Electric Drives Note

Electric Drives Principles: From Basics to Advanced Vector Control and

Encoder-less Operation April 24-25, 2004

Day 2 Advanced: Dynamic Operation and Control of AC Drives (8:00AM – 5:00PM) 1. Representation in Phase Quantities 2. Dynamic Analysis by d-q Representation 3. Vector Control 4. Voltage Space-Vector PWM 5. Encoder-less Direct Torque Control 6. Vector Control of PMAC Drives 7. Switched-Reluctance Motor Drives 8. Synchronous-Reluctance Motor Drives

Reference Book:

Textbook Advanced Electric Drives: Analysis, Control and Modeling using Simulink® by Ned Mohan, year 2001, published by MNPERE. (See www.MNPERE.com for details.)

Chapter 1

Introduction to Advanced Electric Drives

Graduate Course in Electric Drives

Objectives:• Basics to Advanced Topics in 2 Semesters• Seamless Continuation of the First Course• Topics: Dynamic Modeling and Control• Approach/Tools

– dq-Windings based Analysis– Design Examples Using Simulink– Verification 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 Representation• Magnetics – Transformers• Feedback Controller Design• Space Vector Representation for AC Machines• Basic Calculations for Electromagnetic Torque• PMAC Drives – Space Vector Based Steady State

Operation• Induction Motor Drives – Space Vector based

Steady State Analysis

Design Examples• Induction Machine Initially Operating in Steady

state– Load Torque Disturbance at t=0.1 s– Control 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) Drives• Switched-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:

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

R R XX X

J kg m

= Ω = Ω = Ω= Ω = Ω

2-1© Copyright, Ned Mohan, 2001 Reference: Advanced Electric Drives www.MNPERE.com

Chapter 2

Induction Machine Equations in Phase Quantities: Assisted by Space

Vectors

c axis

a axis

b axis

magneticaxis of phase a

′4′3

65 4 3

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

θ θ= ( )(4) ( ) ( ) i coso sa o a a

NB Hpµθ µ θ θ= =

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

θ θ θ= =

Figure 2-2 Three-phase windings.

axisa −

axisb −

axisc −/j4 3e π

/j2 3e π

(a) (b)

Three-Phase Sinusoidally-Distributed Windings

a ax is−

leak age

m agn etiz in g

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

inductance.

Single-Phase Magnetizing Inductance ,1m phaseL −

,(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

πµ− =

Figure 2-4 Mutual inductance .

Stator Mutual Inductance mutualL

(1) bmutual

a i i rotor openb c

,11(3)2mutual m phaseL L −= −

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

a axis−

b axis−

c axis−

A axis−

B axis−

C axis−

−− +

−−

vA = 0

vB = 0

vC = 0

Stator circuit Rotor circuit

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

Equivalent Windings in A Squirrel-Cage Rotor

,13(2)2m m phaseL L −=

23(3) ( )2

(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

( )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.

axisa −

axisb −

axisc −

/j4 3e π

/j2 3e π

at time tmagnetic axis of hypothetical winding

ˆwith current sI

axisa −

(a) (b)

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=

axisa −αα

ˆa aI I α= ∠−

Relationship between space vector and phasor in sinusoidal steady state

as ati I

3ˆ ˆ2s aI I=

ic+−vc

+−vb

+− va

λa t( )

λ b t( )

λ c t( )

a axis−

c axis−

b axis−

e j4 3π /

e j2 3π / i s s,i

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−

A axis−

B axis−

C axis−

, Br r irFr r ir

λat time t' '

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

a axis−

b axis−

c axis−

A axis−

B axis−

C axis−

, Br r irFr r ir

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−

A axis−

B axis−

C axis−

−− +

−−

vA = 0

vB = 0

vC = 0

Stator circuit Rotor circuit

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

(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 tdtλ= +

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+ =

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

(3 / 2)

(using Eq. 2-12)

m phase

axisa −

axisb −

axisc −

axisa −

axisb −

axisc −

axisd −

axisq −

axisa −

axisb −

axisc −

axisd −

axisq −

projection

projectionsq2i3

= ×projection

projectionsd2i3

(b) (c)

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= + +

( )( )/

F ti tN p

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

dAθrqi

a axis−

dωd axis−

q axis−

B axis− A axis−mω

daθ mθ

••

projection2 projection3rqi = ×

projection2 projection3rdi = ×

a axis−stator

A axis−rotor

d axis−

q axis−at t

irq 32

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

irq 32

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 →

(a) stator

[ ]ABC dqrT →

(b) rotordaθ dAθ

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

[ ]abc dqsT →

(a) stator

[ ]ABC dqrT →

(b) rotordaθ dAθ

[ ]abc dqsT →

(a) stator

[ ]ABC dqrT →

(b) rotordaθ dAθ

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

ada da dasdb

sqda da da c

Ts abc dq

i ti t

π πθ θ θ

− − = − − − − − [ ]

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

AdA dA dArdB

rqdA dA dA C

Tr ABC dq

i ti t

π πθ θ θ

− − = − − − − −

Mathematical Relationship between dq and phase Winding Variables

-axisa

-axisd

-axisα

-axisβ

sdisqi

si β- axisq

Figure 3-5 Stator and equivalent windings.dqαβ

(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 jdtλ ω λ= + +

(3) sd s sd sd d sqdv R idtλ ω λ= + −

(4) sq s sq sq d sddv R idtλ ω λ= + +

[M ]rotate

0 1(5)

1 0sd sd sd sd

s dsq sq sq sq

v i dRv i dt

λ λω

λ 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

rdirqi

- axisq

Figure 3-6 Rotor and equivalent windings.dqαβ

-axisA

Derivation of Rotor Voltages in dq Windings

(1) rd r rd rd dA rqdv R idtλ ω λ= + −

(2) rq r rq rq dA rddv R idtλ 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 dt

λ λω

λ λ −

[M ]rotate

0 11 0

sd sd sd sds d

sq sq sq sq

v id Rv idt

λ λω

λ λ −

= − − [M ]rotate

0 11 0

rd rd rd rdr dA

rq rq rq rq

v id Rv idt

λ λω

λ 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 ]

dω ×

_s dqddtλ

_r dqddtλ

1[ ]M −

Eq. 3-62

_s dqi

_r dqi

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

a axis−

d axis−q axis−

subtract

due to and sq rqi i

due to rqileakage flux

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

a axis−

d axis−q axis−

subtract

due to and sq rqi i

a axis−

d axis−q axis−

subtract

due to and sq rqi 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

N LB i ip L

µ = +

,3 / 2 ˆ(2)

d rotor rq rdNpT r B ip

,3 / 2(3) ( )

d rotor sq rq rdg m

Np LT r i i ip L

,3(4) ( )

2 2s r

d rotor sq rq rdg m

Np LT r i i ip L

µπ = +

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

p pT L i L i i i

λ= + =

subtract

due to and sd rdi i

due to rdileakage flux

q axis−d axis−

a axis−

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

subtract

due to and sd rdi i

q axis−d axis−

a axis−

subtract

due to and sd rdi i

q axis−d axis−

a axis−

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

p pT L i L i i i

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

T Tddt Jω −

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= + + + +

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

ω λ=

= − + + +

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

ω λ=

= + + + +

−−

− −

d sdω λ

d sqω λ

dA rqω λ

dA rdω λ

sdddtλ

sqddtλ

rdddtλ

rqddtλ

(a) -axisd

(b) -axisq

Figure 3-5 -winding equivalent circuits.dq

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

( )m mj L jXω =

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

sq sqs m

m rrq rqL

iL LL L i

0 00 0

sd sds msq sqs m

m rrd rd

m rrq rqM

iL LiL L

L L iL L i

Induction Motor Model

_s dqv

_r dqv

_s dqλ

_r dqλ

rotate[M ]

dω ×

_s dqddtλ

_r dqddtλ

1[ ]M −

Eq. 3-62

_s dqi

_r dqi

Eq. 3-47

eqsJmechω

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

Calculation of Steady State Initial Conditions Using Phasors

3ˆ ˆ(0)2

j ia a i s a

I I i I e θθ= ∠ ⇒ =

2 2 3 ˆ(0) (0) - = cos( )3 3 2sd s i

i projection of i on d axis I θ = ×

2 2 3 ˆ(0) (0) - = sin( )3 3 2sq s i

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= + [ ]

s syn s syn m sdsd

syn s s syn m sqsq

syn m r syn r rd

rqsyn m syn r r

R L L ivL R L iv

s L R s L iis L s L R

− − = − −

i vi v

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

− aeav

em T aT k i=emT

Tem = kT ia

Brushless DC Motor Drive

( )rB t

( )si t

o90 NS axisa −

em T sT k I=emT

ˆ em T ST k I=

Vector-Controlled Induction Motor Drive

axisa −'( )ri t

'( )rF t( )rB t

( )rF t

( )msB t( )lrB t

( )sv t

speed feedback

currentfeedback

( ) 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

shown explicitly2R

net = 0

2v 0=2R

+ -2 2(0 ) = (0 ) = 0λ λ

+ + +m,i m,i 2(0 ) = (0 ) + (0 )φ φ φ

+ +m2 1

Li (0 ) = i (0 )L

Analogy to a Current-Excited Transformer With a Shorted

Secondary

(0 ) (0 )mLi iL

22 2( ) (0 ) ti t i e τ−+=

φmi, 2 φmi, 1

net =0

∴ = +φ φ φmi mi, ,1 2 2

t = +0

R shown licitly2 exp

L didt

dtL di

mφ , 2 2=

dtmi: ,φ

21 0+d i

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

1 2N N=

2 2mL L L= +

R1 R2i2

00 tt im

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

L Li i iL L L

+ + += =+

2 21 1

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

L Li i iL L L

+ + += =+

22 2( ) (0 ) ti t i e τ−+=

d- and q- Axis Winding Representation

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

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

a axis−stator

A axis−rotor

d axis−

q axis−at t

irq 32

d axis−d axis−

ωmωm

isdisd

isqisq

i rqi rq

irdird

θmθm

irq 32

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=

t−∞ 0

, sdm iφ

t 0−=

- - -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

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

LˆT α B , iL

-t = 0

Flux Densities at +t = 0

lrBmsB

axisa −

Transformer Analogy – Voltage Needed to Prevent the Decay of

Secondary Current

( ) ( )2 2 2v R i 0 u t+′ ′ ′=

l2L′l1L

( )1i u t⋅

1R t 0>

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

m 0ω =

net = 0

axisa −

slipωlrφ

, rm iφ, sqm iφsqi

slipω

axisd −

axisq −

r m r sqslip

R , (L /L )i

Bω α

Vector-Controlled Condition With a Rotor Speed ωm

net = 0

axisa −

synωlrφ

, rm iφ, sqm iφsqi

axisd −

axisq −

syn m slipω = ω + ω

Torque, Speed, and Position Control

(measured)/d dt

encoder

PPUregulated

current

to abcdq

to dqabc

*ai*bi*ci

aibici

rBθcalculations

mθmω

(measured)mθ mω emT

(measured) (calculated)

P PI PI

∑∑ ∑

−−+ +

+ PImω(measured)

ˆrB (calculated)

(measured)mωmotor mathematical model

syn m slipω (t) = ω (t) + ω (t)

B syn0θ (t) = 0 + ω (τ) dτ∫ ⋅

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

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

Σ ( )Aq t

phasea

( )ai t

dV+−

( )m tθ

reference current

actual current

.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−

q axis−at t

irq 32

d axis−d axis−

ωmωm

isdisd

isqisq

i rqi rq

irdird

θmθm

irq 32

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

= −( ) 0rqd tdtλ =

−−

− −

d sdω λ

d sqω λ

sqidA rdω λ

0rdv =

0rqv =

sdddtλ

sqddtλ

rdddtλ

0rqddtλ =

(a) -axisd

(b) -axisq

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

em rd rq rd sqr

Lp pT i iL

= − =

rdddtλ rdi

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

D-Axis Rotor Flux Dynamics

(1) ( ) ( )mrd sd

sLi s i sR sL

= −+

(2) rd r rd m sdL i L iλ = +

(3) ( ) ( )(1 )

mrd sd

Ls i ss

(4) rd mrd sd

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

rdλrdλ

N/D1/ sdaθ

emTmω

mechω

dω dAωmL

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.

Current −Regulated

( )mech measuredωmechθ

mod elMotor

of Fig.49

iaibic

Encoder

Transformations

Eq. 107

Eq. 103

Motor Model(Fig. 5-4)

abctodq

dqtoabc

Speed and Position Loops for Vector Control

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

(measured)/d dt

PPUregulated

current

to abcdq

to dqabc

*ai*bi*ci

aibici

sqiemTrdλdaθ

Fig. 5-4

mechθmechω

(measured)

mechθ mechω emT

(calculated)

P PI PI

∑∑ ∑

−−+ +

+ PImechω *

rdλ*rdλ

mechω

rdλ (calculated)

*mechθ

*mechω *

(measured)

mechωEstimated Motor Model

Design of Speed Loop

rd m sdL iλ =

em sd sqrk

LpT i iL

mechωk

( )sqi si

*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

σ = −

(2) msd s sd rd

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.

∑ ∑

Motorvsd*

Transformations

M o to rM o d e l

Encodermechθ( )mech measuredω

Transformations

Modulated

SpaceVector

rdλemT

,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

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

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

,da estθdaθ

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

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

,da estθdaθ

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

,da estθdaθ

Effect of Detuning due to Incorrect Rotor Time Constant

,(1) r

r estkτ

( ) ( ), ,,_ _ _(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

,da estθdaθ

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

,da estθdaθ

-axisd

-axisestd

-axisa

-axisq

-axisestq

_s dqi

,da estθdaθ

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)

dqtoabc

CR-PWMInv.

abctodq

dAω 1s

,rd estλ,1

,r estτ

,dA estθ

abctodq

D,dA estω

*,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

daθ ,da estθ

,i estsθ

Calculations of Steady State Errors

1 sqdA

dA estsd

ir estω

1 1sq sq

r est r sdsddAdA est

τ τ=

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

-axisd

-axisestd

-axisa

-axisq-axisestq _s dqi

daθ ,da estθ

,i estsθ

Calculations of Steady State Errors (continued) *

* 2 21

i mi k mτ

* *sq sd

i iki i

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

,err i est is sθ θ θ= −

1 ( )em

T mkT k m

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

+−av

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

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

12 / 3

34 / 3

45 / 3

(000) 0

(111) 0

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

v v V e

Synthesis of Voltage Vector in Sector 1

jdv V e=

jdv V e π=

ˆ j ss sv V e θ=

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= +

/ 2sT7z

0 / 2z 0 / 2z

/ 2y/ 2y

/ 2x / 2xaNv

,control avtriv

,control bv

,control cv

ˆ / 2

control a a k

control b b k

control c c k

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

ˆFigure 7-7 Limit on amplitude .sV

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

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

aibici

∑ ∑PI+ +− −

+∑−

*mechω

mechω

*ˆsλ

sθ∠

Estimator

Selectionof

Figure 8-1 Block diagram of DTC.

Principle of DTC Operation

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

( )s tλ

( )s t Tλ −∆sv T∆i

( ) ( )r rt t Tλ λ≈ −∆

-axisa

Rotor -axisAArθ

2ˆ ˆ sin

em s r srLpTLσ

λ λ θ=

sr s rθ θ θ= −

slip rr

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

θ θω θ − − ∆= =

Im( )2

conjem s s

pT iλ=

σ = −

Calculation of Stator Flux:

Calculation of Rotor Flux:

Estimating Torque:

Estimating Mechanical Speed:

s s s sdv R idtλ= + ⇒

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

Figure 8-3 Inverter basic vectors and sectors.

Stator Voltage Vector Selection in Sector 1

sector 1sλ

4v 5v6v

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

Selection of the Stator Voltage Space Vector

sector 1sλ

4v 5v6v

sector 1sλ

4v 5v6vsector 1sλ

4v 5v6v

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

increasedecrease

decreasedecrease

decreaseincrease

increaseincreasesv

emT ˆsλ

a-axis1(001)v

3(011)v2(010)v

6(110)v

4(100)v 5(101)v

b-axis

c-axis

Figure 8-3 Inverter basic vectors and sectors.

Effect of Zero Stator Voltage Space Vector

( ) ( )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−

axisc −

axisa −

axisb −

-axisq

-axisd

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

sd s sd sd m sqdv R idtλ ω λ= + −

sq s sq sq m sddv R idtλ 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

T Tddt Jω −

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

v R i j L i jω ω λ= + +

23a s a m s a m fd

V R I j L I jω ω λ= + +

2ˆ3fa fd m E m

E kλ ω ω= =

23E fdk λ=

m mLωm sLω

m lsLω

ˆ( )fa fa E mE E k ω=

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

Controller in the dq Reference Frame

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

ω= + + −

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

ω λ= + + +

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

Figure 9-3 Controller in the dq reference frame.

Σ*sqi

( )2m mechpθ θ=

sensorposition

abctodq

( )m s sd fdL iω λ+

m s sqL iω−

*sqv Inverter

mωsqi

decouplingterms

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

rdisdi

Figure 9-6 Salient-pole machine.

sd s sd sd m sqdv R idtλ ω λ= + −

sq s sq sq m sddv R idtλ ω λ= + + ( 0)

rd rd rd rddv R idtλ

( 0)rq rq rq rq

dv R idtλ

= +fd fd fd fd

dv R idtλ= +

Winding Voltages

( ) -axisa d

( ) -axisb q

fdvsR m sqω λ sL rdL rdR

sR m sdω λ sL rqL rqR

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

( )a ( )b

Chapter 10

Switched-Reluctance Motor (SRM) Drives

mechθo0mechθ =

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

Cross-Section of a Switched-Reluctance Machine

mechθo0mech

mechθ

o0mechθ =

(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.

1λ2λ

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)

T area area areaarea area area

areaλ

θ λ λ λ λλ λ λ λ

− − − −

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

= − − −

(0 1 2 0)em

areaTθ− − −

=∆ constant

emmech ia

WTθ =

′∂=∂

Figure 10-5 Performance assuming idealized current waveform.

( )ai A

( sec)a

Vλ−

( )em aTNm

( )ae V

(deg)mechθ

unθalθ

Waveforms Assuming Ideal Current Waveforms

a a av Ri e= +

( , )a a a mechde idtλ 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∂ ∂

= =∂ ∂

(V-sec)aλ

(Nm)em aT

onθ unθ

onθ unθ offθ alθ

offθ alθ

offθ alθonθ unθ

,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.

unalignedposition

alignedposition

Role of Magnetic Saturation

WEnergyConversion FactorW W

Da1 Db1

Sd2Sb2

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

Power Processing Unit

−+ Σ

i( , )iθ λ

mechθ

Figure 10-9 Estimation of rotor position.

Determining Rotor Position for Encoder-less Operation

positions

*mechω

refIPI

currentcontroller

powerconverter

signals

mechω

Figure 10-10 Control block diagram for motoring.

Control in Motoring Mode

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