Lecture 9: Elementary Single-Phase Machines and
Lossless Magnetic Field
ELEC-E8402 Control of Electric Drives and Power Converters
Marko Hinkkanen
Spring 2019
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Single-Phase Machines
I Single-phase machines are not used in real applications
I Why should we study them?
I To get more thorough understanding of fundamental concepts
I
I
I
I
Flux linkages
Conservative magnetic field systems
Selection of state variables
Modeling concepts introduced are very general and powerful
I For simplicity, 2-pole machines will be considered here
I More poles can be easily taken into account
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Outline
Permanent-Magnet Machine
Salient-Pole Machine
Lossless Magnetic Field
Voltage Equations
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Airgap Field Produced by PMs
I Ferrite PMs are cheap and easy to manufacture
I Rare-earth magnets (e.g. NdFeB, SmCo) provide much higher energy
products, but they are expensive and difficult to manufacture and handle
ϑ
B
ϑ
π/2
π
π/2
Assumed uniform radial air-gap flux density
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Equivalent Current Source
I PMs can be modeled with an equivalent field winding
I Permeability of the magnets is approximately the same as that of air
I Current if is constant
ϑ
B
if
ϑ
π/2
π
π/2
Assumed uniform radial air-gap flux density
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Induced Voltage
Uniform flux density B
I Rotor rotates at ωm = dϑm /dt
I Magnetic field moves at v = r ωm with
respect to the conductors
I Voltage B`v is induced in each
conductor
I N turns in the coil
ϑm
ea
I Induced voltage ea = 2Nr `Bωm is
proportional to the angular speed
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Phase Voltage
I Phase voltage
ua = Ra ia +
ia
Ra
dψa
dt
I Phase flux linkage
ϑm
ψa = La ia + ψaf (ϑm )
ua
where La is the self-inductance and ψaf
is the flux linkage due to the PMs
I Based on the geometry
π
ψaf = 2Nr `B
− ϑm
when
2
0 ≤ ϑm ≤ π
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Phase Voltage
ia
Ra
I Voltage can be expressed as
ua = Ria + La
dia
+ ea
dt
I Voltage induced by the magnets
ϑm
ua
dψaf dϑm
dψaf
=
dt
dϑm dt
= −2Nr `Bωm
ea =
which is the same result as earlier
The sign in the expression for ea is valid at 0 < ϑm < π, cf. the following waveforms.
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Torque Production
ia
I Forces F = NB`ia act on both coil sides
I Opposite counterforces act on the rotor
F
I Electromagnetic torque
Tm = −2Nr `Bia
ϑm
dψa
dt
is proportional to the current
I Mechanical power
F
pm = Tm ωm = ea ia
The sign in the expression for TM is valid at 0 < ϑm < π, cf. the following waveforms.
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Example Waveforms
ψaf
ia
ϑm
ea
ϑm
ϑm
ia
dψa
dt
ϑm
Tm
0
π/2
π
3π/2
2π ϑm
The waveform of the current ia is chosen such that the torque Tm becomes constant.
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Simple Distributed Winding
ψaf
π/6
2π ϑm
π
ea
ϑm
π/2
3π/2
ϑm
ea
Fundamental
I Waveforms approximately sinusoidal
ψaf = ψ1 cos(ϑm )
ea = −ωm ψ1 sin(ϑm )
I PM-flux constant ψ1 ∝ Nr `B
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Example Waveforms: Ideal Sinusoidally Distributed Winding
ψaf
ia
ψ1
ϑm
ea
ωm ψ1
ϑm
ϑm
ia
dψa
dt
ϑm
Tm
0
π/2
π
3π/2
2π ϑm
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Outline
Permanent-Magnet Machine
Salient-Pole Machine
Lossless Magnetic Field
Voltage Equations
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Reluctance Machine
ia
La
0
π/2
π
3π/2
ϑm
2π ϑm
dψa
dt
I Ideal anisotropy assumed above
(with constant leakage inductance)
I Inductance La depends on
the rotor position
I Flux linkage ψa = La (ϑm )ia
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Simple Distributed Winding
ia
La
0
π/2
π
3π/2
ϑm
2π ϑm
dψa
dt
I Ideal anisotropy assumed above
(with constant leakage inductance)
I If spatial harmonics are omitted
ψa = [L0 + L2 cos(2ϑm )]ia
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PM Reluctance Machine
ia
ψaf
ψ1
ϑm
ϑm
La
dψa
dt
0
π/2
π
3π/2
2π ϑm
I Approximate flux linkage
ψa = [L0 −L2 cos(2ϑm )]ia +ψ1 cos(ϑm )
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Salient-Pole Machine With Field Winding
I Field winding in the rotor
I Flux linkages
ia
ψa = La (ϑm )ia + Laf (ϑm )if
if
ψf = Laf (ϑm )ia + Lf if
where
dψa
dt
ϑm
dψf
dt
La (ϑm ) = L0 + L2 cos(2ϑm )
Laf (ϑm ) = M cos(ϑm )
I How to determine torque Tm of
salient-pole machines?
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Outline
Permanent-Magnet Machine
Salient-Pole Machine
Lossless Magnetic Field
Voltage Equations
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Lossless Magnetic Field
Understanding lossless magnetic field systems often helps in developing machine
models for control purposes
I Very general and powerful concept
I Only assumption is that the magnetic field is lossless (conservative)
I Forces and torques in complex electromechanical systems can be determined
I Independent of machine type, number of terminals, number of poles, etc.
I Most lumped-parameter electric machine models are based on it
I Magnetic saturation and spatial harmonics can be taken into account
I Core losses can be modeled outside the lossless field system
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Lossless Magnetic Field
Stored magnetic field energy Wm
I is a state function, depending only on its
independent state variables
I is independent of the path used to reach
the state
I can be determined completely if the
electrical port relations are known
I can be evaluated by means of numerical
techniques (e.g. FEM) or measurements
F1
i1
dψ1
dt
..
.
dψN
dt
iN
Lossless
magnetic
field
FM
dx1
dt
..
.
dxM
dt
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Example System: Single-Phase Machine With Field Winding
Rotor
magnetic
axis
ia
I Two electrical ports and one
mechanical port
I Stored field energy
Wm = Wm (ψa , ψf , ϑm )
where ψa , ψf , and ϑm are
independent state variables
if
dψa
dt
dψf
dt
ϑm
Stator
magnetic
axis
I This example system is
considered in the following
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I Power balance
dWm
dψa
dψ
dϑm
= ia
+if f −Tm
dt
dt
dt
dt
ia
dψa
dt
I Currents
∂Wm (ψa , ψf , ϑm )
∂ψa
∂Wm (ψa , ψf , ϑm )
if =
∂ψf
ia =
Lossless
if
ia = ia (ψa , ψf , ϑm )
if = if (ψa , ψf , ϑm )
dψf
dt
Tm
magnetic field
dϑm
dt
Tm = Tm (ψa , ψf , ϑm )
I Torque
Tm = −
∂Wm (ψa , ψf , ϑm )
∂ϑm
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Lossless Field System Should Satisfy Reciprocity Conditions
I Incremental mutual inductances should be equal in any operating point
∂ia
∂if
=
∂ψf
∂ψa
I If multiple mechanical ports, analogous conditions hold for them as well
I Conditions between electrical and mechanical ports
∂Tm
∂ia
=−
∂ϑm
∂ψa
∂if
∂Tm
=−
∂ϑm
∂ψf
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Integration Path for the Field Energy Can Be Chosen Freely
I For illustration purposes
ψf = 0 assumed
I Integration along Path 1
Z
ψa
Wm (ψa , ϑm ) =
0
Z ϑm
−
0
ia (ψa0 , 0)dψa0
Path 2b
ia = ia (ψa0 , ϑm )
ϑ0m
ϑm
Wm (ψa , ϑm )
Path 3
Path 2a
Path 1b
Tm = Tm (ψa , ϑ0m )
Tm = 0
Tm (ψa , ϑ0m )dϑ0m
I We should know Tm (ψa , ϑm )
0
Path 1a
ia = ia (ψa0 , 0)
ψa
ψa0
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Path 2b
ia = ia (ψa0 , ϑm )
ϑ0m
I Integration along Path 2
Z
ψa
Wm (ψa , ϑm ) =
0
ia (ψa0 , ϑm )dψa0
ϑm
Wm (ψa , ϑm )
Path 3
Path 2a
Path 1b
Tm = Tm (ψa , ϑ0m )
Tm = 0
since Tm (0, ϑm ) = 0
I Torque is not needed in Path 2
0
Path 1a
ia = ia (ψa0 , 0)
ψa
ψa0
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Illustration of Field Energy and Coenergy
I For illustration purposes
ψf = 0 assumed
I Area of the rectangle ψa ia
ψa0
Z
Wm =
ψa
ia0 dψa0
0
ψa0 = ψa0 (ia0 , ϑm )
ψa
Energy
I Relation of coenergy to field energy
Wm + Wm0 = ψa ia
I Magnetically linear case: Wm = Wm0
Coenergy
Z ia
0
Wm =
ψa0 dia0
0
ia
ia0
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Field Energy and Coenergy
I Field energy
Z
ψa
Wm (ψa , ψf , ϑm ) =
0
ψf
Z
ia (ψa0 , ψf , ϑm )dψa0
+
ψa (ia0 , if , ϑm )dia0
Z
0
if (0, ψf0 , ϑm )dψf0
I Coenergy
Wm0 (ia , if , ϑm )
Z
=
0
ia
if
+
0
ψf (0, if0 , ϑm )dif0
I Relation of coenergy to field energy
Wm + Wm0 = ψa ia + ψf if
I Torque is typically easier to calculate from coenergy
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Torque from Coenergy
I Power balance
dWm0
dia
di
dϑm
= ψa +ψf f +Tm
dt
dt
dt
dt
I Flux linkages
∂Wm0 (ia , if , ϑm )
∂ia
∂Wm0 (ia , if , ϑm )
ψf =
∂if
ia
dψa
dt
ψa =
Lossless
if
ψa = ψa (ia , if , ϑm )
ψf = ψf (ia , if , ϑm )
dψf
dt
Tm
magnetic field
dϑm
dt
Tm = Tm (ia , if , ϑm )
I Torque
Tm =
∂Wm0 (ia , if , ϑm )
∂ϑm
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Analytical Example
I Assume a magnetically linear machine with the flux linkages
ψa = La (ϑm )ia + Laf (ϑm )if
ψf = Laf (ϑm )ia + Lf if
where the inductances are
La (ϑm ) = L0 + L2 cos(2ϑm )
Laf (ϑm ) = M cos(ϑm )
I Coenergy
Wm0 (ia , if , ϑm ) =
1
1
[L0 + L2 cos(2ϑm )] ia2 + M cos(ϑm )ia if + Lf if2
2
2
I Torque
Tm = −M sin(ϑm )ia if − L2 sin(2ϑm )ia2
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Torque
Tm
Total
torque
Tm
Reluctance
torque
0
π
ϑm
−π
0
π
ϑm
ia
Mutual
interaction
torque
Currents ia and if are constant
Field current if is constant
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Outline
Permanent-Magnet Machine
Salient-Pole Machine
Lossless Magnetic Field
Voltage Equations
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Inclusion of Voltage Equations
Ra
ua (t)
ia
dψa
dt
Rf
Lossless
if
ia = ia (ψa , ψf , ϑm )
if = if (ψa , ψf , ϑm )
uf (t)
dψf
dt
Tm
magnetic field
dϑm
dt
Mechanical
subsystem
Tm = Tm (ψa , ψf , ϑm )
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Voltage Equations: Flux Linkages as State Variables
I Voltage equations
dψa
= ua − Ra ia
dt
dψf
= uf − Rf if
dt
where the currents are known static functions of the state variables
ia = ia (ψa , ψf , ϑm )
if = ia (ψa , ψf , ϑm )
I Electromagnetic torque is the input for the mechanical subsystem
Tm = Tm (ψa , ψf , ϑm )
and the state variable ϑm is the output of the mechanical subsystem
I This set of equations is very simple to implement
The torque expression Tm = Tm (ia , if , ϑm ) could be used as well since the currents ia = ia (ψa , ψf , ϑm ) and if = if (ψa , ψf , ϑm ) are known.
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Voltage Equations: Currents as State Variables
I If the currents are used the state variables, the representation of the voltage
equations becomes complex, for example
dψa
∂ψa dia ∂ψa dif
∂ψa dϑm
=
+
+
= ua − Ra ia
dt
∂ia dt
∂if dt
∂ϑm dt
I In general case, all the partial derivatives are functions of ia , if , and ϑm
I In the magnetically linear example case
∂La (ϑm )
dia
dif
∂Laf (ϑm ) dϑm
La (ϑm )
+ Laf (ϑm ) +
ia +
i
= ua − Ra ia
dt
dt
∂ϑm
∂ϑm f dt
and similarly for the rotor voltage equation
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Further Reading
I H. H. Woodson and J. R. Melcher, Electromechanical Dynamics, John Wiley & Sons,
1968.
I A. E. Fitzgerald, C. Kingsley, Jr., and S. D. Umans, Electric Machinery, 6th ed.,
McGraw-Hill, 2003.
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