Matrix Converter Based Five-Phase Series-Connected Two-Motor Drive System
Matrix Converter Based Five-Phase Series-Connected Two-Motor Drive System
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techniques and are further supported by simulation results. v INV = [v A vB vC vD v E ]T
The major aim of the modulation is to produce two (4)
fundamental frequency output from the matrix converter that i INV = [i A iB iC iD i E ]T
can be used to control two series-connected five-phase
machines. i r1 = [iar1 ibr1 icr1 idr1 ier1 ]T
(5)
i r 2 = [iar 2 ibr 2 icr 2 idr 2 ier 2 ]T
II. FIVE-PHASE SERIES-CONNECTED TWO MOTOR DRIVE
The basic topology of a five-phase series-connected two- In order to simplify the phase-domain model, the decoupling
motor drive system is shown in Fig. 1. The variable frequency transformation is applied. The Clark’s decoupling
(VF) source is supplying a five-phase induction machine transformation matrix in power invarient form is [21]:
(Motor 1) whose stator windings are connected to another α ⎡ 1 cos α cos 2α cos 3α cos 4α ⎤
five-phase induction machine (Motor 2) through appropriate ⎢ 0 sin α sin 2α sin 3α sin 4α ⎥⎥
β ⎢
phase transposition. The rotor of the two machines are 2 ⎢
C= x 1 cos 2α cos 4α cos 6α cos8α ⎥ (6)
independent and they are connected to different mechanical 5 ⎢ ⎥
loads [20]. y ⎢ 0 sin 2α sin 4α sin 6α sin 8α ⎥
0 ⎢⎣⎢1/ 2 1/ 2 1/ 2 1/ 2 1/ 2 ⎥⎦⎥
The new variables are defined as:
IA INV INV r1 r1 r2 r2 INV INV
vαβ = Cv vαβ = Cv vαβ = Cv i = Ciαβ
IB i r1 = Ciαβ
r1
i r 2 = Ciαβ
r2
(7)
By omitting the x-y and zero-sequence equation for rotor
IC windings and the zero-sequence equation of the inverter, the
complete d-q model in stationary reference frame for the two
ID five-phase series-connected machines can be written in
developed form as:
IE di dINV di di INV
v dINV = R s1i dINV + ( Lls1 + L m1 ) + L m1 dr1 + R s 2 i dINV + Lls 2 d
dt dt dt
ω1 ω2 v qINV = R s1i qINV + ( Lls1 + Lm1 )
di qINV
+ Lm1
di qr1
+ R s 2 i qINV + Lls 2
di qINV
dt dt dt
Fig. 1. A five-phase drive system block diagram.
di xINV di INV di
v xINV = R s1i xINV + Lls1 + R s 2 i xINV + ( Lls 2 + L m 2 ) x + Lm 2 dr 2
Due to the series connection of two stator windings according dt dt dt
to Fig. 4.1 the following holds true: di yINV di yINV di qr 2
v yINV = R s1i yINV + Lls1 + R s 2 i yINV + ( Lls 2 + L m 2 ) + Lm 2
v A = vas1 + vas 2 i A = i as1 = i as 2 dt dt dt
vB = vbs1 + vcs 2 i B = ibs1 = i cs 2 (8a)
didINV di
vC = vcs1 + ves 2 iC = i cs1 = i es 2 (1) 0 = Rr1idr1 + Lm1
dt
(
+ ( Llr1 + Lm1 ) dr1 + ω1 Lm1iqINV + ( Llr1 + Lm1 ) iqr1
dt
)
vD = vds1 + vbs 2 i D = i ds1 = ibs 2
diqINV diqr1
vE = ves1 + vds 2 i E = i es1 = i ds 2 0 = Rr1iqr1 + Lm1
dt
+ ( Llr1 + Lm1 )
dt
(
− ω1 Lm1idINV + ( Llr1 + Lm1 ) idr1 )
Capital letters stand for inverter phase-to-neutral voltages and (8b)
inverter phase currents in equation (1).
In a general case the two machines, although both five-phase, dixINV di
may be different and therefore may be with different 0 = Rr 2 idr 2 + Lm 2
dt dt
(
+ ( Llr 2 + Lm 2 ) dr 2 + ω 2 Lm 2 i yINV + ( Llr 2 + Lm 2 ) iqr 2 )
parameters. Let the index ‘1’ denote induction machine
di yINV diqr 2
directly connected to the five-phase inverter and let the index
‘2’ stand for the second induction machine, connected after
0 = Rr 2 iqr 2 + Lm 2
dt
+ ( Llr 2 + Lm 2 )
dt
(
− ω 2 Lm 2 ixINV + ( Llr 2 + Lm 2 ) idr 2 )
the first machine through phase transposition. (8c)
Voltage equation for the complete system can be written in a
compact matrix form as
d (Li ) Te1 = PL ⎡ INV − idINV iqr1 ⎤⎦
v = Ri + (2) 1 m1 ⎣ idr 1iq
dt
th
where the system is of the 15 order and (9)
⎡v INV ⎤ ⎡i INV ⎤ Te 2 = P2 Lm 2 ⎡⎣idr 2i yINV − ixINV iqr 2 ⎤⎦
⎢ ⎥ ⎢ ⎥
v=⎢ 0 ⎥ i = ⎢ i r1 ⎥ (3)
⎢ 0 ⎥ ⎢i ⎥
⎣ ⎦ ⎣ r2 ⎦
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J dω elimination of bulky dc link capacitor. The power factor at the
Te1 − TL1 = 1 1 source side is also controllable.
P1 dt
The control decoupling is possible due to decoupling of the d-
P
∫ dω1 = J11 ∫ (Te1 − TL1 )dt q and x-y components. The d-q components of one machine
become the x-y to the other and vice-versa. The independent
(10)
J dω2 control is achieved of the two five-phase motors using vector
Te2 − TL 2 = controllers. However, in this paper, open-loop operation of the
P dt
two-motor drive system is presented. The control decoupling
P
∫ dω2 = J22 ∫ (Te2 − TL2 )dt is proved by showing the two frequency components in the
output voltage and the independent speed and torques. Five-
phase Inverter model
In five-phase system two set of orthogonal voltage/current
components are produced namely d-q and x-y. In single-motor III. CARRIER-BASED PULSE WIDTH MODULATION TECHNIQUE
drive system, only d-q components are utilised and the x-y FOR TWO-MOTOR DRIVES
components are free to flow creating losses. Thus concept of The power circuit topology of a 3 to n phase matrix converter
two-motor five-phase drive system is developed where both is shown in Fig. 3. The input is three-phase fixed voltage and
these components are utilised, d-q by one machine and x-y by fixed frequency supply from the grid system (50 Hz, 220 V
other machine. The extra set of current components (x-y) rms). The output is n-phase with variable voltage and variable
available in a five-phase system is effectively utilised in frequency. A small filter is needed at the input source side and
independently controlling an additional five-phase machine the switches are bidirectional for allowing regenerative
when the stator windings of two five-phase machines are operation of the load. The matrix converter is modulated either
connected in series (Fig. 1). As such the two five-phase using carrier-based PWM [32-34] and space vector PWM [35-
machines are supplied from one five-phase variable frequency 36].
source but are controlled independently. More detail on this
configuration of the drive system is available in references
isa ia
[22-26]. The supply options for the series-connected five-
phase two-motor drive are shown in Fig. 2. v S 11 S
a 1n
isb ib
vb S 21 S 2n
isc ic
vc S 31 S 3n
iA in
Fig. 3. Three to n-phase matrix converter.
v a = V cos(ωt )
v b = V cos(ωt − 2π / 3) (11)
v b = V cos(ωt − 4π / 3)
Since the matrix converter outputs voltages with frequency
b. decoupled from the input voltages, the duty ratios of the
Fig. 2. Supply options for five-phase series-connected two-motor drive: a.
switches are to be calculated accordingly. The five-phase
Inverter based solution, b. matrix converter based solution.
output voltage duty ratios should be calculated in such a way
The two possible solutions shown in Fig. 2 can be used. The that output voltages remains independent of input frequency.
available literature discusses the first option. However, this In other words, the five-phase output voltages can be
paper analyses the second option of supplying the two-motor considered in synchronous reference frame and the three-
drive using Matrix converter or direct AC-AC converter. The phase input voltages can be considered to be in stationary
major advantage is the sinusoidal source side current and reference frame, so that the input frequency term will be
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absent in output voltages. Considering the above, duty ratios k A = k A1 + k A2
of output phase j is chosen as
k B = k B1 + kC 2
δ aj= k j cos(ωt − ρ ), kC = kC1 + k E 2 (18)
k D = k D1 + k B 2
δ bj = k j cos(ωt − 2π / 3 − ρ ), (12)
k E = k E1 + k D 2
δ cj = k j cos(ωt − 4π / 3 − ρ )
Therefore, from (15), the output voltages are obtained as;
Where ρ is the phase shift at the input side. The input and ⎡3 ⎤ ⎡3 ⎤
V A= ⎢ k A1 V cos( ρ ) ⎥ cos(ωo1t ) + ⎢ k A2 V cos( ρ ) ⎥ cos(ωo 2t )
output voltages are related as: ⎣2 ⎦ ⎣2 ⎦
⎡V A ⎤ ⎡δ aA δ bA δ cA ⎤ ⎡3 ⎤ π ⎡3 ⎤ π
V B= ⎢ k B1 V cos( ρ ) ⎥ cos(ωo1t − 2 ) + ⎢ kC 2 V cos( ρ ) ⎥ cos(ωo 2t − 4 )
⎢ ⎥ ⎢δ ⎥ ⎣2 ⎦ 5 ⎣2 ⎦ 5
⎢VB ⎥ ⎢ aB δ bB δ cB ⎥ ⎡Va ⎤ ⎡3 ⎤ π ⎡3 ⎤ π
⎢VC ⎥ = ⎢δ aC δ bC δ cC ⎥ ⎢Vb ⎥ (13) V C= ⎢ kC1 V cos( ρ ) ⎥ cos(ωo1t − 4 ) + ⎢ k E 2 V cos( ρ ) ⎥ cos(ωo 2t − 8 )
⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣2 ⎦ 5 ⎣2 ⎦ 5
⎢VD ⎥ ⎢δ aD δ bD δ cD ⎥ ⎢⎣Vc ⎥⎦ ⎡3 ⎤ π ⎡3 ⎤ π
V D= ⎢ k D1 V cos( ρ )⎥ cos(ωo1t − 6 ) + ⎢ k D 2 V cos( ρ )⎥ cos(ωo 2t − 2 )
⎢V ⎥ ⎢δ aE δ bE δ cE ⎥ ⎣2 ⎦ 5 ⎣2 ⎦ 5
⎣ E⎦ ⎣ ⎦
⎡3 ⎤ π ⎡3 ⎤ π
V E= ⎢ k E1 V cos( ρ )⎥ cos(ωo1t − 8 ) + ⎢ k D 2 V cos( ρ )⎥ cos(ωo 2t − 6 )
Therefore the phase A output voltage can be obtained by ⎣2 ⎦ 5 ⎣2 ⎦ 5
using the above duty ratios as
V A= k A V [cos(ωt ) • cos(ωt − ρ ) + cos(ωt − 2π / 3) • cos(ωt − 2π / 3 − ρ ) (19)
The discussion on the common mode voltage addition and
+ cos(ωt − 4π / 3) • cos(ωt − 4π / 3 − ρ )] subsequent enhancement in the modulation index is presented
(14) in [37].
3
V A= k A V cos( ρ ) (15)
2 IV. SIMULATION RESULTS
In eq (15), cos( ρ ) term indicates that the output voltage is
affected by ρ . The term kA is defined in equation (18). Thus, The simulation model is developed in Matlab/Simulink for the
the output voltage V A is independent of the input frequency whole drive system. Three-phase grid supply is assumed as 50
Hz 440 V rms phase voltage (double voltage is assumed since
and only depends on the amplitude V of the input voltage two-motor drive is considered). Five-phase reference voltage
and k A is a reference output voltage time-varying modulating is chosen for the first motor and another set of five-phase
signal for the output phase A with the desired output reference is assumed for the second motor. The five-phase
modulating signals is formulated by adding the two five-phase
frequency ωo1 + ωo 2 , ωo1 is the operating frequency of
references according to the transposition rule (equation (1)).
machine-1 or the first fundamental output frequency and ωo 2 The parameter of the simulation is given in Table 1.
is the operating frequency of machine-2 or the second The simulation condition is taken as;
fundamental output frequency. The fundamental output Motor-1 operating at rated speed of 1500 rpm (reference
voltage magnitude corresponding to ωo1 is given as m1 and frequency of 50 Hz)
corresponding to ωo 2 is given as m2. The five-phase reference Motor-2 operating at half rated speed of 750 rpm (reference
frequency of 5 Hz)
output voltages can then be represented as Load (half rated) applied to motor-1 at t = 1.2 sec
k A1 = m1 cos(ωo1t ), Load (one quarter of rated value) applied to motor-2 at t = 1.1
k B1 = m1 cos(ωo1t − 2π / 5) Switching frequency of the Matrix converter is kept at 6 kHz.
kC1 = m1 cos(ωo1t − 4π / 5) (16) The resulting waveforms for motor side and matrix converter
sides are shown in Fig. 4 and Fig. 5, respectively.
k D1 = m1 cos(ωo1t − 6π / 5)
k E1 = m1 cos(ωo1t − 8π / 5) Table I: Simulation Parameters
Parameters Name Parameters Values
Source side resistance Rs 0.05 Ω
k A2 = m2 cos(ωo 2t ) Source side inductance Ls 8 mH
DC link capacitor 2000 µF
k B 2 = m2 cos(ωo 2t − 2π / 5) Stator resistance 10 Ω
Stator leakage inductance 40 mH
kC 2 = m2 cos(ωo 2t − 4π / 5) (17) Mutual inductance 420 mH
k D 2 = m2 cos(ωo 2t − 6π / 5) Inertia J 0.03 kg sq m
Number of Poles 4
k E 2 = m2 cos(ωo 2t − 8π / 5) Rated Torque 8.33 Nm
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2000 5
-10
1.92 1.93 1.94 1.95 1.96 1.97 1.98 1.99 2
1000 Time [s]
0 0
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0 50 100 150 200 250 300 350 400 450 500
Time [s] Frequency [Hz]
a. a.
0
0
0 Fundamental = 313.3556
200
-10
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
Time [s] 0
0 50 100 150 200 250 300 350 400 450 500
b. Frequency [Hz]
20
b.
15 Fig. 6. Matrix converter output current and voltage time domain and frequency domain waveform.
1
Source current [A]
Phase 'A' MC current [A]
10
5 0
0
-5 -1
1.94 1.95 1.96 1.97 1.98 1.99 2
Time [s]
Spectrum source current [A]
-10
-20
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0.2
Time [S]
c. 0
0 50 100 150 200 250 300 350 400 450 500
Fig. 4. Response of two-motor drive, a. speeds, b. torques, c. phase ‘a’ current Frequency (Hz)
from matrix converter. Fig. 7. Filtered source side current spectrum.
The spectrum for the output side current and voltage is shown
The source side current is sinusoidal and working at unity
in Fig. 6 and that of source side current unfiltered is shown in
power factor. This is the distinct feature of the matrix
Fig. 7.
converter based drives. The total harmonic distortion (THD) is
3 computed for the voltage and current as follows;
2 Voltage
2
∞
⎛ vn ⎞
Source side current and
∑
Voltage 150:1 [A, V]
1
THD = ⎜⎜ ⎟⎟ (25)
0 n =3, 5, 7.. ⎝ v1 ⎠
-1
Current Where vn is the nth harmonic component and v1 is the
fundamental component magnitude. For the computation of
-2
THD, upto 10th lower order harmonic components are taken.
-3 The THD for the source side current is calculated as 1.66%,
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1
Time [s] the output side current THD is 4.55% while the output voltage
Fig. 5. Source side voltage and current, voltage is reduced to 150 times. THD is 4.48%. These values are well within the specified
limit.
The output voltage and current waveform shows two
frequency components at the two operating frequencies of the V. CONCLUSION
two motors. The two machines shows acceleration at the
initial response. When load is applied the speeds drops and the A three to five-phase matrix converter based five-phase series-
motor settles at the same speed. The speed is not corrected as connected two-motor drive system is presented in this paper.
no closed-loop controller is employed in this analysis. The Simple carrier-based PWM technique is used to control the
motor torque is typical of a five-phase induction machine. matrix converter. The matrix converter successfully drive two
five-phase series-connected induction machine. This solution
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has advantage of higher power factor and sinusoidal source [21] M. Jones, E. Levi, and S.N. Vukosavic, “Independent control of two
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