JPH11275828A - Rotatary machine - Google Patents

Abstract

PROBLEM TO BE SOLVED: To prevent the magnetic pole of a rotor and a revolving magnetic field produced for driving the rotor from affecting the rotation of the other rotor, when two rotors are provided. SOLUTION: Having a body 1, this rotating machine is constituted of two rotors 3, 4 and a stator 2 structured in three layers on the same shaft. Then, a single coil 6 is formed on the stator 2 and compound currents are caused 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. In this case, the ratio of the number of magnetic poles of the two rotors 3, 4 is expressed as a combination of K:L (where K is an even number, and L is an odd number).

Description

DETAILED DESCRIPTION OF THE INVENTION

[0001]

The present invention relates to a rotating electric machine.

[0002]

[Prior Art] Synchronous motors with the same rated torque
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.

Therefore, it is conceivable to prevent a loss due to a current by forming a single coil in order to share the coil and supplying a composite current to the coil so as to generate a plurality of rotating magnetic fields.

For example, as shown in FIG. 1, rotors 3 and 4 made of permanent magnets are arranged at a predetermined gap on the outer peripheral side and the inner peripheral side of a cylindrical stator 2. A single coil 6 is formed on the circumference.

In this case, depending on the ratio of the number of magnetic poles between the outer rotor 3 and the inner rotor 4, the rotation of the outer rotor 3 may be affected by the inner rotor 4 (or the inner rotating magnetic field) to cause a torque fluctuation, or vice versa. It is conceivable that the rotation of the inner rotor 4 may undergo torque fluctuations under the influence of the outer rotor 3 (or the outer rotating magnetic field). If such torque fluctuations occur, the motor cannot be operated at a constant rotation.

Therefore, the present invention is made to drive the magnetic poles of one of the rotors and the other by setting the ratio of the number of magnetic poles of the two rotors to a combination of K: L (K is an even number and L is an odd number). An object is to prevent a rotating magnetic field from affecting the rotation of another rotor.

[0010]

According to a first aspect of the present invention, two rotors and one stator are formed in a three-layer structure on the same axis, and a single coil is formed on the stator.
In a rotating electric machine in which a composite current flows so as to generate the same number of rotating magnetic fields as the number of rotors in this single coil, the ratio of the number of magnetic poles of the two rotors is K: L (K is an even number,
L is an odd number).

According to a second aspect, in the first aspect, the rotor drive current on the side of the composite current having the smaller number of magnetic poles is 2K.
× 3-phase alternating current.

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

[0013]

According to the first aspect of the present invention, the magnetic poles of one of the rotors and the rotating magnetic field generated for driving the rotor do not affect the rotation of the other rotor. It becomes possible to drive at

When the rotor drive current on the side with the smaller number of magnetic poles is a three-phase alternating current, the current concentrates on a specific coil. According to the second invention, a 2K × 3 phase alternating current (for example, Since K is 2 and 12-phase alternating current), current does not concentrate on a specific coil, thereby suppressing copper loss.

According to the theoretical analysis, the drive current for generating the same drive torque on the rotor with the smaller number of magnetic poles is smaller than that when the rotor drive current on the side with the smaller number of magnetic poles is three-phase AC. When the current is, for example, K = 2, 1/4 is sufficient, so that the current rectifier can be further miniaturized.

Also in the third invention, the current is not concentrated on a specific coil, thereby suppressing copper loss.

[0017]

FIG. 1 is a sectional view of a rotary electric machine main body 1. FIG. In the figure, rotors 3 and 4 are arranged at predetermined intervals 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 S pole and another half circumference of N pole, while 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.

By arranging the magnetic poles of the rotors 3 and 4 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 as a component, the meaning is different.

The following composite currents I _{1 to} I ₁₂ flow through 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 flow a current (three-phase alternating current) for generating a rotating magnetic field for 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 only need to flow through the 12 coils.

[0029] _{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 that, for comparison with FIG. 1, different rotating magnetic fields are applied to the inner and outer peripheral sides of the stator 2 for each rotor. 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 spite of 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, as described later,
Proven by theoretical analysis.

The above Id, If, and Ie current settings are based on the inner rotor 4
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.

Since the ratio of the number of magnetic poles between the outer rotor 3 and the inner rotor 4 (hereinafter referred to simply as the ratio of the number of magnetic poles) is a combination of 2: 1, the magnetic poles of one of the rotors and the rotating magnetic field generated for driving the rotor are combined. However, it does not affect the rotation of the other rotors, which allows any rotor to operate at a constant rotation.

On the other hand, for comparison, FIGS. 6 and 7 show combinations having a magnetic pole ratio of 3: 1. The pole ratio is
In the case of the 3: 1 combination, the magnetic pole ratio is different from that of the 2: 1 combination, and the magnet of the outer rotor 3 and the inner rotor 4 are different.
Some effect occurs between the magnets. That is, the outer rotor
The magnet 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 applied to the outer rotor 3. It rotates with fluctuations. This point is proved by the theoretical analysis described later.

Incidentally, the interference of the outer rotor 3 with the rotation of the inner rotor 4, that is, the constant torque fluctuation generated 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.

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 production steps is reduced.

In the first and second embodiments, when the ratio of the number of magnetic poles is a combination of 2: 1, the magnetic pole of one rotor and the rotating magnetic field generated for driving the rotor are different from each other. 6 and 7 for comparison, when the ratio of the number of magnetic poles is a combination of 3: 1, the magnetic poles of the outer rotor and the rotor are driven. It has been described that the rotating magnetic field created for this influences the rotation of the inner rotor. The theoretical analysis on which this is based will be explained in the following sections.

<1> N (2p-2p) basic type This is the case of a combination in which 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 an 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.

Assuming that 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, 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), a constant torque term (second term) The term of torque fluctuation (the first term) due to the influence of the magnetic field of the inner magnet is added to the above.

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, respectively, 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 / 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 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 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 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.

[0073] _{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, from 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.

[0076] Thus, in both of the magnet to a constant rotation may be solved following simultaneous binary equations Ic, the Ic _2.

[0077] _{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 becomes 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 generated 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 outer and inner stator magnet coils are used. 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) Assuming that the torque acting on the outer magnet m ₁ is τ ₁ , τ ₁ = f ₁
× r ₁ (r ₁ is a 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) indicates 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) + 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) 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> Coil when inner rotating magnetic field is applied
To allow a phase difference current of γ to flow through d, e, and f according to the outer magnet, Icd (t), Ice (t), and Icf (t) are calculated as follows: 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 coil 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 the outer rotating magnetic field and a dedicated coil for generating the 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 coils for passing the 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 other words, in other words, the coil is divided into six (see broken lines in FIGS. 1 and 2) so that three phases are formed per one pole of the inner rotor (the rotor having the smaller number of magnetic poles). In other words, each of the separated coils a and c (or each of the coils a and c ) has a drive current I of the inner rotor.
d '(or Id' ) is the number of coils separated by 2
To flow the current value (1/2) Id ' (or (1/2) Id ' ) divided by.

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.

[0099] 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> When an inner rotating magnetic field is applied by 12-phase alternating current <2-6-1> Considering that an inner rotating magnetic field is created by 12-phase alternating current, 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 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 is 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 the equations (36a) to (36l) into the 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 (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)) = 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> Inside rotation The obtained when given field in 12-phase AC (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.

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 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 in line symmetry with respect to the diameter, τ ₂ = 2f ₂ × r ₂ when f ₂ is a force corresponding to a half turn. 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 influence of the magnetic field other than the calculated magnetic field on the rotation of the inner magnet ( It can be seen that there are 2π / 3 and 4π / 3) in relative phase angles.In order 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 each effect, 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 outer and inner rotating magnetic fields are given: 3 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 by the equation (54), it is possible to cancel the constant torque fluctuation, and the inner magnet can be driven with the 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 and inner 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 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 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 are
FIG. 18 shows the positional relationship between I _ix and I _i to I _ix .

<3-5> Case of Giving 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 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 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 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)

[0132] (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 third to eighth embodiments. These also have the rotors 3 and 4 arranged inside and outside the stator similarly to the above-described four embodiments. 19 and 20, the magnetic pole ratio is 2: 1, FIG. 21 is magnetic pole ratio 2: 3, FIG. 22 is magnetic pole ratio 4: 3, and FIG.
3. FIG. 24 shows a combination in which the ratio of the number of magnetic poles is 2: 1. In summary, as long as the ratio of the number of magnetic poles is a combination of even: odd or odd: even, the number of magnetic poles of the outer magnet is not limited to the number of magnetic poles of the inner magnet. It may be less than the number. 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.
5 is a structure in which the intermediate rotor 42 and the inner rotor 43
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.

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

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

In the above embodiment, the electric machine has a radial gap type (a gap between the rotor and the stator in the radial direction), but an axial gap type (a gap between the rotor and the stator in the axial direction). 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 cross-sectional view of a rotating electrical machine main body in a case of a combination having a magnetic pole ratio of 3: 1 shown for comparison.

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

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 rotating electric machine main body according to a third embodiment.

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

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

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

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

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

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

[Explanation of symbols]

 2 Stator 3 Outer rotor 4 Inner rotor 6 Coil

Claims (3)

[Claims] 1. A three-layer structure comprising two rotors and one stator on the same axis, a single coil formed on the stator, and the single coil having the same number as the number of rotors. In a rotating electric machine in which a composite current is caused to flow so as to generate a rotating magnetic field, the ratio of the number of magnetic poles of the two rotors is a combination of K: L (K is an even number and L is an odd number). Electric machine. 2. The rotating electric machine according to claim 1, wherein the rotor drive current on the side of the composite current having the smaller number of magnetic poles is a 2K × 3 phase alternating current. 3. The method according to claim 1, wherein the coil is divided into three phases per pole of a rotor having a smaller number of magnetic poles, and a rotor drive current having a smaller number of magnetic poles is divided into each of the divided coils. 2. The rotating electric machine according to claim 1, wherein a current value divided by the number of coils is applied.