JP3480302B2 - Rotating electric machine - Google Patents

Description

Detailed Description of the Invention

[0001]

TECHNICAL FIELD The present invention relates to a rotary electric machine.

[0002]

2. Description of the Related Art Two synchronous motors with the same rated torque are independently
It has been proposed to provide three of them and rotate them synchronously (see Japanese Patent Laid-Open No. 9-275673).

[0003]

By the way, in order to make the structure compact, it is conceivable to construct two rotors and one stator on the same shaft with a three-layer structure (see Japanese Patent Laid-Open No. 8-340663). ).

In this case, in order to rotate the two rotors separately and synchronously, a dedicated coil is prepared for each rotor in the stator, and two inverters (current controllers) for controlling the current flowing in each dedicated coil are provided. You have to be prepared.

However, if a current is passed through each coil and each inverter, the loss due to the current (copper loss, switching loss) cannot be avoided.

Therefore, in the present invention, the coils are commonly used.
A common coil, by passing a double coupling current to the coil, and to prevent the loss due to current.

For example, as shown in FIGS. 1 and 5, the rotors 3 and 4 are arranged with a predetermined gap between the outer peripheral side and the inner peripheral side of the cylindrical stator 2, and the outer rotor 3 and the inner rotor 4 are arranged. When the magnetic pole number ratio of is set to 2: 1,
There are 12 common coils in total.

In this case, depending on the ratio of the number of magnetic poles of the outer rotor 3 and the inner rotor 4, the inner rotor 4 (or the inner rotating magnetic field) may affect the rotation of the outer rotor 3 to cause a torque fluctuation, or vice versa. It is conceivable that torque fluctuation may occur in the rotation of the inner rotor 4 under the influence of the outer rotor 3 (or the outer rotating magnetic field). If such torque fluctuation occurs, it is not possible to operate at a constant rotation.

Therefore, in the present invention, the ratio of the numbers of magnetic poles of the two rotors is set to a combination of K: L (K is an even number, L is an odd number), so that the magnetic poles of one rotor and the rotor are driven. The purpose is to prevent the rotating magnetic field from affecting the rotation of other rotors.

[0010]

According to a first aspect of the present invention, two rotors and one stator are constructed in a three-layer structure and on the same shaft, and the two rotors have different rotating magnetic fields.
Forming a common coil in the stator to generate a field ,
Apply the current corresponding to each rotor to this common coil
In a rotating electrical machine that is designed to pass combined currents,
The ratio of the numbers of magnetic poles of the two rotors is a combination of K: L (K is an even number, L is an odd number).

According to a second invention, in the first invention, the rotor drive current on the side having a smaller number of magnetic poles is 2K.
× Three-phase AC.

According to a third aspect of the present invention, the coils are divided into three phases per one pole of the rotor on the side having the smaller number of magnetic poles in the first aspect, and each divided coil has a small number of magnetic poles. The current value obtained by dividing the side rotor drive current by the number of the divided coils is passed.

[0013]

According to the first aspect of the present invention, the magnetic poles of one rotor and the rotating magnetic field created to drive that rotor do not affect the rotation of the other rotor, and as a result, a constant rotation occurs for both rotors. It becomes possible to drive in.

When the rotor drive current on the side having a small number of magnetic poles is a three-phase alternating current, the current concentrates on a specific coil. Since 12-phase AC when K is 2), the current does not concentrate in a specific coil, which can suppress copper loss.

Further, according to theoretical analysis, compared with the case where the rotor drive current on the side with a smaller number of magnetic poles is a three-phase AC, the rotor on the side with a smaller number of magnetic poles is driven to generate the same drive torque. When the current is, for example, K = 2, 1/4 is sufficient, so that the current rectifier can be further downsized.

Also in the third aspect of the invention, the electric current does not concentrate on a specific coil, and thus copper loss can be suppressed.

[0017]

1 is a sectional view of a rotary electric machine main body 1. As shown in FIG. In the figure, rotors 3 and 4 are arranged on the outer side and inner side of a cylindrical stator 2 with a predetermined space (three-layer structure), and the outer and inner rotors 3 and 4 respectively cover an outer frame.
It is rotatably and coaxially provided with respect to 5 (see FIG. 3).

The inner rotor 4 is formed of a pair of permanent magnets having an S pole on one half and an N pole on the other half, whereas the outer rotor 3 has twice the number of poles per pole of the inner rotor 4. So that the permanent magnet poles are arranged. That is, S of the outer rotor 3
There are two poles and two north poles, and the south pole and the north pole are switched every 90 degrees.

When the magnetic poles of the rotors 3 and 4 are arranged in this manner, the magnet of the inner rotor 4 is not given a rotational force by the magnet of the outer rotor 3, and conversely, the magnet of the outer rotor 3 does not. No rotating force is applied by the magnet.

For example, consider the effect of the magnets of the inner rotor 4 on the outer rotor 3. Inner rotor 4 for simplicity
Think fixed. First, in the relationship between the S pole of the inner rotor 4 and the upper magnet SN of the outer rotor 3 facing the S pole, the magnetic force generated by the S pole of the inner rotor 4 in the illustrated state is received,
If the upper magnet SN of the outer rotor tries to rotate in the clockwise direction, in the relationship between the N pole of the inner rotor 4 and the lower magnet SN of the outer rotor 3 facing it, the N of the inner rotor 4 is
The poles cause the lower magnet SN of the outer rotor 3 to rotate counterclockwise. That is, the S pole of the inner rotor 4 is the outer rotor
The magnetic force exerted on the upper magnet of Fig. 3 and the magnetic force exerted by the N pole of the inner rotor 4 on the lower magnet of the outer rotor 3 cancel each other out, and the outer rotor 3 has no relation to the inner rotor 4 and has a relationship with the stator 2. It can be controlled only by itself. This is the same between the rotating magnetic field generated in the stator coil and the rotor as described later.

The stator 2 has 3 poles per magnetic pole of the outer rotor 3.
12 coils (= 3 x 4) composed of 6 coils
Are evenly distributed on the same circumference. Reference numeral 7 denotes a core around which a coil is wound, and the same number of cores 7 as the coils 6 are arranged on the circumference at equal intervals with a predetermined gap (gap) 8.

As will be described later, twelve coils are distinguished by numbers, and in this case, the coil 6 comes out to mean the sixth coil. It is confusing with the expression coil 6 as a component, but the meaning is different.

The following composite currents I _{1 to} I ₁₂ are passed through these 12 coils.

First, since a current (three-phase alternating current) for generating a rotating magnetic field for the inner rotor 4 is passed, [1, 2] = [ 7 ,
8 ], [ 3 , 4 ] = [9, 10], [5, 6] = [ 11 , 12 ] 3
Currents Id, If, and Ie that are 120 degrees out of phase with each other are set in the pair of coils.

Here, the underline added below the number means that the current flows in the opposite direction. For example, letting a current Id flow through a pair of coils [1, 2] = [ 7 , 8 ] means that a current half of Id is directed from coil 1 to coil 7 and a current Id is transferred from coil 2 to coil 8. The other half is to pass the current. Since 1 and 2, 7 and 8 are located close to each other on the circumference, this current supply makes it possible to generate rotating magnetic fields of the same number (2 poles) as the magnetic poles of the inner rotor 4.

Next, since a current (three-phase alternating current) for generating a rotating magnetic field for the outer rotor 3 is flown, [1] = [ 4 ] =
[7] = [ 10 ], [ 2 ] = [5] = [ 8 ] = [11], [3]
The currents Ia, Ic, and Ib, which are out of phase by 120 degrees, are set in the three sets of coils of = [ 6 ] = [9] = [ 12 ].

For example, a set of coils [1] = [ 4 ] =
[7] = flowing current Ia in [ 10 ] means coil 1 to coil
The current of Ia is applied to 4 and the current of Ia is also applied from coil 7 to coil 10 . Since the coils 1 and 7 and the coils 4 and 10 are located 180 degrees apart from each other on the circumference, this current supply can generate a rotating magnetic field of the same number as the magnetic poles of the outer rotor 3 (4 poles). .

Each of the 12 coils has the following respective wires.
The combined current (hereinafter simply referred to as
It is called "composite current". ) Flow I _{1 to} I ₁₂ .

I ₁ = (1/2) Id + Ia I ₂ = (1/2) Id + Ic I ₃ = (1/2) If + Ib I ₄ = (1/2) If + Ia I ₅ = (1/2 ) Ie + Ic I ₆ = (1/2) Ie + Ib I ₇ = (1/2) Id + Ia I ₈ = (1/2) Id + Ic I ₉ = (1/2) If + Ib I ₁₀ = (1/2) If + Ia I ₁₁ = (1/2) Ie + Ic I ₁₂ = (1/2) Ie + Ib However, the underline below the current symbol indicates that the current flows in the opposite direction.

The setting of the composite current will be further described with reference to FIG. 2. In FIG. 2, for comparison with FIG. It has a dedicated coil to generate it. That is,
The inner side coils d, f, e generate a rotating magnetic field for the inner rotor, and the outer side coils a, c, b generate a rotating magnetic field for the outer rotor. In this case, the common two dedicated coils, to reconfigure a common coil shown in FIG. 1, of the inner circumferential side coil, certain halves of the current flowing through the coil d near the coil d In the same way, the coils a and c are loaded, and similarly, half of the current flowing in the coil f is placed in the coils b and a near the coil f, and half of the current flowing in the coil e is placed in the coil near the coil e. You just have to burden c and b. The formulas of the composite currents I _{1 to} I ₁₂ described above are expressed as mathematical formulas. The current setting method is not limited to this, and other current setting methods may be used, as will be described later.

When the current is set in this manner, two magnetic fields, that is, a rotating magnetic field for the inner rotor 4 and a rotating magnetic field for the outer rotor 3 are simultaneously generated, even though they are common coils, but the magnet of the inner rotor 4 is outside. The rotating magnetic field for the rotor 3 does not give a rotating force, and the magnet of the outer rotor 3 does not give a rotating force for the inner rotor 4 by a rotating magnetic field. This point has been proved by theoretical analysis, as will be described later.

The current settings of Id, If, and Ie are set to the inner rotor 4
The rotation speed of the outer rotor 3 is set in synchronization with the rotation of the outer rotor 3, and the currents Ia, Ic and Ib are set in synchronization with the rotation of the outer rotor 3. The phase lead and lag are set with respect to the torque direction, which is the same as for the synchronous motor.

FIG. 3 is a block diagram for controlling the rotating electric machine.

In order to supply the composite currents I _{1 to} I ₁₂ to the stator coil, an inverter 12 for converting a direct current from a power source 11 such as a battery into an alternating current is provided. Since the sum of all the instantaneous currents becomes 0, this inverter 12 is the same as a normal three-phase bridge type inverter with 12 phases, as shown in detail in FIG. 4, and has 24 transistors Tr1 to T1.
It consists of r24 and the same number of diodes as this transistor.

The ON and OFF signals given to each gate (base of the transistor) of the inverter 12 are PWM signals.

In order to rotate the rotors 3 and 4 synchronously, rotation angle sensors 13 and 14 for detecting the phase of the rotors 3 and 4 are provided, and the control circuit 15 to which the signals from these sensors 13 and 14 are input. , A PWM signal is generated based on data (necessary torque command) of required torque (positive or negative) with respect to the outer rotor 3 and the inner rotor 4.

As described above, in the embodiment of the present invention, 2
The three rotors 3 and 4 and the one stator 2 are configured in a three-layer structure and on the same shaft, and the coil 6 common to the stator 2 is used.
Forming a, since it has to pass a double coupling current to the common coil 6, one of the rotor as a motor, when operating the remainder as a generator, the minute current difference of the motor drive power and generated power Since it only has to flow to the common coil, the efficiency can be greatly improved.

Further, one inverter is provided for two rotors.
When operating one of the rotors as a motor and the other as a generator, it suffices to flow a current corresponding to the difference between the motor drive power and the generated power to a common coil as described above. Therefore, the capacitance of the power switching transistor of the inverter can be reduced, which improves the switching efficiency,
Overall efficiency is improved.

Also, since the magnetic pole number ratio of the outer rotor 3 and the inner rotor 4 (hereinafter simply referred to as the magnetic pole number ratio) is a combination of 2: 1, the magnetic pole of one rotor and the rotating magnetic field created to drive that rotor However, it does not affect the rotation of the other rotors, and this allows any rotor to operate at a constant rotation.

On the other hand, for comparison, FIG. 6 and FIG. 7 show combinations in which the magnetic pole number ratio is 3: 1. Magnetic pole ratio
The 3: 1 combination differs from the 2: 1 magnetic pole ratio combination in that the magnets of the outer rotor 3 and inner rotor 4
Some influence occurs between the magnets. That is, the outer rotor
The magnet of No. 3 is not given a rotational force by the rotating magnetic field to the inner rotor 4, but the magnet of the inner rotor 4 is affected by the rotating magnetic field given to the outer rotor 3, so that the inner rotor 4 has a torque. It rotates with fluctuations. This point will be proved by theoretical analysis described later.

The interference of the outer rotor 3 with respect to the rotation of the inner rotor 4, that is, the constant torque fluctuation generated in the inner rotor 4, is a phase difference between the outer rotor 3 and the inner rotor 4, as will be understood from theoretical analysis described later. Since it becomes a function of (ω ₁ −ω ₂ ), amplitude modulation is applied to the alternating current for generating the rotating magnetic field for the outer coil so that the constant torque fluctuation is canceled in advance. It is possible to cancel the torque fluctuation that occurs in the.

FIG. 5 shows a second embodiment, which corresponds to FIG. 1 of the first embodiment.

In FIG. 1, the core 7 around which the coil is wound is the coil 6.
There are 12 cores, which is the same as the total number of cores, but in the second embodiment, one core 21 is provided for every two coils.
However, in order to avoid the interference between the magnetic fluxes generated in the two coils 6, a slit 22 is provided at the center of the core 21 in the circumferential direction.

In the second embodiment, the total number of cores 21 is six, which is a half of that in the first embodiment, and therefore the number of production steps is reduced.

In the first and second embodiments described above, when the magnetic pole number ratio is a combination of 2: 1, the magnetic pole of one rotor and the rotating magnetic field generated to drive the rotor are 6 does not affect the rotation of the rotor, and for comparison, the magnetic poles of the outer rotor and the rotor are driven when the magnetic pole ratio is a combination of 3: 1 using FIGS. It was described that the rotating magnetic field created for that purpose influences the rotation of the inner rotor. The theoretical analysis which is the basis for this is explained below in terms of sections.

<1> N (2p-2p) basic type This is the case where the ratio of the number of magnetic poles is 1: 1.

Here, the notation of N (2p-2p) will be explained. 2p on the left side is the number of magnetic poles of the outer magnet (outer rotor),
2p on the right side represents the number of magnetic poles of the inner magnet (inner rotor). In addition, N is an integer, which means that the same thing can be obtained by expanding (2p-2p) and multiplying it by an integer to form a ring.

The simplest one with a magnetic pole number ratio of 1: 1 is
When the number of magnetic poles of the outer magnet is 2, and the number of magnetic poles of the inner magnet is 2,
This is shown in FIG.

<1-1> In FIG. 8, when the outer magnet m ₁ and the inner magnet m ₂ are replaced with equivalent coils, the magnetic flux densities B ₁ and B ₂ generated in each magnet can be expressed as follows. .

B ₁ = Bm ₁ sin (ω ₁ t-θ) = μ Im ₁ sin (ω ₁ t-θ) (1) B ₂ = Bm ₂ sin (ω ₂ t + α-θ) = μ Im ₂ sin (ω ₂ t + α-θ) (2) where Bm ₁ and Bm ₂ : Amplitude μ: Permeability Im ₁ : Equivalent DC current of outer magnet Im ₂ : Equivalent DC current of inner magnet ω ₁ : Outer magnet Rotational angular velocity ω ₂ : Rotational angular velocity of inner magnet α: Phase difference between two magnets (when t = 0) However, in FIG. 8, the time when the outer magnet and the coil are in phase is considered to be 0.

If the current flowing through the stator coil is a three-phase alternating current, the magnetic flux density Bc due to the stator coil is Bc = μn (Ica (t) sin (θ) + Icb (t) sin (θ-2π / 3) + Icc (t) sin (θ-4π / 3)) (3) However, it can be given by the formula of n: coil constant.

In equation (3), Ica (t), Icb (t), Icc (t)
Are currents that are 120 degrees out of phase.

FIG. 9 shows the changes in the magnetic flux densities B ₁ , B ₂ and Bc shown in FIG.

The total magnetic flux density B at the angle θ is as follows.

B = B ₁ + B ₂ + Bc = μIm ₁ sin (ω ₁ t-θ) + μIm ₂ sin (ω ₂ t + α-θ) + μn (Ica (t) sin (θ) + Icb ( t) sin (θ-2π / 3) + Icc (t) sin (θ-4π / 3)) (4) Here, if the torque acting on the outer magnet m ₁ is τ ₁ ,
The generated torques are equal to each other in line symmetry about the diameter. Therefore, if f ₁ is the force for half a circle, the total driving force is 2f ₁ , so τ ₁ = 2f ₁ × r ₁ (r ₁ is the radius).

To consider the torque τ ₁ , it is sufficient to consider f ₁ (that is, the driving force generated by the influence of the magnetic field (magnetic flux density B) on the equivalent DC current Im ₁ ). Since only one equivalent DC current flows in the half circle, f ₁ becomes as follows.

F ₁ = Im ₁ × B (θ = ω ₁ t) = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + μn (Ica (t) sin (ω ₁ t) + Icb (t) sin (ω ₁ t-2π / 3) + Icc (t) sin (ω ₁ t-4π / 3))) (5) Similarly, the torque τ ₂ acting on the inner magnet m ₂ is also the diameter. Generated torque is equal in line symmetry about
If f ₂ is a force for half a circle, then τ ₂ = 2f ₂ × r ₂ (r ₂ is a radius). Since only one equivalent DC current flows in the half circle, f ₂ is as follows.

F ₂ = Im ₂ × B (θ = ω ₂ t + α) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) + μn (Ica (t) sin (ω ₂ t + α) + Icb (t) sin (ω ₂ t + α-2π / 3) + Icc (t) sin (ω ₂ t + α-4π / 3))) (6) <1-2> Outer rotating magnetic field When the current is given, the phase difference current of β is made to flow in the coil in accordance with the phase of the outer magnet. Therefore, the three-phase AC Ica (t), Icb (t), and Icc (t) in Eq. Ic cos (ω ₁ t-β)… (7a) Icb (t) ＝ Ic cos (ω ₁ t-β-2π / 3)… (7b) Icc (t) ＝ Ic cos (ω ₁ t-β-4π / 3) (7c) However, Ic: amplitude β: phase shift.

The driving forces f ₁ and f ₂ are calculated by substituting the equations (7a) to (7c) into the equations (5) and (6).

F ₁ = Im ₁ × B (θ = ω ₁ t) = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t)) + μn Ic (cos (ω 1t-β) sin (ω ₁ t) + cos (ω ₁ t-β-2π / 3) sin (ω ₁ t-2π / 3) + cos (ω ₁ t-β-4π / 3) sin (ω ₁ t-4π / 3))) Here, using the formula of cos (a + b) = 1/2 (sin (2a + b) -sin (b)), f ₁ = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t ) + μn Ic (1/2 (sin (2ω ₁ t-β) + sin (β)) +1/2 (sin (2 (ω ₁ t-2π / 3) -β) + sin (β)) + 1/2 (sin (2 (ω ₁ t-4π / 3) -β) + sin (β))) = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) +1/2 μn Ic ( 3sin (β) + sin (2 (ω ₁ t-2π / 3) -β) + sin (2 (ω ₁ t-4π / 3) -β))) = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + 1 / 2μn Ic (3sin (β) + sin (2ω ₁ t-4π / 3-β) + sin (2ω ₁ t-8π / 3-β))) = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) +1/2 μn Ic (3sin (β) + sin (2ω ₁ t-β-2π / 3) + sin (2ω ₁ t-β-4π / 3))) = -Im ₁ (μIm ₂ sin ((ω ₂ -ω ₁ ) t-α) -3 / 2μn Ic sin (β)) (8) According to Eqs. (8), the constant torque term (second term) The term of the torque fluctuation (first term) due to the influence of the magnetic field of the inner magnet is added to.

F ₂ = Im ₂ × B (θ = ω ₂ t + α) Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) + μn Ic (cos (ω ₁ t-β) sin ( ω ₂ t + α) + cos (ω ₁ t-2π / 3-β) sin (ω ₂ t-2π / 3 + α) + cos (ω ₁ t-4π / 3-β) sin (ω ₂ t- 4π / 3 + α)) where, using the formula cos (a) sin (b) = 1/2 (sin (a + b) -sin (a−b)), f ₂ = Im ₂ (μIm ₁ sin ( ω ₁ t-ω ₂ t-α) + μn Ic 1/2 (sin (ω ₁ t-β + ω ₂ t + α) -sin (ω ₁ t-β-ω ₂ t-α) + sin (ω ₁ t-2π / 3-β + ω ₂ t-2π / 3 + α) -sin (ω ₁ t-2π / 3-β-ω ₂ t + 2π / 3-α) + sin (ω ₁ t-4π / 3-β + ω ₂ t-4π / 3 + α) -sin (ω ₁ t-4π / 3-β-ω ₂ t + 4π / 3-α)) = Im ₂ (μIm ₁ sin (ω ₁ t -ω ₂ t-α) + μn Ic 1/2 (sin ((ω ₁ + ω ₂ ) t + α-β) -sin ((ω ₁ -ω ₂ ) t-α-β) + sin ((ω ₁ + ω ₂ ) t-4π / 3 + α-β) -sin ((ω ₁ -ω ₂ ) t-α-β) + sin ((ω ₁ + ω ₂ ) t-8π / 3 + α-β ) -sin ((ω ₁ -ω ₂ ) t-α-β) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) -3/2 μn Ic sin ((ω ₁ -ω ₂ ) t -α-β) + μn Ic 1/2 (sin ((ω ₁ + ω ₂ ) t + α-β) + sin ((ω ₁ + ω ₂ ) t + α-β-2π / 3) + sin ( (ω ₁ + ω ₂ ) t + α-β-4π / 3)) = μ Im ₂ (Im ₁ sin ((ω ₁ -ω ₂ ) t-α) -3 / 2n Ic sin ((ω ₁ -ω ₂ ) t-α-β))… (9) <1-3> When an inner rotating magnetic field is applied, the phase difference current of γ flows in the coil in accordance with the phase of the inner magnet. Phase exchange Ica (t), Icb (t), Icc (t)
Ica (t) ＝ Ic cos (ω ₂ t-γ)… (10a) Icb (t) ＝ Ic cos (ω ₂ t-γ-2π / 3)… (10b) Icc (t) ＝ Ic cos (ω ₂ t-γ-4π / 3) (10c) where Ic: amplitude γ: phase shift.

The driving forces f ₁ and f ₂ of the outer magnet and the inner magnet are calculated by substituting the equations (10a) to (10c) into the equations (5) and (6).

F ₁ = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + μn Ic (cos (ω ₂ t-γ) sin (ω ₁ t) + cos (ω ₂ t-γ- 2π / 3) sin (ω ₁ t-2π / 3) + cos (ω ₂ t-γ-4π / 3) sin (ω ₁ t-4π / 3)) Again, cos (a) sin (b) ＝ Using the formula of 1/2 (sin (a + b) -sin (ab)), f ₁ = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) +1/2 μn Ic (sin (ω ₂ t-γ + ω ₁ t) -sin (ω ₂ t-γ-ω ₁ t) + sin (ω ₂ t-γ-2π / 3 + ω ₁ t-2π / 3) -sin (ω ₂ t- γ-2π / 3-ω ₁ t + 2π / 3) + sin (ω ₂ t-γ-4π / 3 + ω ₁ t-4π / 3) -sin (ω ₂ t-γ-4π / 3 + ω ₁ t + 4π / 3)) = Im ₁ (μIm ₂ sin ((ω ₂ −ω ₁ ) t + α) +1/2 μn Ic (sin ((ω ₂ + ω ₁ ) t-γ) -sin ((ω ₂ -ω ₁ ) t-γ) + sin ((ω ₂ + ω ₁ ) t-γ-4π / 3) -sin ((ω ₂ -ω ₁ ) t-γ) + sin ((ω ₂ + ω ₁ ) t -γ-8π / 3) -sin ((ω ₂ -ω ₁ ) t-γ))) = Im ₁ (μIm ₂ sin ((ω ₂ -ω ₁ ) t + α) -3 / 2μn Ic sin (( ω ₂ -ω ₁ ) t-γ) +1/2 μn Ic (sin ((ω ₂ + ω ₁ ) t-γ) + sin ((ω ₂ + ω ₁ ) t-γ-2π / 3) + sin (( ω ₂ t + ω ₁ ) t-γ-4π / 3))) ＝ -μIm ₁ (Im ₂ sin ((ω ₂ -ω ₁ ) t-α) -3/2 n Ic sin ((ω ₁ -ω _{2) t + γ)) ...} (11) (11) equation only torque fluctuation outside magnet generator Which indicates that.

F ₂ = Im ₂ (μIm ₁ sin (ω ₂ t-ω ₁ t-α) + μn Ic (cos (ω ₂ t-γ) sin (ω ₂ t + α) + cos (ω ₂ t- γ-2π / 3) sin ((ω ₂ t + α-2π / 3) + cos (ω ₂ t-γ-4π / 3) sin ((ω ₂ t + α-4π / 3))) where Using cos (a) sin (b) ＝ 1/2 (sin (a + b) -sin (ab)) f ₂ ＝ Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) -3 / 2 μn Ic sin (-α-γ) +1/2 μn Ic (sin (2ω ₂ t + α-γ) + sin (2ω ₂ t + α-γ-2π / 3) + sin (2ω ₂ t + α- γ-4π / 3))) = μIm ₂ (Im ₁ sin ((ω ₁ -ω ₂ ) t-α) +3/2 n Ic sin (α + γ))… (12) (12) For example, the term is a sum of the constant torque term (second term) and the torque fluctuation term (first term) due to the influence of the magnetic field of the inner magnet.

<1-4> When Both Outer Rotating Magnetic Field and Inner Rotating Magnetic Field are Given Since currents flowing in the coil are synchronized with the outer magnet and the inner magnet, respectively, Ica (t), Icb (t), Icc ( t) as Ica (t) ＝ Ic cos (ω ₁ t-β) + Ic ₂ cos (ω ₂ t-γ)… (13a) Icb (t) ＝ Ic cos (ω ₁ t-β-2π / 3) + Ic ₂ cos (ω ₂ t-γ-2π / 3) (13b) Icc (t) ＝ Ic cos (ω ₁ t-β-4π / 3) + Ic ₂ cos (ω ₂ t-γ-4π / 3)… (13c).

F ₁ = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + μn ((Ic cos (ω ₁ t-β) + Ic ₂ cos (ω ₂ t-γ)) sin ( ω ₁ t) + (Ic cos (ω ₁ t-β-2π / 3) + Ic ₂ cos (ω ₂ t-γ-2π / 3)) sin (ω ₁ t-2π / 3) + (Ic cos ( ω ₁ t-β-4π / 3) + Ic ₂ cos (ω ₂ t-γ-4π / 3)) sin (ω ₁ t-4π / 3))) = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + μn (Ic cos (ω ₁ t-β) sin (ω ₁ t) + Ic ₂ cos (ω ₂ t-γ) sin (ω ₁ t) + Ic cos (ω ₁ t-β -2π / 3) sin (ω ₁ t-2π / 3) + Ic ₂ cos (ω ₂ t-γ-2π / 3) sin (ω ₁ t-2π / 3) + Ic cos (ω ₁ t-β- 4π / 3) sin (ω ₁ t-4π / 3) + Ic ₂ cos (ω ₂ t-γ-4π / 3) sin (ω ₁ t-4π / 3))) = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + μn (Ic (cos (ω ₁ t-β) sin (ω ₁ t) + cos (ω ₁ t-β-2π / 3) sin (ω ₁ t-2π / 3 ) + cos (ω ₁ t-β-4π / 3) sin (ω ₁ t-4π / 3)) + Ic ₂ (cos (ω ₂ t-γ) sin (ω ₁ t) + cos (ω ₂ t- γ-2π / 3) sin (ω ₁ t-2π / 3) + cos (ω ₂ t-γ-4π / 3) sin (ω ₁ t-4π / 3)))) = Im ₁ (μIm ₂ sin ( ω ₂ t + α-ω ₁ t) + μn (Ic (3 / 2sin (β)) + Ic ₂ (3 / 2sin ((ω ₁ -ω ₂ ) t + γ))))… (14) (14 ), A constant torque is generated according to the rotation phase difference (β) with respect to the outer magnet. The form that change was riding.

F ₂ = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) + μn ((Ic cos (ω ₁ t-β) + Ic ₂ cos (ω ₂ t-γ)) sin ( ω ₂ t + α) + (Ic cos (ω ₁ t-β-2π / 3) + Ic ₂ cos (ω ₂ t-γ-2π / 3)) sin (ω ₂ t + α-2π / 3) + (Ic cos (ω ₁ t-β-4π / 3) + Ic ₂ cos (ω ₂ t-γ-4π / 3)) sin (ω ₂ t + α-4π / 3))) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) + μn (Ic cos (ω ₁ t-β) sin (ω ₂ t + α) + Ic ₂ cos (ω ₂ t-γ) sin (ω ₂ t + α ) + Ic cos (ω ₁ t-β-2π / 3) sin (ω ₂ t + α-2π / 3) + Ic ₂ cos (ω ₂ t-γ-2π / 3) sin (ω ₂ t + α- 2π / 3) + Ic cos (ω ₁ t-β-4π / 3) sin (ω ₂ t + α-4π / 3) + Ic ₂ cos (ω ₂ t-γ-4π / 3) sin (ω ₂ t + α-4π / 3)) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) + μn (Ic (cos (ω ₁ t-β) sin (ω ₂ t + α) + cos ( ω ₁ t-β-2π / 3) sin (ω ₂ t + α-2π / 3) + cos (ω ₁ t-β-4π / 3) sin (ω ₂ t + α-4π / 3)) + Ic ₂ (cos (ω ₂ t-γ) sin (ω ₂ t + α) + cos (ω ₂ t-γ-2π / 3) sin (ω ₂ t + α-2π / 3) + cos (ω ₂ t- γ-4π / 3) sin (ω ₂ t + α-4π / 3))) where cos (a) sin (b) ＝ 1/2 (sin (a + b) -sin (ab)) Te _{f 2 = Im 2 (μIm 1} sin (ω 1 t-ω 2 t-α) + μn (Ic (1 / 2sin (ω 1 t-β + ω 2 t + α) -sin (ω 1 t- _{-ω 2 t-α)) +} 1 / 2sin (ω 1 t-β-2π / 3 + ω 2 t + α-2π / 3) -sin (ω 1 t-β-2π / 3-ω 2 t- α + 2π / 3)) +1/2 sin (ω ₁ t-β-4π / 3 + ω ₂ t + α-4π / 3) -sin (ω ₁ t-β-4π / 3-ω ₂ t-α + 4π / 3))) + Ic ₂ (1 / 2sin (ω ₂ t-γ + ω ₂ t + α) -sin (ω ₂ t-γ-ω ₂ t-α)) + 1 / 2sin (ω ₂ t-γ-2π / 3 + ω ₂ t + α-2π / 3) -sin (ω ₂ t-γ-2π / 3-ω ₂ t-α + 2π / 3)) + 1 / 2sin (ω ₂ t -γ-4π / 3 + ω ₂ t + α-4π / 3) -sin (ω ₂ t-γ-4π / 3-ω ₂ t-α + 4π / 3)))) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) + μn (Ic (1 / 2sin (ω ₁ t-β + ω ₂ t + α) -sin (ω ₁ t-β-ω ₂ t-α)) +1 / 2sin (ω ₁ t-β + ω ₂ t + α-4π / 3) -sin (ω ₁ t-β-ω ₂ t-α)) + 1 / 2sin (ω ₁ t-β + ω ₂ t + α-8π / 3) -sin (ω ₁ t-β-ω ₂ t-α))) + Ic ₂ (1 / 2sin (ω ₂ t-γ + ω ₂ t + α) -sin (ω ₂ t- γ-ω ₂ t-α)) + 1 / 2sin (ω ₂ t-γ + ω ₂ t + α-4π / 3) -sin (ω ₂ t-γ-ω ₂ t-α)) + 1 / 2sin (ω ₂ t-γ + ω ₂ t + α-8π / 3) -sin (ω ₂ t-γ-ω ₂ t-α)))) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t -α) +1/2 μn Ic (sin (ω ₁ t-β + ω ₂ t + α) -sin (ω ₁ t-β-ω ₂ t-α) + sin (ω ₁ t-β + ω ₂ t + α-4π / 3) -sin (ω ₁ t-β-ω ₂ t-α) + sin (ω ₁ t-β + ω ₂ t + α-8π / 3) -sin (ω ₁ t-β-ω ₂ t-α)) +1/2 μn Ic ₂ (sin (ω ₂ t-γ + ω ₂ t + α) -sin (ω ₂ t-γ-ω ₂ t- α)) + sin (ω ₂ t-γ + ω ₂ t + α-4π / 3) -sin (ω ₂ t-γ-ω ₂ t-α) + sin (ω ₂ t-γ + ω ₂ t + α-8π / 3) -sin (ω ₂ t-γ-ω ₂ t-α)) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) + 1/2 μn Ic (-3sin (( ω ₂ -ω ₁ ) t-α-β) + 1 / 2μn Ic ₂ (-3sin (-α-γ)) = Im ₂ (μIm ₁ sin (ω ₁ t-ω ₂ t-α) -3 / 2μn Ic sin ((ω ₂ -ω ₁ ) t-α-β) + 3 / 2μn Ic ₂ 3 sin (α + γ) (15) (15) also depends on the rotational phase difference (α + γ) with respect to the inner magnet. In addition, the rotation fluctuation is added to the constant torque.

<1-5> Summary The above (8), (9), (11), (12) and (1
4) and (15) are arranged next.

When an outer rotating magnetic field is applied f ₁ = -μIm ₁ (Im ₂ sin ((ω ₂ -ω ₁ ) t-α) -3 / 2n Ic sin (β)) (8) f ₂ = μIm ₂ (Im ₁ sin ((ω ₁ -ω ₂ ) t-α) -3 / 2n Ic sin ((ω ₁ -ω ₂ ) t-α-β)) (9) When an inner rotating magnetic field is applied f ₁ = -μIm ₁ (Im ₂ sin ((ω ₂ -ω ₁ ) t-α) -3/2 n Ic sin ((ω ₁ -ω ₂ ) t + γ)) (11) f ₂ = μIm ₂ (Im ₁ sin ((ω ₁ -ω ₂ ) t-α) +3/2 n Ic sin (α + γ)) (12) When both outer and inner rotating magnetic fields are given f ₁ = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + μn (Ic (3 / 2sin (β)) + Ic ₂ (3 / 2sin ((ω ₁ -ω ₂ ) t + γ)))) ... (14) f ₂ = μIm ₂ (Im ₁ sin (ω ₁ t-ω ₂ t-α) + 3 / 2n Ic sin ((ω ₁ -ω ₂ ) t-α-β) + 3 / 2n Ic ₂ sin (α + γ)) (15) The meanings of these equations are as follows. The second term on the right side of equation (8), the second term on the right side of equation (12), the second term on the right side of equation (14), and (15)
Only the third term on the right side of the equation is a fixed term (constant value), and the rotational torque is generated only when the fixed term is included. On the other hand, since the terms other than the fixed term are trigonometric functions, the average value of the driving force f is zero, and therefore the rotational torque is not generated depending on the terms other than the fixed term. That is, when a current is passed through the stator coil in synchronization with the outer magnet, rotational torque is generated only in the outer magnet, and when a current is passed through the stator coil in synchronization with the inner magnet, rotational torque is generated only in the inner magnet. When a current is passed through the stator coil in synchronization with each of the inner magnets, rotational torque is generated in each of the magnets.

From this, it was proved that it is possible to work as a rotating electric machine when the magnetic pole number ratio is a combination of 1: 1. By analogy with this, even when the number of magnetic poles is an arbitrary combination, it is possible to work as a rotating electric machine.

<1-6> Suppression of Torque Fluctuation On the other hand, in the equation including the fixed term, the remaining term of the fixed term, that is, the first term on the right side of the equation (8), the first term and the third term on the right side of the equation (14). The constant torque fluctuation according to the phase difference (ω ₁ -ω ₂ ) between the two magnets causes the rotation of the outer magnet, and the first term on the right side of Eq. (12), (1
Due to the first term and the second term on the right side of the equation (5), a constant torque fluctuation corresponding to the phase difference (ω ₁ −ω ₂ ) between the two magnets also occurs in the rotation of the inner magnet.

Therefore, it is considered to suppress the torque fluctuation when both the outer rotating magnetic field and the inner rotating magnetic field are applied. From the above formula (14), f ₁ = μIm ₁ Im ₂ sin (ω ₂ t + α-ω ₁ t) + Ic μn Im ₁ Ic (3 / 2sin
Since (β)) + Ic ₂ Im ₁ 3/2 sin ((ω ₁ −ω ₂ ) t + γ), f ₁ is set as follows.

F ₁ = A + Ic C + Ic ₂ V (16) where A = μIm ₁ Im ₂ sin (ω ₂ t + α-ω ₁ t) V = Im ₁ 3/2 sin ((ω _1- ω ₂ ) t + γ) C = μn Im ₁ Ic (3 / 2sin (β)) where, if the modulation Ic = (C1-A-Ic ₂ V) / C is added, f
₁ = C1 (constant), and torque fluctuation is eliminated from the rotation of the outer magnet.

Similarly, from the above formula (15), f ₂ = μIm ₂ Im ₁ sin (ω ₁ t-ω ₂ t-α) + Ic 3/2 μIm ₂ n sin
Since ((ω ₁ -ω ₂ ) t-α-β) + Ic ₂ 3/2 μIm ₂ n sin (α + γ), f ₂ is set as follows.

F ₂ = -A + Ic D + Ic ₂ E (17) where D = 3/2 μIm ₂ n sin ((ω ₁ -ω ₂ ) t-α-β) E = 3/2 μIm ₂ n sin (α + γ) Here, if the modulation of Ic ₂ = (C 2 + A-IcD) / E is added, then f ₂ = C 2 (constant), and the torque fluctuation is eliminated from the rotation of the inner magnet.

Therefore, in order to make both magnets rotate at a constant value, the following simultaneous binary equations should be solved for Ic and Ic ₂ .

C1 = A + Ic C + Ic ₂ V (18) C2 = -A + Ic D + Ic ₂ E (19) <2> N (2 (2p) -2p) basic form <2-1> FIG. 10 Consider a case where the magnetic pole number ratio is 2: 1 (in FIG. 10, the outer magnet has four magnetic poles and the inner magnet has two magnetic poles).

When each magnet is replaced by an equivalent coil, the magnetic flux density B ₁ generated in the outer magnet is B ₁ = Bm ₁ sin (2ω ₁ t-2θ) = μIm ₁ sin (2ω ₁ t-2θ) (21) On the other hand, the magnetic flux density B ₂ generated in the inner magnet is the same as the above equation (2), that is, B ₂ = Bm ₂ sin (ω ₂ t + α-θ) = μIm ₂ sin (ω ₂ t + α-θ) (22)

Since the magnetic field produced by the stator coil is calculated separately for the outer rotating magnetic field and the inner rotating magnetic field, the coils are arranged as shown in FIG. 10, and the stator coils for the outer and inner magnets are used. The magnetic flux densities Bc ₁ and Bc ₂ can be calculated as Bc ₁ = μn (Ica (t) sin (2θ) + Icb (t) sin (2θ-2π / 3) + Icc (t) sin (2θ-4π / 3)) ... (23) Bc ₂ = μn (Icd (t) sin (θ) + Ice (t) sin (θ-2π / 3) + Icf (t) sin (θ-4π / 3)) (24).

However, in addition to Ica (t), Icb (t), Icc (t),
Icd (t), Ice (t), and Icf (t) are also currents that are 120 degrees out of phase.

FIG. 11 shows a model of changes in the magnetic flux densities B ₁ , B ₂ , Bc ₁ , and Bc ₂ described above.

The magnetic flux density B at the angle θ is the sum of the above four magnetic flux densities.

B = B ₁ + B ₂ + Bc ₁ + Bc ₂ = μIm ₁ sin (2ω ₁ t-2θ) + μIm ₂ sin (ω ₂ t + α-θ) + μn (Ica (t) sin (2θ ) + Icb (t) sin (2θ-2π / 3) + Icc (t) sin (2θ-4π / 3)) + μn (Icd (t) sin (θ) + Ice (t) sin (θ-2π / 3) + Icf (t) sin (θ-4π / 3)) (25) If the torque acting on the outer magnet m ₁ is τ ₁ , τ ₁ = f ₁
× r ₁ (r ₁ is the radius). In FIG. 10, since the generated torques are not equal to each other in line symmetry with the diameter as the center, consideration is made for the entire circumference. Since four equivalent DC currents flow around one round, the sum of the forces acting on these four currents is f ₁ .

F ₁ = Im ₁ × B (θ = ω ₁ t) + Im ₁ × B (θ = ω ₁ t + π) -Im ₁ × B (θ = ω ₁ t + π / 2) -Im ₁ × B (θ = ω ₁ t + 3π / 2) = μIm ₁ (Im ₁ sin (2ω ₁ t-2ω ₁ t) + Im ₁ sin (2ω ₁ t-2ω ₁ t-2π) -Im ₁ sin (2ω ₁ t-2ω ₁ t-π) -Im ₁ sin (2ω ₁ t-2ω ₁ t + 3π) + Im ₂ sin (ω ₂ t + α-ω ₁ t) + Im ₂ sin (ω ₂ t + α- ω ₁ t + π) -Im ₂ sin (ω ₂ t + α-ω ₁ t + π / 2) -Im ₂ sin (ω ₂ t + α-ω ₁ t + π / 2) + n (Ica (t ) sin (2ω ₁ t) + Icb (t) sin (2ω ₁ t-2π / 3) + Icc (t) sin (2ω ₁ t-4π / 3)) + n (Ica (t) sin (2ω ₁ t + 2π) + Icb (t) sin (2ω ₁ t + 2π-2π / 3) + Icc (t) sin (2ω ₁ t + 2π-4π / 3)) -n (Ica (t) sin (2ω ₁ t + π) + Icb (t) sin (2ω ₁ t + π / 3) + Icc (t) sin (2ω ₁ t-π / 3) -n (Ica (t) sin (2ω ₁ t + π) + Icb (t) sin (2ω ₁ t + π / 3) + Icc (t) sin (2ω ₁ t-π / 3)) + n (Icd (t) sin (ω ₁ t) + Ice (t) sin (ω ₁ t-2π / 3) + Icf (t) sin (ω ₁ t-4π / 3)) + n (Icd (t) sin (ω ₁ t + π) + Ice (t) sin (ω ₁ t + π -2π / 3) + Icf (t) sin (ω ₁ t + π-4π / 3)) -n (Icd (t) sin (ω ₁ t + π / 2) + Ice (t) sin (ω ₁ t + π / 2-2π / 3) + Icf (t) sin (ω ₁ t + π / 2-4π / 3) -n (Icd (t) sin (ω ₁ t + 3π / 2) + Ice (t) sin (ω ₁ t + 3π / 2-2π / 3) + Icf (t) sin (ω ₁ t + 3π / 2-4π / 3) = 4μIm ₁ n (Ica (t) sin (2ω ₁ t) + Icb (t) sin (2ω ₁ t-2π / 3) + Icc (t) sin (2ω ₁ t -4π / 3)) (26) According to the formula (26), it is shown that the torque acting on the outer magnet can be controlled by the exciting currents of the coils a, b, and c. It also shows that it is not affected by the exciting currents of the coils d, e, and f.

Next, the torque acting on the inner magnet m ₂ is set to τ ₂
Then, τ ₁₂ = f ₂ × r ₂ (r ₂ is radius). Since two equivalent direct currents flow around one round, the sum of the forces acting on these two currents is f ₂ .

F ₂ = Im ₂ × B (θ = ω ₂ t + α) -Im ₂ × B (θ = ω ₂ t + π + α) = μI m ₂ (Im ₁ sin (2ω ₁ t-2ω ₂ t -2α) -Im ₁ sin (2ω ₁ t-2ω ₂ t-2α-2π) + Im ₂ sin (2ω ₂ t + 2α-2ω ₂ t-2α) -Im ₂ sin (2ω ₂ t + 2α-2ω ₂ t-2α-2π) + n (Ica (t) sin (2ω ₂ t + 2α) + Icb (t) sin (2ω ₂ t + 2α-2π / 3) + Icc (t) sin (2ω ₂ t + 2α -4π / 3) -n (Ica (t) sin (2ω ₂ t + 2π + 2α) + Icb (t) sin (2ω ₂ t + 2π + 2α-2π / 3) + Icc (t) sin (2ω ₂ t + 2π + 2α-4π / 3) + n (Icd (t) sin (ω ₂ t + α) + Ice (t) sin (ω ₂ t + α-2π / 3) + Icf (t) sin (ω ₂ t + α-4π / 3)) -n (Icd (t) sin (ω ₂ t + π + α) + Ice (t) sin (ω ₂ t + π + α-2π / 3) + Icf (t ) sin (ω ₂ t + π + α-4π / 3))) = ₂ μIm ₂ n (Icd (t) sin (ω ₂ t + α) + Ice (t) sin (ω ₂ t + α-2π / 3 ) + Icf (t) sin (ω ₂ t + α-4π / 3)) ... (27) According to Eq. (27), the torque acting on the inner magnet can be controlled by the exciting currents of the coils d, e, and f. Moreover, it is shown that it is not affected by the exciting currents of the coils a, b, and c.

<2-2> When an outer rotating magnetic field is applied A current having a phase difference of β is passed through the coils a, b, and c in accordance with the outer magnet. In other words, the above three-phase AC Ica (t), Icb (t), Icc
(t) is Ica (t) ＝ Ic cos (2ω ₁ t-2β)… (28a) Icb (t) ＝ Ic cos (2ω ₁ t-2β-2π / 3)… (28b) Icc (t) ＝ Ic cos (2ω ₁ t-2β-4π / 3) (28c). The (28a) ~ (28c) (26), to calculate the f ₁ are substituted into the expression (27).

F ₁ = 4 μIm ₁ n Ic (cos (2ω ₁ t-2β) sin (2ω ₁ t) + cos (2ω ₁ t-2β-2π / 3) sin (2ω ₁ t-2π / 3) + cos (2ω ₁ t-2β-4π / 3) sin (2ω ₁ t-4π / 3)) where cos (a) sin (b) ＝ 1/2 (sin (a + b) -sin (ab)) F ₁ = 4μIm ₁ n Ic (1/2 (sin (2ω ₁ t-2β + 2ω ₁ t) -sin (2ω ₁ t-2β-2ω ₁ t)) +1/2 (sin ( 2ω ₁ t-2β-2π / 3 + 2ω ₁ t-2π / 3) -sin (2ω ₁ t-2β-2π / 3-2ω ₁ t + 2π / 3)) +1/2 (sin (2ω ₁ t -2β-4π / 3 + 2ω ₁ t-4π / 3) -sin (2ω ₁ t-2β-4π / 3-2ω ₁ t + 4π / 3))) = 2μIm ₁ n Ic (sin (4ω ₁ t- 2β) + sin (2β) + sin (4ω ₁ t-2β-4π / 3) + sin (2β) + sin (4ω ₁ t-2β-8π / 3) + sin (2β)) = 2μIm ₁ n Ic ( sin (4ω ₁ t-2β) + sin (4ω ₁ t-2β-4π / 3) + sin (4ω ₁ t-2β-4π / 3) + 3sin (2β)) = 6μIm ₁ n Ic sin (2β)… (29) Equation (29) shows that the torque of the outer magnet changes according to the phase difference (β). Therefore, it is understood that it is sufficient to measure the rotation angle of the outer magnet and shift the phase by β with respect to the rotation angle to supply the exciting current to the coils a, b, and c.

<2-3> When an inner rotating magnetic field is applied Coil
Icd (t), Ice (t), Icf (t) are expressed as Icd (t) = Ic cos (ω ₂ t-γ) ... (30a) Ice (t) = Ic cos (ω ₂ t-γ-2π / 3) (30b) Icf (t) = Ic cos (ω ₂ t-γ-4π / 3) (30c)

Substituting these into the equation (27), f ₂ is calculated.

F ₂ = ₂ μIm ₂ n (Ic cos (ω ₂ t-γ) sin (ω ₂ t + α) + Ic cos (ω ₂ t-γ-2π / 3) sin (ω ₂ t + α-2π / 3) + Ic cos (ω ₂ t-γ-4π / 3) sin (ω ₂ t + α-4π / 3) where cos (a) sin (b) ＝ 1/2 (sin (a + b ) -sin (ab)) formula, f ₂ = 2μIm ₂ n Ic (1/2 (sin (ω ₂ t-γ + ω ₂ t + α) -sin (ω ₂ t-γ-ω ₂ t -α)) +1/2 (sin (ω ₂ t-γ-2π / 3 + ω ₂ t + α-2π / 3) -sin (ω ₂ t-γ-2π / 3-ω ₂ t-α + 2π / 3)) +1/2 (sin (ω ₂ t-γ-4π / 3 + ω ₂ t + α-4π / 3) -sin (ω _2t -γ-4π / 3-ω ₂ t-α + 4π / 3)) = μIm ₂ n Ic (sin (2ω ₂ t-γ + α) + sin (γ + α) + sin (2ω ₂ t-γ-4π / 3 + α) + sin (γ + α) + sin (2ω ₂ t-γ-8π / 3 + α) + sin (γ + α)) = μIm ₂ n Ic (sin (2ω ₂ t-γ + α) + sin (2ω ₂ t-γ-4π / 3 + α) + sin (2ω ₂ t-γ-8π / 3 + α) + 3sin (γ + α)) = 3μIm ₂ n Ic sin (γ + α)… (31) (31) It is shown that the torque of the inner magnet changes due to the phase difference (γ + α) .Therefore, the rotation angle of the inner magnet is measured, and the phase is shifted by (γ + α) to the coils d, e, f. It is enough to supply the exciting current. It is seen.

<2-4> Summary Expression (29) is applied only to the outer magnet when current is supplied to the stator coil in synchronization with the outer magnet, and expression (31) is applied to the stator coil in synchronization with the inner magnet. When flowing, the rotational torque is generated only in the inner magnet. Since each magnetic field corresponds only to each phase current, we did not calculate it, but when current is passed through the stator coil in synchronization with each of the outer magnet and the inner magnet, rotational torque is generated in both magnets. .

From this, it was proved that it is possible to work as a rotating electric machine even when the magnetic pole number ratio is a combination of 2: 1.

<2-5> Setting of Current Flowing in Stator Coil In FIG. 10, for the theoretical calculation, a dedicated coil for generating the outer rotating magnetic field and a dedicated coil for generating the inner rotating magnetic field were considered. , As shown in FIG.
Consider sharing a coil. In FIG. 10, the coils a and d, the coils b and f , the coils c and e, the coils a and d , the coils b and f, and the coils c and e can be combined. Therefore,
Comparing the coils of FIGS. 10 and 12, the combined currents I _{1 to} I ₁₂ flowing in the coils 1 to 12 of FIG. ₁₂ are I ₁ = Ia + Id I ₂ = Ic I ₃ = Ib + If I ₄ = Ia I It can be seen that ₅ = Ic + Ie I ₆ = Ib I ₇ = Ia + Id I ₈ = Ic I ₉ = Ib + If I ₁₀ = Ia I ₁₁ = Ic + Ie I ₁₂ = Ib .

In this case, the load of the coils for passing the respective currents I ₁ , I ₃ , I ₅ , I ₇ , I ₉ , I ₁₁ is I ₂ , I ₄ , I ₆ , I ₈ , I ₁₀ , I _12.
It is considered that the inner rotating magnetic field is formed by distributing the load also to the remaining coils because the currents are larger than the remaining coils for passing the respective currents.

For example, comparing FIG. 2 with FIG.
2 , the parts corresponding to 1, 1 , 2 , and 2 are a, a , c , and c of the outer coil and d and d of the inner coil in FIG. In this case, consider a state where the phases of the coils d and d are equivalently shifted,
New coil d'what was its shifting, when d', the halves of the current Id' flowing Among the coil d'coil a
Coil and half of the current Id 'flowing in the coil d '
Allocate to a and c . The rest is the same.

In other words, this means that the coils are divided into 6 (see the broken lines in FIGS. 1 and 2) so that there are 3 phases per pole of the inner rotor (the rotor with the smaller number of magnetic poles). each coil of the coils a and c in which the separated speaking (or each coil of the coil a and c), the driving current I of the inner rotor
d '(or Id' ) is the number of separated coils 2
The current value divided by is (1/2) Id ' (or (1/2) Id ' ).

By doing so, as another current setting, I ₁ = Ia + (1/2) Id 'I ₂ = Ic + (1/2) Id' I ₃ = Ib + (1/2) If ' I ₄ ＝ Ia + (1/2) If´ I ₅ ＝ Ic + (1/2) Ie´ I ₆ ＝ Ib + (1/2) Ie´ I ₇ ＝ Ia + (1/2) Id´ I ₈ ＝ Ic + (1/2 ) Id ' I ₉ = Ib + (1/2) If' I ₁₀ = Ia + (1/2) If 'I ₁₁ = Ic + (1/2) Ie' I ₁₂ = Ib + (1/2) Ie ' . However, the coils e ′ and f ′ are also equivalently displaced from the coils e and f.

Further considering, I ₁ = Ia + I _i I ₂ = Ic + I _ii I ₃ = Ib + I _iii I ₄ = Ia + I _iv I ₅ = Ic + I _v I ₆ = Ib + I _vi I _{7 = Ia + I vii I 8} = Ic + I viii I 9 = Ib + I ix I 10 = Ia + I x I 11 = Ic + I xi I 12 = Ib + I xii may even. That is, the currents I _{i to} I _{xii in} the second term on the right side of the equations I _{1 to} I ₁₂ are 12-phase alternating current as shown in FIG. 13, and the 12-phase alternating current forms an inner rotating magnetic field. You can do this.

<2-6> When applying an inner rotating magnetic field with 12-phase AC <2-6-1> Considering that an inner rotating magnetic field is created with 12-phase AC, the magnetic flux density Bc _{2 at} this time is as follows. become.

Bc ₂ = μn (Ic _i (t) sin (θ) + Ic _ii (t) sin (θ-2π / 12) + Ic _iii (t) sin (θ-4π / 12) + Ic _iv (t ) sin (θ-6π / 12) + Ic _v (t) sin (θ-8π / 12) + Ic _vi (t) sin (θ-10π / 12) + Ic _vii (t) sin (θ-12π / 12 ) + Ic _viii (t) sin (θ-14π / 12) + Ic _ix (t) sin (θ-16π / 12) + Ic _x (t) sin (θ-18π / 12) + Ic _xi (t) sin (θ-20π / 12) + Ic _xii (t) sin (θ-22π / 12)) (32) At this time, the overall magnetic flux density B is as follows.

B = B ₁ + B ₂ + Bc ₁ + Bc ₂ = μIm ₁ sin (3ω ₁ t-3θ) + μIm ₂ sin (ω ₂ t + α-θ) + μn (Ica (t) sin (3θ ) + Icb (t) sin (3θ-2π / 3) + Icc (t) sin (3θ-4π / 3) + μn (Ic _i (t) sin (θ) + Ic _ii (t) sin (θ-2π / 12) + Ic _iii (t) sin (θ-4π / 12) + Ic _iv (t) sin (θ-6π / 12) + Ic _v (t) sin (θ-8π / 12) + Ic _vi (t ) sin (θ-10π / 12) + Ic _vii (t) sin (θ-12π / 12) + Ic _viii (t) sin (θ-14π / 12) + Ic _ix (t) sin (θ-16π / 12) ) + Ic _x (t) sin (θ-18π / 12) + Ic _xi (t) sin (θ-20π / 12) + Ic _xii (t) sin (θ-22π / 12))… (33) When the f ₁ will be _{calculated, f 1 = Im 1 × B} (θ = ω 1 t) + Im 1 × B (θ = ω 1 t + π) -Im 1 × B (θ = ω 1 t + π / 2) -Im ₁ × B (θ = ω ₁ t + 3π / 2) = μIm ₁ (Im ₁ sin (2ω ₁ t-2ω ₁ t) + Im ₁ sin (2ω ₁ t-2ω ₁ t-2π ) -Im ₁ sin (2ω ₁ t−2ω ₁ t-π) -Im ₁ sin (2ω ₁ t-2ω ₁ t + 3π) + Im ₂ sin (ω ₂ t + α-ω ₁ t) + Im ₂ sin (ω ₂ t + α-ω ₁ t + π) -Im ₂ sin (ω ₂ t + α-ω ₁ t + π / 2) -Im ₂ sin (ω ₂ t + α-ω ₁ t + π / 2 ) + n (Ica (t) sin (2ω ₁ t) + Icb (t) sin (2ω ₁ t-2π / 3) + Icc (t) sin (2ω ₁ t-4π / 3)) + n (Ica ( t) sin (2ω ₁ t + 2π) + Icb (t) sin (2ω ₁ t + 2π-2π / 3) + Icc (t) sin (2ω ₁ t + 2π-4π / 3)) -n (Ica (t) sin (2ω ₁ t + π) + Icb (t) sin (2ω ₁ t + π / 3) + Icc (t) sin (2ω ₁ t + 2π-π / 3)) -n (Ica (t) sin (2ω ₁ t + π) + Icb (t) sin (2ω ₁ t + π / 3) + Icc (t) sin (2ω ₁ t + 2π-π / 3)) + n (Ic _i (t) (sin (ω ₁ t) + sin (ω ₁ t + π) -sin (ω ₁ t + π / 2) -sin (ω ₁ t + 3π / 2)) + Ic _ii (t) (sin (ω ₁ t-2π / 12) + sin (ω ₁ t-2π / 12 + π)- sin (ω ₁ t-2π / 12 + π / 2) -sin (ω ₁ t-2π / 12 + 3π / 2)) + Ic _iii (t) (sin (ω ₁ t-4π / 12) + sin ( ω ₁ t-4π / 12 + π) -sin (ω ₁ t-4π / 12 + π / 2) -sin (ω ₁ t-4π / 12 + 3π / 2)) + Ic _iv (t) (sin ( ω ₁ t-6π / 12) + sin (ω ₁ t-6π / 12 + π) -sin (ω ₁ t-6π / 12 + π / 2) -sin (ω ₁ t-6π / 12 + 3π / 2 )) + Ic _v (t) (sin (ω ₁ t-8π / 12) + sin (ω ₁ t-8π / 12 + π) -sin (ω ₁ t-8π / 12 + π / 2) -sin ( ω ₁ t-8π / 12 + 3π / 2)) + Ic _vi (t) (sin (ω ₁ t-10π / 12) + sin (ω ₁ t-10π / 12 + π) -sin (ω ₁ t- 10π / 12 + π / 2) -sin (ω ₁ t-10π / 12 + 3π / 2)) + Ic _vii (t) (sin (ω ₁ t-12π / 12) + sin (ω ₁ t-12π / 12 + π) -sin (ω ₁ t-12π / 12 + π / 2) -sin (ω ₁ t-12π / 12 + 3π / 2)) + Ic _viii (t) (sin (ω ₁ t-14π / 12) + sin (ω ₁ t-14π / 12 + π) -sin (ω ₁ t-14π / 12 + π / 2) -sin ( ω ₁ t-14π / 12 + 3π / 2)) + Ic _ix (t) (sin (ω ₁ t-16π / 12) + sin (ω ₁ t-16π / 12 + π) -sin (ω ₁ t- 16π / 12 + π / 2) -sin (ω ₁ t-16π / 12 + 3π / 2)) + Ic _x (t) (sin (ω ₁ t-18π / 12) + sin (ω ₁ t-18π / 12 + π) -sin (ω ₁ t-18π / 12 + π / 2) -sin (ω ₁ t-18π / 12 + 3π / 2)) + Ic _xi (t) (sin (ω ₁ t-20π / 12) + sin (ω ₁ t-20π / 12 + π) -sin (ω ₁ t-20π / 12 + π / 2) -sin (ω ₁ t-20π / 12 + 3π / 2)) + Ic _xii ( t) (sin (ω ₁ t-22π / 12) + sin (ω ₁ t-22π / 12 + π) -sin (ω ₁ t-22π / 12 + π / 2) -sin (ω ₁ t-22π / 12 + 3π / 2)) = 4μn Im ₁ (Ica (t) sin (2ω ₁ t) + Icb (t) sin (2ω ₁ t-2π / 3) + Icc (t) sin (2ω ₁ t-4π / 3))… (34), which is the same as Eq. (26) when the inner rotating magnetic field is created by three-phase alternating current.

On the other hand, the calculation of f ₂ is as follows.

F ₂ = Im ₂ × B (θ = ω ₂ t + α) -Im ₂ × B (θ = ω ₂ t + π + α) = μI m ₂ (Im ₁ sin (2ω ₁ t-2ω ₂ t -2α) -Im ₁ sin (2ω ₁ t-2ω ₂ t-2α-2π) + Im ₂ sin (2ω ₂ t + 2α-2ω ₂ t-2α) -Im ₂ sin (2ω ₂ t + 2α-2ω ₂ t-2α-2π) + n (Ica (t) sin (2ω ₂ t + 2α) + Icb (t) sin (2ω ₂ t + 2α-2π / 3) + Icc (t) sin (2ω ₂ t + 2α -4π / 3) -n (Ica (t) sin (2ω ₂ t + 2π + 2α) + Icb (t) sin (2ω ₂ t + 2π + 2α-2π / 3) + Icc (t) sin (2ω ₂ t + 2π + 2α-4π / 3) + n (Ic _i (t) (sin (ω ₂ t + α) -sin (ω ₂ t + π + α)) + Ic _ii (t) (sin (ω ₂ t + α-2π / 12) -sin (ω ₂ t + π + α-2π / 12)) + Ic _iii (t) (sin (ω ₂ t + α-4π / 12) -sin (ω ₂ t + π + α-4π / 12)) + Ic _iv (t) (sin (ω ₂ t + α-6π / 12) -sin (ω ₂ t + π + α-6π / 12)) + Ic _v (t) (sin (ω ₂ t + α-8π / 12) -sin (ω ₂ t + π + α-8π / 12)) + Ic _vi (t) (sin (ω ₂ t + α-10π / 12) -sin (ω ₂ t + π + α-10π / 12)) + Ic _vii (t) (sin (ω ₂ t + α-12π / 12) -sin (ω ₂ t + π + α-12π / 12)) + Ic _viii (t) (sin (ω ₂ t + α-14π / 12) -sin (ω ₂ t + π + α-14π / 12)) + Ic _ix (t) (sin (ω ₂ t + α-16π / 12) -sin (ω ₂ t + π + α-16π / 12)) + Ic _x (t) (sin (ω ₂ t + α-18π / 12) -sin (ω ₂ t + π + α-18π / 12)) + I c _xi (t) (sin (ω ₂ t + α-20π / 12) -sin (ω ₂ t + π + α-20π / 12)) + Ic _xii (t) (sin (ω ₂ t + α-22π / 12) -sin (ω ₂ t + π + α-22π / 12))) = ₂ μIm ₂ n (Ic _i (t) sin (ω ₂ t + α) + Ic _ii (t) sin (ω ₂ t + α-2π / 12) + Ic _iii (t) sin (ω ₂ t + α-4π / 12) + Ic _iv (t) sin (ω ₂ t + α-6π / 12) + Ic _v (t) sin ( ω ₂ t + α-8π / 12) + Ic _vi (t) sin (ω ₂ t + α-10π / 12) + Ic _vii (t) sin (ω ₂ t + α-12π / 12) + Ic _viii ( t) sin (ω ₂ t + α-14π / 12) + Ic _ix (t) sin (ω ₂ t + α-16π / 12) + Ic _x (t) sin (ω ₂ t + α-18π / 12) + Ic _xi (t) sin (ω ₂ t + α-20π / 12) + Ic _xii (t) sin (ω ₂ t + α-22π / 12))… (35) <2-6-2> Inward rotation When a magnetic field is applied, the above 12-phase AC Ic _i (t) to Ic _xii (t) is Ic _i (t) = Ic ₂ (t) cos (ω ₂ -γ) (36a) Ic _ii (t) = Ic ₂ (t) cos (ω ₂ t-γ-2π / 12)… (36b) Ic _iii (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-4π / 12)… (36c) Ic _iv ( t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-6π / 12)… (36d) Ic _v (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-8π / 12)… (36e ) Ic _vi (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-10π / 12)… (36f) Ic _vii (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-12π / 12 )… (36g) Ic _viii (t) ＝ Ic ₂ ( t) cos (ω ₂ t-γ-14π / 12)… (36h) Ic _ix (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-16π / 12)… (36i) Ic _x (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-18π / 12)… (36j) Ic _xi (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-20π / 12)… (36k) Ic _xii (t) = Ic ₂ (t) cos (ω ₂ t-γ-22π / 12) (36l).

Substituting the expressions (36a) to (36l) into the expression (35), f ₂
To calculate.

F ₂ = ₂ μIm ₂ n Ic ₂ (t) (cos (ω ₂ t-γ) sin (ω ₂ t + α) + cos (ω ₂ t-γ-2π / 12) sin (ω ₂ t + α-2π / 12) + cos (ω ₂ t-γ-4π / 12) sin (ω ₂ t + α-4π / 12) + cos (ω ₂ t-γ-6π / 12) sin (ω ₂ t + α-6π / 12) + cos (ω ₂ t-γ-8π / 12) sin (ω ₂ t + α-8π / 12) + cos (ω ₂ t-γ-10π / 12) sin (ω ₂ t + α-10π / 12) + cos (ω ₂ t-γ-12π / 12) sin (ω ₂ t + α-12π / 12) + cos (ω ₂ t-γ-14π / 12) sin (ω ₂ t + α-14π / 12) + cos (ω ₂ t-γ-16π / 12) sin (ω ₂ t + α-16π / 12) + cos (ω ₂ t-γ-18π / 12) sin (ω ₂ t + α-18π / 12) + cos (ω ₂ t-γ-20π / 12) sin (ω ₂ t + α-20π / 12) + cos (ω ₂ t-γ-22π / 12) sin (ω ₂ t + α-22π / 12)) where, using the formula cos (a) sin (b) = 1/2 (sin (a + b) -sin (ab)), f ₂ = 2μIm ₂ n Ic ₂ (t ) (1/2 (sin (ω ₂ t-γ + ω ₂ t + α) -sin (ω ₂ t-γ-ω ₂ t-α)) +1/2 (sin (ω ₂ t-γ-2π / 12 + ω ₂ t + α-2π / 12) -sin (ω ₂ t-γ-2π / 12-ω ₂ t-α + 2π / 12)) +1/2 (sin (ω ₂ t-γ- 4π / 12 + ω ₂ t + α-4π / 12) -sin (ω ₂ t-γ-4π / 12-ω ₂ t-α + 4π / 12)) +1/2 (sin (ω ₂ t-γ -6π / 12 + ω ₂ t + α-6π / 12) -sin (ω ₂ t-γ-6π / 12-ω ₂ t-α + 6π / 12)) +1/2 (sin (ω ₂ t- γ-8 π / 12 + ω ₂ t + α-8π / 12) -sin (ω ₂ t-γ-8π / 12-ω ₂ t-α + 8π / 12)) +1/2 (sin (ω ₂ t-γ -10π / 12 + ω ₂ t + α-10π / 12) -sin (ω ₂ t-γ-10π / 12-ω ₂ t-α + 10π / 12)) +1/2 (sin (ω ₂ t- γ-12π / 12 + ω ₂ t + α-12π / 12) -sin (ω ₂ t-γ-12π / 12-ω ₂ t-α + 12π / 12)) +1/2 (sin (ω ₂ t -γ-14π / 12 + ω ₂ t + α-14π / 12) -sin (ω ₂ t-γ-14π / 12-ω ₂ t-α + 14π / 12)) +1/2 (sin (ω ₂ t-γ-16π / 12 + ω ₂ t + α-16π / 12) -sin (ω ₂ t-γ-16π / 12-ω ₂ t-α + 16π / 12)) +1/2 (sin (ω ₂ t-γ-18π / 12 + ω ₂ t + α-18π / 12) -sin (ω ₂ t-γ-18π / 12-ω ₂ t-α + 18π / 12)) +1/2 (sin ( ω ₂ t-γ-20π / 12 + ω ₂ t + α-20π / 12) -sin (ω ₂ t-γ-20π / 12-ω ₂ t-α + 20π / 12)) +1/2 (sin (ω ₂ t-γ-22π / 12 + ω ₂ t + α-22π / 12) -sin (ω ₂ t-γ-22π / 12-ω ₂ t-α + 22π / 12)) = 2μIm ₂ n Ic ₂ (t) (1/2 (sin (2ω ₂ t-γ + α) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-4π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-8π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-12π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-16π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α- 20π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t- γ +
α-24π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-28π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-32π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-36π / 12) + sin (γ + α)) +1/2 ( sin (2ω ₂ t-γ + α-40π / 12) + sin (γ + α)) +1/2 (sin (2ω ₂ t-γ + α-44π / 12) + sin (γ + α)) = μIm ₂ n Ic ₂ (t) (sin (2ω ₂ t-γ + α) + sin (2ω ₂ t-γ + α-4π / 12) + sin (2ω ₂ t-γ + α-8π / 12) + sin (2ω ₂ t-γ + α-12π / 12) + sin (2ω ₂ t-γ + α-16π / 12) + sin (2ω ₂ t-γ + α-20π / 12) + sin (2ω ₂ t -γ + α-24π / 12) + sin (2ω ₂ t-γ + α-28π / 12) + sin (2ω ₂ t-γ + α-32π / 12) + sin (2ω ₂ t-γ + α- 36π / 12) + sin (2ω ₂ t-γ + α-40π / 12) + sin (2ω ₂ t-γ + α-44π / 12) + 12sin (γ + α)) = μ Im ₂ n Ic ₂ (t ) (sin (2ω ₂ t-γ + α) + sin (2ω ₂ t-γ + α-π / 3) + sin (2ω ₂ t-γ + α-2π / 3) -sin (2ω ₂ t-γ + α) -sin (2ω ₂ t-γ + α-π / 3) -sin (2ω ₂ t-γ + α-2π / 3) + sin (2ω ₂ t-γ + α) + sin (2ω ₂ t -γ + α-π / 3) + sin (2ω ₂ t-γ + α-2π / 3) -sin (2ω ₂ t-γ + α) -sin (2ω ₂ t-γ + α-π / 3) -sin (2ω ₂ t-γ + α-2π / 3) + 12sin (γ + α)) = 12μIm ₂ n Ic ₂ (t) sin (γ + α)… (37) <2-6-3> Summary Inner rotating magnet This is obtained when the field is given by 12-phase alternating current (3
Comparing Eq. (7) with Eq. (31) obtained when the inner rotating magnetic field is given by three-phase alternating current, Eq. (37) has a fixed term (last term) of Has doubled. That is,
If the driving current of the inner magnet is 12-phase alternating current (Ii ~ Ixii), four times more driving force can be obtained than when the driving current of the inner magnet is 3-phase alternating current. This means, conversely, to generate the same driving force in the inner magnet, the inner driving current only needs to be 1/4 of that in the three phases.

<3> N (3 (2p) -2p) basic form <3-1> Referring to FIG. 14, the magnetic pole number ratio is 3: 1 (for example, the outer magnet has 6 magnetic poles, the inner magnet has 5 magnetic poles). Consider the case 2).

The magnetic flux densities B ₁ and B ₂ generated in the outer and inner magnets in this case are as follows.

B ₁ = Bm ₁ sin (3ω ₁ t-3θ) = μIm ₁ sin (3ω ₁ t-3θ) (41) B ₂ = Bm ₂ sin (ω ₂ t + α-θ) = μIm ₂ sin (ω ₂ t + α-θ) (42) Since the rotating magnetic field created by the stator coil is also calculated separately,
Magnetic flux density due to stator coils for outer and inner magnets
Let Bc1 and Bc2 be Bc ₁ = μn (Ica (t) sin (3θ) + Icb (t) sin (3θ-2π / 3) + Icc (t) sin (3θ-4π / 3)) (43) Bc ₂ = μn (Icd (t) sin (θ) + Ice (t) sin (θ-2π / 3) + Icf (t) sin (θ-4π / 3)) (44).

FIG. 15 shows changes in the magnetic flux densities B ₁ , B ₂ , Bc ₁ and Bc ₂ described above.

The overall magnetic flux density B is as follows.

B = B ₁ + B ₂ + Bc ₁ + Bc ₂ = μIm ₁ sin (3ω ₁ t-3θ) + μIm ₂ sin (ω ₂ t + α-θ) + μn (Ica (t) sin (3θ ) + Icb (t) sin (3θ-2π / 3) + Icc (t) sin (3θ-4π / 3)) + μn (Icd (t) sin (θ) + Ice (t) sin (θ-2π / 3) + Icf (t) sin (θ-4π / 3)) ... (45) torque tau ₁ acting on the outer magnet m _1, since occurs in line symmetry about a diameter, the f ₁ of half round force Then, τ ₁ = 2f ₁ × r ₁ (r ₁ is radius). Since three equivalent DC currents flow in half a circle, the sum of the forces acting on these three currents is f ₁ .

F ₁ = Im ₁ × B (θ = ω ₁ t) + Im ₁ × B (θ = ω ₁ t + 3π / 2) -Im ₁ × B (θ = ω ₁ t + π / 3) = μIm ₁ (Im ₁ sin (3ω ₁ t-3ω ₁ t) + Im ₁ sin (3ω ₁ t-3ω ₁ t-2π) -Im ₁ sin (3ω ₁ t-3ω ₁ t-π) + Im ₂ sin ( ω ₂ t + α-ω ₁ t) + Im ₂ sin (ω ₂ t + α-ω ₁ t-2π / 3) -Im ₂ sin (ω ₂ t + α-ω ₁ t-π / 3) + n (Ica (t) sin (3ω ₁ t) + Icb (t) sin (3ω ₁ t-2π / 3) + Icc (t) sin (3ω ₁ t-4π / 3)) + n (Ica (t) sin (3ω ₁ t + 2π) + Icb (t) sin (3ω ₁ t + 2π-2π / 3) + Icc (t) sin (3ω ₁ t + 2π-4π / 3)) -n (Ica (t) sin (3ω ₁ t + π) + Icb (t) sin (3ω ₁ t + π-2π / 3) + Icc (t) sin (3ω ₁ t + π-4π / 3)) + n (Icd (t) sin (ω ₁ t) + Ice (t) sin (ω ₁ t-2π / 3) + Icf (t) sin (ω ₁ t-4π / 3)) + n (Icd (t) sin (ω ₁ t + 2π / 3) + Ice (t) sin (ω ₁ t + 2π / 3-2π / 3) + Icf (t) sin (ω ₁ t + 2π / 3-4π / 3)) -n (Icd (t) sin (ω ₁ t + π / 3) + Ice (t) sin (ω ₁ t + π / 3-2π / 3) + Icf (t) sin (ω ₁ t + π / 3-4π / 3)) = μIm ₁ (n (Ica (t) sin (3ω ₁ t) + Icb (t) sin (3ω ₁ t-2π / 3) + Icc (t) sin (3ω ₁ t-4π / 3)) + n (Ica ( t) sin (3ω ₁ t) + Icb (t) sin (3ω ₁ t-2π / 3) + Icc (t) sin (3ω ₁ t-4π / 3)) + n (Ica (t) sin (3ω ₁ t) + Icb (t) sin (3ω ₁ t-2π / 3) + Icc (t) si n (3ω ₁ t-4π / 3)) + n (Icd (t) sin (ω ₁ t) + Ice (t) sin (ω ₁ t-2π / 3) + Icf (t) sin (ω ₁ t- 4π / 3)) + n (Icd (t) sin (ω ₁ t + 2π / 3) + Ice (t) sin (ω ₁ t) + Icf (t) sin (ω ₁ t-2π / 3)) + n (Icd (t) sin (ω ₁ t + 4π / 3) + Ice (t) sin (ω ₁ t + 2π / 3) + Icf (t) sin (ω ₁ t))) = μn Im ₁ (3 (Ica (t) sin (3ω ₁ t) + Icb (t) sin (3ω ₁ t-2π / 3) + Icc (t) sin (3ω ₁ t-4π / 3)) + Icd (t) sin (ω ₁ t) + Icd (t) sin (ω ₁ t + 2π / 3) + Icd (t) sin (ω ₁ t + 4π / 3) + Ice (t) sin (ω ₁ t) + Ice (t) sin (ω ₁ t + 2π / 3) + Ice (t) sin (ω ₁ t + 4π / 3) + Icf (t) sin (ω ₁ t) + Icf (t) sin (ω ₁ t + 2π / 3) + Icf (t) sin (ω ₁ t + 4π / 3)) = 3 μIm ₁ n (Ica (t) sin (3ω ₁ t) + Icb (t) sin (3ω ₁ t-2π / 3) + Icc (t) sin (3ω ₁ t-4π / 3)) (46) According to the equation (46), when the outer magnet is approximated by a sine wave, the torque acting on the outer magnet is controlled by the exciting currents of the coils a, b, and c. It shows that you can do it. It also shows that it is not affected by the exciting currents of the coils d, e, and f.

Next, since the torque τ ₂ acting on the inner magnet m ₂ is also generated in line symmetry about the diameter, if f ₂ is a force for half a circumference, then τ ₂ = 2f ₂ × r ₂ . Since one equivalent DC current flows in the half circumference, the force acting on this one equivalent DC current is f ₂ .

F ₂ = Im ₂ × B (θ = ω ₂ t + α) = μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) + μIm ₂ sin (ω ₂ t + α-ω ₂ t-α) + n (Ica (t) sin (3ω ₂ t + 3α) + Icb (t) sin (3ω ₂ t + 3α-2π / 3) + Icc (t) sin (3ω ₂ t + 3α-4π / 3) + n (Icd (t) sin (ω ₂ t + α) + Ice (t) sin (ω ₂ t + α-2π / 3) + Icf (t) sin (ω ₂ t + α-4π / 3)) = μIm ₂ (Im ₁ sin (3 (ω ₁ −ω ₂ ) t-3α) + n (Ica (t) sin (3ω ₂ t + 3α) + Icb (t) sin (3ω ₂ t + 3α -2π / 3) + Icc (t) sin (3ω ₂ t + 3α-4π / 3) + n (Icd (t) sin (ω ₂ t + α) + Ice (t) sin (ω ₂ t + α-2π / 3) + Icf (t) sin (ω ₂ t + α-4π / 3)) ... (47) Looking at Eq. (47), the effects of other than the calculated magnetic field on the rotation of the inner magnet ( It can be seen that there are 2π / 3, 4π / 3 in relative phase angle.To make this effect easier to understand, the position of each outer magnet at the peak time t is φ1 = ωt + π / 6, φ
2 = ωt + 5π / 6 and φ3 = ωt + 9π / 6.

Considering each influence, the magnetic field of the rotation angle θ is B ₁ = Bm ₁ (cos (ω ₁ t + π / 6-θ) + cos (ω ₁ t + 5π / 6-θ) + cos (ω ₁ t + 9π / 6-θ)) = μIm ₁ (cos (ω ₁ t + π / 6-θ) + cos (ω ₁ t + 5π / 6-θ) + cos (ω ₁ t + 9π / 6-θ)) = 0 This indicates that the magnetic poles with the crossing angle of 120 degrees cancel each other out on the inner coil. That is,
The number of magnetic poles of the outer magnet does not affect the inner magnet. Similarly, the total magnetic field produced by the outer coil is zero. Therefore, the driving force f _{2 at} this time is as follows.

F ₂ = μIm ₂ (n (Icd (t) sin (ω ₂ t + α) + Ice (t) sin (ω ₂ t + α-2π / 3) + Icf (t) sin (ω ₂ t + α-4π / 3)) (48) <3-2> When both outer and inner rotating magnetic fields are applied: Same as above three-phase alternating current Ica (t), Icb (t), Icc (t) 3 Exchange
Icd (t), Ice (t), Icf (t) are Ica (t) = Ic ₁ cos (3ω ₁ t-3β)… (49a) Icb (t) = Ic ₁ cos (3ω ₁ t-3β-2π / 3)… (49b) Icc (t) ＝ Ic ₁ cos (3ω ₁ t-3β-4π / 3)… (49c) Icd (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ)… ( 50a) Ice (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-2π / 3)… (50b) Icf (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-4π / 3) … (50c)

However, in equations (50a) to (50c), since amplitude modulation is possible, it is set as Ic ₂ (t) which is a function of time.

Expressions (49a) to (49c) are changed to expressions (46) and (49a) to
Substituting equation (49c) and equations (50a) to (50c) into equation (47), f
Calculate ₁ and f ₂ .

F ₁ = 3 μIm ₁ n Ic ₁ (cos (3ω ₁ t-3β) sin (3ω ₁ t) + cos (3ω ₁ t-3β-2π / 3) sin (3ω ₁ t-2π / 3) + cos (3ω ₁ t-3β-4π / 3) sin (3ω ₁ t-4π / 3)) where cos (a) sin (b) ＝ 1/2 (sin (a + b) -sin (ab) ) Formula, f ₁ = 3μIm ₁ n Ic ₁ (1/2 (sin (3ω ₁ t-3β + 3ω ₁ t) -sin (3ω ₁ t-3β-3ω ₁ t)) +1/2 ( sin (3ω ₁ t-3β-2π / 3 + 3ω ₁ t-2π / 3) -sin (3ω ₁ t-3β-2π / 3-3ω ₁ t + 2π / 3)) +1/2 (sin (3ω ₁ t-3β-4π / 3 + 3ω ₁ t-4π / 3) -sin (3ω ₁ t-3β-4π / 3-3ω ₁ t + 4π / 3))) = 3 / 2μIm ₁ n Ic ₁ (sin (6ω ₁ t-3β) + sin (3β) + sin (6ω ₁ t-3β-4π / 3) + sin (3β) + sin (6ω ₁ t-3β-8π / 3) + sin (3β)) = 3/2 μIm ₁ n Ic ₁ (sin (6ω ₁ t-3β) + sin (6ω ₁ t-3β-4π / 3) + sin (6ω ₁ t-3β-8π / 3) + 3sin (3β)) = 9 / 2 μIm ₁ n Ic ₁ sin (3β)… (51) f ₂ = μIm ₂ (Im ₁ sin (3 (ω ₁ -ω ₂ ) t-3α) + n Ic ₁ (cos (3ω ₁ t-3β) sin (3ω ₂ t + 3α) + cos (3ω ₁ t-3β-2π / 3) sin (3ω ₂ t + 3α-2π / 3) + cos (3ω ₁ t-3β-4π / 3) sin (3ω ₂ t + 3α-4π / 3)) + n Ic ₂ (t) (cos (ω ₂ t-γ) sin (ω ₂ t + α) + cos (ω ₂ t-γ-2π / 3) sin (ω ₂ t + α-2π / 3) + cos (ω ₂ t-γ-4π / 3) s in (ω ₂ t + α-4π / 3)) where f ₂ = μIm using the formula cos (a) sin (b) ＝ 1/2 (sin (a + b) -sin (ab)) ₂ (Im ₁ sin (3 (ω ₁ -ω ₂ ) t-3α) + n Ic ₁ (1/2 (sin (3ω ₁ t-3β + 3ω ₂ t + 3α) -sin (3ω ₁ t-3β- 3ω ₂ t-3α)) +1/2 (sin (3ω ₁ t-3β-2π / 3 + 3ω ₂ t + 3α-2π / 3) -sin (3ω ₁ t-3β-2π / 3-3ω ₂ t -3α + 2π / 3)) +1/2 (sin (3ω ₁ t-3β-4π / 3 + 3ω ₂ t + 3α-4π / 3) -sin (3ω ₁ t-3β-4π / 3-3ω ₂ t-3α + 4π / 3))) + n Ic ₂ (t) (1/2 (sin (ω ₂ t-γ + ω ₂ t + α) -sin (ω ₁ t-γ-ω ₂ t-α )) +1/2 (sin (ω ₂ t-γ-2π / 3 + ω ₂ t + α-2π / 3) -sin (ω ₂ t-γ-2π / 3-ω ₂ t-α + 2π / 3)) +1/2 (sin (ω ₂ t-γ-4π / 3 + ω ₂ t + α-4π / 3) -sin (ω ₂ t-γ-4π / 3-ω ₂ t-α + 4π / 3)))) = μIm ₂ (Im ₁ sin (3 (ω ₁ -ω ₂ ) t-3α) +1/2 n Ic ₁ (sin (3ω ₁ t + 3ω ₂ t-3β + 3α) + sin (3ω ₁ t + 3ω ₂ t-3β + 3α-4π / 3) + sin (3ω ₁ t + 3ω ₂ t-3β + 3α-8π / 3) -3sin (3ω ₁ t-3β + 3ω ₂ t-3α )) +1/2 n Ic ₂ (t) (sin (2ω ₂ t-γ + α) + sin (2ω ₂ t-γ + α-4π / 3) + sin (2ω ₂ t-γ + α-8π / 3) + 3sin (γ + α))) = μIm ₂ (Im ₁ sin (3 (ω ₁ -ω ₂ ) t-3α) +1/2 n Ic ₁ (sin (3ω ₁ t + 3ω ₂ t- 3β + 3α) + sin (3 _{1 t + 3ω 2 t-3β} + 3α-2π / 3) + sin (3ω 1 t + 3ω 2 t-3β + 3α-4π / 3) -3sin (3ω 1 t-3β-3ω 2 t-3α)) +1/2 n Ic ₂ (t) (sin (2ω ₂ t-γ + α) + sin (2ω ₂ t-γ + α-2π / 3) + sin (2ω ₂ t-γ + α-4π / 3 ) + 3sin (γ + α))) = μIm ₂ (Im ₁ sin (3 (ω ₁ -ω ₂ ) t-3α) -3/2 n Ic ₁ sin (3ω ₁ t-3β-3ω ₂ t-3α ) +3/2 n Ic ₂ (t) sin (γ + α)) (52) Here, for f ₂ , as seen in the equation (48),
When there is no influence of the magnetic field created by the outer magnet and outer coil, f ₂ = 3/2 n Ic ₂ (t) sin (γ + α)) (53), and it can be driven with a constant torque.

On the other hand, when the influence of the magnetic field generated by the outer magnet or the outer coil remains, Ic ₂ (t) = (2 / 3C / μIm ₂ -Im ₁ sin (3 (ω ₁ -ω ₂ ) t-3α) + n Ic ₁ sin (3ω ₁ t-3β-3ω ₂ t-3α)) / (n sin (γ + α)) (54) However, if C: constant, Since f ₂ = C, it becomes possible to drive with a constant torque. That is, when the magnetic pole ratio is 3: 1, it means that the outer magnet slightly affects the rotation of the inner magnet according to the equation (52). More precisely, the phase difference (ω _1-
A constant torque fluctuation according to ω ₂ ) occurs in the rotation of the inner magnet. This is shown in FIG. When the rectangular wave model is used, the influence of magnetic force interference between the outer magnet and the inner magnet is remarkably expressed. Considering state A now, state B rather than this state
Is more stable, a torque that tends to shift to the state B is generated. This torque becomes an intermittent torque and the phase difference
It is generated by (ω ₁ -ω ₂ ). Furthermore, in reality, it may not be possible to completely cancel the influence of the outer magnet because the influence of the distance between the coils or the perfect sine wave cannot be realized, and in this case, the most extreme case is It is expressed by equation (52).

However, by performing the amplitude modulation by the equation (54), it becomes possible to cancel the constant torque fluctuation, and the inner magnet can be driven with a constant torque even when the magnetic pole number ratio is 3: 1. is there.

<3-3> Summary According to the equations (51) and (52), when a current is passed through the stator coil in synchronization with each of the outer magnet and the inner magnet, the rotational torque is applied to both magnets. You can see that it occurs. Although not calculated, the rotational torque is applied only to the outer magnet when the current is applied to the stator coil in synchronization with the outer magnet, and only to the inner magnet when the current is applied to the stator coil in synchronization with the inner magnet. It goes without saying that it will occur. From this, it was proved that it is possible to work as a rotating electric machine even when the magnetic pole number ratio is a combination of 3: 1.

<3-4> Current Setting Consider that the outer circumference side coil and the inner circumference side coil shown in FIG. 14 are commonly used as shown in FIG. In FIG. 14, the coils a and d, the coils a and f , the coils a and e, the coils a and d , the coils a and f, and the coils a and e can be combined. I ₁ = Ia + Id I ₁₀ = I ₁ = Ia + Id I ₂ = Ic I ₁₁ = I ₂ = Ic I ₃ = Ib I ₁₂ = I ₃ = Ib I ₄ = Ia + If I ₁₃ = I ₄ = Ia + If I ₅ = Ic I ₁₄ = I ₅ = Ic I ₆ = Ib I ₁₅ = I ₆ = Ib I ₇ = Ia + Ie I ₁₆ = Ia + Ie I ₈ = Ic I ₁₇ = I _It can be seen that ₈ = Ic I ₉ = Ib I ₁₈ = I ₉ = Ib . That is, in a combination with a magnetic pole number ratio of 3: 1 it can be represented by 9-phase currents.
Compared with the combination with a magnetic pole number ratio of 2: 1, this means that a combination with a magnetic pole number ratio of 3: 1 requires 18-phase alternating current, but a combination with a magnetic pole number ratio of 3: 1 This is because only in the case, since the phase is inverted in half a circle, it can be represented by an alternating current of 9 phases, which is half of 18 phases.

However, since the load on the coils 1 , 4 , 7, 1 , 1 , 4 , 7 becomes large, considering that the remaining rotating coils are also used to form the inner rotating magnetic field, I ₁ = Ia + I _i I ₁₀ = I ₁ = Ia + I _i I ₂ = Ic + I _vi I ₁₁ = I ₂ = Ic + I _vi I ₃ = Ib + I _ii I ₁₂ = I ₃ = Ib + I _ii I ₄ = Ia + I _vii I ₁₃ = I ₄ = Ia + I _vii I ₅ = Ic + I _iii I ₁₄ = I ₅ = Ic + I _iii I ₆ = Ib + I _viii I ₁₅ = I ₆ = Ib + I _viii I ₇ = Ia + I _iv I ₁₆ = I ₇ = Ia + I _iv I ₈ = Ic + I _ix I ₁₇ = I ₈ = Ic + I _ix I ₉ = Ib + I _v I ₁₈ = I ₉ = Ib + I _v Good.

Current I _i for Forming Inner Rotating Magnetic Field
The positional relationship between I _ix and I _i to I _ix is shown in FIG.

<3-5> When applying an inner rotating magnetic field with 9-phase AC <3-5-1> Considering that an inner rotating magnetic field is created with 9-phase AC, the magnetic flux density Bc _{2 at} this time is as follows. become.

Bc ₂ = μn (Ic _i (t) sin (θ) + Ic _ii (t) sin (θ-2π / 9) + Ic _iii (t) sin (θ-4π / 9) + Ic _iv (t ) sin (θ-6π / 9) + Ic _v (t) sin (θ-8π / 9) + Ic _vi (t) sin (θ-10π / 9) + Ic _vii (t) sin (θ-12π / 9) ) + Ic _viii (t) sin (θ-14π / 9) + Ic _ix (t) sin (θ-16π / 9) (55) Therefore, the total magnetic flux density B is as follows.

B = B ₁ + B ₂ + Bc ₁ + Bc ₂ = μIm ₁ sin (3ω ₁ t-3θ) + μIm ₂ sin (ω ₂ t + α-θ) + μn (Ica (t) sin (3θ ) + Icb (t) sin (3θ-2π / 3) + Icc (t) sin (3θ-4π / 3) + μn (Ic _i (t) sin (θ) + Ic _ii (t) sin (θ-2π / 9) + Ic _iii (t) sin (θ-4π / 9) + Ic _iv (t) sin (θ-6π / 9) + Ic _v (t) sin (θ-8π / 9) + Ic _vi (t ) sin (θ-10π / 9) + Ic _vii (t) sin (θ-12π / 9) + Ic _viii (t) sin (θ-14π / 9) + Ic _ix (t) sin (θ-16π / 9) ) (56) Calculating f ₁ at this time, f ₁ = Im ₁ × B (θ ＝ ω ₁ t) + Im ₁ × B (θ ＝ ω ₁ t + 2π / 3) -Im ₁ × B (θ = ω ₁ t + π / 3) = μIm ₁ (Im ₁ (sin (3ω ₁ t-3ω ₁ t) + sin (3ω ₁ t−3ω ₁ t + 2π) -sin (3ω ₁ t− 3ω ₁ t + π)) + Im ₂ (sin (ω ₂ t + α-ω ₁ t) + sin (ω ₂ t + α-ω ₁ t-2π / 3) -sin (ω ₂ t + α-ω ₁ t + π / 3) + n (Ica (t) (sin (3ω ₁ t) + sin (3ω ₁ t + 2π) -sin (3ω ₁ t + π)) + Icb (t) (sin (3ω ₁ t-2π / 3) + sin (3ω ₁ t + 2π-2π / 3) -sin (3ω ₁ t + π-2π / 3)) + Icc (t) (sin (3ω ₁ t-4π / 3) + sin (3ω ₁ t + 2π-4π / 3) -sin (3ω ₁ t + π-4π / 3))) + n (Ic _i (t) (sin (ω ₁ t) + sin (ω ₁ t + 2π / 3) + _{sin (ω 1 t + π /} 3)) + Ic ii (t) (sin (ω 1 t-2π / 9) + sin ( _{1 t-2π / 9 + 2π} / 3)) + sin (ω 1 t-2π / 9 + π / 3)) + Ic iii (t) (sin (ω 1 t-4π / 9) + sin (ω 1 t-4π / 9 + 2π / 3)) + sin (ω ₁ t-4π / 9 + π / 3)) + Ic _iv (t) (sin (ω ₁ t-6π / 9) + sin (ω ₁ t -6π / 9 + 2π / 3) + sin (ω ₁ t-6π / 9 + π / 3)) + Ic _v (t) (sin (ω ₁ t-8π / 9) + sin (ω ₁ t-8π / 9 + 2π / 3)) + sin (ω ₁ t-8π / 9 + π / 3)) + Ic _vi (t) (sin (ω ₁ t-10π / 9) + sin (ω ₁ t-10π / 9 + 2π / 3)) + sin (ω ₁ t-10π / 9 + π / 3)) + Ic _vii (t) (sin (ω ₁ t-12π / 9) + sin (ω ₁ t-12π / 9 + 2π / 3) + sin (ω ₁ t-12π / 9 + π / 3)) + Ic _viii (t) (sin (ω ₁ t-14π / 9) + sin (ω ₁ t-14π / 9 + 2π / 3) + sin (ω ₁ t-14π / 9 + π / 3)) + Ic _ix (t) (sin (ω ₁ t-16π / 9) + sin (ω ₁ t-16π / 9 + 2π / 3 )) + sin (ω ₁ t-16π / 9 + π / 3))) = μIm ₁ (Im ₁ (sin (3ω ₁ t-3ω ₁ t) + sin (3ω ₁ t−3ω ₁ t + 2π)- sin (3ω ₁ t−3ω ₁ t + π)) (= 0) + Im ₂ (sin (ω ₂ t + α-ω ₁ t) + sin (ω ₂ t + α-ω ₁ t-2π / 3) -sin (ω ₂ t + α-ω ₁ t + π / 3) (= 0) + n (Ica (t) (sin (3ω ₁ t) + sin (3ω ₁ t + 2π) -sin (3ω ₁ t + π)) + Icb (t) (sin (3ω ₁ t-2π / 3) + sin (3ω ₁ t + 2π-2π / 3) -sin (3ω ₁ t + π-2π / 3)) + Icc ( t) (sin (3ω ₁ t-4π / 3) + sin (3ω ₁ t + 2π-4π / 3) -sin (3ω ₁ t + π-4π / 3))) + n (Ic _i (t) (sin (ω ₁ t) + sin (ω ₁ t + 2π / 3) + sin (ω ₁ t + π / 3)) (= 0) + Ic _ii (t) (sin (ω ₁ t-2π / 9) + sin (ω ₁ t-2π / 9 + 2π / 3)) + sin (ω ₁ t-2π / 9 + π / 3)) (= 0) + Ic _iii (t) (sin (ω ₁ t-4π / 9) + sin (ω ₁ t-4π / 9 + 2π / 3)) + sin (ω ₁ t-4π / 9 + π / 3)) ( = 0) + Ic _iv (t) (sin (ω ₁ t-6π / 9) + sin (ω ₁ t-6π / 9 + 2π / 3) + sin (ω ₁ t-6π / 9 + π / 3) ) (= 0) + Ic _v (t) (sin (ω ₁ t-8π / 9) + sin (ω ₁ t-8π / 9 + 2π / 3)) + sin (ω ₁ t-8π / 9 + π / 3)) (= 0) + Ic _vi (t) (sin (ω ₁ t-10π / 9) + sin (ω ₁ t-10π / 9 + 2π / 3)) + sin (ω ₁ t-10π / 9 + π / 3)) (= 0) + Ic _vii (t) (sin (ω ₁ t-12π / 9) + sin (ω ₁ t-12π / 9 + 2π / 3) + sin (ω ₁ t- 12π / 9 + π / 3)) (= 0) + Ic _viii (t) (sin (ω ₁ t-14π / 9) + sin (ω ₁ t-14π / 9 + 2π / 3) + sin (ω ₁ t-14π / 9 + π / 3)) (= 0) + Ic _ix (t) (sin (ω ₁ t-16π / 9) + sin (ω ₁ t-16π / 9 + 2π / 3)) + sin (ω ₁ t-16π / 9 + π / 3))) (= 0) ＝ 3μn Im ₁ (Ica (t) sin (3ω ₁ t) + Icb (t) sin (3ω ₁ t-2π / 3) + Icc (t) sin (3ω ₁ t-4π / 3)) (57), which is the same as the equation (46) obtained when the inner rotating magnetic field is given by three-phase alternating current.

On the other hand, the calculation of f ₂ is as follows.

F ₂ = Im ₂ × B (θ = ω ₂ t + α) = μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) + Im ₂ sin (ω ₂ t + α-ω ₂ t -α) + n (Ica (t) sin (3ω ₂ t + 3α) + Icb (t) sin (3ω ₂ t + 3α-2π / 3) + Icc (t) sin (3ω ₂ t + 3α-4π / 3) + n (Ic _i (t) sin (ω ₂ t + α) + Ic _ii (t) sin (ω ₂ t + α-2π / 9) + Ic _iii (t) sin (ω ₂ t + α- 4π / 9) + Ic _iv (t) sin (ω ₂ t + α-6π / 9) + Ic _v (t) sin (ω ₂ t + α-8π / 9) + Ic _vi (t) sin (ω ₂ t + α-10π / 9) + Ic _vii (t) sin (ω ₂ t + α-12π / 9) + Ic _viii (t) sin (ω ₂ t + α-14π / 9) + Ic _ix (t) sin (ω ₂ t + α-16π / 9))) = μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) + n (Ica (t) sin (3ω ₂ t + 3α) + Icb ( t) sin (3ω ₂ t + 3α-2π / 3) + Icc (t) sin (3ω ₂ t + 3α-4π / 3) + n (Ic _i (t) sin (ω ₂ t + α) + Ic _ii (t) sin (ω ₂ t + α-2π / 9) + Ic _iii (t) sin (ω ₂ t + α-4π / 9) + Ic _iv (t) sin (ω ₂ t + α-6π / 9) ) + Ic _v (t) sin (ω ₂ t + α-8π / 9) + Ic _vi (t) sin (ω ₂ t + α-10π / 9) + Ic _vii (t) sin (ω ₂ t + α -12π / 9) + Ic _viii (t) sin (ω ₂ t + α-14π / 9) + Ic _ix (t) sin (ω ₂ t + α-16π / 9)))… (58) <3- 5-2> When both outer and inner rotating magnetic fields are applied AC Ica (t), Icb (t ), Icc (t) is _{Ica (t) = Ic 1 cos} (3ω 1 t-3β) ... (59a) Icb (t) = Ic 1 cos (3ω 1 t-3β- 2π / 3)… (59b) Icc (t) ＝ Ic ₁ cos (3ω ₁ t-3β-4π / 3)… (59c) and the above 9-phase alternating current Ic _i (t) to Ic _ix (t) Ic _i (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ)… (60a) Ic _ii (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-2π / 9)… (60b ) Ic _iii (t) = Ic ₂ (t) cos (ω ₂ t-γ-4π / 9)… (60c) Ic _iv (t) = Ic ₂ (t) cos (ω ₂ t-γ-6 π / 9) )… (60d) Ic _v (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-8π / 9)… (60e) Ic _vi (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ -10π / 9)… (60f) Ic _vii (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-12π / 9)… (60g) Ic _viii (t) ＝ Ic ₂ (t) cos (ω ₂ t-γ-14π / 9) (60h) Ic _ix (t) = Ic ₂ (t) cos (ω ₂ t-γ-16π / 9) (60i).

The expressions (59a) to (59c) and the expressions (60a) to (60i) are substituted into the expression (58) to calculate f ₂ .

F ₂ = μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) + n (Ic ₁ cos (3ω ₁ t-3β) sin (3ω ₂ t-3α) + Ic ₁ cos (3ω ₁ t-3β-2π / 3) sin (3ω ₂ t + 3α-2π / 3) + Ic ₁ cos (3ω ₁ t-3β-4π / 3) sin (3ω ₂ t + 3α-4π / 3) + n (Ic ₂ (t) cos (ω ₂ t-γ) sin (ω ₁ t + α) + Ic ₂ (t) cos (ω ₂ t-γ-2π / 9) sin (ω ₁ t + α-2π / 9) + Ic ₂ (t) cos (ω ₂ t-γ-4π / 9) sin (ω ₁ t + α-4π / 9) + Ic ₂ (t) cos (ω ₂ t-γ-6π / 9) sin (ω ₁ t + α-6π / 9) + Ic ₂ (t) cos (ω ₂ t-γ-8π / 9) sin (ω ₁ t + α-8π / 9) + Ic ₂ (t) cos ( _{ω 2 t-γ-10π /} 9) sin (ω 1 t + α-10π / 9) + Ic 2 (t) cos (ω 2 t-γ-12π / 9) sin (ω 1 t + α-12π / 9) + Ic ₂ (t) cos (ω ₂ t-γ-14π / 9) sin (ω ₁ t + α-14π / 9) + Ic ₂ (t) cos (ω ₂ t-γ-16π / 9) sin (ω ₁ t + α-16π / 9))) where f ₂ = using the formula cos (a) sin (b) ＝ 1/2 (sin (a + b) -sin (ab)) μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) + n Ic ₁ (1/2 (sin (3ω ₁ t-3β + 3ω ₂ t + 3α) -sin (3ω ₁ t-3β-3ω ₂ t-3α) +1/2 (sin (3ω ₁ t-3β-2π / 3 + 3ω ₂ t + 3α-2π / 3) -sin (3ω ₁ t-3β-2π / 3-3ω ₂ t-3α + 2π / 3)) +1/2 (sin (3ω ₁ t-3β-4π / 3 + 3ω ₂ t + 3α-4π / 3) -sin (3ω ₁ t -3β-4π / 3-3ω ₂ t-3α + 4π / 3))) + n Ic ₂ (t) (1/2 (sin (ω ₂ t-γ + ω ₂ t + α) -sin (ω ₂ t-γ-ω ₂ t-α)) +1/2 (sin (ω ₂ t-γ-2π / 9 + ω ₂ t + α) -sin (ω ₂ t-γ-2π / 9-ω ₂ t -α)) +1/2 (sin (ω ₂ t-γ-4π / 9 + ω ₂ t + α) -sin (ω ₂ t-γ-4π / 9-ω ₂ t-α)) + 1 / 2 (sin (ω ₂ t-γ-8π / 9 + ω ₂ t + α) -sin (ω ₂ t-γ-6π / 9-ω ₂ t-α)) +1/2 (sin (ω ₂ t -γ-10π / 9 + ω ₂ t + α) -sin (ω ₂ t-γ-8π / 9-ω ₂ t-α)) +1/2 (sin (ω ₂ t-γ-12π / 9 + ω ₂ t + α) -sin (ω ₂ t-γ-10π / 9-ω ₂ t-α)) +1/2 (sin (ω ₂ t-γ-14π / 9 + ω ₂ t + α)- sin (ω ₂ t-γ-12π / 9-ω ₂ t-α)) +1/2 (sin (ω ₂ t-γ-16π / 9 + ω ₂ t + α) -sin (ω ₂ t-γ -14π / 9-ω ₂ t-α))) = μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) + 1 / 2n Ic ₁ (sin (3ω ₁ t + 3ω ₂ t-3β + 3α) -sin (3ω ₁ t-3ω ₂ t-3α-3β) + sin (3ω ₁ t + 3ω ₂ t-3β + 3α-4π / 3) -sin (3ω ₁ t-3ω ₂ t-3α-3β ) + sin (3ω ₁ t + 3ω ₂ t-3β + 3α-2π / 3) -sin (3ω ₁ t-3ω ₂ t-3α-3β)) + n Ic ₂ (t) (1/2 (sin ( ω ₂ t-γ + ω ₂ t + α) + sin (γ + α)) +1/2 (sin (ω ₂ t-γ + ω ₂ t + α-4π / 9) + sin (γ + α) ) +1/2 (sin (ω ₂ t-γ + ω ₂ t + α-8π / 9) + sin (γ + α)) +1/2 (sin (ω ₂ t -γ + ω ₂ t + α-12π / 9) + sin (γ + α)) +1/2 (sin (ω ₂ t-γ + ω ₂ t + α-16π / 9) + sin (γ + α )) +1/2 (sin (ω ₂ t-γ + ω ₂ t + α-2π / 9) + sin (γ + α)) +1/2 (sin (ω ₂ t-γ + ω ₂ t + α-6π / 9) + sin (γ + α)) +1/2 (sin (ω ₂ t-γ + ω ₂ t + α-10π / 9) + sin (γ + α)) +1/2 ( sin (ω ₂ t-γ + ω ₂ t + α-14π / 9) + sin (γ + α)))) = μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) -3/2 n Ic ₁ sin (3ω ₁ t + 3ω ₂ t-3α-3β) + 9 / 2n Ic ₂ (t) sin (γ + α)) (61) <3-5-3> Summary (61) Right side The first and second terms of are canceled out when the other phases are taken into account, as in the case of Eq. (48), as in the case of three-phase alternating current.

On the other hand, comparing this equation (61) obtained when the inner rotating magnetic field is applied with 9-phase alternating current with the above equation (52) obtained when applying the inner rotating magnetic field with 3-phase alternating current, (61)
The equation has three times as many fixed terms (the last term) as equation (52). That is, the drive current of the inner magnet is changed to a 9-phase alternating current (I
_i ~ I _ix ), three times as much electromagnetic force (driving torque) can be obtained as when the driving current of the inner magnet is three-phase alternating current. This means that, conversely, the drive current need only be 1/3 to generate the same drive torque in the inner magnet.

This completes the theoretical analysis.

Next, FIGS. 19 to 25 show the third to eighth embodiments. Similar to the four embodiments described above, the rotors 3 and 4 are arranged inside and outside the stator. However, the magnetic pole number ratio is 2: 1 in FIGS. 19 and 20, the magnetic pole number ratio is 2: 3 in FIG. 21, and the magnetic pole number ratio is 4: 3 in FIG.
3 and FIG. 24 show a combination in which the magnetic pole number ratio is 2: 1. In summary, as long as the number of magnetic poles is a combination of even: odd number or odd: even number, it is not limited to the case where the number of magnetic poles of the outer magnet is larger than that of the inner magnet, and the number of magnetic poles of the outer magnet is the magnetic pole of the inner magnet. It may be less than the number. Further, the rotor can be treated in the same manner as that before being expanded even if it is constructed in a cylindrical shape by expanding one round described in each of the first to fourth embodiments and connecting a plurality of rotors.

The arrangement of the stator and the two rotors may basically be any arrangement. For example, in Figure 2.
5 is an intermediate rotor 42 and an inner rotor 43 inside the stator 41.
Two rotors are arranged. In this case, if the intermediate rotor 42 is covered with the same iron frame as the outer frame 44,
Since the magnetic flux generated in 41 does not reach the inner rotor 43,
The intermediate rotor 42 is not covered with the iron frame. Although not shown, the same applies when two rotors are arranged outside the stator.

In the embodiment, the case where the two rotors are made of permanent magnets has been described, but it goes without saying that each rotor can be made of electromagnets.

The motor drive current circuit is not limited to the case of using the PWM signal, but may be the case of using the PAM signal or other signals.

In the embodiment, the structure of the electric machine is a radial gap type (there is a gap between the rotor and the stator in the radial direction), but the structure of the axial gap type (there is a gap between the rotor and the stator in the axial direction). The present invention can be applied also to.

[Brief description of drawings]

FIG. 1 is a schematic cross-sectional view of a rotary electric machine body according to a first embodiment.

FIG. 2 is a schematic cross-sectional view of a rotating electric machine body in which dedicated coils are arranged on an inner peripheral side and an outer peripheral side of a stator 2.

FIG. 3 is a control system diagram.

FIG. 4 is a circuit diagram of an inverter.

FIG. 5 is a schematic sectional view of a rotary electric machine body according to a second embodiment.

FIG. 6 is a schematic cross-sectional view of a rotating electric machine main body in the case of a combination with a magnetic pole number ratio of 3: 1 shown for comparison.

FIG. 7 is a schematic cross-sectional view of a rotating electric machine main body in the case of a combination with a magnetic pole number ratio of 3: 1 shown for comparison.

FIG. 8 is a model diagram to be referred to when considering an N (2p-2p) basic form.

FIG. 9 is a model diagram showing changes in magnetic flux density.

FIG. 10 is a model diagram to be referred to when considering an N (2 (2p) -2p) basic form.

FIG. 11 is a model diagram showing a change in magnetic flux density.

FIG. 12 is a model diagram to be referred to when considering an N (2 (2p) -2p) basic form.

FIG. 13 is a waveform diagram showing a distribution of 12-phase alternating current.

FIG. 14 is a model diagram that is referred to when considering an N (3 (2p) -2p) basic form.

FIG. 15 is a model diagram showing changes in magnetic flux density.

FIG. 16 is an explanatory diagram of magnetic interference between an outer magnet and an inner magnet.

FIG. 17 is a model diagram to be referred to when considering an N (3 (2p) -2p) basic form.

FIG. 18 is a waveform chart showing the distribution of 9-phase alternating current.

FIG. 19 is a schematic sectional view of a rotary electric machine body according to a third embodiment.

FIG. 20 is a schematic sectional view of a rotary electric machine body according to a fourth embodiment.

FIG. 21 is a schematic sectional view of a rotary electric machine body according to a fifth embodiment.

FIG. 22 is a schematic sectional view of a rotary electric machine body according to a sixth embodiment.

FIG. 23 is a schematic sectional view of a rotary electric machine body according to a seventh embodiment.

FIG. 24 is a schematic sectional view of a rotating electrical machine body according to an eighth embodiment.

FIG. 25 is a schematic sectional view of a rotary electric machine body according to a ninth embodiment.

[Explanation of symbols]

2 stator 3 outer rotor 4 Inner rotor 6 coils

─────────────────────────────────────────────────── ─── Continuation of the front page (58) Fields surveyed (Int.Cl. ⁷ , DB name) H02K 16/02 H02K 21/12

Claims (3)

(57) [Claims] 1. Two rotors and one stator are constructed in a three-layer structure and on the same shaft, and the two rotors are
A common coil for generating different rotating magnetic fields is formed on the stator, and the common coil is provided on each rotor.
A rotating electric machine configured to flow a composite current obtained by adding corresponding currents, characterized in that a ratio of the numbers of magnetic poles of the two rotors is a combination of K: L (K is an even number, L is an odd number). Electric machinery. 2. The rotating electric machine according to claim 1, wherein the rotor driving current on the side having a smaller number of magnetic poles in the composite current is 2K × 3 phase alternating current. 3. The coils are divided into three phases per one pole of the rotor having the smaller number of magnetic poles, and the rotor driving current on the side having the smaller number of magnetic poles is divided into each of the divided coils. The rotary electric machine according to claim 1, wherein a current value divided by the number of the coils is passed.