Lecture 3: The Synchronous Machine - a-lab.ee · Lecture 3: The Synchronous Machine 2 Mechanical...

Yoash Levron Juri Belikov½ The Andrew and Erna Viterbi Faculty of ElectricalEngineering, Technion—Israel Institute of Technology,Haifa 3200003, Israel

½ Department of Software Science, Tallinn Uni-versity of Technology, Akadeemia tee 15a, 12618Tallinn, Estonia

R [email protected] R [email protected]

Lecture 3: The Synchronous Machine

This lecture presents a dynamic model of the synchronous machine. We demonstrate howto use this model in power system simulations, and explain the relations between the ma-chine’s dq0 model and time-varying phasor model.Synchronous machines are often operated as generators, and are a major source of energyin electric power systems. In several applications synchronous machines are also operatedas motors. A basic diagram of the machine is shown in Fig. 1. The key mechanical com-ponents of the machine are the rotor and stator. There are also two key electric parts: thefield winding on the rotor, and the three-phase winding on the stator. The field winding istypically connected to a DC source, which creates a magnetic field with alternating northand south polarities, as illustrated in Fig. 1. As the rotor rotates AC voltages are inducedin the stator windings. In addition, the interaction between magnetic fields and currents inthe machine produces a torque that acts to decelerate the rotor. These two processes resultin energy conversion, from mechanical energy to electromagnetic energy as a generator, orvice-versa as a motor.The machine is named “synchronous” since at steady-state the rotor speed is proportionalto the frequency of currents and voltages in the stator. This is not necessarily the case inother machines. For instance in induction motors the frequency of the rotor must be slightlylower than the frequency of the AC current.

b′

a

c′

b

a′

cif

S

N

Rotor

Stator

Air gapStator windings

Magnetic field line

Figure 1: Schematic diagram of a three-phase synchronous machine with two poles.

Remark 1: Synchronous machines have complex dynamics and control mechanisms that can-not be fully addressed in a short lecture. Therefore several topics such as the accurate primemover dynamics or the structure of excitation systems are not fully explained. Additionalinformation may be found in [1, 2].Remark 2: For the convenience of the reader a list of symbols is provided in Appendix A.

Series of lectures on power system dynamics. The lectures are freely available onhttps://a-lab.ee/projects/dq0-dynamics

https://a-lab.ee/projects/dq0-dynamics

Lecture 3: The Synchronous Machine 2

Mechanical model

Synchronous machines often have more than two magnetic poles on the rotor. Since quanti-ties associated with any pair of poles are identical, it is usually convenient to define electricalquantities which are related to a single pair of poles. For this reason the rotor angle is oftenexpressed in electrical degrees or electrical radians, rather than in mechanical units.One pair of poles is mapped to 360 electrical degrees or 2π electrical radians. Since there arepoles/2 electrical cycles in one mechanical cycle it follows that

θ =pf2θm, (1)

where

3 θ denotes the rotor electrical angle;

3 θm denotes the rotor mechanical angle, with respect to a fixed point on the stator;

3 pf denotes the number of magnetic poles on the rotor (an even integer).

In addition frequencies are defined as follows:

3 ω = dθ/dt denotes the rotor electrical frequency;

3 ωm = dθm/dt denotes the rotor mechanical frequency;

3 ωs denotes the nominal grid frequency, for instance ωs = 2π50 or 2π60 rad/s.

Following (1) we haveω =

pf2ωm. (2)

The frequencies are typically measured in rad/s.

The angular acceleration of the rotor is given by

d

dtωm =

1

J(Tm − Te), (3)

where

3 J denotes the rotor moment of inertia.

3 Tm denotes the mechanical torque, accelerating the rotor.

3 Te denotes the electromagnetic torque, decelerating the rotor.

In addition powers are defined as

pm = Tmωm =2

pfTmω,

pe = Teωm =2

pfTeω,

(4)

where

3 pm denotes the mechanical power accelerating the rotor,

3 pe denotes the electromagnetic power decelerating the rotor.


Equations (2) and (3) yieldd

dtω =

pf2J

(Tm − Te), (5)

and following (4) and (5) we have

d

dtω =

(pf2

)2 1

Jω(pm − pe) = K

ωsω(pm − pe), (6)

where K =(pf

2

)2 1Jωs

.Under the approximation ω ≈ ωs this is the classic swing equation. However in this lecturewe do not use this approximation, and work directly with the exact expressions in (5) and(6). It is also assumed that the mechanical torque is governed by a droop mechanism whichis described as

Tm =pf2ωs

(3Pref −

1

D(ω − ωs)

), (7)

where

3 D denotes the damping coefficient.

3 Pref denotes the reference power.

As we have seen in Lecture 1 this type of control is crucial for regulating the frequency andmaintaining stability.

Remark 1: The linear relationship in (7) ignores the complex dynamics of the prime-mover,and holds only for slow transients. A more accurate model may be found in [2].Remark 2: The droop mechanism in (7) is defined with respect to torque instead of withrespect to power, as done in Lecture 1. This form is chosen to simplify the dynamic equationswithout using the approximation ω ≈ ωs.

Substituting (7) into (5) yields

d

dtω = K

(3Pref −

2

pfωsTe −

1

D(ω − ωs)

). (8)

This equation defines the dynamics of the frequency ω with respect to the electromagnetictorque Te. Recall that in Lecture 1 we have used a time-varying phasor model and definedthe swing equation as

d

dtω = K

(3Pref − 3P − 1

D(ω − ωs)

). (9)

Equations (8) and (9) are similar except for the more accurate term 2pfωsTe that replaces the

active power 3P .

Electrical model

This section presents an electrical model of the machine which is based on dq0 quantities.The model is an extension of the time-varying phasor model presented in Lecture 1, andmay be used to describe fast transients. However, as all models, the model presented hereis an approximation. Specifically, it is assumed that

3 The machine is a magneto-quasi-static device,


3 Saturation of the magnetic materials and other sources of imbalance and harmonicdistortion are ignored,

3 Self- and mutual inductances are composed of a constant term, in addition to a sinu-soidal term varying with 2θ.

Although these assumptions may seem restrictive, they form the basis of the classic dq0model, and have been found to give excellent results in a wide variety of applications [1].The machine is described as a system of coupled inductors

λaλbλcλf

=

àa àb àc àf`ba `bb `bc `bf`ca `cb `cc `cf`fa `fb `fc `ff

−ia−ib−icif

, (10)

where

3 λa, λb, λc denote the stator flux linkages;

3 ia, ib, ic denote the stator currents (generator output currents). The negative signs havebeen included since currents are positive when flowing out of the generator;

3 λf denotes the field winding flux linkage;

3 if denotes the field winding current.

Following these definitions we have

va = −Raia +d

dtλa,

vb = −Raib +d

dtλb,

vc = −Raic +d

dtλc,

vf = Rf if +d

dtλf ,

(11)

where

3 va, vb, vc denote the stator terminal voltages (generator output voltages);

3 vf denotes the field winding voltage;

3 Ra denotes the resistance of each winding on the stator;

3 Rf denotes the field winding resistance.

The self- and mutual inductances in (10) depend on the rotor position. The stator self-inductances are given by

àa = Laa + Lg2 cos(2θ),

`bb = Laa + Lg2 cos(2θ + 120◦),

`cc = Laa + Lg2 cos(2θ − 120◦),

(12)

and the stator-to-stator mutual inductances are

àb = `ba = Lab + Lg2 cos(2θ − 120◦),

`bc = `cb = Lab + Lg2 cos(2θ),

àc = `ca = Lab + Lg2 cos(2θ + 120◦),

(13)


where Laa is a positive constant and Lab is a negative constant. These inductances are com-posed of a constant term, in addition to a sinusoidal term varying with 2θ. This additionalterm is required in case the rotor is not perfectly round (which causes “saliency effects”).In addition, the stator-to-rotor mutual inductances vary according to the rotor position andare given by

`af = `fa = Laf cos(θ),

`bf = `fb = Laf cos(θ − 120◦),

`cf = `fc = Laf cos(θ + 120◦),

(14)

and the field winding self-inductance is

`ff = Lff = constant. (15)

The model defined by the above equations typically does not have an equilibrium pointsince the inductances depend on the rotor angle θ, and thus vary with time. For this reasonwe will now develop an equivalent model which is based on dq0 quantities. Such a modelmay be obtained by applying the dq0 transformation to the equations above. Omitting thealgebraic details, the resulting dq0-based model is given by

λd = −Ldid + Laf if ,

λq = −Lqiq,λ0 = −L0i0,

λf = −3

2Laf id + Lff if ,

(16)

andvd = −Raid +

d

dtλd − ωλq,

vq = −Raiq +d

dtλq + ωλd,

v0 = −Rai0 +d

dtλ0,

vf = Rf if +d

dtλf ,

(17)

where

3 vd, vq, v0 are the dq0 transformation of va, vb, vc (stator terminal voltages).

3 λd, λq, λ0 are the dq0 transformation of λa, λb, λc (stator flux linkages).

3 Ld is the direct-axis synchronous inductance.

3 Lq is the quadrature-axis synchronous inductance.

3 L0 is the zero-sequence inductance.

Notes:

1. The reference angle for the dq0 transformation is the rotor electrical angle θ.

2. The dq0 variables do not depend directly on θ.


The inductances in (16) are given by

Ld = Laa − Lab +3

2Lg2,

Lq = Laa − Lab −3

2Lg2,

L0 = Laa + 2Lab.

(18)

We will usually work directly with these inductances.We also denote

Ls =1

2(Ld + Lq). (19)

For a perfectly round rotor with Lg2 = 0 (no "saliency effects") the synchronous inductancesare equal:

Ls = Ld = Lq. (20)

In this case Ls is the synchronous inductance of the machine, as in Lecture 1.To complete the basic model, the machine’s output power (flowing from the stator terminalsinto the grid) is

ps =3

2(vdid + vqiq + 2v0i0), (21)

and the electromagnetic power decelerating the rotor is

pe =3

2ω(λdiq − λqid). (22)

Following (22) and using (2) the electromagnetic torque is given by

Te =peωm

=3

2

pf2(λdiq − λqid). (23)

A complete state-space model of the synchronous machine is obtained by merging the equa-tions above. Using equations (8), (16), (17) and (23) and again omitting the algebraic detailswe obtain

d

dtθ = ω,

d

dtω = K

(3Pref −

1

D(ω − ωs) +

(3βωs − 6LffLqωs

2βLq

)λdλq +

3Lafωsβ

λqλf

),

d

dtλd = −

2RaLffβ

λd + ωλq +2RaLafβ

λf + vd,

d

dtλq = −ωλd −

Ra

Lqλq + vq,

d

dtλ0 = −

Ra

L0

λ0 + v0,

d

dtλf =

3RfLafβ

λd −2RfLdβ

λf + vf ,

(24)

The inputs of this model are vd, vq, v0, vf , Pref , and the constants are defined as

K =(pf2

)2 1

Jωs,

β = 2LdLff − 3L2af .

(25)


Several optional outputs (in addition to the state variables) are

id =2Lafβ

λf −2Lffβ

λd, iq = −1

Lqλq, i0 = −

1

L0

λ0,

if =2Ldβλf −

3Lafβ

λd,

Te =3

2

pf2(λdiq − λqid), pe =

3

2ω(λdiq − λqid),

Tm =pf2ωs

(3Pref −

1

D(ω − ωs)

), pm =

ω

ωs

(3Pref −

1

D(ω − ωs)

),

ps =3

2(vdid + vqiq + 2v0i0).

(26)

Following are several examples demonstrating how to use the above model in simulation.Consider first a synchronous generator feeding a symmetrically configured resistive loadRL. The objective here is to simulate the transient that follows a sudden 3-phase short circuitat the stator terminals. A possible signal flow diagram is shown in Fig. 2. Under normaloperating conditions the ratio between the voltages and currents is RL. When the shortoccurs the voltages vd, vq, v0 are zeroed, and a transient takes place.

Vf

Pref sync.

machine

model

if ps

idq0RL

vdq0

0short-circuit

normaloperation

switch

Figure 2: Simulating a sudden 3-phase short circuit at the stator terminals.

As a second example, consider the system shown in Fig. 3. A signal flow diagram is shownin Fig. 4.

∼medium length transmission line

(a) One-line diagram

sync.machine

i v

C2

i′

R L

v′

C2

i′′

RL

load

(b) Single-phase diagram

Figure 3: Example: synchronous machine connected to a medium length transmission line and load.


Vf

Pref sync.

machine

model

ω ps

idq0+ −

capacitordq0 model

vdq0

R-Ldq0 model

+ −

i′dq0+ −

capacitordq0 model

v′dq0

1RL

i′′dq0

Figure 4: Signal flow diagram for a synchronous machine connected to a medium length transmissionline and load.

The capacitors, inductor and resistor are modeled as in Lecture 3. The left capacitor is mod-eled as

d

dtvd = ωvq +

1

C/2(id − i′d),

d

dtvq = −ωvd +

1

C/2(iq − i′q),

d

dtv0 =

1

C/2(i0 − i′0),

(27)

the right capacitor is modeled as

d

dtv′d = ωv′q +

1

C/2(i′d − i′′d),

d

dtv′q = −ωv′d +

1

C/2(i′q − i′′q),

d

dtv′0 =

1

C/2(i′0 − i′′0),

(28)

and the series resistor and inductor are modeled as

d

dt

i′di′qi′0

=

−RL

ω 0−ω −R

L0

0 0 −RL

i′di′qi′0

+1

L

vd − v′dvq − v′qv0 − v′0

. (29)

Note that these equations depend on ω, which is an output of the machine’s model. It is alsopossible to use the approximation ω ≈ ωs to eliminate this dependency.

Simplified machine model

This section presents a simplified dynamic model of the machine. The model is based on thefollowing assumptions:

3 Round rotor: Lg2 = 0, or equivalently Ld = Lq = Ls,

3 Constant field current: if = If = const,

3 Balanced voltages and currents: v0 = 0, i0 = 0.

We will now see that under these assumptions

3 The machine may be described as an internal voltage source behind a series impedance;

3 The magnitude of the internal voltage is proportional to the frequency ω;


3 The electromagnetic torque is proportional to the current iq.

Based on (16) and (17) and using Ld = Lq = Ls, if = If , v0 = 0, i0 = 0 we have

λd = −Lsid + LafIf ,

λq = −Lsiq,

λf = −3

2Laf id + LffIf ,

(30)

andvd = −Raid +

d

dtλd − ωλq,

vq = −Raiq +d

dtλq + ωλd,

vf = RfIf +d

dtλf .

(31)

In addition, substituting (30) into (31) yields

vd = −Raid − Lsd

dtid + ωLsiq,

vq = −Raiq − Lsd

dtiq − ωLsid + ωLafIf ,

vf = RfIf −3

2Laf

d

dtid.

(32)

Note that vf does not affect the other quantities, and may be considered an output of themodel. Using the definition

VE = ωsLafIf = constant (33)

These equations may be written as

vd = −Raid − Lsd

dtid + ωLsiq,

vq = −Raiq − Lsd

dtiq − ωLsid +

ω

ωsVE,

(34)

or

vd = 0 −Raid −(Ls

d

dtid − ωLsiq

),

vq︸︷︷︸stator

voltages

=ω

ωsVE︸︷︷︸

internal voltage

−Raiq︸︷︷︸voltage drop

on series resistance

−(Ls

d

dtiq + ωLsid

).︸︷︷︸

voltage dropon series inductance

(35)

Based on these expressions the machine’s simplified model may be compactly described asan internal voltage source (induced EMF) behind a series impedance, as shown in Fig. 5.

ea−+

Ra Lsva

eb−+

Ra Lsvb

ec−+

Ra Lsvc

Figure 5: The simplified machine model: an internal voltage source behind a series impedance.


In this equivalent circuit the internal voltage source is given byedeqe0

=

0ωωsVE0

, (36)

where [ed, eq, e0]T is the dq0 transformation of [ea, eb, ec]

T. It can be seen that the internalvoltage is proportional to the frequency, where at steady-state and at nominal frequency thepeak voltage is VE = ωsLafIf .In addition, using (23), (30) and (33), the electromagnetic torque acting on the rotor (in thesimplified model) is

Te =3

2

pf2(λdiq − λqid) =

3

2

pf2LafIf iq =

3

2

pf2ωs

VEiq. (37)

The electromagnetic torque is proportional to the stator current iq.Based on these results, a simplified model of the machine may be obtained by separatingthe internal voltage source from the series impedance (Ra and Ls). This series impedance,although physically a part of the machine, can be modeled as if it is a part of the transmissionnetwork. The dynamic equations of the internal voltage source are

d

dtω = K

(3Pref −

2ωspf

Te −1

D(ω − ωs)

),

Te =3

2

pf2ωs

VEiq,

(38)

and direct substitution of Te provides the simplified model

d

dtθ = ω,

d

dtω = K

(3Pref −

1

D(ω − ωs)−

3

2VEiq

).

(39)

Here the inputs are Pref and iq, and several outputs (in addition to the state variables) are

ed = 0, eq =ω

ωsVE, e0 = 0,

Te =3

2

pf2ωs

VEiq, pe =3

2

ω

ωsVEiq,

Tm =pf2ωs

(3Pref −

1

D(ω − ωs)

), pm =

ω

ωs

(3Pref −

1

D(ω − ωs)

).

(40)

All the dq0 quantities are defined here with respect to the rotor electrical angle θ.

Energy conversion in the machine

We will now use the simplified machine model to explain the flow of energy in the machine.The total energy stored in the machine is composed of two parts:

3 The kinetic energy of the rotor: Erot = 12Jω2

m = 12J(

2pf

)2ω2,

3 The magnetic energy, represented by the energy stored in the synchronous inductance:EL = 3

4Ls(i2d + i2q

).


The kinetic energy derivative is

d

dtErot =

d

dt

(1

2Jω2

m

)= Jωm

dωmdt

, (41)

and dωm

dt= 1

J(Tm − Te) yields

d

dtErot = ωmTm − ωmTe = pm − pe. (42)

This is an intuitive result: the change in kinetic energy is equal to the difference between themechanical and electromagnetic powers. In addition, the magnetic energy derivative is

d

dtEL =

3

2Ls

(iddiddt

+ iqdiqdt

), (43)

and using (35) we have

d

dtEL =

3

2

(id (−Raid + ωLsiq − vd) + iq

(−Raiq − ωLsid +

ω

ωsVE − vq

))=

3

2

ω

ωsVEiq︸︷︷︸pe

− 3

2(vdid + vqiq)︸︷︷︸

ps

− 3

2Ra

(i2d + i2q

)︸︷︷︸

ohmic loss

= pe − ps −3

2Ra

(i2d + i2q

).

(44)

The flow of power in the machine is summarized in Fig. 6.

rotor

energy storage Erot

dErotdt

= pm − pe

energyconversion

(no storage, no loss)

energyloss

energy storage EL

dELdt

= pe − ploss − ps

ea−+

Ra Ls

eb−+

Ra Ls

ec−+

Ra Ls

⇒pm

mechanicalpower ⇒

pe

⇒pe

⇒pe − ploss⇑

ploss

⇒ps

electromagnetic power

to grid

Figure 6: Energy conversion in the machine, based on the simplified model.

The central energy conversion process is described by the internal voltage source, whichconverts mechanical energy to electrical energy (or vice-versa).

From a mechanical perspective, based on (4): pe = Teωm.


From an electrical perspective, based on (40):

pe =3

2(edid + eqiq + 2e0i0)

=3

2

(0 · id +

ω

ωsVE · iq + 2 · 0 · 0

)=

3

2

ω

ωsVEiq

= Te2

pfω

= Teωm.

(45)

Note that both these forms of pe are identical, and therefore the internal voltage source is anideal converter of mechanical energy to electromagnetic energy.

Transformation from one reference frame to another

The dynamic models presented above are defined in the rotor reference frame, with respectto the angle θ. A typical question is how to construct a dynamic model of a system withmore than one natural reference frame. Such a need arises when the system includes morethan one machine, or when a machine is connected to an infinite bus.A typical solution is to transform the dq0 variables from one reference frame to another.Assume a system with two sets of signals:

3 xdq0 is defined with respect to a reference angle θr.

3 xdq0 is defined with respect to a reference angle θ.

The relations between these signals can be found by transforming the signals to the abcreference frame and back, which may be written as xdq0 = TθrT

−1θ xdq0 or xdq0 = TθT

−1θrxdq0.

Using the dq0 identities in Lecture 2 this yields

θ → θr : xdq0 =

cos(θ − θr) − sin(θ − θr) 0sin(θ − θr) cos(θ − θr) 0

0 0 1

xdq0, (46)

θ ← θr : xdq0 =

cos(θ − θr) sin(θ − θr) 0− sin(θ − θr) cos(θ − θr) 0

0 0 1

xdq0. (47)

Consider now a general unit, which dq0 model is described with respect to the angle θ. Thismodel can be linked to a system with a reference angle θr as described in Fig. 7.As an important special case, consider the internal voltage source ed = 0, eq = ω

ωsVE , e0 = 0,

with a reference angle θ. Direct transformation from θ to θr yieldsedeqe0

=

cos(θ − θr) − sin(θ − θr) 0sin(θ − θr) cos(θ − θr) 0

0 0 1

0ωωsVE0

=ω

ωsVE

− sin(θ − θr)cos(θ − θr)

0

(48)

or edeqe0

=ω

ωsVE

cos(δ)sin(δ)0

with δ = θ − θr +π

2, (49)


+ −

θ − θr

θ → θr

θ ← θr

θ − θr

idq0

vdq0

Systemreference

angleθr

θr

idq0

vdq0

Unitreference

angleθ

θ

Figure 7: Transformation from one reference frame to another.

where ed, eq, e0 are defined with respect to θr. The reader may recognize δ as the powerangle of the machine, as discussed in Lecture 1.We will now use this result to show that the simplified machine model of Section 3 leadsto the time-varying phasor model of Lecture 1. Consider a slow transient for which time-varying phasors may be used instead of dq0 quantities. In this case the internal voltage sourcemay be described by a phasor:

E =1√2(ed + jeq) =

ωVE√2ωs

(cos(δ) + j sin(δ)) . (50)

Let |E| denote the amplitude of this phasor,

|E| = ωVE√2ωs

= ωLafIf/√2, (51)

and substitute (51) in (50) to obtainE = |E|∠δ. (52)

Note that |E| depends on ω, however in a time-varying phasor model it is typically assumedthat ω ≈ ωs, and therefore |E| is constant.Also recall that in a time-varying phasor model pe = 3P . This relation holds since

3P = 3Re{EI∗a} = 3Re

{1√2(ed + jeq)

1√2(id + jiq)

∗}

=3

2Re{edid + eqiq} = pe.

(53)

An equivalent circuit of the machine which is based on time-varying phasors is shown inFig. 8. Recall that this is the same circuit presented in Lecture 1. It is valid only under theassumptions mentioned above, which are

3 Round rotor: Lg2 = 0, or equivalently Ld = Lq = Ls;

3 Constant field current: if = If = const;

3 Balanced voltages and currents: v0 = 0, i0 = 0;

3 Slow transients: voltages and currents are nearly sinusoidal over a single line cycle, sotime-varying phasors may be used instead of dq0 quantities.

E = |E|ejδ +− =⇒P

Ra jXs = jωLs Ia

Figure 8: A time-varying phasor model of the synchronous machine, single-phase diagram.


Example: machine connected to an infinite bus

Consider a synchronous machine connected to an infinite bus. The machine is described bythe simplified model presented in Section 3, and the infinite bus is modeled as

vd,∞ =√2Vg = const,

vq,∞ = 0,

v0,∞ = 0,

(54)

with respect to a reference angle θr = ωst. A signal flow diagram is shown in Fig. 9.

+ − ωst

θ − ωst

θ ← ωst

θ − ωst

idq0

R-Ldq0 model

+ −

vdq0,∞θ → ωstedq0

internalvoltagesource

idq0

θ

edq0

Pref

referenceangle θ

referenceangle ωst

Figure 9: Signal-flow diagram: synchronous machine connected to an infinite bus.

The internal voltage source is modeled with respect to the reference angle ωst asedeqe0

=ω

ωsVE

cos(δ)sin(δ)0

(55)

with δ = θ − ωst+ π2. In addition, based on (39) we have

d

dtθ = ω,

d

dtω = K

(3Pref −

1

D(ω − ωs)−

3

2VE iq

),

(56)

where iq is referenced to θ. Using (47) this variable may be expressed in terms of the referenceangle ωst as idiq

i0

=

cos(θ − ωst) sin(θ − ωst) 0− sin(θ − ωst) cos(θ − ωst) 0

0 0 1

idiqi0

, (57)

and since δ = θ − ωst+ π2

we have

iq = cos(δ)id + sin(δ)iq. (58)


Moreover, the machine’s series impedance is modeled (with respect to ωst) as

d

dt

idiqi0

=

−Ra

Lsωs 0

−ωs −Ra

Ls0

0 0 −Ra

Ls

idiqi0

+1

Ls

ωωsVE

cos(δ)sin(δ)0

−√2Vg0

0

. (59)

Combining these equations, the resulting state-space model is

d

dtδ = ω − ωs,

d

dtω = K

(3Pref −

1

D(ω − ωs)−

3

2VEid cos(δ)−

3

2VEiq sin(δ)

),

d

dtid = −

Ra

Lsid + ωsiq +

1

Ls

(ω

ωsVE cos(δ)−

√2Vg

),

d

dtiq = −ωsid −

Ra

Lsiq +

1

Ls

ω

ωsVE sin(δ),

d

dti0 = −

Ra

Lsi0.

(60)

Here δ is used instead of θ as a state variable.

Example: two machines connected to each other, and feeding a resistive load

Consider two synchronous machines connected to each other, and feeding a resistive loadRL. The machines are described using the detailed model in (24). The reference angle ofthe first machine is θ1, and the reference angle of the second machine is θ2. A signal flowdiagram is shown in Fig. 10.

− +

θ2 − θ1

θ1 → θ2vdq0,2

mac

hine

2

θ2

Pref,2

Vf,2

θ1 ← θ2

θ2 − θ1

idq0,2

+ +

RL

vdq0,1

mac

hine

1

θ1

Pref,1

Vf,1

idq0,1

Figure 10: Signal-flow diagram: two machines connected to each other, and feeding a resistive load.

Permanent Magnet Synchronous Motor (PMSM)

The Permanent Magnet Synchronous Motor (PMSM) is a synchronous machine in whichpermanent magnets are embedded in the rotor to create a constant magnetic field. As inall synchronous machines, at steady state the rotor speed is proportional to the frequencyof currents and voltages in the stator. For this reason such motors are especially useful inapplications that require precise speed or position control. This section presents a dq0 modelof a PMSM with sinusoidal EMF (as opposed to motors with trapezoidal EMF).PMSMs may be modeled similar to synchronous generators, with three key modifications:


3 The term Laf if is replaced with λ, which is the amplitude of the flux induced in thestator by the permanent magnets on the rotor.

3 The stator currents are defined positive when flowing into the machine.

3 The electromagnetic torque accelerates the rotor, and the mechanical torque deceler-ates the rotor. The angular acceleration is defined as d

dtωm = 1

J(Te − Tm).

Based on these assumptions the resulting model is

d

dtθ =

pf2ωm,

d

dtωm =

1

J(Te − Tm) ,

d

dtid =

1

Ldvd −

Ra

Ldid +

LqLd

pf2ωmiq,

d

dtiq =

1

Lqvq −

Ra

Lqiq −

LdLq

pf2ωmid −

λpfωm2Lq

,

d

dti0 =

1

L0

v0 −Ra

L0

i0.

(61)

Here the reference angle for the dq0 transformation is the electrical angle θ.The symbols are defined as follows:

3 θ denotes the rotor electrical angle;

3 pf denotes the number of magnetic poles on the rotor (even integer);

3 Ld, Lq, L0 denote the direct axis, quadrature axis, and zero sequence inductances;

3 Ra denotes the resistance of the stator windings;

3 id, iq, i0 denote the stator currents (positive when flowing into the machine);

3 vd, vq, v0 denote the stator terminal voltages;

3 ωm denotes the angular velocity of the rotor;

3 λ denotes the amplitude of the flux induced in the stator phases by the permanentmagnets on the rotor;

3 J denotes the rotor moment of inertia;

3 Tm, Te denote the mechanical and electromagnetic torques;

3 pm, pe denote the mechanical and electromagnetic powers.

The inputs are vd, vq, v0, Tm, and several outputs are

Te =3pf4

(λiq + (Ld − Lq) idiq) ,

pm = Tmωm, pe = Teωm.(62)

Similar to synchronous generators this model may be simplified by assuming a round rotor,so that Ld = Lq = Ls. In this case the motor may be described by the equivalent circuitshown in Fig. 11. In this simplified model the internal voltage source is described asedeq

e0

=

0λpf2ωm0

, (63)


vaia

∼

ea+ −

vbib

∼

eb+ −

vcic

Ls Ra

∼

ec+ −

energystorage

energyloss

energyconversion

Figure 11: Equivalent circuit for a permanent magnet synchronous motor with a round rotor (assum-ing Ld = Lq = Ls).

where[ed eq e0

]T is the dq0 transformation of[ea eb ec

]T. In addition, for Ld = Lq = Lsthe electromagnetic torque is given by

Te =3

4pfλiq. (64)

As we have seen before, the magnitude of the internal voltage is proportional to the angularvelocity, and the electromagnetic torque is proportional to the quadrature axis current.The heart of the energy conversion process is described by the internal voltage source, whichconverts electrical energy to mechanical energy. From a mechanical perspective, the electro-magnetic power is

pe = Teωm =3

4pfλωmiq, (65)

and from an electrical perspective,

pe =3

2(edid + eqiq + 2e0i0)

=3

2

(0 · id + λ

pf2ωm · iq + 2 · 0 · 0

)=

3

4pfλωmiq.

(66)

Both expressions are identical.A basic control scheme for the PMSM is shown in Fig. 12. The design consists of two loops:an inner current loop and an outer speed loop. The inner loop regulates the currents such thatid ≈ i∗d and iq ≈ i∗q , by adjusting the inverter duty cycles. The objective of the outer loop is toregulate the speed, such that in steady state ωm ≈ ω∗m. This is implemented by controlling i∗qbased on the approximated relation between torque and current Te = 3

4pfλiq. If the speed ωm

is too low then i∗q increases to produce more torque, and if ωm is too high then i∗q decreasesto produce less torque. Two sensors are placed on the motor to measure the speed ωm andthe electrical angle θ. The latter is used as a reference angle for the dq0 transformation.


dd

dq

0d0

dq0

↘

abc

θ

da

db

dcPWM

3-phase

inverter

Vdc

c b a

motor

abc↘dq0

θ

iabc currentsensors

currentcontroller

i∗d 0

idq

i∗q

speed

sensor

ωm+−

ω∗m

speed

controller

position

sensor

θ

Figure 12: A basic control scheme for a permanent magnet synchronous motor (PMSM).


Appendix A: summary of symbols

pf - number ofmagnetic poles on therotor (even integer)

θ - rotor electrical angle VE = ωsLafIf - internalvoltage at nominalfrequency, peak value

J - rotor moment ofinertia

ω = dθ/dt - rotor electricalfrequency

|E| = ωVE√2ωs

= ωLafIf/√2

- internal voltage at rotorfrequency, RMS value

D - damping coefficient ωm = 2pfω - rotor mechanical

frequencyid, iq, i0 - stator currents

Ld, Lq - direct axis andquadrature axissynchronousinductances

ωs - nominal frequency, 2π50or 2π60 rad/sec

vd, vq, v0 - stator terminalvoltages

L0 - zero sequenceinductance

Tm - mechanical torque(accelerating the rotor forgenerator)

K =(pf

2

)2 1Jωs

- swingequation constant

Ls =12(Ld + Lq)

synchronousinductance in thesimplified model

Te - electromagnetic torque(decelerating the rotor forgenerator)

ed, eq, e0 - internalvoltages in the simplifiedmachine model

Laf - stator to rotormutual inductance(maximum value)

Pref - reference power for thedroop controller

vf - field winding voltage

Lff - field windingself-inductance

pm = Tmωm - mechanicalpower

if - field winding current

Ra - stator windingresistance

pe = Teωm - electromagneticpower

λd, λq, λ0 - stator fluxlinkages

Rf - field windingresistance

ps - output power λf - field winding fluxlinkage

All the dq0 quantities in this table are defined in the rotor reference frame (with respect to θ).

References

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