Electronic Power Control

ELECTRONIC POWER CONTROL

The textbook Electronic Power Control consists of two volumes: Power Electronics and ElectronicMotor Control. The fundamentals and applications of power electronics are thoroughly discussed withmathematical derivations, a lot of numerical examples and clarifying images. This enables bothstudents and professionals to study power electronics in depth. The author has more than 40 years ofexperience in teaching on the subject. Thousands of students (engineering, electronics) found a usefulcompanion in this book. The seventh edition of the Dutch textbook, which is the result of continualupdating, is now also available in English. The book is in full colour and includes a glossary containingmore than 660 technical terms translated in Dutch, German and Spanish.

An indispensable companion for engineers and other specialists in the field of power electronics

Volume 1 - Power Electronics

ISBN 9789038217918 504 pages ! 33.00 (VAT incl., shipping and handling excl.)

! Click here to order

Volume 2 - Electronic Motor Control

ISBN 9789038219110 404 pages ! 30.00 (VAT incl., shipping and handling excl.)

! Click here to order

TABLE OF CONTENTS

Volume 1 - Power Electronics

PART 1 - SEMICONDUCTOR SWITCHES

1. PHILOSOPHY OF POWER CONTROL1. Controlling of electrical energy using switches2. Switching matrix3. Controllable semiconductors4. Properties of switches5. Commutation6. Power converters7. Power frequency domain8. Evaluation

2. POWER DIODES1. Semiconductors2. I-V characteristic of a junction diode3. Power diodes4. Data of a power diode5. Excerpts from data books6. Evaluation

3. TRANSISTOR POWER SWITCHES1. Bipolar junction transistor (BJT)2. Power MOSFET3. IGBT4. Evaluation

4. THYRISTORS1. Shockley diode2. Silicon controlled rectifier (SCR)3. Diac4. Triac5. Gate turn off thyristor (GTO)6. Mos controlled thyristor (MCT)7. Integrated gate commutated thyristor (IGCT)8. Evaluation

5. NOTES1. Electrical and mathematical notation2. Opto-electronics3. Hall effect sensors4. Heat dissipation from semiconductors5. Colour code for resistors / Non-inductive resistors6. Normalisation of resistor and capacitor values7. Colour code for capacitors8. Type indication of semiconductors and integrated circuits

9. The skin effect10. EMC-EMI11. Si-surface area of power switches12. Size winding wire (AWG = American wire gauge)

6. COMPUTER SIMULATIONS1. Introduction2. History: analogous simulations3. SIMULINK: simulation with block diagrams4. SPICE: simulation of an electric network5. CASPOC: the multilevel model

PART 2 - POWER CONVERTERS

7. UNCONTROLLED RECTIFIERS WITH INDUCTIVE LOAD1. Current flow of an inductive circuit with a sinusoidal supply2. Current and voltage waveforms of an inductively loaded single-phase rectifier3. Approximations in the study of rectifiers4. Half-wave three-phase rectifier. Resistive load5. Three-phase bridge circuit6. Table 7-3: single-phase rectifiers7. Table 7-4: three-phase rectifiers8. Evaluation

8. LINE-FREQUENCY PHASE-CONTROLLED RECTIFIERS1. Half-wave single-phase controlled rectifier (E 1 ). Resistive load2. Resistive and inductive loaded E 1 -controller3. Table 8-1: Single-phase power control4. Harmonics in the case of phase-controlled sine wave5. Distortion power6. Full wave B2-controlled rectifier . Resistive inductive load7. Fully controlled B6-rectifier. Resistive load8. Fully controlled B6-rectifier. Resistive and inductive load9. Full controlled B6-rectifier: dead time, harmonics

10. Twelve-pulse controllers11. Evaluation

9. AC-CONTROLLERS1. AC-controller with phase-control2. AC-controller with integral cycle control3. Turn-off snubber for thyristors4. Solid state relays5. Radio interference suppression (RFI) of thyristors6. Evaluation

10. CYCLOCONVERTERS1. Continuous cycloconverter2. Trapezium cycloconverter

11. CONTROL OF THYRISTORS1. Firing pulses2. Pulse transformers3. IC for phase control of thyristors and triacs4. Triac control with a diac5. Evaluation

12. CHOPPERS1. Operating principle of a chopper2. Control modes3. Resistive and resistive-inductive loaded choppers4. Chopped resistor5. Chopper control IC’s6. Evaluation

13. SWITCH-MODE POWER SUPPLIES1. Basic principles of switch-mode power supplies2. Basic converter configurations3. Isolated switch-mode power supplies4. Flyback converter5. Forward converter6. Converter control strategies7. Two-transistor SMPS of the forward type8. Forward with multiple outputs9. Full-bridge of the buck type

10. Synchronous SMPS11. SMPS components12. Overview of SMPS up to 2500 W13. Non ideal waveform14. Digital control of an SMPS15. Evaluation16. The design of switch-mode power supplies

14. INVERTERSA. THREE-PHASE INVERTERS

1. Voltage source and current source inverter2. Switching matrix of a voltage source inverter3. 180°-type inverter4. Pulse frequency inverter with a constant DC-voltage5. Pulse width modulation (PWM)6. PWM-strategies

7. Harmonics in a PWM-wave8. 120°-type inverter9. Common switches used in inverters

10. Control circuit for three-phase inverter bridge of the 180°-typeB. SINGLE-PHASE INVERTER

1. Basic circuit of full bridge inverter2. Unipolar and bipolar PWM3. Full bridge with unipolar PWM4. Harmonics with unipolar PWM5. Evaluation

15. APPLICATIONS OF POWER ELECTRONICS1. Uninterruptible power supplies2. High-frequency inductive heating3. Power factor correction (PFC)4. Lighting5. Renewable energy

A. Wind turbinesB. Photo voltaic solar panels

6. Drive technology7. Motion control8. High frequency induction cooking plate

Volume 2 - Electronic Motor Control

16. ELECTRIC MACHINES1. Static transformers2. DC commutator machines3. Three-phase asynchronous motors4. Synchronous machines5. Small appliance motors

17. DRIVE SYSTEMS1. History2. Control theory3. Types of drive systems4. Electronic drive technology5. Useful mechanical formulas6. Moment (of torque) and power of a motor7. Run out test to determine moment of inertia of a load8. Numeric examples

18. CURRENT-, ANGULAR POSITION-, SPEED TRANSDUCERS1. Current sensors2. Angular position sensors3. Speed sensors

19. SPEED and (or) TORQUE CONTROL of a DC-MOTORA. DC-MOTOR supplied from an AC POWER GRID

1. Control of an independently excited motor2. M-n curves3. Regulated single quadrant drive4. Two quadrant and four quadrant operation5. Functional control diagram of a single quadrant drive6. Optimising controllers7. Numeric example

B. DC-MOTOR supplied from a DC SOURCE1. Chopper controlled drive2. Chopper control of a series motor

Traction serviceLine filter

20. SPEED- and (or) TORQUE-control of THREE-PHASE ASYNCHRONOUS MOTOR1. Three-phase asynchronous motor2. Electronic control of an induction motor3. Scalar regulation of induction motor speed4. Slip control5. Scalar frequency converters6. Indirect frequency converter of the VSI-type7. Vector control8. Microelectronics with power electronics9. Softstarters

10. Indirect frequency converters with a current DC link (CSI)21. ELECTRONIC CONTROL of appliance motors, switched reluctance motor, synchronous three-phase motor, induction

servo motor1. Appliance motors

Universal motorSingle phase induction motor

2. Switched reluctance motor3. Synchronous AC motor4. Three-phase induction servo motor

22. ELECTRICAL POSITIONING SYSTEMS1. Servomechanisms2. Electrical positioning systems. Definitions3. Position control with DC servomotor4. Position control with brushless DC motor

5. Position control with stepper motor6. Position control with AC servomotor7. Position control with linear motor8. Computer guided motion control9. An integrated system (SIMOTION from Siemens)

23. e-MOBILITY1. Renewable energy2. Electrical traction3. Electrical automobiles4. Electrical boats5. Electrical bicycles

THE PHILOSOPHY OF POWER CONTROL 1.77. POWER FREQUENCY DOMAIN

Fig.1.1illustratesthestateoftheartinpowerswitches.CurrentlyweseethatwiththeexceptionofanSCRof8.5kVthatthyristorsarelimitedto4kV-4kA.Thisisamaximumof16MWperswitch.Thisisconsideredthemidpowerdomain. These days an SCR is also known as Phase Controlled Thyristor (PCT). ThereareforexamplesingleIGBT’sof1200V-3600A,1700V-2400Aand3300V-1500A. InahalfbridgeusingIGBT’satypicalratingwouldbe1700V-1000A. Stateoftheartcomponentsuse125mmSiwafers.

Fig 1-11: Propertiesofpowerswitches

SCR

IGCT

IGBT - module

4 kA

3 kA

2 kA

1 kA

2 kV 4 kV 6 kV 8 kV

MOS-FET

100 Hz

1 kHz

10 kHz

100 kHz

1 MHz

10 MHz

GTO

Intensity kA

VoltagekV

Frequency

Hz

TransportofelectricalenergyoverlongdistancescanbemoreeconomicalusingDCtransmission.Forexamplewhen the distance is greater than 250 km then transmission of 400kV-1200MW is more economical using HVDC (highvoltagedirectcurrent).Withsubmarinecablesthedistanceisevenshorter.InChinaandBrazilwherelargehydroelectricpowerstationsaremorethana1000kmfromthebigcitiesHVDCisused. Example:theUltraHighVoltageDC(UHVDC)of800kVbetweenXiangjialaandShanghai.ThefirmABBdevelopeda6-inch8.5kVthyristorforthisapplication. AnotherapplicationofHVDCistheconnectionbetweenAC-gridsofdifferentfrequencies.AnexampleistheGarabibacktobackstationwhichconnectsthe60HzgridofBrazilandthe50HzgridofArgentina.

3.14 TRANSISTOR POWER SWITCHES

0 10 20 30 40 50 0

ID

(A)

10

8

6

4

2

10

8

6

4

2

9V

8V

7V

6V

5V

4V

(a) (b)

VDS

VGS

(V) (V)

VGS

ID

(A)

VDS

= 30V

1 2 3 4 5 6 7 8 9

a

Vthreshold

DID

DVGS

7,754,75

VGS

(a)

G

D

S

RD

VDD

N+

N+ P

depletion layer

ID

induced N-channel

– +– +– +– +– +– +– +– +

D

S

G

RD

ID

VDDV

GS

(b)

2.3 Operation

IfapositivevoltageV GS isapplied(fig.3-15a)thenasaresultofelectrostaticinductionanN-zoneiscreatedinthePstructurebetweenthetwoexistingN+-zones.SincetheN-zoneconsistsofminoritychargecarriers(fromthePsubstrate)werefertothisasaninversionlayer.Source(S)anddrain(D)arenowconnectedviatheN-channelandwithapositivevoltageV DS a current I D will flowfromDtoS.ThelargerV GS , the wider the N-channel and the larger the current I Dwillbe.Infig.3-15baschematicrepresentationoftheconfigurationoffig.3-15aisshown.

Fig. 3-15: OperationandprincipalconfigurationofaN-channelenhancementtypeMOSFET

2.4 Mosfet characteristics

Thecharacteristicinfig.3-16aisanalogoustotheI C - V CE-characteristicofabipolartransistor.InthecaseofaBJTwecontrolthetransistorwiththebasecurrent,whilewithaMOSFETweworkwithvoltagecontrol.Thedraincurrentconsistsofonetypeofchargecarrierthereforewerefer to a MOSFET as a unipolar transistor in contrast to the BJT.

Fig.3-16(aandb):Characteristics of an N-channel enhancement MOSFET

TRANSISTOR POWER SWITCHES 3.15

g fs =( ∆I D

_____ ∆V GS

)with V DS = C te ; g fs =tgα

Fromtheoutputcharacteristicwecanderivethetransfercharacteristic(fig.3-16b).Thistransferprovidesuswithaveryimportantspecification,namelythetransconductance:

(3-6)

Infig.3-16bweseethatthegatevoltagemustexceedacertainminimum V GS(th) beforethe MOSFET conducts.The SiO 2layerbetweenthegateandsourceisresponsiblefortheveryhighinputresistanceofthe MOSFET ( 10 10 to 10 15Ω).Asingleelementaryunitasshowninfig.3-14isonlycapableofhandlingadraincurrentof100µA.Theparallelcombinationofforexample100,000 similar elementary units results in a MOSFETwithadraincurrentofseveralampsasshowninfig.3-16a.

Fig.3-16cshowsthecompleteI - V characteristic with the three distinct regions: 1. Cut-off:Withthegatevoltagebelowtheturn-offvoltageV GS(th.)thereisnocurrentflowand the MOSFETandfunctionsasanopenswitch.Inthecharacteristicthebreakdownvoltage BV DSS of the MOSFETisshown.Thismeansthattheapplieddrainsourcevoltagemustbeless than this BV DSS BV=Breakdownvoltage;DSS=Drain-Sourceinthecommonsourceconfiguration. 2. Active region:Inthisregionwehaveanalmostlinearrelationshipbetweendraincurrentand gate source voltage ( i D = g fs . u GS ) 3. Ohmic area: WithaBJTtheborderbetweentheactiveareaandsaturationisgiven byV CE = V BE or V CB=0,seefig.3-5.InsimilarfashionforaMOSFETtheborderbetweenthe activeareaandsaturationisgivenby:V DG =0 or V GS − V GS(th.) = V DS . When V DS is low, every value of V GS endsupontheborderlineintheI D − V DS characteristic. Theseborderlines(fig.3-16c)allhavethesamerampirrespectiveofthevalueV GS . On the borderlinetherelationshipbetweenI D and V DS ispracticallyconstant,theMOSFETbehaves as a resistor R DS . We refer to an OhMicoperatingarea.InthisareaorregiontheMOSFET functionsasaclosedswitch.Werefertotheresistanceofthis“closed”(MOSFET) switch as R DS(ON) . The value of R DS(ON)followsfromtherampoftheborderlinein fig.3-16c,butthisisalsotobefoundinthedatasheetoftheMOSFET.

Fig. 3-16c : OperatingregionsofanN-channelMOSFET

0

10

8

6

4

2VDS

(V)

VGS

10 20 30 40 50 60 70 80 90 100

limit curves

Ohmic

CUT - OFF

9V

8V

7V

6V

5V

4V

ID

(A) [ vGS

- VGS(th)

= vDS ]

VGS(th)

Acitve

area

BVDSS(c)

3.36 TRANSISTOR POWER SWITCHES

N+N+

N-

collector C

emitter E

AlSiO2

gate G

P

P P

N+N+

N-

collector C

emitter E

Al

CGE

CGC

T2

T1

RD

R1

SiO2

RBE

CCE

gate G

P

PP

3 IGBT

AsweknowabipolartransistorhasalowV CEsat even with a large current (also with high voltage transistors).ApowerMOSFETincontrastrequiresonlylowgatepowerandcanoperateathighfrequency.IntheconstructionofanIGBT(InsulatedGateBipolarTransistor)themanufacturerssoughttocombinethebenefitsofthebipolartransistorandtheMOSFET.ThefirstdesignofanIGBT dates from 1980.

3.1 ConstructionFig 3.35a shows the construction of a single IGBT cell. Here in the N +−substratefromtheMOSFEToffig.3-20hasbeenreplacedwithaP-substrate.AnIGBTismadeupofthousands ofsuchcellswhichareallconnectedinparallel.Forfurtherstudyadiagramofasinglecell

isshownandthisisconsideredasbeingrepresenta-tiveofthecompleteIGBT.Analogoustofig.3-20theinternalparasiticcomponents(T1 , T 2 , etc) are representedinfig.3-35b. Infig.3-35band3-36notethatanN-channelMOS-FETispresentandaPNPtransistorT1.TheparasiticNPNtransistorfromfig.3-20isalsopresent. If T 2anditscorrespondingR BE are neglected then fig.3-37aandbarearudimentaryequivalentdiagramoffig.3-36.

(a) Construction

(b)IGBTwithinternalparasiticelements

Fig. 3-35: Construction of an IGBT


N-

T2

T1

RBE

CCE

RD

collector C

R1

CGC

CGE

gate G

emitter E

T1

collector C

CCE

CGE

CCG T1

G

emitter E

gate G

C

(b) (c)(a)E

C

C

E

E

G

G

C

E

G

I

ID

R1

R

Theequivalentcircuitasdrawninfig.3-37bconsistsofaPNPtransistorcontrolledbyanN-channelMOSFETinapseudo-Darlingtonconfiguration. SincethereisaMOSFETintheinputandatransistorintheoutputthesymbolsusedinfig.3-37careeasilyunderstood. Inthetextthebottomsymbolofthethreewillbeused. AttheinputtheIGBTbehavesasaMOSFETandattheoutputapproximatelythebehaviourofaBJT.

3.2 Operation

Fig. 3-36: Equivalentdiagramforfig.3-35b

Fig. 3-37: ReducedequivalentdiagramofanIGBTandsymbols

ThephysicaloperationofanIGBTmorecloselyresemblesthatofaBJTthanapowerMOSFET.ThereasonisthattheP-substrateoftheIGBTisresponsibleforinjectingtheminoritycharge carriers in the N −-layersothatconductionismuchbetterthanwithaMOSFETthatworkswithmajoritychargecarriers.SincenohighvoltageexistsacrosstheMOSFETpartwecanforcon-structionpurposesusealowvoltagetype.LowvoltagetypeshavealowR DS(ON) asimportantchar-acteristic.ThebaseofthePNPhasnoexternalconnectionwhichisexpressedintheuppersymboloftheIGBTinfig.3-37c.Intheequivalentdiagramoffig.3-37aitisclearthatthevoltagedropiscomposedoftwoparts:ononesidethediodevoltagedropacrosstheemitter-basejunctionofT1 andontheothersidethevoltagedropacrossR D and the gating MOSFET. Since it is a low voltage MOSFETthevoltagedropacrossthisFETisdependentonthegatecontrolvoltage(thisisnotthecase with high voltage MOSFETS !). We write: (3-8) with I D =

I C

___ h FE V CEsat = V BE(T1) + I D .( R D + R DS(ON) )


N

P

N

P

Emitter

Collector

Emitter

Gate

N-

P

E

x

0

N

P

N

P

Emitter

Collector

Emitter

Gate

N-

P

E

x

N - buffer FS

0

3.5.2 Trench NPT-IGBT The NPT-trench structure was designed to reduce V CEsat and the switch-off losses E off (fig.3-48).IncontrastwithaplanarNPT-IGBTthetrenchtypehasalowerV CEsat , lower switching losses( E off ) and a lower thermal resistance R th . As a result of the trench structure the R DS(ON) iseliminatedfromtheparasiticMOSFEToffig.3-37asothat(3-8)becomes: (3-12) The trench has therefore a lower V CEsat thantheNPTshowninfig.3-46.

3.5.3 Trench-fieldstop (FS) structure Fig.3-49showstheimplementation.Incomparisontothetrenchimplementationoffig.3-48thethickness of the N −-baseismuchreducedandthereisalightlydopedN-bufferadded.Fromthegraphofthefieldstrengthwecanseewherethenamefieldstopcomesfrom.SincetheN−-baseof a trench FS is much thinner than a trench the drift resistance R D islowerandexpression(3-12)shows that V CEsat isreducedincomparisontoatrenchIGBT.

V CEsat = V BE(T1) + I D . R D

Fig. 3-48: Trench structure of an IGBT Fig. 3-49: TrenchfieldstopIGBT

5.16 NOTES

2. OPTO-ELECTRONICS

2.1 LED (Light Emitting Diode) Thelightemittingdiode(LED)hasbeenusedforyearsasasignallamp,asnumberindicatorandasalightemittingtransmitterinanopto-coupler.InrecentyearstheLEDisbecomingevermorepopularasalightsource.Thisaspectisdealtwithinchapter15intheapplicationsofpowerelectronics.TheoperatingprincipleoftheLEDrestsonthereleaseofenergybydirectrecombina-tion of electrons and holes in a PN-semiconductor. This energy is released in the form of electro-magnetic radiation. In general this is: (5-29) with: W =energy(eV) h =Planckconstant=4.133x.10−15 eVs f =frequencyofelectromagneticwavesinHz.Inthespecificcaseofasemiconductorwefind:W= W g =energygaporbandgapbetweenthevalenceandconductionband.Thisenergygapisalsotemperaturedependent.SihasaW=1.12eVat0K(=−273°C)andthisbecomes1.1eVat300K(roomtemperature).Thewavelengthofthereleasedradiationisgivenby: λ= c __

f = 300 . 10 6

__________

f = 300 . 10 6

__________ W g .h= 300. 10

6 x 4.133 x 10 −15

_________________________

W g = 1239 ______ W g

(nm) Herein c=300,000km/s=speedoflight.Tohavecontinuousradiationwerequirecontinuousrecombinationtooccurbetweenelectronsfromtheconductionbandandholesfromthevalenceband.Thiscontinuousprocesscanbeob-tainedbyaforwardpolarisedPN-junction(e.g.withI F =20mA).Toobtainvisiblelight(λ=400to700nm,seefig.5-36)W g mustliebetween1.77and3.1eV.ThesemiconductormaterialmostoftenusedfortheconstructionofLED’sis3-5crystalswiththefollowingconfiguration:.group3:Ga,Al. group5:As,P,N. Thedifferentcombinationsandimpuritiesleadtodifferentwavelengthsoftheradiatedlight.Thespectrumofagalliumphosphide-LED(GaP)canbeyellow(λ=575nm)togreen(λ=560nm).Gallium-arsenidephoshide(GaAsP)emitsyellow(λ=590nm)toredlight(λ=650nm).GaAshasanenergygapofabout1.43eVsothatλ=900nmandGaAscanoperateasaninfraredtransmitter(IRED=infraredemitter).Fig5-35showsthecurvesthataccompanyaGaAsP-LED.

Remarks 1. WithnominalcurrenttheforwardvoltagedropofaLEDisbetween1.3to3Vdepending onthethetypeofsemiconductorandtheimpurities. 2. WiththepassingoftimeLED’sstarttoage:theradiatedpowerdecreases.Highcurrentsand highambienttemperatureshaveadisastrouseffectontheagingprocess.Aloadoperates classically with I F =20mA,butifalowerlightintensityisrequiredanI F =5or10mAisused.3. Fig. 5-35d shows a LED as indicator. With V=12V;I F =10mAandV F =1.6Vthevalueof thecurrentlimitingresistorbecomes:R V =

V− V F ______ I F = 12 − 1.6 ________ 10 ≈1kΩ

W=h.f

!

NOTES 5.17

intensity

100

590(a) Spectral curve

(%)

l (nm)

mW

100

10(b) Radiated power

IF (mA) VF (V)VF

IF (mA)

10

1,6(c) forward

characteristic (d) LED-indicator

circuit

IF RV

V

700680600

500

400

300

200

100

0

VIOLET

CIE CURVE

RED

human eye (receiver)(Lm/W)

BLUE

GREE

N

YELL

OW

AMBE

R

Ga As (transmitter)100%

75

50

25

300 400 500 600 700 800 900 1000 1100 1200

λ (nm)

% re

lativ

e se

nsiti

vity

ultraviolet infrared400

420 460 525 595 650 725

500 600 700 800 nm

violet green orange

blue yellow red

infrared

Ga As Zn

Ga As P6

Ga As P4Ga P:NLED-materials

Lum

ens

/ Wat

t tra

nsfo

rmat

ion

Ga AsP82:N

silicon

Fig. 5-35: Curves of a GaAsP-LED

2.2 Spectral sensitivity

Infig.5-36wedraw: 1. ThestandardhumaneyecurvepreparedbytheCommissionInternationaledel'Eclair (theCIE-curve).Thiscurvenotonlyprovidesthesensitivityoftheeyeatdifferentwavelengths butalsothelumens/wattconversionofthelightsourceatthatparticularfrequency. 2. ThespectralcurveofaGa-Astransmitter(LED). 3. The wavelength of maximum sensitivity of other LED-materials. 4. Thesensitivityofasilicondetector(diode,transistor).ItisclearwhyGa-ASLED’sareused withan(Si-)transistorordiodeasopto-coupleelements.

Fig. 5-36: HumaneyecurvesandspectralsensitivityofLEDsandSi-detectors

NOTES 5.19

Fig. 5-38: Si-photo-voltaiccell

Fromfig.5-36itfollowsthatSi-photo-voltaiccellsaresensitiveforlightwithawavelengthofbe-tween400and1100nm.Thiscellsareespeciallysensitiveforredandinfraredlight.Alogarithmicscalelinksphoto-emf(V ) and illuminance (lux). If we connect a resistor R b acrossaphoto-voltaiccellthenwecandeterminetheoperatingpointofthiscircuit(fig.5-39).Fromtheproductofthephoto-emfandthecurrentproducedwefindthepowerthatthecellgives.Intheexampleoffig.5-39wefindamaximumpowerof27.8mWwithaloadresistorof2.52kΩ.WealwaysattempttoloadaPV-cellwithanoptimalresistorsothatmaximumpowerisobtainedattheexistingillumination.ThisoperatingpointisindicatedastheMPP(maximumpowerpoint).

NotethePV-cellcanbeshortcircuitedandthenhasacurrentofI SC . Infig.5-29thisshortcircuitcurrent I SC =0.15mAandtheopencircuitphotovoltaicemfisabout0.45V.Toincreasethevoltageandrequiredcurrentweformbatteriesofphotovoltaiccells.Thesearecalledsolarbatteriesorsunpanels.Thesedaystheyarebeingmoreandmorepromotedasacleanenergysource.Thiswillbedealtwithinchapter15.Photovoltaiccellsareusedincontrolcircuits,lightmeasurementof visibleandalmostinfraredlight,topowercalculatorsandphoto-electricrelays,etc.

Fig. 5-39: Optimumoperatingpointofaloadedcell

P PN Nm

etal

met

al

VV

dd

I Rb

E

(lux)

( a ) ( b )

V F

Rb= 2,52 kΩ

Rb= 4 kΩ

Rb= 10 kΩ

Rb=1kΩ

E=5000lux

16mW

26mW

27,8mW

19mW

I F

(mA)

0,05

0,10

0,10 0,2 0,3 0,4 0,5

0,15

(V)

MPP

5.20 NOTES

Photo Ixys (KWx): SolarBIT'saremonocrytallinePV-cellswithahighdegreeofefficiency(17%). Spectralsensitivity300to1100nm.Thesolarbitexistsindifferentvoltage-currentconfigurations(see www.ixys.com).Examples:XOB17-12x1:0.69V-39mAwithanMPPof0.51V-39mA. XOB17-04x3:MPPof1.51V-11.7mA.

3. HALL-EFFECT SENSORS

3.1 Hall-effect

IfweplaceacurrentcarryingconductororsemiconductorinaperpendicularmagneticfieldB (fig.5-40a)thenanelectricfieldarisesperpendiculartotheI-B surface. This effect is known astheHall-effect.Thiseffectwasdiscoveredin1879bytheAmericanphysicistEdwinHerbertHall.

Fig 5-40: Hall-effect

y

Ib

d

1

2

I(A)

X

ZB(Wb/m2)

bI I

IB

+

-

VH

VHX

5.40 NOTES

A second method involves a metal enclosurewhichincludesabridgingcapacitor(linebypasscapacitor) C y asshowninfig.5–55.Thedistur-banceflowsviathesignallinesandvia C y backtothecasingandviathestraycapacitancetothesourceofthedisturbance.

Fig. 5-55: Applicationofametalenclosure

Fig.5–56showsanexampleofanEMI-filtercommonlyusedinthesupplycableontheAC-inputsideofequipment(forexampleacomputer).

Remarks 1. Inthecaseofasinglephasemainsthedifferentialmodecurrentconsistsofthe50Hz-current togetherwiththedisturbance.Thecommonmodedisturbancesnormallyhavehigher frequencies(10kHzto10MHz).Theyflowviaparasiticcapacitancesbacktothesourceof noise. 2. A C y isplacedbetweeneverylineandthemetalenclosure.This"linebypasscapacitor"is usually only a few nF, and this is to limit the leakage current V.ω. C y .3. Practicalvaluesforthecommonmodechokeareforexample500µHupto2mH. 4. C x (across-the-linecapacitor)suppressesthedifferentialmodedisturbance.Apracticalvalue isforexample0.015µF. 5. Inthecaseofafrequencyconverter,dependingonwhetheritisasinglephaseorthree-phase supplyanappropriatefilterisincludedintheinput.

Fig. 5-56:EMI-filteratthesupply-sideofequipment

!

SMPSx xC

C

C

C

y

y

load0.015µF 0.015µF

2mH

2mH

2200pF

2200pF

mains

N

ZL

reference ground

noise

signal source

metal package

C Cy y(line bypass capacitor)(stray-

capacitance)

COMPUTER SIMULATIONS 6.5

Fig. 6-5: Classificationofanaloguesimulators

K=compliance;K=rotationcompliance;M a=acousticinertia;C a =acousticcapacitance;f=translationfriction;f=rotationalfriction;R H=hydraulicallyresistance;r a =acousticresis-tance; J=momentofinertia;M =momentum(oftorque);v=speed;v=voltage

ANALOGUE SIMULATORS

INDIRECT

MECHANICAL HYDRAULICALLY ELECTRIC MECHANICAL HYDRAULICALLY ELECTRIC

DIRECT

planimeternomograph

...

unusual opamp(dierentialamplier)

(analogue computer)...

wind tunnel...

hydraulical labs

river beddam model

...

equivalentecircuits

networkanalyser

...

scale models

ParameterMechanical

Electric Hydraulically AcousticTranslation Rotation

Appliedforce F M i p P

Speed v ω= dϴ ___ dt v Q(massflow) c= dx __ dt

Displacement s= ∫ 0

t

v.dt ϴ ∫ 0

t

v.dt Q(mass) x

Acceleration a= dv __ dt dω ___ dt dv __ dt dc __ dt

Pulse ∫ 0

t

F.dt ∫ 0

t

M.dt ∫ 0

t

i.dt ∫ 0

t

p.dt ∫ 0

t

P.dt

Power F . v M.ω v . i q.p P.c

Inertia m . dv __ dt J. dω ___ dt C. dv __ dt M a . dc __ dt

Elasticity K K L C H C a Coefficientoffriction f f R R H r a

Momentum (of torque) J. dω ___ dt ; f. dϴ ___ dt ;K.ϴ

Kinetic energy 1 _ 2 . m. v 2 1 _ 2 . J. ω 2 1 _ 2 . C. v 2 1 _ 2 . M a . c 2

Potential energy 1 _ 2 . K. s 2 1 _ 2 . K. ϴ 2 1 _ 2 . L. i 2 1 _ 2 C H . p 2 M a .xDissipatedenergy 1 _ 2 . f . v 2 1 _ 2 . f. ω 2 1 _ 2 . R. i 2 1 _ 2 . R H . p 2 1 _ 2 . r a . P 2

2.2 Analogies between physical systems

2.2.1 Table 6-1: Similarities between different physical quantities

LINE-FREQUENCY PHASE-CONTROLLED RECTIFIERS 8.34. Disadvantagesofphasecontrol: .poorpowerfactor; . thyristor has to deal with a large di/dt ; .resultsinharmonicscausingradioandTVdisturbance.1.5 Voltage form across the thyristor

0

0

0

180 360 540 720

ωt°

ωt

ωt

ωt

ωt

vs (v)

5

(b)

(a)

(c)

i0

0

0

vRb

vT

α

α

α α

α

(d)

(e)

IH IH

≈ 1,6V

v0

vT

i0

vs Rbv

SCR

v0

^

v

v

^

^

β

v

2

2

2

2

4

+

Aslongasthethyristorisnotfired(between0°and α)wecancompareitsbehaviourtoanopenswitch. The voltage across its terminals follows faithfullythesupplyvoltage.Seefig.8-1e. TheforwardvoltagedropacrossaconductingSCRisapproximately1.6V.Afterthethyristorswitchesoff(isextinguished)becausethei o has fallenbelowI H , the voltage again follows the supplyvoltage.

Fig. 8-1: Circuit and waveforms of a resistive loaded E 1 - controller.

LINE-FREQUENCY PHASE-CONTROLLED RECTIFIERS 8.17 Numeric example 8-4: Given:Fig. 8-1 with R b =10Ω ; V line =230V-50Hz;α=90°;V T=0. Required:Determine P, S,displacementfactor,distortionfactor,powerfactor,Q 1 , D . Solution:. Fundamental harmonic of current p.8-11:A 1 =81.32 V ; B 1=− 51.77 V ; X 1=96.4V;V 1=68.16V I 1 =6.816A;φ 1=bgtg

B 1 __ A 1 =−32°48’;I 1P = I 1 . cos φ 1=5.75A

. RMS value of current Table8-1:V RMS=0.5x230=115V;I RMS=

115 ___ R b =11.5A . Activepower: P=V RMS . I RMS=115x11.5=1322.5W

or: P= V line . I 1P=230x5.75=1322.5W . Totalapparentpower S=V.I=230x11.5=2645VA . Apparentpoweroffundamentalharmonic S 1 =V. I 1=230x6.816=1567.68VA . Displacementfactor cos φ 1 = P __ S 1 =

1322.5 ________ 1567.68 =0.8436 . Distortion factor

cos δ= S 1 ___ S =1567.68 _______ 2645 =0.5926

. Power factor

λ=P __ S =1322.5 ______ 2645 =0.5 . Reactivepoweroffundamentalharmonic

Q 1 = √__________

S 1 2 − P 2 =841.8var . Distortionpower

D= √__________

S 2 − S 1 2 = 2130.35 var

LINE-FREQUENCY PHASE-CONTROLLED RECTIFIERS 8.57

11. EVALUATION

8.1 InwhichspecificsituationdoesanSCRrequireadoublepulse?Why? 8.2 Whatarethedisadvantagesofusingphasecontrolwiththyristors? 8.3 ForanSKT600(Semikron)weassumethatthemaximumambienttemperatureis50°C and that R thj-a =0.11°C/W.WhatisthemaximumpermissibleI T(AV) ? What is then thepowerdissipationinthethyristor?TheSKT600datacanbefoundonp.4.15and p.4.24/25. 8.4 Whyaresomepowerelectroniccircuitscallednaturalcommutatingcircuits? 8.5 WhatisthemaximumRMSanodevoltagethatisallowablewithanSKT10/08Dconfigured asacontrolledrectifier(E1 -controller)?8.6 Intheconfigurationshowninfig.8-10V line =400V.Whatistheaveragerectifiedoutput voltage when α=135°andforα=100°.8.7 Whatismeantby“miss-firing”ofaninverter?Howcanthisbeprevented? 8.8 Whatdoestheexpression“distortionpower"means? 8.9 Whatisthemaximumpowerfactorofafullcontrolledbridge? 8.10 Infig.8-20thefollowinginformationisprovided:R b =10Ω,L b =150mH,transformer andthyristorareideal,linevoltage230V-50Hz.Determinewithafiringangleofα =0° and60°:theoutputDCvoltage;outputcurrent,powerfactor,displacementfactor,distortion factorandreactivepower. 8.11 Whichspecificationsshouldthethyristorsinfig.8-20havegiventheinformationinthe previousquestion. 8.12 Determine the maximum inverse voltage of the semiconductors in the situations shown in fig.8-19and8-21. 8.13 DeterminetheDCoutputvoltageforthecircuitshowninfig.8-9ifthesupplyvoltageis 230V-50Hzandthefiringangleisrespectively60°and105°. 8.14 Infig.8-3thesupplyvoltageis230V-50Hzandtheloadiscomprisedof7.255Ωand 40mH.Determinetheaverageoutputvoltageifthefiringangleis60°.Inadditiondetermine thespecificationsthethyristorneedstocomplywith. 8.15 The full controlled B 2 -controlleroffig.8-9isconnectedtoa230V-50Hzsupply.Theload iscomprisedof7.255Ωand40mH.Calculatetheaverageoutputvoltageifthefiringangle isrespectively30°,60°,90°. 8.16 Determinethelargestinversevoltageacrossthethyristorinfig.8-1.Whatisthelargest possibleblockingvoltageacrosstheSCR?Forwhichvalueofαdoes this occur? 8.17 Set α=120°intheconfigurationoffig.8-20.Canthebridgebebroughtintoconduction? Confirmthiswithagraphicsuchasfig.8-21or8-23. 8.18 Considerthespeedcontrolofanindependentlyexcited100kWmotor. Thecurrentprotectionissetat1.5xdenominalarmaturecurrent.Howmuchreactivepower (kvars)doesthismotordrawfromthenetatstart-up? 8.19 Weconsiderahalfwaveandfullwavecontrolledrectifier,bothwithresistiveload.Let α=30°and150°respectively.Determinefortherectifierstheoutputpowerwithrespectto themaximumpossibleoutputpower(whenα=0°).Usetable8-1.

8.58 LINE-FREQUENCY PHASE-CONTROLLED RECTIFIERS

+- v0

RbLb

i3v3

i2v2

L1

L2

L3

(v1 = v.cosωt)

i1v1

Th3

Th2

Th1

i0

Fig. 8-40: M 3 -controller(halfwavefullcontrolledthreephasebridge)

8.20 Determine for the data used in question 8.10 with α=60°:theRMSvalueofthefundamen- talharmonicoftheinputlinecurrentofthebridge,RMSvaluesofthefifthandseventh harmoniccurrentcomponentsofthelinecurrent,RMSvaluesofthe300Hz-and600Hz- voltagecomponentsoftheoutput. 8.21 An E 1-controllerconnectedtoa230V-50HzhasaresistiveloadofR b =100Ω. Calculatetheamplitudeofthe150Hzcurrentintheloadwhenthefiringangleis60°and 90°respectively. 8.22 A full controlled B 6-controllerisconnectedtoathreephasesupplyof3x400V-50Hzwitha highlyinductiveloadof5Ω-280mH.CalculatetheRMSvalueofthe350Hzcurrentinthe supplylinewhenthefiringangleis30°.

8.23 Sketchtheoutputvoltagewaveformwithafiringangleαfor a highly inductive load in the following situations: a) fullwavefullcontrolledsinglephasebridge b)halfwavefullcontrolledthreephasebridge(M3,seefig.8-40lowerdown). Sketchinasimilarfashiontofig.8-34,aperiodicvoltagev o = f(x)and derive an expressionforωt=f(x).8-24 DeterminethemaximumoutputvoltagesforaB2 - controller, M 3 - and B 6 -controller. Thestartingpointisexpression(8-25). 8.25 Given a full control highly inductive loaded B 6 -controller. Calculate the ratio of the RMS valuesoftheharmonicsintheoutputwhenα=90°inrelationtoα=0°.Checkyouranswer withthevaluesfromtable8-4.

CONTROL OF THYRISTORS 11.7

3. CONTROL IC FOR SCR AND TRIAC

AnumberofcompaniesselllinearIC’sthatcanformpartofacontrolmoduleforthyristors.Itispossibletobuildatemperaturecontrollerforanoven,aspeedcontrollerforamotor,etc...withaminimumnumberofcomponents. WeconsiderjustsuchacontrolIC,theindustrystandardTCA785fromInfineon.

3.1 AC controller using a TCA785

TheoperationoftheTCA785isexplainedunderheading3.2. TheIChas16pins.TheICissuppliedwith15VDCviapin16andpin1. Infig.11-5theDCsupplyiscreatedwithrectification(1N4005),filtering(470µF)andzenerstabi-lisation (BZY97-C15). Todetectzerocrossoverofthesupply,pin5isconnectedtothesupplyvoltageviaaresistorR sync =220kΩ.The value of R 9(resistorconnectedbetweenpin9andground!)determinesthechargingcurrentI 10 thatlinearlychargescapacitorC 10 .Atpin10wehavearampvoltageinwhichtheslopeisdeterminedbyR 9 and C 10 . Thevoltageatpin11isregulatedwithapotmeterof10kΩandasaresultthefiringangleα can varybetween0°and180°. Pin15deliversafiringpulseduringthepositivehalfperiodofthesupplyvoltageandpin14doesthesameduringthenegativehalfperiod. Bypasscapacitorsareconnectedtopins1,11and13.InthiscircuitanACcurrentisphasecon-trolledasshowninfig.9-1.

230V

TCA 785

0,22µF

250V

4,7k

220k9W

1N4005

470µF

16V

15V

10k

0,47µF

RSYNC

BAY61

BAY61

1

2

3

4

5

6

7

8

16

15

14

13

12

11

10

9

BAY 61

BAY 61

22k

100k

4.7k

(bulb als a load)

0,1µF47nF

150pF

C10R9C12

2.2k

2.2µF (MKH)

150

L

triac

10k

R

Mp

BZY

97-C

15

inhibit

Fig. 11-5: PhasecontrolwithaTCA785fortriacswithagatecurrentupto50mA

12.12 CHOPPERS

5.CHOPPERCONTROLIC’S

5.1 Description of block diagram

470

470k

1k

150

150

100µF+-

Compensation

shutdown

oscillator (out)t

tv7

voltage regulator

Oscillator

Vref.

1

2

3

-+5V

8

10

2

1

7

6

3

16

15

9

5

4

14

13

11

12

faultamplifier

+-

comparator

+5V

7+-

+5V

6

5

-+

+5V

0,1µ 1k

10k

NORFLIP-FLOP

8

NOR9

4 Q

Q

270

470

470

2N2904

BC140

68-5W

Rb

RbD1

D2

Lb9,1V 3W

S

IRFN530

M

+24V

T1

T2

Fig. 12-14: Adidacticmodelofachopperwitha“1524”controlIC.ThisICworkswithasupplybetween8and40V(max), hasanoscillatorfrequencybetween100Hzand500kHzandhastwooutputtransistors(100mA)thatcanhandle a maximum collector voltage of 60V

The“1524”anditsderivativesisamuchusedIC-controller.Itfirstappearedin1976asthe“SG1524”(SG=SiliconGeneral)andquicklybecametheindustrystandardandtothisdayisof-feredbymanychipproducers(CA1524,LM1524,UC1524,XR1524,…). Fig.12-14showsadidacticmodelwhichallowsustostudyachoppercircuitwhenPWMisused.WithswitchSwecanchoosebetweenaresistiveload,aninductiveloadoramotorasload. TheinterfacebetweenICandswitchSwerecogniseasacontrolcircuitfromthestudyofthepowerMOSFET. The“1524”isapulsewidthmodulatedcontrollerforvoltagecontrollers.Thefrequencyisdeter-minedfromthefollowingexpression: (12-10) Here in is f inkHzwithR 6 inkΩandC 7 in µF. With R 6 and C 7 wemeanthediscretecomponentsthatareconnectedbetweenpin6and7oftheIC.Practicalvaluesare: 1.8kΩ< R 6 <500kΩ 0.001µF < C 7 < 0.1µF The value of R 6 determines the constant current which linearly charges C 7 . The sawtooth of +1V to+3.5Visappliedtothe(+)terminalofcomparator3. AflipflopmakessurethatonlyoneoftheNORgates8or9receivesalowinput.TheNORgatewhichhasallinputslowwillsaturateitsrespectiveoutputtransistorsT1 or T 2 . For this to occur thefollowingmustbethecase:V 3 =0Vandtheoutputofcomparator3shouldalsobe0V.

f= 1.18 _____ R 6 . C 7

CHOPPERS 12.13

v3

t (ms)

t (ms)

t (ms)

t (ms)

t (ms)

t (ms)

t (ms)

3

0 8 18

v7

1

0 8 1810 20

3.5

1.5

v9

0

1.5

vGS

0

8

8 10 18 20

v11/14

v11

0

22

8 10 18 20

v14

v11/14

v11

0

22

8 10 18 20

v14

vRb

22

80 10 18 20

vDS

0

24

1

8 10 18 20

(V)

(V)

(V)

(V)

(V)

(V)

(V)

An internal voltage controller 1 regulates the 5V(±1%)supplyoftheICandwhichcanbeex-ternally loaded to 20mA. TheoutputpulseV 3 ofoscillator2isa“blank-ing”pulsewho’sfunctionistopreventT1 and T 2 conducting at the same time. This is neces-saryifthetransistorsarenotinparallelasshowninfig.12-14,butwhereforexampletheyoperateseparatelytocontrolabalancecircuit.Inthiscasethefrequencyoftheoutputvoltageishalfofthesawtooth frequency.

5.2 Operation of comparator 3

C 7 providesasawtoothvoltageof+1to3.5Vtothenoninvertingterminal(+)ofcomparator3.Ontheinvertinginput(−)ofthecomparatoravoltageisappliedwithfourpossibilities: A. Theinvertinginputissmallerthan+1Vso thattheoutputofthecomparatorishigh (+5V)andbothoutputtransistorsare blocking. Thiscanbeachievedby: 1) saturatingtransistor5viapin10sothat V CE < 1VandtheICblockswithin 200ns(shutdownprotection). 2) usingopamp7asacurrentlimiter. B. Avoltageofbetween+1Vand+3.5Vis addedtotheinvertinginputofthecompara- tor.AsaresultthepulsewidthattheIC outputiscontrolled. Wecanachievethisby: 3) linearly controlling the duty cycle via thefaultamplifier(6). 4) adding a voltage V 9 as shown in fig.12-14whichalsoregulatestheduty cycle. Inthismannerweseeinfig.12-15that with V 9 =+1.5Vthereisconduction between8and10ms(V 11 ) and between18and20ms(V 14 )

Fig. 12-15: Waveformassociatedwithfig.12-14

SWITCH-MODE POWER SUPPLIES 13.11

2.3 Buck-boost converter

2.3.1 Operation of buck-boost converter

Fig.13-9showsthebasicconfiguration.ClosingScausesacurrenti L toflowandmagneticenergyis stored in the coil ( 1 __ 2 . L . i L 2 ). Whentheswitchopens,themagneticenergycanonlybedischargedintheloadandthisresultsinareversepolarityoftheoutputvoltageV o withrespecttoV i .Thecapacitoronceagainfunctionsasafiltercapacitor. Aswithaboostconverterwefirststoreenergyinthecoilandpump this energy to the load when the switch is open.Thisexplainswhybothconvertersareclassifiedasflyback converters.

vi

is

iL vL RbC

D+

-

L

S

vi

vs

iD

iDvD

iL Rb VoIo

C

D+

-

L

S

+

-vL

Onceagainwedistinguishbetweentwooperatingmodes,namelycontinuousanddiscontinuouscurrent mode in the coil. 2.3.2 Continuous current in the coil

Fig. 13-10a shows the associated waveforms. When the switch closes a voltage results across the

coil: v L = V i =L. ∆i __ ∆t =L.

(I 1 − I 2 ) _______ δ.T or:L.( I 1 − I 2 )=δ.T. V i (13-14).

Whentheswitchopensafterneglectingv D : v L = V o =L ∆i ___ ∆t =L. ( I 2 − I 1 )

_________ (1−δ).T or:

L.( I 2 − I 1 )= V o .(1−δ).T (13-15)

From (13-14) and (13-15) it follows: (13-16)

Fig. 13-9: Basicprincipleofbuck-boostconverter

V o = − V i . δ ____ 1 − δ

13.12 SWITCH-MODE POWER SUPPLIES

2.3.3 Discontinuous current in the coil

Inthiscasetheenergyinthecoiliscompletelydischargedbeforethetransistorswitchrecloses.Once again there is a dead time t d (discontinuouscurrent).Fig.13-10bshowstheassociatedwave-forms.ClosingtheswitchSresultsinavoltageacrossthecoil,whereby V i = v L =L. ∆i __ ∆t =L.

I 1 ___ δ.T or: L . I 1 =δ.T. V i (13-17)

Aftertheswitchopensduringthetimeinterval(1−δ).T− t d almost all the energy in the coil is gone. The voltage across the coil is now V i and the current changes from I 1 tozero,sothat: V i = v L =L.

(0− I 1 ) _______________ [ (1−δ).T− t d ]

→→ L . I 1 =− V o . [ (1−δ).T− t d ] (13-18) From (13-17) and (13-18) it follows that: δ.T. V i =− V o . [ (1−δ).T− t d ] from which: (13-19) 2.3.4 Remarks

1. From(13-16)and(13-19)itfollowsthatthebuck-boostconverterresultsinvoltage inversion. 2. Dependingonthevalueofδwe have | V o |≤or ≥| V i | .Thisisthereforeastep-up/step- down converter. 3. The maximum voltage across the (transistor) switch is V i + V o ascanbeseenin fig.13-10atthebottom. 4. The maximum reverse voltage across the diode is V i + V o . 2.3.5 Numeric example 13-3: Forabuckboostconverterwithcontinuouscurrentmodewearegiven: .inputvoltage:V i =3to15V .max.outputcurrent:I o =3A.outputvoltage:V o =9V± 0.1% .chopperfrequency:f=100kHz Required: determine the values of L and C Solution:

From (13-16) →→ V o = V imax . δ min _________ (1− δ min ) →→ δ min =

V o _________ V imax + V o =

9 ______ 15 + 9 =0.375.

Assumption∆ i Lmax =0.2x I o =0.6A

From (13-15) →→ L max = V o . (1− δ min ).T

____________ ∆ i L =9 . (1 − 0.375) . 10 − 5

__________________ 0.6 =93.75µH

From (13-7) →→ C min = ∆ i L ___________ 8.f.∆ v Cmax

= 0.6 ______________ 8 . 10 5 . 9 . 10 − 3

=83.4µF

and: ESR max = ∆ v omax _______ ∆ i L =0.1% . 9 _______ 0.6 =0.015Ω

V o =− V i . δ _________ 1−δ− t d /T

!


Fig. 13-10: Waveformsoffig.13-9

t

iS

I1I2

t

iLI1I2

t

vL

-Vo

I2

I1

iD

Vi

∆ iL

t

Vi + Vo

vS

t

Vi

T

(a) Continuous current

t

iS

I1

t

iL

I1

t

vL

iDI1

Vi

-Vo

tvS

t

Vi

T

(b) Discontinuous current

td

Vi + Vo

δ.Tδ.T(1 - δ).T

(1 - δ).T


13. NON IDEAL WAVEFORM

Infig.13-26weredrawtheflybackconverter,butnowanumberofparasiticelementsareinclud-ed.Thesehavebeendrawningreen. Where: L cb : parasiticselfinductionoftheprintedcircuitboard; L 0 : magnetizinginductanceofthetransformer; L lk : leakage inductance of transformer; C prim: capacitanceoftheprimarywindings; C sec : capacitanceofthesecondarywindings; C tr : feedbackcapacitancebetweensecondaryandprimarytransformer; C D : diodecapacitance; C o : outputcapacitanceoftransistor. Fig.13-26isanapproximateddrawingsincetheseparasiticelementsdonothaveaconstantvalueandareoftenvoltageandfrequencydependent. Fig.13-27ashowsforexamplethevoltageacrosstheMOSFET.ItisclearthattheoscillationscausethevoltageacrosstheMOSFETtoincrease.AsnubberacrosstheMOSFETcanpreventitbeingdestroyed. Ifwetakealookatthefirsttwoperiodsofoscillation(fig.13-27b),thenweseeinthisexampleaperiodof100ns.Thiscorrespondsto10MHz. Aruleofthumbindicatesthattheoscillationfrequencyisapproximately100timestheswitchingfrequency. Fig.13-27bistheresultofresonancebetweentheleakageinductanceofthetransformerontheonehandandtheselfinductanceofthecurrentloopontheprintedcircuitandontheotherhandtheoutputcapacitanceoftheFETandthetransformercapacitances.ThedistortionofthesinusoidsistheresultofthevoltagedependentoutputcapacitanceoftheMOSFET.

cb

0

lk

0

prim.

sec.

tr. D

C

C

L

L

LC C

C

Fig. 13-26: Flybackconverterwithparasiticelements


0

V

2 4 6 8 10 12t (µs)

t (ns)

DS (V)

140

120

100

80

60

40

20

v (V)

40 80 120 140

80

60

40

20

20

40

60

0

(a) (b)

Fig.13-27a: WaveformontheFETinfig.13.26 Fig.13-27b:Firstperiodoftheoscillationsinfig.13-27a

14. DIGITAL CONTROL OF AN SMPS

Withanewtypeofdigitalsignalcontroller(DSC)thereisnowenoughcalculationpoweronboardtoexecutethesoftwarequicklyenoughtohandleclosedloopcontrol.ThisDSChasin additionbuilt-inintelligentmodulesforPWM,andananaloguetodigitalconverter(ADC).

IN OUT

T

T

1

2

L

C RPWM

PWM

1

2

IN OUTPID PBM

T

T

1

2

S&H

S&HADC

+

DSCFig. 13-29: Digitallycontrolledbuckconverter

Welookattheexampleofthesynchronousbuckconverterfromfig.13-24andapplyavoltagebetweenthepointsAandB.Thisbringsustofig.13-28.Theanaloguecontrolcircuitfromfig.13-24hasnowbeenreplacedwith a digital control system with a DSC (fig.13-29).

Fig. 13-28: Basicconfigurationofasynchronousbuckconverter


Theprocessinginthedigitalcontrolloopresultsindelays.APIDcontrollerthatrunsintheDSChasatypicalcalculationtimeof1µs.ThePIDcontrolitselftakesabout2µs.Theanalogue-digitalconverter(ADC)takessamplesandconvertsinabout500ns.Considerinadditionatransistorswitching time of 100ns, then we arrive at a total delay of 3.6µs. From this we can determine an optimumsamplingrateof277kHz.Inordernottolooseanyinformation,theShannontheoremtellsusthesamplingfrequencyhastobeatleasttwicethehighestfrequencythatFourieranalysisrevealsintheanaloguesignal.Withthis“twice”samplingratethephasedisplacementis180°.Due to the associated delay in the system we exceed the maximum allowed 180° required for a stablesystem.Forthisreasonindigitalcontrolsystemstheanaloguesignalisoversampledwithafactor ten.

15. EVALUATION

13.1 Withwhichisolatedconverteristhesupplyvoltageacrossthetransistorandforwhich converteristhedoublesupplyvoltageacrossthetransistor. 13.2 What does discontinuous current mean for a converter? 13.3 Whichtypesofflybackconverterareyoufamiliarwith? 13.4 Whatisthedifferencebetweenaflybackandaforwardtype? 13.5 Whatdoes“feedforward”controlofanSMPSinvolve? 13.6 WhattypeofoutputdiodeisusedforanSMPSwithahighoutputvoltage? 13.7 Whatisthebigadvantageofferritecoretransformerscomparedtostandardironcore transformers? 13.8 What does off-the-line mean? 13.9 Whyshouldflybackconvertersneverhaveanopenoutput,orbewithoutacontrolloop? 13.10 Uptowhichpowerlevelarelinearpowerregulatorsused?Whatistheefficiencyofsuch a regulator? 13.11 a) What does EMI stands for? b)Whatforthemostpartdeterminestheweightandvolumeofalinearstabilisedpower supply? 13.12 Provethatexpression(13-21)isequivalenttoexpression(13-19).

16. THE DESIGN OF SWITCH-MODE POWER SUPPLIES

ThiscanbefoundonthewebsiteofCASPOC.www.caspoc.com/educationundertheheadingbooks“Pollefliet”.

14.16 INVERTERS

Fig. 14-14: Outputlinevoltageasafunctionofmodulationdepthm(forN=15)

6. PWM STRATEGIES

Upuntil1980thepulsepatternsforPWMwereproducedusinganalogtechniques.Between1980and1990digitaltechnologystartedtobeusedtoproducethePWMwaveformsinthree-phaseinvertersformotorcontrol.Intheearlyyears8bitmicroprocessorswereused.AfterthatDSP’s(digitalsignalprocessors)appearedonthescene.

6.1 Analog technology ( system with modulated output voltage)

Thissystemwithamodulatedoutputvoltagewascommoninearlyanalogtechnology.Areferencevoltagewascomparedwithacarrierwaveofhigherfrequency.Thisiscomparabletothedidacticexampleoffig.14-11.

6.2. Digital technology (system with calculated voltage pattern)

The desired condition of the inverter switches is calculated using a DSP. For the three voltages thereisavaryingsinglestepshapedreference,thiscanforexamplebederivedfroma“look-up”table.This(hypothetical)sinusoidalvariablereferenceiscorrectedduringeverypulsewitha“zerosequence”value.Thiszerosequencevalueisdeterminedperpulsebydividingbytwothesumofthemostpositivevalueandthemostnegativevalueandaddingthiszerosequencevaluetothethreebasicreferences. Inthiswaythezeropausesatthestartandendofeverypulsearesynchronised. 6.3 Value of the output voltage of a PWM inverter

Fig.14-5showstheoutputvoltageofaninverterwithaquasisquarewaveintheoutput.

From(8-28)theamplitudeofsuchawaveformcanbefound:

v 1 = 2 . √

__ 3 . V t _________ π ,

so that the RMS value is: V 1 eff. = √

__ 6 ____ π . V t . Here in : V t ≈ ^

v = √

__ 2 . V net .

v LL 2 Vt1

m0

0,612

0,78

1 3,2

over-modulationsquare wave

linear area

APPLICATIONS OF POWER ELECTRONICS 15.7

. softwarecontrol:viaRS232andalocalnetwork.ViathebuiltinwebservertheUPScanbe remotely monitored and controlled.

1.3 Static UPS without auxiliary network

1.3.1 Passive stand-by topology

Whenthereisnoauxiliarynetworkavailablewhenfaultsoccurthebatterieshavetosupplytheenergy.Thesebatteriesarereferredtoastheback-uporredundantcurrentsource.Forcomputerpowersuppliesextrarequirementsmaybemade. Inthesupplynetworkanumberofdisturbancesmaybepresentduetoatmosphericinfluences(lightning,…)faultsinthesupplynetwork(shortcircuits,…)industrialdisturbance(weldingmachines,motors,lifts,fluorescentlamps,…).Overvoltageandundervoltagecanoccuraswellasmicrointerruptionsorlonginterruptions.Thesevarietyofdisturbancesarenotcompatiblewithgoodoperationof(micro)computers. Thesupplynetworkofacomputershouldbefreefromthefollowing: . frequency and voltage variation. voltagedrop. shortpowerinterruptions. electromagnetic faults.Inadditionthereshouldbesuitableautonomytoenablesavingofdataandthecompletionofcal-culationsintheeventofapowerfailure.Anautonomyofbetween10to15minutesisusuallythemaximum.Ittakesseveralmsfortheautonomytokickin.Fig.15-5showsthebasicschematic. Afilterneutralizesthehighfrequencypeaks.Sometimesanultraisolatorisincluded,Thisisatransformerwithoneormorescreenssothatshortdurationhighfrequencyfaultsmaybefiltered.

Fig. 15-5: UPSofthepassivestand-bytype

Innormalservicetheconsumerissuppliedviaafilterandultra-isolator.Whenavoltagedistur-bancebecomestooextremetheUPSswitchestoaninverteroutput.Thisinverteroutputisincontinuousservicetogetherwiththerectifierandbattery. ThepositionsoftheswitchareindicatedwithS.B.(stand-by)andO.L.(online). Under2kWUPSareusedalmostexclusivelytosupplycomputers.

RECTIFIER INVERTER

24V+

-

230V

GRID

FILTERULTRA-ISOLATORSWITCH OVER

CONSUMER

O.L.

S.B.


Asalightsource,LEDshaveanumberofinterestingproperties: 1. Incontrasttootherlightsources,LEDsdonotsuddenlyfailorburnout.Theluminous intensity of LEDs reduces gradually. They still emit 70% of their initial luminous intensity after50,000hours(onfor12hoursperdayfor11years!).ThelifetimeofLEDsisbetween 50,000and100,000hours.Bycontrast,anincandescentlamphasalifetimeofabout3000 hoursandaCFL(compactfluorescentlamp)lastsforupto10,000hours.Thereistherefore nonecessitytoreplaceaLEDlamp,whichisimportantforcriticalapplicationssuchas beacons,buoys,lighthouses,emergencyexits,etc. 2. Highefficiency.WithcurrenttechnologywehaveUHB-LEDs(ultrahighbrightnessLED) whichhave10timesthelumen/wattratioofahalogenlamp.ALEDheadlightof50Wina caristhereforeequivalenttoa500Whalogenlamp.AUHB-LEDcurrentlyprovidesupto 100lm/W(whitelight),andprototypesexistwhichprovide200lm/W.TheoriginalLED- indicator in the early 70s delivered 0.1 lm/W. 3. LEDs emit little heat. 4. Theyareflatelementswithsmalldimensions. 5. Canwithstandshocksandvibrations,whichisimportantincars. 6. Theyaresemiconductors,whichmeansthereisnogas,mercuryorfilamentpresent.The LED therefore has an environmental advantage. 7. OftenmultipleLEDsareconnectedinseries,withaconstantcurrent.IfoneLEDgets short-circuitedtherestcontinuetooperatesincetheyaresuppliedwithaconstantcurrentsource. 8. ByusingacontrolIC,itispossibletocontroltheluminousintensityusingPWM. Inpractice,itispossibletocontrolfrom0tothefullluminousintensity.UsingPWMitis possibletocontroltheluminousintensity,whilemaintainingfullcolourintegrity. 9. Operationisatlowvoltagelevelsandhighvoltageconnectionsarethereforeavoided. ALEDrequiresabout3V(between2.5and4.47Vdependingonthetype)sothataseries connectionoftensofLEDscaneasilybecontrolledfromanICthatissuppliedwithan inputvoltageof40to50V.TheLT3595(LinearTechnology)isa16channelLED-driver capableofhandling160LEDsfromaDCsourceof45V.ThisICcanhaveadutycycle (dimming) of 5000:1.

Remarks 1. AnotherinterestingapplicationofLEDsisthatanRGB-colouredlightsourcecanbebuilt. ToachievethisoftentwogreenLEDswithoneblueandoneredLEDareused.Thiscom- binationtakesthesensitivityofthehumaneyetocolourintoaccount.ALEDhasanomi- nalwavelengthandcolourwithaspecificcurrent.That’swhyweuseacurrentsourcewith theLEDs.UsingcolourmixingLEDscancreateacontinuouscolourspectrum.Toachieve this,theluminousintensityofeachLEDiscontrolled.Thisisnotachievedbylinearly controlling the current of each LED since this would change the colour. The solution is PWM of the current. FourchannelLED-driversexistinICformtorealizethisRGB-luminousflux(forexample theSTP04CM596fromthefirmST,seewww.st.comunder“lighting”).

!

15.22 APPLICATIONS OF POWER ELECTRONICS

2. AnotherexampleoflightingcontrolusingLEDsistheapplicationofthecontrolIC IRS2540fromInternationalRectifier.ThisICcanoperatewithasupplyof200V.APWM- controlsignalusesburstmodecontrolat175kHz.ThedutycycleofthePWMcontrolis100%. 3. AnimportantnicheforLEDsisthecarindustry.Europeancarswerethefirsttouseblue/ green/white/ambercolouredLEDsintheinstrumentpanel.AmericanandJapaneseluxury carbrandsfollowedthistrend.AbrandsuchasLexususes50WUHB-LEDSforthehead- lights.LEDshavestartedtoreplacehalogenandXenonlamps.Advantagesincludethe streamlineddesignsandthelongerlifetimeofthelamp.LEDtechnologyisnolongerjust usedintheluxurycarmarketbutisalsousedinthemiddleclasssegment.Applications vary from head lights to interior lighting. The LT3755 from Linear Technology is a control-ICintendedfor50Wheadlights.ThisICcanboostthe12Vbatteryvoltageto60V to control a series circuit of 14 LEDS drawing 1A. 4. AnotherimportantterrainforLEDSis“backlighting”ofHDTVLCDscreens.Forback- lightinga46”LCDTVrequiresabouttenLT3595’sperHDTV. 5. IC-manufacturersproducefamiliesofLED-driversthatcomplywithspecificrequirements forspecificapplications.InfineonforexamplehasalowcostLED-driver(BCR450)for controlofhighpowerLEDs.IncombinationwithanexternaltransistorthisBCR450is capableofprecisecontrolofthecurrent.Inadditionthereisovercurrentandovervoltage protection,etc. 6. ALEDcannotbedimmedusingatriacdimmer.NationalhasdevelopedanIC(LM3445) thatmakesitpossibletodimaLEDincombinationwithatriacdimmer.(www.national. com/powerwise) 7. MoreandmorepublicauthoritiesareconsideringinstallingLEDstreetlighting.Astreet- lightisroutinelyrequiredtodeliver10,000lumens.CurrentLEDstreetlightsusebetween 50 and 200 LEDs with a current of 350mA. A solution for street lighting is to connect a numberofseriesLEDcircuitsinparallel.AccordingtoIECspecificationsatransformer needstobeplacedbetweentheLEDsandthesupplynetandthesecondaryvoltagemay notbegreaterthan120V.Often48Visused.Inthismanner12InGaN-LEDscanbecon- nectedinseries(12x3.3V=39.6V).Abuckconverterisrecommendedforthesupply. Itisthecheapest,simplestandmostefficientoftheclassicconverters.Ifaconverteris usedforeveryseriescircuitofLEDsthentheycanbedimmedindependently.Adisadvan- tageofthissetupthoughisthecostprice.Whenthebuckconvertersareconnectedin parallelthenEMCproblemsresult.Tocombatthis,filterscontainingatleastacoilanda capacitorarerequiredintheinputofeachconverter.Thisincreasesthecostpriceeven further.Topreservethequalityofthesupply,networkpowerfactorcorrectionisalso implemented.StreetlightingisusuallyclassifiedasHPWA(highpowerwidearealight- ing).Inadditiontoenergyefficientsolutionsforinteriorlighting,streetsandtouristattrac- tionsattheWorldExpo2010inShanghai,OSRAMtogetherwithitsmothercompany Siemens,installed150,000LEDsinthepavilionsandstreetsaroundtheExpo.


12

7

9

54

3

6 8

5. RENEWABLE ENERGY

A. WIND TURBINES 5.1 History CharlesBrush(1849-1929)wasoneofthepioneersoftheAmericanelectricalindustry.Inthewinterof1887-1888Brushbuiltthefirstautomaticwindturbinetoproduceelectricity.Arotorof17mdiameterwith144rotorbladesmadeofcedarwooddroveaDCgeneratorof12kW.Theturbineworkedfor20years. PoulLaCour(1846-1908)wasthepioneerofmodernaerodynamicsandbuilthisownwindtun-neltouseforexperiments.In1897LaCourhadexperimentalwindturbinesintheAskovFolkHighschool(Denmark).In1918therewereabout120localwindturbinesofbetween20and35kWinDenmark.AllthegeneratorswereDCmachines.In1951aDCgeneratorwasreplacedwitha35kWasynchronousmachine.In1980-81a55kWgeneratorwasdevelopedthatultimate-lywasresponsibleforthebreakthroughofthemodernindustrialwindturbine. Inspecialcaseswindturbinesupto6MWarepossible.Ontheotherextremeofthepowerrangethereisalsoamarketforsmallwindturbines(100kW)forprivateuseonfarmsetc. Themajorityofthewindturbineshaveapowercapabilitybetween500kWand1MW. The6MWturbinefromREpowerinstalledin2009ontheGermanDanishborderhasarotor- diameterof126m,rotorbladeswithalengthof61.5mandahubheightof100m. Currently1%ofalltheworldselectricityisproducedfromwindenergy,16%fromnuclearenergy and 16% from hydro-electricity. 5.2 Construction of a wind turbine

Fig. 15-22: Crosssectionofa2MWwindturbine.CourtesyofVestasWindTechnologyA/S

Thepartsare: 1:rotor 4:alternator 7:measurementofwindspeedanddirection 2: main shaft 5: radiator 8: tower 3:gearbox 6:mechanicalbrake 9:gondola

16.6 ELECTRIC MACHINES

1.4 Impedance transformation

1.4.1 Transformation formula Fig.16-7ashowsanidealtransformer,loadedwithaseriesR-L-C circuit. An ideal transformer is a transformer without losses.

Fig.16-7:Ideal transformer, loaded with a series R-L-Ccircuit:impedancetransformation

Secondary: v s = R s . i s + L s . dis __ dt + 1 __ C s

∫ 0

t

i p.dt Applicationof(16-8)gives: v p=k

2 . R s . i p+k2 . L s .

dip ___ dt +k

2 . 1 ___ C s ∫

0

t

i p.dt Foranidealtransformer,thesecondaryloadcanberepresentedasanequivalentcircuitseenfromtheprimaryside(fig.16-7b),aslongas:R’=k2 . R s;L’=k

2 . L s ; C’ = C S/k2 .

Moregenerally:anidealtransformerwithsecondaryimpedanceZ sec . maybeseenasa primaryimpedance: (16-11) From(16-11)itfollowsthatwecantransformaprimaryimpedancetoanequivalentsecondary impedance: (16-12)

1.4.2 Numeric example 16-1:

1. Anelectricalovenissuppliedwith46voltandhasapowerof4kW.Thesupplynetworkis 230V-50Hz.Ifwehadanidealtransformeravailable,whatisthen: a) the transformation ratio b)theprimaryandsecondarycurrent c)impedanceseenfromthe230V-50Hznet?

Z’ prim. = ( N p

______ N s ) 2 . Z sec. (Ω)

Z’ sec. = ( N s _______

N p ) 2 . Z prim. (Ω)

Vp

IpIp

Vp Vs

Is

NP NS

RS R’ = k2 . RS

L’ = k2 . LS

C’ = CS / k2

LS

CS

(a) (b)

ELECTRIC MACHINES 16.7 Solution:

a) k= N P

___ N S = 230 ___ 46 =5

b) secondarycurrent:I S = P S

___ V S = 4000 ____ 46 =86.95A

primarycurrent:I P = I S

__ k = 86.95 _____ 5 =17.39A

c) load impedance:Z S = V S

__ I S = 46 _____ 86.95 =0.529+j.0Ω

Transformedimpedanceseenfromthesource:Z prim.=k2 . Z sec. = 52x0.529=13.23Ω

Proof: Z prim.xIP = 13.23x17.39=230V!! 2. Ifmaximumpowertransferisrequiredfromthegeneratortotheconsumer,thentheconsumers impedanceshouldbethecomplexconjugatevalueofthegeneratorimpedance. Wehaveapoweramplifierwithanoutputresistanceof48Ωandwishtoconnectaloudspeaker withthefollowingcharacteristics:30W-4Ω. Maximumpowertransferispossiblebyplacinganimpedancetransformerbetweenamplifier

andloudspeaker.Theturnsratioshouldbek= N P

___ N S = √_____

Z prim.

____ Z sec. = √

___

48 ___ 4 = 3.46.

1.5 Magnetizing inductance

Withamagnetisingcurrent I µ themagneticfieldstrengthinthecoreisH = N P . I µ

_____ l k and the

magnetic induction is B = µ 0 . µ r . H sothatthefluxinthecoreofthetransformeris: Φ 0 = B . A k =

µ 0 . µ r . A k ________ l k . N p . I µ

Whereby: . Φ0 (Wb): no-loadflux≈resultingflux(Φ 1 − Φ 2)with load . B(Wb/m²): magneticinductioninthecore . µ r : relativepermeabilityofcorematerial . µ 0 : = 4 . π . 10 −7H/m . l k (m): averagelengthoffieldlineinthecore . A k (m²): cross-sectionalareaofcore . I µ (A): magnetising current of the transformer. If we call L 0 theselfinductanceoftheprimarywithrespecttothefluxΦ 0 in the core, then we may

write: N p.Φ0 = L 0 . I µ so that: N p.Φ0 = µ r . µ 0 . A k ________ l k

. N p 2 . I µ = L 0 . I µ

from which follows: (H) (16-13) L 0 = N p

2 . µ r . µ 0 . A k ________ l k


Numeric example 16-2:

1.Aringcoretransformer(fig.16-8)consistsof:

Core: averageradius60mm;cross-sectionoftorus45mm; µ r =1600.

Insulation layer: 1 mm thick

Primary: 3layers:respectively201,189and140windingsAWG18.

Eachlayerisseparatedby1mmthickinsulation.

Insulation layer: 4 mm thick

Secondary: twolayers:50and22windingsAWG10,separatedby1mmofinsulation

2. Extractfromwindingwiretable(AWG=Americanwiregauge)

Fig. 16-8:Ringcoretransformer(a)cross-section(b)coreandwindings

Question:

1. Resistanceofprimaryandsecondarycoil

2. Magnetising inductance

Solution:

1.1 Primary resistance

LAYER1: lengthofonewinding:πx0.04811=0.15114m

totallength: 201x0.15114=30.38m

AWGdiameter (with insulation) in mm

min. max.

resistance (per 100 m)

Ω

admitted current(on base of 2A/mm²)

A

10 2.64 2.69 0.3276 10.38

18 1.08 1.11 2.095 1.624

seco

ndar

y

60 mmprimary

ø 45 mm

ø 45 mm

core

insulation

1mm1,11

1,11

1,111

1mm

4mm1

2,69

2,69

(a) (b)

ELECTRIC MACHINES 16.9 LAYER2: lengthofonewinding: πx0.05233=0.1644m totallength: 189x0.1644=31.07m LAYER3: lengthofonewinding: πx0.05655=0.1776m totallength: 140x0.1776=24.87m Totallengthprimarywinding:86.32m Resistance: R p = 86.32 _____ 100 x2.095=1.8Ω 1.2 Secondary resistance LAYER1: lengthofonewinding: πx0.06835=0.2147m totallength: 50x0.2147=10.73m LAYER2: lengthofonewinding: πx0.07537=0.238m totallength: 22x0.238=5.234m Totallengthofsecondarywinding:15.96m

Resistance: R s = 15.96 _____ 100 x0.3276=0.0523Ω 2.Magnetizinginductance

L 0 = N p 2 .

µ 0 . µ r . A k ________ l k = 5302 . 4xπx10

−7x1600xπx0.02252 __________________________ 2xπx0.06 = 2.38 H

1.6 Leakage inductance

Fromtheviewpointofvoltagelossleakageinductanceisundesirable.Transformersarethereforeconstructedtominimisetheleakagefluxes.Fig.16-9shows,forexample,howacoaxialimple-mentationofprimaryandsecondarycoilsminimisestheleakagereactancebyminimisingthedistancebetweenconsecutivecoils.Ontheotherhand,possibleshortcircuitcurrentsarelimitedbytheleakagereactance,whichcanformaprotectionforthetransformer.Inpracticedistribu-tiontransformersareconstructedwithsufficientleakagereactance,sothatshort-circuitcurrentislimited to 8 or 10 times the full load current. Inelectronicpowersuppliesringcoretransformersarefrequentlyused.Duetotheconstructionmethodtheyhaveaminimumleakagereactance.Electronictechnicianstalkabout“hard”trans-formerssincelargevariationsintheloadcoupledwithlowleakageinductancecanproducelargecurrentspikes.Thesevaryingloadconditionsoccurforexampleduringcommutationofone rectifierelementtoanotheronthesecondarysideofthree-phasetransformers.


Fig. 16-9:Leakagefluxesbyacoaxiallywoundtransformer

Todeterminetheleakageinductance,weconsidertheprimaryleakageflux(thesamereasoningisvalidforthesecondaryside).Wecannotmakeanaccuratecalculationsincethecross-sectionalareathroughwhichthefluxflowscannotbeaccuratelydetermined.Itispossibletomakeanapproximatecalculation.Ifthecross-sectionalareawhereintheleakagefluxflowsisA lp and the averagelengthofthefieldlineisl lp thensimilartoexpression(16-13),itmaybewrittenas: (16-14) Thefieldlinesoftheleakagefluxcompletetheircircuitthroughtheair(µ r = 1) instead of through the ferromagnetic core ( µ r ),whichexplainsthedifferencewithexpression(16-13).

Numeric example 16-3: Wereusethedataofnumericexample16-2ensurethepossiblepartsoftheleakagefluxesinfig.16-10aandfig.16-10b.

llp

lls

Alp

Als

ls

0

0

0

lp lp ls

primary

secondary

core

Ø

Ø ØØØ

Ø

Ø

lp

ls lp

ls

Als

Alp

23,5

23,5

22,5

22,5

CORE

ø 45primary

28,83

32,83

50 x

2,6

9

ø 4 mm

Ø

Ø

Ø Ø

L lp = sP = N P 2 . µ 0 . A lp

___ l lp

Fig. 16-10a:Primaryleakagefluxoftransformerinfig.16-8a Fig.16-10b:Secondaryleakagefluxfig.16-8b

ELECTRIC MACHINES 16.11

Primary leakage inductance

sp = N p 2 .

µ 0 . A lp ______ l lp

=5302 . 4.π.10−7 . π.(23.52−22.52 ) . 10 −6

_____________________________ 201 x 1.11 x 10 −3

= 228 µH

Secondary leakage inductance

ss = N s 2

. µ 0 . A ls _______ l ls =722 .

4.π.10−7.π.(32.832−28.832 ) . 10 −6 _______________________________

50x2.69x10−3 =37.42µH

If we realise that the magnetising inductance for this transformer is 2.38 H then we see that the leakage inductance is indeed minimal. Itisclearthatthepathoftheleakagefluxesdependsuponthepracticalimplementationofthetransformerwindings.Thepresentnumericexamplegivesusaroughideaoftherelativemagni-tude of the leakage inductance.

1.7 Energy losses 1.7.1. Copper lossesIntheprimaryandsecondarywindingsenergylossesoccur.If R P and R S aretherespectiveresis-tancesofthewindingsthenthelossesmaybewrittenas R P . I P 2 and R S . I S

2 .Thesumofbothisthetotalenergyloss.Thisisreferredtoasthecopperlossesofthetransformer.

1.7.2 Iron lossesInferromagneticmaterials,subjectedtoavaryingmagneticfield,hysteresislossesoccur:

W/kg (16-15)

whereby: kh = material constant of the ferromagnetic material used in relation to hysteresis losses. f = frequency(Hz) ^

B = amplitudeofthemagneticinduction(T=Wb/m²)

n = empiricalconstantforthemagneticmaterial(1<n<3).Sincethemagneticcircuitofatransformerisconstructedfrommetalplates,hysteresislosses occur.Tolimittheselosses,itisdesirablethatthematerialconstantbeassmallaspossible. Apossibilityinthiscaseisanironalloyusingsilicon(e.g.3%silicon).Ifthecorewasmadefromsolidiron,thenconsiderableeddycurrentswouldoccur.Thesecanbedramaticallylimitedbymakingthemagneticcircuitfromplateswhichareinsulatedfromeachotherandthesurfaceofwhich is in thedirectionoftheflux.Asresultofthisthepathoftheeddycurrentsislimited.The eddy current losses P w canbedeterminedwithaformulainthefollowingform: W/kg (16-16)

Here in : kw = materialconstantwithrespecttotheeddycurrentlosses δ = platethicknessinmm.By adding silicon the electrical resistance is also increased as a result of which the eddy current lossesarereduced.Accordingtothelastformula,itisadvantageoustohavetheplatesasthinaspossible.Typicalplatethicknessliesbetween0.3and1mmfor50Hzoperation.Theplatescanbe0.02mmforhighfrequencies.Forbandwoundcoresthicknessesof0.003to0.3mmarepossible.

P h=kh . f . ^

B n

P w =kw.δ2 . f 2 . ^

B 2



Wereconsiderastatorwith12slots(fig.16-49).ThecoilsN1 to N 6 are in series and form one phase.Assumethatonesuchcompletephasewindinghas1200windings.Wewishtodistributethesewindingsassinusoidalaspossible.Determinethenumberofwindingsforthesesixcoils, aswellasthemmfintheairgapifthecurrentis1A.

Fig.16-49: StatorironwoundwithonephasecoilS1 ( = U 1 U 2 )

Solution:Inastatorwithmslotsweareabletodeterminethenumberofwindingsinslotnbytakingtheidealsinusoidaldistributionoveranangle2.π ___ m around slot n: N k = ∫

2π(n−1)/m

2πn/m

n S1 .dα = N S

___ 2 ∫ 2π(n−1)/m

2πn/m

sin α.dα with n = 1 , 2 , ..., 6 (16-43) Withm=12andNS=1200wefind: N 3 = N 6 = 1200 _____ 2 ∫

0

π/6

sin α.dα = 80 ; N 2 = N 5 = 1200 _____ 2 ∫ π/6

π/3

sin α.dα = 220 ; N 1 = N 4 = 1200 _____ 2 ∫

π/3

π/2

sin α.dα = 300 Withi S1 =1AwefindasaresultofN1 and N 4ineachairgapanmmfof300Awacrossawidthof5.π/6.ThecoilsN2 and N 5produceafieldof220Awacrossπ/2radians.N6 and N 3arerespon-sibleforanextra800Awacrossπ/6radians.Thishasbeendrawninfig.16-50.

N6

N4

N5

N1

N2

N3α


Fig.16-50:Magnetomotiveforceintheairgapoftheconfigurationoffig.16-49

ApplicationoftheFourierseriesshowsthatonlyoddcosinetermscanexist. WecalculatetheseFouriercomponents: F S1,α = 4 __ π ∫

0

π/2

F . coskα.dα with k=1,3,5,... Thefundamentalharmonic(k=1)hasanamplitude: F 1 (S1) = 4 __ π . [ 300.sinα| 0

5.π/12

+220.sinα| 0 π/4

+80.sinα| 0 π/12

]=593.39Aw Theamplitudeofthethirdharmonic(k=3)isgivenby: F 3 (S1) = 4 ____ 3 . π . [ 300.sinα| o

15.π/12

+220.sinα| o

3.π/4

+80.sinα| o π/4

] = 0 Aw Fifth,seventhandninthharmonicsprovetobezero. F 11 (S1) = 4 _____ 11 . π . [ 300.sinα| 0

55.π/12

+220.sinα| 0

11.π/4

+80.sinα| 0

11.π/12

]=53.94Aw F 13 (S1) =35.84Aw The eleventh harmonic is 1 __ 11 ≈9%ofthefundamental. Accordingtofig.16-50themaximummmfis600Aw.Theamplitudeofthefundamental accordingtoourcalculationis593.39Aw.

Usuallywedefineanequivalentsinusoidalwindingas: N Se = 1200 . 593.39 ______ 600 =1184.78Aw

100200

400500

600

02 2 2

3 α

F


F µ,α =

N Se .

i µ _______ 2 . [ 3 __ 2 . cos( ω S . t + φ µ − α)]

(16-44) Expression16-44indicatesthatthemagnetomotiveforceissinusoidallydistributedinspaceandrotates with an angular velocity of ω S .TheamplitudeisdeterminedbyNSe .

i µ .

Themaximumofthemagneticfieldateachinstantoccursatanangleα,givenby: α=ωS.t+φµ (fig.16-52).

Fig.16-52:Spacevectorofthemmfintheairgapofanasynchronousmotor

The angle φµ isthedisplacementanglebetweentheappliedphasevoltagev S1 =

v S.cosωS .t and themagnetizingcurrenti µ .WiththechoiceofphaseorderU-V-Wweobtainacounterclockwiserotatingstatorfield.Iftheconnectiontoanytwowindingsarereversed(e.g.U-W-V)thenthestatorfieldrotatesclockwise. 3.1.4 Magnetising inductance Theinductionintheairgapiscalculatedwith:

→

B ag =

µ 0 . →

F µ,α _________ l ag

From (16-44) we can calculate the effective value of F µ,α so that we can write for B ag : (Tesla) (16-45) Weconsiderfig.16-51. Wearelookingfortheinstantatwhich i S1 =

i µ . cos ( ωS.t+φµ ), this occurs at ωS . t = −φµ .

At that instant the magnetic induction (vector) is directed according to the reference shaft of the stator (this is coil U 1 U 2 ). The value of the mmf follows from (16-44): F µ,α =

3 . N Se . ^

i µ __________ 4 . cos α so that: B ag,α =

µ 0 . F µ,α ________ l ag

= 3 . N Se . µ 0

__________ 4. l ag .

i µ .cosα= ^

B ag .cosα

Themagneticinductionissinusoidallydistributedinspacejustlikethewinding.

F µ,α = 3 . N Se .

i µ _________ 4 . cos( ωS.t+φµ−α )

→

B ag = 3 . N Se . µ 0

________ 4.lag .

→

I µ

(t = 0)

Fµ,α

Iµ

φµ

ωs

α

reference axis stator


WewillnowdeterminethefluxinthecoilS1andthatleadsustothemagnetizinginductance.AssumeinthefirstcaseawindingofcoilU1 - U 2atanangleα(andα+π) to the stator. For a stator with internal radius r and axial length lthefluxofonewindingis:

Φ winding = 2 . ^

B ag . l . r . sinα TofindthemaximumofallthelinkedfluxofawindingU1 - U 2 we have to integrate Φwinding across thewindingdistribution: ^

Φ S1 = ∫ −π/2

+π/2

n S1 . Φ winding .dα Applying(16-39):^

Φ S1 = ∫

−π/2

+π/2

N Se

___ 2 . sin α.( 2 . ^

B ag . l . r . sin α).dα (Wb) (16-46) Ifwereplace(16-45)inexpression(16-46),thenwefindtheeffectivevalueofthestatorflux,generated in one coil ( S 1 ): (16-47) Thelinkedflux

→

Φ S1 isproportionaltothemagnetizingcurrent.Fluxandcurrentareinphase.We

refer to →

Φ S1 ____

→

I µ = L 0 asthemagnetizinginductance:

(16-48) Fig.16-53showsthemagnetizinginductance.Theinducedemfinonephasewindingisgivenby: (16-49)

Fig.16-53:Magnetising curve of an asynchronous machine

^

ΦS1 = π __ 2 . N Se . ^

B ag . l . r

→

ΦS1 = 3.π ____ 8 . N Se 2 .

µ 0 . l . r ______ l ag

. →

I µ

L 0 = 3.π ____ 8 . N Se 2 .

µ 0 . l . r ______ l ag

→

E S1=j.ωS . →

ΦS1=j.ωS . L 0 . →

I µ

0

tg β = L0β

Øs1

Iµ



Atwopolethree-phasemachinehasarotorwithalengthof100mm.Theradiusoftherotoris50mmwhiletheairgaphasaneffectivelengthof0.6mm.Theeffectivenumberofsinusoidallydistributedwindingsis160.Themotorisconnectedinstartoa400V-50Hzpowergrid.

Question:

1. Thenominalinductionintheairgap2. The magnetising inductance and the magnetising current.

Solution:If we neglect the resistance R S of the stator coil, then: E S1 = V S1 = 400 ___

√__

3 = 230V

(16-49): Φ S1 = V S1

___ ωS = 230 _______ 2 . π.50 =0.732Wb

(16-46): ^

B l = 2 . √

__ 2 . Φ S1 _________ π. N Se . l . r = 2 . √

__ 2 .0.732 _________________ πx160x0.1x0.05 = 0.824Tesla

A value of 0.8 to 1 T is normal. The maximum inductance in the stator teeth is usually 1.6 to 1.8T. If the stator teeth are as wide as the slots, then the average inductance is 0.8 to 0.9T. Magnetising inductance (16-48): L 0 = 3.π ____ 8 . 160 2 . 4xπx 10 −7 x0.1x0.05 ____________________ 0.0006 =315mH

Magnetising current (16-49): I µ = V S1 _____ ωS . L 0 = 230 _________________ 2 . π.50.315. 10 −3 = 2.324A

Photo Siemens: Cross section of a 1LE1 asynchronous motor

1LE1 Standard induction motor 0.55to200kW 9.9to1546Nm 50/60Hz 2-4-6poles USA(also8poles) Generalpurposeinaluminiumand "Severe Duty" in cast iron for more demanding industry. Rotorincastcopperforhigherefficiency(IE1toIE3) Modulardesign:anumberofoptionsareeaslytoadd(seep.20.15).

18.8 CURRENT - , ANGULAR POSITION - , SPEED TRANSDUCERS

1.6.2 Closed loop sensor

Infacttheconstructionofaclosedloopsensorwithflux-gateiscomparablewiththemagneticcir-cuitinfig.18-4inwhichaHallsensorisnowreplacedbyaflux-gatesensor.Thisisshowninfig.18-9.ThecurrenttobemeasuredI 1 producesafluxΦ 1 and a current I 2 is sent through the second-arycoil(fluxΦ 2 ) in such a way that Φ 2=−Φ 1 . Here N 1 . I 1 = N 2 . I 2 . It follows that: I 1 =

N 2 __ N 1 . I 2 .

Wearecontinuallytryingtobringtheflux-gateoutofsaturationtothepointofsymmetryofthehysteresisloopsothattheresultingflux(Φ 1 + Φ 2 )iszero.Aswasthecasewiththezerofluxtransducer(fig.18-4),thevoltagedropofI 2 overaresistorproducestheoutputvoltageofthecurrentsensor.Sincethisisafloatingoutputitissometimesfedintoadifferentialamplifierwithopamps. ThesignalgeneratorwhichsuppliesthewindingN f iscomprisedofacomparatorcircuitwithhysteresis (Schmitt-trigger). The current change in I f producesnoiseintheprimaryofthecurrentsensorasaresultofthetransformeroperation.Afilterisrequiredtoremovethenoise.

Fig. 18-9: Adjustingtheclosedloop

ThefirmLEMsuppliesthemarketwiththefollowingflux-gatesensorsCAS/CASR,CFSRandCTSR(seephoto’sonp.20.13andp.20.82).Thephotoatthetopofp.18-9showsadidacticmodelofaflux-gatecurrentsensor.

I1

N1

N2 I2

flux-gate

Φ

Φ

ΦΦ1+

2

2 = 0

1

YES

NO

I2

NN

11 22 II = .

Nf

adjust

CURRENT - , ANGULAR POSITION - , SPEED TRANSDUCERS 18.11

1.6.3 Current measurement in a photovoltaic installation

Asdiscussedinchapter5and15wewantaPV-installationtooperateattheMPP(maximumpowerpoint).ThismeansforeveryPVtopologyweneedtomeasuretheDC-output(currentandvoltage)ofthesolarpanel.Inadditionacurrentmeasurementintheinputofthecontrolloopisnecessaryforprotectionagainstshortcircuitandovercurrent.IninstallationswithouttransformerthemaximumDC-currentthatthePVinstallationcansendintothepowergridisamaximumof10mA to 1A. Thevaluedependsonthestandardusedinthedifferentcountries.(IEC61727,…VDE0126-1).From all these requirements current sensors used need to have: an accuracy better than 1%, low offset, low drift in amplification, operate well with DC and low frequencies. Asensorwithclosedloopflux-gatetechnologymeetsthespecifiedrequirements.ThefirmLEMhasdevelopedtheirCTSR-seriesforthistypeofapplication. Inadditiontothementionedrequirements(accuracy,lowdrift,…)theCTSR-serieshasanumberofadditionalfunctionssuchas:referencepin(forselftesting),demagnetizingfunction(alsoviathereferencepin).TheCTSRcanbeusedforsingleandthree-phasevoltages.TheCTSRalsohasaversionwithfourindividualprimaryconductors.Threeconductorsareusedforthree-phasesystemsandthefourthconductorisusedfortestingtheoperationorasneutralconductorofthethree-phasenet. InadditionthemagneticcoreoftheCTSRhastwomagneticscreenstoprotecttheflux-gate fromexternalmagneticfields.Thereferencepinprovidesaccesstothe2.5Vreferencevoltage.This V ref. canbeusedasareferenceforanAD-converter.

Fig.18-10(photoLEM):Exampleofaleakagecurrent

Fig.18-10showsanexampleofhowaleakagecurrentcanflowinaPV-installationwithouttrans-former.Thecapacitancebetweensolarpanelandtheroofcanbethecauseoftheleakagecurrentwhichcanresultinthesolarpanelrisingtothenetpotential.Itisthereforenecessarythatany leakage currents are measured contact free and in a galvanically isolated manner.

CURRENT - , ANGULAR POSITION - , SPEED TRANSDUCERS 18.17

ThephotobelowfromthefirmHeidenhainshowsonceagainthetransparentprincipleofan opticalencoder.

Photo Heidenhain: Principleconfigurationopticalencoder

ThelightbeamemittedbyL(e.g.aLED)isconcentratedbyalensKandthereaftersplitbyade-tectionplateintofivefields.IfaglassrulerMwithanincrementalscaleC(e.g.20µm)comprisedofchromestrips(C/2wide)moveswithrespecttothedetectionplateAthelightwhichshinesthroughissinusoidallymodulated(duetothespecialconstructionofthedetectionplate).Fourofthefivefieldsproduceviaphoto-transistorssinusoidalsignalsdisplacedby90°withrespecttoeachother.Sincethefoursinusoidsarenotsymmetricalwithrespecttothezerolinetworesult-ingsymmetricalsinusoidsS1andS2areproducedviaapushpullcircuitwhichare90°outofphasewitheachother.ThisistheSIN/COSoutput.Atypicalamplitudeis1Vpeaktopeak.Thesinusoidscanalsobeconvertedinternallytoblockwaves(asinfig.18-15)whichthenappearontheoutputatTTLorHTLlevel.ThefifthfieldservesasareferenceR(oneperrevolutionoftheshaft).Encodersmaybesingleturnandmultiturntypes.Singleturnsystemsprovidetheactualpositionwithinonerevolution.Multi-turnpulsesensorsoperatewithmultiplerevolutions.

19.12 SPEED- AND (OR) TORQUE-CONTROL OF A DC-MOTOR

3. CONTROLLED SINGLE QUADRANT DRIVE

3.1 Operation

Agoodspeedcontrollermustmeettworequirements: 1. nshouldbeindependentoftheload2. I a shouldbelimitedtoanadjustablemaximum.Theblockdiagramofthecontrolcircuitforthespeedcontrolwitharmaturecontrolisshowninfig.19-11.

Fig. 19-11: Block diagram of a controlled single quadrant drive

Withapotmeterthedesiredspeedisset.Atachometerproducesavoltageproportionalwiththeactualspeedofthemotor.Set-pointandprocessvalueintheformofvoltagesarecomparedwitheachotherinacomparator.Attheoutputofthiscomparatorthedifferenceεappears.If ε≠ 0 thespeedofthemotorisnotequaltotheset-point,inotherwordsthereisanerror.The fault ε is referred to as the error signal. The error εistransferredtothefirstcontroller.Ifthiscontrollerisaproportionalamplifier,thentheoutputsignalwillbeproportionaltothedifferencebetweenset-pointandtheactualspeedofthemotor.Inotherwordstheoutputofthespeed controllerindicatesifthemotorshouldaccelerateordecelerate.Forapositiveε the motor should accelerate and for a negative ε the motor should decelerate. An acceleration or deceleration correspondswithanincreasedordecreasedmotortorque.Thisisachievedbyanincreasedordecreased current in the motor.

Ia referenceIa actual

nactual

speed reference

(set value)speed regulator

+

-

+

-

current regulator

control unit

PI (D) PI (D)

tachon M+

-

Im

controlledrectifier

GRID3

Φ

ε

currentlimiting circuit

SPEED- AND (OR) TORQUE-CONTROL OF A DC-MOTOR 19.13Theoutputofthespeedregulatoristhereforeameasureofthedesiredarmaturecurrentofthemo-tor.Wecomparethisdesiredarmaturecurrentwiththeactualarmaturecurrentinacurrentcom-parator. Theoutputofthiscomparatorisnowtheinputforasecondcontrollertoachieveanasexactaspossiblecurrentcontrol.Theoutputofthissecondcontrollersendsapulse(controlsignal),whichforitspartcontrolsthethyristorbridge.Theoutputvoltageofthebridgeiscontrolledasafunctionoftheinputvoltageofthepulsegenerator. AsweshallseelaterthecontrollersarenotjustproportionalcontrollersbutnormallyPI-control-lers.InexceptionalcasesPID-controllersmaybeencountered.

3.2 Advantages of internal current control

The current control circuitisespeciallyactivewithchanges in the motor load torque. The shortcontroltime(7or10ms)inthecurrentcircuitdonotallowtheloadvariationstoaffectthespeedcontrolsincethearmaturecurrentisquicklyadjustedtotherequiredvaluebythenewloadtorque. Withthisinternalcurrentcontrolthemaximummotorcurrentiseasilycontrolled.Ifwelimittheinputcurrentcontrollertoforexample120%ofthenominalvalueofthearmaturecurrent,thenthyristorbridgeandmotorareprotectedagainstoverloadandagainstahighstartcurrentsurge of the motor. Current limitationisespeciallyimportantwhenquick speed changes are required from themachine.Ifthemotorloadhasalargemomentofinertiathenlarge(required)speedchangescannotbefollowed.Ifthespeedset-pointissuddenlyraised,thenthecontrolleroutputwillwanta much higher armature current than the nominal armature current. The current limitation sees to itthatthenominalvalueisnotexceeded.Withconstantfluxandnominalarmaturecurrentduringacceleration the motor will maintain an almost constant torque M ≈ M em=k2 . I a . Φ.If the counter torque of the mechanical load is M t , then the difference M v = M − M t will cause the machine to accelerate according to M v = J m.(dω/dt). Here dω/dt = acceleration; J m = moment of inertia. If M v is constant then the machine will acceleratelinearlytothedesiredspeed. Ifthedesiredspeedislessthantheactualspeedthenwithsinglequadrantoperationthemotor willruntillthedesiredspeedisreachedandthiswithzeroarmaturecurrent.

Remark

Whyisthearmaturecurrentlimitsetto120%ofthenominalarmaturecurrent? Verysimplysinceafullyloadedmotor(I a = I a nominal ) could no longer accelerate if we should limitthecurrentto100%.

!


Numeric example 19-2: Given: . DC-motor:220V/10A/1.76kW/1000rpm/R i=1Ω. tachogenerator:constantis10V/1000rpm. B 6-controlledrectifierwithimpulsemodule:V control = 0 to 10V gives V diα = V a = 0 to 220V . controlcharacteristic:fig.19-12 Toenablemaximumuseofthisnumericexampleanumberofvaluesareindicatedinthis characteristic.

Fig. 19-12: Control characteristic of a B 6 - Bridge

Required: 1. Withn=250rpmandwith50%motorloaddeterminetheinputandoutputvaluesofthe different“blocks“infig.19-11.Theinternalcurrentcontrolandcurrentlimitationareinactive.

2. Thesamequestionbutnowwithn=500rpm,50%motorloadandactivecurrentcontrolasin fig.19-11.

Vdi = Va

220

162,5

110

57,5

180150120906030 42 74°50'0

01,6833,335,310

Vcontrol (V)

α

α

SPEED- AND (OR) TORQUE-CONTROL OF A DC-MOTOR 19.15

Solution:Tomakethesolutionasshortaspossiblewewillignorethetransientbehaviour,andonlyconcen-trateonthefinalsteadystatevalue.Thetransientbehaviourofasinglequadrantdriveisstudiedunderoptimisingthecontrollers(p.19.34). Fromthenameplateofthemotorweobtainthefollowing:

at full load, V a = 220V and n =1000rpm:

I a = 10 = V a − E

______ R i = 220−E _______ 1 or: E = 210Vat1000rpm

The normalised emf e N is: e N =0.21V/rpm The nominal torque is: M = 9550 . P _______ n = 9550x1.76 _________ 1000 = 16.8 Nm 1. n=250rpm-50%motorload . M t=50%.16.8=8.4Nm . I a=50%.10=5A . n=250rpm→→E=250.e N=250.0.21=52.5V . V a = E + I a . R i=52.5+5x1=57.5V . Fig. 19-12: V diα=57.5=220.cosα→→α=74°50’→→V control = 1.683V . Thesinglequadrantcontrollerwithoutinternalcurrentloopisdrawninfig.19-13.

Fig. 19-13: SpeedcontrolledDC-motorat250rpm

+ 10 V

(n*)

+

-

pulse unit

T

M

Im

GRID3

2,5V74°50'

2,5V

1,683V0V

(n)

Vcontrol

B6 - controlled rectifier

E = 52,5V

Va = 57,5V

Ia

F1

F2

5A

5A5V

2,5V

Ri1W

VI

8,4 Nm

8,4 Nm

250 tr/min

MMt

n

PI

0 - 10A0 - 10V

Φ

α

speedregulator

ε


2. n=500rpm-50%motorloadTosetthedesiredspeedn * = 500rpmweadjustthewiperofthepotmeterfrom2.5Vto5V.Inthefirstinstantthemotorremainsrotatingataspeedof250rpmandthetachometerproduces2.5V.Theerrorvoltage εbecomes5−2.5=2.5V.The n-controllerimmediatelymakesajumpatitsoutput(proportionalpartofthePIcontroller)andatthesametimethecontrollerbeginsto integrate(I-part)sothattheoutputvoltageofthe n-controllerreaches10Vinanumberofms.WithV control =10Vthenα=0°andV diα = V a = 220 V. Due to the inertia of the motor with its driveinthefirstinstantEremainspracticallyunchangedatE=52.5Vsothat: I a =

V a −E ______ R i

= 220−52.5 ________ 1 =167.5A! Toomuchforour10AmotorandalsofortheassociatedSCRbridge.Wearecompelledtolimitthearmaturecurrentandmaintainitwithinsafeoperatingvalues.Theonlywaytodothisistomeasure the current I a withacurrentsensorandtocontrolthiscurrenttobelowanadjustable(acceptable)maximum.Forthisreasontheinternalcurrentloopoffig.19-11isnowactive.Thisisshowninfig.19-14withthecurrentlimitedto9A,andspeedset-pointn*=500rpmwiththemotor50%loaded.ThecurrentsensorisfollowedbyanI/Vconverterthatforexampleconverts0-10 A to 0-10 V.

Fig. 19-14: Single quadrant DC-motor at n=500rpm,M t = 0.5xM=8.4 Nm and I amax = 9A

E = 500x e N =500x0.21=105V; V a = E + I a . R i = 105+5x1=110V V diα = 110 = 220 .cosα→→α= 60°→→ V control = 3.33V

+ 10 V

(n*)

speed regulator

+

-

+

-

1

currentregulator

control unit

T

M

Im

GRID3

5V 5V 5V 260°

5V 5V

0V 3,33V

(n)

Vcontrol

B6 - controlled rectifier

E = 105V

Va = 110V

Ia

F1

F2

5A

5A5V

5V

Ri1W

VI

Mt

8,4 Nm8,4 Nm

500 tr/minM

n

limited at 9V(Ia = 9A!)

α

Φ

ε ε

20.18 SPEED-ANDTORQUE-CONTROL:3-PHASEASYNCHRONOUSMOTOR6.3 Control range

Wedefinethecontrolrange= (20-11)

. Electromechanical torque

IntheassumptionthatthenominalmotorcurrentI n isflowing,itfollowsfrom(20-6)thatthetorque M em proportionallychangeswithΦ .

In thefieldweakeningareaΦ=k1 . V S / f S=k1 . constant _______ f S , so that the torque: M em ~ 1/ f S ~ 1/n. Thisisindicatedbythehyperboleinfig.20-17.

. Maximum torque

In thefieldweakeningareaV S is constant. From (20-8) it follows that in this area the maximum torque quadratically decreases with increasing frequency f S .AboveF 7 the motor cannot run with constantpower,seefig.20-17.Therangewhereinthemotorpowerremainsconstant(uptof 7 ) dependsuponM n / M po intheareawithnominalfluxΦ n . Often for a motor M n /M po = 0.5, so that : f 7 ≈ 2 . f 4 . From f 7 the motor can only deliver a certain percentageofitsmaximumtorquebecauseofthelosses.Inthisareaof“higherspeed”M em is inverselyproportionaltof S 2 .

Fig.20-17:M-ncurvesofathree-phaseasynchronousmotor

smallestspeedbynominalflux

_____________________________ largestspeed(infieldweakeningarea)

0 f1 f2 f3 f4 f5 f6 f7

M

Mpo

(In = constant)

Mn

Mmax.

nominal flux field weakening

1fs

fs

2

2

Mem1fs

Mem1

fs (Hz)

n (rpm)constant torque constant power higherspeed

20.46 SPEED-ANDTORQUE-CONTROL:3-PHASEASYNCHRONOUSMOTOR

7.5 Coordinate transformation

7.5.1 Transformation formulas

ThefluxcreatingandtorquecreatingcomponentsofthefictivecoilsDandQareinrealityob-tainedfromtwovariable(e.g.sinusoidal)currentsi α and i β intherealcoilsofourtwo-phasemotor.Thesecoilsarefixedinthestator.Wedrawtheα-β coordinate system that is related to the stator insuchawaythatthecoilαproducesafluxaccordingtotheα-axisandcoilβ according to the β-axis.(fig.20-42).

Fig. 20-42:Equivalenttwo-phasestatorcoordinatesystem

Weknowfromfig.20-31thattherotatingfieldinthestatorhasanangularvelocityωS . As with thethree-phasemotortherotorispulledalongwithaspeedn < n S . TherotorhasarotorfluxΦ R which also rotates at ωS rad/s,similartothethree-phasemotor.IfwenowallowthefictivecoilsDandQ(andthereforethed-qcoordinatesystem)torotatewiththerotorfluxΦ R asinfig.20-41,thenwecandrawfig.20-43.

β

α

α

β β

α

ΦR

ωS

λ(t)

SPEED-ANDTORQUE-CONTROL:3-PHASEASYNCHRONOUSMOTOR20.47

Fig. 20-43: d-qcoordinatesystem,linkedtotherotorflux

The current i α producesthecomponentsin: . d-direction: i dα = i α .cosλ . q-direction: i qα = − i α .sinλ The current i β producesthecomponentsin: . d-direction: i dβ = i β.sinλ . q-direction: i qβ = i β .cosλ The currents i α and i β fromthestatorcoordinatesystem(α,β)providecurrentsi d and i q in a coordinatesystemthatrotateswiththerotorfluxΦ R : . i d = i dα + i dβ . i q = i qα + i qβ or: (20-20) (20-21) Conclusion:Theexpressions(20-13)and(20-14)giveforathree-phaseinductionmotorthecurrents i α and i β astransformedcomponentsofthethree-phaserotorcurrent(i S1 , i S2 , i S3 ). Thisequivalentsystemα-βisfixedtothestator.Wecallittheα-β stator coordinate system. The coordinatesystemd-qwithitscomponentsi d and i q arecalledthefieldcoordinatesystemsincetheyarefixedtothe(rotor)flux. Theexpressions(20-20)and(20-21)describethetransformationfromstatorcoordinatesystemto(rotor)fieldcoordinatesystem.ThisisknownastheParktransformation.

i d = i α.cosλ+iβ.sinλ i q = i β.cosλ− i α.sinλ

β

α

q

iqβiβ

iα

idβ

ia

iβ

idα

iqα

d

0

ΦR

λ

ωs

α

β

20.56 SPEED-ANDTORQUE-CONTROL:3-PHASEASYNCHRONOUSMOTOR

Fig.20-54showsfinallyablockdiagramwithassociatedPI-controllersforfieldorientatedspeedcontrolofaninductionmotorwithimposedstatorcurrents.

Fig.20-54:Speedcontrolofinductionmotorinfieldcoordinates(indirectmethod)

7.6.5 Applications with closed loop control

Frequency controllers with vector control are used in: cranes,elevators,liftingequipment,extruders,papermachines,rollingmillsintheproductionofsteel,CNC-machines,cablelayers(shipbased),drumsetc.

n - c

ontro

ller

field

wea

keni

ng

i - c

ontro

ller

limite

rn*

+ -

+ -

+ -

+ -

Mn

i q*

i α* i β*

Φ -

conr

olle

ri d*

i s3* i s2* i s1*

ΦR

L 0k 2

ΦR

M

*

M*

ejλ2/

3

cosλ

sinλ i q i µ

iµ

calc

ulat

ion

mod

ule

ωi s

i s1i s2

CR

-PW

Min

verte

r

M T3

GR

ID

20.66 SPEED-ANDTORQUE-CONTROL:3-PHASEASYNCHRONOUSMOTOR 7.8.3 Moment of work torque

Asshowninfig.20-65theworktorqueinfig.20-69isproportionaltotheareaofthetriangle

formedby→

i S , L R

__ L 0 . →

i R . e jδ and →

i µ .Thisareacanalsobedescribedinfig.20-69as:

1/2 . i q . i µ so that: M em=k2 . i µ . i q (20-39)

With Φr = L 0 . i µ is M em=k3 . i q.ΦR (20-40)

Another method is to assume that M em isproportionalwiththeproductoftherotorfluxwiththecomponentof

→

i S thatisperpendiculartothisrotorflux.Infig.20-69thisleadsto:

M em=k3 . i S.sinε . Φ R , so that: (20-40)

Calculation leads to : k3 = 3.p

____ 2 . L 0

___ L R with p=numberofpolepairspermachine.

7.8.4 Mathematical model in field coordinates With

→

i µ = i µ . e jλ (20-37) becomes:τ R . d __ dt (iµ . e jλ)+(1−jω.τ R).iµ . e jλ =

→

i S

τ R . e jλ . d→

i µ ____ dt +τR . i µ . j . dλ __ dt . e jλ+(1−jω.τ R).iµ . e jλ =

→

i S

τ R . d→

i µ ____ dt + τ R . i µ.j.ωS+(1−jω.τ R).iµ =

→

i S . e −jλ

From (20-24):

→

i S(t)=iS . e jθ it follows:

→

i S . e −jλ = i S . e j(θ−λ) = i S . e jε (20-41)

Thismaybeexpressedinthecomplexd-qcoordinatessystemas:

→

i S . e −jλ = i S . e jε = i d + j. i q

so that: τ R .

d→

i µ ____ dt + τ R . i µ.j.ωS+(1−jω.τ R).iµ = i d + j . i q

Takingtherealandimaginarypartsseparately:

. τ R . d→

i µ ____ dt + i µ = i d (20-42)

. (ωS −ω).iµ . τ R = i q (20-43) Multiplying(20-42)with L 0 : τ R .

d→

ΦR ____ dt +ΦR = L 0 . i d (20-44)

M em=k3 . i q.ΦR

SPEED-ANDTORQUE-CONTROL:3-PHASEASYNCHRONOUSMOTOR20.67

Rewritingof(20-44)usingLaplacenotations: τ R.s.ΦR = L 0 . I d

ΦR ___ I d = L 0 . 1 ________ 1+s.τ R

ThisisthetransferfunctionofafirstordersystemwithtimeconstantτR =

L R __ R R (fig.20-70)

ThefluxΦ R isdeterminedbyi d !

Fig.20-70:Transferfunctionofrotorflux

From(20-43)theslipfollows:g=ωS−ω ______ ωS =

i q ________ i µ . τ R.ωS

Withi µ =

ΦR ___ L 0 then g=

L 0 _____ ΦR . τ R .

i q ___ ωS

From (20.43) we can derive: (20-45) Themathematicalmodelofathree-phasesquirrelcageinductionmotorinfieldcoordinates (referencecoordinatesasfieldcoordinatesrotateswithrotorflux)isgivenby: (20-40) (20-44) (20-45) Ifwenowdeterminethestatorcurrentcomponentsateveryinstantinacoordinatessystemthatrotateswiththerotorfluxthenthetorqueoftheinductionmotormaybecontrolledexactlyas withaDC-motor.InfacttheexpressionM em=k3 . i q . Φ R iscomparablewiththeformulaforthemoment of work torque of a DC-machine: M em=k2 . I a . Φ (19-2). WithaninductionmotorΦ R canbekeptconstantorcontrolledviai d as (20-44) suggests. Thespeednoftheinductionmotorcanbecontrolledwith(20-45)byinfluencingωS and i q .Infactwithfieldorientationviamathematicaltransformationthewayisopentocontrolthetorqueof an induction motor in the same way as with a DC-motor.

ωS −ω= L 0 . i q

______ ΦR . τ R

M=k3 . i q.ΦR τ R .

dΦR ____ dt +ΦR = L 0 . i d

ωS −ω=

L 0 . i q ______ ΦR . τ R

Id ΦRL0 τR

21.12 CONTROLAPPLIANCE,SR,3-PHSYNCHR.,INDUCTIONSERVOMOTOR

Fig.21-17showsatypicalthree-phaseSR-converter.SincethestatorcoilsofanSRMaretotallyinsulatedfromeachotherafaultinonestatorcoilwillhavenoeffectonother“legs”ofthecon-verter.Themotorcanremainoperatingevenwithafaultycoil,althoughwithcopperlosses.Notealsothatthecurrentthroughaphasewindingalwaysflowsinthesamedirection.Opposingpolesareprovidedwithwindingswhichareinparallelorinseriesandeachformsastatorcoil. Thepolarityofthemotortorqueisindependentofthecurrentdirectionthroughthestatorcoil. Itisthepositionoftherotorteethwithrespecttothestatorthatdeterminesthedirectionof rotation. Infactthediodesoperateasfreewheeldiodes.Thetransistorsarealwaysinserieswithacoilinwhichthecurrentalwaysflowsinthesamedirection.Theproblemofprotectionisalotsimplercomparedtothethree-phaseinverterofaninductionmotor.TheseSRM-convertersaremorerobustthanclassicthree-phaseinverters.

Fig.21-17:Typicalthree-phaseinverterusedtosupplyanSRM

2.5 Speed-, position- and torque control

ThetopologyofaclosedcontrolloopiscomparabletothatofaDC-motor.WithanSRM inadditionfeedbackoftherotorpositionisrequiredforcommutationoftheinverter. Fig.21-18showstheclosedloopforapositioncontroller.

Fig. 21-18: Position control using an SR-motor

D2

D1

T1

T2

S1

D4

D3

T3

T4

S2

D6

D5

T5

T6

S3

+Vcc

positioncontroller

velocitycontroller

currentcontroller inverter

PG

SRMθ* n* i* i

i

+

-

θ θ θ

θ / n

rotor position

ELECTRICALPOSITIONINGSYSTEMS22.3

Remarks

Thename“servomechanism”isoftenreservedintheliteratureforapositioningsystem.

Forhigh-dynamicpositioningandtorquecontrolelectricalservo-systemsusedonlyDC-servomo-torsorpermanentmagnetsynchronousmotors. Duetothetechnologicaldevelopmentsinthepreviousyearsthefluxvectorcontrolofanasyn-chronousmotorhasdramaticallyimproved.Theresultisthatthisvectorcontrolisnowcapableinmanyapplicationsofcompetingwithtraditionalservo-systems.Theborderbetweendrivesystemandservo-systemisbecomingmorevague.Certaincompany’ssuchasYaskawausea“radar”cardonwhichthedifferencesinperformancebetweenservo-systemsandfluxvectorcontrol(VC)areindicated.Thisisshowninfig.22-1.

Fig. 22-1: Points of difference servo-system with VC-system Source : YaskawaandControlEngineering

10

110 10 0.1

110

1001,000

1

0.1

0.01

1001000

10

101

0.1

0.01

100

1,000

10,000

1

0.1

0.01

speed regulation range (X : 1)

acceleration time(s)

cycles/min

position accuracy (°)

frequency respons (Hz)

speed regulation (%)

SERVO

VECTOR

CONTROL

!

22.16 ELECTRICALPOSITIONINGSYSTEMS

Fig.22-12showstheprincipleconfigurationofaDC-servo-system.

Fig. 22-12: DC-servo-systemwiththethreecontrolloops

3.4.2 Controlled speed

Once the value ωSC isreachedthespeedcontrollercomesintoaction(SC=SpeedControl).Dependingonthesettingfor ωSC *

, the ωofthemotorwillbecontrolledsothatω = ωSC * .

3.4.3 Delay

Inordertostopatthedesiredpositionatime t 3 hastobedeterminedatwhichthemotorshouldstarttodecelerate.Inrealitywedeterminethenumber(k)ofpulsesthattheencoderstillneedstogiveuntilthemotorshaftisattheset-pointlocation. Expression(22-11)allowsforcalculationofthenegativevalue I a required to decelerate with a

value γ=−dω ___ dt Fig. 22-13a shows that ωSC=(t4−t3).γ.Fromfig.22-13bitfollowsthattheshaftofthemotoris

rotatedbyanangleθ = ( t 4−t3 ) . ωSC

___ 2 between t 3 and t 4 . Elimination of ( t 4 − t 3 )inbothofthepreviousformulasgives θ=(t4 − t 3).

ωSC ___ 2 =

ωSC ___ γ . ωSC

___ 2 = ωSC 2

____ 2.γ

Assumethatthespeediscontrolledat3000rpm,then: ωSC = 2 . π . n ______ 60 =314.1593rad/sIfforexample(accordingtothemicro-controllercountdown)wearestillk= 3800 encoderpulsesremovedfromtheset-pointpositionandtheencodergenerates2000pulses/rev,then: θ=11.938 rad.

µprocessor-board

µprocessor

D / A+

-

waveshaper

+

-

ωsc

ω

* Ia

Ia

*

MOTOR

servoamplifier

TAC

HO

ENC

OD

ER


The required delay is then: γ=

ωSC 2 ____ 2.θ = 4133.6925rad/s2

WithJ R + J eq = 3.9 x 10 −5 kg m 2 and K M = 5.4x10−2Nm/Abecomes: I a =

γ.(JR + J eq) __________ K M =−2.985A

Fig. 22-13: Decelerationoftheservomotortoastop

3.4.4 Stopping

Itisclearthatthecalculationaboveisdependentonthelinearoperationduringdeceleration.Thisisdifficulttoknowforamongstotherreasonsbecauseofthedigitalcontrolofthespeedwhichisalwaysin“steps”.Normallyapositioningsystemiscontrolledinsuchawaythattherotorwithonlyafewoscillationscanbebroughttoahalt.Thetimet S of these oscillations is known as the settlingtime.Mostlythespeedcontrollerisswitchedoffintheregionclosetotheangularpositionset-pointandonlythepositioncontrollerisoperatingsothatwhenthestoppositionisachievedagood holding torque is maintained.

realSTOP

ω

ωSCωSC

-

ω = ωSC

γ = dωdt

0 t3

t = t3

t = t4STOP

t4 ts

t

encoder pulsesk pulses

(a) (b)

θ

(motor shaft)


B. M-n curves

1. Static torque

ThetorquethatisrequiredforastationarySM(zerostepsandnominalcurrentthroughonephase)tomoveonestepiscalledtheholdingtorque.

Fig.22-23: Statictorqueofastepper-motor

2. Dynamic torque

Fig. 22-24: Path of the dynamic torque of a SM

Withahigherstepfrequencytheaveragestatorcurrent(andflux)willdecrease.Thetorquede-creases as a result.

steps

0 1

M

Mh

2

- Mh

∆α

holding torque

pull-out torque

pull-in torque

30

25

20

15

10

5

100

pull - outrate

pull - inrate

maximumslew - rate

200 300 400 500

START-STOP RANGE

pulse rate (Hz)

steps/s

slew range

running

start without error

M (mNm)


F. Micro-stepping

A typicalsmallstepangleofahybridstepperis1.8°(sometimes0.9°).Thiscorrespondsto200(or400)stepsperrevolution.Toobtainahigherprecisionareductionmechanismorgearboxcanbeused.Anelectronicsolutionforobtaininghigherprecisionconsistsof“micro-stepping”. AfullstepoftheSMisdividedintosmallersteps. Assume an SM with ppolepairsontherotor.Forthefullstepdrivewevarythephasecurrentbetween+I f andzeroorbetweenzeroand−I f .Ifwenowchangethephasecurrentinksmall stepsthenitispossibletoorientatethefluxinm=k.p different directions resulting in the rotor assumingmpossiblepositions.Thiscanbeappliedtoathree-phaseSM.Fig.22-38ashowsthebasiccrosssectionofsuchastepper.

Fig. 22-38: Three-phaseSMwiththestatorfluxvector

Infig.22-38bthethreestatorfluxcomponentsareshown.Thesecomponentsarecalledpositive ifforexampletheyinduceamagneticsouthpoleontherotorsideofastatortooth. WithmaximumpositiveandI f2 = I f3 = − 0.5xIf1 weobtainaresultingstatorfluxasshowninfig.22-38c. Apracticalapplicationistochangethestatorcurrentinsmallstepsandthereforethestatorflux,andsuperimposethisonasinusoidalsignalasshowninfig.22-39a. Witha(p=)50-polemachineandk=40wefindthatm=k.p=2000stepsperrevolution.Thisisshowninfig.22-39.Fig.22-39bshowsthefirst11statesofthestatorflux. Themarking1,2,...ofthestatorfluxcorrespondtotheindexes1,2...ofthestatorcurrentchangesinfig.22-39a.Position11infig.22-39bcorrespondswithfig.22-38c.Thismechanismisknownasmicro-stepping.Inthiswaypractically10,000to25,000stepsperrevolutionarepossible.

N

Z

1

2

3

32

1

(a) (b) (c)

1

23

ΦS


Fig. 22-39a: Statorcurrentsduringmicro-stepping

Fig.22-39b:Statorfluxduringmicro-stepping

If1

θ

If2

θ

0

If3

θ

0

1 2 3 4 5 6 7 8 9 10

11 109

87

6

5

4

3

2

1

ΦS


9.INTEGRATEDSYSTEM(SIMOTIONFROMSIEMENS)

AsanexampleofanintegratedpositioningsystemweconsiderSimotion(Siemensmotioncontrolsystem).Integratedsystemshavemultipletools.Theuseofstandardmoduleseachwithitsownintelligenceresultsinnotonlytimesavingduringdevelopmentandimplementationofaproject butinadditionthereisamultitudeofapplicationpossibilities.ThecommunicationbetweenthedifferentmodulesoccursviaProfibusandProfinet. Intheupperpartoffig.22-46theSiemensphilosophybehindSimotionisshownandinthelowerfigurethepossibilitiesareshown(hardwareplatforms). Therearethreeplatforms(SimotionC,DandP).

Simotion Disaverycompactsolution.Hereclosedloopcontrol,PLC-functionsandmotioncontrolareintegratedintothemodule.Thisiscomparabletoacertainextenttothe“stand-alone”solutionoffig.22-45. ConnectionoftheSimotionD(fig.22-48)occursviatheET200.TheET200(ET=externalter-minal)isadecentralized(periphery)moduleforcommunication(ProfibusorProfinet)viaalocalnetworkcabletoavoidlongsignalcables.ConnectionoftheSimotionDwithProfibus,Profinet orEthernetresultsinvisualisationandoperatorcontrol. HMI(fig.22-48)means“humanmachineinterface”.

Fig.22-47: Photo of the Simotion D module, clicked ontoaSinamicsS120powermodule

Fig. 22-46: SiemensSimotionphilosophyandhardwareplatforms

PLC functionality

Motion control(positioning, ...)

Technological functions (cams, temperature control,..)

The system philosophy of

SIMOTIONMotion control, PLC andtechnological functions are merged

Engineering system. programming. parameterizing. graphic or textbased programming. testing and diagnostics

Runtime software. PLC. Positioning. Synchronous operation. Camming. Interpolation

Hardware platforms. Controller-based. PC-based. Drive-based


Fig. 22-48: SimotionDtopology(drivebased)

Machineswithmultipleshaftsareachallengeforeverypositioncontrolsystem.Thereasonisthateveryshaftincreasestheloadonthesystemandonthebus.WithSimotionthemachinepartsaresplitintodifferentmodules,eachcontrolledbyit’sownSimotionsystem.Theindividualsystemsare then connected via PROFIBUS or PROFINET. The Simotion D version is highly suited for thistypeofapplications,ontheonehandbecauseofit’scompactdesignandontheotherhandbecauseofit’sfastcommunicationwithinamultipleshaftsystem.

Simotion Cisamodularsolutionandhasthelargestfieldofapplication.Therearefourinter-facesonboardforanalogueandstepperdriversandalsomultipledigitalinputandoutputs. Duetothesegregatedarchitecture,drivesandI-OportscanbeconnectedviaPROFIBUSor PROFINET.

Simotion Pisan(open)systemforapplicationswhereanopenPC-environmentisrequired.TobenefitfullyfromtheoptionsofSimotionthereisanengineeringsystem(SCOUT)thatallowsmotioncontrol,PLC,technicalfunctionsanddrivestodealtwithinonetool.Configuring,pro-gramming,testingandcommissioningcanallbegraphicallydealtwithfromone“work”space.Afterthedrivehasbeencommissionedalldatacanberequestedviathecentraladministration. InadditionviaLAD(ladderdiagram),FBD(functionblockdiagram)andtextbasedST(struc-turedtext)thepositioningsequencescanbeprogrammedviaMCC(motioncontrolchart).MoreextensivecontroltaskscanbeimplementedbyusingDCC(drivecontrolchart).

SIMATIC HMI

PROFINETIndustrial Ethernet

ET200 ET200

SIMOTION DSIMOTION D

E-MOBILITY23.1

23 e - MOBILITY CONTENTS

1.Renewableenergy 2. Electric traction 3.Electricautomobile 4.Electricboats 5.Electricbicycles

Atthestartofthetwentiethcenturytherewerealreadyelectriccarsinexistence.Duetothecheapavailabilityoffossilfuelstheelectriccarwasunabletocompetewiththebenzineengine.Inad-ditiontheactionradiusofelectriccarswassmallduetotheproblemoflimitedenergythatcouldbestoredinbatteries.Aproblemthatwestillhavetoday.Goodbatterytechnologyisprobablythemostimportantconditionforthemassbreakthroughofelectricauto-mobiles. Inadditiontoroadtransporttherehasalwaysbeenelectrictraction(train/tram/metro).Inaddi-tionelectricbicycles,scootersandsailingvesselsarecandidatesforelectricdrives.Thecompletepacketofelectricaltransportisreferredtoaselectro-mobilityorinshorte-mobility. E-mobilitydealswithspeedsfrom25km/h(bicycle)tomorethan200km/h(train)andthiswithpowersof250W(bicycle)tomorethan9MW(train).Electriccarsaresomewhereinbetweenwithpowersofbetween40and200kWwithmaximumspeedsbetween80and200km/h. Inthischapterwelookatanumberofexamplesoftraction,carsandboatsasapplicationsofelec-tronic motor control.

23.2 E-MOBILITY

1.RENEWABLEENERGY

Sincefossilfuels(oil,gas,coal)arequicklybecomingscarce,inthefuturee-mobilitywillbe suppliedbyrenewableenergy.Theenergyonearthcomesforalargepartfromthesun.There-maindercomesfromtheheatoftheearth,tidepowerandnuclearenergy.Thesunproduces3.89.10 26watt.Theearthisatadistanceof150millionkmfromthesunandabout174petawattreachestheearth.Rightnowhumanityconsumesabout15terawatt,thisislessthanonetenthousandthof174petawatt!Thefollowingtableisamemoryaidinconnectionwithmultiplesandsubdivisionsofunits.

Table 23.1

Canwesatisfyourenergyneedswithsolarpanelsandwindmills?Asmallcalculationshows: . 174petawattofsolarenergyreachestheearth . 25petawatthitsthegroundasradiationenergy.Ifwecouldconvertonethousandofonepercenttoelectricalenergythenwewouldhave25TW,morethanthecurrentconsumptionoftheentirepopulationoftheplanet . 0.5%ofthesunsenergyisconvertedtoaircurrents.Ifwecouldconvert2%ofthiswithwindmillsintoelectricalenergythenwewouldhave174x1015x0.5%x2%=17.4TW.Thisisstillmorethantherequired15TWoftheentireworld. Wewillgeneraterenewableenergyinthemostfavourablelocations(solarpanelsintheSaharaandwindmillsintheseaandoffthecoastofMexico!).Thentheproblempresentsitselfofanelectri-cal-networkcapableoftransportingtheenergyovervastdistances. Suchanetworkiscalledasuper-grid. AstransmissionsystemHVDC(highvoltagedirectcurrent)presentsitselfasacandidate. DC-cablesarecheaperthanAC-cablesandthelossesofHVDCarelessthan3%per1000m. Thedisadvantageisthehighinvestmentcostsoftheconvertersatthebeginningandendoftheline.Thisisonlyviableforlargedistancesandheavyloads.Asuper-gridrequiresinternationalcooperationandenormousinvestment.

SUBDIVISIONS MULTIPLES

yocto = 10 −24 kilo = 10 3

zepto 10−21 mega 10 6

atto 10 −18 giga 10 9

femto 10 −15 tera 10 12

pico 10−12 peta 1015

nano 10 −9 exa 10 18

micro 10 −6 zetta 1021

milli 10 −3 yotta 10 24

E-MOBILITY23.3 2. ELECTRICAL TRACTION

Electrical traction (tram, train, metro) occurs over a rail network (train) or over a local city net-work(tram,metro).Inadditionanumberoflocaltracksexistforexampleinfactories.

Therollingstockismostlycomprisedofamotorisedpart(locomotive),followedbypassengerorfreight carriages. Electriclocomotiveshavespeedsofupto230km/handpowersofmorethan9MW.

Thereareinternationaldifferencesinthesupplyviatheoverheadlinesandrail.Wefind1.5kVDC,3kVDC,15kV-162 __ 3 Hz,25kV-50Hz,mixed3kVDC/25kV-50Hz.Inadditionthereisasystem with a DC third rail. This is usually in metro systems. Anumberofexamplesofstandardvoltages:Germany(15kV-16.2 __ 3 Hz),France(25kV-50Hz),Belgium(3kVDCand25kV-50Hz),Netherlands(1.5kVDC). Turkey,IndiaandtheCongoalsoworkwith25kV-50Hz.ThenewelectrificationnetworksinWesternEurope90%ofmakeuseof25kV-50Hz.Foreaseoftransportoninternationallinesmultiplevoltagelocomotivesareused. UrbantransportoperateswithDC:400V/600V/650V/750V/900V. Mostcommonare600Vand750V.

Inadditiontothe“classic”train/tram/metroanothertermhasbecomepopularinrecentyearsnamely“lightrail”.Thereisnocleardefinitionforthistermsince“lightrail”ismoreacollectivetermforfasttrams(interlocal),railvehiclesthatoperateontramandtraintracksandlightweighttrains(lightandcheaptrains).ThereisforexampletheDocklandsLightRailwayinLondoninwhich you can connect from a metro and ride to the Docklands. SpecialmetrolinesareforexampletheThais“SpecialSkytrainBangkok”withamaximumspeedof80km/handoperatingon750VDC-thirdrail,ortheautomaticmetroinNuremberg-Germanywhichoperateswithoutdriver.

In additiontospeedthereisalsothefactoracceleration/deceleration.Itisacceptedthatthecom-fortvalueofaccelerationanddecelerationis1m/s2 .

Withregenerativebrakingthemotoroperatesasgenerator.Theenergyiseitherreturnedtothepowergridorstoredinbatteriesorsupercaps. Supercapsarecapacitorswhichcanstorealargeamountofenergy.Table23-2providesanidea oftheenergydensityandlifespanofbatteriesandsupercaps.

Table 23-2

COMPONENT ENERGY DENSITY LIFE SPAN

Leadacidbattery 30-50Wh/kg 3-5years

Li-ion 90-150Wh/kg 5-10years

Supercap 3-5Wh/kg unlimited

23.4 E-MOBILITY

2.1 The Flexity Outlook Tram (Brussels) by Bombardier Transportation

Fig.23-1showstheblockdiagramoftheprimaryelectronicsoftheBrusselstram.TheFlexityOutlooktrambyBombardierTransportationhastwomodels.The“long”tramhasalengthof43.2mandthe“short”modelis31.8mlong. Thetramisconstructedwithfour(six)asynchronousmotorsdependingonthemodel. Everymotor(photolowerrightonp.23.5)isdesignedfor3x430V-60Hzandhasanominal powerof112kW.ViapantographPT,powerswitchDJandinputfilterL 10 -C,the650Vsupplyisappliedtotheinverter.ThisinverterisconstructedwithIGBT’s,operatesaccordingtothePWMmethod,hasaswitchingfrequencyof2.5kHzandisdrivenbyacomputerboard(withmicropro-cessorMC68360,DSP56203andFPGAXC4013). Theoutputfrequencyoftheinverteris0to300Hz. At150Hzthetramspeedis70km/h. Duringbrakingenergyisregeneratedaslongasthecatenarywirevoltageislessthan920V. Above920Vitisswitchedovertodissipationbrakingviathebrakechoppertoresistorsof2.6Ω.

Onthelowerleftofp.23.5thereisaphotoofatractionconverter.

Fig. 23-1: BlockdiagramoftheprimaryelectronicsofaFlexityOutlooktram(BombardierTransportation)

E-MOBILITY23.5

Photo Flexity Outlook:TraminBrussels(Bombardiertransportation)

PhotoBombardiertransportation: PhotoBombardiertransportation: Tractionconverterofthetramshownabove Motorofthetramshownabove.Thegreyparthas (Dimensions:b=1600mm;h=450mm; dimensions850x385mm d=1255mm;Weight424kg)

EV.16 EVALUATION: SOLUTIONS

8.12 Fig. 8-19:

v ;fig.8-21:

v

8.13 Φ=72°20’; α=60°→→ (8-13): V diα =103.5V α=105°→→ (8-11): V diα =26.52V 8.14 Φ=60°;fig. 8-4 →→ δ=240°;(8-5)→→ V diα =51.76V (8-5) →→ V di =

v ___ 2 π .(1+0.438)=74.46V;I T(AV)max =10.26A and V RRM ≥325V

8.15 Φ=60°→→ α=30°→→ (8-13) →→ V diα = 179.29V α=60°→→ (8-13) →→ V diα = 103.5V α=90°→→ δ=233°→→ (8-11) →→ V diα =62.30V8.16 a)

v

b)

v with 90° ≤ α < 180° 8.17 No.ConsiderforexampleTh1 and Th 6 with there driving voltage v 12 . At ωt=120°,v 12 is negative so that Th 6cannotfire.Thesameconclusioncanbemadefortheothersemi- conductorpairs

8.18 Start-up:α= φ 1 =90°→→ Q 1α = V di . I d (x1.5)=150kvar

8.19 P= V RMS 2

_____ R b

HALF WAVE: α=0°:P max =0.5/ R b ; α=30° : P 1 =0.4855/ R b α =150°:P 2 =0.0144/R b for the ratios: P 1 / P max=97%and P 2 / P max =3% FULL WAVE: α=0°:P max =1/ R b ; α =30° : P 1 =0.97/ R b α =150°:P 2 =0.03/R b for the ratios: P 1 / P max=97%and P 2 / P max =3% 8.20 I 1 =

√___

6 . I dα _______ π =12.1A;I k =

I 1 __ k →→ I 5=2.42A; I 7 =1.73A

Table8-4→→ 300Hz:V kα =21 ___ 100 . V di = 65.22V;

600Hz:V kα =10.29 _____ 100 . V di = 31.96V; 8.21 Fig. 8-6: α=60° →→ X 3 =0.115x230x√

__ 2 →→

i 3=0.374A

α=90° →→ X 3 =0.153x230x√__

2 →→

i 3=0.497A 8.22 (8-29) and (8-30) →→ I 7 =

√___

6 . I d . cosα ____________ π.7 with I d =

V di ___ R (8-25) →→ V di =540V→→ I 7 =10.42A

EVALUATION: SOLUTIONS EV.17

8.23 Fig. 8-41

Fig. 8-41

The v o =f(x)axesisthesameaspreviouslyfoundinfig.8-34,sothattheexpressionforωt=f(x)in fig.8-34isalsovalidhere 8.24 B 2 -controller: p=2→→

v . p __ π sin π __ p =

v . 2 __ π . sin π __ 2 = 2 .

v ____ π

M 3 -controller: p=3→→

v . p __ π sin π __ p =

v . 3 __ π . sin π __ 3 = 3 √

__ 3 .

v _______ 2.π

B 6 –controller: p=6→→

v . p __ π sin π __ p =

v . 6 __ π . sin π __ 6 = 3 .

v ____ π

8.25 α =0°→→ cosα=1→→ V k0° ____ V di

= √__

2 _______ ( k 2 −1)

. √______________

k 2 −( k 2 −1) = √__

2 _______ ( k 2 −1)

α =90°→→ cosα=0→→

V k90° ____ V di = √

__ 2 _______

( k 2 −1) . √

____ k 2 = k . √

__ 2 _______

( k 2 −1)

V k90° ____ V k0° =k. Table8-4:examplek=2→→ α=0°→→

V kα ____ V di =47.14%

α=90°→→

V kα ____ V di =94.2%

0 0

0 0

v0 v0

v0v0

v1 v2 v1 v2 v3 v1

XX

(a) (b)

/p/p

2.2.

2./p2./pα α

α - /p α - /p

ωt ωt

VOCABULARY VO1

ENGLISH DUTCH GERMAN SPANISHabscissa abscis Abszisse abscisaAC controller wisselstroominsteller Wechselstromsteller Controlador ACaccelerationtorque versnellingskoppel Beschleunigungsmoment torquedeaceleraciónAC-controller AC controller AC-Kontroller Controlador ACactivecomponent actievecomponente aktiveKomponente componenteactivoactivepower aktiefvermogen Wirkleistung potenciaactivaaerodynamic aërodynamisch aerodynamisch aerodinámicoairgap luchtspleet Luftspalte entrehierroalgorithm algoritme Algorithmus algoritmoalternating current (AC) wisselstroom Wechselstrom corriente alternaambienttemperature omgevingstemperatuur Umgebungstemperatur temperaturaambienteamplifier versterker Verstärker amplificadoranalogons analogons analogesModell modeloanalógicoanaloguecomputer analogecomputer Analogrechner ordenadoranalógicoanalogue analoog analog análogoangulardisplacementsensor hoekstandopnemer Winkelencoder sensordeposición angularangularposition hoekpositie Winkelposition posiciónangularanode anode Anode ánodoanodizing geanodiseerd eloxiert anodizadoantistaticpackaging antistatischeverpakking antistatischeVerpackung Empaqueantiestaticoapparentpower schijnbaarvermogen Scheinleistung potenciaaparenteappliancemotor huishoudelijkemotor Haushaltmotor motordomésticoapplications toepassingen Anwendungen aplicacionesarcfurnace boogoven Lichtbogenofen hornodearcoarmaturereaction ankerreactie Ankerrückwirkung reaccióndearmaduraarmature anker Anker armaduraasynchronous motor asynchrone motor Asynchronmaschine motor asincronoauxiliarypole hulppool Hilfspol poloauxiliarauxiliary thyristor doofthyristor Löschthyristor tiristor auxiliarauxiliarywinding hulpwikkeling Hilfswicklung arrollamientoauxiliaravalanchediode zenerdiode Zenerdiode diodozeneraverageresistivity soortelijkeweerstand spezifischerWiderstand resistividadmediaaverage value gemiddelde waarde (arithmetischer) Mittelwert valor medioback-to-back(SCRs) antiparallel(SCRs) antiparalleleVentilen SCRsenantiparalelosballast ballast Vorschaltgerät balastbandgap(energygap) bandafstand Energielücke(Bandabstand) bandaprohibidabandwidth bandbreedte Bandbreite anchuradebandabankofaccumulators accubatterij Akkumulatorenbatterie bateriadeacumuladoresbarrierlayer sperlaag Sperrschicht capadebarrerabasevoltage basisspanning Basisspannung tensióndebasebase basis Basis basebatterycharger batterijlader Batterieladegerät cargadordebateríabiasing voormagnetisatie Vormagnetisierung polarizaciónmagnéticabidirectional bidirectioneel bidirektional bidireccionalbipolartransistor bipolairejunctietransistor bipolarerTransistor transistorbipolarbistable bistabiel bistabil biestableblockdiagram blokschema Blockschaltbild diagramaduncionalblockorientated blokgeoriënteerd blockorientiert bloqueorientación

Documents

Electronic Power Control