JPH11275826A - Rotating machine - Google Patents

Abstract

PROBLEM TO BE SOLVED: To prevent a loss caused by a stator coil current, when two rotors are provided. SOLUTION: In a rotating machine having a body 1, two rotors 3, 4 and a stator 2 are structured in three layers on the same shaft. Then, a single coil 6 is formed on the stator 2 and compound currents are made to flow in this single coil 6, in such a way that the same number of revolving magnetic fields with the number of rotors 3, 4 are produced. When one of the rotor 3, 4 is operated as a motor and the other as a generator, it is sufficient merely for a current corresponding to the difference between motor driving power and generating power to be caused to flow in the single coil 6. As a result, the efficiency is improved.

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

The present invention relates to a rotating electric machine.

[0002]

2. Description of the Related Art Two synchronous motors of the same rated torque are independently controlled.
There has been proposed a configuration in which a plurality is provided and each is rotated synchronously (see Japanese Patent Application Laid-Open No. 9-275573).

[0003]

By the way, in order to make the structure compact, it is conceivable that two rotors and one stator are formed in a three-layer structure on the same shaft (see Japanese Patent Application Laid-Open No. 8-34063). ).

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

However, when current flows through each coil and each inverter, losses due to the current (copper loss and switching loss) cannot be avoided.

SUMMARY OF THE INVENTION Accordingly, it is an object of the present invention to prevent a loss due to a current by forming a single coil to share a coil and supplying a composite current to the coil so as to generate a plurality of rotating magnetic fields. .

[0007]

According to a first aspect of the present invention, two rotors and one stator are formed in a three-layer structure on the same shaft, and a single coil is formed on the stator.
A composite current is applied to this single coil so as to generate the same number of rotating magnetic fields as the number of rotors.

According to a second aspect, in the first aspect, the rotational phase of each rotor is detected, and the composite current is controlled according to the detected rotational phase.

In a third aspect, in the second aspect, the means for flowing the composite current through the single coil is an inverter.

According to a fourth aspect of the present invention, in any one of the first to third aspects, the rotor is constituted by a permanent magnet.

According to a fifth aspect of the present invention, in any one of the first to third aspects, the rotor is disposed at predetermined intervals on the outside and inside of the cylindrical stator.

According to a sixth aspect of the present invention, in any one of the first to third aspects, when the stator is arranged at the outermost side or the innermost side, the rotor closer to the stator than the rotor farther from the stator. Make the magnetic flux reach through the rotor.

[0013]

According to the first and second aspects of the invention, when one of the rotors is operated as a motor and the other is operated as a generator, a current corresponding to the difference between the motor driving power and the generated power is supplied to a single coil. The efficiency is greatly improved because it is only necessary to flow to

According to the third aspect, when one of the rotors is operated as a motor and the other is operated as a generator,
Since only the current corresponding to the difference between the motor driving power and the generated power needs to flow through a single coil, the capacitance of the power switching transistor of the inverter can be reduced, thereby improving the switching efficiency and increasing the overall efficiency. improves.

According to the fourth aspect of the present invention, since the rotating body is only a permanent magnet, a slip ring is not required.

According to the fifth aspect, since the distance between the two rotors from the stator is the shortest, when the same current is applied to the stator coil, compared with the case where one rotor is located farther than the stator. The driving torque increases.

According to the sixth aspect, when it is necessary to cool the coils of the stator, the cooling is facilitated.

[0018]

FIG. 1 is a sectional view of a rotary electric machine main body 1. FIG. In the figure, rotors 3 and 4 are arranged with a predetermined gap on the outside and inside of a cylindrical stator 2 (three-layer structure), and each of the outside and inside rotors 3 and 4 is an outer frame that covers the whole.
5 (see FIG. 3) and rotatably and coaxially.

The inner rotor 4 is formed of a pair of permanent magnets having a half circumference of the S pole and another half circumference of the N pole, whereas the outer rotor 3 has twice the number of poles per one pole of the inner rotor 4. The permanent magnet poles are arranged as follows. That is, S of the outer rotor 3
There are two poles and two N poles, and the S pole and the N 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 is No rotational force is given by the magnet.

For example, consider the effect of the magnet of the inner rotor 4 on the outer rotor 3. Inner rotor 4 for simplicity
Is fixed. First, in the relationship between the S pole of the inner rotor 4 and the upper magnet SN of the outer rotor 3 opposed thereto, temporarily receive the magnetic force generated by the S pole of the inner rotor 4 in the illustrated state,
If the upper magnet SN of the outer rotor attempts to rotate in the clockwise direction, the relationship between the N pole of the inner rotor 4 and the lower magnet SN of the outer rotor 3 facing the N pole indicates that
The pole causes the lower magnet SN of the outer rotor 3 to rotate counterclockwise. In other words, the south pole of the inner rotor 4 is
The magnetic force exerted on the upper magnet 3 and the magnetic force exerted by the N pole of the inner rotor 4 on the lower magnet of the outer rotor 3 exactly cancel each other, so that the outer rotor 3 has no relation to the inner rotor 4 but has a relation to the stator 2. It is possible to control only by. This is the same between the rotating magnetic field generated in the stator coil and the rotor as described later.

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

As will be described later, the twelve coils are distinguished by numbers, and in this case, the coil 6 comes out in the meaning of the sixth coil. Although confusing with the expression coil 6 above, the meaning is different.

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

First, in order to supply a current (three-phase alternating current) for generating a rotating magnetic field to the inner rotor 4, [1, 2] = [ 7 ,
8 ], [ 3 , 4 ] = [9, 10], [5, 6] = [ 11 , 12 ]
The currents Id, If, and Ie, which are out of phase by 120 degrees, are set in the set coils.

Here, an underline below the number means that a current flows in the opposite direction. For example, flowing a current Id through a set of coils [1, 2] = [ 7 , 8 ] means that half of the current Id flows from coil 1 to coil 7 , and Id flows from coil 2 to coil 8. Is to pass the other half of the current. Since 1 and 2, 7 and 8 are close to each other on the circumference, it is possible to generate the same number (two poles) of rotating magnetic fields as the number of magnetic poles of the inner rotor 4 by this current supply.

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

For example, one set of coils [1] = [ 4 ] =
Flowing the current Ia at [7] = [ 10 ] means that coil 1
4 is to pass the current of Ia even when the current of Ia is directed from the coil 7 to the coil 10 . Since the coils 1 and 7 and the coils 4 and 10 are located 180 degrees apart on the circumference, the current supply can generate the same number (4 poles) of rotating magnetic fields as the magnetic poles of the outer rotor 3 .

As a result, the following composite currents I _{1 to} I ₁₂ may be applied to the twelve coils.

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, underline attached under the current symbol represents that a reverse current.

Further, the setting of the composite current will be described with reference to FIG. 2. FIG. 2 shows, for comparison with FIG. 1, a separate rotating magnetic field for each rotor on the inner circumference and the outer circumference of the stator 2. A dedicated coil to be generated is arranged. That is,
The arrangement of the inner peripheral coils d, f, and e generates a rotating magnetic field for the inner rotor, and the arrangement of the outer coils a, c, and b generates a rotating magnetic field for the outer rotor. In this case, in order to share the two dedicated coils and reconstruct the single coil shown in FIG. 1, half of the current flowing through the coil d in the inner peripheral side coil is placed near the coil d. In the same way, half of the current flowing through coil f is placed on coils b and a near coil f, and half of the current flowing through coil e is placed near coil e. That is, the coils c and b may be charged. The expressions of the composite currents I _{1 to} I ₁₂ express such a concept in a mathematical expression. The current setting method is not limited to this, and another current setting method may be used as described later.

When the current is set in this manner, two magnetic fields, a rotating magnetic field for the inner rotor 4 and a rotating magnetic field for the outer rotor 3, are simultaneously generated in a single coil. Neither is the rotating force applied to the outer rotor 3 by the rotating magnetic field, nor is the magnet of the outer rotor 3 applied to the inner rotor 4 by the rotating magnetic field. This point has been proved by theoretical analysis as described later.

The above Id, If, Ie current settings are
The current setting of Ia, Ic, and Ib is performed in synchronization with the rotation of the outer rotor 3, respectively. The phase lead / lag is set for the direction of the torque, 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 DC current from a power source 11 such as a battery into an AC current is provided. Since the sum of all instantaneous currents becomes zero, this inverter 12 is the same as a normal three-phase bridge type inverter having 12 phases, as shown in detail in FIG. 4, and has 24 transistors Tr1 to T
It consists of r24 and the same number of diodes as this transistor.

The ON / OFF signal applied to each gate (base of the transistor) of the inverter 12 is a PWM signal.

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

As described above, in the embodiment of the present invention, 2
One rotor 3 and 4 and one stator 2 have a three-layer structure on the same axis, and a single coil 6
Is formed, and a composite current is caused to flow through the single coil 6 so that the same number of rotating magnetic fields as the number of rotors is generated.When one of the rotors is operated as a motor and the other is operated as a generator, Since only the current corresponding to the difference between the motor drive power and the generated power needs to flow through a single coil, the efficiency can be greatly improved.

An inverter is provided for each of the two rotors.
In the case where one of the rotors is operated as a motor and the other is operated as a generator, as described above, it suffices to flow only a current corresponding to the difference between the motor driving power and the generated power to a single coil. Therefore, the capacitance of the power switching transistor of the inverter can be reduced, thereby improving the switching efficiency,
Overall efficiency is improved.

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

In FIG. 1, the core 7 for winding the coil is the coil 6
In the second embodiment, one core 21 is provided for each two coils, whereas the total number of twelve coils is equal to twelve.
However, in order to avoid interference between magnetic fluxes generated in the two coils 6, a slit 22 is provided in the center of the core 21 in the circumferential direction.

In the second embodiment, since the total number of cores 21 is six, which is half that in the first embodiment, the number of manufacturing steps is reduced.

FIG. 6 shows a third embodiment, which corresponds to FIG. 5 of the second embodiment.

In FIG. 5, the ratio of the number of magnetic poles between the outer rotor 3 and the inner rotor 4 (hereinafter simply referred to as the ratio of the number of magnetic poles) is a combination of 2: 1. : 1 combination.

In the case of the combination having the magnetic pole ratio of 3: 1, unlike the case of the combination having the magnetic pole ratio of 2: 1, there is a slight effect between the magnet of the outer rotor 3 and the magnet of the inner rotor 4.
That is, although the magnet of the outer rotor 3 is not given a rotational force by the rotating magnetic field with respect to the inner rotor 4,
Since the magnet of the inner rotor 4 is affected by the rotating magnetic field applied to the outer rotor 3, the inner rotor 4 rotates while generating torque fluctuation.

However, the interference of the outer rotor 3 with the rotation of the inner rotor 4, that is, the constant torque fluctuation occurring in the inner rotor 4 is, as understood from the theoretical analysis described later, the phase difference between the outer rotor 3 and the inner rotor 4. Since it becomes a function of (ω ₁ −ω ₂ ), amplitude modulation is applied to an alternating current for generating a rotating magnetic field for the outer coil so as to cancel the constant torque fluctuation in advance, so that the inner rotor 4 Can cancel the torque fluctuations that occur.

Therefore, even with this combination of the magnetic pole ratio of 3: 1, basically the same operation and effect as the combination of the magnetic pole number ratio of 2: 1 can be obtained.

On the other hand, the point that three coils are provided per magnetic pole of the outer rotor 3 is the same as in FIG. 5, so that the total number of the stator coils 6 is 18 (= 3 × 6). Therefore, an inverter for passing an 18-phase alternating current through the stator coil 6 is required. However, since the current of the 18-phase alternating current is inverted every 180 degrees, any inverter that generates 9-phase alternating current, which is half of the 18-phase AC, may be used. That is, the inverter may be constituted by 18 transistors and the same number of diodes as the transistors, and the number of transistors and diodes constituting the inverter can be reduced as compared with the first and second embodiments.

Also, one core 25 is provided for each of the three coils, and slits 2 are formed at positions where the core 25 is divided into three in the circumferential direction.
By forming 6, as in the first and second embodiments, the total number of cores is reduced.

FIG. 7 shows a fourth embodiment, which corresponds to FIG. 6 of the third embodiment.

In this embodiment, the core 31 around which the stator coil 6 is wound is formed integrally, thereby further reducing the number of manufacturing steps as compared with the case of FIG.

It is needless to say that wide slits 32 and 33 are provided in every third coil 6 to increase the magnetic resistance.

In the first and second embodiments, the combination of magnetic pole ratios is 2: 1. In the third and fourth embodiments, the combination of magnetic pole ratios is 3: 1. In fact, the combination of the magnetic pole ratios is not limited to this, and the following theoretical analysis shows that any combination can be used as a rotating electric machine.

Hereinafter, this theoretical analysis will be described with respect to each term.

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

Here, the notation of N (2p-2p) will be described. The left 2p is the number of magnetic poles of the outer magnet (outer rotor),
2p on the right side indicates the number of magnetic poles of the inner magnet (inner rotor). Also, N is a positive integer, which indicates that the same applies to a ring obtained by expanding (2p-2p) and multiplying it by an integer.

The simplest one having a magnetic pole ratio of 1: 1 is
When the outer magnet has two magnetic poles and the inner magnet has two magnetic poles,
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 the respective magnets can be expressed as follows. .

B ₁ = Bm ₁ sin (ω ₁ t-θ) = μIm ₁ sin (ω ₁ t-θ) (1) B ₂ = Bm ₂ sin (ω ₂ t + α-θ) = μIm ₂ sin (ω ₂ t + α-θ)… (2) where Bm ₁ , Bm ₂ : amplitude μ: magnetic 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 phases of the outer magnet and the coil match are considered to be 0.

If the current flowing through the stator coil is a three-phase alternating current, the magnetic flux density Bc of the stator coil is Bc = μn (Ica (t) sin (θ) + Icb (t) sin (θ-2π / 3) + Icc (t) sin (θ-4π / 3)) (3) where n: coil constant.

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

FIG. 9 shows changes in the magnetic flux densities B ₁ , B ₂ , and Bc. Each magnetic flux density changes as a sine wave.

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 ( in t) sin (θ-2π / 3) + Icc (t) sin (θ-4π / 3)) ... (4) here, when the torque applied to the outer magnet m ₁ and tau _1,
The generated torques are equal in line symmetry about the diameter. Therefore, when the half circumference of the force of f _1, the whole driving force from becoming a _{2f 1, τ 1 = 2f 1} × r 1 (r 1 is the radius) is.

To consider the torque τ ₁ , it is sufficient to consider f ₁ (that is, the driving force generated when the equivalent DC current Im ₁ is affected by the magnetic field (magnetic flux density B)). Since only flows one equivalent DC current to the semicircle, f ₁ is 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. The generated torques are equal in line symmetry with respect to
Assuming that f ₂ is a force for a half turn, τ ₂ = 2f ₂ × r ₂ (r ₂ is a radius). Since only one equivalent DC current flows in a half circumference, 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 a phase difference current of β flows through the coil in accordance with the phase of the outer magnet, the three-phase alternating currents Ica (t), Icb (t), and Icc (t) in equation (3) are calculated as Ica (t) = 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) where 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 (ω1t-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 Then, 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 / 2μn 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 / 2μn 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 equation (8), the constant torque term (second term) The term of torque variation (first term) due to the influence of the magnetic field of the inner magnet is added.

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)) Here, using the formula of 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 / 2μn 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)) = μIm 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 A phase difference current of γ flows through 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) however, Ic: amplitude gamma: the shift of the phase.

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 / 2μn 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 / 2μn 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 / 2μn 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 / 2μn Ic sin (-α-γ) + 1 / 2μn 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) For example, it has a form in which a term of torque variation (first term) due to the influence of the magnetic field of the inner magnet is added to a term of constant torque (second term).

<1-4> When Both Outer Rotating Magnetic Field and Inner Rotating Magnetic Field Are Applied In order to supply currents synchronized with the outer magnet and the inner magnet to the coil, the above Ica (t), Icb (t), Icc ( t) is 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 According to the expression (3), the motor is turned to a constant torque corresponding to the rotational 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 / 2sin (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 / 2μn 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 / 2μn 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 / 2μn 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 depending on _{Ic sin ((ω 2 -ω 1} ) t-α-β) + 3 / 2μn Ic 2 3sin (α + γ) ... (15) (15) rotating the phase difference also with respect to the inner magnet (alpha + gamma) This is a form in which the rotation fluctuation is superimposed on the constant torque.

<1-5> Conclusion The above (8), (9), (11), (12), (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 the outer rotating magnetic field and the inner rotating magnetic field are given, f ₁ = Im ₁ (μIm ₂ sin (ω ₂ t + α-ω ₁ t) + μn (Ic (3 / 2sin (β)) + Ic ₂ (3/2 sin ((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 expressions 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), (15)
Only the third term on the right side of the equation is a fixed term (constant value), and a 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, no rotational torque is generated depending on the term other than the fixed term. In other words, when a current flows through the stator coil in synchronization with the outer magnet, a rotational torque is generated only in the outer magnet, and when a current flows through the stator coil in synchronization with the inner magnet, a rotational torque is generated only in the inner magnet. When a current is supplied to the stator coil in synchronization with each of the inner magnets, a rotational torque is generated in both magnets.

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

<1-6> Suppression of Torque Fluctuation On the other hand, in the equation including the fixed term, the remaining terms 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 corresponding to the phase difference (ω ₁ −ω ₂ ) between the two magnets due to the term causes the rotation of the outer magnet, and the first term on the right side of the equation (12), (1
According to the first and second terms on the right side of 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 that the torque fluctuation is suppressed when both the outer rotating magnetic field and the inner rotating magnetic field are applied. From the above equation (14), f ₁ = μIm ₁ Im ₂ sin (ω ₂ t + α−ω ₁ t) + Icμn Im ₁ Ic (3 / 2sin
(β)) + because Ic is a _{2 Im 1 3/2 sin ((ω} 1 -ω 2) t + γ), put the f ₁ in the following manner.

[0082] _{f 1 = A + IcC + 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, Ic = (C1-a- Ic 2 V) / C that f be added modulation
₁ = C1 (constant), and the torque fluctuation is eliminated from the rotation of the outer magnet.

Similarly, according to the above equation (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 μm ₂ n sin (α + γ), f ₂ is set as follows.

F ₂ = −A + IcD + 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, f ₂ = C 2 (constant), and the torque fluctuation is eliminated from the rotation of the inner magnet.

[0085] Thus, in both of the magnet to a constant rotation may be solved following simultaneous binary equations Ic, the Ic _2.

[0086] _{C1 = A + IcC + Ic 2} V ... (18) C2 = -A + IcD + Ic 2 E ... (19) <2> N (2 (2p) -2p) Basic <2-1> 10 Let us consider a case where the ratio of the number of magnetic poles is 2: 1 (the number of magnetic poles of the outer magnet is 4 and the number of magnetic poles of the inner magnet is 2 in FIG. 10).

When each magnet is replaced with 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)

The magnetic field generated by the stator coil is calculated separately for the outer rotating magnetic field and the inner rotating magnetic field. Therefore, the coils are arranged as shown in FIG. The magnetic flux densities Bc ₁ and Bc ₂ are calculated as follows: 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) and Icc (t),
Icd (t), Ice (t) and Icf (t) are also currents that are 120 degrees out of phase.

FIG. 11 shows a model change in the magnetic flux densities B ₁ , B ₂ , Bc ₁ , and Bc ₂ .

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) when the torque applied to the outer magnet m ₁ and _{τ 1, τ 1 = f 1} × r 1 (r 1 is the radius) . In FIG. 10, since the generated torques are not equal symmetrically with respect to the diameter, the entire circumference is considered. Flows through four equivalent direct current round, the sum of the forces acting on these four current 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) Equation (26) shows that the torque acting on the outer magnet can be controlled by the exciting current of the coils a, b, and c. It also shows that it is not affected by the exciting current of the coils d, e, and f.

Next, the torque acting on the inner magnet m ₂ is given by τ ₂
Then, τ ₁₂ = f ₂ × r ₂ (r ₂ is a radius). Since two equivalent direct current flows in 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) = μIm 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))) = 2μIm 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 equation (27), the torque acting on the inner magnet can be controlled by the exciting current of coils d, e, and f. , And that they are 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 applied to the coils a, b, and c in accordance with the outer magnet. That is, the above three-phase alternating currents 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-4) + sin (4ω ₁ t-2β-4π / 3) + sin (4ω ₁ t-2β-4π / 3) +3 sin (2β)) = 6μIm ₁ n Ic sin (2β)… (29) Equation (29) indicates that the torque of the outer magnet changes according to the phase difference (β). Therefore, it is understood that the rotation angle of the outer magnet is measured, and the excitation current is supplied to the coils a, b, and c by shifting the phase by β.

<2-3> When an inner rotating magnetic field is applied Since a phase difference current of γ flows through the coils d, e, and f in accordance with the outer magnet, Icd (t), Ice (t), Icf (t) 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).

These are substituted into equation (27) to calculate f ₂ .

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) According to equation (31), This indicates 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, and f. It is sufficient to supply the exciting current. It is seen.

<2-4> Conclusion Equation (29) applies the current to the outer magnet only when the current flows through the stator coil in synchronization with the outer magnet, and Equation (31) applies the current to the stator coil in synchronization with the inner magnet. , A rotational torque is generated only in the inner magnet. Since each magnetic field corresponds only to each phase current, it was not calculated, but when a current is applied to the stator coil in synchronization with the outer magnet and the inner magnet, a rotating torque is generated in both magnets .

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

<2-5> Setting of Current Flowing in Stator Coil In FIG. 10, for theoretical calculation, a dedicated coil for generating an outer rotating magnetic field and a dedicated coil for generating an inner rotating magnetic field were considered. Now, as shown in FIG.
Consider sharing a coil. In FIG. 10, coils a and d, coils b and f , coils c and e, coils a and d , coils b and f, and coils c and e can be summarized. Therefore,
When comparing the coils of FIGS. 10 and 12 with each other, the composite currents I _{1 to} I ₁₂ flowing through the coils 1 to 12 of FIG. 12 are as follows: I ₁ = Ia + Id I ₂ = Ic I ₃ = Ib + If I ₄ = Ia I _{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 = it can be seen that if Ib.

In this case, the load of the coil 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
Therefore, it is considered that the load is distributed to the remaining coils to form an inner rotating magnetic field because the current is larger than the remaining coils through which the respective currents flow.

For example, comparing FIG. 2 with FIG. 1, FIG.
In FIG. 2, the portions corresponding to 1, 1 , 2 , and 2 are a, a , c , c of the outer peripheral coil and d, d of the inner peripheral coil. In this case, consider a state in which the phases of coils d and d are equivalently shifted,
Assuming that the shifted coils are new coils d ′ and d ′, half of the current Id ′ flowing through the coil d ′ is a coil a ′ .
c and half of the current Id ' flowing through the coil d '
Assign to a and c . The rest is the same.

In this way, 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 shifted from the coils e and f.

[0107] Further _{consider, 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. In other words, 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 currents as shown in FIG. 13, so that the 12-phase alternating currents form an inner rotating magnetic field. You just have to do it.

<2-6> Case of Giving Inner Rotating Magnetic Field with 12-Phase AC <2-6-1> Considering that inner rotating magnetic field is created with 12-phase AC, the magnetic flux density Bc _{2 at} this time is as follows. become.

[0109] _{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 entire 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 equation (26) when an inner rotating magnetic field is created by three-phase alternating current.

On the other hand, when f ₂ is calculated, it becomes as follows.

F ₂ = Im ₂ × B (θ = ω ₂ t + α) −Im ₂ × B (θ = ω ₂ t + π + α) = μIm 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))) = 2μIm 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> Inner rotation 12-phase AC when the providing a magnetic field Ic _i (t) ~Ic _xii the _{(t) 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).

By substituting equations (36a) to (36l) into equation (35), f ₂
Is calculated.

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 f ₂ = ₂ μIm ₂ n Ic ₂ (t) using the formula cos (a) sin (b) = 1/2 (sin (a + b) −sin (ab)). ) (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 ( + 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)) = μIm 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 magnetic field 12 Ai交In obtained when given this (3
Comparing equation (7) with equation (31) obtained when the inner rotating magnetic field is given by three-phase alternating current, equation (37) has a fixed term (last term) of 4 compared to equation (31). Doubled. That is,
If the driving current of the inner magnet is 12-phase alternating current (Ii to Ixii), four times as much driving force can be obtained as when the driving current of the inner magnet is 3 phase alternating current. Conversely, this means that generating the same driving force on the inner magnet requires only one-fourth of the inner driving current in three phases.

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

In this case, the magnetic flux densities B ₁ and B ₂ generated in the outer and inner magnets 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 generated by the stator coil is also calculated separately,
Magnetic flux density due to stator coils for outer and inner magnets
Bc1, the _{Bc2, 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 .

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 (Icd (t) sin (θ) + Ice (t) sin (θ-2π / 3) + Icf (t) sin (θ-4π / 3)) (45) Since the torque τ ₁ acting on the outer magnet m ₁ is generated in line symmetry with respect to the diameter, a force corresponding to half the circumference of f ₁ Then, τ ₁ = 2f ₁ × r ₁ (r ₁ is a radius). Since the three equivalent direct current to half flows, the sum of the forces acting on the three current 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)) = 3μIm 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 equation (46), when the outer magnet is approximated by a sine wave, the torque acting on the outer magnet is controlled by the exciting current of coils a, b, and c. Indicates that you can do it. It also shows that it is not affected by the exciting current of the coils d, e, and f.

Next, since the torque τ ₂ acting on the inner magnet m ₂ is also generated symmetrically with respect to the diameter, τ ₂ = 2f ₂ × r ₂ where f ₂ is the force for one half circumference. Since one equivalent DC current flows through the half, the force acting on the 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 equation (47), the effect of the magnetic field other than the calculated magnetic field on the rotation of the inner magnet (Relative phase angle is 2π / 3, 4π / 3.) 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, φ3 = ωt + 9π / 6.

Considering the respective effects, the magnetic field at 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 having a crossing angle of every 120 degrees cancel each other out on the inner coil. That is,
The number of poles of the outer magnet does not affect the inner magnet. Similarly, the magnetic field generated by the outer coil becomes zero in total. 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 the outer rotating magnetic field and the inner rotating magnetic field are applied, the same as the three-phase alternating currents Ica (t), Icb (t), and Icc (t) described above. Phase exchange
Icd (t), Ice (t) and Icf (t) are expressed as 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), Ic ₂ (t), which is a function of time, is set to enable amplitude modulation.

Formulas (49a) to (49c) are replaced by formulas (46) and (49a) to
(49c) and Equations (50a) to (50c) are substituted into Equation (47) to obtain f
_1, to calculate the 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) 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 / 2μIm 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 / 2μIm 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 of 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) where f ₂ is, as seen from equation (48),
When there is no influence of the magnetic field generated by the outer magnet and the outer coil, f ₂ = 3/2 n Ic ₂ (t) sin (γ + α)) (53), and the motor 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, in equation (52), 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) where C: constant f ₂ = C, and driving with a constant torque becomes possible. That is, when the ratio of the number of magnetic poles is 3: 1 according to the equation (52), it means that the rotation of the inner magnet is slightly affected by the outer magnet. More precisely, the phase difference (ω _1-
ω ₂ ), a constant torque fluctuation occurs in the rotation of the inner magnet. This is shown in FIG. When a rectangular wave model is used, the influence of magnetic interference between the outer magnet and the inner magnet is remarkably exhibited. Now, considering state A, state B is better than state B.
Is more stable, a torque is generated to shift to the state B. This torque becomes intermittent torque, and the phase difference
(ω ₁ -ω ₂ ). In addition, since the effect of the distance between the coils and the complete sine wave cannot be realized in reality, it may not be possible to completely cancel the effect of the outer magnet, and the most extreme case in this case is this. It is expressed by equation (52).

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

<3-3> Conclusion According to the equations (51) and (52), when a current is applied to the stator coil in synchronization with the outer magnet and the inner magnet, the rotational torque is applied to both magnets. It can be seen that this occurs. Although the calculation was not performed, when the current was passed through the stator coil in synchronization with the outer magnet, the rotation torque was applied only to the outer magnet, and when the current was passed through the stator coil in synchronization with the inner magnet, the rotation torque was applied only to the inner magnet. Needless to say. From this, it was proved that even when the ratio of the number of magnetic poles was 3: 1, it was possible to function as a rotating electric machine.

<3-4> Current Setting Consider sharing the outer peripheral side and inner peripheral side coils shown in FIG. 14 as shown in FIG. In FIG. 14, coils a and d, coils a and f , coils a and e, coils a and d , coils a and f, and coils a and e may be put together. the composite current supplied _{to, 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 _{8 = Ic I 9 = Ib I} 18 = I 9 = or it can be seen that if Ib. In other words, a combination having a magnetic pole number ratio of 3: 1 can be represented by a nine-phase current.
This means that in terms of contrast with the combination with a magnetic pole ratio of 2: 1, a combination with a magnetic pole ratio of 3: 1 requires 18-phase alternating current, but a combination with a magnetic pole ratio of 3: 1. Only in the case of, since the phase is inverted in half the circumference, it can be represented by the AC of 9 phases, which is half of the 18 phases.

However, since the load on the coils 1 , 4 , 7, 1 , 4 , and 7 becomes large, considering that the remaining coils are 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 if _{+ 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.

The currents I _i to generate the inner rotating magnetic field
FIG. 18 shows the positional relationship between I _ix and I _i to I _ix .

<3-5> When an inner rotating magnetic field is applied by 9-phase AC <3-5-1> Considering that an inner rotating magnetic field is created by 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 ) + Ic
c (t) sin (3ω ₁ t−4π / 3)) (57), which is the same as the above equation (46) obtained when the inner rotating magnetic field is given by three-phase alternating current.

On the other hand, when f ₂ is calculated, it becomes as follows.

F ₂ = Im ₂ × B (θ = ω ₂ t + α) = μIm ₂ (Im ₁ sin (3ω ₁ t-3ω ₂ t-3α) + Im2 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 rotating magnetic field and inner rotating magnetic field are given Phase exchange 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) is _{Icc (t) = Ic 1 cos} (3ω 1 t-3β-4π / 3) ... (59c), 9 -phase AC of the _{Ic i (t) ~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-6π / 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)

[0140] (59a) to Expression and (59c) and formula (60a) formula - (60i) Equation (58) are substituted into equation to calculate the 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))) Here, using the formula of cos (a) sin (b) = 1/2 (sin (a + b) -sin (ab)), f ₂ = 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 are canceled out in consideration of other phases, as seen from equation (48), as in the case of three-phase alternating current.

On the other hand, when this equation (61) obtained when the inner rotating magnetic field is given by the nine-phase alternating current is compared with the above equation (52) obtained when the inner rotating magnetic field is given by the three-phase alternating current, (61)
In the equation, the fixed term (last term) is three times that of the equation (52). That is, the drive current of the inner magnet is changed to 9-phase alternating current (I
_{If i to Iix} ), an electromagnetic force (drive torque) that is three times that of the case where the drive current of the inner magnet is a three-phase alternating current can be obtained. Conversely, this means that the drive current may be よ い to generate the same drive torque for the inner magnet.

This completes the theoretical analysis.

Next, FIGS. 19 to 25 show the fifth to eleventh embodiments. These also have the rotors 3 and 4 arranged inside and outside the stator similarly to the above-described four embodiments. However, FIGS. 19 and 20 show that the ratio of the number of magnetic poles is 2: 1,
21 shows a magnetic pole ratio of 2: 3, FIG. 22 shows a magnetic pole ratio of 4: 3, FIGS. 23 and 25 show a magnetic pole ratio of 2: 1, and FIG. 24 shows a magnetic pole ratio.
9: 1 combination. In summary, the present invention is not limited to the case where the number of magnetic poles of the outer magnet is larger than the number of magnetic poles of the inner magnet, and may be the case where the number of magnetic poles of the outer magnet is smaller than the number of magnetic poles of the inner magnet. In addition, even if the rotor is expanded for one round as described in each of the first to fourth embodiments and a plurality of rotors are connected to form a cylindrical shape, the rotor can be handled in the same manner as before the expansion.

The arrangement of the stator and the two rotors may be basically any arrangement. For example, FIG.
6 has an intermediate rotor 42 and an inner rotor 43 inside the stator 41.
It has two rotors. In this case, if the intermediate rotor 42 is covered with an iron frame in the same manner 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. As described above, an advantage of disposing the stator at the outermost or innermost position is that cooling is facilitated when the coils of the stator need to be cooled.

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

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

In the embodiment, the electric motor has a radial gap type (a gap between the rotor and the stator in the radial direction). However, an axial gap type (a gap between the rotor and the stator in the axial direction) has been described. The present invention can also be applied to

[Brief description of the drawings]

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

FIG. 2 is a schematic cross-sectional view of a rotating electric machine main 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 rotating electric machine main body according to a second embodiment.

FIG. 6 is a schematic sectional view of a rotating electric machine main body according to a third embodiment.

FIG. 7 is a schematic sectional view of a rotary electric machine main body according to a fourth embodiment.

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

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

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

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

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

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

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

FIG. 15 is a model diagram showing a change 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 referred to when considering an N (3 (2p) -2p) basic form.

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

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

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

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

FIG. 22 is a schematic sectional view of a rotating electric machine main body according to an eighth embodiment;

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

FIG. 24 is a schematic sectional view of a rotary electric machine main body according to a tenth embodiment.

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

FIG. 26 is a schematic sectional view of a rotary electric machine main body according to a twelfth embodiment.

[Explanation of symbols]

 2 Stator 3 Outer rotor 4 Inner rotor 6 Coil

Claims (6)

[Claims] The present invention comprises two rotors and one stator having a three-layer structure on the same axis, a single coil formed on the stator, and the single coil having the same number as the number of rotors. A rotating electric machine characterized by flowing a composite current so as to generate a rotating magnetic field. 2. The rotating electric machine according to claim 1, wherein a rotational phase of each rotor is detected, and the composite current is controlled according to each detected rotational phase. 3. The rotating electric machine according to claim 2, wherein the means for flowing the composite current through the single coil is an inverter. 4. The rotating electric machine according to claim 1, wherein said rotor is constituted by a permanent magnet. 5. The rotor according to claim 1, wherein a rotor is arranged outside and inside the cylindrical stator at a predetermined interval.
4. The rotating electric machine according to any one of items 1 to 3. 6. When the stator is arranged at the outermost or innermost position, the magnetic flux reaches the rotor farther from the stator through the rotor closer to the stator. Item 4. The rotating electric machine according to any one of Items 1 to 3.