recap on theory of synchronous machines. variable … section 4... · recap on theory of...

§ Recap on theory of synchronous machines.§ Variable-speed (frequency) performance

using the steady-state equivalent circuit.

1

Section 4 - Synchronous Motor Drive

ELEC4613 – Electric Drive Systems

Synchronous motor drive applications

§ Used in large drive applications at constant speed such as oil rigpumping, cement mills, ship propulsion and so on. (no slip)

§ More recently they are used in many other variable speedapplications such as disk drives, robotics, machine tool drives andother automation applications requiring high dynamicperformance.

§ They are very suitable for applications where a number of motorshave to be operated with fixed speed ratio of a reference speedsetting. Numerous examples exist in paper, textile and metalcontinuous strip rolling mills.

§ Used in many wind power generators, large and small. Most HEVsand EVs employ PMSMs.

2ELEC4613 – Electric Drive Systems

SM types

• Non salient-pole (NSP) synchronous machine • Salient-pole (SP) synchronous machine

3

Rotor with non-salient and salient poles


Rotors of various type of synchronous motors

DC current terminals

Shaft

Steelretaining

ring

DC currentterminals

Wedges

Shaft

Pole

DC excitationwinding

Fan

Sliprings


Main features

• The rotor speed is given by

ssyn

fNp

= rev/sec or ssyn

2 fp

πω = mech rad/sec

• Average speed regulation = 0

• Voltage control has no effect on the steady-state speed. Thestator voltage and the rotor excitation determine the maximumtorque which the machine will develop.


Stator winding and airgap field

a’

b’ c’

b c

Axis of phase a

a


Non-salient pole SM with sine distributed stator winding

θsin2

Nn s=


lg

Ns is the total number of turns in aa′

Non-salient pole SM with sine distributed stator winding

o s ag o g

g

N IB / 2l cos2l

µµ θ= ℑ =

sa s a

N I sin d N I cos2

θ π

θθ θ θ

+ℑ = =∫

At any position θ, mmf is found by applying Amp’s law along a closed semi-circular path A-B-C-A (amp’s law contour)

The close path includes air gap twice, so

Note: (i). Since the iron core of stator has infinite permeability, the mmf in the core can be neglected.

(ii).The air-gap B field due to sinusoidal distributed winding varies with θ along the air-gap.

(iii). r 0B Hµ µ=MMF I N= × H MMF / l=

MMF – magnetomotive force, ampere-turns.H – magnetic field strength. MMF per unit lengthµ0 , µr – free space, and relative permeability.

MMF and B field of sine distributed windings

11

sa a s a

N I sin d N I cos2

θ π

θℑ θ θ θ

+= =∫

g g c c s aH 2l H l N I cosθ θ× + =

o s aga o g

g

N IB H cos2lθ

µµ θ= = assuming infinite permeability in iron


For DC current I in Phases B and C,

( ) ( )cos cos 2 / 3 cos 4 / 3 =02o s

gg

N IBl

µ θ θ π θ π= + − + −

The resultant field B is given by

cos( 2 / 3)2o s

gbg

N IBl

µ θ π= − cos( 4 / 3)2o s

gcg

N IBl

µ θ π= −

a b ci i i I= = =

MMF and B field of sine distributed windings

0gB =

Will the AC machine rotate?

Air-gap field B distribution of each phase

With sinusoidal excitation, the B-field of each phase is sinusoidally distributed in space (θ) around the axes of the respective windings, which is the same as DC excited field. However, the amplitude of each phase B at a given θ is alternating in time with frequency ωs , but stationary in space.

( )a m si I cos tω=

( )b m si I cos t 2 / 3ω π= −

( )c m si I cos t 4 / 3ω π= −

cos( ) cos( )2

o s mga s

g

N IB tl

µ ω θ=

cos( 2 / 3)cos( 2 / 3)2

o s mgb s

g

N IB tl

µ ω π θ π= − −

cos( 4 / 3) cos( 4 / 3)2

o s mgc s

g

N IB tl

µ ω π θ π= − −

Bm

Magnetic Field Axes of Stator Windings

Stationary: The peaks of field B in the air gap coincide with magnetic field axes at any instant of time. Eg. Bbg is aligned with axis-B (120 degree), its amplitude is alternating with the time. ( )scos t 2 / 3ω π−

Rotating air-gap field due to stator currents

( )

2 2 4 4Acos cos Acos cos Acos cos3 3 3 3

3 Acos2

π π π πα β α β α β

α β

+ − − + − −

= −

Using the following identity,

For balanced three-phase currents of amplitude Im in windings aa’, bb’, and cc’, the resultant field B in the air-gap is given by

( ) ( )o s mg ag bg cg s

g

3 N IB B B B cos t4l

µθ ω θ= + + = −

o s ms

g

N It , , A2l

µω α θ β= = =

• Bg (rotating magnetic field) retains its sinusoidal waveform and magnitude but rotates progressively around the air-gap.

• It rotates at a constant angular speed of ωs. The direction of rotation is from the leading phase axis to lagging phase axis (A-B-C).

• The peak value is 1.5 times Bmax of each phase whenever ωst=θ.

( ) ( )o s mg s

g

3 N IB cos t T4l

µθ ω θ= −

( )s mg s

3N I cos t4

ω θℑ = − A-T


Physical Interpretation

Rotating air-gap fieldωt =0, ia=Im, ib= ic= -0.5ImThe resultant flux mag. 1.5Bm is on the axis a (θ= 0);

ωt =60⁰, ic=-Im, ia= ib= 0.5Im, The resultant flux mag. 1.5Bm is on the magnetic axis of phase c but in opposite direction (θ= 60);

ωt =120⁰, ib=Im, ia= ic= -0.5Im, The resultant flux mag. 1.5Bm is on the positive direction of magnetic axis of phase b (θ= 120⁰).

Pole numbers and synchronous speed§ The stator and the rotor may have a

higher number (always an eveninteger) of poles.

§ The winding structure of a 2p polemotor (p is the number of polepairs) is similar to two-polewinding except that sinusoidallydistributed winding is repeated ptimes in 360 mechanical degrees

§ For a 2p pole machine, each cycleof 2π electrical radians of supplycurrent causes the stator field torotate by 2π/p mechanical radians.

rev/secssyn

fNp

=

electrical mechanicalpθ θ=


a1

a1' a2'

a2

Exercise 1: Synchronous speed

21

What is the base speed of a 300kW, 6600 V, 8-pole, 50Hz Y-connected non-salient pole synchronous machine in mechanical rad/sec and rev/min? What about the base speed for 60Hz?

rev/mins60 f 60 50N 750p 8 / 2× ×

= = =

( ) rev/secsfN 50 8 2 12.5p

= = =

rad/secs2 f 2 3.14 50N 78.5p 8 / 2

π × ×= = =

rev/mins60

60 f 60 60N 900p 8 / 2× ×

= = =

Summary

§ Stator has sinusoidally distributed winding.§ Balance three phase supply to such winding

produces a revolving magnetic field which rotates at synchronous speed and causes the rotor to turn.

§ Two types of rotor : Non-salient pole (uniform air gap) and salient pole.

§ Rotor field is produced either by electro-magnets supplied from a separate DC supply or by permanent magnets.

Machine representation: stator flux linkages


Laao = Lbbo = Lcco

Machine representation: stator flux linkages


va ea

Ra La

vf

ia Rf Lf if

Equivalent Circuit of DC Machine

The flux linkage of the aa’ winding due to the co-sinusoidallydistributed rotor field around the rotor poles is

Flux linkage of aa’ winding due to rotor fieldaf af f f

ˆL I cos cosλ θ λ θ= =

ˆaf fλ λ=

0dt

d af =λ

θ = 0°

θ = 90°

af 0λ =

25

afdmax

dtλ

=


a

a’

We assume sine distribution of rotor field:

Stator flux linkage contd.Each stator winding links with field from the other two windingsand the rotor. Fields from other two stator windings are displaced by120° and 240° (mechanical).

( )a aao aal a ab b ac c afL L i L i L iλ λ= + + + + a

b’

b

θ

120 °

λbb

a

aao a aal a aao b

aao c af

L i L i L i cos 120

L i cos 240 λ

= + + °

+ ° +

s a af s a af f s a fˆL i L i L i cos L i cosλ θ λ θ= + = + = +


aao a aal a aao b aao c af

aao a b c aal a af

1 1L i L i L i L i2 2

1 1L i i i L i2 2

λ

λ

= + − − +

= − − + + alaao2

3s LLL +=

Phase voltage equations

dtd

dtdi

LRidt

dRiv afa

saa

aaλλ

++=+=

This equation is very similar to the voltage equation of a DCmachine. Note that induced voltage due to mutual flux betweenwindings are included in the second term in the RHS.

where Laao = air-gap inductance of each winding

Lal = leakage inductance of each winding

alaao23

s LLL += = Synchronous inductance, H

bfb bb b b s

dd div Ri Ri Ldt dt dt

λλ= + = + +

dtd

dtdi

LRidt

dRiv cfc

scac

ccλλ

++=+=


Back emf due to rotor field at steady speed

m sp dt dtθ ω δ ω δ= + = +∫ ∫For a fixed or constant speed of rotation, rotor field angle,

electrical radians

( ) ( )af f s af f sˆ cos t L I cos tλ λ ω δ ω δ= + = +

( )afaf s f s s af f s

d ˆe sin t L I cos tdt 2λ πω λ ω δ ω ω δ = = − + = + +

δ

eaf

90°

λaf va

E af

λaf

28

ˆˆ.f

af f s s2 f

E 4 44 N f2

π λϕ= =


B field with DC current (Review)

29

ga mB B cosθ=


( ) ( )cos cos 2 / 3 cos 4 / 3 =02o s

gg

N IBl

µ θ θ π θ π= + − + −

The resultant field B is given by

cos( 2 / 3)gb mB B θ π= −

cos( 4 / 3)gc mB B θ π= −

2o s

mg

N IBl

µ=where

Bga

Bgb

Bgc

Air-gap field B distribution of each phase

cos( ) cos( )ga m sB B tω θ=

cos( 2 / 3) cos( 2 / 3)gb m sB B tω π θ π= − −

cos( 4 / 3)cos( 4 / 3)gc m sB B tω π θ π= − −

B field with three-phase sinusoidal excitation (Review)

( ) ( )g m sB 1.5B cos tθ ω θ= −

gaBgbB gcB

1.5 mB

mB

sω

θ

0stω = 2stω π=

Synchronous speed and rotor speed (Review)

• The rotor speed is given by

ssyn

fNp

= rev/sec or ssyn

2 fp

πω = mech rad/sec

• Average speed regulation = 0

• Voltage control has no effect on the steady-state speed. Thestator voltage and the rotor excitation determine the maximumtorque which the machine will develop.


Stator flux linkage (Review)

aa aa aL iλ =

a

b’

b

θ

120 °

λbb

a

ab bbo b aao bL i cos120 0.5L iλ = ° = −


a

a’

θ

120°

ic

c c’

ic λcc

ac cco c aao cL i cos 240 0.5L iλ = ° = −

a’

a θ

+ia

ia

λaa

af f cosλ λ θ=

Per-phase equivalent circuit of SM (Review)


va ea

Ra Ls

vf

ia Rf Lf if

ˆˆ.f

af f s s2 f

E 4 44 N f2

π λϕ= =

E af

λaf alaao2

3s LLL +=

dtd

dtdi

LRidt

dRiv afa

saa

aaλλ

++=+=

Steady-state equivalent circuit with R 0 and no-load

Ef∠δ°

jXs=jωLs I

V∠0°

Eaf

I

jIXs V

λaf

δ = 0° γ or φ

Eaf

I

jIX s

V

λa f

δ = 0°

γ or φ

(a) Under-excited rotor (Φ=90) (b) Over-excited rotor (Φ=-90)

dev afP E I cos 0γ= =

90γ φ= = ± oinP VI cos 0φ= =

Phasor diagram with load and R > 0

(a) Under-excited motor, R > 0 (b) Over-excited motor, R > 0

A

35

f

S

V 0 EI

R jXδ∠ − ∠

=+

o o

γ

φ

Ef I

IR

jIXs

V

λf

δ γ φ

Ef

I

IR jIXs

V

λ f

δ


Ef∠δ°

jXs=jωsLs I

V∠0°

R

Steady-state load characteristic, R = 0

0 0f f

s s

V E V EI

R jX Zδ δ

θ∠ − ∠ ∠ − ∠

= =+ ∠

Ef∠δ°

jXs=jωLs I

V∠0°

R

( )/ 2/ 2

s s

EVIX X

δ ππ ∠ −∠ −= −

[ ]Re cos( ) sin2

f f

s s

E EI

X Xπδ δ= − − = −

[ ] f

s

VEP V Re I sin

Xδ= × = −

Developed power per-phase

ft

s

3VEP sin

Xδ= − Watts

f

s

VEP VI cos sin

Xφ δ= = −

Watts/ph EfI

jIXs

V

f

( )f sE sin X I cosδ φ= −

Or using the phasor diagram

Exercise 2: Steady-state equivalent circuit with R 0

37

300kW, 6600 V, 50Hz, Y-connected non-salient pole synchronous motor: Xs=72Ω/ph at 50Hz, R≈0Ω/ph, and Efo=3200V/ph at base speed. Calculate δwhen the motor delivers rated load at base speed.

( )

sin

/sin ,

sin .

f 0o dev

S0

3VEP P

X

3 6600 3 3200300 000

72

0 59

δ

δ

δ

= =

× ×= =

∴ =

Assuming R=0 and no other losses,

.36 2δ = o

Developed power and torque in non salient-pole machines

Developed power3

sinf

s

VEP

Xδ= − W

3 sinf

syn s s s

VEP P pTp X

δω ω ω

= = = − ×Developed torque

90°

Genera tor δ

P , Watts T, Nm

M otor 180°

− 180° − 90°

− δ

The developed torque is +ve when δ is negative, Ef lags V.


Nm


39

Calculate the input current and power factor (PF) of the motor used in Exercise 2 when it delivers rated load at base speed.

cos sinSX I Vγ δ=

sin cosS foX I V Eγ δ= −

( ) cos2 2 2S f 0 f 0IX V 2VE Eδ= − +

, ..

P 300 000PF 0 837753VI 3 3810 31 33

= = =× ×

cos

cos .

. A

2 2f 0 f 0 S

2 2

I V 2VE E X

3810 2 3810 3200 36 2 320072

31 3

δ= − +

− × × × +=

=

o

( ) ( )2 21 2+

(1)

(2)

Method I:


40

Method II (equivalent circuit):

0

3810 3200 36 272 90

31 33 33 1

o fo

so

V EI

jX.

. .

δϕ

∠ ° − ∠ −∠ =

− ∠ − °=

∠ °= ∠ − °

cos . .PF 33 1 0 83775∴ = =o lagging.

Phasor diagram and the stator reference frame

Ef

jIdXs

Iq I

Id λaf

jIqXs

V

δ

q-axis

d-axis

jIXs

γ

φ

Id and Iq are currents in two fictitious windings in the stator whichproduce mmf along the d and q axes of the rotor.


Salient-Pole Synchronous Motor

Xd = ωs Ld

Xq = ωsLq

In general, Xd > Xq.

Xd and Xq are synchronousreactances of the statorwindings on the d and q axes,respectively.


•The SP machine does not have an uniform air-gap. •The air-gap length . Magnetic reluctance is low along the poles and high in the inter-pole regions. Hence, more flux can pass the air-gap along d-axis than q-axis.

ad aql l<

Developed power in Salient-Pole machines, R ≈ 0.

δsinVIX qq =

δcosVIXE ddf =+

cos sinq dP V I V Iδ δ= ⋅ − ⋅

2

sin sin 22

f d q

d d q

VE X XVPX X X

δ δ −

= +

Watts/phase


cos cos sinq dI I Iφ δ δ= −

δ φ

γ

q- axis

d- axis

Id

I V

jIdXd Ef

jIqXq

Iq

Developed power & torque of the salient-pole SM

23 3sin sin22

f d q

d d q

VE X XVPX X X

δ δ −

= +

The total developed power,

Nm

m

syn s

2f d q

s d d q

3P 3PT/ p

VE X X3 p Vsin sin 2 NX 2 X X

ω ω

δ δω

= =

− = × +

The developed torque,

The second term, the reluctance torque, is because of the difference between the d and q axes reluctances (or saliency ratio Ld/Lq) of the rotor. Higher saliency implies higher reluctance torque.


T - δ characteristic of the SP SM with separately excited rotor

0 0.5

1.0 1.5

− 20° − 40° − 60° 0° − 80°

− 100°

− 140° − 180°

Ef/V = 2.0

T, Nm

δ


For such machines, Lq < Ld , the max torque occurs for -δ<90°

SP permanent-magnet synchronous motor

Torque due to permanent

magnet

Reluctance torque

Total torque

90° 180° − δ

T, Nm

For such machines, Lq > Ld

46

q- axis d- axis


m2

f q d

s d d q

VE X X3 p VT sin sin 2 NX 2 X X

δ δω

− = × + −

Synchronous machine with no excitation in the rotor

Nmδω

2sinXX

XX2pV3T

qd

qd

s

2

−=

D-axis

Q-axis

Air flux Barriers

Narrow ironbridge

Shaft

T, Nm

δ

− 45° − 90° − 180°


Summary

§ Synchronous speed syn s2 f pω π= mech rad/sec

Ef∠δ°

jXs=jωLs I

V∠0°

R=0

EfI

jIXs

V

f

§ Equivalent circuit and phasor diagram (non-SP)

3 sinf

syn s s s

VEP P pTp X

δω ω ω

= = = − ×

Summary

Generator operation Motor operationI lags V Ef >V, over-excitation Ef <V, under-excitation

I leads V Ef <V, under-excitation Ef >V, over-excitationAngle δ Positive, Ef leads V Negative, Ef lags V

γ

φ

Ef I

IR

jIXs

V

λf

δ γ φ

Ef

I

IR jIXs

V

λ f

δ

q qI V sin Xδ=

fd

d

V cos EI

Xδ −

=


δ φ

γ

q- axis

d- axis

Id

I V

jIdXd Ef

jIqXq

Iq

Summary§ Phasor diagram of salient pole SM

m2

f d q

s d d q

VE X X3 p VT sin sin 2 NX 2 X X

δ δω

− = × +


Summary

Non-SP SM Reluctance SM

SP PM SP SM (Wound Rotor)

inductance Lq = Ld = Ls Lq < Ld Lq > Ld Lq < Ld

excitation Ef Ef =0 Ef Ef

-δ|Tmax (deg) 90 45 >90 <90

Exercise 4: Salient-Pole machines, R ≈ 0.


A 4-pole, Y-connected, SP permanent magnet synchronous

motor has the following parameters at 50 Hz:

Xd=7.5 Ω/ph, Xq =15.7Ω/ph, R ≈0Ω/ph.

Ea=150 Vrms at 1200 rev/min.

The inverter supplies 50 Hz 200Vrms/ph to the motor,

(i) Calculate the developed torque at the base speed.

o= 120δ

Solution of Exercise 4

f 1 f 0 1 sE / E N / N=

4-pole, 50 Hz / / ( / ) rev / minS 0N f p 50 4 2 1500= = =

. V/phase;f 0150E 1500 187 51200

∴ = × =

1 150V@1500rpmfE =

oV/ph ; = 1200V 200 @ 50Hz δ=

20 f 0 d0 q00

devs d0 s d0 q0

3 pV E X X3 pVT sin sin 2 105.57X 2 X X

δ δω ω

−= + =

Tdev = ? at 1200rpm, 600rpm, 300rpm …?

VSI driven NSP synchronous motor

The inverter supplies three-phase AC sinusoidal voltage of variable RMS value V1 and frequency f1.

1 f 1

1 s1

3 pV ET sin

Xδ

ω= Nm, where &s1 1 s 1 1X 2 f L 2 fπ ω π= =

Tmax occurs at δ = 90° (electrical).54

ˆˆ.1 f

f 1 f s 12 f

E 4 44 N f2

π λϕ= = where is the peak flux/pole

from rotor. ˆ fϕ

Ef∠δ°

jXs=jωsLs I

V∠0°


3-phase VSI VVVf inverter

55

dAn,1 d

VV m 0.354m V2 2

= ⋅ = ⋅

where m is the depth of modulationELEC4613 – Electric Drive Systems

1

6

3

4

T - ω characteristic of a NSP SM with constant V/f ratio

ˆ.

ˆ ˆ.

s0 0 s

0 0

f 0 f s 0

f 0 f 0f s f

0 0

X 2 f L2 f

E 4 44 N f

E E4 44 N K

f

πω π

ϕ

ϕ ϕω

= = = = ⇒ =

f 1 f 0

s1 s0

1 0

E E

X X

λ

λω λω

= = =

We define, 1

0

ff

λ =

where

f 1 1 fo 1

syn o s1 o so

3 pE V E3P VT sin 3 p sin/ p X X

δ δω λω ω λ

= = =

( )ˆ ˆsin sin1 1f f

so s 1

V V3pT 3p K KX 2 L f

ϕ δ ϕ δλ π

= =

56

where f0 is the base frequency, say 50 Hz, at which rated voltage Vo /phase is applied


Nm

Nm

Speed rpm

Torque, N m

V 1/ f1 = C

V 2/ f2 = C V 3/f3= C

V 4/f4 = C V 5/f5 = C

T m ax

f1 > f2 > f3 > f4 > f5

fo

57


Torque characteristic with limited V1

T = Constant

Speed rpm

o

Torque, Nm

V1/ f1 = C

V2/ f2 = C V3/f3= C

V4/f4 = C V5/f5 = C

Tmax

Vo, f6

Vo, f7

Vo, f8

1 1 s sˆV 4.44 f N ϕ≈

58

If voltage drop in stator resistance is negligible,

V1 = Vo beyond fo, the stator flux linkage and torque falls proportionately when f1 > fo, resulting in constant maximum-power operation.


ϕs is the rotating fieldproduced by the 3-phasestator windings suppliedwith V1/phase.

( )ˆ ˆsin sin1 1f f

so s 1

V V3 pT 3 p K KX 2 L f

ϕ δ ϕ δλ π

= =

T-ω characteristic with constrained V1 and δmax(Constant maximum power)

1f max

s 1

3p VˆT K sin2 L f

ϕ δπ

=

syn1

K 'Tω

=


δ 2 ma x, V1

δ 1 ma x, V1 = cons tan t

δ 3 ma x ,V1

Speed

Torque

Variable-frequency performance of NSP SM at low speed

jXs=jωLs I

Ef1 V1

R

δsinVRIIX 1dq1s =−

1 1 1 coss d q fX I RI E V δ+ + =

jIqXs1

Ef1

RIq

V1

IdR

Id

λf

q-axis

d-axis

jIdXs1

I

Iq δ

60

The IR drop is not negligible at low speed when V1 is also low.


T-ω characteristic of NSP SM at low speed

( )21s

211f11s

dXR

sinRVEcosVXI

+

−−=

δδ

( )s1 1 1 f 1q 2 2

s1

X V sin R V cos EI

R Xδ δ+ −

=+

( )2 2dev q 1 d 1 d q f 1 qP I V cos I V sin R I I E Iδ δ= − − + =

( ) ( )1 1 1 1 1 12 2

1

sin cosf s f f

s

E V X RE V E

R X

δ δ+ −=

+W/phase

1

3PT/ pω

=( ) ( )f 1 1 s1 f 1 1 f 1

2 21 s1

E V sin X RE V cos E3 pR X

δ δω

+ −=

+Nm


(1) * Xs+(2) * R

(1) * R - (2) * Xs

Condition for Tmax

( )

foso

fo 1 122

o so

EX sin R cos

3 pE V VT

R X

λλ δ δ

ω λ

+ −

=+

oo1 f2πλλωω ==o1 ff λ=

f 1 foE Eλ= sosos11s Lf2XLX πλλω ===

Nm

1 somT

XtanR

λδ − =

For maximum torque, 0dTdδ

=


Tmax at low frequency

Tmax for fo

ωo fo

f1

f3

f2

Tmax for f3

fo>f1>f2>f3…


for a given mT maxTδ λ⇒

Low frequency voltage boostFrom 4.3.9 and 4.3.11, for λ = 0,

1d

V sinI

Rδ−

= RcosV

I 1q

δ= 2 2 1 ;d q

VI I IR

= + =

Thus, V1 for f1 = 0 = Irated × R

10

3 cosfo

o

pE VT

Rλ

δω≈ =

64


Neglecting the terms containing frequency

0 @ 0maxT λ δ≈

=

Salient Pole SM under VSI drive, R ≈0

−+= δδ

ω2sin

XXXX

2Vsin

XVEp3T

1q1d

1q1d2

1

1d

11f

1

1 ;of fλ= fo1f EE λ=

1 ;oω λω= do1d XX λ= qo1q XX λ=

2fo 1 do qo1

2o do do qo

E V X XV3 pT sin sin 2X X X2

δ δω λ λ

−= +

Nm

Thus, if V1/f1 ratio is kept constant, the same maximum torquewill occur at all speeds (frequency).


Salient-pole motor under VSI drive, R ≈ 0With V1 and δ constrained, thetor-speed characteristic isgiven by

1 22

K KTλ λ

= +

fo o1 1

o do

E V3 pK sin ;X

δω

=

1qodo

qodo2

o

o2 2sin

XXXX

2Vp3K δ

ω

−=

f3

f2

f1

fo ωo

Speed rpm

Tmax T, Nm


Torque with constrained V1 and δmax(constant maximum power)

221 KKT

λλ+=

fo o1 max

o do

E V3 pK sinX

δω

=

If the load angle δ is limited to some arbitrary value δ1max, and the supply voltage is also kept constant, say at the rated value Vo

maxsin2

do qoo2

o do qo

X X3 p VK 22 X X

δω

−=

δ2, V2

δ1, V1 = constant

δ3 ,V3

Speed

Torque


A constant power like T-ω characteristics

The salient pole motor at low frequency

q1 q d 1X I RI V sinδ− =

d 1 d q f 1 1X I RI E V cosδ+ + =

68

q - a x i s

E f V

δ

I q

I d

I

λ f

d - a x i s

R I d

j I q X q

R I q

j I d X d

φ

γ


Torque at low speed with R not negligible( )q1 1 f 1 1

d 2d q

X V cos E RV sinI

R X Xδ δ− −

=+

( )d 1 1 1 f 1q 2

d q

X V sin R V cos EI

R X Xδ δ− −

=+

( )2 2dev 1 q 1 d d qP V cos I V sin I R I Iδ δ= × − × − +

( )d q d q f 1 qX X I I E I= − + W/phase

( )2 2

1 11 fo qo do qo 1 fo

2 2o do qo

V VR V E X sin X X sin2 V RE cos3p 2TR X X

λ δ δ δλ

ω λ

+ + − −

= +

Tmax at fo

ωo fo

f1

f3 f2

Tmax for f3

fo>f1>f2>f3…


Torque at low speed with R not negligible

Tmax at fo

ωo fo

f1

f3 f2

Tmax for f3

fo>f1>f2>f3…


( ) for a given mT maxdT 0 f cos ,cos 2 ,sin Td

δ δ δ δ λδ

= ⇒ = ⇒

Torque at low and high speeds (Review)

( )2 2

1 11 fo qo do qo 1 fo

2 2o do qo

V VR V E X sin X X sin2 V RE cos3p 2TR X X

λ δ δ δλ

ω λ

+ + − −

= +

Tmax at fo

ωo fo

f1

f3 f2

Tmax for f3

fo>f1>f2>f3…


f3

f2

f1

fo ωo

Speed rpm

Tmax T, Nm

2fo 1 do qo1

2o do do qo

E V X XV3 pT sin sin 2X X X2

δ δω λ λ

−= +

1V Cλ

=

Torque characteristic with limited V1 (review)T = Constant

Speed rpm

o

Torque, Nm

V1/ f1 = C

V2/ f2 = C V3/f3= C

V4/f4 = C V5/f5 = C

Tmax

Vo, f6

Vo, f7

Vo, f8

1 1 s sˆV 4.44 f N ϕ≈


syn1

K 'Tω

=

1 22

K KTλ λ

= +

Non SP SM

SP SM

Low frequency voltage boost (Review)

V1 for f1 = 0 = Irated × R

10

3 cosfo

o

pE VT

Rλ

δω≈ =

73


0 @ 0maxT λ δ≈

=

CSI synchronous motor drive§ Higher dynamic response and better reliability due

to high dynamics of current control.

§ Automatic current limiting feature in a CSI drive.

§ In variable-speed applications, the synchronousmotor is normally driven from stiff current sourcesrather than voltage sources.

§ A shaft position sensor is normally required tosynchronise the supply current waveforms with rotorposition.


PWM current regulated VSI drive scheme

+ DC-Link

− DC-Link

−

E

−

−

LOOK

UP

TABLE

Sinusoidal Current

References (iaref, ibref &

icref ) of amplitude

= m

ia

ic

ib

iaref

ibref

icref

P W M

P W M

P W M

iaref

ibref

icref

T1

T1

T3 T5

T6 T4 T2

T1

T4

T3

T6

T5

T2

AC supply ia

ic

m γ


Torque reference = m

CSI driven synchronous motor in BLDC operation

~ ~ ~

θ Converter Switching

Circuit

Current controller

T1-T6

T1

T4 T2

T5

T6

T3

α

DC Link Inductor IDCLlink

Iref +

−

IDCLink

E


CSI drive contd.

IDCLink = Id

60° 120°

120° 60° ea ia

ia1

γ − Id

γ +Id


Torque characteristic with CSI drive for a NSP SM

γcosIE3P f=

3 cosˆ cosf

fE I

T K Iγ

ϕ γω

= = Nm

γ

efa

ia

t

78

jIqXs

Ef V

Id λaf

q-axis

d-axis

jIdXs

I Iq δ γ

IqR IdR

φ


Maximum Torque per Ampere (MTPA) drive

79

V E f

λ a f

IqX s

φ

γ = 0 °

I= Iq

IqR


Operation with field weakening

ˆo o fV E Kϕ ω≈ =

Above base speed

Vo can be higher than the rated voltage, which should be avoided.When the I phasor leads Ef , it has –ve Id, component which mayreduce ϕf by armature reaction.Thus, operation above base speed can be achieved by forcing thephase current waveform to lead the ef. If Ld is not relatively larger,e.g. in Surface PMSM, the armature reaction may not be effective.

80

d-axis λaf

Ef I

Id

Iq

q-axis


s sd d d s fN L I Nϕ ϕ= +

Power factor improvement using CSI drive

va efa ia

γ

φ δ

I lags Ef

81

V E f

λ a f

I q X s

φ

γ

j I d X s

R I q R I d

I

I d

I q


PF improvement using CSI driven over-excited SM

I leads Ef

82

V E f

λ af

jIqX s

φ

I

IqR

γ

Id

Iq

jIdX s

IdR

I leading E f

γ

δ

φ

va

efa ia


recap on theory of synchronous machines. variable … section 4... · recap on theory of...

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