Advanced Course on Chapter 1 Electric Drives Introduction ... · Title: Microsoft PowerPoint -...

© Copyright Ned Mohan 2001 1

Ned Mohan Oscar A. Schott Professor of Power Electronics and Systems Department of Electrical and Computer Engineering University of Minnesota Minneapolis, MN 55455 USA

Advanced Course onElectric Drives


Chapter 1

Introduction to Advanced Electric Drives

3

Graduate Course in Electric Drives

Objectives:Basics to Advanced Topics in 2 SemestersSeamless Continuation of the First CourseTopics: Dynamic Modeling and ControlApproach/Tools

dq-Windings based AnalysisDesign Examples Using SimulinkVerification in the Hardware Lab using dSPACE


Topics (Lectures)1. Introduction to Advanced Electric Drive Systems (2)2. Induction Machine Equations in Phase Quantities: Assisted by Space

Vectors (5)3. Dynamic Analysis of Ind. Mach. in terms of dq-Windings (7)4. Vector Control of IM Drives: A Qual. Examination (4)5. Mathematical Description of Vector Control (5)6. Detuning Effects in Induction Motor Vector Control (3)7. Space Vector PWM (SV-PWM) Inverters (3)8. Direct Torque Control (DTC) and Encoder-Less Operation of

Induction Motor Drives (5)9. Vector Control of Perm-Magnet Syn. Motor Drives (3)10. Switched-Reluctance Motor (SRM) Drives (3)


Continuation of Topics Discussed in the First Course

Switch-Mode Converters Average RepresentationMagnetics – TransformersFeedback Controller DesignSpace Vector Representation for AC MachinesBasic Calculations for Electromagnetic TorquePMAC Drives – Space Vector Based Steady State OperationInduction Motor Drives – Space Vector based Steady State Analysis


Design ExamplesInduction Machine Initially Operating in Steady state

Load Torque Disturbance at t=0.1 sControl Objective is to keep Speed Constant (design speed loop with a bandwidth of 25 rad/s and a phase margin of 60 degrees)

Permanent Magnet AC (PMAC) DrivesSwitched-Reluctance Motor (SRM) Drives


“TEST” INDUCTION MOTORNameplate Data:Power: 3 HP/2.4 kWVoltage: 460 V (L-L, rms)Frequency: 60 HzPhases: 3Full Load Current: 4 AFull-Load Speed: 1750 RPMFull-Load Efficiency: 88.5 %Power Factor: 80.0 %Number of Poles: 4

Per-Phase Motor Circuit Parameters:

2

1.77 , 1.34 , 5.25 (at 60 Hz)4.57 (at 60 Hz), 139.0 (at 60 Hz)

Full-Load Slip = 1.72 %, 0.025

s r s

r m

eq

R R XX X

J kg m

= Ω = Ω = Ω= Ω = Ω

= ⋅© Copyright Ned Mohan 2001 8

Chapter 2

Induction Machine Equations in Phase Quantities:

Assisted by Space Vectors

9

c axis

a axis

b axis

i c

i b

i a

θ

magneticaxis of phase a

ia

θ=0

′4′3

′2

′1

7

65 4 3

2

1

′7

′6

′5

θ π=

(a) (b)

Figure 2-1 Stator windings.

Sinusoidally-Distributed Stator Windings

(1) ( ) sin2s

sNn θ θ= 0 θ π≤ ≤

(2) ( ) i cossa a

g

NHp

θ θ= ( )(4) ( ) ( ) i coso sa o a a

g

NB Hp

µθ µ θ θ= =

( ) ( )(3) i cossa g a a

NF Hp

θ θ θ= =


Figure 2-2 Three-phase windings.

a'a

b

'b

c

'c

ai

bi

ci

axisa −

axisb−

axisc −/j4 3e π

/j2 3e π

j0e

θb

bi

ai

cic

a

(a) (b)

Three-Phase Sinusoidally-Distributed Windings


n

b

c

a

ia

a ax is−

ia

ia

r

leak age

m agn etiz in g

(a)

(b)Figure 2-3 Single-phase magnetizing inductance and leakage

inductance.

Single-Phase Magnetizing Inductance ,1m phaseL −

,,

,1

,(1) a leakagea as self

a a ai onlyaL Ls m phase

magnetizingLi i i

λλ λ

−

= = +

, ,1(2) s self s m phaseL L L −= +

2,1(3) ( )o sm phase

g

r NLp

πµ− =


n

b

c

a

ia

λ b

Figure 2-4 Mutual inductance .

Stator Mutual Inductance mutualL

0,,

(1) bmutual

a i i rotor openb c

Liλ

==

,11(3)2mutual m phaseL L −= −

, ,(2) cos(120 )ob duetoi a magnetizing dueto ia aλ λ=


a axis−

b axis−

c axis−

θm

ia

ib

ic

iA

iC

iB

A axis−

B axis−

C axis−

ωm

( )a

ia

ib

ic

−− +

+

+

va

vb

vc

Rs

−−

−

+

+

vA = 0

vB = 0

vC = 0

RriA

iB

iC

Stator circuit Rotor circuit

( )b

− +

Figure 2-5 Rotor circuit represented by three-phase windings.

(b)

Equivalent Windings in A Squirrel-Cage Rotor

,13(2)2m m phaseL L −=

23(3) ( )2

o sm

g

r NLp

πµ=

(4) s s mL L L= +

(1) ( ) ( ) ( ) 0A B Ci t i t i t+ + =

(1) ( ) ( ) ( ) 0a b ci t i t i t+ + =

,13(2)2m m phaseL L −=

(3) r r mL L L= +

,1(4) cosaA m phase mL L θ−= ⋅

Stator

Rotor


( )sF t( )

sF tθ

( )aF t( )bF t

( )cF t

axisc −

axisb −

axisa −

ci = −

bi =−

ai = +

( )aF t

( )bF t

( )cF t

axisc −

axisb −

axisa −

(a) (b)

Figure 2-6 Space vector representation of mmf.

Review of Space Vectors

( ) ( ) ( ) ( )a a a as a b cF t F t F t F t= + +


Figure 2-7 Physical interpretation of stator current space vector.

a'a

b'b

c

'cai

bi

ci

axisa −

axisb−

axisc −

/j4 3e π

/j2 3e π

j0e

sI

at time tmagnetic axis of hypothetical winding

ˆwith current sI

axisa −

(a) (b)

si

Physical Interpretation of Current Space Vector

( )0 2 / 3 4 / 3 ˆ(1) ( ) ( ) ( ) ( ) ( )j tia j j j ss a b c si t i t e i t e i t e I t eθπ π= + + =

(2) ( ) ( / ) ( )a as s sF t N p i t=


( )0 2 / 3 4 / 3 ˆ( ) ( ) ( ) ( ) ( )j tva j j j ss a b c sv t v t e v t e v t e V t eθπ π= + + =

( )0 2 / 3 4 / 3 ˆ( ) ( ) ( ) ( ) ( )j ta j j j ss a b c st t e t e t e t eθλπ πλ λ λ λ λ= + + =

Voltage and Flux-Linkage Space Vectors


Figure 2-8 Relationship between space vector and phasor in sinusoidal steady state.

ˆa js si I e α−=

@ t 0=

(b)

axisa −αα

ref

ˆa aI I α= ∠ −

(a)

Relationship between space vector and phasor in sinusoidal steady state

0

32

as ati I

== 3ˆ ˆ

2s aI I=


i a

ib

ic+−vc

+−vb

+− va

λ a t( )

λ b t( )

λ c t( )

a axis−

c axis−

b axis−

e j0

e j4 3π/

e j2 3π/ i s s,i

s s,i

B

Fs

s

,, λ

at time t' '

Figure 2-9 All stator space vectors are collinear (Rotor open-circuited).

All stator space vectors are collinear (Rotor open-circuited)

,due to leakage flux due to magnetizing flux

( ) ( ) ( ) ( )a a a as s m s s ss is

t L i t L i t L i tλ = + =


a axis−

b axis−

c axis−

θm

iA

iC

iB

A axis−

B axis−

C axis−

ωm

,

, ,

, Br r irFr r ir

i

λat time t' '

Figure 2-10 All rotor space vectors are collinear (Stator open-circuited).

a axis−

b axis−

c axis−

θm

iA

iC

iB

A axis−

B axis−

C axis−

ωm

,

, ,

, Br r irFr r ir

i

λat time t' '

Figure 2-10 All rotor space vectors are collinear (Stator open-circuited).

All rotor space vectors are collinear (Stator open-circuited)

,due to leakage flux due to magnetizing flux

( ) ( ) ( ) ( )A A A Ar r m r r rr ir

t L i t L i t L i tλ = + =


Making the case for dq-axis analysis

(3) ( ) ( ) ( ) ja a A ms s s m rt L i t L i t e θλ = +

(2) ( ) ( ) ja A mr ri t i t e θ=

a axis−

b axis−

c axis−

θm

ia

ib

ic

iA

iC

iB

A axis−

B axis−

C axis−

ωm

( )a

ia

ib

ic

−− +

+

+

va

vb

vc

Rs

−−

−

+

+

vA = 0

vB = 0

vC = 0

RriA

iB

iC

Stator circuit Rotor circuit

( )b

− +

Figure 2-5 Rotor circuit represented by three-phase windings.

(b)

(1) ( ) ( )a a as s s m rt L i t L iλ = +

(4) ( ) ( ) ( )a a as s s s

dv t R i t tdt

λ= +


Chapter 3

Dynamic Analysis of Induction Machines in Terms of dq-Windings


Representation of Stator MMF by Equivalent dq Windings

2 / 3 4 / 3(2) ( ) ( ) ( ) ( )a j js a b ci t i t i t e i t eπ π= + +

(3) ( ) ( )a ass s

NF t i tp

=

( )3/ 2(4) ds ssd sq s

N Ni ji ip p

+ =

( ) 2(5)3d

sd sq si ji i+ =

2,1

,1

(1) winding magnetizing inductance =( 3 / 2)

(3 / 2)

(using Eq. 2-12)

m phase

m phase

m

dq L

L

L

−

−=

=

axisa −

axisb −

axisc −

si

ai

bi

ci

at t

axisa −

axisb −

axisc −

si

at t

axisd −

axisq −

sdi

sqi

axisa −

axisb −

axisc −

si

at t

axisd −

axisq −

projection

projectionsq2i3

= ×projection

projectionsd2i3

= ×

(a)

(b) (c)

3

2 sN

Figure 3-1 Representation of stator mmf by equivalent dq winding currents.

daθ daθ


Representation of Rotor MMF by Equivalent dq Windings

2 / 3 4 / 3( ) ( ) ( ) ( )A j jr A B Ci t i t i t e i t eπ π= + +

( )( )/

AA rr

s

F ti tN p

=

Figure 3-2 Representation of rotor mmf by equivalent dq winding currents.

Ai

dAθrqi

a axis−

dωd axis−

q axis−

B axis− A axis−mω

daθ mθ

ri

••

rdi

projection2 projection3rqi = ×

projection2 projection3rdi = ×


a axis−stator

A axis−rotor

d axis−

ωm

isd

isq

q axis−at t

i rq

ird

θm

is

ir

32

isd

32

isq

32

irq 32

ird

dω

dω

daθ

dAθ

32 sN

32 sN

32 sN

Figure 3-3 Stator and rotor mmf representation by equivalent dq winding currents.

a axis−a axis−stator

A axis−A axis−rotor

d axis−d axis−

ωmωm

isdisd

isqisq

q axis−q axis−at tat t

i rqi rq

irdird

θmθm

isis

irir

32

isd32

isd

32

isq32

isq

32

irq32

irq 32

ird32

ird

dω

dω

daθ

dAθ

32 sN

32 sN

32 sN


Figure 3-3 Stator and rotor representation by equivalent dq winding currents. The dq winding voltages are defined as positive at the dotted terminals. Note that the relative positions of the stator and the rotor current space vectors are not actual, rather only for definition purposes.

Mutual Inductance between dq Windings on the Stator and the Rotor

sd s sd m rdL i L iλ = +

sq s sq m rqL i L iλ = +

rd r rd m sdL i L iλ = +

rq r rq m sqL i L iλ = +


Figure 3-4 Transformation of phase quantities into dq winding quantities.

[ ]abc dqsT →

aibi

ci

sdi

sqi

(a) stator

[ ]ABC dqrT →

Ai

Bi

Ci

rdi

rqi

(b) rotordaθ dAθ

Figure 3-4 Transformation of phase quantities into dq winding quantities.

[ ]abc dqsT →

aibi

ci

sdi

sqi

(a) stator

[ ]ABC dqrT →

Ai

Bi

Ci

rdi

rqi

(b) rotordaθ dAθ

[ ]abc dqsT →

aibi

ci

sdi

sqi

(a) stator

[ ]ABC dqrT →

Ai

Bi

Ci

rdi

rqi

(b) rotordaθ dAθ

[ ]

2 4 ( )cos( ) cos( ) cos( )( ) 2 3 3 ( )( ) 2 43 sin( ) sin( ) sin( ) ( )

3 3

ada da dasdb

sqda da da c

Ts abc dq

i ti t

i ti t

i t

π πθ θ θ

π πθ θ θ

→

− − = − − − − − [ ]

2 4 ( )cos( ) cos( ) cos( )( ) 2 3 3 ( )( ) 2 43 sin( ) sin( ) sin( ) ( )

3 3

AdA dA dArdB

rqdA dA dA C

Tr ABC dq

i ti t

i ti t

i t

π πθ θ θ

π πθ θ θ

→

− − = − − − − −

Mathematical Relationship between dq and phase Winding Variables


-axisa

-axisd

-axisα

-axisβ

daθ

i

ii

i

sλ

sdisqi

si α

si β- axisq

Figure 3-5 Stator and equivalent windings.dqαβ

dω

(1) s s s sdv R idtα α αλ= +

(2) s s s sdv R idtβ β βλ= +

_ _ _(4) s s s sdv R idt

α α ααβ αβ αβλ= +

_ _(6) ; etc.j das s dqv v e θααβ = ⋅

_(5) ; etc.s dq sd sqv v jv= +

Derivation of Stator Voltages in dq Windings

_(3) ; etc.s s sv v jvααβ α β= +


_ _ _ _(2) s dq s s dq s dq d s dqdv R i jdt

λ ω λ= + +

(3) sd s sd sd d sqdv R idt

λ ω λ= + −

(4) sq s sq sq d sddv R idt

λ ω λ= + +

[M ]rotate

0 1(5)

1 0sd sd sd sd

s dsq sq sq sq

v i dRv i dt

λ λωλ λ

− = + +

Derivation of Stator Voltages in dq Windings (continued)

_ _ _(1) s s s sdv R idt

α α ααβ αβ αβλ= +


-axisa

-axisd

-axisα

-axisβ

daθ

ii i

i

rλ

rdirqi

ri α

ri β

- axisq

Figure 3-6 Rotor and equivalent windings.dqαβ

dω

-axisA

dAθ

mθ

mω

Derivation of Rotor Voltages in dq Windings

(1) rd r rd rd dA rqdv R idt

λ ω λ= + −

(2) rq r rq rq dA rddv R idt

λ ω λ= + +

(3) dA d mω ω ω= −

(4) ( / 2)m mechpω ω=

[M ]rotate

0 1(5)

1 0rd rd rd rd

r dArq rq rq rq

v i dRv i dtλ λ

ωλ λ −

= + +


[M ]rotate

0 11 0

sd sd sd sds d

sq sq sq sq

v id Rv idt

λ λωλ λ

− = − −

[M ]rotate

0 11 0

rd rd rd rdr dA

rq rq rq rq

v id Rv idt

λ λωλ λ

− = − −

Obtaining Flux Linkages: Voltages as Inputs

[ ]_ _ _ rotate _[ ] [ ] [ ] M [ ]s dq s dq s s dq d s dqd v R idt

λ ω λ= − −

[ ]_ _ _ rotate _[ ] [ ] [ ] M [ ]r dq r dq r r dq dA r dqd v R idt

λ ω λ= − −

_s dqv

_r dqv

_s dqλ

_r dqλ

rotate[M ]

rotate[M ]

1s

1s

sR

rR

dAω

dω×

×

∑

∑

+

−

−

+

−

−

_s dqddt

λ

_r dqddt

λ

1[ ]M −

Eq. 3-62

dqi

_s dqi

_r dqi

Figure 3-7 Calculating winding flux linkages and currents.dq


a axis−

d axis−q axis−

rqi

rdi

sqi

d

d

add

subtract

due to and sq rqi i

due to rqileakage flux

i

i i

Figure 3-8 Torque on the rotor -axis.d

a axis−

d axis−q axis−

rqi

rdi

sqi

d

d

add

subtract

due to and sq rqi i


i

i i

a axis−

d axis−q axis−

rqi

rdi

sqi

d

d

add

subtract

due to and sq rqi i


i

i i

Figure 3-8 Torque on the rotor -axis.d

Electromagnetic Torque on the Rotor d-Axis

0 3/ 2ˆ(1) ( )s rrq sq rq

g m

mmf

N LB i ip L

µ = +

,3 / 2 ˆ(2)

2s

d rotor rq rdNpT r B ip

π

=

20

,3 / 2(3) ( )

2s r

d rotor sq rq rdg m

Np LT r i i ip L

µπ

= +

20

,3(4) ( )

2 2s r

d rotor sq rq rdg m

Lm

Np LT r i i ip L

µπ = +

,(5) ( )2 2d rotor m sq r rq rd rq rd

rq

p pT L i L i i i

λ

λ= + =


rqi

rdi

q

add

subtract

due to and sd rdi i

due to rdileakage flux

q axis−d axis−

a axis−

sdi

q

i i

i

Figure 3-9 Torque on the rotor -axis.q

rqi

rdi

q

add

subtract

due to and sd rdi i


q axis−d axis−

a axis−

sdi

q

i i

i

rqi

rdi

q

add

subtract

due to and sd rdi i


q axis−d axis−

a axis−

sdi

q

i i

i

Figure 3-9 Torque on the rotor -axis.q

Electromagnetic Torque on the Rotor q-Axis

, ( )2 2q rotor m sq r rq rq rd rq

rd

p pT L i L i i iλ

λ= − + = −

, ,(1) em d rotor q rotorT T T= +

(2) ( )2em rq rd rd rqpT i iλ λ= −

(3) ( )2em m sq rd sd rqpT L i i i i= −

(4) em Lmech

eq

T Tddt J

ω −=

Net Electromagnetic Torque


Equivalent d and q Axis Equivalent Circuits

( )sd s sd d sq s sd m sd rdd dv R i L i L i idt dt

ω λ= − + + +

( )sq s sq d sd s sq m sq rqd dv R i L i L i idt dt

ω λ= + + + +

0

( )rd r rd dA rq r rd m sd rdd dv R i L i L i idt dt

ω λ=

= − + + +

0

( )rq r rq dA rd r rq m sq rqd dv R i L i L i idt dt

ω λ=

= + + + +

sR

+

+

+

+

−−

− −

sR

sL

sL rL

rL

mL

mL

sdv

sqv

d sdω λ

d sqω λ

sdi

sqi

dA rqω λ

dA rdω λ

rdv

rqv

+

−

+

−

sdddt

λ

sqddt

λ

+

−

+

−

rdddt

λ

rqddt

λ

(a) -axisd

(b) -axisq

1 2

1 2

1 2

1 2

1

2

1

2

Figure 3-5 -winding equivalent circuits.dq

rdi

rqi

rR

rR

Figure 3-10 dq-winding equivalent circuits.


Figure 3-11 Per-phase equivalent circuit in steady state.

+

−

(at )ωaV

+

−

sj Lω

maImaE

raI ′aIsR

rj Lω

rRs

( )m mj L jXω =

AI

Per-Phase Equivalent Circuit

_ _ _(1) s dq s s dq syn s dqv R i jω λ= +

_ _(4) 0 rdq syr n r dq

R i js

ω λ= +

_ _ _ _ _(2) ( )s dq s s dq syn s s dq syn m s dq r dqv R i j L i j L i iω ω= + + +

_ _ _ _(5) 0 ( )rr dq syn r r dq syn m s dq r dq

R i j L i j L i is

ω ω= + + +

(3) ( )a s a syn s a syn m a AV R I j L I j L I Iω ω= + + +

(6) 0 ( )ra syn r A syn m a A

R I j L I j L I Is

ω ω= + + +


Obtaining Currents from Flux Linkages

[ ]

sd s m sd

rd m r rdL

L L iL L i

λλ

=

[ ]

sq sqs m

m rrq rqL

iL LL L i

λλ

=

[ ]

0 00 0

0 00 0

sd sds msq sqs m

m rrd rd

m rrq rqM

iL LiL L

L L iL L i

λλλλ

=

1[ ]

sd sd

sq sq

rd rd

rq rq

ii

Mii

λλλλ

−

=


Induction Motor Model

_s dqv

_r dqv

_s dqλ

_r dqλ

rotate[M ]

rotate[M ]

1s

1s

sR

rR

dAω

dω×

×

∑

∑

+

−

−

+

−

−

_s dqddt

λ

_r dqddt

λ

1[ ]M −

Eq. 3-62

dqi

_s dqi

_r dqi

emT

LT

∑

Eq. 3-47

1

eqsJmechω

2p

∑

mω

+

−

+−

Figure 3-12 Induction motor model in terms of windings.dq


Calculation of Steady State Initial Conditions Using Phasors

ˆ

3ˆ ˆ(0)2

j ia a i s a

Is

I I i I e θθ= ∠ ⇒ =

ˆ

2 2 3 ˆ(0) (0) - = cos( )3 3 2sd s i

Is

i projection of i on d axis I θ = ×

ˆ

2 2 3 ˆ(0) (0) - = sin( )3 3 2sq s i

Is

i projection of i on q axis I θ = ×


(1) sd s sd syn sqv R i ω λ= −

(2) sq s sq syn sdv R i ω λ= +

(3) 0 r rd syn rqR i sω λ= −

(4) 0 r rq syn rdR i sω λ= + [ ]

0

0

000 0

s syn s syn m sdsd

syn s s syn m sqsq

syn m r syn r rd

rqsyn m syn r r

A

R L L ivL R L iv

s L R s L iis L s L R

ω ωω ω

ω ωω ω

− − = − −

[ ] 1

00

sd sd

sq sq

rd

rq

i vi v

Aii

−

=

Calculation of Steady State Initial Conditions Using Voltage Equations


Simlink-based dq-Axis Simulation of Induction Motor


Simulation Results


Chapter 4

Vector Control of Induction-Motor Drives: A Qualitative Examination


DC Motor Drive

aS

SN

aNemT

fφ

aφ

PPU

+

− aeav

aL

aR

ai

−

+

ai

0

0

t

em T aT k i=emT

t

Tem = kT ia


Brushless DC Motor Drive

( )rB t

( )si t

o90 NS axisa −

mθ

sI

0

0

tˆ

em T sT k I=emT

t

ˆ em T ST k I=


Vector-Controlled Induction Motor Drive

axisa −'( )ri t

'( )rF t( )rB t

( )rF t

( )msB t( )lrB t

rθ

rθ

( )sv t

speed feedback

currentfeedback

Tacho

IMPPU

( ) perpendicuar to ( ) and ( )r r rB t F t F t′'ˆ ˆ (keeping constant)em T r rT k I B=


Analogy to a Current-Excited Transformer With a Shorted Secondary

, 2m iφ , 1m iφ

1i l2φ

N:N

shown explicitly2R

short

net = 0

2v 0=2R

2i

0 t1i

+ -2 2(0 ) = (0 ) = 0λ λ

1 2

+ + +m,i m,i 2(0 ) = (0 ) + (0 )φ φ φ

+ +m2 1

2

Li (0 ) = i (0 )L


Analogy to a Current-Excited Transformer With a Shorted

Secondary

2 12

(0 ) (0 )mLi iL

+ +=

22 2( ) (0 ) ti t i e τ−+=

22

2

LR

τ =

φmi, 2 φmi, 1

i1 i1

i2

N1 N2

net =0

R2φ2

∴ = +φ φ φmi mi, ,1 2 2

0 t

t = +0

short

R shown licitly2 exp

( )a

i2

−

−+

+Lm

L2

+

−

e2

Nddt

L didt

φ2

2

2=

Nd

dtL di

dtmi

mφ , 2 2=

Noted

dtmi: ,φ

1 0=

t >0

( )b

0

0i2

i1

t

t

LL

im

21 0+d i

( )c

Figure 4-5 Analogy of A Current-Excited Transformer with a Short-Circuited Secondary;

0

1 2N N=

2 2mL L L= +


R1 R2i2

i2

00 tt im

i1i1

L1

L2

Lm

N N N1 2= =

Figure 4-6 Equivalent-Circuit Representation of the Current-Excited Transformer with a Short-Circuited Secondary.

Using the Transformer Equivalent Circuit

2 1 12 2

(0 ) (0 ) (0 )m m

m

L Li i iL L L

+ + += =+

2 21 1

2 2(0 ) (0 ) (0 )m

m

L Li i iL L L

+ + += =+

22 2( ) (0 ) ti t i e τ−+=


d- and q- Axis Winding Representation

sd s

2i = the projection of i (t) vector along the d-axis3

sq s

2i = the projection of i (t) vector along the q-axis3

a axis−stator

A axis−rotor

d axis−

ωm

isd

isq

q axis−at t

i rq

ird

θm

is

ir

32

isd

32

isq

32

irq 32

ird

dω

dω

daθ

dAθ

32 sN

32 sN

32 sN




d axis−d axis−

ωmωm

isdisd

isqisq


i rqi rq

irdird

θmθm

isis

irir

32

isd32

isd

32

isq32

isq

32

irq32

irq 32

ird32

ird

dω

dω

daθ

dAθ

32 sN

32 sN

32 sN


Figure 3-3 Stator and rotor representation by equivalent dq winding currents. The dq winding voltages are defined as positive at the dotted terminals. Note that the relative positions of the stator and the rotor current space vectors are not actual, rather only for definition purposes.


Initial Flux Buildup Prior to axisq −

axisd −

sqi 0=

sdi

t−∞ 0

, sdm iφ

si

rB

t 0−=

sdi

- - -a m,rated b c m,rated

1ˆ ˆi (0 ) = I and i (0 ) = i (0 ) = - I2

-sd ms,rated m,rated m,rated

2 2 3 3ˆ ˆ ˆi (0 ) = I = ( I ) = I3 3 2 2

sqi = 0

-t = 0


Step Change in Torque at m 0ω = +=t 0

rB sdi

net = 0

axisa −

slipωlrφ

, rm iφ, sqm iφ

slipω

sqi sqi

sdi

0−∞ t

ωm = 0isd unchangedStep-change in isq

Φq,net = 0sq r

+ + +m,i m,i r(0 ) = (0 ) + (0 )φ φ φ

mem r sq

r

LˆT α B , iL

-t = 0


Flux Densities at +t = 0

rB

t 0+=

rθ

lrBmsB

axisa −


Transformer Analogy –Voltage Needed to Prevent the Decay of Secondary Current

−

+

2i′

( ) ( )2 2 2v R i 0 u t+′ ′ ′=

2R′

mimL

l2L′l1L

( )1i u t⋅

1R t 0>


Currents and Fluxes at Sometime Later t > 0, Blocked Rotor

m 0ω =

t 0>

rBsdi

net = 0

axisa −

slipωlrφ

, rm iφ, sqm iφsqi

slipω

sqi

axisd −

axisq −

r m r sqslip

r

R , (L /L )i

Bω α

′


Vector-Controlled Condition With a Rotor Speed ωm

t 0>

rBsdi

net = 0

axisa −

synωlrφ

, rm iφ, sqm iφsqi

synω

sqi

axisd −

axisq −

mω

syn m slipω = ω + ω© Copyright Ned Mohan 2001 54

Torque, Speed, and Position Control

(measured)/d dt

encoder

Motor

PPUregulated current

to abcdq

to dqabc

*ai*bi*ci

aibici

sdisqi

emTˆrB

rBθcalculations

mθmω

(measured)mθ mω emT

(measured) (calculated)

P PI PI

∑

∑∑ ∑

−−+ +

−+

−

+ PImω(measured)

*ˆrB

ˆrB

mω

ˆrB (calculated)

*sdi

*sqi

*mθ *

mω *emT

(measured)mωmotor mathematical model

syn m slipω (t) = ω (t) + ω (t)

r

t

B syn0θ (t) = 0 + ω (τ) dτ∫ ⋅


Figure 4-14 (a) Block diagram representation of hysteresis current control; (b) current waveform.

T

1k

*sI

*emT

*( )ai t*( )bi t*( )ci t

Σ ( )Aq t

phasea

( )ai t

dV+−

( )m tθ

+

−

(a)

(b)

reference current

actual current

t0

.From Eq 10 6−

using Eqs. 10-11 and 10-12

Power-Processing Unit (PPU)


Chapter 5

Mathematical Description of Vector Control


a axis−stator

A axis−rotor

d axis−

ωm

isd

isq

q axis−at t

i rq

ird

θm

is

ir

32

isd

32

isq

32

irq 32

ird

dω

dω

daθ

dAθ

32 sN

32 sN

32 sN




d axis−d axis−

ωmωm

isdisd

isqisq


i rqi rq

irdird

θmθm

isis

irir

32

isd32

isd

32

isq32

isq

32

irq32

irq 32

ird32

ird

dω

dω

daθ

dAθ

32 sN

32 sN

32 sN


Figure 5-1 Stator and rotor mmf representation by equivalent dq winding currents. The d-axis is aligned with rλ .

Motor Model with the d-Axis Aligned with the Rotor Flux Linkage Axis

( ) 0rq tλ = mrq sq

r

Li iL

= −( ) 0rqd tdt

λ =


sR

+

+

+

+

−−

− −

sR

sL

sL rL

rL

mL

mL

sdv

sqv

d sdω λ

d sqω λ

sdi

sqidA rdω λ

0rdv =

0rqv =

+

−

+

−

sdddt

λ

sqddt

λ

+

+

−

rdddt

λ

0rqddt

λ =

(a) -axisd

(b) -axisq

rdi

rqi

rR

rR

−

+

−

rqi

Figure 5-2 Dynamic circuits with the d-axis aligned with rλ .

Dynamic Circuits with the d-Axis Aligned with the Rotor Flux Linkage Axis

Calculation of :dAωrq m

dA r sqrd r rd

i LR iωλ τ λ

= − =

Calculation of Torque

2 2m

em rd rq rd sqr

Lp pT i iL

λ λ = − =

:emT


rL

mLsdi

+

−

rdddt

λ rdi

rR

Figure 5-3 The d-axis circuit simplified with a current excitation.

D-Axis Rotor Flux Dynamics

(1) ( ) ( )mrd sd

r r

sLi s i sR sL

= −+

(2) rd r rd m sdL i L iλ = +

(3) ( ) ( )(1 )

mrd sd

r

Ls i ss

λτ

=+

(4) rd mrd sd

r r

Ld idt

λλτ τ

+ =


Motor Model

0( ) 0 ( )

tda dt dθ ω τ τ= + ∫

rFigure 5-4 Motor model with d-axis aligned with .λ

rdλrdλ1m

r

Lsτ+

rτ

DN

N/D1/ sdaθ

emTmω

mechω

dω dAωmL

sqi

sdi

2m

r

Lp

L⋅

2

p

rdλ

∑

×

++


Simulation with d-Axis Aligned with the Rotor Flux Linkage (line-fed machine; achieving vector control is not an objective)


Figure 5-6 Results of Example 5-1.

Simulation Results match that of Example 3-3


Vector-Controlled Induction Motor with a CR-PWM Inverter

Figure 5-7 Vector-controlled induction motor with a CR-PWM Inverter.

emT

isd*

isq*

ia*

ib*

ic*

Current −Regulated

PPU

Motor

( )mech measuredωmechθ

mod elMotor

of Fig.49

sdi

sqi

iaibic

Encoder

Transformations

Transformations

ddt

Eq. 107

Eq. 103

Motor Model(Fig. 5-4)

abctodq

dqtoabc

rdλ

daθ

daθ


Speed and Position Loops for Vector Control

Figure 5-8 Vector controlled induction motor drive with a current-regulated PPU.

(measured)/d dt

Motor

PPUregulated current

to abcdq

to dqabc

*ai*bi*ci

aibici

sdi

sqiemTrdλdaθ

Fig. 5-4

mechθmechω

(measured)

mechθ mechω emT

(calculated)

P PI PI

∑

∑∑ ∑

−−+ +

−+

−

+ PImechω *

rdλ*rdλ

mechω

rdλ (calculated)

*sdi

*sqi

*mechθ

*mechω *

emT

(measured)

daθ

mechωEstimated Motor Model


Design of Speed Loop

rd m sdL iλ =

2*

2m

em sd sqrk

LpT i iL

=

∑+

−

mechωk

( )sqi si

pkks

+ 1

eqsJ

*mechω emT


Simulation of CR-PWM Vector Controlled Drive using Simulink


Figure 5-11 Simulation results of Example 5-2.

Simulation Results of a Vector Controlled Induction Motor Drive


Calculation of Stator Voltages in Vector Control

2(1) 1 m

s r

LL L

σ = −

(2) msd s sd rd

r

LL iL

λ σ λ= +

(3) sq s sqL iλ σ=,

(5) msq s sq s sq d rd d s sd

rvsq vsq comp

Ldv R i L i L idt L

σ ω λ ω σ

′

= + + +

Figure 5-12 Vector control with applied voltages.

isd*

isq*

PI

PI

∑

∑ ∑

∑+

+

++

+

+

−

−

isd

isq

ia

ic

ib

va*

vb*

vc*

Motorvsd*

vsq*

Transformations

M o to rM o d e l

isd

isq

Encodermechθ( )mech measuredω

Transformations

Modulated

SpaceVector

PPU

rdλemT

daθ

daθ

,sd compv

,sq compv

sdv′

sqv′

,

(4) msd s sd s sd rd d s sq

rvsd vsd comp

Ld dv R i L i L idt L dt

σ λ ω σ

′

= + + −


Figure 5-13 Design of the current-loop controller.

∑+−

( )sdi s( )sdv s′* ( )sdi sPI

1

s sR s Lσ+

Design of the Current-Loop Controller

(1) sd s sd s sddv R i L idt

σ′ = +

(2) sq s sq s sqdv R i L idt

σ′ = +

1(3) ( ) ( )sd sds s

i s v sR s Lσ

′=+

1(4) ( ) ( )sq sqs s

i s v sR s Lσ

′=+


Simulation of Vector Controlled Drive with supplied Voltages



Simulation Results of Vector Controlled Drive with supplied Voltages


Chapter 6

Detuning Effects in Induction Motor Vector Control


sdi

0sqi =

rλ-axisa-axisd

-axisq

0t −=

-Figure 6-1 windings at 0 .dq t =

Initial conditions

0t −=*

sd sdi i=

, 0da da estθ θ= =


-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

sdi

*sdi

*sqi

sqi

sdi

*sdi

*sqi

sqi

,da estθdaθ

0t >

Figure 6-2 windings at 0; drawn for 1.dq t kτ> <

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

sdi

*sdi

*sqi

sqi

sdi

*sdi

*sqi

sqi

,da estθdaθ

0t >

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

sdi

*sdi

*sqi

sqi

sdi

*sdi

*sqi

sqi

,da estθdaθ

0t >


Effect of Detuning due to Incorrect Rotor Time Constant

,(1) r

r estkτ

ττ

=

( ) ( ), ,,_ _ _(3) j jd d est d estda da est errs dq s dq s dqi i e i eθ θ θ− − −= =

,(4) err da da estθ θ θ= −, * *_(2) d est

sd sqs dqi i ji= +


* *(3) cos sinsd sd err sq erri i iθ θ= +

* *(4) cos sinsq sq err sd erri i iθ θ= −

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

sdi

*sdi

*sqi

sqi

sdi

*sdi

*sqi

sqi

,da estθdaθ

0t >


-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

sdi

*sdi

*sqi

sqi

sdi

*sdi

*sqi

sqi

,da estθdaθ

0t >

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

sdi

*sdi

*sqi

sqi

sdi

*sdi

*sqi

sqi

,da estθdaθ

0t >


Effect of Detuning due to Incorrect Rotor Time Constant (continued)

( ) ( ), ,,_ _ _(1) j jd d est d estda da est errs dq s dq s dqi i e i eθ θ θ− − −= =

,err da da estθ θ θ= −

, * *_(2) d est

sd sqs dqi i ji= +


Estimated Motor Model (Rotor Blocked)

*sdi

*sqi

dqtoabc

CR-PWMInv.

*ai

*bi

*ci

ai

bi

ci

abctodq

sdi

sqi

1m

r

Lsτ+

mL

rτ

DN ND

dAω 1s

dAθ

rdλ

,rd estλ,1

m

r est

Lsτ+

,r estτ

,dA estθ

,dA estθ

abctodq

mLN

D,dA estω

ND

1s

*,sd est sdi i=

*,sq est sqi i=

(ideal)

Estimated Motor Model(Rotor Flux and Slip Estimator)

Actual Motor Model

Figure 6-3 Actual and the estimated motor models (blocked-rotor).


Simulation of Vector Control with Estimated Motor Parameters



Effect of Detuning in Dynamic and Steady States


-axisd

-axisestd

-axisa

-axisq-axisestq _s dqi

sdi

*sdi

sqi

*sqi

daθ ,da estθ

0t >


,i estsθ

isθ

Calculations of Steady State Errors

1 sqdA

r sd

ii

ωτ

=

*

, *1

,sq

dA estsd

i

ir estω

τ=

*

*,

,

1 1sq sq

r est r sdsddAdA est

i iii

ωω

τ τ=

2 22 2 * *_ˆsd sq sd sq s dqi i i i I+ = + =


-axisd

-axisestd

-axisa

-axisq-axisestq _s dqi

sdi

*sdi

sqi

*sqi

daθ ,da estθ

0t >


,i estsθ

isθ

Calculations of Steady State Errors (continued) *

*sq

sd

im

i=

2

* 2 21

1sd

sd

i mi k mτ

+=+ ⋅

* *sq sd

sq sd

i iki i

τ=

1 1tan ( ) tan ( )err m k mτθ − −= − ⋅

,err i est is sθ θ θ= −

2

* 21

1 ( )em

em

T mkT k m

ττ

+=+ ⋅


Chapter 7

Space-Vector Pulse-Width-Modulated (SV-PWM) Inverters

Advantages

• Full Utilization of the DC Bus Voltage

• Same simplicity as the Carrier-Modulated PWM

• Applicable in Vector Control, DTC and V/f Control


Synthesis of Stator Voltage Space Vector

Figure 7-1 Switch-mode inverter.

aq bq cq

+

−

dV

ab

c

N

+−av

bv+

cv +

ai

bi

ci

0 2 / 3 4 / 3(1) ( ) ( ) ( ) ( )a j j js a b cv t v t e v t e v t eπ π= + +

(2) ; ;a aN N b bN N c cN Nv v v v v v v v v= + = + = +

0 2 / 3 4 / 3(3) 0j j je e eπ π+ + =

0 2 / 3 4 / 3(4) ( )a j j js aN bN cNv t v e v e v eπ π= + +

0 2 / 3 4 / 3(5) ( ) ( )a j j js d a b cv t V q e q e q eπ π= + +


0 7Figure 7-2 Basic voltage vectors ( and not shown).v v

-axisa

sv

1(001)v

3(011)v2(010)v

6(110)v

4(100)v5(101)v

sector 1

sector 2

sector 3

sector 4

sector 5

sector 6

Basic Voltage Vectors

00

12 / 3

2/ 3

34 / 3

45 / 3

5

6

7

(000) 0

(001)

(010)

(011)

(100)

(101)

(110)

(111) 0

asa js da js da js da js da js da js das

v v

v v V e

v v V e

v v V e

v v V e

v v V e

v v V e

v v

π

π

π

π

π

= =

= =

= =

= =

= =

= =

= =

= =


Synthesis of Voltage Vector in Sector 1

01

jdv V e=

/ 33

jdv V e π=

ˆ j ss sv V e θ=

sθ

1xv

3yv

Figure 7-3 Voltage vector in sector 1.

1 31(1) [ 0]a

s s s ss

v xT v yT v zTT

= + + ⋅

1 3(2) asv xv yv= +

(3) 1x y z+ + =

0 / 3ˆ(4) j j jss d dV e xV e yV eθ π= +


Synthesis using Carrier-Modulated PWM

0 7Figure 7-4 Waveforms in sector 1; .z z z= +

sT

/ 2sT7z

0 / 2z 0 / 2z

/ 2y/ 2y

/ 2x / 2xaNv

bNv

cNv

0

0

0

dV

dV

dV

0

,control avtriv

,control bv

,control cv

,

,

,

ˆ / 2

ˆ / 2

ˆ / 2

control a a k

dtri

control b b k

dtri

control c c k

dtri

v v vVV

v v vVV

v v vVV

−=

−=

−=

max( , , ) min( , , )2

a b c a b ck

v v v v v vv +=

( ) ( ) ( ) 0a b cv t v t v t+ + =


Synthesis of Space Vector using Carrier-Modulated PWM in Simulink



Control Waveforms for Carrier Pulse-Width-Modulation


Limit on the Amplitude of the Stator Voltage Space Vector

,max,max

ˆ(4) ( ) 3 0.707

2 2phase d

LL dV VV rms V= = =

,max3(5) ( ) 0.612

2 2LL d dV rms V V= = (sinusoidal PWM)

030,maxsV

dV

dV

ˆFigure 7-7 Limit on amplitude .sV

,max ,maxˆ(1) ( ) j ta syns sv t V e ω=

0,max

60 3ˆ(2) cos( )2 2s d dV V V= =

,max ,max2ˆ ˆ(3)3 3

dphase s

VV V= =


Chapter 8

Direct Torque Control (DTC) and Encoder-less Operation of Induction

Motors


DTC System Overview

Measured Inputs: Stator Voltages and Currents

Estimated Outputs: 1) Torque, 2) Mechanical Speed, 3) Stator Flux Amplitude and 4) its angle

+−dV aq

bq

cq

aibici

IM

∑ ∑PI+ +− −

+∑

−

*mechω

mechω

*emT

*ˆsλ

ˆsλ

emT

sθ∠

Estimator

Selectionof

sv

Figure 8-1 Block diagram of DTC.


Principle of DTC Operation

Figure 8-2 Changing the position of stator flux-linkage vector.

sθ

rθ

mθ

( )s tλ

( )s t Tλ − ∆sv T∆i

( ) ( )r rt t Tλ λ≈ − ∆

-axisa

Rotor -axisAArθ

2ˆ ˆ sin

2m

em s r srLpTLσ

λ λ θ=

sr s rθ θ θ= −


22 3

ˆ2em

slip rr

TRp

ωλ

=

m r slipω ω ω= −

(2 / )mech mpω ω=

ˆ( ) ( ) ( )t j ss s s s s s

t Tt t T v R i d e θλ λ τ λ

−∆= − ∆ + − ⋅ =∫

ˆ( ) jr rr s s s rm

L L i eL

θλ λ σ λ= − =( ) ( )r r

r rt t Td

dt Tω

ω

θ θω θ − − ∆= =∆

Im( )2

conjem s s

pT iλ=

21 m

s r

LL L

σ = −

Calculation of Stator Flux:

Calculation of Rotor Flux:

where

Estimating Torque:

Estimating Mechanical Speed:

s s s sdv R idt

λ= + ⇒


Inverter Basic Vectors and Sectors

a-axis1(001)v

3(011)v2(010)v

6(110)v

4(100)v 5(101)v

b-axis

c-axis

1

23

4

5

6

Figure 8-3 Inverter basic vectors and sectors.


Stator Voltage Vector Selection in Sector 1

sector 1sλ

1v

3v2v

4v 5v6v

Figure 8-4 Stator voltage vector selection in sector 1.


Selection of the Stator Voltage Space Vector

sector 1sλ

1v

3v2v

4v 5v6v


sector 1sλ

1v

3v2v

4v 5v6vsector 1sλ

1v

3v2v

4v 5v6v


Effect of Voltage Vector on the Stator Flux-Linkage Vector in Sector 1.

increasedecrease

decreasedecrease

decreaseincrease

increaseincreasesv

3v

2v

4v

5v

emT ˆsλ

a-axis1(001)v

3(011)v2(010)v

6(110)v

4(100)v 5(101)v

b-axis

c-axis

1

23

4

5

6

Figure 8-3 Inverter basic vectors and sectors.


Effect of Zero Stator Voltage Space Vector

sθrθ

mθ

( ) ( )s st t Tλ λ≈ − ∆

-axisa

Rotor -axis at t- tA ∆

Arθ Rotor -axis at tA

( )r t Tλ − ∆( )r tλ

mθ∆

sin ( )sr s rθ θ θ≈ −

( )em s rT k θ θ−

0sθ∆

0Arθ∆

Ar m r mθ θ θ θ∆ = ∆ + ∆ ∆ ( )em mT k θ∆ − ∆


DTC in Simulink


Fig. 2 Torque Waveforms.


Fig. 3 Speed Waveforms.


Fig. 4 Stator Flux.


Fig. 5 Stator and Rotor Fluxes.


Chapter 9

Vector Control of Permanent-Magnet

Synchronous-Motor Drives


Non-Salient Permanent-Magnet Synchronous Motor

sd s sd fdL iλ λ= + sq s sqL iλ =

axisc−

axisa−

axisb−

N

S

bi

ai

ci

mθ

( )a

axisc −

axisa −

axisb −

N

S

bi

ai

ci

mθ

rB

'a

a

-axisq

( )b

-axisd

Figure 9-1 Permanent-magnet synchronous machine (shown with =2).p


sd s sd sd m sqdv R idt

λ ω λ= + −

sq s sq sq m sddv R idt

λ ω λ= + +

2m mechpω ω=

( )2em sd sq sq sdpT i iλ λ= − [( ) ]

2 2em s sd fd sq s sq sd fd sqp pT L i i L i i iλ λ= + − =

em Lmech

eq

T Tddt J

ω −=

Non-Salient Permanent-Magnet Synchronous Motor (Continued)


sd s sd m s sqv R i L iω= −

Per-Phase Steady Stead Equivalent Circuit

sq s sq m s sd m fdv R i L iω ω λ= + +

3/ 2s s s m s s m fd

e fs

v R i j L i jω ω λ= + +

23a s a m s a m fd

E fa

V R I j L I jω ω λ= + +

2ˆ3fa fd m E m

kE

E kλ ω ω= =

23E fdk λ=

sR

aV

m mLωm sLω

m lsLω

ˆ( )fa fa E mE E k ω=

−

+

aI

−

+

Figure 9-2 Per-phase equivalent circuit in steady state ( in elect. rad/s)mω


Controller in the dq Reference Frame

compd

( )sd s sd s sd m s sqdv R i L i L idt

ω= + + −compq

( )sq s sq s sq m s sd fddv R i L i L idt

ω λ= + + +

22, ,

3ˆ ˆ( )2sd sq dq rated a ratedi i I I+ ≤ =

Figure 9-3 Controller in the dq reference frame.

ai

bi

ci

Motor

+ Σ

+−

Σ

+−

Σ*sqi

*sdi

( )2m mechpθ θ=

sensorposition

abctodq

+ Σ+

+

( )m s sd fdL iω λ+

m s sqL iω−

*sdv

*sqv Inverter

PWMi

pkks

+

mωsqi

sdi

decouplingterms

1s

toabc

dq

*av

ipkks

+

*bv

*cv


Vector Control of a Permanent-Magnet Synchronous-Motor Drive



Simulation Results


Salient-Pole Synchronous Machine

sd md sL L L= +

sq mq sL L L= +

fd md fdL L L= +

rd md rdL L L= +

rq mq rqL L L= +

sd sd sd md rd md fdL i L i L iλ = + +

sq sq sq mq rqL i L iλ = +rd rd rd md sd md fdL i L i L iλ = + +

rq rq sq mq sqL i L iλ = +

fd fd d md sd md rdL i L i L iλ = + +

-axisa

-axisd

-axisb-axisq

-axisc

•

× -axisa

-axisd-axisq

sqi

rqi

rdisdi

fdi

( )a

( )b

Figure 9-6 Salient-pole machine.


sd s sd sd m sqdv R idt

λ ω λ= + −

sq s sq sq m sddv R idt

λ ω λ= + + ( 0)rd rd rd rd

dv R idt

λ=

= +

( 0)rq rq rq rq

dv R idt

λ=

= +fd fd fd fd

dv R idt

λ= +

Winding Voltages

+

−

+

−

+

−

+

−

+

−

( ) -axisa d

( ) -axisb q

sdv

sqv

sqi

sdi

rqi

rdi

fdi

rdv

rqv

fdvsR m sqω λ sL rdL rdR

mdL

sR m sdω λ sL rqL rqR

mqL

fL fdR

Figure 9-7 Equivalent circuits for a salient-pole machine.


field+damper in d-axis saliency

[ ( ) ( ) ]2em md fd rd sq sd sq sd sq rq sdmqpT L i i i L L i i L i i= + + − −

Electromagnetic Torque

( )2em sd sq sq sdpT i iλ λ= −

sd sd sd md rd md fdL i L i L iλ = + +

sq sq sq mq rqL i L iλ = +


Space Vector Diagram in Steady State

sd sd sd md fdL i L iλ = + sq sq sqL iλ =

sd s sd m sq sqv R i L iω= − sq s sq m sd sd m md fdv R i L i L iω ω= + +

sd sq s sd s sq m sd sd m md fd m sq sqv jv R i jR i j L i j L i L iω ω ω+ = + + + −

23sd sq sv jv v+ =

23sd sq si ji i+ =Figure 9-8 Space vector and phasor diagrams.

-axisd

-axisq

sdv 2 / 3 sv

2 / 3 si

sqv

aV

aI

Re

Im

( )a ( )b


Chapter 10

Switched-Reluctance Motor (SRM) Drives


mechθo0mechθ =

da

ad

c c

b

b

Figure 10-1 Cross-section of a four-phase 8/6 switched reluctance machine.

Cross-Section of a Switched-Reluctance Machine


mechθo0mech

al

θθ

==

b

a

a

b

c c

d

d

mechθo0mechθ =

b

a

a

b

c c

d

d

unθ

(a) (b)

Figure 10-2 Aligned position for phase a; (b) Unaligned position for phase a.

Aligned and Unaligned Positions for Phase-a


unaligned position

aligned position

(V-sec)aλ

(A)aiFigure 10-3 Typical flux-linkage characteristics of an SRM.

Typical Flux-Linkage Characteristics


Figure 10-4 Calculation of torque.

aλ

ai1I

1λ2λ

0

12

1θ1 mechθ θ+ ∆

Calculation of Torque

mech em mechW T θ∆ = ∆

1 2(1 2 1)elecW area λ λ∆ = − − − −

2 1(0 2 0) (0 1 0)storageW area areaλ λ∆ = − − − − − − −

mech elec storageW W W∆ = ∆ − ∆

1 2 2 1

1 2 1 2(0 1 2 0)2

(1 2 1) (0 2 0) (0 1 0) (1 2 1) (0 1 0) (0 2 0)

(0 1 2 0)

em

area

T area area areaarea area area

areaλ

θ λ λ λ λλ λ λ λ

− − − −

∆ = − − − − − − − − − − − −= − − − − + − − − − − − −

= − − −

(0 1 2 0)em

mech

areaTθ− − −=

∆ constantem

mech ia

WTθ =

′∂=∂

© Copyright Ned Mohan 2001 118Figure 10-5 Performance assuming idealized current waveform.

( )ai A

( sec)a

Vλ−

,

( )em aTNm

( )ae V

(deg)mechθ

unθ

unθalθ

alθ

Waveforms Assuming Ideal Current Waveforms

a a av Ri e= +

( , )a a a mechde idt

λ θ=

a aa a mech

a mech imech a

d de ii dt dtθ

λ λ θθ

∂ ∂= +∂ ∂

a aa mech mech

mech mechi ia amech

dedtω

λ λθ ωθ θ∂ ∂= =

∂ ∂


(A)ai

(V-sec)aλ

,

(Nm)em aT

onθ unθ

onθ unθ

onθ unθ offθ alθ

offθ alθ

offθ alθ

offθ alθonθ unθ

emT

,em aT,em bT ,em cT ,em dT(Nm)

Figure 10-6 Performance with a power-processing unit.

Performance with a Power Processing Unit


Figure 10-7 Flux-linkage trajectory during motoring.

0

emW

ai

aλ

fW

unalignedposition

alignedposition

offθ

Role of Magnetic Saturation

em

em f

WEnergyConversion FactorW W

=+


Dc1

Sc1

Pha

se a

Dd1

Dc2

Sa1

Sc2

Pha

se d

Sd1

Pha

se b

Sa2

Pha

se c

Db2

Da1 Db1

Da2

Sd2Sb2

Dd2

Sb1

dcV

+

−

Figure 10-8 Power converter for a four-phase switched reluctance drive.

Power Processing Unit


∫v

−+ Σ

λ

i( , )iθ λ

mechθ

iR

Figure 10-9 Estimation of rotor position.

Determining Rotor Position for Encoder-less Operation


positions

*mechω

+−

refIPI

currentcontroller

powerconverter

gate

signals

aλ

Σ

mechω

Figure 10-10 Control block diagram for motoring.

Control in Motoring Mode

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