Optimum use of DC bus by fitting the back-electromotive force of a 7-phase Permanent Magnet Synchronous machine F. Locment, E. Semail, X. Kestelyn

L2EP ENSAM, 8 Bd Louis XIV, 59 046, Lille, France. E-mail: [email protected] URL: http://www.univ-lille1.fr/l2ep

Acknowledgements This work is part of the project ’Futurelec2’ within the ’Centre National de Recherche Technologique (CNRT) de Lille’.

Keywords Multiphase drive, Multi-machine system, Converter machine interactions, Harmonics.

Abstract This paper deals with design constraints of a 7-phase Permanent Magnet Synchronous Machine (PMSM) supplied by a 7-leg Voltage Source Inverter. The optimum back electromotive force waveform is determined in order to get maximum torque for a given DC bus voltage.

Introduction Multi-phase DC brushless machines suffer from an apparent higher number of switching devices than 3-phase ones. Nevertheless, in high power applications such as electrical ship [1]-[2] or low voltage/high current applications such as on-board systems (traction) [3]-[5], this factor is not so obvious: use of high current devices implies high heat dissipation capabilities especially with high frequencies. Then it is common to use either parallel converters or parallel/serie device associations. Moreover, when reliability is required such as in aircraft [6], in marine applications [7]-[8] and in offshore variable speed wind generators, multiphase drives [10]-[11] must be considered as an alternative to 3-phase multi-level converter drives whose reconfiguration in safety mode is not obvious. In this paper, an axial double-rotor 7-phase virtual prototype is considered and has been modelled with 3D-finite element method. The global aim is to fit the machine to its 7-leg Voltage Source Inverter (VSI) in order to optimize the global drive. For 3-phase machines supplied by 3-leg VSI, the optimum use of the DC bus voltage has been widely studied [12]. It consists in injecting a third harmonic component in the voltage references of the VSI when triangle intersection method is employed or to use a space-vector modulation [13]. When the 3phase machine is wye-coupled the injection of a third harmonic component has impact neither on the torque nor on the currents of the machine. For wye-coupled 7-phase machines supplied by 7-leg VSI (Fig. 1), the problem is quite all different. The injection of a third component implies currents and eventually torque components in the machine. A Multi-Machine modelling is used in the paper to prove and explain this difference. Nevertheless the injection of a third harmonic component remains interesting for an optimal use of the 7-leg VSI DC bus Voltage [14].

i1(t) VBUS

ν7(t) Fig. 1: symbolic representation of 7-leg PWM-VSI and wye-coupled 7-phase machine The aim of this paper is to find necessary fitting of the machine in order to be able to inject a maximum third harmonic component in the reference voltages. At first, a Multi-Machine modelling of a 7-phase axial permanent magnet machine is presented: it allows to transform a complex problem into simpler ones. Then, for a given machine and a maximum value of the first harmonic voltage components, effects of injection of a third harmonic voltage component are studied: it appears that results depend on the harmonic spectrum of the backelectromotive force. Finally, thanks to the Multi-Machine modelling, machine design constraints are deduced in order to take maximum advantage of a third harmonic voltage component: extra torque is produced for a given DC bus voltage.

Multi-Machine vectorial characterization Under assumptions of no saturation, no reluctance effects and regularity of design, a vectorial formalism allows to prove that a 7-phase machine is equivalent to a set of three magnetically independent fictitious 2-phase machines [15] named M1, M2 and M3. Each equivalent machine is characterized by its inductance (resp. LM1, LM2 and LM3), resistance (resp. RM1, RM2 and RM3), and backJJJG JJJG JJJG EMF (resp. eM 1 , eM 2 and eM 3 ). The torque of the real machine T, is the sum of the torque of these three machines TM1, TM2 and TM3. The 7-leg VSI can also be decomposed into three fictitious VSI electrically coupled by a mathematical transformation Concordia’s type [15]. A fictitious VSI is characterized by a set of space phasors as it is the case for 3-leg VSI (with the usual hexagonal representation). The equivalence is based on a generalized Concordia transformation characterized by the [C7] matrix:

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ 2⎢ [C7 ] = ⎢ 7⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣⎢

1 2 1 2 1 2 1 2 1 2 1 2 1 2

1

0

1

0

1

2π 7 4π cos 7 6π cos 7 8π cos 7 10π cos 7 12π cos 7

2π 7 4π sin 7 6π sin 7 8π sin 7 10π sin 7 12π sin 7

4π 7 8π cos 7 12π cos 7 16π cos 7 20π cos 7 24π cos 7

4π 7 8π sin 7 12π sin 7 16π sin 7 20π sin 7 24π sin 7

6π 7 12π cos 7 18π cos 7 24π cos 7 30π cos 7 36π cos 7

cos

sin

cos

sin

⎤ ⎥ 6π ⎥ ⎥ sin 7 ⎥ 12π ⎥ sin 7 ⎥ 18π ⎥ ⎥ sin 7 ⎥ 24π ⎥ sin 7 ⎥ 30π ⎥ sin ⎥ 7 ⎥ 36π ⎥ sin 7 ⎥⎦⎥ 0

cos

(1)

Relationships between values of fictitious machines and real values (noted with subscripts 1, 2, …, 7) are then defined by:

JJJJG

imach = ⎡⎣0 iM 1α

JJJJG

vmach = ⎡⎣0 vM 1α

iM 1β vM 1 β

iM 2α

iM 2 β

iM 3α

iM 3 β ⎤⎦ = [C7 ] [i1

vM 2α

vM 2 β

vM 3α

vM 3 β ⎤⎦ = [C7 ] [ v1 v2

t

t

t

t

i2

i3 v3

i4 v4

i5 v5

i7 ]

i6 v6

t

v7 ]

(2) t

(3)

Currents and voltages obtained using this transformation can be decomposed into three subsystems associated with the M1, M2 and M3 machines: t t t ⎧vJJJG = ⎡v ⎧vJJJG = ⎡v ⎧vJJJG = ⎡ v ⎤ ⎤ v v 1 1 α 1 β 2 2 α 2 β M M M M M M ⎦ ⎦ ⎪ ⎣ ⎪ ⎣ ⎪ M 3 ⎣ M 3α vM 3 β ⎤⎦ (4) ⎨ JJG ⎨ JJJG ⎨ JJJG t t t ⎪ iM 1 = ⎡iM 1α iM 1 β ⎤⎦ ⎪ iM 2 = ⎡iM 2α iM 2 β ⎤⎦ ⎪ iM 3 = ⎡iM 3α iM 3 β ⎤⎦ ⎣ ⎣ ⎣ ⎩ ⎩ ⎩ A key of the problem is that each one of the 2-phase fictitious machine is characterized by an harmonic family (Table I) and a vectorial subspace Sk. The three subspaces are orthogonal each to other. It is this orthogonality which allows to introduce the concept of fictitious machine.

Table I: Harmonic characterization of fictitious machines for wye-coupled 7-phase machine Fictitious 2-phase machines M1 M2 M3

Families of odd harmonics 1, 13, 15, …, 7 h ± 1 5, 9, 19, …, 7h ± 2 3, 11, 17, …, 7 h ± 3

To get a synthetic representation, a graphical formalism (Energetic Macroscopic Representation: EMR) is used (see Appendix and Fig. 2). Interleaved triangles traduce a mechanical coupling between the three fictitious machines (T=TM1+TM2+TM3). Interleaved squares traduce an electrical coupling: the three fictitious VSI are supplied by only one DC bus. The voltage equations of these M1 and M3 machines are: JJG JJG ⎧ JJJG diM 1 JJJG + eM 1 ⎪vM 1 = RM 1 iM 1 + LM 1 ⎪ dt JJJG ⎨ JJJG diM 3 JJJG ⎪ JJJG ⎪⎩vM 3 = RM 3 iM 3 + LM 3 dt + eM 3 The electromechanical conversion is traduced by equation (6): JJJG JJG JJJG JJJG JJJG JJJG eM 1 . iM 1 = TM 1Ω eM 2 . iM 2 = TM 2Ω eM 3 . iM 3 = TM 3Ω

JJG diM 1 JJJG JJJG = vM 1 − eM 1 LM 1 dt vM1 VBUS

vM2

SE iBUS

vM3

iM1 iM1 iM2 iM3

DC bus

Electrical coupling

Fictitious inverter

JJJG JJG eM 1.iM 1 = TM 1.Ω

iM2 iM3

M1 eM1

M2 eM2

M3 eM3

Fictitious machine (5)(6)

(5)

(6)

T = TM 1 + TM 2 + TM 3

TM1

Ω

TM2

T

Ω

Ω

Ω

TLoad

TM3

J

Ω Mechanical coupling

SM

dΩ = T − TLoad dt Load

Fig. 2: Multi-Machine Energetic Macroscopic Representation of the 7-phase machine

Characterization of the 7-phase machine prototype The studied machine is an axial flux double rotor permanent magnet machine (Fig. 3) whose windings are toroidal type. It has been modelled with a 3D-finite element method developed in laboratory (L2EP). The harmonic spectrum of the back-EMF (Fig. 4) is presented in Table II. To simplify the study we consider only the first and the third harmonics of back-EMF. Consequently it appears that, according to the Multi-Machine harmonic characterization (Table II), only the M1 and M3 machines must be considered since they will produce torque. Injection of a third harmonic voltage component is thus equivalent to supplying the M3 machine.

Fig. 3: one sixteenth of the studied machine

Table II: Harmonic breakdown of back-electromotive force of one phase Order of harmonic Relative RMS amplitude

fem6

150

1 100% (88.5V)

3 22%

first harmonic of fem6

5 0%

7 8%

9 9%

third harmonic of fem6

100

(V)

50 0 0

0.01

0.02

0.03

0.04

-50 -100 -150

times(s) Fig. 4: back-EMF “fem6” of one phase of the 7-phase machine

Optimal third harmonic injection DC bus voltage determination in case of sinusoidal back-EMF Before injecting a third harmonic voltage component, we search the necessary value of the DC bus voltage by considering only the first harmonic of back-EMF. In this case, the back-EMF of the M3 machine is not taken into account. Only the M1 machine is used to produce the nominal torque at nominal speed with minimum Joule losses. JJG 2 From relation (6), it is easy to deduce that, for a given torque TM1, the Joule losses ( RM 1 iM 1 ) are JJG JJG JJJG JJJG minimum if the two vectors iM 1 and eM 1 are collinear ( iM 1 = k eM 1 ). In steady states, equation (5) becomes then:

JJJG JJJG G JJJG JJJG vM 1 = k RM 1 eM 1 + ω LM 1 z ∧ k eM 1 + eM 1

(7)

JJJG JJJG G with ω pulsation and z , the normal vector orthogonal to eM 1 and iM1 . Besides, as for a 3-leg VSI with no injection of a third harmonic component [9], we have: VBUS = V peak

(8)

with V peak the peak value of the vm ( t ) voltage of the phase n°m. JJJG To link real vm ( t ) voltage of one phase and vM 1 we use a relation obtained by the transformation of Concordia: JJJJG 2 JJJG (9) vmach = vM 1 7

Finally, thanks to (7), (8) and (9), we get: JJJG 2 JJJG 2 2 VBUS = vM 1 = (1 + k RM 1 ) + ( k ω LM 1 ) eM 1 (10) 7 This case is shown in Fig. 5 . The voltage vm ( t ) is then sinusoidal with a 150V maximum value V peak . By (8), we get the particular value of the DC bus voltage: sin1 VBUS = 150V

(11)

Effect of the third-harmonic of back-EMF JJJJG

We want now to study the effect of eM 3 , the third harmonic of the back-EMF. To quantify the effect JJJJG

of eM 3 we use the ratio:

JJJG eM 3 k ′ = JJJG eM 1

(12)

sin1−3 More precisely, we search the necessary new DC bus voltage VBUS in order to use only the M1 machine to produce torque, as previously. JJJG G In this case, iM 3 = 0 , and the relation (5) becomes: JJJG JJJG JG JJJG JJJG ⎧⎪vM 1 = k RM 1 eM 1 + k ω LM 1 z ∧ eM 1 + eM 1 (13) ⎨ JJJG JJJG ⎪⎩vM 3 = eM 3 G JJJG JJJG It is then possible to get vm ( t ) , the waveform of the component of v = vM 1 + vM 3 relative to the phase n°m. It appears in Fig. 5 that the maximum voltage Vpeak is equal to 166V for the studied prototype (with k ′ = 0.22 found thank Table II). Consequently the minimum value of the DC bus voltage is in this case: sin1− 3 VBUS = 166V (14) sin1− 3 is increased More generally, if we suppose a variation of only k ′ from 0 to 0.5, we find that VBUS from 150V to 200V.

Consequently, the impact of a third back-EMF force is harmful for the sizing of the DC bus voltage if the M3 machine is not used to supply torque. The question is: is it possible to use the M3 machine without increasing DC bus voltage by injecting a third harmonic component of voltage ?

k'=0.22

case of sinusoidal emf

200 150 100

(V)

50 0 -50

0.4

0.41

0.42

0.43

0.44

0.45

-100 -150 -200 time(s)

JJJG Fig. 5: ν m (t ) voltage of one phase at nominal load for k ′ = 0 and k ′ = 0.22 with k ′ = eM 3

JJJG eM 1

Fitting of the machine

JJJG We suppose that the machine M1 is unchanged (same inductance, resistance and back-EMF eM 1 ) and JJJG JJJG is supplied by the same voltage vM 1 and current iM 1 as previously (paragraph DC bus voltage sin1 determination in case of sinusoidal back-EMF). The imposed value of VBUS is VBUS , found in (11). The JJJG inductance and resistance of the machine M3 are also unchanged but this machine is supplied by iM 3 JJJG current and voltage vM 3 .

By a modification of the spatial repartition of permanent magnet [17], the aim is to find the back-EMF JJJG vector eM 3 which allows to verify the following constraints: • minimization of extra losses due to the use of M3 machine ; (15) JJJG of the third harmonic voltage for an • insertion of the maximum value vM 3 max

optimum use of

sin1 VBUS .

(16)

The first constraint (15) and relation (6) imply that, for a given torque T, the Joule losses are minimum JJG JJG if the current vector iM is collinear to eM : JJG JJG iM = k eM (17) Using (17), (5) becomes then in steady states: JJJG JJJG G JJJG JJJG ⎧⎪ vM 1 = k RM 1 eM 1 + k ω LM 1 z ∧ eM 1 + eM 1 JJJJG G JJJG JJJG ⎨ JJJG ⎪⎩vM 3 = k RM 3 eM 3 + k 3ω LM 3 z ∧ eM 3 + eM 3

Consequently the following equation (19) must be verified: JJJG 2 2 vM 3 1 + k RM 3 ) + ( k 3ω LM 3 ) ( k ′′ = JJJG = k ′ 2 2 vM 1 (1 + k RM 1 ) + ( k ω LM 1 )

(18)

(19)

where k ′′ (resp. k ′ , k ) depends on the voltages imposed by the VSI (resp. on the design of the ′′ , machine, on the required torque). Thanks to the second constraint (16) an optimal value of k ′′ , kopt can be calculated.

′′ Determination of kopt G JJJG JJJG By projection of the vector v = vM 1 + vM 3 , we get its coordinate relative to the phase n°1: JJJG 2 JJJG ν 1 (t ) = ν M 1 sin(ω t ) + ν M 3 sin(3ω t − ϕ ) 7 Consequently, we have to solve the equation: JJJG 2 sin1 VBUS = ν M1 ( sin θ + k ′′ sin(3θ − ϕ ) ) 7

(

)

(20)

(21)

′′ ≈ 0.4 , ϕ = 0 and θ = 0, 74 rad ≈ 42.6° (Fig. 7) Numerical resolution leads to kopt ′′ . The maximum value of 150V In Fig. 6, we have represented the voltage of the phase n°1 for k ′′ = kopt

is reached at 42.6°. k"=0

k"=0.4

VBUS

200

k ′′ ≈ 0.4

150 100

1.5

(V)

50

1

0 -50

0.4

0.41

0.42

0.43

0.44

0.45

1

0.5

0.75

0

-100

0.5

0

-150

1

θ ≈ 0.74 rad

-200

time(s)

0.25 2 3

Fig. 6: optimum voltage ν 1 (t )

0

Fig. 7: numerical resolution of (21)

JJJJG

Determination of eM 3

Thanks to relations (19) it can be deduced: JJJG ′′ eM 3 = kopt

(1 + k RM 1 ) + ( k ω LM 1 ) 2 2 (1 + k RM 3 ) + ( k 3ω LM 3 ) 2

2

JJJJJG eM 1

(22)

JJJG JJJG To find the α Mh angle between eMh and vMh (for h =1 or 3) we use (18): ⎡ h ω LMh k ⎤ (23) ⎥ ⎣ 1 + RMh k ⎦ JJJG eM 3 For the studied machine we have found: k ′ = JJJG = 0.48 , α M 1 = 0.53 rad and α M 3 = 0.22 rad . eM 1

α Mh = tan −1 ⎢

Relations (22) and (23) are graphically expressed in Fig. 8.

VM1

VM3

EM1

EM3

300 200

(V)

100 0 0.4

0.41

0.42

0.43

0.44

0.45

-100 -200 -300

time(s)

Fig. 8: For one phase, voltages vMh and back-EMF eMh of the M1 and M3 machines JJG JJJG JJJG We get finally from (22) and (23), eM = eM 1 + eM 3 , the total back-EMF of the real machine. Its projection relatively to the phase n°1 gives the back-EMF represented in Fig. 9 200 150 100

(V)

50 0 -50

0.4

0.41

0.42

0.43

0.44

0.45

-100 -150 -200

time(s)

Fig. 9: required optimum back-EMF of phase n°1 Determination of extra torque

The extra torque TM3 developed by the M3 machine is 25N.m, value which represents 20% of the nominal torque (125N.m) developed by the M1 machine. Of course, extra Joule losses appears. JJG JJG Nevertheless, the imposed constraint of minimum losses ( iM = k eM ) allows to reduce their increase to JJJG JJJG 20%. If we impose to work with the same Joule losses, by reducing iM 1 and iM 3 in the same ratio, the torque increase is 9%.

Conclusions It has been shown that, as for 3-phase machines, it was possible to take advantage of injection of a third harmonic component of voltage. The DC bus voltage is then better used, as it was with the 3phase machines. Nevertheless, for optimum use of the DC bus voltage, the 7-phase machine must be fitted to the 7-leg VSI. With improvements of design of PM rotor, such as Halbach arrays, such optimal design can be considered. Of course, the obtained extra torque is a “booster” torque for transient states, unless improved heat dissipation is achieved.

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Appendix: Synoptic of Energetic Macroscopic Representation Source of energy Electrical converter (without energy accumulation)

Electromechanical converter (without energy accumulation) Mechanical Converter (without energy accumulation)

Control block without controller Control block with controller

Element with energy accumulation

Coupling device (distribution of energy)

Control block with coupling criterion