Evaluation of Total Loss for an Inverter andMotor by Applying Modulation Strategies
Jun-ichi Itoh, Takumi Ogura
Nagaoka University of Technology
Department of Electrical, Electronics and Information Engineering,1603-1 Kamitomioka-cho,Nagaoka city Niigata, Japan,
itoh@vos.nagaokaut.ac.jp, takumi94@stn.nagaokaut.ac.jp

Abstract—This paper evaluates total loss of an induction
motor and a permanent magnetic motor drive systemapplying two various modulation strategies on each motor,
which are the PWM operation and the six-step operation. At
first, the loss analysis is implemented based on thetheoretical model of an inverter and an induction motor. As
a result, when the leakage inductance is less than 25%, thePWM drive should be used instead of six-step operation inorder to obtain the minimum loss. On the other hands, when
the leakage inductance is more than 65%, the six-step
operation is suitable for obtaining high efficiency.
Additionally, an optimization point of the switchingfrequency is mentioned according to the leakage inductance.
The validity of the loss analysis is confirmed by simulationand experimental results in which an induction motor of
1.5kW was used. Similarly, the total loss of the permanent
magnet motor drive system is calculated based on the same
method. When the synchronous reactance of the motor islower than 26%, the total loss of the PWM with a 5kHz
carrier is lower than the six-step modulation. On the otherhands, when the synchronous reactance of the motor is morethan 26%, the total loss of the six-step modulation is smaller
than the PWM with a 5kHz carrier.
Keywords— Inverter loss, Six-step modulation, PWM
drive, Motor loss

I.

INTRODUCTION

In recent years, motor drive systems are used in many
applications, such as trains, hybrid electric vehicle (HEV),
home appliances and industrial system. The power
consumption by motors occupies more than 50% of thetotal amount of the power consumption. Therefore, it isvery important to improve the motor drive systemefficiency in consideration of energy saving andsuppression of CO2 emission.
In order to achieve energy saving in recent years,
inverters are widely applied to industry applications andhome appliance. According to the applications, themodulation strategy of the inverter chooses either a sixstep modulation or pulse width modulation (PWM). For
example, the six-step modulation is used in trains and
HEV. The six-step modulation can reduce the switchingloss of the inverter because the number of the switchingtimes is only six in per output period. Additionally, alarger output voltage can be obtained in comparison to
the PWM; however the motor loss will be increased
because the output voltage contains many low frequency

harmonics components the 5th and 7th order of output
frequency. These harmonics current increases the motor
loss. It is noted that the harmonics current generates
ripple in torque, however the amount of ripple does notcontribute to increase average torque.
On the other hand, the PWM is extensively used inhome appliances and industrial drives. The sinusoidal
output current is obtained and the output current is
controlled by an auto current regulator (ACR). Since theharmonics components in the output voltage are higher
than the output frequency, the copper loss and torqueripple of the motor can be reduced by the six-step
modulation. However, in PWM drive the inverter loss islarger than the six-step modulation because the switchingfrequency of the inverter is higher than the six-step
modulation. There is a trade-off relationship between the
inverter loss and the motor loss according to themodulation strategies and its switching frequency.
There are many analyses about the losses among
inverter, induction motor and permanent magnetic motor
regarding to efficiency [1-3]. Almost all studies discussed
the motor structure and losses by using the finite element
method (FEM). The FEM can easily understand the
behavior of flux in the motor. Then, the best modulation
strategy will be decided for each motor based on the FEM.
However, when the motor parameter has changed, thisprocess needs to be repeated to find out the bestmodulation strategy again. Therefore, it is difficult to
discuss the best modulation strategy based on only the
FEM.
This paper discusses the most suitable modulation
strategy for the inverter in order to obtain a highefficiency in the inverter and the motor. The aim of thispaper is to optimize the modulation strategy in the
inverter according to the motor parameter. The model ofthe theoretical loss calculation of the inverter and motor
are provided, respectively.
At first, this paper describes the inverter loss and theinduction motor loss based on the theoretical model.
Second, the harmonics components in the output voltageof the inverter are calculated. After that, the losses of theharmonic components are calculated by an equivalentcircuit of the motor. Then, the relations between the
motor parameters and the total loss are being clarified.
The motor loss, which is caused by the harmonics

components, is dominated by a leakage inductance Lσ ofthe induction motor. Therefore, the modulation strategy ischosen according to the leakage inductance of theinduction motor. Note that in a PM motor thesynchronous reactance will dominate the current
harmonics. The calculation results are confirmed by thesimulation and the experimental results with using a 1.5
kW induction motor. Finally, the permanent magnet
motor losses are calculated in a similar manner. The
copper loss of the PM motor is calculated by theequivalent circuit of the motor. The iron loss of thepermanent magnet motor is calculated by FEM in order tomake a loss table. The motor loss which is generated by
the harmonics components is dominated by asynchronous reactance La of the PM motor.
II.

(a) Switching loss(b) Conduction loss

LOSS ANALYSIS METHODA

Loss analysis of an inverter
Inverter losses consists of the switching loss generates
while switches are turned on and off, and the conduction(c) Recovery loss
loss is generated by saturation voltage of a switching
Fig. 1. Inverter loss.
device.
Figure 1(a) shows a waveform of the current andvoltage in a switching device while the switch is turned
I1IT
on and off. If the switching pattern of the device is ideal,
then the voltage and current interchange with each other
L1
L2
R1
R2
instantaneously. However, the voltage and current
interchange slowly as shown in Figure 1(a). That period
1− sIMis considered as the switching loss which is proportionalRV1
Msto the shaped area in Figure 1(a). The switching loss Psw
is obtained by (1) using turn-on loss Pon and turn-off loss
Poff.
Psw = Pon + Poff
(1) Fig. 2. Equivalent circuit of an induction motor verified by the fundamental
2

The turn-on loss Pon and the turn-off loss Poff are
obtained by (2) and (3), which multiplied by turn-off Eoff,or turn-on Eon and the switching frequency fs from thedata sheet of the switching device.
Pon = Eon × f s(2)

Poff = Eoff × f s(3)

Figure 1(b) shows a current behavior in the switching
device that regards to the conduction loss Pcon. The
conduction loss Pcon can be separated into two losses
PIGBT and PFWD, which are generated either by a switching
device or a free wheel diode (FWD) according to thecurrent direction. The conduction loss Pcon is obtained byPcon = PIGBT + PFWD

(4)
The conduction loss PIGBT of a switching device andconduction loss PFWD from a FWD are obtained
respectively by⎛1 m⎞
PIGBT = I p ×V CE×⎜⎜ + i cos θ ⎟⎟⎝ 8 3π⎠⎛1 m⎞
PFWD = I p ×V F ×⎜⎜ − i cos θ ⎟⎟⎝ 8 3π⎠(5)(6)

where VCE and VF are the saturation voltage of IGBTand FWD, which are obtained from a data sheet, mi is thecomponent.

modulation index, cosθ is the load power factor and Ip isthe peak value of the load current. Equations (4) and (5)indicate that the switching frequency is not an element
will affect on the conduction loss of IGBT and FWD.
Figure 1(c) shows the recovery current in a FWD.
When an IGBT is turned on, a counter FWD, which islocated in a opposite side arm, becomes reverse bias.
Then, the reverse recovery current occurs in the diodeand a recovery loss is generated. The recovery loss of
FWD is obtained byPre =1
× I × Vdc × t rr × f s
8(7),

where I is the RMS value of the load current, reverse trr
is the recovery time, which are obtained from a data sheet,
Vdc is the DC link voltage of inverter.
B

Verification of Copper loss by the fundamental
component of a motor current
Figure 2 shows a one-phase T type equivalent circuit of
an induction motor. The motor loss consists of copper
loss and iron loss. This paper mainly considers the copper
loss according to the different modulation strategies
because the iron loss in an induction motor does notdepends on the modulation strategies as discussed in [1].
The copper loss is calculated by the fundamental
component as follows,
P1 = R1 I 12 + R2 I T

2(8)
where R1 is the primary resistance, R2 is the secondary
resistance, I1 is the fundamental component of the load
current of the inverter and IT is the torque component
current.
C

Verification of Copper Loss by the harmonics components
The harmonic components are contained in the output
voltage when the six-step modulation or PWM is used.
As a result, loss is occurring in the induction motor
because of containing the harmonic components.
Figure 3 shows a one-phase harmonic component
equivalent circuit of an induction motor. The frequencyof the harmonic components is higher than the output
frequency. Therefore, slip and exciting current can beneglected in the harmonic component equivalent circuit.
In addition, the impedance of the motor is dominated by
the leakage inductance Lσ. Then the n-order harmonics
current In is obtained byIn =Vn
ω n Lσ(9),where Vn is the n-order harmonic voltage in the inverter
output and ωn is the n-times angler frequency of theinverter output.
The copper loss by the harmonic current are obtainedbyPn = ( R1 + R2 )I n2(10).

III. QUANTIFICATION OF TOTAL LOSSIn this chapter, the optimization of the modulation
strategy is clarified for the use in an induction motor. At
first, in order to calculate the harmonic components of theoutput voltage, the six-step waveform and the PWM
waveform are implemented with the Fourier series
expansion. Next, the harmonic voltage is calculated, andthe copper loss is calculated based on the harmonics
current according to (10). Then, the converter loss, which
consist of the switching loss and the conduction loss
according to (1), (4) and (7), is calculated. Besides, thetotal loss of the inverter and the motor are obtained fromeach of the modulations. Finally, the leakage inductanceand total loss are normalized by the motor rating. Then,the relations between the leakage inductance and themodulation strategy can be clarified.
Fourier series expansion of the output voltage fromusing a six-step modulation waveform is obtained by (11).
The amplitude Vn of each harmonics voltage is given by
(12).
En =

4 2 E dc
⎛ π⎞ 1cos⎜ n ⎟ ×
nπ⎝ 6⎠ 2

Fig.3. Equivalent circuit of an induction motor verified by harmoniccomponents.

(11)

Fig. 4. Relations between the loss and the leakage inductance.

where Edc is DC link voltage of the inverter.
On the other hand, Fourier series expansion in the
PWM waveform is obtained by (13). Likewise, theamplitude of each harmonics voltage is obtained by (14).
e M (t )= a cos (ω s t + α )E∞⎡⎛ nπ ⎞⎤
4+∑sin ⎢⎜⎟ ⋅ {a cos (ω s t + α ) − 1}⎥ cos nω r t
n =1 nπ⎣⎝ 2 ⎠⎦(13)⎛ 4 E ⎞ ⎛ nπa ⎞ n = 1,3,5K k = 0 ,2 ,4K
Vn = ⎜ dc ⎟ J k ⎜⎟⎝ nπ ⎠ ⎝ 2 ⎠ n = 2 ,4,6K k = 1,3,5K(14)

where, Jk is the n-order Bessel function.
Eq. (9) can obtain the harmonics current by knowing
the harmonics voltage. Then the equation (9) is
normalized by the rated voltage, rated current and rated
angler frequency of the motor, the n-order or thenormalized current In is expressed by
In =

Vn
n% X

(15)

where %X is the normalized reactance of the induction
motor.
From (1)-(6), the conduction loss and the switchingloss are proportional to the RMS value of the current, andthe switching loss is proportional to the switchingfrequency. Therefore, proportionality coefficients are
introduced to the conduction loss and the switching loss
for simplification. Using these coefficients, the total loss
including inverter loss is obtained by (16) substituting(15) for (10).
22 2 E dc
Vn =
nπ

(12)

⎛V ⎞
⎛V ⎞
P = 3R ∑ ⎜⎜ n ⎟⎟ + ( k conVcon + k s f s ) ∑ ⎜⎜ n ⎟⎟nX
⎝ nX ⎠
⎝
⎠

2

(16)

where, n is the harmonics order(n = 1, 5, 7, 11, 13…),TABLE I. PARAMETERS OF INDUCTION MOTOR
kcon is the coefficient of the conduction loss calculated by(4), ks is the coefficient of the switching loss calculated
Items
ValuesRated power
1.5kW
by (1) and (7).
Phases and poles
3phases, 4poles
Figure 4 shows the relations between the losses and
Rated frequency
51Hz
the leakage inductance, which is obtained by (16). Note
Rated voltage
188Vthat the calculation results are normalized with the motor
Rated current
7A
Rated speed
1500r/minparameters in Table 1. In addition, the fundamental
1.09Ω
Primary
resistanceR1
voltage of the six-step modulation is set to rated voltage
Secondary resistance R2
0.79Ω
of the motor. The DC link voltage of the inverter using
Leakage inductance L
7.29mHthe PWM is set to the peak voltage of the rated phase
Mutual inductance Lm
85.3mH
Moment of Inertia
0.009kg m2
voltage of the motor.
In figure 4, the point P is a boundary point between theswitching frequency of 16 kHz and a six-step modulation.
When the leakage inductance of the motor is lower than
25%, the total loss of the PWM with a 16-kHz carrier islower than the six-step modulation. In contrast, the point
Q is a boundary point between the switching frequency of
500 Hz and the six-step modulation. When the leakage
inductance of the motor is more than 65%, the total lossof the six-step modulation is smaller than that of the
PWM with 500-Hz carrier. That is, when the leakage
inductance is small, the total loss is dominated by the
motor loss because each harmonics current is increased.
The suitable modulation strategy is obtained by (17)and (18) substituting losses of the point P and the point Q
Fig. 5. Relations between the loss and the switching frequency.
into (16), which are 0.03p.u. and 0.01p.u., respectively.
The conditions for which the PWM drive system issuitable are obtained by (17).
6 R ∑ (V / n )%X <− (k V + k f ) ∑ (V / n ) + {k V + k f } ∑ (V / n ) + (0.36 R ∑ (V / n ) )2

n2

con con

s

s

n

2

con con

s

s

2

2

n

6 R ∑ (Vn / n )

n

(17)

2

%X >

− (kconVcon + k s f s ) ∑ (Vn / n ) +2{kconVcon + ks f s }2 ∑ (Vn / n )2 + (0.12 R∑ (Vn / n )2 )(18)
The PWM is better than the six-step modulation when
the leakage inductance is constrained by (17). Likewise,
the six-step modulation is better than the PWM when theleakage inductance is constrained by (18).
IV. VERIFICATION OF THE LOSS ANALISYS BASED ON

Fig. 6. Loss comparison for various modulation strategies.

SIMULATION AND EXPERIMANTAL RESULTS

A

Simulation verifications
At first in order to evaluate the validity of Figure 4,
loss analyses for the motor and inverter were simulated.
Table 1 shows the parameters of an induction motor. The
simulation condition, where %X = 14.7%, is used.
Figure 5 shows a relationship between the total loss
and the switching frequency. The simulation conditions
are shown in Table 1. From the results, the copper loss of
the induction motor becomes larger when the switchingfrequency is lower because the large low-order harmonics
components are contained in the motor current. In
contrast, when the switching frequency is higher, theinverter loss is increased because of higher switching loss
is occurring. As a result, the total loss is reducing to aminimum at PWM 5kHz where the motor loss isdecreasing to balance with the increased inverter loss.
Figure 6 shows the loss comparison among various
modulation strategies. The harmonics loss in the motor isincreased by the six-step modulation. Thus, the total loss
is increased. On the other hand, the PWM with 16-kHz
carrier generates little loss depending on harmonics
component though the switching loss of the inverter
increases. As a result, the switching frequency of 5 kHz is
optimal modulation strategy for the motor as shown in
Table 1.
B

Experimental verifications
The total loss was evaluated by experimental results
using the induction motor of Table 1. The vector control

is used in PWM in order to keep the same current without
any compensation for the dead time voltage error. It isnoted that the V/f control is used in six-step modulation;
however the fundamental voltage of the inverter output is
set to the same as other conditions.
Figure 7 shows the output voltage and current
waveform of the six-step modulation. The output current
has large distortion due to the six-step modulation. The
total harmonic distortion (THD) of the output current is
62.9 %.
Figure 8 show the harmonics analysis results of the sixstep modulation. The harmonic components of 5th and 7thorder are appeared largely. The average torque is notincreased by these harmonic components because theharmonic current generates torque ripple of 6th order inthe output frequency.
Figure 9 shows the output voltage and current
waveforms using the PWM with a 16-kHz carrier. The
motor current becomes a sinusoidal waveform by using
the PWM. The THD of the motor current is 2.59 %.
Likewise, when the switching frequency of 5kHz is used
the motor current can obtain a sinusoidal waveform as
well.
Figure 10 shows the harmonics analysis results of
PWM with a 16-kHz carrier. Since the current feedback
is applied in vector control, the output current contains
few of the low-order harmonics components.
Figures 11 (a) and (b) show the losses about theinverter and motor at 50% and 25 %load condition,
respectively. The minimum total loss is obtained when
the switching frequency is 5 kHz. This result well agrees
with the simulation result. It is noted that the loss in theexperimental results is different from the simulation
because of the iron loss of tan induction motor isFig. 9 Experimental results using the PWM with a 16-kHz carrier.

100[%]

fout = 51[Hz]20[dB/div]

10[%]
1[%]

500[Hz/div]

0.1[%]

Frequency [Hz]Fig. 10. Harmonics analysis results using the PWM with a 16-kHz carrier.
Total loss ( 50% Load )

400

300

200

100

Motor loss ( 50% Load )Inverter loss ( 50% Load )

0
10

100

1k

10k

100k

Switching frequency [Hz](a) 50% load condition.

Fig.7 Experimental results with the six-step modulation.
100[%]

fout = 51[Hz]20 [dB/div]

10[%]

500 [Hz/div]1[%]
0.1[%](b) 25% load condition.
Fig.11. Losses about the inverter and motor under various conditions.
Frequency [Hz]Fig. 8. Harmonics analysis results of the six-step modulation.

neglected in the simulation analysis.
Then, the modulation is optimized by considering to
the motor characteristic. When the leakage inductance isless than 25%, the PWM should be used in order toobtain the minimum loss. On the other hands, when theleakage inductance is more than 65%, the six-step
modulation is suitable to obtain the highest efficiency.
From this, since the leakage inductance of a general
purpose induction motor is approximately 10%-20%, thePWM is suitable for its drive. On the other hand, theadvantage of the six-step modulation is appeared in the
motor, which has high fundamental frequency, small
primary and secondary resistance, and a large leakage
inductance. That is, for a high speed motor that has large
normalized leakage reactance, is suitable for the six-step
modulation.
V. LOSS ANALYSIS OF PERMANENT MAGNET MOTOR
Verification of Copper loss by the fundamental
component of the motor current
Table 2 shows the parameters of a permanent magnet
motor that used as an example in this paper. It is notedthat the inverter loss analysis was evaluated by the same
method as discussed in Chapter II.
Figure 12 shows a single-phase equivalent circuit of a
PM motor. The motor loss composed of the copper loss
and the iron loss. The copper loss of the permanent
magnet motor is calculated by the fundamental
component of the motor as shown in (19).

TABLE II. PARAMETERS OF PERMANENT MAGNET MOTOR
Items
Rated power
Phases and poles
Rated frequency
Rated voltage
Rated current
Rated speed
Back electromotive force
Winding resistance Ra
Synchronous reactance La
Number of poles
Stator outer diameter
Stator inner diameter
Winding configuration
Rotor outer diameter
Rotor shaft diameter
Electricalresistivityρ(magnet)Values
1.5kW
3phases, 6poles
90Hz
180V
6.1A
1800r/min
147V
0.783Ω
11.5mH
36slots
130mm
83mm
138turn,series per phase
82.2mm30mm
1.4×10-6Ωm

A

Fig 12 Equivalent circuit of a PM motor verified by the fundamental
component.

P1 = Ra I 12

(19)
where Ra is a winding resistance, I1 is a fundamental
component of the load current of the inverter.
B

Copper loss by the harmonics components
The harmonic voltage is applied to the PM motor when
the six-step modulation and the PWM are applied. As aresult, the loss which is caused by the harmoniccomponents occurs in the permanent magnet motor.
Figure 13 shows the single-phase equivalent circuit ofa PM motor for high frequency components. The backelectromotive force is disappeared in the equivalentcircuit because the back electromotive force does notcontain high frequency components. The impedance ofthe motor is dominated by the synchronous reactance La
from figure 13. Then the n-order harmonics current In is
obtained by
In =

Vn

ω n La(20),

where Vn is the n-order harmonic voltage of theinverter output voltage and ωn is the n-times naturalangular frequency of the inverter output.
The copper loss is calculated based on the harmonic
current is obtained by
Pn = Ra I n2

C

(21),

Iron loss analysis by FEM
The iron loss of the permanent magnet motor occurs inthe stator, the rotor and the magnet. An eddy current loss

Fig 13 Equivalent circuit of a PM motor verified by the harmoniccomponents.

in the magnet is changed largely by the switchingfrequency. Therefore, the eddy current loss should beanalyzed by a simplified model using mathematics
equation; however the eddy current loss in the magnet ischanged by the motor shape. That is, it is difficult toanalysis only mathematic model. Therefore, in the firststep, the analysis using FEM is introduced for the eddy
current loss analysis in this chapter.
Figure 14 shows a motor model by using FEM. The
specification of the magnet is shown in Figure 14 andTable 2. Internal magnet motor (IPM) of 1.5kW, 180V is
selected as one of example. The number of pole is 6. This
motor is used for industrial applications in 200V-power
grid lines.
Figure 15 shows the current density of the magnet foreach of the input voltage waveforms. The current density
in the sinusoidal wave drive, which is an ideal condition,is distributed equally. On the other hands, in the six-step
modulation and the PWM with a 450Hz carrier, a current

31density becomes greatly higher than the other modulation
conditions. That is, the current density of the PM motor
becomes larger when the switching frequency is lower
because the low order harmonics components of the
motor are becoming larger.
Figure 16 shows the loss from the eddy current byusing FEM analysis. From the results, we know that theeddy current loss of the PM motor is proportional to thecurrent density, therefore the loss becomes larger when
the switching frequency is lower.
Figure 17 concludes the relations between the eddy
current loss and the synchronous reactance. The eddy
current loss for each of the modulation strategies wasanalyzed by the FEM according to various values ofsynchronous reactance. The eddy current loss of FEM
result is approximated by

Fig. 14. The model of a FEM motor.

Pe = aX − b(22),where a and b are the constant parameters, which are
fitting according to each simulation results. Table 3
shows the constant parameter a and b for each results.
The eddy current loss is estimated under various
conditions by (22) with Table 3.
VI. QUANTIFICATION OF TOTAL LOSS

Loss [W]Fig. 15. Magnetic current density.

Fig. 16. Eddy current loss by the FEM analysis.

Eddy current loss [W]

In order to discuss which of the modulation strategy isbest for the PM motor, first the motor loss needs toquantify according to the modulation strategy. The sixstep drive waveform and the PWM drive waveform isimplemented with the Fourier series expansion to obtain
the harmonic component of the inverter output. As aresult, the harmonic voltage and current are obtained,
respectively. Then, the copper loss is calculated from theharmonics current according to (21). After that, the eddy
current loss is calculated by (22), and the total loss ofboth the inverter and the motor is sum-up from eachvarious modulation. Finally, the synchronous reactanceand the total loss are normalized by the motor rating.
Then, the relations between the synchronous reactanceand the modulation strategy are clarified.
As discussed in the previous chapter, Fourier series
expansion of the inverter output voltage using a six-step
modulation waveform is obtained by (11). The amplitude
Vn of each harmonics voltage is given by (12). In addition,
Fourier series expansion of the PWM waveform isobtained by (13). Therefore, the amplitude of eachharmonics voltage is obtained by (14).
The harmonics current which is generated by eachharmonics voltage is obtained by (20). When the equation(20) is normalized by the rated voltage, current, angler
frequency of the motor and the n-order normalized
current In is obtained by (15).
The total loss is quantified by the expression of the loss.
From (1)-(6), the conduction loss and the switching loss
are proportional to the RMS value of the current; theswitching loss is proportional to the switching frequency.
Therefore, the proportionality coefficient kcon and ks are
introduced to the conduction loss and the switching loss,
respectively. The copper loss of the motor is obtained by

Fig. 17. Relations between the eddy current loss and the synchronous
reactancesubstituting (15) in (19). In addition, the eddy current lossof the motor is obtained by (22). The total loss which
consists of the inverter loss, the copper loss of the motor
and the eddy current loss of the motor can be obtained by
2

TABLE III.

WAVEFORM

Input waveform
Six-step
PWM450Hz
PWM990Hz
PWM5kHzPWM10kHzPWM16kHz

2

VII. CONCLUSION
This paper described the total loss among the inverter
loss, the induction motor and PM motor. The motor loss
and the inverter loss are calculated by simple equations.
Besides, the optimization of modulation strategy for the
inverter is clarified.
As a result, when the leakage inductance in the
induction motor is less than 25%, the PWM should be
used in order to obtain the minimum loss. On the otherhands, when the leakage inductance is more than 65%,
the six-step modulation is suitable to obtain highest
efficiency. The total loss is at minimum for a 1.5-kW
general purpose induction motor when the switchingfrequency is 5 kHz on the PWM drive. The simulationand experiment results confirmed the validity of thetheoretical analysis.
On the other hands, when the synchronous reactance in
the PM motor is lower than 26%, the total loss of the
PWM with a 5kHz carrier is lower than the six-step
modulation. When the synchronous reactance is morethan 26%, the total loss of the six-step modulation issmaller than the PWM with a 5-kHz carrier.
In future study, the iron loss will be considered and the
optimization of the modulation strategy PM motor will be
discussed in the term of experiment.
REFERENCES[1]

K. Yamazaki, Y. Seto and M. Tanida: “Iron Loss Analysis of IPM
Motor Considering Carrier Harmonics”, IEEJ Trans., Vol.125-D,
No.7, pp.758-766 (2005) In Japanese.

a
560
2300
3200
1200
220
110

b
1.5
1.5
1.6
1.7
1.4
1.3

Loss [pu]

⎛V ⎞
⎛V ⎞
P = 3Ra ∑ ⎜ n ⎟ + ( k conVcon + k s f s ) ∑ ⎜ n ⎟ + aX −bnX⎝ nX ⎠⎝⎠(23).
Figure 18 shows relations between the loss and thesynchronous reactance, which is obtained by (23). Notethat the calculation results are normalized with the motor
parameters in Table 2.
In figure 18, the point Q is the boundary point between
the PWM drive with a 5kHz carrier and the six-step
modulation. When the synchronous reactance of the
motor is lower than 26%, the total loss of the PWM witha 5kHz carrier is lower than the six-step modulation. On
the other hands, when the synchronous reactance is morethan 26%, the total loss of the six-step modulation issmaller than the PWM with a 5kHz carrier. The copper
loss of the PM motor is lower than the induction motor
because PM motor does not contain the secondary
resistance. In the case of synchronization reactance isabout 30%, the eddy current loss is almost equally on all
the control strategies. Therefore, the range which issuitable for the six-step drive expands in comparison tothat of the induction motor.

THE CONSTANT NUMBER OF EACH INPUT

Fig 18 Relations between the loss and the synchronous reactance.
[2]

K. Yamazaki and A. Abe: “Loss Analysis of IPM Motors
Considering Carrier Harmonics –Calculation of Eddy Current
Loss in Permanent Magnet Using 3-D Finite Element Method–”,
IEEJ Trans., Vol.127-D, No.1, pp.87-94 (2007) In Japanese.
[3] K. Yamazaki and N. Fukushima: “Carrier Loss of Induction
Motors Driven by Inverters –Comparison between Results
Separated by Experiment and Field Analysis–”, IEEJ Trans.,
Vol.129-D, No.11, pp.1068-1074 (2009) In Japanese.
[4] J. Itoh, J. Toyosaki, and H. Ohsawa: “High performance V/fcontrol method for PM Motor”, IEEJ Trans., Vol.122-D, No.3,
pp.253-259 (2002) In Japanese.
[5] H. Yonezawa, K. Taniguchi, T. Morizane and N. Kimura:“Modified Trapezoidal Modulating Signal suitable for PM
Synchronous Motor Drives”, IEEJ Trans., Vol.125-D, No.1,
pp.46-53 (2005) In Japanese.
[6] I. Takahashi and S. Miyairi: “Relation between the Output Voltage
Waveform of the PWM Inverter and its Gate Control Signals”,
IEEJ Trans., Vol.95-B, No.2, pp.73-80 (1975) In Japanese.
[7] I. Takahashi, T. Sekiguchi and S. Miyairi: “Control Method of
PWM Inverter Output Wave with Reduced Low Order Harmonic
Components”, IEEJ Trans., Vol.97-B, No.5, pp.302-308 (1977) In
Japanese.
[8] T. Ogura and J. Itoh: “Open Loop Control with a Square Wave
Operation and a Total Loss for a Permanent Magnet Motor”,
JIASC IEEJ, pp.1-101 (2009) In Japanese.
[9] R. Liu, C. Chunting Mi and D. Wenzhong Gao: “Modeling of
Eddy-Current Loss of Electrical Machines and Transformers
Operated by Pulsewidth-Modulated Inverters”, IEEE Trans. Magn.,
pp.2021-2028 (2008)[10] M. OGAWA : An Analysis of Transient Phenomena of
Magnetizing Current of Transformer Generated by DC Interrupted
Excitations[11] K. Itoh, Y. Hashiba, K. Sakai and T. yagisawa: “The a.c. losses of
the rare-earth permanent magnets”, IEEJ Trans., Vol.118-A, No.2,
pp.182-187 (1998) In Japanese.
[12] R. A. NEWBURY: “Prediction of Loss in Silicon Steel from
Distorted Waveforms”, IEEE Trans. Magn, pp.262- (1977) In
Japanese.