JPH06133582A - Inverter device for driving induction motor - Google Patents

Abstract

PURPOSE:To suppress the ripple of a torque generated by an induction motor to a minimum value by operating a DC voltage generating section and inverter section in accordance with a value found from a specific calculating formula. CONSTITUTION:A control section decides voltage vector time widths TW1 and TW2 to be impressed based on a formula and, at the same time, decides the DC stage voltage Vdc1 and Vdc2 of an inverter based on formulae, Vdc1= SQR(3/2).omega1.phi2/cos(pi/6+theta) and Vdc2=SQR(3/2).omega1.phi2/cos(pi/6-theta) (where, phi2, omega1 and theta respectively represent a secondary cross flux, the angular velocity of the secondary cross flux phi2, and the angle indicating the position of the secondary cross flux phi2) and operates a DC voltage generating section and inverter section in accordance with the decided values so that the rotating speed and generating torque of an induction motor can always coincide with commanded values.

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention relates to a device for variable speed control of an induction motor using an inverter, and more particularly, it controls the speed and generated torque of the induction motor instantaneously and instantaneously so that the induction motor always operates according to a command value. The present invention relates to an inverter device for driving an induction motor.

[0002]

2. Description of the Related Art There are various kinds of inverter devices for controlling an induction motor at a variable speed. In a conventional device, the voltage of a DC stage of an inverter unit is fixed as a DC unstable voltage obtained as a rectifier output. Value or
By keeping the voltage corresponding to the desired motor speed,
I was trying to get the required inverter output frequency.

In such a device, when the induction motor is to be rotated at an arbitrary speed, the optimum one is selected from the voltage vectors (six kinds other than the zero vector) and output, but the voltage vector is always There is a problem that the vector continues for a long time in order to keep outputting, and the secondary interlinkage magnetic flux passes over the position originally required, and excessive torque is generated.

In order to solve such a problem, a zero vector is inserted to add a standby state of secondary interlinkage magnetic flux.
In this case, the generated torque has a drawback that the generated torque may drop below the command torque, causing ripples in the generated torque, which makes it impossible to control the speed and torque with high accuracy.

[0005]

In the case of the conventional device described above, the average speed and torque can be maintained, but the instantaneous speed and torque do not match the desired values. . That is, the speed and the like become larger than the desired value at a certain time, but become smaller at the next time, and the actual time integral value of the vector does not rotate at a uniform speed. This causes the generated torque ripple and speed ripple without any correction.
This is an important issue in controlling the induction motor's variable speed,
In particular, when it is desired to control the rotor angular velocity ω _m at a low speed, the influence is great.

A method of adjusting the temporal position of the voltage vector so as to minimize the torque ripple evaluation function and a method of increasing the switching frequency of the constituent elements of the inverter to reduce the torque ripple are also available. Although proposed, none of them is a fundamental measure. The present invention has been made in view of this problem, and it is possible to control the speed and the generated torque of the induction motor at each instant according to the command value, and to suppress the ripple of the generated torque to the minimum value. An object is to provide an inverter device for driving an electric motor.

[0007]

In order to achieve the above-mentioned object, an induction motor driving inverter device according to the present invention supplies an inverter section for controlling the induction motor at a variable speed and a DC stage voltage to this inverter section. A DC voltage generation unit, an operation state detection unit that detects the primary current of the induction motor, the rotor speed, and the like, a signal from the operation state detection unit, and a command value such as an externally supplied rotor speed, Based on [Equation 1] (Equation 9) (Equation 10), the voltage vector time widths T _w1 and T _w2 and the DC stage voltages V _dc1 and V _{dC 2} of the inverter are determined, and the DC voltage generating unit is determined according to these values. And a control unit for operating the inverter unit.

[0008]

The controller determines the voltage vector time widths T _w1 and T _w2 based on [Equation 1], and at the same time, (Equation 9) and (Equation 10)
Based on the DC stage voltage V _dc1 , V _{dC of} the inverter
₂ is determined, and the DC voltage generator and the inverter are operated according to these determined values, so that the speed and the generated torque of the induction motor are always in accordance with the command values.

Hereinafter, the operation of this control unit will be shown [Equation 1],
The theoretical basis of (Equation 9) and (Equation 10) will be described. When a three-phase induction motor is driven by a three-phase inverter, the voltage vectors that can be applied to the motor except for the multiple type are V ₁ , V ₂ , V ₃ , V ₄ , V ₅ , and V ₆ except for the zero vector. Limited to types.

FIG. 2 shows the six types of voltage vectors V ₁ to
V ₆ is illustrated. For example, voltage vector V
₁ indicates that the DC stage voltage of the inverter is applied only to the U phase when the phases of the induction motor are U phase, V phase, and W phase. , 0)
], A voltage vector that causes a secondary magnetic flux φ ₂ to flow in the direction shown by flowing a current from the U phase to the V phase and the W phase.

The present invention intends to realize an inverter system in which torque ripple is minimized by optimally designing the type of vector and its application time using these six types of voltage vectors. Therefore, next, a method of selecting a voltage vector and a method of determining the application time of the voltage vector will be described. [Selection of voltage vector and determination of application time of the voltage vector] When the secondary interlinking magnetic flux φ ₂ is rotated to generate the power running torque (or regenerative torque) in the induction motor, (1) The secondary flux linkage φ ₂ is always on a circular orbit, and (2) the secondary flux linkage φ ₂ is always rotated at a constant speed. However, it is assumed that the load torque and the rotation speed of the electric motor are constant.

Now, in order to satisfy the above conditions by applying a computer-controlled inverter, the secondary interlinkage magnetic flux φ ₂ is on a constant circular orbit at every sample time T _S , and the sample time is The distance traveled during T _S (or the central angle spanned by this travel) must be constant. Therefore, in order to move the secondary interlinkage magnetic flux φ ₂ from one point on the circular orbit to the next point after one sampling time T _S , two adjacent magnetic fluxes corresponding to the current position of the secondary interlinkage magnetic flux φ ₂ are adjacent to each other. It _suffices to appropriately select the two voltage vectors V _x1 and V _x2 and control the application times T _w1 and T _w2 of the selected voltage vectors to appropriate values to satisfy the above two conditions. .

FIG. 3 shows that the positions of the secondary interlinkage magnetic flux φ ₂ before and after one sample time T _S are P ₁ and P ₂ , and the secondary interlinkage magnetic flux φ ₂ is moved from the position of P _{1 to} the position of P _2. If you want to
The voltage vectors V _x1 and V _x2 are appropriately selected, the voltage vector V _x1 is applied for the time T _W1 , and then the voltage vector V _x2.
Is applied for a time T _W2 . That is, T
_S = T _W1 + T _W2 , and the vector T _W1 V _x1 and the vector T
It is shown that the vector sum with _W2 V _x2 is the vector P ₁ P ₂ = Δλ.

The d and q coordinate values of the voltage vector V _xi (where V _xi is any of V _{1 to} V ₆ ) in the stator coordinate system d and q (see FIG. 2) are represented as V _dxi and V _qxi. , The d and q coordinate values of the voltage vector time integration value Δλ are Δλ _d and Δλ _q
If expressed, Δλ _d = V _dX1 · T _w1 + V _dx2 · T _w2 (Equation 1) Δλ _q = V _qx1 · T _w1 + V _qx2 · T _w2 (Equation 2) . By solving (Equation 1) and (Equation 2) and calculating the application times T _w1 and T _w2 of each voltage vector V _xi , the above [Equation 1] is obtained.

1 sample time T_SElectric power needed between
If the pressure vector time integration value Δλ is determined, the d and q coordinates of Δλ
Standard value Δλ_d, Δλ_qIs determined at each sample time,
In addition, Δλ (= vector P₁P₂) Two phases sandwiching
Adjacent voltage vector V_x1, V _x2Choose to
, The selected voltage vector V_x1, V_x2D and q coordinate values of
V_dx1, V_qx1And V_dx2, V_qx2Will also be decided
It

Therefore, the control unit controls the voltage vector time integration.
After determining the value Δλ, V_dx1, V_{qx 1}, V_dx2, V_qx2
Substituting the value of the above into [Equation 1] above, each voltage vector selected
Toru V_x1, V_x2Application time T_w1, T_w2To calculate
It [Determination of DC stage voltage] Next, the DC voltage of the inverter
A method of determining the tage voltage will be described with reference to FIG. Figure
In 4, the vector OP₁Is the secondary flux linkage φ₂Shows
And the vector P ₁P₂Is the secondary flux linkage φ₂A sun
Pull time T_SLater P₁P from the point₂Needed to move to a point
The voltage vector time product Δλ is shown. Also, vector
Le P₁P'represents the first voltage vector time product,
Cutle P₁P ″ represents the second voltage vector time product
It In the case of FIG. 4, the first voltage vector is V
₂Is selected and V is set as the second voltage vector.₃Is selected
ing.

[0017] Secondary interlinkage magnetic flux phi ₂ is the vector OP ₀ when the angle formed by the (consistent with the voltage vector V ₁ in FIG. 4) and theta, time product P ₁ P ^'and the voltage of the first voltage vector The angle formed by the vector time product P ₁ P ₂ is π / 6 + θ + δ
θ / 2. Here, δθ represents the displacement of the angle θ in one sample time T _S, but since the displacement δθ is almost zero, the direction of the voltage vector time product P ₁ P ₂ substantially coincides with the tangential direction at the point P ₁ . Therefore, the secondary flux linkage φ ₂
Consider the movement distance and movement speed in the tangential direction (circumferential direction) of.

According to the first voltage vector time product P ₁ P ',
The moving distance of the secondary interlinkage magnetic flux φ _{2 in} the circumferential direction is P ₁ P ^'
-It is given by COS (π / 6 + θ). And the first
Since the application time of the voltage vector is T _w1 , the velocity of the secondary interlinkage flux φ ₂ in the circumferential direction at the point P ₁ is P ₁ P ^′ · COS.
(π / 6 + θ) / T _w1 . On the other hand, the second voltage vector time product P ₁ P ^″ is π / 6 with the tangent to the circumference at the point P _1.
Since the angle is −θ, the moving distance in the circumferential direction of the secondary interlinkage magnetic flux φ ₂ is P ₁ P ^″ · COS (π / 6−θ),
The velocity in the circumferential direction at the point P _{1 of the} secondary flux linkage φ ₂ is P ₁ P
^″ · COS (π / 6−θ) / T _w2 .

By the way, when the DC stage voltage of the inverter when the first voltage vector is applied is V _dc1 and the DC stage voltage of the inverter when the second voltage vector is applied is V _dc2 , Vector p
₁ P ^′ and P ₁ P ^″ are respectively P ₁ P ^′ = SQR (2/3) · V _dc1 · T _w1 (Equation 3) P ₁ P ^″ = SQR (2/3) · V _dc2・ T _w2 (Equation 4)
Given in.

In order to prevent torque ripple, the circumferential velocity of the secondary interlinkage magnetic flux φ ₂ is different between when the first voltage vector is applied and when the second voltage vector is applied. Since they must match, P ₁ P ^' · COS (π / 6 + θ) / T _w1 = P ₁ P ^″ · COS (π / 6−θ) / T _w2 (Equation 5) Substituting (Equation 3) and (Equation 4) into this equation, SQR (2/3) * _Vdc1 * COS (π / 6 + θ) = SQR (2/3) * _Vdc2 * COS (π / 6-θ) ... (Equation 6) is obtained.

[0021] than the speed of the circumferential-direction twin given by both sides primary interlinkage magnetic flux phi ₂ of the equation (6), when the angular velocity of the secondary flux linkage phi ₂ and _{ω 1, ω 1 = SQR (} 2/3 ) ・ V _dc1・ COS (π / 6 + θ) / φ ₂ …… (Equation 7) ω ₁ = SQR (2/3) ・ V _dc2・ COS (π / 6−θ) / φ ₂ …… (Equation 8) Is asked. Therefore, if the angular velocity ω _{1 of the} secondary interlinkage magnetic flux φ ₂ is determined, the DC stage voltages V _dc1 and V _{dc2 of the} inverter are determined _.
Is V _dc1 = SQR (3/2) · ω ₁ · φ ₂ / COS (π / 6 + θ) (Equation 9) V _dc2 = SQR (3/2) · ω ₁ · φ ₂ / COS (π / 6-θ) (Equation 10) is given.

That is, the controller determines the voltage of the DC stage of the inverter according to the position of the secondary interlinkage magnetic flux φ ₂ (angle θ from the voltage vector V _{1 in} FIG. 4) (Equation 9),
By controlling to the value given by (Equation 10), the torque ripple of the induction motor is suppressed to the minimum value. [Others] By the way, the fact that the switching from the first voltage vector to the second vector is performed at the point P ′ in FIG. 4 means that when moving from the point P _{1 to the} point P ₂ , the secondary interlinkage magnetic flux φ ₂ Means deviating from the circumference. However, since the magnitude of the secondary interlinkage magnetic flux φ ₂ does not change instantaneously due to its time constant, it is necessary to apply the second voltage vector after applying the first voltage vector at the first sample time. , If the first voltage vector is applied after the second voltage vector is applied at the next sample time, the secondary interlinkage magnetic flux φ ₂ is moved on the circumference with almost no change. be able to. That is, if the order of applying the first voltage vector and the second voltage vector is reversed one after another, it is possible to generate almost no torque ripple.

In the above description, the angle θ representing the position of the secondary interlinkage magnetic flux is based on the V ₁ vector, but θ = −
In the range of π / 6 to θ = + π / 6, the V ₁ vector is used as the reference, but when θ exceeds + π / 6, the reference vector is V ₂
To V _{3 to} V ₆ depending on the value of θ.
Since the standard is shifted to, the above description can be applied as it is.

[0024]

The present invention will be described in more detail based on the following examples. FIG. 1 is a block diagram of an induction motor driving inverter device according to an embodiment of the present invention. This device includes a converter 1, a DC / DC converter 2, a brake circuit 3, an inverter 4, an induction motor 5, a rotary encoder 6, a DC stage voltage determination unit 7, an arithmetic processing unit 8, and an applied voltage. It is composed of a vector decision unit 9. The DC stage voltage determination unit 7, the arithmetic processing unit 8, and the applied voltage vector determination unit 9 are
As a whole, the control device 10 using a computer is configured.

In FIG. 1, a converter 1 is a circuit for converting a commercial AC power supply into a direct current, and a DC / DC converter 2 receives a direct current output of the converter 1 and receives an inverter 4
DC most suitable for the speed and load torque
It is a circuit that outputs a stage voltage. Brake circuit 3
Is a circuit for generating the regenerative torque and absorbing the braking force of the induction motor 5 when the induction motor 5 is requested to generate the regenerative torque by the load. It is for actually measuring the position or rotational movement distance.

The DC stage voltage determination unit 7 receives the speed command ω _m ^* of the electric motor applied from the outside, the slip angular speed command value ω _S ^* calculated by the arithmetic processing unit 8, and the secondary interlinkage magnetic flux φ _2.
And a signal from the rotary encoder 6 to determine the optimum values V _dc1 and V _dc2 of the DC stage voltage of the inverter 4. The arithmetic processing unit 8 determines the secondary flux linkage command φ.
₂ ^* , the speed command ω _m ^{* of} the induction motor, the signal from the rotary encoder 6, and the detected value i _{1 of the} primary current flowing in the induction motor 5, the command value φ of the secondary interlinkage magnetic flux. ₂ ^* , the current value φ ₂ of the secondary interlinkage magnetic flux, the generated torque command value τ ^{* of} the induction motor, the slip angular velocity command value ω _S ^* ,
The primary side current command value i ₁ ^* of the induction motor necessary to comply with these command values is calculated, and finally these are used to apply the vector voltage product Δλ ^{* of the} voltage applied to the induction motor at the next sampling time ^. Is a part for calculating.

The applied voltage vector determination unit 9 is a DC
After receiving the calculation results of the voltage determination unit 7 and the calculation processing unit 8,
Voltage vector V_xiHow long T_w1, T_w2, Induction
This is a part for determining whether to apply to the electric motor 5. More than
About the induction motor drive inverter device consisting of
The operation contents of each unit will be described below.

When the speed of the induction motor 5 is controlled, the speed command value ω _m ^* in the operation of the motor is determined by the control device 1.
Given to 0. On the other hand, the actual speed ω _m of the electric motor can be obtained by actually measuring the number of pulses transmitted by the rotary encoder 6 and calculating the number of pulses per unit time. Therefore, the speed command value ω _m ^* and the actual speed ω _m are The deviation ω _m ^* −ω _m can be calculated naturally. The output obtained by PI-amplifying this value is the torque required for the electric motor 5 at that time, and this becomes the torque command value τ ^* . When controlling the torque of the electric motor 5, the torque command value τ ^* is given to the control device.

In the arithmetic processing unit 8, this torque command value τ^*
Based on the slip angular velocity command value ω_S ^*To ω_S ^*= R₂
/ Φ₂ ²・ Τ^*… Calculated from (Equation 11). R here₂Is the secondary resistance of the induction motor 5.
It is a value. Calculated slip angular velocity command value ω_S ^*Than two
Next flux linkage φ₂Angular velocity command value ω₁ ^*Is ω₁ ^*= Ω_m ^*+ Ω_S ^*...... (Equation 12) is obtained, so each sample time T is calculated based on this value._Severy
Secondary flux of φ₂The position θ can be obtained.

On the other hand, the value φ _{2m (n-1) in} the rotor coordinate system of the secondary interlinkage magnetic flux φ _{2 (n-1)} at a certain sampling time _(n-1) is
Secondary interlinkage magnetic flux φ at the previous sampling time (n-2)
_{2 (n-2)} and (n-1) motor primary current point in time i _{1 (n-1)} and the value of the rotor coordinate system obtained from _{φ 2m (n-2),} i 1m (n-1)
Φ _{2m (n-1)} = T ₂ / (T ₂ + T _S ) ・ φ _{2m (n-2)} + M ・ T _S / (T ₂ + T _S ) ・ i _{1m (n-1)} …… ( It is calculated as in Expression 13). However, T ₂ is the secondary circuit time constant, M
Is the mutual inductance. In _order to use φ _{2m (n-1)} obtained by (Equation 13) for control, it is necessary to convert this into a value φ _{2 (n-1)} in the stator coordinate system. If the position θ of the secondary interlinkage magnetic flux φ ₂ is obtained by integrating the output, conversion from the rotor coordinate system to the stator coordinate system can be performed by a well-known method, so a detailed description will be omitted.

In this way, the secondary interlinkage magnetic flux φ _{2 (n-1) at time (n-1)} is obtained, while the torque command value τ described above is obtained.
^* And the command value φ ₂ ^* of the secondary interlinkage magnetic flux given to the arithmetic processing unit 8 from the outside by the calculation of [Equation 2] (n +
1) The motor primary current command value i ₁ ^* _{(n + 1) at the time point} is obtained.

[0032]

[Equation 2]

However, L₂Is the secondary inductance.
For example, the current time point is between the (n-1) time point and the n time point.
Then, the quantities required for control at the next time are n hours
Motor primary current i at the point₁The actual value of i_{1 (n)}(This value is
(Estimated from the detected value of the next current) and the secondary at time n
Interlink magnetic flux φ₂Actual value of φ_{2 (n)}And the power at (n + 1)
Motivation primary current command value i₁ ^* _{(n + 1)}And the secondary flux linkage φ₂
^* _{(n + 1)}And. The secondary flux linkage φ at time n₂of
Actual value φ _{2 (n)}Is i in (Equation 13)_{1m (n-1)}Instead of
Rini_{1m (n)}Using φ_{2m (n-2)}Instead of φ_{2m (n-1)}To
Use the secondary flux linkage φ_{2m (n)}And find this in the stator coordinates
It was converted to.

The above variables i _{1 (n)} , φ _{2 (n)} ,
i ₁ ^* _{(n + 1)} , φ ₂ ^* _{(n + 1)} , the primary voltage vector time integral value Δλ ^* _{(n) of the} motor required at time _n
Is calculated as in [Equation 3].

[0035]

[Equation 3]

However, Lσ = L ₁ -M ² / L ₂ and R
₁ is a primary resistance and L ₁ is a primary inductance. still,
The magnitude of φ ₂ ^* _{(n + 1)} is externally given to the control device 10 as a command value, and its angular velocity ω ₁ ^* is given by (Equation 12). The size and position of the next interlinkage magnetic flux φ ₂ ^* _{(n + 1)} can be easily obtained. The following [Equation 4] and [Equation 5] are basic equations that are the starting points for deriving [Equation 3], and p is a differential operator.

[0037]

[Equation 4]

[0038]

[Equation 5]

The above is the contents of the calculation in the calculation processing section 8. [Equation 3] expresses the time integral value of the primary voltage vector by converting it to a discrete value system, and gives a command value of the time integral value of the primary voltage vector required at the time point n.
Therefore, [Formula 3] is added to Δλ _d and Δλ _q in (Formula 1) and (Formula 2).
By substituting Δλ _d ^* _(n) and Δλ _q ^* _(n) , the time widths T _w1 and T _{w2 of the} voltage vector required at that time can be obtained from [Equation 1].

[0040] [Delta] [lambda] ^* of d, at the stage of Motoma' is q coordinates, this [Delta] [lambda] ^* also becomes clear whether the between which the neighbors are vectors in the six voltage vectors, applied voltage vector determining unit 9 Specifically determines the type of voltage vector V _xi to be applied. Further, the calculation in the applied voltage vector determination unit 9 requires the voltages V _dc1 and V _dc2 of the DC stage when each of the first voltage vector and the second voltage vector is applied to the motor. voltage determining unit 7, (equation 9) omega ₁ of the omega ₁ (equation 10) (equation 12)
By substituting ^* , the DC stage voltages V _dc1 and V _dc2 are calculated.

V _dc1 calculated in this way,
V _dc2 , T _w1 and T _w2 are given as commands to the DC / DC converter 2, and the inverter 4 is given a voltage vector V _xi to be generated and its time widths T _w1 and T _w2 as command values. Speed control is performed. When regenerative torque is required, these signals are also given to the brake circuit 3. In the case of a small power system, a brake resistor is installed to absorb the regenerative energy of the electric motor, and in the case of a large power system, the power is supplied to the converter 1. It is necessary to equip an electric power regeneration device to regenerate energy into a power source.

[0042] As mentioned above, the DC stage of the inverter
If the voltage is controlled according to (Equation 9) and (Equation 10), the
Minimizes torque ripple in the barter system
You can do it. However, with this method, one sample
Longing time T_sOf which, T_w1Time is the DC of the inverter
Stage voltage is V_dc1Control the remaining time T_w2(= T
_S-T_w1) Is V_dc2I have to control
The power supply design of the inverter is slightly complicated. There
The power supply design of the inverter can be simplified in consideration of
The method is described below.

Inverters given by (Equation 9) and (Equation 10)
DC stage voltage V_dc1, V _dc2Is the secondary interlinkage
Bundle φ₂Since the value varies depending on the position θ of,
Will have different values. However, the absolute value of θ is expressed as ABS (θ).
V_dC1≒ V_dc2≒ V_dc= SQR (3/2) ・ ω₁・ Φ₂/ COS (π / 6-ABS (θ)) (Equation 14) does not cause a large error. So 1 sun
Pull time T_sWhile the voltage of the DC stage is constant value V_dcTo
I will control it.

In the case of this simplification method,
DC stage voltage V of Maru inverter_dcOn the basis of,
From [Equation 1], the application time T of the first and second voltage vectors_w1
And T_w2And the calculated application time T_w1, T_w2
Is always 1 sample time T_STo not match
Become. The reason is that originally, from (Equation 9) and (Equation 10),
The DC stage voltage of the inverter is set to the first vector and the second vector.
The results were obtained after individually calculating the size of the kutor.
V in (Equation 3) and (Equation 4)_dx1, V_qx1, V_dx2, V_qx2
And apply time T _w1, T_w2And need to ask
Nevertheless, the DC stage voltage is set to the same value V_dcAs (expression
3) The application time is calculated from (Equation 4).

Further, when the simplified expression of (Equation 14) is used, the equation of (Equation 6) does not hold, and therefore, the secondary flux linkage of the first voltage vector time product P ₁ P ^' The circumferential angular velocity and the circumferential velocity of the secondary interlinkage magnetic flux of the second voltage vector time product P ₁ P ^″ do not match, and as a result, the slip angular velocity ω _s changes and the generated torque τ changes. A slight torque ripple may occur.

Therefore, by adopting the following method
Torque ripple is suppressed without using the zero vector.
I will control. (Equation 14), that is, SQR (3/2)
・ Ω₁・ Φ₂/ COS (π / 6-ABS (θ)) V_dc1=
V_dc2= V_dC Decide this V_dCBased on the value of
1] the first vector V calculated from_x1And the second vector V
_x2And apply them to T_{w1 S}, T_w2SAnd
Sum of calculated application times T_w1S+ T_w2SWhen 1 sample
Interval T _SAs mentioned above, it does not always match
Assuming that the current time is between (n-1) time point and n time point,
DC stage voltage V between (n + 1) points_dCSTo V_dcS= (T_w1S+ T_w2S) ・ V_dc/ T_S…… (Equation 15) If corrected, insert zero vector at each sample time
Torque ripple of the induction motor
It becomes possible to control.

As described above, according to the present invention, since it is not necessary to transmit the zero vector at each sample time T _S , including the case of the simplified method, the switching frequency of the inverter can be reduced to about 1/3. There is an advantage. Further, since the voltage of the DC stage of the inverter is applied over the entire sampling time, the deterioration of the characteristics due to the dead time for preventing the short circuit between the positive arm portion and the negative arm portion of the inverter is prevented. You can also

[0048]

As described above, in the induction motor driving inverter device according to the present invention, the control unit is represented by [Equation 1].
Since it operates based on (Equation 9) and (Equation 10), it follows the speed command and the torque command extremely faithfully, and realizes smooth control in which actual speed ripples and torque ripples are minimized.

[Brief description of drawings]

FIG. 1 is a block diagram of an inverter device for driving an induction motor that is an embodiment of the present invention.

FIG. 2 is a diagram showing voltage vectors for driving an induction motor.

FIG. 3 is a conceptual diagram showing a method of selecting a voltage vector when moving a secondary interlinkage magnetic flux φ ₂ .

FIG. 4 is a more detailed illustration of FIG.

[Explanation of symbols]

 1 Converter 2 DC / DC Converter 3 Brake Circuit 4 Inverter 5 Induction Motor 6 Rotary Encoder 10 Controller

Claims (1)

[Claims] 1. An inverter section for controlling the induction motor at a variable speed, a DC voltage generating section for supplying a DC stage voltage to the inverter section, and an operation state detecting section for detecting the primary current of the induction motor and the rotor speed. And the rotor speed command value supplied from the outside and the signal from the operation state detection unit, [Equation 1] (Equation 9) (Equation 1)
0), the voltage vector time widths T _w1 and T _w2 and the DC stage voltages V _dc1 and V _dC2 of the inverter are determined, and the DC voltage generating unit and the control unit for operating the inverter unit are provided according to these values. An inverter device for driving an induction motor, which is characterized in that [Equation 1] V _dc1 = SQR (3/2) ・ ω ₁ · φ ₂ / COS (π / 6 + θ) (Equation 9) V _dc2 = SQR (3/2) ・ ω ₁・ φ ₂ / COS (π / 6- θ) …… (Equation 10)