section 6. dynamics of electric drive systems section 6 - dynamics... · dcm has two windings...

Section 6. Dynamics of Electric Drive Systems

1

2

Review of steady state analysis of DC machine Dynamics of Separately Excited (SE) DC machine

Dynamics of Synchronous Machine (SM)

Rotor flux oriented control (RFOC) of SM

Dynamics of Induction Machine (IM)

Rotor Flux Oriented (FOC) Vector Control of IM

Outlines

3

Review of steady state analysis of DC machine Review of dynamics of Separately Excited (SE) DC machine





Outlines

Review of DC Machine Steady-States

4

Steady-state analysis (Va, ωm, Ia)

a a a E mV R I K

e T aT K I

c

a am e

E E T

V RT

K K K

If = CONST

Slope-Intercept Form

Review of DC Machine Steady-States

5

Steady states for field weakening (Va=CONST)

' ' ' 2

a am e

E E T

V RT

K K K

'a a a E f f mV R I K K I

'

e t f f aT K K I I

CONSTaV

f fK I

A

B

TL

6

Review of steady state analysis of DC machine Dynamics of Separately Excited (SE) DC machine





Outlines

7

aa a a a E m

div R i L K

dt

Lmm

TaT TDdt

dJiKT

Electric equation in the armature circuit

Motion equation

Dynamics of the Separately Excited DC Motor

8

aa a a a E m

div R i L K

dt

τa=La/Ra

d/dt=>s

a aa a a

a

V ( s ) E ( s )s I ( s ) I ( s )

R

Va(s)

Ea(s)

Ia(s) +

a

a

1 / R

1 s

a

a a a

a

1/ RI ( s ) V ( s ) E ( s )

1 s


9

Lmm

TaT TDdt

dJiKT

)s(T)s(D)s(sJ)s(IK)s(T LmmTaT

d/dt=>s

DsJ

)s(T)s(IK

DsJ

)s(T)s(T)s(

T

LaT

T

Lm


10

KT DsJ

1

T

KE

Va(s)

Ea(s)

Ia(s)

T (s)

TL(s)

m(s) a

a

1 / R

1 s

DsJ

)s(T)s(IK

DsJ

)s(T)s(T)s(

T

LaT

T

Lm

a

a a a

a

1/ RI ( s ) V ( s ) E ( s )

1 s


11

a

1

1 s

TJ

Ds

1

ref(s) Ia(s)

TL(s)/JT(s)

m(s)

E

1

K

m(s)

aT

TE

RJ

KK Va

+

a Tm

E T

R J

K K a

a

a

L

R

τm >> τa Current changes much faster than the speed

aV s

Case A: Purely inertia load with D 0 and TL 0

Case B: Inertia load with viscous friction and TL 0

Case C: Effect of load torque on motor dynamics



Va(s)

ref

m

a

1 /

1 s s

m

E

1

K

m

m E a mE

2

a a m a a m

1 / K1 / K

V 1 s s 1 s s / 1 /

0s2s 2nn

2 n

a m

1

m

a

1

2

12

Natural frequency Damping coefficient

Step Response of a Second-Order System

13

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.5

0

0.5

1

1.5

2

2.5

Time(sec)

Input

output

n=20 =0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.80

0.5

1

1.5

2

=0.1 =0.5 =1 =2 =4

n=20

Undamped Natural frequency 10Hz

2

1 2

2 2

1 1 1, 1 4

2 2

1 1 1 1 1 1 1 21 4 2

2 2 2 2

1 1,

a a m

a

a a m m

a a a m

m a

s s


14

T a Tm E

a a m

a m

K / L J 1 / K

V s 1 s 11 1s s

Closed-loop poles are obtained by solving the

characteristic equation (denominator of TF = 0)

a m

using the indentity if x

1 x 1 x 12

Case B: Inertia load with viscous friction and TL 0

=

m T

a a a T E T

T T a

2

T a a T E T T a a a

K

V sL R J s D K K

K J L

s D J R L s D J K K J R R L

1 1n

a m T

D

J

1 1

2n

a T

D

J

15

Standard second order transfer characteristic

T

mTa

K

DsJI

mmTaTm DsJIKT

mEaaaa KIRsLV

1 a1 m

Case C: Effect of load torque on motor dynamics

TL m

Va = 0

+

+

Ea

KE Ia 1 /

1

a

a

R

s

KT

T

T

1

J s D

16

Va = CONST

Small perturbations in TL

1

1

m a

E TLa T

a

s

K KT s J s DR

17

Va=50V; 1 Nm load applied @ 1s

0 0.5 1 1.5 20

1

2

Load t

orq

ue(N

m)

0 0.5 1 1.5 20

200

400

time(sec)

Speed(r

pm

)Case C: Effect of load torque on motor dynamics

Dynamics of field controlled separately excited DC motor

if

Rf Lf

vf Va

Ia

m

(s)Tf fmf f a Tf f T m m

T

K IdT K i i K I J D

dt J s D

( )( )

f f

f f f f f

f f

di V sv R i L I s

dt R sL

Tfm

f f f T

Ks

V s R L s J s D

18

E T f fK K K i

m

f

s?

V s


19

dt

diLiRv

fffff

fv fi


20

Va =200V, If =0.2A 0.1A at 10 sec. wm = 160 rpm 300 rpm

Position response with inner current control

a

1

1 sKT

T

1

J s D

1

s

q

KE

Va

Ea

+

-

Ia T

21

Dynamics of separately excited DC motor

22

• The nonlinear model results in a two-input (Va, If), one-output (wm) map, having a disturbance input TL.

• Field (If) control with constant armature current (Ia); torque decreases with the increase of speed (field weakening).

• Armature current control by adjusting the armature voltage with a constant field (constant torque).

• Exceeding rated voltages accelerates ageing of conductor insulation, field weakening is used to increase the speed further.

• Open loop control. Exact values of armature voltage and field need to be known in order to obtain the expected speed. They are load dependent.

Example I: Compare the dynamics of two SE DC motors

23

Parameters Motor 1 Motor 2

Rated power (kW) 0.75 75

Rated speed (r/min) 500 1750

Ra (Ohm) 7.56 0.0114

La (H) 0.055 0.0011

KE (V/rad/s) 4.23 1.27

JL, Jm (kgm^2) 0.068 1.82

The torque-speed characteristic of the load is a straight line through the origin and the rated load point.


24

Compare the un-damped natural frequencies ωn, the damping ratio ξ and the time constants Ta (or τa) and Tm

(or τm ) of the two motors.

TL

ω Nm / rad / sT 14.3

D 0.27352.3

rated

P 750T 14.3 Nm

52.3

2T m LJ J J 2 0.068 0.136 kgm

aa

a

L0.0073

R

a Tm

E T

R J0.0575

K K

n

a m T

1 1 D51.54 rad/s

J

n a T

1 1 D1.348

2 J

For Motor 1:


25

Parameters Motor 1 Motor 2

Electrical time constant, τa (s) 0.0073 0.0965

Mechanical time constant, τm (s) 0.0575 0.246

Natural frequency, ωn (rad/s) 69.49 9.11

damping ratio, ξ 1.348 0.582

0 0.5 1 1.5

0

4000

8000

Torq

ue(N

m)

0 0.5 1 1.5

0

1000

2000

Time(sec)

Speed(r

pm

)

0 0.5 1 1.5

0

50

100

150

200

Am

atu

re V

oltage(V

)

speed

Armature voltage

0 0.5 1 1.5

0

20

40

60

80

Torq

ue(N

m)

0 0.5 1 1.5

0

100

200

300

400

Time(sec)

Speed(r

pm

)

0 0.5 1 1.5

0

50

100

150

200

Am

atu

re V

oltage(V

)

speed

armature voltage

Under-damped response Over-damped response

Closed loop control of separately excited DC motor

26

Speed control (200rpm),

If=0.2A The armature current exceeds the current limit of the DC motor during the transient (10-10.2 sec).


27

Speed control (200rpm) Field current control (0.20.1A) The armature current exceeds the limit during the speed transient.


28

Speed control (200rpm) Field current control (0.2A) Armature current control (maximum armature is limited within 20A)

29






Outlines

Representation of SM in the rotor reference frame

33

• The steady-state equivalent circuit of SM does not address the dynamics of the motor in the transient state.

• The equation of torque developed earlier does not have a direct or linear relationship with instantaneous values of stator current.

Ef

jXs=jLs I

V0

3sin

f

s s

pVET

X

Recall Questions

34

• Write the equations of the balanced three-phase currents in the stator windings of SM? Draw the waveform of 3-ph currents (amplitude=1, 50Hz).

• How are the 3-ph windings placed physically?

The 3-ph windings are displaced physically by 120° from each other

Representation of the synchronous machine in the rotor reference frame

Purpose: We seek to express the developed torque and flux in terms

of some instantaneous currents, as for the DC machine. This is not

possible with steady-state equivalent circuit models.

a

a’

b

b’

c

c’θ

ia

ib

ic

ea

eb

ec

Axis of

phase ‘a’

mmf

d-axis;

axis of

rotor mmf

q-axis;

orthogonal

to d-axis

35


36

What are the differences between a DC motor and a SM? 1. DCM has two windings (field and armature in a SE DC motor) SM has three-phase stator windings. Solution: Convert three-phase quantities of SM into two phase quadrature quantities (fictitious). 2. DCM driven by DC and the AC machine is by AC (three-phase SINE balance) We need to convert vectors in two-phase orthogonal stationary system into orthogonal rotating reference frame, where the quantities are DC values. abc (3-ph AC)α-β (2-phase AC)d-q (2-ph DC)

Warm-up Question I (Clarke Transformation)

37

Consider the case of balanced three-phase stator currents

1 11

2 2

3 30

2 2

1 1 1

2 2 2

sa

b s

c s

cos( t)i

i cos( t 120 )

i cos( t 120 )

Find two-phase orthogonal currents using the following

transformation matrix

Warm-up Question I

38

Three-phase to two-phase transformation

sa b c

a

b b c s

c0

a b c

1 1 31 cos( t)i 0.5i 0.5i2 2 2i i3 3 3 3

i 0 i i i sin( t)2 2 2 2

ii 1 1 1 1 0i i i

2 2 2 2

Warm-up Question II

39

Find a factor in front of the transformation matrix to make the 2-ph currents have the same amplitude as 3-ph currents?

β

α

2/3 2 2 3 2i i

Warm-up Question III (Park Transformation)

40

• Transform the two-phase (α-β) currents to another 2-ph (d-q) currents by the following matrix (θ is the rotor position).

d

q

0 0a b c

0 0a b c

m s

m s

i icos sin

i isin cos

i cos i cos 120 i cos 2402

3 i sin i sin 120 i sin 240

I cos( t )

I sin( t )

q q

q q

q q q

q q q

q

q

stq

d m

q m

i I cos( )

i I sin( )

cos sin

sin cos

q q

q q

2 2d q mi i I

Check the amplitude

Warm-up Question III

41

• α-β: 2-ph AC currents on stationary reference frame • d-q: 2-ph DC currents on rotor reference frame.

30


42

abcαβ0 αβ0dq0

abc αβ0 dq0

Clarke Transformation Park Transformation


43

0 0

ds as

0 0qs bs

cs0s

cos cos 120 cos 240f f

2f sin sin 120 sin 240 f

3ff 1 1 1

2 2 2

q q q

q q q

Any three-phase sinusoidal set of quantities in the stator can be transformed to an orthogonal reference frame by

magnetic axis of the rotor

Assumptions

stator windings are sine-distributed, balanced windings around the respective winding axes, (no zero or negative sequence)

the rotor field is cosine distributed about the rotor axis, (mutual inductance between stator and rotor varies as cosθ)

stator and rotor slots do not affect the mutual windings appreciably, (only consider mutual inductances between phase windings and stator and rotor windings)

magnetic saturation is negligible.

44

What do we need to find for SM Dynamics

45

Voltage equations in rotor (d-q) ref. frame. Expectation: One voltage equ. for torque current control; the other for field control, like Va and Vf in DC machine.

Torque equation . A direct or linear relationship with instantaneous stator current, like Te=Kt·Ia in DC machine.

Steps to Find Torque and Flux of SM

46

Rotor and stator flux linkages of SM on 3-ph ref. frame;

Transform the stator flux linkages to the rotor (d-q) ref. frame;

Machine voltage equations derived from flux linkage

equations;

Derive the equation of output power from instantaneous input

power;

Derivation of torque equation from output power and

mechanical speed.

Step I:

53

Rotor and stator flux linkages of SM on abc ref. frame;



equations;


power;


mechanical speed.

Winding self-inductances

aa A BL L L cos 2q

bb A BL L L cos 2 120q

cc A BL L L cos 2 240q

• Self inductance of each winding is λ/I of its own current; • It varies between max and min values according to the

position of the rotor. • Self-inductance goes through two cycles of variations for each

rev. of the rotor (2θ);

54

Self inductance expressions [1]

q

gmin

gmax

2A s

min max

C 1 1L N

2 g g

2B s

min max

C 1 1L N

2 g g

2aas,min l s

max

1L L CN

g

2aas,max l s

min

1L L CN

g

oC rl4

[1] P. C. Krause, O. Wasynczuk, S. D. Sudhoff, S. Pekarek, Ch1, Analysis of Electric Machinery and Drive Systems, 3rd ed., Wiley-IEEE, 2013.

Self-inductances

56

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04

Time (s)

0

1

2

3

Laa

LA

LB

Laas

0° 90° 180° 270° 360°

0

q

aas A BL L L cos 2q

bbs A BL L L cos 2 120q

ccs A BL L L cos 2 120q

LA is independent on rotor position

LB is max value of inductance varying with rotor position

57

Mutual inductances between phase windings

ab ba A B

1L L L L cos 2 30

2q

bc cb A B

1L L L L cos 2 90

2q

ca ac A B

1L L L L cos 2 150

2q

ab ba A BL L L cos(120 ) L cos 2 120q

bc cb A BL L L cos(120 ) L cos 2 120 120q

ca ac A BL L L cos( 240 ) L cos 2 240q

Mutual inductance between stator and rotor windings

58

af fa mL L L cosq

0bf fb mL L L cos 120q

0cf fc mL L L cos 240q

1m s r

min

L CN Ng

• mutual inductances between a stator and the rotor winding varies as a cosine function of θ.

• The mutual inductance between phase a winding and the rotor becomes maximum when the rotor aligns with the axis of this winding

0 0d a b ci K i cos i cos 120 i cos 240q q q

0 0q a b ci K i sin i sin 120 i sin 240q q q

fd f f af a bf b cf c

0 0f f m a b c

L i L i L i L i

L i L i cos i cos 120 i cos 240

q q q

Total rotor flux linkage along the d-axis (rotor)

59

Machine rotor flux linkages

fd f f m d f m d

3 3L i L i L i

2 2

Let

Expressed with one current rather than 3 currents

2 2

amax bmax cmax d qI I I i i

0 a b c

1i i i i

3

0 0

d a

0 0q b

c0

cos cos 120 cos 240i i

2i sin sin 120 sin 240 i

3ii 1 1 1

2 2 2

q q q

q q q

60

We choose K = 2/3, so that,

61

0 0

0 0

cos cos 120 cos 2401 0.5 0.5cos sin 02 2

A sin cos 0 0 3 2 3 2 sin sin 120 sin 2403 3

0 0 1 1 / 2 1 / 2 1 / 2 1 1 1

2 2 2

q q qq q

q q q q q

Stay Awake Question I

d

q

0 0

i cos sin 0 i

i sin cos 0 i

i 0 0 1 i

q q

q q

a

b

c0

1 11

2 2i i

2 3 3i 0 i

3 2 2ii 1 1 1

2 2 2

Find the matrix A, which can convert 3-ph (abc) currents to 2-ph currents in the rotor ref. frame (dq)

62

0 0

0 0

cos cos 120 cos 240

2A sin sin 120 sin 240

31 1 1

2 2 2

q q q

q q q

Stay Awake Question II

Find the reverse transformation, which can convert dq currents to 3-ph (abc) currents

1 0 0

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q

63

Stay Awake Question II

1AA E

a d

1

b q

c 0

i i

i A i

i i

1 0 0

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q

0 0

0 0

cos cos 120 cos 240


31 1 1

2 2 2

q q q

q q q

64

Inverse transformation

Stator flux linkages a aa a ab b ac c af fL i L i L i L i

65

0

a A B a A B b

0

A B c af f

1L L cos 2 i L L cos 2 30 i

2

1L L cos 2 150 i L i

2

q q

q

0 0

b A B a A B b

0

A B c b f f

1L L cos 2 30 i L L cos 2 120 i

2

1L L cos 2 90 i L i

2

q q

q

0 0

c A B a A B b

0

A B c c f f

1 1L L cos 2 150 i L L cos 2 90 i

2 2

L L cos 2 240 i L i

q q

q

b ba a bb b bc c bf fL i L i L i L i

c ca a cb b cc c cf fL i L i L i L i

Step II:

66




equations;


power;


mechanical speed.

Stator flux linkages

67

1

0 0

1 0.5 0.5

0.5 1 0.5

0.5 0.5 1

cos(2 ) cos(2 120) cos(2 240)

cos(2 120) cos(2 240) cos(2 )

cos(2 240) cos(2 ) cos(2 120)

d a d

q b A q

c

B

i

A L A A i

i

L A

q q q

q q q

q q q

1

0

cos( )

cos( 120)

cos( 240)

d

q m f

i

A i L i A

i

q

q

q

0 0

0 0

cos cos 120 cos 240


31 1 1

2 2 2

q q q

q q q

1 0 0

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q


68

cos 0.5cos 120 0.5cos 1202

1 term sin 0.5sin 120 0.5sin 1203

0

0.5cos cos 120 0.5cos 120

0.5sin sin 120 0.5sin 120

0

0.5cos 0.5

Ast L

q q q

q q q

q q q

q q q

q

1

cos 120 cos 120

0.5sin 0.5sin 120 sin 120

0

A

q q

q q q

cos cos 120 cos 120q q q

sin sin 120 sin 120q q q

Using the following identity

0.5cosq

0.5sinq

0.5cos 120q

0.5sin 120q

0.5cos 120q

0.5sin 120q


69

1

1 1

1.5cos 1.5cos 120 1.5cos 1202

1 term 1.5sin 1.5sin 120 1.5sin 1203

0 0 0

0 0 0 1 0 03 3

= 0 0 0 0 1 02 2

0.5 0.5 0.5 0 0 0

A

A A A

st L A

L AA L A L

q q q

q q q

1 0 03

2 term 0 1 02

0 0 0

Bnd L

0 0

0 0

cos cos 120 cos 240


31 1 1

2 2 2

q q q

q q q


70

2 2 2cos cos 120 cos 1202

3rd term sin cos sin 120 cos 120 sin 120 cos 1203

0.5 cos cos 120 cos 120

1

= 0

0

m f

m f

L i

L i

q q q

q q q q q q

q q q

2cos 1 cos 2 2q q

2cos 120 1 cos 2 240 2q q

2cos 120 1 cos 2 240 2q q

cos 2 cos 2 240 cos 2 240 0q q q

sin 2 sin 2 240 sin 2 240 0q q q

71

d A A B d a f f

d d m f d d f

1 3L L L i L i

2 2

L i L i L i

q A A B q

q q

1 3( L L L )i

2 2

L i

d A B

3L L L

2

q A B

3L L L

2

Stator flux linkages on Rotor Reference Frame

Ld and Lq are independent of position now

Step III:

72




equations;


power;


mechanical speed.

0 0

d a

0 0

q b

c0

cos cos 120 cos 240

2sin sin 120 sin 240

31 1 1

2 2 2

q q q

q q q

73

0 0

d a a

0 0

q b b

c c0

sin sin 120 sin 240

d 2 d dcos cos 120 cos 240 A

dt 3 dt dt0 0 0

q q q q

q q q

qa d

b q d

c 00

d dA

dt dt

' ' 'uv u v uv

Stator voltage equations

74

aa a

dv Ri

dt

bb b

dv Ri

dt

cc c

dv Ri

dt

1

0 0

1 0 0

0 1 0

0 0 1

d a d a

q b q b

c c

v v id

v A v R A A i Adt

v v i

qdd d

q q q d

0 0 00

v id

v R idt

v i

1 0 0

0 1 0

0 0 1

a a a

b b b

c c c

v id

v R idt

v i

d d d d m f e q q

dv Ri L i L i L i

dt

fd dd d m e q q d d e q q

didi diRi L L L i Ri L L i

dt dt dt

q

q q q e d d f

div Ri L L i

dt

00 0

dv Ri 0

dt

75

e r

dp

dt

q

Review of Dynamics of SM

76

77

0 0

0 0

cos cos 120 cos 240


31 1 1

2 2 2

q q q

q q q

Forward(abcdq) and reverse (dqabc) transforms

1 0 0

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q

a

a’

b

b’

c

c’θ

ia

ib

ic

ea

eb

ec

Axis of

phase ‘a’

mmf

d-axis;

axis of

rotor mmf

q-axis;

orthogonal

to d-axis

78

fd f f m d f m d

3 3L i L i L i

2 2

Machine rotor and stator flux linkages

d d d fL i

q q qL i

d A B

3L L L

2

q A B

3L L L

2

79

Recall dynamic model of DC machine

dt

diLiRv

fffff

dd d d e q q

div Ri L L i

dt

q

q q q e d d f

div Ri L L i

dt

Voltage equ. of SM machine

mffa

aaaa iKdt

diLiRV

Step IV:

80




equations;


power;


mechanical speed.

The developed power

81

a a b b c c

d d q q

P v i v i v i

3v i v i

2

Power input to the machine and the developed torque

The factor 3/2 in the power is because of the factor 2/3 in the abc-dq transformation for both voltage and current.

The developed power

82

a a b b c c

2 2 2d d

d q

q d

2 2 2q q

P v i v i v i

=v i cos cos 120 cos 240

-v i cos sin cos 120 sin 120 cos 240 sin 240

-v i cos sin cos 120 sin 120 cos 240 sin 240

+v i sin sin 120 sin 240

q q q

q q q q q q

q q q q q q

q q q

d d

d q q d

q q

cos 2 240 1 cos 2 480 1cos2 1 =v i

2 2 2

- v i v i 0.5sin2 0.5sin 2 240 0.5sin 2 480

1-cos 2 240 1-cos 2 4801-cos2 +v i

2 2 2

q qq

q q q

q qq

0

3/2

3/2

The developed power

qdd q q d d q

2 2d q

d3 d 3 dP i i i i

2 dt dt 2 dt

3 i i R

2

q

We expect to see only current in the torque equation.

q 2 2dd d q q d d q q

didi d 1 d 11st term L i L i L i L i

dt dt dt 2 dt 2

Represent rate of change of stored energy in the inductors

Copper loss

q

q q e d

dv Ri

dt

dd d e q

dv Ri

dt

d d d fL i

q q qL i

The developed power

qdd q q d d q

2 2d q

d3 d 3 dP i i i i

2 dt dt 2 dt

3 i i R

2

q

q d d q

32nd term i i

2

This term being the product of the back emf and currents of the d- and q-windings, is the developed power of the machine.

back emf

Step V:

85




equations;


power;


mechanical speed.

dev dev q d d q

f q d q q d

3 pT P / i i

2

3 pi L L i i Nm

2

For a non-salient pole synchronous machine, Ld = Lq

dev f q

3T p i

2

86

For a machine with p pole pairs.

The developed torque

dev q d d q

3 dP i i

2 dt

q

aT K iCompare to DC machine torque

Steps to Model Dynamics of SM

89

Clarke and Park Transformation, A

2. Obtain the dq currents from the state equations.

dd d e q d

e fq q qe dq

dL

i v R L i 0dt

i v id L RL

dt

3. Obtain the developed torque from

dev f q

3T p i

2

4. Obtain speed by integrating dev L

d 1T D T sign

dt J

5. dq flux linkages d d d fL i

q q qL i

Steps to Model Dynamics of SM

90

6. Find θ by integrating ω

7. Use the θ to recalculate dq voltage for the next integration step.

8. Repeat Step 2-7

9. Continue until the end of the duration of modeling.

10. Obtain actual voltages, currents and flux linkages in the abc

reference frame with the inverse of A.

1 0 0

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q

91






Outlines

Torque control using id and iq controls

d

** * *d

d d e q q

div Ri L L i

dt

*

q* * *

q q q e d d f

div Ri L L i

dt

These equations are coupled and non-linear.

Back emf

dev f q d q q d

3pT i L L i i Nm

2

dev f q

3T p i Nm

2

d d d fL i q q qL i 2 2

s d q 92

for a SP SM

for a NSP SM

Rotor flux oriented MTPA control of IPM

93

2

f s q d s

dT 3 3I sin p( L L )I cos 2 0

d 2 2

2

f s q d s

3 3T I cos p( L L )I sin 2

2 4

2 2 2

f f q d q

d

q d

2

f f 2

q2

q d q d

4( L L ) ii

2( L L )

i2( L L ) 4( L L )

2 2

s d qI i i

Vector or rotor flux oriented control of the SM

dq-1

*qi

*di

*av

*bv

*cv

d/dt

PWM

Inverter

q

q

*ai

*bi

*ci

* T *

s

Current

Reference

Generator

ia - ic

SMSpeed

Controller

Current

Controllers

94

• In servo drive, id is maintained at a constant value corresponding to the rated flux.

• In PM and large air gap motor, id is maintained at 0 (λf>>idLd). • In field weakening, id is maintained at negative values.

AC dev f q

3T p i Nm

2


s d q


95

** * dd d d e q q

div Ri L L i

dt

*

q* *

q q q e d d f

div Ri L L i

dt

De-coupling compensation results in faster dynamics of current (and hence torque) control.

dq

*qi

*di

*av

*bv

*cv

w

q

w*

T *

Current

Ref

Gen

ia

SMSpeed

Controller

dq current

Controllers

w

ic

ib

d/dt

+

-

+

-

iq

id

PWM

Inverter

*qv

*dv

*

s

dq-1

e q qL i

-

e d d fL i

+


96

With de-coupling compensation

Without de-coupling compensation

Example of RFOC SM

97

A 10-pole, 50Hz, Surface mounted PMSM, Ef= 115 Vrmsl-l/krpm, Irated=4.17 Arms, Imax=6 A (peak) within the rated speed. The developed torque and stator flux linkage of such a motor under rotor flux oriented control (without field weakening)

1.5 f qsT p i

ds d ds fL i

Example of RFOC SM

98

(a) Calculate the rotor flux linkage, λf.

115 2 30.18 V/rad/s

5 1000 30f

(b) Calculate the stator current reference ids in the rotor reference frame, so that the stator flux linkage d along the rotor d-axis remains constant at all times.

* 0 so that 0.18ds ds fi

(c) Calculate the stator current reference for iqs if the motor is to develop 8 Nm of torque. And the maximum developed torque?

1.5 5 5 Nmf qsT i 5 1.5 5 0.18 3.7 Aqsi

max ,max1.5 5 1.5 5 0.18 6 8.1 Nmf qsT i

Example of RFOC SM

99

(d) Sketch the torque and flux-linkage control schemes in which the stator flux-linkage must be maintained at f.

Voltage

Source

Inverter

PI

*

qi

ˆqi

* 0di *

dsv

*

qsv

PIˆdi ˆ

req

-+

*

anv

*

bnv

-+

abc

dq

d−q

abc

EPMSM

ˆreq

*

cnv

asi

bsi

csi

Pp

Simulation of Vector or RFOC of PMSM

100

Vector control: Stator currents are identified as two orthogonal components (dq), representing a vector. RFOC: The d component defines the magnetic flux and q the torque. d axis is aligned with the electrical position of the rotor flux.


101

vd

Current controller step response


102

Speed controller step response And load disturbance response

Review of Dynamics of SM

103

104

0 0

0 0

cos cos 120 cos 240


31 1 1

2 2 2

q q q

q q q

Forward(abcdq) and reverse (dqabc) transforms

1 0 0

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q

a

a’

b

b’

c

c’θ

ia

ib

ic

ea

eb

ec

Axis of

phase ‘a’

mmf

d-axis;

axis of

rotor mmf

q-axis;

orthogonal

to d-axis

fi

Torque control using id and iq controls

** * *dd d d e q q

div Ri L L i

dt

*

q* * *

q q q e d d f

div Ri L L i

dt

These equations are coupled and non-linear.

Back emf

dev f q d q q d

3pT i L L i i Nm

2

dev f q

3T p i Nm

2


s d q 105

for a SP SM

for a NSP SM


106

** * dd d d e q q

div Ri L L i

dt

*

q* *

q q q e d d f

div Ri L L i

dt

De-coupling compensation results in faster dynamics of current (and hence torque) control.

dq

*qi

*di

*av

*bv

*cv

w

q

w*

T *

Current

Ref

Gen

ia

SMSpeed

Controller

dq current

Controllers

w

ic

ib

d/dt

+

-

+

-

iq

id

PWM

Inverter

*qv

*dv

*

s

dq-1

e q qL i

-

e d d fL i

+

107






Outlines

Dynamics of Induction Machines

108

Review of SM Working Principle

109

The rotating magnetic field rotates at synchronous speed, like a N-S poles rotating at the same speed.

Rotor&stator poles repel each other get aligned with the opposite poles

Rotor&stator poles interlocked Rotating at synchronous speed

Working Principle of IM

110

• Rotating magnetic field at the synchronous speed • Closed rotor windings (wound, squirrel cage), emf I F • Rotor cutting rotating magnetic field relative speed, slip s

http://www.electrical4u.com/what-is-magnetic-field/

Why Do we Need the Dynamic Model of an IM

111

• Scalar control methods such as VVVF and slip power recovery

schemes are not instantaneous control of torque .

• VSD/VFD drives require high dynamic performance (fast acceleration

and deceleration), like in a servo system

• The vector control of an IM is based on the decoupling between the

current components used for generating magnetizing flux and torque.

• The decoupling allows the induction motor to be controlled as a simple

DC motor and high dynamic performance, comparable to that of a DC

motor.

• Vector control uses machine dynamic model rather than steady-state

model (scalar control) .

Reference Frame Transformation in an IM

112

• Transform 3-ph sinusoidal quantities of stator windings and rotor windings to an orthogonal reference frame separately (abcsdqs, abcrdqr).

• Pick a rotating reference frame synchronously rotating reference frame (ωs, ωs≠ωr);

• For IM, the steps of the transformation are the same as SM except for inductance expression (much simpler).

• The stator voltage (dq) equations of IM should be the same as SM, but the rotor voltage equations are a little different.

Dynamic model of an IM in the synchronously rotating reference frame

0

t

s odtq q

q is the angle of the

synchronously

rotating d-axis with

respect to the ‘a’

stator winding axis.

113

iar

idr

iqr

ibr

icr

ids

ibs

ics

iqs

ias

r

s

s

qr

ids /idr

q

iqs /iqr

Axes transformation

120 240

120 240

ds as

qs bs

cso

cos cos cosf f2

f sin sin sin f3

f1 1 1f

2 2 2

q q q

q q q

s as

s bs

csos

1 11

2 2f f

2 3 3f = 0 f

3 2 2ff 1 1 1

2 2 2

114

Assumptions & machine inductances

Balanced windings, uniform air-gap and linear magnetic field.

Stator winding self-inductances, Laas= Lbbs=Lccs=L1 (LB=0), independent of

the angular position of the rotor.

Rotor winding self inductances, Laar = Lbbr = Lccr = L2

Mutual inductances between stator windings, Labs = Lbcs = Lcas = M1

Mutual inductances between rotor windings, Labr = Lbcr = Lcar = M2

The mutual inductances between stator and rotor windings (M) have

cosine variation with qr.

Thus, Lasar = Mcosqr, ; Lbsar = Mcos(qr −120°) and so on.

115

Compare Self Winding Inductances

aa A BL L L cos 2q

bb A BL L L cos 2 120q

cc A BL L L cos 2 240q

116

1aas bbs ccsL L L L

2aar bbr ccrL L L L

Salient Pole SM IM

Dependent on rotor position Independent of rotor position Constant

fL

Compare Mutual Inductances between Stator

117

Salient Pole SM IM

Dependent on rotor position Independent of rotor position Constant

ab ba A B

1L L L L cos 2 30

2q

bc cb A B

1L L L L cos 2 90

2q

ca ac A B

1L L L L cos 2 150

2q

1abs bcs casL L L M

Compare Mutual Inductances between Rotor

118

Salient Pole SM IM

No mutual inductance between the rotor windings

Independent of rotor position Constant

2abr bcr carL L L M

Compare Mutual Inductances Stator- Rotor

119

Salient Pole SM IM

Dependent on rotor position Dependent of rotor position

;

120

240

( )

( );

( )

asar bsbr cscr r

bsar csbr ascr r

csar asbr bscr r

L L L Mcos

L L L Mcos

L L L Mcos

q

q

q



af fa mL L L cosq

Compare Mutual Inductances between Rotor-Stator

120

Salient Pole SM IM

Dependent on rotor position Dependent of rotor position

;

240

120

( )

( );

( )

aras brbs crcs r

bras crbs arcs r

cras arbs brcs r

L L L Mcos

L L L Mcos

L L L Mcos

q

q

q



af fa mL L L cosq


121

iar

idr

iqr

ibr

icr

ids

ibs

ics

iqs

ias

r

s

s

qr

ids /idr

q

iqs /iqr

Flux linkages

For stator:

1 1

cos cos 240 cos 120

as as bs cs

ar r br r cr r

L i M i i

M i i i

q q q

1 1

cos 120 cos cos 240

bs bs as cs

ar r br r cr r

L i M i i

M i i i

q q q

1 1

cos 240 cos 120 cos

cs cs as bs

ar r br r cr r

L i M i i

M i i i

q q q

qr is the angular displacement of the rotor a-axis with respect to stator a-

axis.

r r edt 1 s dtq 122

0as bs csi i i

2 2

cos cos 120 cos 240

ar ar br cr

as r bs r cs r

L i M i i

M i i i

q q q

2 2

cos 240 cos cos 120

br br ar cr

as r bs r cs r

L i M i i

M i i i

q q q

2 2

cos 120 cos 240 cos

cr cr ar br

as r bs r cs r

L i M i i

M i i i

q q q

For rotor:

1 1sL L M

2 2rL L M

123

Flux linkage contd.

0ar br cri i i

cos cos 240 cos 120as s as ar r br r cr rL i M i i i q q q

cos 120 cos cos 240bs s bs ar r br r cr rL i M i i i q q q

cos 240 cos 120 coscs s cs ar r br r cr rL i M i i i q q q

cos cos 120 cos 240ar r ar as r bs r cs rL i M i i i q q q

cos 240 cos cos 120br r br as r bs r cs rL i M i i i q q q

cos 120 cos 240 coscr r cr as r bs r cs rL i M i i i q q q

124

Rotor flux linkages


Stay-awake Questions

q

q

axis

d

axisq

cos cos 120 cos 240


3

1/ 2 1 / 2 1 / 2

sA

q q q

q q q

125

Transform 3-ph stator currents to the synchronously rotating reference

frame (iabcs idqs )

1 0 0

s

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q

0

t

s odtq q


126

Transform 3-ph rotor currents to the synchronously rotating reference frame

(iabcr idqr ) sl

e e r

ds

dt

q

=

sl e

e r slo

s dt

dt

q

q

0ar br cr ori i i i

cos cos 120 cos 240


3

1/ 2 1 / 2 1 / 2

sl sl sl

r sl sl slA

q q q

q q q

0ar br cr ori i i i

127

sle e r

ds

dt

q

=

sl e

e r slo

s dt

dt

q

q


Transform rotor currents from the synchronously rotating to the 3-ph

reference frame (iabcr idqr )

1

cos sin 1

cos 120 sin 120 1

cos 240 sin 240 1

sl sl

r sl sl

sl sl

A

q q

q q

q q

128


1

0

cos cos 240 cos 120

cos 120 cos cos 240

cos 240 cos 120 cos

r r r dr

ds ds

s s r r r r qr

qs qs

r r r r

ii

L MA A ii

i

q q q

q q q

q q q

cos( ) cos 120 cos 240 cos sin 1

sin( ) sin( 120 ) sin( 240 ) cos 120 sin 120 1

0 0 0 cos 240 sin 240 1

sl sl sl sl sl

sl sl sl sl sl

sl sl

q q q q q

q q q q q

q q

cos cos 240 cos 120

cos 120 cos cos 240

cos 240 cos 120 cos

as as r r r ar

bs s bs r r r br

cs cs r r r cr

i i

L i M i

i i

q q q

q q q

q q q

3 / 2 0 0

0 3 / 2 0

0 0 0

3

2

ds ds dr

s

qs qs qr

i iL M

i i

Machine Stator voltages

as s as as

dv R i

dt

bs s bs bs

dv R i

dt

cs s cs cs

dv R i

dt

Non-linear differential equations because

of coefficients which have trigonometric

terms of qr.

Can these equations have constant

coefficients by transforming current and

voltage from abcdq0?

129

0 0

ds as ds as

qs s bs s qs s bs

s cs s cs

v v id

v A v R i Adt

v v i

1 1 1

0 0 0

as ds ds ds

s bs s s qs s s qs s s qs

cs s s s

d d d dA A A A A A A

dt dt dt dt

' ' 'AB A B AB

Machine Stator voltages

130

0 0

sin cos 0 1 0 0

sin 120 cos 120 0 0 1 0

sin 240 cos 240 0 0 0 1

as ds ds

s bs e s qs qs

cs s s

d dA A

dt dt

q q

q q

q q

cos ' sind

dt

qq q

0 0

0 1 0 1 0 0

1 0 0 0 1 0

0 0 0 0 0 1

as ds ds

s bs e qs qs

cs s s

d dA

dt dt

0 0 0 0

1 0 0 0 1 0

0 1 0 1 0 0

0 0 1 0 0 0

ds ds ds ds

qs s qs qs e qs

s s s s

v id

v R idt

v i

ds s ds ds e qs

dv R i

dt

qs s qs qs e ds

dv R i

dt

131

Rotor flux linkages

1

0

cos cos 120 cos 240

cos 240 cos cos 120

cos 120 cos 240 cos

r r r ds

dr dr

r r r r r s qs

qr qr

r r r s

ii

L MA A ii

i

q q q

q q q

q q q

cos( ) cos 120 cos 240 cos sin 1

sin( ) sin( 120 ) sin( 240 ) cos 120 sin 120 1

0 0 0 cos 240 sin 240 1

q q q q q

q q q q q

q q

cos cos 120 cos 240

cos 240 cos cos 120

cos 120 cos 240 cos

ar ar r r r as

br r br r r r bs

cr cr r r r cs

i i

L i M i

i i

q q q

q q q

q q q

3 / 2 0 0

0 3 / 2 0

0 0 0

3

2

dr dr ds

r

qr qr qs

i iL M

i i

Machine Rotor voltages

ar r ar ar

dv R i

dt

br r br br

dv R i

dt

cr r cr cr

dv R i

dt

Non-linear differential equations because

of coefficients which have trigonometric

terms of qr.

Can these equations have constant

coefficients by transforming current and

voltage from abcdq0?

132

0 0

dr ar dr ar

qr r br r qr r br

r cr r cr

v v id

v A v R i Adt

v v i

1 1 1

0 0 0

ar dr dr dr

r br r r qr r r qr r r qr

cr r r r

d d d dA A A A A A A

dt dt dt dt

' ' 'AB A B AB

Machine Rotor voltages

133

0 0

sin cos 0 1 0 0

sin 120 cos 120 0 0 1 0

sin 240 cos 240 0 0 0 1

ar sl sl dr dr

r br e r r sl sl qr qr

cr sl sl r r

d dA A

dt dt

q q

q q

q q

cos ' sin sinslsl sl sl e r

d

dt

qq q q

0 0

0 1 0 1 0 0

1 0 0 0 1 0

0 0 0 0 0 1

ar dr dr

r br e r qr qr

cr r r

d dA

dt dt

0 0 0 0

1 0 0 0 1 0

0 1 0 1 0 0

0 0 1 0 0 0

dr dr dr dr

qr r qr qr e r qr

r r r r

v id

v R idt

v i

dr r dr dr e r qr

dv R i

dt

qr r qr qr e r dr

dv R i

dt

Flux linkages and voltages in synchronous dq frame

ds s ds m drL i L i

qs s qs m qrL i L i

2 2s ds qs

dr r dr m dsL i L i

qr r qr m qsL i L i

2 2r dr qr

ds s ds ds e qs

dv R i

dt

qs s qs qs e ds

dv R i

dt

dr r dr dr e r qr

dv R i

dt

qr r qr qr e r dr

dv R i

dt

After a lot of trigonometric manipulations, and using 3

2mL M

134

For stator For rotor

Flux linkages and voltages in synchronous dq frame

ds s ds s ds m dr e s qs m qr

dv R i L i L i L i L i

dt

qs s qs s qs m qr e s ds m dr


dt

dr r dr r dr m ds e r r qr m qs


dt

qr r qr r qr m qs e r r dr m ds


dt

135

For stator:

For rotor

Replacing fluxes in terms of current

Dynamic equivalent circuits in dq-axes

s ls mL L L r lr mL L L

+ +

Rs Llr =Lr - Lm Lls =Ls - Lm

Lm

(e - r)dr eds Rr

iqs iqr

qs qr

vqr vqs

137

Part of the self flux linkage of each winding is leakage flux that does not cross the air gap. Thus,

qr r qr e r dr r m qr m qr qs

d dv R i L L i L i i

dt dt

qs s qs e ds s m qs m qs qr


dt dt

qsd

dt

qrd

dt

Dynamic equivalent circuits in dq-axes

s ls mL L L r lr mL L L

+ +

Rs Llr Lls

Lm

(e - r)qr eqs Rr

ids idr

ds dr

vdr vds

138

Part of the self flux linkage of each winding is leakage flux that does not cross the air gap. Thus,

ds s ds e qs s m ds m ds dr


dt dt

dr r dr e r qr r m dr m dr ds


dt dt

dsd

dt

drd

dt

The developed power and torque

Ws a a b b c cp v i v i v i

139

The power input to the 3-ph stator is given by

With abcdq0 transformation, it becomes

3

W2

s ds ds qs qsp v i v i

Substituting stator dq voltages yields

2 23

2

s s ds qs s ds ds qs qs

m ds dr qs qr e ds qs qs ds

d dp R i i L i i i i

dt dt

d dL i i i i i i

dt dt


140

3

mr dr ds qr qs m ds ds qs qs

r

e r r qr ds m qs ds r dr qs m ds qs

L d drd term R i i i i L i i i i

L dt dt

L i i L i i L i i L i i

For a squirrel-cage rotor, 0dr qrv v

0 r dr r dr m ds e r r qr m qs

dR i L i L i L i L i

dt

0 r qr r qr m qs e r r dr m ds

dR i L i L i L i L i

dt

1

dr r dr m ds e r r qr m qs

r

d di R i L i L i L i

dt L dt

1

qr r qr m qs e r r dr m ds

r

d di R i L i L i L i

dt L dt

mp 4 term= e m dr qs qr dsth L i i i i


3

2

3

2

mm e r r qr ds m qs ds r dr qs m ds qs

r

e m dr qs qr ds

Lp L i i L i i L i i L i i

L

L i i i i

141

The electromechanical power is given by

3

2m m r dr qs qr dsp L i i i i

3 3

2 2

m mdev m dr qs qr ds dr qs qr ds

r r

p LT pL i i i i p i i

p L

dr r dr m dsL i L i

qr r qr m qsL i L i


142

A simple way to find the power associated with speed (motional)

voltages is from the equivalent dynamic circuit.

3

2m e ds qs e r dr qr e qs ds e r qr drp i i i i

+ +


Lm

(e - r)dr eds Rr

iqs iqr

qs qr

vqr vqs

+ +

Rs Llr Lls

Lm

(e - r)qr eqs Rr

ids idr

ds dr

vdr vds

dr r dr m dsL i L i

qr r qr m qsL i L i


143

3

2m e ds qs e qs ds e r ds m qr e r qs m drp i i i L i i L i

3

2m e ds qs e qs ds e r ds qs s qs e r qs ds s dsp i i i L i i L i

m qr qs s qsL i L i

m dr ds s dsL i L i

3

2m r ds qs r qs dsp i i

3

2

mdev ds qs qs ds

r

p pT i i

p

The mechanical output power is given by

The developed electromagnetic torque is given by

In terms of stator variables


144

3

2m e dr qr r qr e qr dr r dr e r dr qr e r qr drp i L i i L i i i

m qs qr r qrL i L i

m ds dr r drL i L i

3

2m r qr dr r dr qrp i i

3

2

mdev qr dr dr qr

r

p pT i i

p

3

2m e ds qs e r dr qr e qs ds e r qr drp i i i i

ds s ds m drL i L i

3

2m e qs s ds m dr e ds s qs m qr e r dr qr e r qr drp i L i L i i L i L i i i

In terms of rotor variables

qs s qs m qrL i L i

145






Outlines

Review of Dynamics of IM

146

147

Self and Mutual Inductances of IM

1aas bbs ccsL L L L 2aar bbr ccrL L L L

1abs bcs casL L L M 2abr bcr carL L L M

;

120

240

( )

( );

( )

asar bsbr cscr r

bsar csbr ascr r

csar asbr bscr r

L L L Mcos

L L L Mcos

L L L Mcos

q

q

q

;

240

120

( )

( );

( )

aras brbs crcs r

bras crbs arcs r

cras arbs brcs r

L L L Mcos

L L L Mcos

L L L Mcos

q

q

q

Stator windings Rotor windings


148

iar

idr

iqr

ibr

icr

ids

ibs

ics

iqs

ias

r

s

s

qr

ids /idr

q

iqs /iqr

149

cos cos 120 cos 240


3

1/ 2 1/ 2 1/ 2

sA

q q q

q q q

1 0 0

s

0 0

cos sin 1

A cos 120 sin 120 1

cos 240 sin 240 1

q q

q q

q q


cos cos 120 cos 240


3

1/ 2 1/ 2 1/ 2

sl sl sl

r sl sl slA

q q q

q q q

1

cos sin 1

cos 120 sin 120 1

cos 240 sin 240 1

sl sl

r sl sl

sl sl

A

q q

q q

q q

abcsdq0s dq0sabcs

abcrdq0r dq0rabcr

Flux Linkages and Voltages of IM

ds s ds m drL i L i

qs s qs m qrL i L i

2 2s ds qs

dr r dr m dsL i L i

qr r qr m qsL i L i

2 2r dr qr

ds s ds ds e qs

dv R i

dt

qs s qs qs e ds

dv R i

dt

dr r dr dr e r qr

dv R i

dt

qr r qr qr e r dr

dv R i

dt

After a lot of trigonometric manipulations,

150

For stator For rotor

151

3 3

2 2

m mdev m dr qs qr ds dr qs qr ds

r r

p LT pL i i i i p i i

p L

Developed Torque of IM

+ +


Lm

(e - r)dr eds Rr

iqs iqr

qs qr

vqr vqs

+ +

Rs Llr Lls

Lm

(e - r)qr eqs Rr

ids idr

ds dr

vdr vds

152

RFOC (Rotor Flux Oriented Control)

3

2

mdev dr qs qr ds

r

LT p i i

L

1.5dev f qsT p i

Recall developed torque of non-SP SM

0qr

r dr

Comparing to the developed torque of IM

3

2

mdev r qs

r

LT p i

L

Conditions for rotor flux oriented control (RFOC)

For the squirrel-cage motor, vqr and vdr are both zero.

0qr

r qr e r dr

dR i ( )

dt

0drr dr e r qr

dR i ( )

dt

1 mqr qr qs

r r

Li i

L L

1 mdr dr ds

r r

Li i

L L

Eliminating iqr and idr (induced currents), which are not accessible

and hence cannot be controlled, using

153

(a)

(b)

qr mrqr r qs e r dr

r r

d LLR i ( ) 0

dt R L

dr mrdr r ds e r qr

r r

d LRR i ( ) 0

dt L L

qr 0 r dr̂

We require all the rotor flux to be in the rotor d-axis.

so that

Also, the rotor flux should remain constant. Thus

qrdrdd

0dt dt

154

mr qs e r r

r

L ˆR i ( ) 0L

m re r sl qs

rr

L Ri

ˆ L

The first rotor voltage equation becomes

(1)

(b)(a)

2

m mdev dr qs qr ds r qs

r r

L L3P 3 P ˆT i i i2 L L

r rr m ds

r

ˆL d ˆ L iR dt

155

mr rr r ds

r r

ˆ Ld R ˆ R i 0dt L L

The second rotor voltage equation becomes

Rotor flux λr is determined by ids, subject to a time delay Tr which is the rotor time constant (Lr/Rr).

Current iqs controls the developed torque T without delay.

2m m

qs s qs m qr s qs qrr r

L LL i L i L i

L L

2m m

ds s ds m dr s ds drr r

L LL i L i L i

L L

3

2dev ds qs qs ds

pT i i

qr 0 r dr̂ v

v v

(2)

(3)

156

• ids and iqs are orthogonal to each other and called the flux and

torque-producing currents.

• RFOC is achieved by aligning the d-axis of the synchronous

reference frame with the rotor flux vector λr (λdr=|λr|, λqr=0

and dλdr/dt= dλqr/dt=0).

• Ids=CONST for operation up to the base speed. It is reduced in

order to weaken the rotor flux (constant-power-like

characteristic).

• Based on how the rotor flux is detected and regulated, indirect

rotor flux oriented control (IFOC) method and the direct vector

control method.

Direct RFOC

157

In direct FOC, rotor flux angle θ is obtained by using flux-sensing

devices embedded inside the motor or using measured motor

terminal voltages and currents.

The requirement for accurate air-gap flux sensors with ability to

reject the effects of slot ripples are significant problems.

Also, the necessity for mounting additional search coils in the air-

gap which requires machine modifications have precluded the

adoption of the direct method.

Estimation Error in the rotor flux and angle comprises of a drift

produced by the integrator because of initial conditions and

terminal voltage and/or phase current sensor measurement offset

error.

Indirect RFOC

160

In the indirect RFOC, the rotor flux angle θ is obtained from

detected rotor position angle θr and calculated slip angle θsl :

θ=θr+θsl

Terminal voltages sensors or flux sensors are not needed.

Please note the terminal voltages are PWM pulse trains (amplitude

is ±Vdc) and the flux calculation only needs the fundamental

component of the terminal voltage, which normally requires a low

pass filter or is estimated using DC bus voltage and duty cycle of

the switching signals.

Indirect RFOC scheme 1

161

r rr m ds

r

ˆL d ˆ L iR dt

* *m rsl qs

rr

L Ri

ˆ L

Voltage

Source

Inverter

PI

*

qi

ˆqi

*

di*

dsv

*

qsv

PIˆdi ˆ

eq

-+

*

anv

*

bnv

-+

abc

dq

d−q

abc

EIM

ˆeq

*

cnv

asi

bsi

csi

1/Lm

*

dr

*

2

3

r

m dr

L

PL

*

eT

*

r m

r dr

R L

L +

+

0

t

ed *

sl

ˆre

Ppˆ

rm

e

Indirect RFOC scheme 2

163

* m rsl qs*

rdr

L Ri

L

2

* * *mdev dr qs

r

L3 PT i

L * *

rd m dsL i

Indirect RFOC scheme

164

Indirect RFOC scheme

165

vds

dq Rotor Flux in RFOC scheme

166

Psi_qr

ωs

ωr

Dynamic Model

167

( )ABC dqo sv v v A

Solve diff equations (6.3.22) for ids and iqs

1.

2.

mdev m L m

d 1T D T si gn

dt J

3.

r = pm . 4.

r sl oq q 5.

6. Repeat steps 2-5, taking values of variables in the current step as initial values for the next step.

Dynamic Model

168

s s e s m e mds ds

e s s s e m mqs qs

dr drm e r r r e r r

qr qre r m m e r r r r

d dR L L L Lv i

dt dt

d dL R L L Lv i

dt dt

d dv iL R L Ldt dt

d dv iL L L R L

dt dt

(6.3.22)

Example of RFOC IM

170

50Hz parameters of a 415V/50Hz, 4-pole, squirrel-cage IM are: 𝑅𝑠 = 3.257 Ω; 𝑅𝑟 = 3.392 Ω; 𝐿𝑠 = 𝐿𝑟 = 589.9 mH; 𝐿𝑚 =569.5 mH; 𝑁𝑟 = 1450 rpm; 𝐼𝑟𝑎𝑡𝑒𝑑 = 5.8 Arms. The maximum current allowed in the machine is 2*𝐼𝑟𝑎𝑡𝑒𝑑. Equations of IM under Indirect RFOC (without field weakening), are given by

drrdr m ds

r

dLL i

R dt

m rsl e re qs

dr r

L Ri

L

2

mdev dr qs

r

L3 PT i

L

Example of RFOC SM

171

(a) Calculate the approximated value of the rated rotor flux linkage and stator current reference of 𝑖𝑑𝑠.

ph,* * 415 2 31.08 Wb-t

50 2

peak

r dr

e

V

(b) Calculate the stator current reference for 𝑖𝑞𝑠 if the motor is to

develop 20 Nm of torque. And maximum developed torque under rated rotor flux linkage.

* *

*

2 2*589.9*20 6.39A

3*2*569.5*1.083P

rqs e

m dr

Li T

L

*max max

3P1.5*2*569.5 / 589.9*1.08*16.3 51Nm

2

me qsdr

r

LT i

L

** 1.08

1.9 A0.5695

drds

m

iL

2

* 2 2 2max s max ds 2*5.8* 2 1.9 16.3Aqsi i i

Voltage

Source

Inverter

PI

*

qi

ˆqi

*

di*

dsv

*

qsv

PIˆdi ˆ

eq

-+

*

anv

*

bnv

-+

abc

dq

d−q

abc

EIM

ˆeq

*

cnv

asi

bsi

csi

1/Lm

*

dr

*

2

3

r

m dr

L

PL

*

eT

*

r m

r dr

R L

L +

+

0

t

ed *

sl

ˆre

2ˆ

rm

e

-+

ˆrm

*

rm

Block Diagram of IRFOC IM Drive

172

(c) Complete the following block diagram for RFOC IM drive.

section 6. dynamics of electric drive systems section 6 - dynamics... · dcm has two windings...

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