbine inertia
and the inertia of the generator. The turbine data used is the same as in [2]. In [3] the gear box losses are
estimated in 3%. This means:
12
2
43.2157
03.02%3 −=
×=⇔== Nmsrad
MWB
P
B
P
Pm
nom
rotorm
nom
friction ω (2.17)
In [2] the speed ratio of the gear box used is ng = 77. This value will be used in this thesis.
2.2.5. Characteristics of the Transformers
The three phase transformers that will be used in this thesis will have in each side:
Rtrans=0.002pu
Ltrans=0.05pu
The single phase transformers will have leakage inductance 5% total. The frequency can be 100Hz or
400Hz.
23
Chapter 3 – Control Fundamentals
In this thesis the control of power electronics equipment such as IEGTs converters is going to be used.
For this reason in this chapter a brief description of the control methods will be given.
In this thesis the current, the voltage or the speed will be controlled, depending of the used converter.
Considering the block diagram that describes the dynamics of the system and the problem, different control
methods need to be used. The methods that will be used in this thesis will be three: the method of dominant pole,
the method of symmetric criterion and the Ziegler-Nichols method.
3.1. – Method of dominant pole
The method of dominant pole is the simplest method, and it is used only for the current control in
Chapter 6. In this case the block diagram of the system is presented in Fig.3. 1 :
Ctrl(s) Mod(s) F(s)
P(s)
Vref(s) V(s)
+
-
-
+Ctrl(s) Mod(s) F(s)
P(s)
Vref(s) V(s)
+
-
-
+
Fig.3. 1 – Block diagram used in the method of the dominant pole.
V(s) is the variable to control (in the current control of chapter 6 Id and Iq), and Ctrl(s) is the function
transfer of the controller, in this case a PI controller. Mod(s) is the function transfer of the modulator which is all
cases in this thesis a first order function. F(s) is the function transfer between the perturbation and the variable to
control. P(s) is the perturbation (the voltage in the grid). As this voltage is constant, it can be removed from the
block diagram. The block diagram will be as in Fig.3. 2 :
Ctrl(s).Mod(s).F(s)Vref(s) V(s)
-
+Ctrl(s).Mod(s).F(s)
Vref(s) V(s)
-
+
Fig.3. 2 – Block diagram simplified.
F(s) in the figure is a first order function where the pole is not close to the origin. Otherwise, this
method cannot be used, as it will give unstable responses. The method of dominant pole is to place the zero of
the PI controller in the dominant pole, which will be in the case of Chapter 6 s=-Rtrans/Ltrans, where Rtrans is the
transformer onshore resistance and Ltrans is the transformer onshore inductance. This will give:
24
trans
trans
p
i
L
R
K
K=
. (3.1)
With one equation and two parameters to compute there are two strategies to determine the parameters.
3.1.1. – Setting ξ
The close looped system will be a second order system. The ξ of the system will be 0.707 for the
optimum transient in overshoot and rise time, and with this Kp and Ki are computed.
3.1.2. – Setting the bandwidth of the system
The other method is to neglect the pole from the modulator dynamics, and thus consider the system to
be a first order system. However this can only be done if the poles are not to far away from the real axis. The cut
off frequency of the system in close loop shall be chosen, usually 7 to 10 times longer than the frequency of the
system. In this way Kp will be computed. From the equation above, Ki will be computed.
3.2. Symmetry criterion method
More in-depth information about this method in the applications of electric motor control can be found
in [4]. In this thesis it is not motors which are being controlled, but power electronic converters. However, the
block diagram of both will be the same, so the same method can be used. For example, the speed control of the
PMSG with ATR in Chapter 5 will be used, and the voltage control of the voltage in the submarine cable in the
Chapter 6 will be used as well. Generally, the block diagram of the system to control will be as presented in
Fig.3. 3 :
Fig.3. 3 – Block diagram using the symmetry criterion.
Mod(s) in Section 4.1.3. is the modulator dynamics, in Chapter 6 and in Chapter 5 will be the current
control dynamics. F(s) will be 1/ (sLboost) in Section 4.1.3. 1/(sJeq) in Chapter 5 and 1/(sConshore) in Chapter 6, but
it will be similar mathematically. Only when P(s) can be considered to vary too slowly in the time scale of the
dynamics of the system, P(s) can be considered to be zero. In this case the method of dominant pole can be used.
Ctrl(s) Mod(s)
F(s)
P(s)
Vref(s)+
-
-
+F(s)Ctrl(s) Mod(s)
F(s)F(s)
P(s)
Vref(s)+
-
-
+F(s)F(s)
25
However, if large perturbations arise, P(s) needs to be accounted for (for example in Chapter 6). In
these cases the method of symmetric criterion should be used instead. The function transfer of the close looped
system will be:
( ) ( ) ( )( ) ( )
( ) ( )( ) ( )
( )sPsFsModsCtrl
sFsV
sFsModsCtrl
sFsModsCtrlsV ref )(1)(1
)(
−+
−−= (3.2)
Because the minus sign is in the action loop in Fig.3.3. Kp and Ki should be negative for the system to
be stable (only in the speed control this does not happen and so the parameters will be positive). Renaming:
( ) ( )sHsFsModsCtrl −=)()( .(3.3)
Then:
( ) ( )( )
( ) ( )( )
( )sPsH
sFsV
sH
sHsV ref
++
+=
11 (3.4)
( )
( )
( ) ( )( )
( )sPsH
sFsV
sH
sVref
++
+
=11
1
1. (3.5)
As -H(s) has already a pole in the origin, a P controller was used [4] to guarantee steady state error
equal to zero, so Ki=0. In this case the zero of the PI regulator will cancel the pole of G(s). Doing the
calculations the function transfer between V(s) and Vref(s) will be a second order system:
( )( )
( )sVss
sV ref
nn
n
sP 22
2
0 2 ωξω
ω
++=
= (3.6)
Doing the calculations with ξ=0.707, the following equation will be derived:
τZ
Kp 2
1= . (3.7)
Where τ is the time constant of Mod(s) and Z is the gain of the cascaded Mod(s) and F(s). It will be the
inductance in the boost converter in Section 4.1.3., the inertia seen from the PMSG in Chapter 5, the Capacitance
of the capacitor between the submarine cable and the VSC on-shore in Chapter 6.
When s=0, V(s) follows the reference but it will have a error from the disturbance, because
F(s)/(H(s)+1) doesn’t lead to zero when s=0. A PI controller is needed to solve this matter [2]. In this case, it can
be seen doing the computations that V(s) will follow the reference and the perturbation won’t have influence on
it (it will appear a zero in the origin in F(s)/(H(s)+1), which is the goal. Doing various simulations with many
values of Ki, the best value that will give fast response to the perturbation is with Tz/τ =4, where Tz is the time
26
constant of the zero from the PI controller. However this parameter will give an overshoot of 43%. A first order
filter in the reference with time constant equal to Tz will reduce the overshoot to acceptable values while
maintaining good responses. With this value doing the calculations, Ki will be:
28
1
τZK
i= (3.8)
3.3. Method of Ziegler Nichols
For this method the information was taken from [5]. The method of Ziegler Nichols is a set of rules for
obtaining the values of the regulators, obtained empirically from simulation studies. In this thesis it was used
only for the current control of the full bridge converter. From the two possible Ziegler Nichols methods the
method of stability margin was selected.
The first step is to consider the regulator to be a P controller, and continuously increasing the gain until
the response is unstable. The gain where the response will become unstable will be called KPcrit. Then the period
of the oscillations when the gain is KPcrit will be measured, and it will be called TCcrit.
With these two values, depending on if the regulator will be a PI or PID, or the method used is the
Ziegler Nichols or the Tyreus-Luyben, the parameters Kp, Ti and Td will be computed using Table3. 1:
Table3. 1 – Ziegler-Nichols and Tyreus-Luyben tuning rules.
Controller parameters (Ziegler-Nichols) Controller parameters (Tyreus-Luyben) Type
of controller Kp Ti Td Kp Ti Td
PI 0.45KCcrit 0.85TCcrit - KCcrit/3.2 2.2TCcrit -
PID 0.6KCcrit 0.5TCcrit 0.12TCcrit KCcrit/2.2 2.2TCcrit TCcrit/6.3
Ki and Kd can be computed from Kp, Ti and Td:
=
=
dpd
i
p
i
TKK
T
KK
(3.9)
It’s important to know that the response obtained by the Ziegler Nichols Method is acceptable, but not
optimal. It means that there can be other responses that are even faster and with less overshoot than the one
obtained in the Ziegler Nichols Method.
27
Chapter 4 – Torque control solutions
In this chapter the sizing, control, simulation and efficiency analysis of several solutions for torque
control of the PMSG are presented. The techniques for the determination of the control parameters were
announced in Chapter 3. The calculation of losses will be presented to determine the efficiency of each
converter.
The three DC/DC converters analysed here are in order of appearance: the boost converter, the full
bridge converter, and the ATR (Active Three-Phase Rectifier) with IEGTs.
These converters were chosen to raise the voltage from the low level in the PMSG to a high level
suitable for transmission. Because of this, all the following converters will have an output voltage higher than
the input voltage. Also, in this thesis the electrical grid will be as a constant voltage grid.
The rated voltage in the PMSG can be 690V or 5kV, but the level of the output voltage in the three
converters is always the same: 60kV. This value will be used later in Chapter 6 for the determination of losses in
the whole park in option 2. The output voltage in the simulations for the three converters will be considered
constant, because it is controlled by the DC/DC converter that will raise the voltage from 60kV to 200kV, a
voltage level suitable to transport the whole power from the park (200MW) across 300 km until the shore.
The IEGTs are chosen to be used as switches because they can withstand high voltage (4.5kV) and high
currents (750A) when the converter doesn’t need a lot of current it is assumed the IEGT with the same voltage
but lower current, has shorter switching times. The data sheet is in [18]. For the diodes the data sheet is in [21].
The first converter described in this section is the simplest of the kind of voltage source converter, for
the purpose to raise the voltage. It is called boost converter (step-up converter).
4.1. Boost converter connected to PMSG
The first application is to use a PMSG of 690V, followed by a three phase diode rectifier, and a boost
converter that raises the voltage from 931V to 60kV. The schematic of this system is shown in Fig. 4.1.:
wrotor
rated speed
Continuous
pow ergui
A
B
C
+
-
Universal BridgeRload
Tm
mA
B
C
Permanent Magnet
Synchronous Machine
L
Iref
g CE
IEGT
Diode
Idc
Iref
Out
Current_Control
i+
-
Current Measurement1
Cout
iL
Fig 4.1 – Diagram with the first implementation using the boost converter.
28
However is not possible to raise the voltage from 690V to 60kV because the duty cycle of the converter
will be very close to 1. This will lead to very low efficiency, close to 3.5%. This is unacceptable.
4.1.1. Description
The maximum ratio for the voltage raise in the boost converter is 4 times approximately [7], therefore a
three phase transformer next to the PMSG to raise the voltage must be used.
In all configurations the speed in the PMSG is considered to be constant (and equal to the rated speed)
because in the time scale of the dynamics of the system, the speed is constant due to the fact the mechanical time
constants are much higher than the electric ones.
4.1.2. Sizing
The PMSG can have a nominal voltage of 5kV or 690V. The transformer, depending of the case, raises
it from 5kV or 690V (the values used in this thesis, in Section 2.2.3.) to 35kV. It has 0.002pu of resistance and
0.05pu of inductance on the primary and secondary side. In this case, the boost converter will raise the voltage
from 35kV*1.35=47kV (due to the diode rectifier) to 60kV.
The dimensioning of the inductor Lboost must be done in order to guarantee that the current ripple will be
less than 10% of the nominal current. From [7] the following formula can be extracted:
( ) ( )
HMWfU
UUU
ifU
UUUL
como
inoin
Lcomo
inoin
boost 3.221.0
2
=×
−=
∆
−= . (4.1)
Where Uin=47kV, Uo =60kV, fcom=1000Hz.
4.1.3. Control
Usually it is desired to control the current in the inductor, or the input voltage of the converter. In this
case it is necessary to control the input current to control the torque of the generator as they are proportional (as
large and sudden variations in the torque can provoke too much mechanical stresses in the shaft of the PMSG)
and to protect the semiconductors in case of short-circuits.
The output of the regulator, the control voltage Uc determines the right duty cycle δ in order for the
inductor current to achieve the desired result. Uc is compared with a carrier wave having a much higher
frequency. It can be triangular, or saw-tooth type. This comparison can be done digitally in a microprocessor or
analogically by an AMPOP.
A saw-tooth signal from 0 to 10V with frequency 1 kHz will be used as a carrier. The frequency could
be higher, but that would lead to too high switching losses. The current control block is presented in Fig 4.2. :
29
1
OutModulator
PI
Discrete
PI Control ler
2
Iref
1
Idc
Fig 4.2 – Current control block used in the simulation of the boost converter.
Writing the equation for the inductor gives:
dt
diLiRUU L
boostLLpin+=− . (4.2)
Where Uin is the voltage after the diode rectifier bridge, Up is the voltage across the IEGT, RL is the
resistance of the inductor, Lboost the inductance, iL is the current in the inductor. Using the Laplace transformation
it is found:
sLR
UUI
boostL
pin
L+
−= . (4.3)
Uc enters the modulator. It has a statistic delay, which normally is considered to be half of the switching
period. It has also a gain. This gain can be computed in [7] and for the boost converter it is equal to:
600010
60000
10max
====∂
∂=
V
V
V
U
u
U
u
UK o
c
o
c
pav
v. (4.4)
With the modulator dynamics together with (4.3) the complete block diagram representing the dynamics
of the system is presented in Fig 4.3. – Block diagram representing the dynamics of the system. :
Fig 4.3. – Block diagram representing the dynamics of the system.
30
In Fig 4.3. the pole s=-RL/Lboost is very close to the origin, because RL is dimensioned as small as
possible in order for the inductor to have as low losses as possible. If the technique to place the zero of the PI
controller near this pole is used, this can lead to instabilities problems as mentioned in Chapter 3. For this reason
the symmetry criterion will be used. The low-pass filter in the reference is used to decrease the overshoot and
(3.7) and (3.8) are used to compute the parameters of the controller, presented again in (4.7.).τ is the time
constant of the modulator dynamics, in this case half of the switching period. Remembering the minus sign in the
action loop in Fig 4.3. the following expressions are found:
−=−=
−=−=
4008
395.02
2v
boost
i
v
boost
p
K
LK
K
LK
τ
τ . (4.5)
4.1.4. Simulation Results:
The simulation is shown below. The step time for the current reference is at 0.03s. The result is shown
in Fig :
0.02 0.025 0.03 0.035 0.04 0.045 0.0510
15
20
25
30
35
40
45
50
55
60
inducto
r curr
ent(
A)
time(s)
Fig 4.4 – Inductor current Response.
Looking at Fig 4. the response takes 8 ms to follow the reference without steady state error, as the
current in the inductor in the nominal power should be 2MW/47kV=42A. The overshoot is 10%. The current
ripple is 10% as desired in the inductor sizing. The voltage and current for the Diode and IEGT are in Fig 4.5.
and Fig 4.6.
31
0.025 0.03 0.035 0.04 0.045 0.05-10
0
10
20
30
40
50
dio
de c
urr
ent(
A)
0.025 0.03 0.035 0.04 0.045 0.05
-60
-40
-20
0
dio
de v
olt
ag
e(k
V)
time(s)
Fig 4.5. – Voltage and Current in the diode.
0.025 0.03 0.035 0.04 0.045 0.05
0
20
40
60
80
IEG
T c
urr
ent(
A)
0.025 0.03 0.035 0.04 0.045 0.05
0
20
40
60
IEG
T v
olta
ge(k
V)
time(s)
Fig 4.6. – Voltage and Current in the IEGT.
Looking at Fig 4.5. and Fig 4.6. the IEGTs and diodes have a current of 42A and voltage of 60kV as
expected. Fig. 4.7. presents the evolution of the most important generator variables:
32
0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-50
0
50
Ia(A
)
0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-60
-40
-20
Iq(A
)
0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-40
-20
0
20
40
Id(A
)
0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-20
-10
0
time(s)
Torq
ue(k
Nm
)
Fig 4.7. – Currents Ia, Iq and Id in the stator in the PMSG and electromagnetic torque.
It is seen that at 30ms, when the current reference has a step, all generator variables change. It means
that the current controller is controlling the generator currents and therefore the generator torque. It can be seen
that Iq and the torque are proportional as expected. The current is not sinusoidal; Id is not constant and has a
small average value that is due to the diode bridge conduction overlap. This will cause additional losses in the
PMSG and lowers the power factor, which means that the PMSG will be more expensive.
4.2. Full Bridge Converter connected to the PMSG
4.2.1. Description
In this section the full bridge converter is going to be described for the application of raising the 690V
from the PMSG to 60kV, like in the boost converter section.
The full bridge converter can implement voltage or current sources. In the voltage case it should have a
capacitor on the input and an inductor on the output. The purpose of the inductor is to smooth the current in the
output and consequently in the transformer and in the IEGTs. In the current source case, it is the opposite: an
inductor is placed in series with the input and a capacitor in the output. In this thesis the first case was
investigated. The schematic of the full bridge in this application is given in Fig 4.8. :
33
wrotor
rated speed
Uo=60kV1 2
Single Phase Transformer
A
B
+
-
Rectifier2
A
B
C
+
-
Rectifier1
Tm
mA
B
C
Permanent Magnet
Synchronous Machine
Lload
Iref
g
A
B
+
-
Inverter
Idc
Iref
Out
Current_Control
i+
-
Current Measurement
Cin
current
Fig 4.8. – Full bridge converter used in this application.
The full bridge converter is connected at the input capacitor Ci. The inverter converts DC to AC and
controls the current and speed of the PMSG. The inverter output connects to a single phase transformer that will
raise the voltage. This transformer usually is made of a special type of iron giving to acceptable iron losses so
that it can operate at higher frequencies, like 400 Hz or even 1 kHz. In this thesis the frequency used is 400 Hz.
Adopting 400 Hz leads to small transformers that are cheaper and lighter than the usual grid transformers of
50Hz. The transformer will isolate the load galvanically from the generator that will protect it against short-
circuits from the grid, which is an advantage for the full bridge converter [9].
The output of the single-phase transformer is connected to a diode rectifier bridge to convert to DC. The
output inductor filter Lload is displayed and at the end there is a voltage source of 60kV that represents the DC
grid, since as usual this voltage is controlled to be constant.
4.2.2. PMSG, transformer and full bridge converter
The usual value for the voltage of the PMSG is 690V. However this will lead to very high ratios for the
single phase transformer. For this a 10kV PSMG was used instead.
The first parameter of the converter to be dimensioned is the minimum duty cycle (at rated conditions).
It will be set to 30%. Knowing [9]:
in
o
nU
U=δ . (4.6)
It is known Uo is 60kV and Uin 13.5kV, so n will be 14.5. The minimum duty cycle cannot be higher
than 30% because the transformer ratio would be too small and the duty cycle at lower wind speeds would be
higher than 1.
The flux of the PMSG was computed in order to obtain the rated voltage at rated conditions.
The current in the IEGTs is [9]:
34
AU
MWI
in
IEGT478
2==
δ. (4.7)
4.2.3. Sizing the output inductor and input capacitor
Applying the inductor equation in the interval when the voltage in the output of Rectifier2 is zero, and
setting the current ripple to 10% the output inductor will be:
( )
HIf
UL
loadcom
o
load77.15
2.0
1=
−=
δ (4.8)
fcom is the switching frequency, equal to 400Hz. The current in the load will be 2MW/60kV=33A, and
δ=30%. From [9] it is known that the input capacitor needed for 10% of voltage ripple will be:
( ) ( ) FkV
kVInI
nUf
UC
inload
incom
o
inµ961485.1433
5.145.13400
6055
22=−×
××=−= (4.9)
Iin is the current in the input, equal to P/Uin.
4.2.4. Control and determination of the controller parameters
The current in the inductor and the input voltage can be controlled in the full bridge converter. In this
approach only the current needs to be controlled, in order to prevent uncontrolled currents that can destroy the
semiconductors, and also to control the current in the generator, this way controlling the torque. Usually the
speed and current control of the generator are cascaded; the output of the PI controller of the speed is the current
reference. In this study of the full bridge, only the current control is studied, the speed control will be explained
in the next section.
The current is always compared with the reference, the error passes through a PI controller, and the
output will be a control voltage Uc. This control voltage is approximately equal to the duty cycle of the converter
(the ratio between the active period1 and the switching period). From this control voltage to the gate signals of
the switches there are some alternatives: unipolar and bipolar switching, phase shift, duty cycle control [10]. For
the beginning of the study the phase shift control was chosen, because it seems to be the best option for this
application [9]. The schematic of the inverter with the input and output voltage and currents is depicted in Fig
4.9:
1 Active period – The period when power flows from the source to the load. The voltage in the transformer is not zero.
35
S1
S2
S3
S4
Vi
It
Ii
Vp
Fig 4.9 – Schematic of the inverter.
The phase shift control is based on two square waves with 50% pulse width; one square wave
commands switches 1 and 2 in Fig 4.9 and the other commands switches 3 and 4. Switch 2 is instructed to be the
logical negative of switch 1 and switch 4 to be the logical negative of switch 3 in order to prevent short circuits
of the capacitor Ci. The command voltage Uc that leaves the regulator is multiplied by half the switching period.
This value is the delay between these 2 square waves. Uc varies between 0 and 1; this means the delay will vary
between 0 and T/2. For a better understanding of this control, the simulink model used for simulation is shown in
Fig 4.10 :
1/(2Fcommutation)
1
Pulses
t
Pulse
Generator1
Pulse
Generator1/800
boolean
boolean
double
Clock
NOT
NOT
Add
1
Uc
Fig 4.10 – Phase shift control block.
The square waves “Pulse Generator” and “Pulse Generator1” in Fig 4.10 and the voltage Up in Fig 4.9
are shown in Fig 4.11 :
Fig 4.11 – Waveforms for the control of switch 1 and switch 3 (switches 2 and 4 are the negative).
36
This way, there are four intervals in one period:
1. Switches 1&4 ON
2. Switches 1&3 ON
3. Switches 2&3 ON
4. Switches 2&4 ON
It can be seen that when Uc=0 there is no intervals 1 and 3, which means the voltage at the primary of
the transformer (Up) will always be zero. This gives the lowest Up RMS (0). If Uc=1 intervals 2 and 4 don’t exist,
it means Up will never be zero, it will be a square wave with pulse width 50% between Uin and – Uin. This gives
the highest Up RMS (Uin).
The parameters for the current control regulator of the converter will be computed using the tuning rules
of Ziegler Nichols explained in Chapter 3. The goal as was said before is to control the torque of the PMSG. The
output current was controlled. In the simulation it can be seen that the input and output current steady state
values are proportional, so controlling the output current is the same as controlling the torque of the PMSG.
KPcrit was found to be 0.1 and TPcrit to be 15 ms. The response of the output current was simulated for PI
and PID controllers. The reference changed from 5 to 15A. The responses are presented in Fig 4.12:
0.03 0.035 0.04 0.045 0.05 0.055 0.060
2
4
6
8
10
12
14
16
18
Iout(
A)
time(s)
PI Ziegler
PID Ziegler
PI Tyreus
PID Tyreus
Fig 4.12 – Step response of the full bridge with different controllers.
By comparing in Fig 4.12. the four responses it can be seen that the red and blue curves have steady
state error, and the black curve doesn’t start in 5A. For this it was concluded that the best curve is the green one,
which represents the PI controller using the Ziegler-Nichols tuning rules.
4.2.5. Simulation results
The controller presented above was used in simulations. The speed is constant and equal to its rated
value, and the reference current changes from 5 to 33A at 35ms. The voltage in the capacitor Ci, the control
voltage Uc and the input and output currents are in Fig 4.13 :
37
0.03 0.035 0.04 0.045 0.05 0.055 0.06
10
15
Uin
(kV
)0.03 0.035 0.04 0.045 0.05 0.055 0.06
200
400
Iin(A
)
0.03 0.035 0.04 0.045 0.05 0.055 0.060
20
40
Iou
t(A
)
0.03 0.035 0.04 0.045 0.05 0.055 0.060
0.5
1
Uc
(V)
time(s)
Fig 4.13 – Input voltage, control voltage Uc and input and output currents of the full bridge converter.
When the reference current changes the control voltage increases immediately for the current to
increase faster. The ripple in the output current and in the input voltage is 10% as expected. The input and output
currents are proportional apart from the transient state which means controlling the output current will control
the torque of the PMSG. It takes approximately 20 ms to stabilize.
0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06
-400
-200
0
200
400
Ia(A
)
0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06-400
-200
0
Iq(A
)
0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06-300
-200
-100
0
Id(A
)
0.02 0.025 0.03 0.035 0.04 0.045 0.05 0.055 0.06
-30
-20
-10
0
time(s)
To
rque(k
Nm
)
Fig 4.14 – Stator Ia, Iq, Id and electromagnetic torque of the PMSG.
38
Fig. 4.14. shows the variables in the PMSG. The current is not sinusoidal, because of the Rectifier1. For
this the current Id is not zero as in the boost converter. Iq and the torque are proportional. The input speed value
is ωerated/np=628rad/s/4=157rad/s. The torque is 2MW/157rad/s=13 kNm as expected. It doesn’t have
oscillations, which is essential for a good speed control of the PMSG. In Fig 4.15 the variables in the
transformer, the current and the voltage in the primary side are displayed:
0.03 0.032 0.034 0.036 0.038 0.04 0.042 0.044 0.046 0.048 0.05-600
-400
-200
0
200
400
600C
urr
ent
in t
he
Prim
ary
(A)
0.03 0.032 0.034 0.036 0.038 0.04 0.042 0.044 0.046 0.048 0.05-15
-10
-5
0
5
10
15
Volt
age
in t
he P
rim
ary
(kV
)
Fig 4.15 – Voltage Vp and the current in the primary side of the transformer.
The current is approximately constant. The voltage is controlled by the phase shift control. The current
in the IEGTs is displayed in Fig. 4.16. :
0.03 0.032 0.034 0.036 0.038 0.04 0.042 0.044 0.046 0.048 0.05-600
-400
-200
0
200
400
600
Curr
ent
IEG
T(A
)
time(s)
Fig 4.16 – Current in the Rectifier1 and in the IEGTs.
It is seen the maximum current is 470A as in (4.7.). Changing the reference current from 5 to 15A it was
seen that the output current took 10ms to stabilize, half of the time to stabilize from 5 to 33A. This is due to the
fact that the converter has a maximum duty cycle to increase the output current as fast as possible. That
39
maximum is 1. So if the current only changes from 5 to 15A, it is normal it takes less time to get there.
Resuming, for large perturbations it takes 20ms, for small perturbations 10ms.
4.3. PMSG with a ATR
The electric circuit of this implementation is in Fig 4.17 The first step is to determine the line to line
voltage in the AC side, and the inductance between the generator and the ATR:
wrotor
rated speed
generator_measurements
Ud
Uq
teta
Ualpha
Ubeta
dq alpha_beta
Uo=60kV
w
mA
B
C
Permanent Magnet
Synchronous Machine
Iqref
g
A
B
C
+
-
Inverter
Idref
Ualpha
Ubeta
Pulses
Discrete SV PWM
Generator
Id
Idref
Iq
Iqref
wm
Ud
Uq
Current_Control
<Stator current is_a (A)>
<Stator current is_q (A)>
<Stator current is_d (A)>
<Electromagnetic torque Te (N*m)>
<Rotor angle thetam (rad)>
<Rotor speed wm (rad/s)>
Fig 4.17 – Equivalent electric circuit.
In this converter, IEGTs will be used. A PMSG of 690V could be used in this simulation. But then the
problem is that the DC voltage Uo would be very low and couldn’t reach 60 kV. Using the formulas below it is
concluded that a good value for the line to line voltage in the AC side of the ATR can be 10 kV.
4.3.1. Determination of the LATR and Udc
The control for the ATR is made using PWM (Pulse Width Modulation) modulation. In the next figure,
the three phase voltage sources represent the open voltage from the generator, the impedance represents the
impedance of the generator plus the impedance introduced between the generator and the ATR in order to reduce
the current harmonics. The voltage in the DC link is going to be controlled by the main inverter onshore, so this
voltage is assumed to be constant.
In order to determine the voltage harmonics caused by the PWM in the ATR another simulation model
was used, which is displayed in Fig 4.18:
40
Ts=1/20000
Discrete,
Ts = 1e-006 s.
pow ergui
v+-
Voltage Measurement
Vdc =100V
Vab (av)
Vab
g
A
B
C
+
-
Universal Bridge
3 arms
1
0.0001s+1
Transfer Fcn
Idc
To Workspace1
Vab
To Workspace
A B C
Three-Phase
Series RLC Load
Scope
3
Multimeter
Pulses
Discrete
PWM Generator
i+
-
Current Measurement
Fig 4.18 – Simulation model to determine the maximum percentage of voltage harmonics.
In this model for the PWM modulation the most significant voltage harmonic was for index modulation
equal to 0.5, where the largest harmonic is 31% of Udc. The commutation frequency used is 1 kHz. LATR can be
found such as the current harmonic will be 10% of the nominal current. Quoting [16] is known:
mHMW
kV
If
UL
nomcom
ATR256
310000
21.010002
6031.0
20 =
××
×==
ππ. (4.10)
( )[ ]223 dgridtransmDC iLEU ω+> . (4.11)
Where Em is the peak line to ground voltage, ωe the electric speed of the generator and Iq is the nominal
current. In [16] instead of iq it is id. However in [16] the d-axis is aligned with the voltage. As in this case we
have a synchronous generator the voltage will be aligned with the flux of the machine. So instead of id it will be
iq. In this case:
AkV
MW
U
PI
qnom
nom
qnom200
10
2===
(4.12)
sradf
ratederated/62810022 === ππω (4.13)
kVEm 1.8
3
210000 == . (4.14)
Inserting these values gives:
kVU
o6.57> . (4.15)
41
This means that the value 60kV can be used. Quoting also [16]:
mHi
EU
Lqe
m
ATR268
200628
81003
60000
32
22
20
=×
−
=
−
<ω
. (4.16)
The required LATR of 256mH fulfils the above condition.
4.3.2. Control and Simulation
As it can be seen in Fig. 4.18., id, iq, the speed and the position of the rotor are measured from the
PMSG in order to control iq and id. The current control is in Fig. 4.19:
2
Uq
1
Ud
Product1
Product
1/Uo
Gain5
1/Uo
Gain4
Lq
Gain3
Ld
Gain2
np
Gain1
PI
Discrete
PI Controller1
PI
Discrete
PI Controller
5
wm
4
Iqref
3
Iq
2
Idref
1
Id
Fig. 4. 19. – Current control for Id and Iq in the ATR.
Wm is the mechanical speed of the PMSG, and is multiplied by the number of pair of poles to get the
electric speed. It is multiplied with Id and Iq and added in the end to decouple the currents. This means they will
be independent. This will be explained latter exactly the same way in chapter 6. Ud and Uq are divided by 60kV
because the MATLAB block that creates the SVPWM (Space Vector Pulse Width Modulation) pulses needs Uα
and Uβ between -1 and 1.
In this simulation the reference for Id was zero, and Iq changes from -100 to -200A at 0.02s. The
simulation file is described in the appendix. The PMSG variables are in Fig. 4.20. :
42
0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
-200
0
200
Ia(A
)
0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
-200
-100
0
Iq(A
)
0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-50
0
50
100
Id(A
)
0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-15
-10
-5
0
time(s)
Torq
ue(k
Nm
)
Fig. 4.20. – PMSG measurements with the ATR control.
The torque and Iq are proportional. Iq follows the reference of 200A and the torque 12kNm which is the
right torque for the mechanical power to be 2MW. It takes 5ms to reach the reference. It can be seen in the
Figure that the switching frequency is the double of what it should be. In the MATLAB block SVPWM there are
two patterns to use, one uses the switching frequency and the other uses the double, reducing harmonics without
increasing the switching losses. This is one of the advantages of the SVPWM.
4.4. Efficiency calculation of the three previous converters
Following are the formulas to derive the losses in the three previous converters. These formulas are
valid for 12kV or 60kV converters, but in order to avoid repetition only the results for the 60kV converters are
shown. The comparison between them is made, and later in Chapter 7 the results for the 12kV converters are
shown in order to choose the converter that has the minimum of losses using a two level voltage DC grid in the
wind park. The values for the resistor and core losses for inductors and single phase transformers were computed
in order to minimize the total losses at rated power, and those computations are given in appendix. The three
phase transformers were assumed to have 1% losses at rated power and core losses to be 1/5 of the copper losses.
4.4.1. 60kV Boost converter
The duty cycle in the boost converter is:
o
in
U
U−= 1δ . (4.17)
43
The data sheet of the diodes used is in [21]. Using a safety margin of 50% for the voltage the number of
semiconductors in series should be minimum:
o
IEGT
sU
Vn 5.1= (4.18)
In this case, the number of diodes in series will be ns=1.5x47/4.5=15.66 ≈ 16. The conduction losses in
the diode rectifier bridge will be:
diodeinslossesDRc
UInP 2= . (4.19)
As the current will be though 2 diodes in all period, where Udiode is the on state voltage in each diode
and Iin is the current DC in the output of the diode rectifier. The switching losses in the IEGT are:
com
fr
inIEGTlossesDRs fK
ttIVP
2
+= . (4.20)
fcom is the switching frequency, VIEGT is the voltage in the off state, Iin is the current in the on state which
is equal to the current in the inductor, tr is the rise time and tf the fall time, and K is a scale factor to take into
account that in many converters in this thesis the maximum current required is 166A, other cases it is 33A, much
lower than the maximum allowed for the IEGT used which is 750A. Thus the fall and rise times are smaller than
the ones in the catalogue, proportional to the decrease of current. In [21] it is stated that the reverse recovery
energy of the diodes (Err) depends on the current and the current derivative. For each current the derivative is
considered to give the largest Err for pessimistic loss accountability.
In [18] dv/dt=4500V/µs and di/dt=3600A/µs. The fall time is constant because the voltage in the IEGT
in the off state is constant for every wind speed. However the rise time will increase with the current. The
conduction losses for the IEGT and for the diode are:
inavIEGTslossesDRc
IVnP = (4.21)
VIEGT is the on voltage that depends on the current according to the current-voltage characteristic of the
IEGT. In the boost converter the average current for the IEGT is δIin, for the diode is (1-δ)Iin. Iin is the maximum
current in the IEGT and the diode, which is equal to the current in the inductor. It is computed as the power
divided by the input voltage 47kV. The losses in the inductor are the core losses and the ohmic losses (RL*Iin2).
The losses are presented as percentage of transmitted power for every wind speed in Table 4.1 :
44
Table 4.1 – Losses for all wind speeds for the diode rectifier plus boost converter to 60kV.
DR is Diode Rectifier and WS is Wind Speed. The switching frequency is 1 kHz as in the simulation. It
is seen that the main losses in this converter are in the inductor and in the three phase transformer. This is
because very good IEGT and diodes are used that give low conduction and switching losses.
4.4.2. 60kV Full Bridge
The efficiency was computed for two different control types: the phase shift and the duty cycle control.
Reference [19] presents a capacitive snubber that can strongly reduce the IEGTs switching losses, and so these
won’t be considered here. The switching losses in the DR are computed the same way as for the boost converter,
for DR1 is 100Hz and for DR2 is 400Hz.
For each control type the conduction losses will be different. The formulas for the conduction losses in
the IEGTs, Output Diodes, freewheeling diodes and transformer ohmic losses for the phase shift control are
presented next:
( )
( )2secsec
1
12
2
IRP
IVnP
IVnIVnP
IVnP
trans
diodeDiodesCFWDiodes
IEGTIEGTsIEGTIEGTsCIEGT
diodeDiodesCDiodes
=
−=
−+=
=
δ
δδ (4.22)
For the duty cycle:
( )
2secsec
0
22
142
IRP
P
IVnP
IVnIVnP
trans
CFWDiodes
IEGTIEGTsCIEGT
diode
DiodesdiodeDiodesCDiodes
δ
δ
δδ
=
=
=
−+=
(4.23)
WS(m/s)
Power(kW)
Iin(A) Duty cycle
3phase Transf
losses(%)
DR cond losses(%
)
DR switch losses(
%)
Lboost Losses(
%)
IEGT cond
losses(%)
IEGT switch losses(%)
Diode cond
losses(%)
Diode switch losses
(%)
Total (%)
4 74 1,58 0,74 5,43 0,12 0,04 3,27 0,07 0,13 0,02 0,07 9,15
5 145 3,08 0,67 2,83 0,12 0,03 1,70 0,07 0,13 0,02 0,05 4,94
6 250 5,32 0,61 1,71 0,12 0,02 1,02 0,06 0,13 0,03 0,04 3,13
7 397 8,45 0,54 1,18 0,12 0,03 0,69 0,05 0,13 0,03 0,05 2,29
8 593 12,61 0,48 0,94 0,12 0,03 0,53 0,05 0,13 0,04 0,05 1,88
9 844 17,96 0,41 0,85 0,12 0,03 0,46 0,04 0,13 0,04 0,05 1,72
10 1158 24,64 0,35 0,86 0,12 0,03 0,45 0,04 0,13 0,05 0,04 1,71
11 1541 32,79 0,28 0,94 0,12 0,02 0,48 0,03 0,13 0,05 0,04 1,81
12 2001 42,57 0,22 1,08 0,12 0,02 0,54 0,02 0,13 0,06 0,03 2,01
45
These formulas are in [20]. The only losses left are the output inductor losses and the core losses in the
transformer. The duty cycle is computed with (4.17.) and the current in the IEGTs with . (4.7).
The summary of all the losses for the phase shift is in Table 4.2. The duty cycle is the same for both
type of control:
Table 4.2 – Losses for the full bridge converter to 60kV using phase shift control.
T1 is Single Phase Transformer and FD is Freewheeling Diode. The fourth column is the current in the
output, the power divided by 60kV. For the duty cycle control the losses are in Table 4. 3
Table 4. 3 – Losses for the full bridge to 60kV using duty cycle control.
Comparing Table 4.2. and Table 4. 3 it is seen the duty cycle control will have lower losses than the
phase shift. For both cases the IEGT losses are larger than the diode ones, and the main losses are the output
inductor, single transformer and IEGT conduction losses.
4.4.3. 60kV ATR
WS(m/s)
Power(kW)
Duty cycle
Iin(A) Iout(A)
DR1 cond
losses(%)
DR1 switch losses(
%)
IEGT cond
losses(%)
T1 losses
(%)
DR2 cond
losses(%)
DR2 switch losses
(%)
Losses
FD(%)
Lload Losses(%)
Total (%)
4 74 1,00 16,41 1,24 0,39 0,54 0,54 5,60 0,11 0,11 0,00 5,25 12,01
5 145 0,80 25,64 2,41 0,31 0,28 0,51 2,91 0,11 0,08 0,05 2,71 6,70
6 250 0,66 36,92 4,17 0,26 0,16 0,51 1,74 0,11 0,06 0,07 1,62 4,38
7 397 0,57 50,26 6,62 0,23 0,10 0,53 1,15 0,11 0,08 0,08 1,08 3,27
8 593 0,50 65,64 9,88 0,20 0,07 0,56 0,82 0,11 0,08 0,09 0,81 2,69
9 844 0,44 83,08 14,07 0,18 0,05 0,60 0,63 0,11 0,08 0,10 0,68 2,39
10 1158 0,40 102,6 19,30 0,16 0,03 0,64 0,52 0,11 0,07 0,11 0,64 2,27
11 1541 0,36 124,1 25,69 0,15 0,03 0,69 0,45 0,11 0,06 0,12 0,66 2,25
12 2001 0,33 147,7 33,35 0,14 0,02 0,75 0,40 0,11 0,06 0,12 0,73 2,32
WS(m/s)
Power(kW)
Iin(A) Iout(A)
DR1 cond
losses(%)
DR1 switch losses(
%)
IEGT cond
losses(%)
T1 losses
(%)
DR2 cond
losses(%)
DR2 switch losses
(%)
Lload Losses(%)
Total (%)
4 74 16,41 1,24 0,39 0,54 0,53 5,53 0,11 0,11 5,25 11,94
5 145 25,64 2,41 0,31 0,28 0,45 2,83 0,11 0,08 2,71 6,51
6 250 36,92 4,17 0,26 0,16 0,40 1,64 0,11 0,06 1,62 4,11
7 397 50,26 6,62 0,23 0,10 0,38 1,03 0,11 0,08 1,08 2,93
8 593 65,64 9,88 0,20 0,07 0,37 0,69 0,11 0,08 0,81 2,28
9 844 83,08 14,07 0,18 0,05 0,36 0,49 0,11 0,08 0,68 1,91
10 1158 102,6 19,30 0,16 0,03 0,36 0,36 0,11 0,07 0,64 1,72
11 1541 124,1 25,69 0,15 0,03 0,37 0,27 0,11 0,06 0,66 1,63
12 2001 147,7 33,35 0,14 0,02 0,37 0,21 0,11 0,06 0,73 1,62
46
The formulas to compute the conduction and switching losses are the same except the current used is
the RMS current in the AC side, and the power electronic losses are multiplied by 3 as there are 3 legs. The
switching frequency is 1000Hz. For all wind speeds Table 4.4 show all the losses:
Table 4.4 – Losses for the 60kV ATR.
WS(m/s) Power(kW) Rise Time
IEGT(s)
IRMS_AC (A)
IEGT cond losses(%)
IEGT switch
losses(%)
Inverse Diode cond
losses(%)
Inverse Diode switch
losses(%)
3phase Transf Losses
(%)
Losses total(%)
4 74 7,81e-7 4,28 0,27 0,24 0,20 0,27 5,43 8,38
5 145 7,82e-7 6,69 0,22 0,20 0,16 0,21 2,82 5,16
6 250 7,83e-7 9,63 0,18 0,16 0,14 0,16 1,70 3,62
7 397 7,84e-7 13,10 0,16 0,14 0,12 0,13 1,17 2,79
8 593 7,85e-7 17,12 0,14 0,12 0,10 0,10 0,91 2,31
9 844 7,86e-7 21,66 0,13 0,11 0,09 0,09 0,81 2,07
10 1158 7,87e-7 26,74 0,12 0,10 0,08 0,09 0,81 1,96
11 1541 7,89e-7 32,36 0,11 0,09 0,07 0,06 0,88 1,89
12 2001 7,91e-7 38,51 0,10 0,08 0,07 0,05 1,00 1,90
It is seen the IEGT switching losses are the same as the conduction losses, which indicates the
switching frequency was well chosen. The transformer gives 50% of the total losses, while the power electronics
give the other 50%.
4.5. Conclusion:
The implementation of using the boost converter to raise the voltage from 690V to 60kV is impossible
to do, because the duty cycle will be close to 1 and it will have very low efficiency.
The implementation with a transformer to raise the voltage from 690V to 35kV is more acceptable. The
efficiency rises from 3.5% to 98.17%. The symmetry criterion was used to compute the parameters of the
regulator for the current control, and consequently the torque control. With this implementation and these
parameters a simulation was done in MATLAB/SIMULINK. There are no problems in the PMSG variables. All
the PMSG variables in the ATR don’t have problem as well.
This latest implementation takes 8ms to reach steady state, the full bridge takes 20ms and the ATR
takes 5ms.
The purpose of introducing an output filter Lload in the full bridge was to smoother the currents in the
transformer and in the IEGTs. As it can be seen in Fig 4.15 and Fig 4.16 this is accomplished.
Fig. 4.21 shows the losses for the converters:
47
Losses of 60kV 2MW converters
0,00
2,00
4,00
6,00
8,00
10,00
12,00
14,00
4 5 6 7 8 9 10 11 12
Wind speed(m/s)
Lo
ss
es
(%
)
boost
full bridge phase shift
control
full bridge duty cycle
ATR
Fig. 4.21. – Losses for the three converters discussed in this chapter in percentage of transmitted power.
It is seen that the losses decrease with the increasing transmitted power due to the core losses in the
transformers and inductors. For low wind speeds the ATR and the boost are better and for high wind speeds the
boost and the full bridge with duty cycle control have a small advantage. However, other factors need to be taken
into account. For example the ATR uses the 3 phases of the generator while the others need to converter to one
phase, and three phases is superior economically comparing with one phase [22]. Also the current in the boost
and in the full bridge is not sinusoidal, giving additional losses in the generator. Considering this and the small
difference between losses in the converters, the ATR is clearly superior in efficiency compared with the other
converters, for all wind speeds.
48
Chapter 5 – Speed and pitch control of the turbine
The torque control is necessary in order to avoid mechanical stresses in the wind turbine, and also to
avoid overcurrents in the system. As was said in Chapter 2 the speed control is necessary for the wind turbine
rotor to rotate at a speed where maximum wind power extraction can be done. These two controls are cascaded;
the output of the speed controller is the reference for the Iq current, which in the PMSG is proportional to the
torque. The current control works in a time scale 1000 times quicker than the speed control. Because of this, in
this chapter only the speed control will be simulated, otherwise it would lead to unnecessary large simulation
times. This means that the current is considered to be equal to its reference value.
The speed control is done in the same way for the three converters studied in the previous chapter, so
the speed control will be used only in the ATR. It is controlled with PWM modulation. The main goal of the
speed control is to extract the maximum power from the wind for all operating conditions. The output power
from the generator should be designed like in Fig. 2.2 .
For reasons of simplification and to lead to short simulation times, the turbine and PMSG are going to
be modelled with their steady state equations, not the dynamic model. Only the differential equation representing
the mechanical dynamics is maintained. The differential equations representing the voltage balance (2.15.) are
replaced by steady state equations (current derivatives equal to zero), as the current control is considered to be
ideal.
5.1. Description: Presentation of the matlab model used
The model used to simulate the turbine, and to compute the efficiency is in Fig. 5.1 :
we
Wind speed
Tw_g
wturbine
Vw
Tw_g
weref
Pt
Turbine Model
Te and turbine speed
weref
weIqref
Speed Control
wmG
Iqref
Te
Pinelectric
we
PMSG and ATR
Iqref
1/(ng)
G4
Te
Tw_g
wmG
DriveTrain2.501e+005
Display
Fig. 5.1 – Simulink model for the simulation of the one turbine.
The wind speed and turbine speed enter the block “Turbine Model” and it calculates the Cp, λ and the
reference generator frequency so the maximum power from the wind is extracted, as it is explained in the
Chapter 2, Section 2.1. It calculates also the turbine torque and the power extracted from the wind, Section 2.1.5.
The “block “speed control” calculates the Iq so that the generator will have the right torque for the speed
to follow the reference. Id will be zero all the time.
49
The block “Drive Train” with the current, flux of the generator and turbine torque computes the speed
of the turbine and the torque of the turbine from the generator side and the electromechanical torque, with the
mechanical equation from the turbine (2.9) and the equation of the torque (2.16).
The block “PMSG and ATR” contains the blocks “Voltage Cal” and “Determination Losses”. The block
“Voltage cal” computes Ud and Uq of the generator with Iq, we and the flux of the generator, using (2.15.). Now
the sub models are discussed:
5.1.2. Block “Turbine Model”
This block chooses the best Cp depending of the wind speed. If the wind speed is lower than 12m/s,
Cp=0.4. If not, it is calculated in (2.8.). If the wind speed is lower than 12m/s, λ=7. Otherwise it is calculated in
(2.5.).
Using (2.11.), (2.12.) and (2.13.) the torque and power of the turbine and the reference frequency of the
generator are computed.
5.1.3. Block “Drive Train”
This block is just the mechanical equation of the turbine (2.9), with just one detail. This equation is in
motor convention, it assumes the turbine is working as a motor. Actually the turbine is working as a generator,
so the torque from the turbine will need to change the sign. Jeq is the inertia seen from the PMSG and the
damping coefficient Bm is computed so the PMSG will have 3% of mechanical losses, as it was computed in
Section 2.2.4.
5.1.4. Block “PMSG and ATR”
In this block the electromagnetic torque is computed through (2.16). The reference for Id is zero as is
usual in this type of control.
Considering the current control to be much faster than the speed control, the derivative of the currents is
zero. With the flux, the electric speed and (2.15.) Uq is computed. (2.15.) is in motor reference. Iqref comes from
the speed control block in motor reference as well. As Id=0 the electric power will be UqIq, in motor reference.
So it needs a minus sign to be expressed in generators references.
5.1.5. Block “Speed Control”
Next there is the speed controller. It compares the electrical frequency of the PMSG, but it controls also
the mechanical speed of the turbine as they are proportional. The controller is a PI and the block diagram
representing the dynamics of the system is equal to Fig 3.3, except F(s) doesn’t have the pole in the origin. F(s)
will be 1/(Bm+Jeqs). However the parameters for the controller were computed as in the symmetry criterion and
the results were good. They were:
50
28
2
τ
τ
eq
i
eq
p
JK
JK
=
=
,
(5.1.)
where τ is the time constant of Mod(s) in Chapter 3, in this case, it represents the delay introduced by the torque
controller, i.e. current controller, for example 10 ms. However these formulas were derived in order to obtain
fast control. In reality the speed control is very slow as the mechanical system has a large inertia. For this a low
pass filter with time constant equal to 30s was introduced before the PI speed regulator, for the speed reference
change to vary more slowly and consequently the torque in the PMSG won’t be the maximum and won’t change
very suddenly which could mechanically damage the generator.
5.2. Simulation
The simulation program was used with a 5kV PMSG. The wind speed starts with 5m/s and changes to
6m/s at 3s. The reference frequency, the frequency of the PMSG and Iq are in Fig 5.2 :
0 10 20 30 40 50 60 70 80 90 100
300
310
320
Fre
quency R
efe
rence(r
ad/s
)
time(s)
0 10 20 30 40 50 60 70 80 90 100
300
310
320
Fre
quency(r
ad/s
)
time(s)
0 10 20 30 40 50 60 70 80 90 100-100
-80
-60
-40
Iq(A
)
time(s)
Fig 5.2 – Frequency reference and the frequency of the generator.
In the beginning, the current is negative because the initial speed is adjusted to the reference of the wind
speed 6m/s, which is 271rad/s. At 3s the rotor speed starts to rise and takes 90s to get to the reference of 6m/s
322rad/s. This is very indicative how much time does the turbine take to control its speed. The current will go to
zero at 3s for the torque to be maximum and change speed as fast as possible, but the speed reference is so slow
that the torque from the wind is too large and the current has to increase again to produce more torque and send
more energy to the grid. The higher limit of the controller is zero to prevent the PMSG to work as a motor which
51
could cause sudden power changes in the turbine. The lower limit was set to 100% of the rated current, which is
-400A. When the PMSG reaches its desired speed, Iq will be negative again to produce torque that will balance
the torque from the wind and the sum of all torques will be zero, thus the speed will be constant.
In summary it can be seen that the current and the speed vary very smoothly, which is the goal of the
speed control.
In the next simulation the wind speed changes from 11 to 10.5m/s at 3s in Fig 5.3. :
0 10 20 30 40 50 60 70 80 90 100
570
580
590
Fre
quency R
efe
rence(r
ad/s
)
time(s)
0 10 20 30 40 50 60 70 80 90 100
570
580
590
Fre
quency(r
ad/s
)
time(s)
0 10 20 30 40 50 60 70 80 90 100-400
-350
-300
Iq(A
)
time(s)
Fig 5.3 – Electric power in generator and mechanical power in the turbine.
As can be seen in Fig. 5.2., the speed takes 90s as well to get to the reference value, and the current
won’t change so suddenly as in the previous case, for the same reasons. In both cases the speed follows the
reference perfectly.
5.3. Conclusion
It is concluded that the speed control is much slower than the current control as it should be. The time to
get to steady state is the same if the wind speed rises or falls, because the current doesn’t get to the maximum, so
the system is linear. So the time to get to the reference is due to the characteristics of the system, not the
amplitude of the reference.
The speed control operates well for all condition and the PMSG cannot work as a motor as that gives
much power fluctuations in the wind turbine. For this the upper limit of the speed regulator should be 0, for Iq to
always be negative.
In order for the control to be smooth and not as fast as possible to prevent mechanical stresses in the
PMSG a low pass filter was introduced before the regulator for the speed reference change to be slower, and the
necessary torque to follow it to be smoother.
52
Chapter 6 - Control of the main inverter onshore
The DC voltage in the submarine cable is controlled by the converter onshore. As a consequence its
voltage doesn’t change very much, and the power from the wind turbines can be represented by a current source,
the system can be separated in two parts: from the wind turbines to the submarine cable, and from this point to
the electrical grid. In this chapter only the second part is going to be treated. The approach is geared towards
finding good values for the PI regulators that control the DC voltage in the cable. This controller gives the
reference for an inner current controller, and the active power though Id and reactive power though Iq.
6.1. Block diagram of the Plant
The system in study in this chapter is therefore depicted in Fig 6.1:
Fig 6.1 – Wind park from the submarine cable to the grid.
The inverter uses bidirectional current switches; the most used ones are IEGTs with diodes in anti-
parallel. The DC voltage is always positive; which means that the active power can be positive or negative. The
method for controlling this inverter is the SVPWM.
It permits that the amplitude and phase of the three voltages can be adjusted freely under the physical
limits of the system. The voltage can be in delay or advance to the current, which means that the reactive power
that goes to the grid can be positive or negative.
Considering that all wind turbines are represented by a current source that delivers current to the
submarine cable, they can be represented by a circuit in Fig 6.2. :
g
A
B
C
+
-
VSC
A
B
C
a
b
c
Three-Phase
Transformer A B C
Grid
Current Source C(cable)
Fig 6.2 – Electric circuit that represents the wind park from the submarine cable to the grid.
Let us neglect the harmonics created by the VSC, that is, consider that the inverter can create 3 phase
voltages perfectly sinusoidal, so it can be represented by a 3 phase voltage source. The amplitude of these
voltages can be controlled and the phase can be regulated as well. The circuit is represented in Fig 6.3:
53
u3u2u1e3 e2 e1
R and L (phase3)
R and L (phase2)
R and L (phase1)
Fig 6.3 – Simplified electric circuit.
Note that the impedance represented in Fig 6.3 is the sum of the leakage impedance of the transformer
in Fig 6.2 with the Thevenin impedance of the grid from the PCC. If the voltage sources from the grid (e1, e2 and
e3 in Fig. 6.3.) are equal to the voltage level on the secondary side of the transformer, the leakage impedance is
referred to the secondary side. Otherwise, it is referred to the primary side. The values used for this impedance
can be found in the appendix.
This simplification is very important because it neglects the delay from the PWM modulation; it
considers that the VSC creates 3 sinusoidal voltage signals instantaneously. This will result in that the current
control transfer function in close loop will be a first order function transfer. This simplification is valid because
the switching frequency is assumed to be high enough, so the delay will be low enough.
As it can be seen in Fig 6.3, in each phase there are 2 sinusoidal voltage sources connected by an
impedance in between. This circuit is similar to the synchronous machine connected to the grid in steady state.
The no load electromotive force of the machine is analogous to the voltage created by the PWM in the inverter.
The impedance in between should be the sum of the impedance of the machine and the Thevenin impedance of
the grid seen from the PCC, is now the leakage impedance of the transformer plus the Thevenin impedance of
the grid seen from the PCC.
As this important conclusion is stated, the equations from the synchronous generator presented in
Chapter 2 can be used again. As it will be seen, the method onward used will be very similar to the control of the
synchronous generator with an ATR presented in Chapter 4.
Looking at Fig 6.3, we can write the Kirchhoff laws of the voltages for each phase, where u is the
voltage created by the VSC and e is the voltage from the grid side.
++=
++=
++=
33
33
22
22
11
11
edt
diLiRu
edt
diLiRu
edt
diLiRu
transtrans
transtrans
transtrans
. (6.1.)
The active power is positive when the wind park sends energy to the AC network. Applying the Park
Transformation to these equations, the following expressions are obtained:
54
+++=
+−+=
qdtransgrid
q
transqtransq
dqtransgrid
d
transdtransd
eiLdt
diLiRu
eiLdt
diLiRu
ω
ω. (6.2.)
Note that these equations are very similar to the PMSG equations in Chapter 2 (2.15), except there is no
flux and ed doesn’t appear in (2.15). The equations are represented in Fig 6.4 as a block diagram:
Fig 6.4 – Block diagram of the Plant, relating Ud and Uq with Id and Iq.
6.2. Inner Current Control
It is now desired to control the current Id (in order to control the voltage of the submarine cable) and Iq,
(in order to control the reactive power that goes to the AC grid). Because the angle for the dq transformation is
the same angle in the alfa-beta plane, the d axis is aligned with the alfa-beta vector at all times. This means that
Uq is always zero. So:
=−=×=
=+==
qddqqd
ddqqdd
iuiuiuIUQ
iuuiiuIUPrr
rr.
. (6.3.)
Since it is desired to control Q to the PCC and not the reactive power that leaves the inverter, the
voltage used instead of Ud is Ed.
The control of Id and Iq is performed in the usual way: the reference is compared with the value, the
error passes though a PI controller, and the result will be ud or uq depending if Id or Iq are controlled, see Fig 6.5,
where Fig 6.4, the block diagram of the plant was put together with the control:
55
Fig 6.5 – Block diagram with the Current Control System and the Plant.
As it can be seen in Fig 6.5, the block diagram for Id depends on Iq and vice-versa. It is concluded that
the systems are coupled. In order to de-couple them, the control block is altered in the following way, see Fig
6.6:
Fig 6.6 – Block diagram with the Plant and the Current Control System altered in order to de-couple Id
and Iq.
It is seen now that the signal that subtracts with ed and eq is now independent of the other current. The
systems are now de-coupled and can be represented in Fig 6.7:
56
Fig 6.7 – Block diagram with Id and Iq de-coupled.
Note that Fig 6.7 is the block diagram representing the dynamics of the system in Fig. 6.6. Fig 6.7
doesn’t have any physical representation it is just used to compute the parameters for the PI controller.
The voltage Eq is always zero as the d axis is aligned with the α-β vector. The voltage Ed is always
constant, as Ea, Eb, Ec are considered to be 3 perfect sinusoids with the same amplitude and phase shifted of 120
degrees. As so Ed can be viewed as a constant perturbation and it will be compensated by the integral part of the
PI controller.
With Ed and Eq neglected, Fig 6.7 is a system in the form of Fig.3.1. The pole s=-Rtrans/Ltrans is not too
close to the origin so the method of dominant pole can be used. From Chapter 3 and considering R to be 20 times
less than ωgridLtrans:
7.1520
314
20=====
grid
transgrid
transgrid
trans
trans
p
i
L
R
L
R
K
K ω
ω
ω. (6.4.)
6.2.1. Determination of the voltage level on the primary side of the transformer
This system (Fig 6.1) is the same as the system as the one in Section 4.5. The section to determine Udc
and Ltrans can be used. The same equation appears. Em is the peak line to ground voltage and Id is the nominal
current. As UDC is 200kV, for the voltage level of the primary side of the transformer the 60kV level is chosen.
This means kVkV
Em
493
602 == . As the nominal power than leaves the inverter is 200MW, the nominal
current Id in the AC side is:
Au
Pi
d
d 333310.60
10.2003
6
=== . (6.5.)
Using these values in the formula:
57
( )[ ] kViLEU dgridtransmDC 873 22 =+> ω . (6.6.)
This is smaller than 200kV. This means the inverter can operate with the voltage level 60kV in the
primary side of the transformer. The other equation is:
mHi
EU
Ldgrid
m
DC
trans9.99
3333314
1009.1310
22
=×
×=
−
<ω
. (6.7.)
The Ltrans used is 11.5mH (calculations in appendix). It doesn’t violate (6.7.).
6.2.2. Determination of the PI parameters for the current control
Writing the system in Fig 6.7 in close loop, using the method of dominant pole, F(s) is Kp/(sLtrans):
( )( )
( )( )
( )p
transdref
d
K
sL
sF
sF
sF
sI
sI
+
=
+
=+
=
1
11
1
1
1. (6.8.)
The system in close loop is a first order system with a pole s=-Kp/Ltrans. Normally the system is designed
for the dynamic response to have a bandwidth 7 to 10 times larger than the 314rad/s [3]. Considering 7 times, the
pole will be: s=-7x314=-2198. So Kp=2198Ltrans=25.3Ω.
Using (6.4.):
18.3963.257.157.15 −Ω=×== radsKK pi . (6.9.)
Ω=
Ω=−18.396
3.25
radsK
K
i
p. (6.10.)
6.3. Simulation of the system with current control
The model above discussed is only valid for small perturbations. For large perturbations, the regulator
will put a very high voltage, until it reaches its limits. If the id (or iq) is lower than its reference, the regulator will
increase the voltage ud (or uq) in order to increase the current. In normal conditions ud is 60kV. Therefore the
higher limit that the regulator could set ud would be good a little bit higher than that, to increase the current
quickly. It is usual to set the limits of the regulators a bit higher than the normal situation. (This will happen
ahead in the voltage controller). The limits for the current control were set to 70kV and -70kV. When it reaches
58
its limits the regulator gives more voltage than usual, and consequently the current changes faster than usual than
the speed predicted in the model for small perturbations. For these reason, the reference for id and iq was set to
small values, 100A and -50A respectively.
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
0
20
40
60
80
100
Id(A
)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
-50
-40
-30
-20
-10
0
Iq(A
)
time(s)
Fig 6.8 – Id and Iq using the Simulation Model.
It can be seen in Fig 6.8 that both Id and Iq have a first order dynamic response, and looking closely in
Fig 6.9 and Fig 6.10;
0.008 0.009 0.01 0.011 0.012 0.013 0.014 0.0150
10
20
30
40
50
60
70
80
90
100
110
Id(A
)
time(s) Fig 6.9 – Transient of Id zoomed in.
59
0.018 0.019 0.02 0.021 0.022 0.023 0.024
-50
-40
-30
-20
-10
0
Iq(A
)
time(s) Fig 6.10 – Transient of Iq zoomed in.
It can be seen that both Id and Iq take 3ms to reach the reference, which is 7 times smaller than 20 ms,
(period of 50Hz) as predicted in the model.
In Fig 6.8 it is clearly seen that Id and Iq are de-coupled, they are completely independent and the
control of each one doesn’t interfere in the other one, as Id as instructed to change at 0.01s and Iq to change at
0.02s. This is why the VSC can control active and reactive power separately and independently.
6.3.1. Test with Id and Iq coupled
A test was now performed to see what would happen if in the control block in Fig 6.6, the ωgridLtrans gain
was set to zero both in the Id and in the Iq. The limits for the PI controller for the Id and Iq were 70kV and -70kV,
for the same reason as with the previous simulation. The results are presented in Fig 6.11 :
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
0
20
40
60
80
100
Id(A
)
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05
-60
-40
-20
0
Iq(A
)
time(s)
Fig 6.11 – Id and Iq with the control system with Id and Iq coupled, Fig 6.5.
As it can be seen, now Id and Iq are not de-coupled. At 0.01s when Id changes, Iq changes also without
be order so. The same happens as at 0.02s. The steady state error in both currents takes a long time to reach zero.
It can be concluded that this type of control is worse in all aspects comparing with the previously one. Especially
nowadays when the parameter ωgridLtrans is easily measured, a de-coupled control should always be used.
60
6.4. Voltage control
From the results in the section of the current control, the following block diagram relating the voltage
Udc with Id can be discovered in Fig 6.12 :
Fig 6.12 – Transfer Function of the Current Control System and the Plant (Fig 6.6) in close loop.
Remembering the system to control, the voltage and currents that are going to be used are represented in
Fig 6.13:
Fig 6.13 – Circuit with the currents and Voltages used in the voltage control system.
The active power that enters the VSC is the same as the active power that leaves the VSC, considering
that there are no losses in the switches. Considering that the resistor in the transformer and in the Thevenin
impedance of the grid is very small; there are no active power losses as well to the grid.
ddDCDCEIUIP == . (6.11.)
Applying Kirchhoff Law for the currents:
ddDC
DC
onshoreg EIUdt
dUCI =
− . (6.12.)
With this equation we can relate UDC with Id in Fig 6.14 :
Fig 6.14 – Block diagram with Voltage control in close loop.
61
Udc in the action block can be considered to vary not much (that’s what the voltage control achieves)
and it can be considered that the system will suffer perturbations in the parameters when Udc varies. This system
is stable and so variations in the parameters won’t cause any problems. The function transfer 1/(sConshore) will
pass for the action loop, and so the gain of the action loop excluding the regulator is:
onshoredc
d
CU
EZ = . (6.13.)
τ is the time constant of the current control. In the current control section it was set to
1/2198=0.455ms. The calculations are in the m file that is presented in appendix. With these two parameters Ki
and Kp for the voltage control are found easily thought the symmetry criterion:
366.02
1−=−=
τZK
p. (6.14.)
2018
12
−=−=τZ
Ki
. (6.15.)
The parameters for the current control are computed in the previous section. Ltrans and Rtrans are
computed in annex. The value used for the capacitance is 100µF.
6.5. Simulation of the voltage control for large perturbations
The voltage control system is shown in Fig. 6.14, the current control and the plant are presented in Fig
6.5. It was considered that large perturbations happen when the voltage varies 50%. Therefore in the simulation
the step considered was from 500 to 1kA at 0.03s. The value of the capacitor is initially 200kV, equal to the
voltage reference. The limits for the PI dc voltage controller were set in a way that the limit current was 4kA, a
bit higher than the rated current 3.33kA. The results are shown below, in Fig. 6.15, and 6.16:
0.03 0.035 0.04 0.045 0.05 0.055 0.061000
1500
2000
2500
3000
3500
4000
4500
Idre
f(A
)
time(s)
Fig 6. 15 – Id response simulated in Simulink (Large perturbations).
62
0.03 0.035 0.04 0.045 0.05 0.055 0.06190
192
194
196
198
200
202
204
206
208
210
UD
C(k
V)
time(s)
Fig 6.16 – UDC response simulated in Simulink (Large Perturbations).
At 0.04s Id reaches the maximum 4kA (it means the voltage regulator got to its upper limit). It stabilizes
at rated current 3.33kA, because at this time the current from the park is nominal (1kA). If the limits of Idref were
higher than 4kA, the system would take less time to follow the reference. The downside is that the IEGTs have a
maximum current they can withstand. If the limit of the regulator is too high the IEGTs can be destroyed.
The analysis of the voltage UDC in Fig 6.16 concludes that the voltage suffers a small perturbation due
to the change in the current Ig but it stabilizes very quickly, in 10ms.
6.6. Simulation of the voltage control for small perturbations
Now small perturbations will be tested. It is considered when the current from the park changes 10%.
For example, when changes from 900 to 800A. The results are in Fig.6.17 and 6.18:
0.03 0.035 0.04 0.045 0.05 0.055 0.06190
192
194
196
198
200
202
204
206
208
210
UD
C(k
V)
time(s)
Fig 6. 17 – UDC response simulated in Simulink for small Perturbations.
63
0.03 0.035 0.04 0.045 0.05 0.055 0.061000
1500
2000
2500
3000
3500
4000
4500
Idre
f(A
)
time(s)
Fig 6.18 – Idref response simulated in Simulink for small Perturbations.
Looking at Fig. 6.17 and Fig. 6.18 it is seen that the voltage has a small perturbation but stabilizes
quickly as in the large perturbation case, and Idref changes quickly also to keep up with the decrease in Ig from
900 to 800A. It takes 8ms to change.
6.7. Conclusion
In this chapter it is concluded that the VSC onshore can control the reactive power that flows to the grid
and the voltage of the submarine cable, independently from each other. In the simulations it was seen that the
current control works really well, as in Fig.6.21 the current follows the reference. The voltage control works well
too and is not sensitive to small, large perturbations or variations in its parameters.
64
Chapter 7 – Analysis of the connection of the wind park
In this Chapter the topology of the DC grid will be investigated in order to obtain the best efficiency
possible. In the first option presented the grid philosophy will follow [8] with two voltage levels: 12kV and
60kV. In the second option there is only one voltage level: 60kV. With the losses the output power for each wind
speed will be known, and integrating it with the wind distribution the average power and yearly energy of the
park is known.
7.1. Best Connection: Parallel or Series?
In [11] many possible choices for connecting a wind park were discussed. Only the options for a DC
park will be addressed, which are:
• Wind turbines connected in parallel, one or two steps to raise the voltage.
• Series connected turbines.
These two options have several advantages and disadvantages. The Series connection has the advantage
that DC/DC converters to raise the voltage are not needed because the turbines are connected in series, so the
voltage will be raised by adding up the output voltage from the turbines directly. The disadvantage is that the
generators have to be insulated for high voltages, a process which is difficult and expensive to do.
The parallel connection doesn’t have the insulation problem (disadvantage) of the series connection, but
instead it needs DC/DC converters to raise the voltage. According to some authors [11] it says that the series
connection is cheaper than the parallel connection. In this report only the parallel connection is going to be
explored because it’s the one which is best developed.
Two steps to raise the voltage are adopted, as it is hard to raise the voltage from 690V AC to 200kV DC
in only one step. It would be possible with one full bridge converter in each turbine, for example, but the output
voltage in the transformer would be too high and this would cause high costs. The best topology for a 200MW
wind park with 2 steps is determined in [8]. It is 5 turbines connected in parallel to a DC/DC converter. This
means there will be 20 DC/DC converters in total in the park.
The approach proposed for the park is in Fig. 7.1:
65
Fig. 7.1 – Proposed connections for the Wind park.
The voltage level 200kV was chosen so the nominal current in the main cable will be
200MW/200kV=1000A, which is an appropriate value. A voltage level of 60kV was chosen as analogy to the
inland grids, because this voltage level is common. The ratio of the main DC/DC converter would be around 3-4,
which is a suitable value. The voltage level of 12kV was chosen in order for the current to be low in the cables
and for the ratio of the 12/60kV converter not to be too high. In this case, this ratio is 5, acceptable also.
Of all the converters seen in Chapters 4 and 5, the only one that can be used in the 12/60kV and
60/200kV converters is the full bridge converter, as the voltage ratio is more than 3 times, the efficiency of the
full bridge is much better than the efficiency of the boost.
There are several ways to raise the voltage to 12kV (DC) after the wind turbines. In Fig. 7.2 the losses
are shown for the 3 converters in this thesis. All the variables in the following tables are in percentage, relative to
the transmitted power of the converter:
Losses for 12kV converters
0,00
1,00
2,00
3,00
4,00
5,00
6,00
7,00
8,00
9,00
4 5 6 7 8 9 10 11 12
Wind speed (m/s)
Lo
sses (
%) boost
full bridge phase shift
full bridge duty cycle
ATR
Fig. 7.2 – Losses for 12kV converters in this thesis.
66
It is seen that the losses in the boost converter decrease with the transmitted power because the inductor
was designed for its copper losses to be very small. The losses in the ATR will decrease to 1% because the
voltage level is very low and consequently the IEGT and inverse diodes will be fast to switch, (only 2.6µs fall
time, leading to low switching losses) and few semiconductors in series are needed (leading to low conduction
losses). The full bridge converter has high losses for low loads due to the core losses. The boost converter option
cannot be used as the current in the PMSG will not be sinusoidal, (id is not zero) leading to extra losses in it, the
torque is oscillating strongly which can deteriorate the shaft, and very few semiconductors are used for very high
powers. For this the ATR will be chosen instead for option1. In option2 it is the full bridge that will be chosen
looking to Fig. 4.21.
7.1.1. Wake effect
The wind speed is never the same for all turbines. This is due to the fact that when the wind passes
thought a wind turbine it losses some kinetic energy to the turbine. This is known as the Wake effect. This effect
is very complex, but a fair approximation to model it is to consider that the wind speed in the turbines in the
edges of the park is 100%, and the wind speed hitting the inside turbines is 80%. The first group will be called
group1 and the second group2. As there are 10*10 turbines in a square, there are 38 turbines in group1 and the
remaining ones in group2, as it is considered the wind hits all edges of the square.
7.2. Option 1: 6kV PMSG with ATR to 12kV
7.2.2. Loss calculation
As the wind speed is different, the power and the current in which PMSG is different also. In Table 7.1
some variables are computed for every wind speed.
Table 7.1 – Losses for the 12kV converter, currents and power in the 2 groups.
WS G1(m/s)
WS G2(m/
s)
Losses of
G1(%)
Losses of G2 (%)
Outpower all turbines (MW)
Iout(A) group1
Iout(A) group2
Pin Cluster1
(kW)
Pin Cluster2 (kW)
4 3,2 2,62 100,00 2,58 5,97 0,00 71,67 358,36
5 4,0 2,07 2,62 9,65 11,73 5,97 427,46 703,87
6 4,8 1,68 1,80 16,79 20,35 10,41 743,79 1221,13
7 5,6 1,40 1,50 26,73 32,41 16,58 1184,64 1944,64
8 6,4 1,20 1,40 39,97 48,48 24,77 1770,71 2908,67
9 7,2 1,10 1,33 56,95 69,09 35,29 2523,23 4145,64
10 8, 1,01 1,20 78,21 94,87 48,48 3465,32 5691,93
11 8,80 0,86 1,10 104,23 126,46 64,59 4617,78 7587,43
12 9,6 0,76 1,00 135,46 164,34 83,94 6001,19 9860,49
13 10,4 0,76 0,90 153,04 164,34 106,83 7099,90 9860,49
14 11,2 0,76 0,80 173,57 164,34 133,56 8383,06 9860,49
15 12,0 0,76 0,76 197,21 164,34 164,34 9860,49 9860,49
67
In Table 7.1 the losses from the 12kV ATR were taken for the respective wind speed. Dividing the
power by the voltage 12kV, the current Iout in both groups is obtained. The power that enters the 12/60kV
converters will be different depending of the cluster of 5 turbines. If the cluster has 5 turbines in the edges
(called Cluster1), the current is 5 times the current in group1. Otherwise it will have 4 turbines inside and 1
turbine in the edge (called Cluster2). The current will be the current in group1 plus 4 times the current in group2.
The output power is the sum of all the power that leaves the turbines, in the beginning of the 12kV cables.
Next there are other variables in Table 7.2 :
Table 7.2 – Losses for cables and 12/60kV full bridge converter.
Losses 12kV
cables(kW)
Losses 60kV
cables(kW)
Losses
(12/60)
1-4(kW)
Losses
(12/60)
5(kW)
Losses
12/60kV
converter
total (kW)
Losses
60/200kV
converter
(kW)
0,30 0,15 0,00 12,97 51,89 92,03
1,50 1,34 15,47 14,36 305,02 196,35
4,54 4,06 15,62 16,49 315,86 243,75
11,50 10,30 14,22 20,03 307,57 317,10
25,72 23,02 20,36 25,31 427,03 355,85
52,25 46,75 25,23 32,75 534,72 411,86
98,51 88,15 30,15 44,40 659,96 504,09
175,01 156,57 36,48 59,94 823,45 651,49
295,58 264,44 47,41 68,04 1030,70 860,35
337,48 330,61 55,38 68,04 1158,22 972,05
399,14 421,95 66,23 68,04 1331,77 1102,33
487,13 545,86 68,04 68,04 1360,75 1253,43
The model of the wake effect has error and so these results are approximated. In order for the
calculations to be easier, the input power of the 12/60kV, 60/200kV, VSC onshore, and cable losses is
considered to be always the power from the wind, not the actual input power of each converter. This
approximation won’t give a lot of error as the losses in all converters are very low for high wind speeds, where
the wind is more frequent in the site considered (10m/s average wind speed). The losses for the 12/60kV and
60/200kV converters were taken from Fig. 7.4 and Fig. 7.4 :
68
Losses for the full bridge converter 12/60kV 10MW
0,00
0,50
1,00
1,50
2,00
2,50
3,00
3,50
4,00
4 5 6 7 8 9 10 11 12
Wind speed (m/s)
Lo
sses (
%)
phase shift
duty cycle
Fig. 7.3 – Losses for the 12/60kV converter.
Losses for 60/200kV 200MW
0,00
0,50
1,00
1,50
2,00
2,50
3,00
3,50
4,00
4 5 6 7 8 9 10 11 12
Wind speed(m/s)
Lo
sses(%
)
phase shift
duty cycle
Fig. 7.4 – Losses for the 60/200kV converter.
It is seen that for both converters the losses are below 1%, this is due to the transformers and inductors
that were chosen in order to have losses as low as 0.15%, 0.25% at rated power. In Table 7.3 the rest of the
variables are presented:
69
Table 7.3 – Losses in the main cable, VSC onshore and output power of the park.
Losses main cable (kW)
Losses VSC onshore(kW)
Pout(MW)
0,55 38,00 2,40
7,71 152,73 8,99
24,22 293,25 15,90
62,71 515,70 25,51
141,25 851,12 38,14
288,59 1381,83 54,24
546,16 1943,17 74,37
971,47 2646,48 98,81
1641,36 3522,49 127,85
2095,20 3973,36 144,17
2694,48 4497,41 163,12
3483,80 5102,68 184,98
The current in the main cable is the power that arrives to the cable dividing by the voltage 200kV. The
square of it times the resistance of the cable gives the losses in the cable. The losses for the VSC onshore were
done the same way for the 12/60kV and 60/200kV converters. They can be seen in Fig. 7.5.:
VSC onshore Losses
0,00
0,50
1,00
1,50
2,00
2,50
3,00
3,50
4,00
4 5 6 7 8 9 10 11 12
Wind speed(m/s)
Lo
sses(%
)
VSC
Fig. 7.5 – Losses in the VSC onshore.
As the margin for the voltage is 50%, the voltage in each IEGT will be 4.5kV/1.5=3kV. Using the
normal voltage derivative in the IEGT catalogue, the time the IEGT takes to go to the off state is 10µs, and to go
on state is 0.78µs. With these results if the switching frequency used is 500 Hz the switching losses are
acceptable (0.52%).
The cable length of the cables is computed to get the resistance.
70
7.3. Calculation of the cable length
7.3.1. Length and resistance for 12kV cables
A program of optimization could be used to find the best way to connect all turbines in order to
minimize the cable length and consequently the losses and costs. However this is outside of the scope of this
work, so an elementary way to interconnect all turbines will be used. It’s not proved that is the best one, it’s
going to be used just to get an idea of what the total efficiency of this park could be.
Looking at Fig. 7.1, 5 turbines will be connected in parallel, in a group which will be called cluster.
These turbines will be connected in the connecting point, where it is the DC/DC converter 12/60kV. It will be
placed on the middle turbine of the 5 group, but possibly it’s not the best choice in order to minimize cable
length. According to [8] it is.
The representation is in Fig. 7.6. :
Fig. 7.6 – Representation of one quarter of the wind park. Each circle is one turbine.
Each turbine will be distanced by 400 meters. Let us assume that each cluster is 5 turbines in line. It is
seen that 4 clusters in the park will have 5 turbines with 100% wind speed (Cluster1) and the others 16 clusters
will have 1 turbine in the edge and the other 4 with 80% wind speed (Cluster2). The total cable length for
turbines with 100% wind speed will be:
( )( ) kmLG
4.2240016242121 =××+××+= . (7.1)
Also for each cluster the length of the hub tower, the depth of the sea bed and the depth which the cable
is buried need to be taken into account. Resuming the total length for each cluster will be:
( )( )( )buriedDepthseaDepthTowerburiedcableLenght
Lcable
__25__2 +++=
= (7.2)
71
Looking at [6] the hub height it could be 100 meters, and looking at [12] the depth which the cable is
buried is 1 meter. In the North Sea the depth of the sea where offshore parks can be built is usually 40meters.
( )( )( ) kmLcable 6.4614021005224002 =+×+×+= (7.3)
This is the length for the turbines of group1. For turbines of group2:
( ) kmLG
6.25164001222 =×××+= . (7.4)
Considering the height of the tower, depth of the buried cable and the sea:
( )( )( ) kmLcable 5314021005256002 =+×+×+= (7.5)
Consulting [12] it is seen that a cable with 95mm2 is appropriate for this current, as its maximum
current is around 370A. The resistance of the cable per unit of length will be:
16
9
05.1811095
102.17 −
−
−
Ω=×
×== m
SR
copper
cable µρ
, (7.6)
Where S is the cross section area and copperρ is the resistivity of the copper (17.2nΩm). The resistance
for the 4 clusters of group1 is Ω=×× − 44.81005.18146600 6 and the resistance for the 16 clusters of
group2 is Ω=×× − 59.91005.18153000 6 .
Concluding the total losses in these cables will be:
2
2
2
112 12
59.912
44.8
+
=
kV
P
kV
PP
groupgroup
kVohmic . (7.7)
Pgroup1 and Pgroup2 are the power of the turbines in group1 and group2 respectively. It is the power from
the wind speed (100% for group1 and 80% for group2) minus the losses in the 12kV ATR.
7.3.2. . Length and resistance for 60kV cables
It is considered that all turbines are in a square aligned 10x10, with a distance of 400 meters from each
other. Each cluster is 5 turbines aligned, and each connecting point of each cluster is on the middle turbine. It
means all the connecting points are aligned as well, in two lines of 10 collecting points, and the lines are
72
separated 400*5=2km (Fig 7.7). It will have 4 clusters with 5 turbines of group1 and 16 clusters with 1 turbine of
group1 and 4 turbines of group2. Joining all the points of these 16 clusters in each line in the middle of the line:
( ) kmnLn
cable 8.442000165.0224003
0
=×++××= ∑=
(7.8)
Fig. 7.7 – Representation of half of the wind park.
Fig 7.7 shows half of the whole park. The other half is symmetric, with the other line (or set of
collecting points) parallel to the one in Fig 7.7. Considering the height of the tower and the depth of the sea the
same way as used for 12kV cables:
( )( )( ) kmLcable 42.9114021005448002 =+×+×+= (7.9)
The length of the 4 clusters with 5 turbines of group1 will be:
kmLcable 2.92000224005.4 =+×××= (7.10)
Considering the height of the tower and the depth of the sea:
( )( )( ) kmLcable 22.201402100592002 =+×+×+= (7.11)
As the nominal current in the 60kV is the same as in the 12kV, the cross section of 95mm2 can be used
as well. The resistance per unit length is the same, and multiplied by 91.42km gives 16.55Ω, the resistance of the
16 clusters with 1turbine of group1. Multiplying by the resistance per unit length by 20.22km gives 3.66Ω, the
resistance of the 4 clusters with 5 turbines of group1. The total losses for the 60kV cables will be:
2122
60 6066.3
6055.16
+
=
kV
P
kV
PP
Cluster
in
Cluster
in
kVohmic (7.12)
73
7.3.3. Length and resistance for the main cable
The main cable is the cable that links the DC/DC converter of 60/200kV that gathers all power from the
wind park to the PCC onshore. It has a voltage of 200kV and a nominal current of 200MW/200kV=1000A.
Consulting [12] the cable chosen has a cross section of 1400mm2, wide spacing, and 100% of armour resistance
as percentage of conductor resistance. This cable has a current rating of 1600A, which has a good safety margin.
The resistance per unit length of this cable will be:
16
9
286.12101400
102.17 −
−
−
Ω=×
×== m
SR
copper
cable µρ
. (7.13)
Multiplied by the distance to onshore location, 300km, the resistance will be 3.68Ω.
7.4. Option2: 10kV PMSG with Full Bridge Converter to 60kV
This converter with a 10kV PMSG will have the lowest losses of all converters studied in this thesis.
Therefore it will be used. The calculations and reasoning are the same as the ones used for option1. Now the
voltage will go immediately to 60kV so there is no need for a 12/60kV converter. The losses in the 60kV cables
are the losses of the 60kV cables in option1 plus the losses in the 12kV cables in option1, except the voltage
used is 60kV not 12kV.
2
2
2
1
2
2_1
2
1_160 60
55.1660
66.360
59.960
44.8
+
+
+
=
kV
P
kV
P
kV
P
kV
PP
in
cluster
in
clustergroupturbinegroupturbine
kVohmic (7.14)
The first variables are presented in Table 7.4 and Table 7.5
Table 7.4 – Losses for 60kV converters and currents in both groups.
WS G1 (m/s) WS G2 (m/s)
Losses of group1
(%)
Losses of group2 (%)
Outpower of all turbines (MW)
Iout(A) group1
Iout(A) group2
4 3,20 11,94 100,00 2,33 1,08 0,00
5 4,00 6,51 11,94 8,99 2,24 1,08
6 4,80 4,11 6,60 16,18 3,97 1,98
7 5,60 2,93 4,70 26,10 6,38 3,21
8 6,40 2,28 3,50 39,33 9,59 4,85
9 7,20 1,91 2,70 56,33 13,71 6,96
10 8,00 1,72 2,28 77,51 18,84 9,59
11 8,80 1,63 2,00 103,36 25,10 12,80
12 9,60 1,62 1,80 134,32 32,58 16,65
13 10,40 1,62 1,66 151,80 32,58 21,20
14 11,20 1,62 1,63 172,10 32,58 26,49
15 12,00 1,62 1,62 195,50 32,58 32,58
74
Table 7.5 – Losses in the 60kV cables, 60/200kV converter and in the main cable.
Losses 60kV cables(kW)
Losses 60/200kV converter (kW)
Losses main cable (kW)
Losses VSC onshore(kW)
Pout(MW)
0,14 84,92 0,46 35,06 2,21
1,23 188,68 7,12 146,78 8,64
3,95 239,36 23,36 287,98 15,62
10,28 313,10 61,14 509,21 25,21
23,32 353,78 139,61 846,17 37,97
47,81 410,89 287,22 1378,57 54,21
90,49 503,24 544,31 1939,90 74,43
160,86 650,14 967,47 2641,08 98,94
271,63 857,94 1632,17 3512,73 128,05
338,51 969,33 2083,49 3962,40 144,44
430,51 1098,67 2676,61 4482,71 163,41
555,59 1247,65 3451,72 5079,56 185,17
The efficiency of both options is in Fig. 7.8 . The efficiency here is the ratio between output power and
output power from the PMSGs, in Fig. 2.2 .
Efficiency of wind park
76,00
78,00
80,00
82,00
84,00
86,00
88,00
90,00
92,00
94,00
96,00
4 5 6 7 8 9 10 11 12 12 12 12 12
Wind speed (m/s)
Eff
icie
nc
y(%
)
option1
option2
Fig. 7.8 – Efficiency of the whole park using option1 and option2.
7.5. Energy production of the park
The expected value of power produced, Pavg, can be computed though the formula:
( ) ( )dvvfvPPavg ∫∞
=0
(7.15)
75
The probability density function of the wind is ( )vf , ( )vP is the output power from the whole park for
each wind speed, and v is the wind speed. The Weibull distribution is described by the following probability
density function:
( ) ( )kcv
k
ec
v
c
kvf
/1
−
−
= (7.16)
Where k is a shape parameter, c is a scale parameter and v is the wind speed. VESTAS specifies the parameters
k=2 and c=11.38. As k=2 the distribution is called the Rayleigh distribution. In these distributions the parameter
c and the average wind speed can be related by:
2
πcv = (7.17)
With c=11.38 the average wind speed would be 10m/s. The columns of Pout in Tables 7.3. and 7.5. are
used for P(v) for options 1 and 2 respectively. They are used until the wind speed of 25m/s. After that the wind
turbines in the edges stop working and the inside turbines continue to work because their wind speed is less than
25m/s. Consequently the output power from 25m/s to 31m/s (31m/s is the wind speed when the inside turbines
stop working) is much less than the rated value. In this range of wind speeds the output power in the whole park
for both options is 122MW. This is accounted in (7.14) to compute the average power for the whole park. The
result was 82MW for both options. The number of hours in a year is 8640. This means the number of hours at
nominal power is 82/200*8640=3500hours. The typical number for offshore wind parks is between 3000 and
4000 hours, concluding that this park has a normal value of hours at nominal power. The energy produced by the
park in one year will be:
GWhPE avgproduced 5.7088640828640 =×== . (7.18)
7.6. Conclusion
It was concluded that if the grid of the wind park will use the voltage level 12kV and 60kV (option1)
then the best converter to get to the voltage level of 12kV is the ATR.
Also looking at Fig. 7.8 it is seen that option1 gives more energy to the grid, which means two voltage
levels of 60kV are better than one voltage level. This is probably due that the 12kV ATR has 1% losses (used in
option1) while the 60kV full bridge has 2% losses (used in option2).
However looking at Fig. 7.8. it is seen that the efficiency of both options is almost the same from 8m/s
onward. At low wind speeds the total power from the park is very low, so this difference is not important.
Consequently the energy produced yearly by both options is the same, 709GWh, which results in around 3500
hours at nominal power, which is normal for a wind park offshore.
76
Conclusion:
In Chapters 2 and 3 the main theoretical topics (aerodynamic principals, modelling of PMSG and wind
turbine, control methods) and the problem specifications were exhibited.
In Chapters 4 and 5 the design and current and speed control of the boost, full bridge and ATR
converters was done. The design is in the Appendix. The comparison of these three converters was done
regarding response time and efficiency.
In the boost converter the control method used was the symmetry criterion as the resistance of the
inductor is designed to be very small to decrease losses, and this approaches the load pole to the origin. This
invalidates the method of the dominant pole. For the full bridge the tuning Rules of Ziegler-Nichols and Tyreus-
Luyben were used, and it was concluded the best controller for this situation was the PI controller using the
Ziegler-Nichols tuning rules, in Fig. 4.13. For the ATR the control method for the torque was setting Id=0 and
controlling Iq. The symmetry criterion was used once again as the resistance in the PMSG is very small.
For the speed control Idref=0 too and the output of the PI controller for the speed is Iq. The method used
was the symmetry criterion as the damping coefficient is designed to be very low to reduce the mechanical losses
in the gear box, and this creates a pole close to the origin. However this criterion is used to make the regulator as
fast as possible. This is not desirable in wind turbines because it creates sudden changes in the PMSG torque.
Placing a low pass filter with a large time constant like 30s forces the speed reference variation to be slow and
thus the PMSG won’t have high torques.
The response time of the ATR is 5ms, for the boost converter around 8ms and for the full bridge is
20ms. The efficiency will depend on if it is a 12kV or 60kV converter, because it will alter the topology. In the
60kV case the best is the ATR and in 12kV is the full bridge with duty cycle control.
These converters were AC/DC converters close to the PMSG. For DC/DC converters in the DC grid,
the ATR will become a full bridge converter. The boost converter is not doable here, as only a transformer can
raise the voltage more than 3 times without large losses. From this comparison it is concluded that the ATR is
the best converter overall for AC/DC converters (low losses, high power factor in the PMSG, fast response), and
the full bridge is the only option doable for DC/DC converters.
In Chapter 6 the design, current (Id and Iq) and voltage control and simulation were done for the
200MW VSC onshore. The control method was the symmetry criterion as the system can be subject to very large
perturbations if the wind speed varies very quickly. From the simulations, the current and voltage control are
very quick, taking 10ms for large perturbations and 8ms for small perturbations. The system is not sensible to
perturbations, and is stable.
In Chapter 7 the energy evaluation of the whole park was done. Here two implementations for the DC
grid were made: Option1 with two voltage levels, 12kV and 60kV, and Option2 with just one voltage level. The
main losses in the whole park for both options are in the VSC onshore, the main cable and in the 60/200kV
converter, mainly in the VSC onshore with 3.5% of the 7.5% total losses in the park. Summing up the efficiency
of both options are approximately the same. Option1 is a bit cheaper because it uses cables of lower voltage. For
this option1 is better. The efficiency is 92.54% for rated power.
77
Future Work
In this thesis only three converters were studied, but other converters could be studied and maybe are
better than the VSC for offshore DC wind parks.
Although the main losses in the full bridge are the conduction losses, [19] presents a snubber circuit
made of capacitors that can reduce greatly the switching losses if the control creates a square voltage wave. This
will happen in the full bridge converter. These snubbers could make possible to use higher switching frequencies
reducing the filter sizing, as the fact that limits this is the iron losses in the transformer. Today special kind of
iron exists that can be used up to 1 kHz, and a possible future work is to use 1kHz instead of 400Hz in the full
bridge and discover how large the output inductor and input capacitor will be.
It is also possible to use a three phase inverter instead; each phase is connected to one single-phase
transformer. The control will create three square voltage waves phase lagged, and as a result the frequency in the
output inductor would be three times larger, and the switching frequency in each transformer would be the same.
In chapter 6 it was concluded that it is best to use one voltage level in the DC grid. A possible future
work is to investigate which voltage level is more suitable in terms of efficiency and cable cost.
References
[1] ‘Wind Turbine Operation in Electric Power Systems’, Z. Lubosny, Springer, 2003
[2] Ming Yin, Gengyin Li, Ming Zhou, Chengyong Zhao, ‘Modeling of the wind turbine with a
Permanent Magnet Synchronous Generator for integration’.
[3] Andreas Petersson, ‘Analysis, Modelling and Control of Doubly-Fed Induction Generators for Wind
Turbines’, Ph.D.Thesis, Chalmers University of Technology, 2005
[4] G. Marques, ‘Controlo de Motores Eléctricos’, 2006
[5] http://www.esr.ruhr-uni-bochum.de/rt1/syscontrol/node64.html
[6] Catalogue V80-2MW VESTAS
[7] J. F. A. Silva, ‘Projecto de Conversores Comutados (ERC_06_07)’, Instituto Superior Técnico
[8] Stefan Lundberg, ‘Configuration study of Large Wind parks’, Licentiate Thesis, Chalmers
University of Technology
[9] Lena Max, ‘Chalmers Energy evaluation for DC_DC converters in DC Based Wind Farms’,
Licentiate Thesis, Chalmers University of Technology
[10] Power Electronics, Converters, Applications and Design, Mohan, Underland, Robbins John Wiley
& Sons
[11] Stefan Lundberg, ‘Evaluation of wind farm layouts’, Chalmers University of Technology, pp. 6 fig
12.
[12] Catalogue for XLPE Submarine Cable Systems, User’s Guide ABB, tables 42, 43.
[13] Brendan Peter McGrath, Member, IEEE, Donald Grahame Holmes, Senior Member, IEEE, Patrick
John McGoldrick, Member, IEEE, and Andrew Douglas McIver, ‘Design of a Soft-Switched 6-kW Battery
Charger for Traction Applications’, pp8 fig 12
[14] Catalogue of Transformer 170MVA Substation Alto Mira, EFACEC
[15] Catalogue of Hermetically Sealed Distribution Transformers, EFACEC
78
[16] Mariusz Malinowski, ‘Sensorless Control Strategies for Three - Phase PWM Rectifiers’, Ph.D.
Thesis, Warsaw University of Technology, 2001.
[17] ‘Design of Rotating Electrical Machines’, Juha Pyrhonen, Tapani Jokinen, Valéria Hrabovcoxá,
2008, John Wiley & Sons, Ltd, ISBN: 978-0-470-69516-6
[18] ‘The 45OOV-750A Planar Gate Press Pack IEGT’, Hironobu Kon, Kazuya Nakayama, Satosi
Yanagisawa, Junichi Miwa and Yosinari Uetake, 2009.
[19] Stephan Meier, ‘System Aspects and Modulation Strategies of an HVDC-based Converter System
for Wind Farms’, KTH Electrical Engineering, 2009.
[20] Lena Max ‘Energy Evaluation for DC/DC Converters in DC-Based’, 2007.
[21] ‘Applying Fast Recovery Diodes’, ABB, December 2008
[22] Olle I. Elgerd, ‘Electric Energy Systems Theory: An Introduction’, TATA McGraw-Hill
Publishing Company Ltd. New Delhi, 1971
79
Appendix
Sizing of the components for the boost converter
PMSG
Nominal Power: 2MW
Nominal Frequency: 100Hz
Nominal Voltage: 690V
Three Phase Transformer
Nominal Power: 2MW
Nominal Frequency: 100Hz
Nominal voltage primary side: 690V
Nominal voltage secondary side: 35kV
Resistance in both sides: 0.002pu
Inductance in both sides: 0.05pu
Three-phase low voltage diode rectifier
The current that flows in this rectifier will be 40A as seen in Fig 4.5. The voltage will be the output voltage
60kV.
Voltage: 60kV*1.5=90kV
Current: 40*1.8=72A
Input Inductor Filter
Current: 40A
Inductance: 2.96H
IGBT
The current that flows in the IGBT can be seen in Fig 4.6. .
Current: 150A*1.8=270A
Voltage: 60kV*1.5=90kV
Sizing of the components for the full bridge converter PMSG
Nominal Power: 2MW
Nominal Frequency: 100Hz
Nominal Voltage: 690V
Three Phase Transformer
Nominal Power: 2MW
Nominal Frequency: 100Hz
Nominal voltage primary side: 690V
Nominal voltage secondary side: 10kV
Resistance in both sides: 0.002pu
80
Inductance in both sides: 0.05pu
Three-phase low voltage diode rectifier
The current that flows in this rectifier will be 150A as seen in Fig 4.16. The voltage will be the voltage of the
capacitor 14kV.
Voltage: 14kV*1.5=21kV
Current: 150*1.8=270A
Input Inductor Filter
Current: 146A
Inductance: 40mH
Input Capacitor
Nominal Voltage 14kV
Capacitance: 93µF
Single-Phase Inverter with IEGTs
The current that flows in the IEGTs can be computed in . (4.7) and seen in Fig 4.16.
Current:533A*1.8=959A
Voltage: 14kV*1.5=21kV
Single Phase Transformer
Nominal Power: 2MW
Nominal Frequency: 400Hz
Nominal voltage primary side: 10kV
Nominal voltage secondary side: 160kV
Resistance in both sides: 0.002pu
Inductance in both sides: 0.05pu
Single Phase high voltage diode rectifier
This high voltage rectifier will support voltages of 60kV and currents of 33A. Using a safety margin of
80% for the current and 50% for the voltage to take into account the value in transient state is higher than in
steady state:
Voltage: 60kV*1.5=90kV
Current: 33A*1.8=60A
Output Inductor Filter
Current: 33A
Inductance: 16H
Sizing of the components for 2MW ATR PMSG
Nominal Power: 2MW
Nominal Frequency: 100Hz
Nominal Voltage: 690V
Three Phase Transformer
Nominal Power: 2MW
81
Nominal Frequency: 100Hz
Nominal voltage primary side: 690V
Nominal voltage secondary side: 35kV
Resistance in both sides: 0.002pu
Inductance in both sides: 0.05pu
Input Inductor Filter
Current: 66.7/ 3 =40A
Inductance: 57mH
6 switches using IEGTs
The current that flows in the IEGTs will be less than 750A, so no IEGTs are needed in parallel. The
nominal voltage is 4500V and nominal current is 750A.
No IEGTs in parallel.
( ) 205.4
605.1__ =×=
kV
kVswitcheachforseriesNumberIEGT
The program for the design and control of the boost converter is the section “Design, Control and
efficiency evaluation of the 2MW 60kV boost converter with transformer” as all was computed in the same
program.
Design and current control of the 2MW 60kV boost converter
P=2e6; Ud=35e3; Uo=60e3; Uin=Ud*1.35; Rload=Uo^2/P;
RPMSGpu=0.008; LPMSGpu=0.25; np=4; frated=100; we=2*pi*frated; wrotor=we/np; Zb=Ud^2/P; RPMSG=RPMSGpu*Zb; LPMSG=LPMSGpu*Zb/we; Flux=Ud/(sqrt(3)*we);
%inductor and capacitor sizing fcom=1000; Lboost=Uin^2*(Uo-Uin)/(Uo*fcom*0.1*P); R=1; C=10e-3;
%control parameters Kv=6000; tau=1/(2*fcom); Kp=-Lboost/(2*tau*Kv); Ki=-Lboost/(8*tau^2*Kv);
82
Design and current control of the 2MW 60kV full bridge converter
vwind=12; P=1156*vwind^3; Ud=10e3; Uo=60e3; Uin=Ud*1.35; Rload=Uo^2/P;
RPMSGpu=0.008; LPMSGpu=0.2; np=4; frated=vwind*100/12; we=2*pi*frated; wrotor=we/np; Zb=Ud^2/P; RPMSG=RPMSGpu*Zb; LPMSG=LPMSGpu*Zb/we; Iin=P/Uin; Udco=Uin+3/pi*we*LPMSG*Iin; Uac=Udco/(1.35*sqrt(3)); Flux=Uac*sqrt(2)/we; %inductor transformer and capacitor sizing fcom=400; delta=0.3; n=Uo/(delta*Uin); Iload=P/Uo; Lload=Uo*(1-delta)/(0.2*fcom*Iload); % Lload=(Uo-Vout^2/(n*Uin))*1/(fcom*0.2*Iload); Cin=5*Uo*(Iload*n-Iin)/(fcom*n*Uin^2); R=1; C=10e-3;
Kcrit=0.1; Tcrit=0.015; %control parameters Ziegler Nichols PI Kp=0.45*Kcrit; Ti=0.85*Tcrit; Td=0; %control parameters Ziegler Nichols PID % Kp=0.6*Kcrit; % Ti=0.5*Tcrit; % Td=0.12*Tcrit;
%control parameters Tyreus-Luyben PI % Kp=Kcrit/3.2; % Ti=2.2*Tcrit; % Td=0; %control parameters Tyreus-Luyben PID % Kp=Kcrit/2.2; % Ti=2.2*Tcrit; % Td=Tcrit/6.3;
Ki=Kp/Ti; Kd=Kp*Td;
83
Design and current control of the 2MW 60kV ATR
P=2e6; Ud=10e3; Uo=60e3; RPMSGpu=0.008; np=4; frated=100; fcom=1000; we=2*pi*frated; wrotor=we/np; Zb=Ud^2/P; RPMSG=RPMSGpu*Zb; LPMSG=0.31*Uo*sqrt(3)*Ud/(2*pi*fcom*0.1*P); Em=Ud*sqrt(2/3); iq=P/Ud; Ld=LPMSG; Lq=Ld; Flux=Ud/(sqrt(3)*we);
%control parameters tau=1/(2*fcom); Z=1/Ld; Kp=1/(2*Z*tau); Ki=1/(8*Z*tau^2);
Parameters for the VSC onshore
This was the values used in the simulations and calculations above. Except the parameters for the
controllers all this values were just considered, and then tested to see if the results were reasonable. This is
particularly true for the capacitance Conshore, as maybe it is not the best capacitance. No calculations were made to
find the best capacitance; a standard value used in the applications was used:
Transformer:
Frequency: 50Hz
Nominal Power: 200MVA
Voltage levels: 60/220kV
Xcc+Zgrid=20%
Xtotal/Rtotal=20
The amplitude of the impedance will be 20%, where the base impedance is:
( )Ω=== 18
10.200
10.606
232
S
UZ l
b
.
So the impedance of the transformer plus the Thevenin impedance of the grid is (viewed from the
primary side):
84
mH
ZLZ t
t 5.11502
6.36.3182.0 ===⇔Ω=×=
πω .
Using (6.9.):
Ω=×== − 18.07.1510.5.117.15 3
VSCLR .
Conshore=100µF
Parameters for the Current control
Ω=
Ω=−18.396
3.25
radsK
K
i
p.
Efficiency of the 2MW 12kV VSC and 200MW 200kV VSC onshore
The program for the 2MW 12kV VSC is the following. For the 200MW 200kV VSC onshore is exactly
the same, except Vo=200kV and Vgen=60kV.
Vwind=11.2; fcom=5000; Vgen=6e3; Pin=1158*Vwind^3; Vo=12e3; IIEGT=Pin/Vgen; tr=0.78e-6; tf=6e-6; Ron=2.5e-3; Von=4; VIGBT=4.5e3; nsIEGT=1.5*Vo/VIGBT;%number of IEGT in series %switching and conduction losses-IEGT Ps=(1.7+1.8)*2*sqrt(2)/(pi*750)*fcom*IIEGT; Pc=Von*2*sqrt(2)/pi*(IIEGT)+Ron*(IIEGT)^2; Ploss=3*(nsIEGT*Pc+Ps); etaVSC2MW=(1-Ploss/Pin)*100; lossVSC2MW=100-etaVSC2MW;
Efficiency of the 10MW 12/60kV and 200MW 60/200kV full bridge converters
As there are not enough data from 400Hz transformers, it is not possible to compute the transformers. It
is considered that the transformer has 99% efficiency in rated power, and the core losses are 1/5 of the total
transformer losses. This is considered to the 10 and 200MW converters. The program for the 10MW converter is
below.
For the 200MW 60/200kV converter, the program is the same as the previous one, except that the core
losses now are 50kW, Vload=200kV and Vin=60kV.
85
Vwind=12; P=1758e3; fcom=400; Vload=60e3; Vin=12e3;%voltage in the capacitor Ci nT2=Vload/(Vin/3);%single phase transformer ratio Iload=P/Vload; Iin=P/Vin;
%converter parameters D=Vload/(nT2*Vin); IIGBT=P/(D*Vin);
%efficiency calculations %transformer losses Pcore=2500; Ptrans=0.004*P/8+Pcore; %semiconductor losses tr=0.78e-6; tf=6e-6; Ron=2.5e-3; Von=3; npIEGT=1.8*IIGBT/750;%number of IEGT in parallel nsIEGT=1.5*Vin/4500;%number of IEGT in series nsdhv=1.5*Vload/5000;%number of diodes in series %conduction losses-IGBT PcIEGT=Von*IIGBT+Ron*IIGBT^2; %switching and conduction losses-Diodes high voltage ts=0.6e-6; trr=1e-6; Vdondb=1.5; Idlv=P/Vin; rond=2/Idlv; VCESS=4000; Vdhv=Vdondb*Vload/VCESS; Idhv=P/Vload; ronhv=rond*Vload/VCESS;
Pshv=Vdhv*Idhv*(trr-ts)*fcom*nsdhv; Pchv=Vdhv*Idhv+ronhv*Idhv^2; PDChv=2*Pchv*nsdhv; PDhv=Pshv+PDChv;
%losses semiconductores total phase shift control Psemi=PcIEGT+PDhv;
eta=(1-(Psemi+Ptrans)/P)*100;
%losses semiconductores total duty cycle control
Pdhv=2*D*nsdhv*Pchv+4*(1-D)*nsdhv*(Vdhv*Idhv/2+ronhv*(Idhv/2)^2); Pdhvduty=Pdhv+Pshv;
PIEGT=2*D*nsIEGT*npIEGT*PcIEGT;
Pduty=PIEGT+Pdhvduty; etaduty=(1-(Pduty+Ptrans)/P)*100; lossduty=100-etaduty;
86
Design, Control and efficiency evaluation of the 2MW 60kV boost converter with transformer Vgen=35e3; Vd=Vgen*sqrt(2)*3/pi; Vwind=11.2; P=1158*Vwind^3; RL=3; %resistance in the input inductance Vo=60e3; Iref=P/Vd; Ro=Vo^2/P; %load resistance fcom=5000; L=Vd^2*(Vo-Vd)/(Vo*fcom*P*0.1); d=1-Vd/Vo; %duty cycle of the converter Kv=6000; tau=1/(2*fcom); %time constant of the modulator used %Symmetrium Criterion Kp=-L/(2*Kv*tau); Ki=-1/(fcom*8*tau^2); Tz=Kp/Ki; %Efficiency computation VIGBT=4.5e3; NumberIGBT=1.5*Vo/VIGBT; Wcom=1.7+1.8; VCE=3*NumberIGBT; VD=VCE; eta=100*(1-RL/(Ro*(1-d)^2+RL)-d*VCE/(Vo*(1-d))-VD/Vo-Wcom*fcom/P); %switching and conduction losses-Diodes low voltage ts=0.6e-6; trr=1e-6; Vdondb=1.5; Idlv=Iref; rond=2.5e-3; nsdlv=1.5*Vd/5000; Pslv=Vdondb*Idlv*(trr-ts)*fcom*nsdlv*6; Pclv=Vdondb*Idlv+rond*Idlv^2; PDlv=Pslv+Pclv; etaD=1-PDlv/P; loss=100-eta*etaD; %PMSG Parameters frated=100; werated=2*pi*frated; Flux=Vgen/werated; Zb=(Vgen)^2/P; we=werated;%conditions of the wind LPMSG=(0.25+0.05*2)*Zb/we; RPMSG=(0.008+0.002*2)*Zb; Np=4; wrotor=we/Np;
Design of the transformer for the full bridge converter
For the design of this transformer the material chosen for the core is VACOFLUX48, as it says by the
manufacturer to be used in special transformers with low losses.
The transformer will be design according to Fig. A.1 :
87
d/2
d/2
h h
Fig. A.1 – Dimensions of the single-phase transformer.
The frequency used is 400Hz, the B is 1T. The maximum current in the transformer is 148 A; the
density current admitted is 4A/mm2. This gives a copper section of 37mm2. In the catalogue of the
VACOFLUX48 it is seen that for 1T the specific losses are 20W/kg. With this data it is known:
NAfBV ironcomef max44.4= ,
Where Vef is the RMS value of the voltage applied to the primary side of the transformer, Bmax is the
flux density in the iron, fcom is the switching frequency, Airon is the cross section area of the iron and N is the
number of turns in the primary side. Knowing for the 60kV converter the voltage in the primary side is 13.5kV,
it is known 6.7=NAiron . For each value of D, the area in the iron is computed and N also consequently. With
N and the copper cross section h is computed as NAh copper3= , where the number 3 is to account both
windings and leaving a margin of 1. With h and D the iron volume is computed. In the catalogue the iron density
is seen as 8.12g/cm3. The iron mass is computed. Consequently the iron losses are computed using the specific
losses. Using D and h to admit a medium radius for the winding and using N the length of the copper wire is
computed. With this and the copper section the copper resistance and consequently the copper losses are
computed.
NhD
Length
+=
22π
LengthS
Rcopper
copper
ρ=
22 MAXcoppercopper IRP =
Note that the Length is just for the primary winding of the transformer. However quoting [21] it is
known that the copper volume is a minimum when the volume of the primary and secondary windings is equal.
As the secondary side has more turns, but its cross section is lower the cross sections of the windings are equal.
88
It is considered that the losses in the primary and secondary sides are equal. Summing the copper and iron losses
the following graph can be obtained:
Losses and Price for the 12kV transformer
0,25
0,26
0,27
0,28
0,29
0,30
0,31
0,32
31 32 33 34 35 36 37 38 39 40
D(cm)
pe
rda
s(%
)
Losses(%)
Price
Fig. A.2 – Total Losses and relative price for the transformer for the best iron area.
Generally the copper price is 20 times larger than the iron price. The price is proportional to the volume
of the material, so the copper volume times 20 plus the iron volume is proportional to the price. That calculation
is the blue curve.
Design of the Inductor
The drawing of the inductor will be as in Fig. A.3 . The iron cross section will be a square with side D:
h
DD
H
Fig. A.3 – Dimensions of the Inductor.
The average leakage flux in the inductor will be:
av
avLinductor
feavfeavLinductoravLB
iLNABNAiL =⇔==ψ
89
Giving several values of D Afe is known. Linductor, iavL and Bav are known too, so N is computed for each
D. Using the Maxwell equations to the magnetic circuit, considering the magnetic permeability of the iron
infinite:
av
avL
avL
av
B
NihNih
B 0
0
µ
µ=⇔= .
The iron volume will be:
( )( )hDHDVcore −+= 42
With the iron density the iron weight is known. In the boost and in the full bridge converters the current
ripple allowed for these inductors is 10%, consequently the ripple in the flux density is 10% as well. The iron has
a saturation flux density of 2.35T. For safety measures an average value of 1.5T is going to be chosen. Its
maximum is 1.65T, much lower than the limit 2.35T. This means the amplitude of the oscillating B that will
cause core losses is 0.15T. Using the switching frequency used in the converter and looking at the catalogue the
core losses are known. The copper losses are computed the same way as in the transformer, except only one
winding needs to be considered. The total losses for the inductor used in the 12/60kV full bridge converter are
shown:
Losses for inductor 12/60kV converter
0,1375
0,1380
0,1385
0,1390
0,1395
0,1400
0,1405
0,1410
0,1415
0,1420
40 41 42 43 44 45 46 47 48 49 50
D(cm)
To
tal L
os
se
s(%
)
Fig. A.4 – Total losses for the inductor in the 12/60kV full bridge converter.
