pmsm foc dtc

UPC

PMSM Control (FOC & DTC) for WECS & Drives

Dr. Antoni Arias. Universitat Politècnica de Catalunya. Catalonia. (Spain)

1/98

UPC

• Introduction & motivation• Permanent Magnet Synchronous Machine (PMSM)

– Dynamic equations– Modeling

• Vector Control Strategies– Field Oriented Control (FOC)

• Current & Speed Loops• PI & IP controllers: root locus, specifications, tuning and DSP C code

Guidelines

–Direct Torque Control (DTC)• VSI review • Flux & Torque Loops• Hysteresis Comparators

• Space Vector Modulation (SVM)

• Case study: Wind Turbine• Experimental approach:

– Platforms– Implementing issues

2/98

UPCIntroduction & Motivation

• An electric drive is an industrial system which performs the conversion of electrical energy to mechanical energy (in monitoring) or vice versa (in generator braking) – Production plants, transportation, home appliances, pumps, air

compressors, computer disk drives, robots, etc…

Regenerative braking

3/98

UPCIntroduction & Motivation

Field Orientated Control (FOC) & Direct Torque Control (DTC) deserve special attention since both are high performance vector controllers.

• Electric drives involves many applications ranging from below 1kW up to several dozens of MW.

High performance means: (i) wide speed range,

(ii) fast and precise response in speed or position.

4/98

UPC

• Electric machines drives absorb a large amount of the electrical energy produced. – 50% of the electrical energy is used in electric drives today (1).– In the industrial countries, EM take around 65% of the entire electrical

energy available (2).(1) I. Boldea, S. A. Nasar. “Electric Drives”. CRC Press. Second Edition. 2006

(2) Kazmierkowski, M. P; Tunia, H. “Automatic Control of Converter-Fed Drives". Elsevier. Studies in Electrical and Electronic Engineering 46. 1994.

• Electric Drives market by 2017: US$ 62.3 billion by 2017 (3). (3) Global Industry Analysts, Inc.

• Sensorless and High Performance Controls are still hot research topics (4).

(4) IEEE Xplore & Elsevier

Introduction & Motivation

5/98

UPC

• AC versus DC: (i) brushless (ii) higher torque density (iii) lower initial and maintenance costs, (iv) power electronics for AC and DC have comparable prices.

AC versus DC electric drives market evolution (1)

(1) I. Boldea, S. A. Nasar. “Electric Drives”. CRC Press. Second Edition. 2006


6/98

UPC

• DC • AC

– Induction– Switched Reluctance. Stepper. Hybrid.– Permanent Magnet

• PMDC – Brushless• PMSM: SMPMSM, IPMSM

• Linear• Nano• etc...

PMSMs are gaining market becoming a key point for the

sustainable development of the present century.


SRMs possess potential market. Second machine most used in China !http://green.autoblog.com/2011/06/28/need-rare-earth-

metals-switched-reluctance-motors-dont/

clas

sific

atio

n

7/98

UPCPermanent Magnet Machines

• Magnetic material to establish the rotor flux.– Neodymium-iron-boron (Nd2Fe14B) magnet is the strongest type of

rare-earth permanent magnet, which was developed in 1982 by GM.

• Advantages:– No rotor currents => no rotor losses => higher efficiency => energy

saving capability. Attractive for Wind applications.– Smaller rotor diameters, higher power density and lower rotor inertia.– Higher torque per ampere constant.– Less weight and volume for the same power. Attractive for aerospace

applications such as aircraft actuators.– Other applications: machine tools, position servomotors (replacing the

DC motors).

• Inconvenience: – Synchronous machines => need of rotor position. Sensorless control.– Price.– Permanent magnet is becoming a scarce raw material.

8/98


Mavilor. Might be used in Electrical Vehicles

1,3 - 31,8 (Nm) 6000 - 9000 (rpm)

Unimotor. Emerson. Control Techniques. 0,72-73,2 (Nm) 2000-6000 (rpm) 230/400(V)

General purpose

9/98


PMSG 13.5kWMulti-pole. External rotorWind turbine laboratory rig

(in collaboration with Robotiker - Tecnalia)

4Q Motor Drive System4kW DC – PMSM 1 kW

General purpose laboratory rig

10/98

UPC

• Considering the shape of the back EMF• PMDC – brushless DC (BLDC) - trapezoidal• PMAC – sinusoidal

Surface Mount Permanent Magnet Synchronous Machine (SMPMSM)

Permanent Magnet Machines

Interior Mount Permanent Magnet Synchronous Machine (IMPMSM)

11/98

UPCPMSM Dynamic Equations

• Combines the individual phase quantities into a single vector in the complex plane

• Same transformation is applied to all other magnitudes such as voltages, fluxes, etc…

invariant power invariantpower non

)()()(

32

32

34

32

c

etietiticij

c

j

ba

• Space vector transformation

12/98


• Basic equation for phase windings voltages

• Total flux linkage

flux produced by the rotor magnet

Leakage inductance

Magnetising inductance

Self inductance

Mutual inductance

c

b

a

c

b

a

s

c

b

a

dtd

iii

Rvvv

)cos()cos(

)cos(

343

2

r

r

r

m

c

b

a

ccbca

bcbba

acaba

c

b

a

ψiii

LMMMLMMML

ψψψ

mψ

lL

mL

mlcba LLLLL

2m

cabcabLMMM baab MM; = cbbc MM; = acca MM; =

a

b

c

i

ib

ic

b

c

a

ava +-

vc+

-

-

+vb

m

r

13/98


• Voltage vector equation in the stationary - frame

)cos(

)cos(23

2

r

rmmls dt

dψii

dtdLL

ii

Rvv

i

v +-

m

r

i

v

+

-

a

b

c

i

ib

ic

b

c

a

ava +-

vc+

-

-

+vb

m

rClarke

transformation


º30cosº30cosº60cosº60cos

32

32

c

xxcxxxxcx

cb

cba

14/98

UPC

i

v +-

m

r

i

v

+

-

a

b

c

i

ib

ic

b

c

a

ava +-

vc+

-

-

+vb

m

r

invariant power

invariantpower non;

23

230

21

211

32

32

cxxx

cxx

c

b

a

invariant power invariantpower non1

;

23

21

23

21

01

32

cxx

cxxx

c

b

a

Clarke-1 Transformation

Clarke Transformation

• Non power invariant versus power invariant. See the comparison.

PMSM Dynamic Equations

15/98


• Saliency: Variation of the stator phase inductance as function of the rotor position.

)2cos(

)2cos(

)2cos(

32

34

rmmla

rmmlb

rmmla

LLLL

LLLL

LLLL

)2cos(2

)2cos(2

)2cos(2

34

32

rmm

ca

rmm

bc

rmm

ab

LLM

LLM

LLM

m

d

q

mml

rmmla

LLL

)θ2cos(LLLL

Δ-+=

=Δ-+=

m

r

dq

ml

rmmla

LL

)θ2cos(LLLL

+=

=Δ-+=

m

r

d

q

mml

rmmla

LLL

)θ2cos(LLLL

Δ++=

=Δ-+=

16/98


• voltage vector equation in the stationary - frame considering saliency

• Where

• If there is no saliency and it is obtained the previous equation.

)cos()cos(

)2cos()2sin()2sin()2cos(

2

r

rm

rssrs

rsrsss

dtdψ

ii

LLLLLL

dtd

ii

Rvv

msmls LLLLL 23;

23

0mL

17/98


• Voltage vector equation in the synchronous reference d/q frame fixed to the rotor • Angle chosen equal to the PM position• Direct axis ( Ld = LS

-ΔLS ) and quadrature ( Lq = LS+ΔLS ) axis inductances

m

r

d

id+

-vd

vq

+

-

iq

q

Parktransformation

i

v +-

m

r

i

v

+

-

10

rmq

d

qrd

rqd

q

ds

q

d ψii

dtdLLLdtdL

ii

Rvv

18/98

UPC

i

v +-

m

r

i

v

+

-

xx

xx

q

d

cossinsincos

m

r

d

id+

-vd

vq

+

-

iq

q

q

d

xx

xx

cossinsincos


Park Transformation

Park-1 Transformation

19/98

UPC

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25

Time [ s ]

[ A ]

ialpha

ibeta

ibiciqid

• Non power invariant • Different constant values. (c=2/3) for Clark & (c=1) for Clarke-1.

• Same amplitudes for (a,b,c) & (α,β) & (d,q).


20/98

UPC

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35-0.4

-0.3

-0.2

-0.1

0

0.1

0.2

0.3

Time [ s ]

[ A ]

ialpha

ibeta

ibiciqidia

• Power invariant • Same constant value. ( c=sqrt(2/3) ) for Clark & Clarke-1.• Different amplitudes for (a,b,c) versus (α,β) & (d,q)


21/98

UPC

c

b

a

q

d

xxx

cxx

23

230

21

211

cossinsincos


;

32sin3

2sinsin3

2cos32coscos

32

32

cwhere

xxx

cxx

c

b

a

q

d

• Direct (Clarke & Park) transformation


22/98

UPC

q

d

c

b

a

xx

cxxx

cossinsincos

23

21

23

21

01

invariant power invariantpower non1

;

32sin3

2cos3

2sin32cos

sincos

32

cwhere

xx

cxxx

q

d

c

b

a

• Direct (Clarke & Park) Inverse transformation


23/98


• Instantaneous torque expression:

qdqdqm

dqqqddm

dqqde

LLiiiPciiLiiLPc

iiPciPcT

fluxmagnet permanent theis psi - andnumber poles'pair theis ,

invariant power 1invariantpower non

where; 23

m

P

c

• First term, usually called as magnet torque, is directly proportional to iq and independent of id .

• Second term, or reluctance torque, is only present in salient machines where ( Ld - Lq ) ≠ 0 and is proportional to the current product ( id ·iq).

24/98

UPC

function transfer;

equation ldiferentia ;

DJsTTw

TTwDdt

dwJ

Lem

Lemm

• Motion equation• Drive Inertia ( J )• Damping constant due to Friction ( D or F )• Load Torque ( TL )


25/98

UPC

Exercise: Sketch a complete PMSM scheme model• Use the d/q model equations• Find out all the required equations involved• Determine where the Clarke (a,b,c) to (α,β) & Park (α,β) to (d,q)

transformations should be applied to get a more realistic model

PMSM Modeling

26/98

UPCPMSM Modeling

27/98

UPCPMSM Modeling

•The adopted parameters for the model are as follows: Lq=0.009165; Ld=0.006774; R=1.85; psi=0.1461; P=3.(specific motor data is on the next page)

•For the motion equation of the whole drive it can be used: J=0.77e-3; F=1e-3.

28/98

UPC

unimotor fmEmerson Control Techniques

Model: 115E2A300BACAA115190

Number of poles: 6

Rated speed: 3000 (rpm)

Rated torque: 3.0 (Nm)

Rated power 0.94 (kW)

KT: 0.93 (Nm/Arms)

KE: 57.0 (Vrms/krpm)

Inertia: 4.4 (kgcm²)

R (ph-ph): 3.7 (Ohms)

L (ph-ph): 15.94 (mH)

Continuous stall: 3.5 (Nm)

Peak: 10.5 (Nm)

PMSM Modeling

• Ld and Lq values?

• ΔLs can be obtained measuring the inductance for different rotor positions.

• The first order system, composed by an L and R, can be identified by supplying a step voltage in the d and q components.

How to obtain the parameters from a commercial catalogue?

ssdssq

dqdqs

LLLLLL

LLLs

LLL

;2

;2

29/98

UPC

• How to obtain the J and F values? • The first order system, composed by a F and F, can be identified by

supplying a step torque. • Usually, the machine manufacturer provides the J value. In a laboratory

rig composed by two machines, the total J value can be then worked out from the manufacturer data.

• Why KT and KE are not equal? • The units must be in the International System.

• Experimental (taken in the laboratory) KE value is 56.22 instead of 57.

rms

rmsANm

radrevs

rpmV

TE

Fm

Lstar

connectedFm

FmFF

KK

ITV

ITVTIV

929.02

1160

100022.563;3

333;3

min

PMSM Modeling

30/98

UPC

•How to obtain the value?

From the torque expression, and the KT (Nm/Arms) parameter, it can be deduced the following expression:

Finally, the numerical value will be:

And the factor is used to convert the rms to the maximum value.

qme iPcT

PcKPcK

iT T

mmTq

e

WbPc

KTm 1461.0

323

93.0 21

m

21

PMSM Modeling

31/98

UPCPMSM Modeling

Lq=9.1e-3 Ld=6.7e-3R=1.85psi=0.1461 P=3J=0.77 F=1e-3

Ld=Lq=0.016416 R=0.656psi=3.184 P=10J=206.05 F=1.5

32/98

• Parameters are PMSM (L, R, psi, P) and whole drive (J, F) dependent.

UPCTypes of Controllers

33/98

UPCField Oriented Control (FOC) for PMSM

• Instantaneous torque for the PM synchronous machine

m

r

dq

Iq>0

m

r

dq

Iq<0

Te > 0 Te < 0

)( 23

qdqdqme LLiiiψPT

• If id = 0, for not demagnetizing the PMSM, the reluctance torque is 0.• Therefore, the electromagnetic torque will be regulated with iq .

• Duality with the DC Machine control.

qm iψP 23

id = 0

34/98

UPCFOC in PMSM

Four quadrant Motor (Q1 & Q3) Generator (Q2 & Q4) 35/98

UPCFOC Scheme

Typical dynamic values (for a PMSM 0.5 / 10 kW)

discretization time / switching frequency

Closed loop time response / band width

Current 100 µs / 10kHz 5ms / 200HzSpeed 5ms / 200Hz 0.5s / 2Hz (inertia dependent)

• 3 PI control loops: • 2 inner (and faster) current loops.• 1 outer (and slower) speed loop.

Limited by the semiconductors

switching frequencySeparation principle will be applied

36/98

UPCCurrent Controller

• The current loop, for controlling design purposes, can be modeled as a first order system.

s

s

s

RLsRL

Rvi

;1

1

10

emq

d

qed

eqd

q

ds

q

d ψii

sLLLsL

ii

Rvv

11

sRLR

sd

s

11

sRLR

sd

s

11

sRLR

sq

s

11

sRLR

sq

s

• The coupling terms are eliminated, since they are considered as a perturbation.

37/98

UPC

• The current loop can be modeled as a first order system, whose transfer function in Laplace domain is:

)(99.01)(;5)(98.01)(;4)(95.01)(;3)(632.01)(;

)()1(1)(

tvRtittvRtittvRtittvRtit

tveRti

s

s

s

s

t

s

4

1)(1

)()(

sRLR

sVsI

s

s

• Its open loop step time response is:

Current Controller

38/98

UPC

0 0.005 0.01 0.015 0.02 0.025 0.030

0.1

0.2

0.3

0.4

0.5

0.6

0.7

System: gTime (seconds): 0.0172Amplitude: 0.531

System: gTime (seconds): 0.0298Amplitude: 0.54

Step Response

Time (seconds)

Am

plitu

de

• Considering the motor parameters given before:

• Time constant:

• Time response at 98%

• Final value:

)(23.174 ms

185.110·97.785.11

)()(

3 ssV

sI

)(31.485.110·97.7 3

ms

54.085.11

Current Controller

39/98

UPC

0 0.01 0.02 0.030

0.2

0.4

0.6

0.8

-1000 -500 0

-500

0

500

Open loop s = -Rs/L = -232 = L/Rs = 4.3(ms)

KP=1s = -(KP+Rs)/L = -358

= L/(KP+Rs) = 2.8(ms) Final value: KP/(RS+KP) = 0.35

KP=10s = -(KP+Rs)/L = -1486 = L/(KP+Rs) = 0.6(ms)

Final value: KP/(RS+KP) = 0.84

Root locus

Step response

i* i 1

1sRL

R

s

s

PK• P controller always introduces the Position Error E0

E0 = Rs / (Rs+KP)

-Rs / L

• Solution: add a PI controller

Current Controller

40/98

UPC

-3000 -2000 -1000 0

-1000

0

1000

KP = 14.09 KI = 15945 Closed loop poles -1000+1000j

• Specifications:Settling time at 2% = 4ms

Overshoot 4.23% Damping =0.707

Root locus

-Rs / L 0-KI / KP

11

sRLR

s

s

sKsK IP

Current Controller

41/ 98

UPC

Step response

The overshoot (17%) is too large ! The settling time is not fulfilled !

• The integrator makes the Error E0 = 0• The specifications are not fulfilled:

• What is the problem ?

Current Controller

42/98

UPC

LK

LK

LRss

sKK

LK

i

iIPs

I

PI

2

1

*

22

2

2 nn

n

ss

• What is the closed loop transfer function?

• Is it a pure 2nd

order term? NO! There is an unwanted zero !

A pre filter is used to cancel the zero 1

1sRL

R

s

s

sKsK IP

11

sKK IP

LK

LK

LRss

sKK

LK

sKK

i

iIPs

I

PI

I

P

2

1

1

1*

LK

LK

LRss

LK

i

iIPs

I

2*

Current Controller

43/98

UPC

0 1 2 3 4 5 6 7x 10-3

0

0.5

1

Time (s)

LK

LK

LRss

LK

LK

LK

LRss

LK

sKK

LK

LK

LRss

sKK

LK

i

iIPs

I

IPs

I

I

P

IPs

I

PI

222

1

*

11

sRLR

s

s

sKsK IP

11

sKK IP 11

sRLR

s

s

sKsK IP

zero

PIPF+PI

Current Controller

44/98

UPC

+

-

i* i 1

1sRL

R

s

s

sK I +

-

PK+

-

i* is

K I

1

1

sKR

LKR

Ps

Ps

Root locus

-(Rs +KP )/ L 0

KP = 14.09 KI = 15945 Closed loop poles

-1000+1000j

Settling time at 2% = 4msOvershoot 4.23% Damping =0.707

Current Controller

45/98

UPCCurrent Controller

Fixing the settling time at 2% (Ts2%) and the damping factor (ζ) the KP and KI can be calculated

nTs

4%2

22

2

2 nn

n

ss

L

KL

KLRss

LK

i

iIPs

I

2*

LKR Ps

n

2

snP RLK 2

sP RTs

LK

%2

8

LKI

n 2

2nI LK

2%2

216

TsLKI

2nd order TF Current loop TF with PI or IP

Identifying coefficients

4.22 instead of 4 when ζ = 0.707 46/98

UPC

11

sRLR

s

s

sK I

PK 1

1sRL

R

s

s

sKsK IP

11

sRLR

s

s

sKsK IP

i 1

1sRL

R

s

s

sK I

+

+PK

+P-

LK

LK

LRss

sLi

IPs

2

1

P0 2 4 6 8

x 10-3

0

0.01

0.02

0.03

0.04

Step Response

Time (seconds)

Ampl

itude

PI IP

In terms of Perturbation, both solutions (PI & IP) have the same performance

Current Controller

47/98

UPC

11

sRLR

sd

s

11

sKK IP

• Decoupling terms have to be added:

11

sRLR

sq

s

• Current loop with PI + PF (d axis) and IP (q axis)

-

Lq·we·iq

Current Controller

48/98

UPC

• Implementing PI in a DSP– From S to Z domain

– Ts=100us

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8

x 10-3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1-z)Ts-1(-z

K)z(PIs

sK)s(PI P

IP

IK

K

P

KK

P =→+

=

1-z,920-z

5,5)z(PIs800s

5,5)s(PI =→+

=

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Root Locus Editor (C)

Real Axis

Imag

Axi

s

0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2Root Locus Editor (C)

Real Axis

Imag

Axi

sCurrent Controller

49/ 98

UPC

// start iq PI controlleriq_error = iq_ref-iq;

vq_ref = vq_ref_last+c_Kp*(iq_error-c_Ki_Kp*iq_error_last);iq_error_last = iq_error;

if (vq_ref > VPI_MAX) vq_ref = VPI_MAX;if (vq_ref < -VPI_MAX) vq_ref = -VPI_MAX;vq_ref_last = vq_ref;// end iq PI controller

• Implementing a PI Controller in a DSP– From Z to discrete time domain

– C code for the TI DSP 6711

1-z)Ts-1(-z

K)z(error_iq

)z(ref_vq)z(PI P

IK

K

P==

))]Ts-1(-z(K)[z(error_iq)1-z)(z(ref_vq PI

KK

P=

))]K_K_c(z-1(K)[z(error_iq)z-1)(z(ref_vq pi-1

P-1 =

))K_K_c(last_error_iq-error_iq(Klast_ref_vqref_vq piP+=

Current Controller

50/98

UPC

• The plant can be simplified as follows• First order with one pole • Mechanical time constant might be 50 times slower than the

electrical one. Current loop can be neglected.• Typical sampling time 5ms.

JD-s =

PITe*

DsJ1

e* TL

Speed Controller

51/98

UPC

s1

• Windup phenomenon– Limits of the real plant (currents and voltages)– Might end up with instability

• Anti Windup PI

Speed and Current Controller

52/98

UPCExercise

• Sketch a complete FOC scheme for a PMSM. It has to include:

Speed and currents control loops (PI + pre filter) or IP Saturations (software protections) Anti Windup Feed forward (decoupling) terms

• Draw the magnitudes (speed reference, speed response, iqreference and iq response) waveforms for the cases: (i) electrical car & (ii) lift. Indicate, in the drawn waveforms, the motoring or generating operating modes.

• Propose a programming code (any generic language) to implement a PI.

53/98

UPC

PMSM model in a b c All control in s domain Speed and currents control loops with: PI + pre filter,

Saturations (software protections) and Anti Windup. Feed forward (decoupling) terms for the current loops.

54/98

UPC

The PMSM model has been simplified to d q

55/98

UPC

PMSM model in a b c All control in z domain. Two different sampling periods: Tsm, Ts

56/98

UPC

• SISO tool. Matlab• Matlab/Simulink PMSM FOC. Waveforms.• FOC in a DSP

57/98

pmsm foc dtc

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