JPS6248479B2 - - Google Patents

Description

[Detailed description of the invention]

(Industrial Application Field) The present invention relates to a torque control device for an induction motor. (Prior Art) Conventionally, shunt-wound DC motors have been used in fields where quick response is required in position control devices, speed control devices, etc.
It has also been used extensively. The reason for this is that the output torque of a shunt-wound DC machine is proportional to the armature current, and as soon as the armature current flows, output torque is generated immediately, making it possible to realize a highly responsive control system.The control system also follows the linear automatic control theory. This is because it is possible to realize a control system with good quick response as intended by the designer. On the other hand, if induction motors, which are widely used as constant-speed motors, can be used in these control systems, they will have several advantages over DC motors.
For example, there are no brushes, which makes maintenance easier;
be robust, free of sparks and electrical noise;
Since there is no problem with rectification, high current and high speed rotation are possible, and furthermore, it is dust-proof and explosion-proof, and is small and inexpensive. (Problem to be solved by the invention) However, conventional control methods for induction motors include constant V/f control that makes the primary voltage V and primary frequency f proportional to the rotational speed; It is impossible to instantaneously generate a normal output torque, and it has not been possible to realize a control system with quick response. Furthermore, it has been impossible to obtain a torque generation mechanism that conforms to the linear automatic control theory. Therefore, it is an object of the present invention to make the torque generation principle of an induction machine exactly the same as that of a DC machine, to provide a system that instantly generates an output torque completely proportional to a command value, and to provide a method for linear automatic control. The aim is to realize a control device with high response as expected in theory. (Example) Principle of torque control of induction motor First, the principle of torque control of induction motor according to the present invention will be explained. Here, in order to simplify the explanation, a two-phase two-pole induction motor will be explained. Figure 1 shows the cage type 2 phase 2
FIG. 2 is a cross-sectional view of a polar induction motor, with orthogonal d-axes,
A q-axis coordinate system is defined as shown, and a cross section 1-1' of one stator winding and a cross section 2-2' of another stator winding orthogonal thereto are shown. When a constant excitation current i _ds =I ₀ is passed through the winding 1-1', the magnetic flux Φ _dr in the d-axis direction within the rotor becomes a constant value Φ ₀ . In this state, in order to generate a constant torque from the two-phase induction machine, if the current i _qs flowing through the stator winding 2-2' is changed from 0 to a certain constant value I ₂ , the second
The magnetic flux Φ _qr in the q-axis direction within the rotor shown in the figure changes, and this magnetic flux change induces a voltage in the rotor winding 4-4'. This induced voltage causes a current i _qr to flow through the winding 4-4' which is short-circuited by the rotor winding resistance r _r .
At this time, the magnetic flux Φ _qr does not interlink with the rotor winding 3-3', so no voltage is induced in the winding 3-3', the current i _dr becomes 0, and the magnetic flux Φ _dr remains at ₀ . It's summery. Assuming that the rotor is not rotating, the following equation holds true at this time. Φ _qr = l _n i _qs − _{l r} i _qr ... (1) dΦ _qr / dt = r _r i _qr ... (2) where, l _n is the mutual inductance (H) between the stator winding and rotor winding , l _r is the self-inductance (H) of the rotor winding. The following equation is obtained from equations (1) and (2) above. l _r di _qr /dt+r _r i _qr = l _n di _qs /dt…
...(3) Here, when i _qs changes stepwise from 0 to I ₂ at time t=0, i _qr is given by the following equation from equation (3). i _qr =l _n /l _r I ₂ e ^- 〓〓 ...(4) Thus, when t=0, i _qr /
t=0=l _n /l _r I ₂ flows, and due to the magnetic flux Φ ₀ perpendicular to the winding 4-4', the force F= in the rotational direction shown in FIG.
K _F Φ ₀ l _n /l _r I ₂ is generated (K _F is a constant coefficient). In this state, i _qr decreases with time according to equation (4), and the force F also decreases. Now, if you want to keep the force F at a constant value of F=K _F Φ ₀ l _n /l _r I ₂ , that is, if you want to keep the torque constant, always i
Keep _qr at a constant value i _qr = l _n /l _r I ₂ , and set Φ ₀ to i _qr
It is necessary to make it perpendicular to . In order to make i _qr =l _n /l _r I ₂ , currents I ₀ and I 2 flow through the stator windings 1-1' and 2-2' as shown in FIG _.
From time t = 0, the stator windings 1-1' and 2-2' are rotated at a constant speed 〓 (rad/sec) while keeping the same value.
1' and 2-2' cannot be physically rotated, but for convenience of explanation, it is assumed that they can be rotated). At a certain point in time t, these windings are rotated by an angle to the positions 1a-1a' and 2a-2a' in FIG. At time t=0, the magnetic flux Φ _dr /t=0 whose magnitude is Φ ₀ shown in FIG. 4', a voltage _〓Φ0 is induced. Due to this induced voltage 〓Φ ₀ , a current i _q flows through the winding 4-4'.
_r = Φ ₀ /r _r flows. The speed 〓 that makes the following equation hold is determined so that this current i _qr becomes l _n /l _r I ₂ . Φ ₀ /r _r =l _n /l _r I ₂ →=l _n /l _r r _r
/Φ ₀ I ₂ ...(5) On the other hand, at time t = 0, i _qr = l _n /l _r I ₂ and the current i _qs flowing through the stator winding 2-2' is I ₂ , so the above ( From formula 1), Φ _qr /t = 0 becomes 0, no voltage is induced in the rotor winding 3-3', i _dr = 0, and the magnitude of the magnetic flux Φ _dr /t = 0 remains as Φ ₀ . It's summery. Next, consider the case where the stator winding is rotated at a speed 〓 from time t = 0, and at a certain time t the stator winding comes to positions 1a-1a' and 2a-2a' in Fig. 3. As you can see, winding 1 has a constant excitation current I ₀ flowing through it.
a-1a' creates a magnetic flux vector Φ _dr /t= _t shown in FIG. 〓 and rotor winding 4a on axis 2a-2a'
-4a', the rotor winding 4a-4
A voltage 〓Φ ₀ is induced in the winding represented by a′. Note that the magnetic flux Φ _dr /t=t is the 3− of the squirrel cage rotor winding.
3', 4-4' and other squirrel cage windings, but since the vector direction of the combined voltage value is the direction of winding 4a-4a', winding 4a-4
It can be considered that a voltage is generated concentratedly in a' and no voltage is induced in other windings. Due to this induced voltage, a current i _qrt =Φ ₀ / _rr flows through the windings 4a-4a'. This i _qrt is l _n /l _r I ₂ from the above equation (5), and the winding 4a-4
Due to the magnetic flux Φ _dr /t=t perpendicular to a′, a rotational force F=K _F Φ ₀ l _n /l _r I ₂ is obtained in this case as well. At this time, the current I ₂ flowing through the stator windings 2a-2a' and the current i _qrt flowing through the rotor windings 4a-4a' are equal in magnitude (i _qrt = l _n /l _r I ₂ ≒ I ₂ , normally In the induction machine, the _direction is _opposite , so the magnetic flux Φ _qr /t=t is set to 0.
Then, the rotor current i _drt of the rotor windings 3a-3a' on the axis 1a-1a' becomes 0, and the magnitude of the magnetic flux Φ _dr /t=t remains at Φ ₀ . Since the above relationship holds true at any time t, the output torque T _e of the motor is always constant as given by the following equation. T _e =K _T Φ ₀ l _n /l _r I ₂ ...(6) However, K _T is a fixed constant. From the above explanation, the current I ₀ creates a magnetic flux Φ _dr /t=t, and the rotor current _{i qrt on} the t-axis is the winding 2
The current I ₂ flowing through a-2a' becomes equal in magnitude and opposite in direction. This means that the current I ₀ corresponds to the excitation current of a DC machine, and the current I ₂ corresponds to the armature current of a DC machine, and it is possible to obtain exactly the same torque generation mechanism with an induction machine as with a DC machine. Become. By the way, the above explanation is for the case where the rotor in Fig. 3 is not rotating (rotor rotational speed θ = 0), but now when the rotor is rotating at the rotational speed θ〓 (rad/sec) The same result can be obtained if the rotor is fixed on a rotating rotor and the d-axis and q-axis are taken as shown in FIG. In this case, 〓 in equation (5) above is d on the rotor.
This is the value for the axis and q-axis coordinates, and in the stationary coordinate system (on the stator), it is given as 〓/on the stator = 〓/on the rotor + θ〓/on the stator, so from equation (5), 〓/on the stator = θ〓/on the stator +l _n r _r /l _r
Φ ₀ I ₂ ...(7) / Stator windings 1-1', 2 on the stator
-2' rotation is sufficient. In the above explanation, θ〓 and I ₂ were explained as constant, but as can be seen from the above explanation, these values do not necessarily have to be constant, and no matter how they change over time, the instantaneous values ( 7) If we make the formula hold,
The output torque T _e of the induction motor is given by equation (6). Furthermore, in the above explanation, it is assumed that the stator windings 1-1' and 2-2' in FIG. 3 can be physically rotated, but in an actual induction machine, the stator windings 1-1' and 2-2'
1' and 2-2' are fixed at the positions shown in FIG.
Therefore, the stator windings are 1a-1a' and 2a in Figure 3.
-2a', the current components _I0 , 1a-1a', 2a-2a',
I ₂ as d-axis component i _ds = I ₀ cos−I ₂ sin, q-axis component i
_qs = I ₀ sin + I ₂ cos, and these i _ds and i _qs are fixed stator windings 1-1', 2-
It is better to let it flow to 2'. i.e. It's fine. Determination of Voltage Control Equation for Induction Motor Next, a description will be given of how to control the stator voltage of the induction motor in order to implement the above-mentioned principle. FIG. 4 is a sectional view of a two-phase induction motor exactly the same as FIG. 1, in which a stator winding 1-1' and another stator winding 2-2' perpendicular thereto are arranged. The current flowing through the winding 1-1' that creates a magnetic flux Φ _dr in the d-axis direction is i _ds , and the current flowing through the winding 2-2' that creates a magnetic flux Φ _qr in the q-axis direction is i _qs . Street. Due to the electromagnetic induction of the stator winding currents i _ds and i _qs , rotor currents flow through all the squirrel-cage rotor windings in Figure 4, but these rotor currents have a cross section on the q-axis as shown in Figure 4. 3-3' and another rotor winding having a cross section 4-4' on the d-axis, the current i _dr flowing through the rotor winding 3-3' and the rotor It is assumed that the current i _qr flowing through the winding 4-4' is represented by orthogonal coordinate components. The currents i _ds and i _dr flowing through the windings 1-1' and 3-3' in FIG. 4 create a magnetic flux Φ _dr in the d-axis direction in the rotor as shown in FIG. 2'
A magnetic flux Φ _qr in the q-axis direction is created by the currents i _qs and i _qr flowing through the terminals 4-4' and 4-4'. Therefore, the magnetic fluxes Φ _dr and Φ _qr are given by the following equations. On the other hand, now the rotor is θ〓 in the rotation direction shown in Figure 4.
If it is rotating at a rotation speed of (rad/sec),
Under the condition that the rotor windings 3-3' and 4-4' are short-circuited to the rotor winding resistance r _r , the following equation holds true. Now, if the magnetic fluxes Φ _dr and Φ _qr are the components of the orthogonal coordinates of the rotating magnetic flux vector Φ _r whose magnetic flux size is Φ ₀ and whose direction is the flux angle as shown in Figure 5, then Φ _dr , Φ _qr is given by the following formula. Here, by substituting equation (11) into equation (10), the rotor current becomes On the other hand, the output torque T _e of the induction machine shown in FIG. 4 is given by the following equation. T _e = K _T (Φ _dr i _qr −Φ _qr i _dr ) ...(13) Here, if we substitute equations (11) and (12) into equation (13), we get This formula (14) means that if the magnitude of the magnetic flux of the rotating magnetic field Φ ₀ is constant, the output torque T _e is completely proportional to the slip frequency (〓−θ〓) [rad/sec]. There is. On the other hand, the stator winding currents i _ds and i _qs for generating this output torque T _e can be calculated using equations (11) and (12).
It can be obtained by substituting into equation (9). Now, in order to simplify equation (15), we will introduce the following variables. Substituting this into equation (16), T _e = K _T l _n / l _r Φ ₀ I ₂ = K _T l _n ² / l _r I ₀ I ₂ ... (
17), and further substituting equation (16) into equation (15), we get becomes. Further, from the above equation (16), the following relationship must be established between I ₂ and I ₀ . I ₂ = l _r /r _r (〓−θ〓)I ₀ ...(19) Here, = d/dt, θ = dθ/dt, and this equation (19) is equivalent to the following equation. It is. -θ=r _r /l _r ∫I ₂ /I ₀ dt ... (20) Here, the above equation (17) matches the equation (16), and the equation (18) is dI ₀ /dt=0 (When I ₀ and Φ ₀ are constant), then it agrees with equation (8), and (19)
The equation is consistent with equation (7). In other words, the current I ₀ in equation (16) is the excitation current for creating magnetic flux Φ ₀ , and the current I ₂
corresponds to the rotor current that causes the current to flow through the rotor winding orthogonal to the magnetic flux Φ ₀ . From the above equation (17), it can be seen that the output torque is proportional to the product of the magnetic flux magnitude Φ ₀ and the rotor current I ₂ . Now, if the magnitude of the magnetic flux Φ ₀ is kept constant, the output torque T _e is completely proportional to the current I ₂ . This is exactly the same as the output torque relationship of a separately excited DC machine. Now, when the desired specified torque T _e for the two-phase induction motor is given, the rotor rotation angle θ
(rad), the excitation current I ₀ and its rate of change dI ₀ /dt, then from the above equations (17), (18), and (20), i _ds , i _qs
is found. Therefore, if these i _ds and i _qs are applied to the stator winding, the output torque of the induction machine becomes the value T _e as the command value. If the stator currents (primary currents) i _ds and i _qs of the induction motor are controlled in this way, the output torque of the induction motor can be controlled in the same way as a direct current machine, but the currents i _ds ,
In order to actually supply i _qs to the stator winding of an induction machine, a current control amplifier is required, and making this current control amplifier highly responsive and highly accurate poses problems in control stability. However, it has the disadvantage of being expensive. Therefore, if the stator winding voltages V _{ds and} V _qs could be controlled instead of controlling the stator currents i _ds and i _qs of the induction machine, a voltage amplifier could be used, which would be economical and solve the stability problem. Ru. Therefore, regarding the above principle, stator voltage V _ds ,
A method for controlling V _qs will be described next. In FIG. 4, magnetic fluxes Φ _ds and Φ _qs within the stator in the d-axis and q-axis directions interlinking with the stator windings are given by the following equations. However, l _s is the self-inductance (H) of the stator winding. Therefore, when voltages V _ds and V _qs are applied to the stator windings, the following equation holds true. However, r _s is the stator winding resistance. Here, if we assume that all of the magnetic flux that is interlinked with the rotor winding is interlinked with the stator winding, this is correct because in a normal induction machine, the leakage magnetic flux is very small compared to the total interlinked magnetic flux. , and the following equation (23) holds true. Furthermore, by substituting equations (18) and (23) into equation (22) to obtain the voltages V _ds and V _qs , become that way. Therefore, when the rotor rotation angle θ〓 (rad/sec), the excitation current I ₀ and its rate of change dI ₀ /dt are given, the desired command torque T _e is entered into the equation (17) above to calculate the current.
At the same time as finding I ₂ , put this current I ₂ into equation (19) to find 〓, and then find from equation (20). Then, by entering these values into equation (24), V _d
If _s and V _qs are calculated and V _ds and V _qs are added to the stator windings 1-1' and 2-2' in FIG. 4, the output torque of the induction motor becomes the value T _e as the command value. Note that the above explanation is about the voltage control equation for a two-phase induction motor, but when applying the above principle to a three-phase two-pole induction motor, the three-phase voltage V applied to the stator winding of the three-phase induction motor _a ,
V _b and V _c should satisfy the following relationship. Note that 〓 in the above equation (25) is given by the above equations (19) and (20). The above explanation has been based on a two-pole induction motor, but in general, in the case of a P-pole motor, if the mechanical angle of the motor is θ, then the electrical angle of the motor θ′ = P/2・θ is given by (19) This invention can be easily applied to multi-polar induction machines other than bipolar machines by calculating by entering the equation (20). The present invention is a torque control method for an induction motor using the above principle, and some embodiments thereof will be described below. Example 1 (Magnetic flux constant torque control) When controlling the magnetic flux size Φ ₀ to always be constant, the current i ₀ is also constant, dΦ ₀ /dt=0, dI ₀ /dt
= 0, the voltage control equation for the three-phase induction machine in equation (25) can be simplified as shown in equation (26) below. Furthermore, since the current I ₀ is constant in the above equation (20), it becomes as follows. =θ+r _r /l _r I ₀ ∫I ₂ dt ...(27) Therefore, from equation (17), when the magnetic flux Φ ₀ is constant,
Noting that the output torque T _e is completely proportional to the current I ₂ , an example of torque control when the magnitude of the magnetic flux is constant will be described below. Figure 6 shows this example 1.
It is a circuit configuration diagram to which the above is applied. That is, 2 poles 3
The rotation axis of the pulse generator 6 is directly connected to the rotation axis of the phase induction motor 5, and when the electric motor 5 rotates, the pulse generator 6 generates an electric signal 6a. Here, it is assumed that the pulse electric signal 6a generates 8192 pulses when the electric motor 5 rotates once. Further, the pulse generator 6 outputs a direction determination signal 6b in response to forward or reverse rotation of the electric motor 5. These signals 6a and 6b are led to a reversible counter 7. Here, reversible counter 7 is binary 13
It is a reversible bit counter that counts up the contents one by one each time the pulse 6a is given from the pulse generator 6 when rotating in the forward direction, and counts down the contents one by one when rotating in the reverse direction. be. Thus, if the rotation angle within one revolution of the electric motor 5 is θ (rad), then the binary 13
The contents of the reversible counter 7 having a value of bits 0 to 8191 are given by 8192×θ/2π. Note that this θ is a quantity corresponding to θ in the above equation (27). Further, the sample pulse generator 8 generates a sampling pulse SP at every sampling period T=1/1000 seconds. This sampling pulse SP serves as an interrupt input to the computer 10 described below. The block 10 surrounded by a dotted line here is composed of a digital computer up to this point. Therefore, the coefficient unit 1 inside the digital computer 10
3, 14, 17, 18, 19, 20, 32, 33
and 34, adders 23, 24, 25, 26 and 2
7. Multipliers 21 and 22, buffer register 1
2, integrator 15, differentiator 16, trigonometric function generator 2
Each of the components 8 is not located at a specific location in the digital computer 10, but is installed in the program control unit 11 within the computer 10 during the operating cycle of the computer 10.
Under the control of the computer 10, the common hardware of the computer 10 is used in a time-sharing manner. However, in order to explain in detail the program processed by the computer 10 according to the present invention, for convenience, it is represented as if it were constituted by individual hardware within the block 10 described above. With this representation, an experienced computer programmer can write an arbitrary computer program to implement this generation. In other embodiments, each hardware element in block 10 in FIG. 6 may be constructed of fixedly connected digital circuits without using a computer. However, in the following explanation, it will be assumed that block 10 is constituted by a digital computer. Now, the sampling pulse generator 8 is T=1/10
When a sampling pulse SP is generated every 00 seconds,
This pulse SP triggers the interrupt input terminal of the program control unit 11 of the computer 10, and the program control unit sequentially executes the program from step 1 to step 8 described below. These programs are executed every T = 1/1000 seconds, and the time to execute the programs from step 1 to step 8 is set to be completed within T = 1/1000 seconds, and then the sampling pulse SP is The program control unit 11 controls the computer 10 until
or execute a program completely unrelated to this invention. When the sampling pulse SP is generated and the program control unit 11 first executes the program in step 1, the computer 10 continues to read the contents 30 of the reversible counter 7 and temporarily stores them in the buffer register 12. Note that a specific memory address of the computer 10 may be used as the buffer register 12. Next, when the program control unit 11 executes the program in step 2, it reads the digital data I ₂ corresponding to the rotor current in equation (17) from the command torque supply means 9, and adds this value I ₂ to the coefficient multiplier 14. Data I ₂ r _r /l _r I ₀ multiplied by a constant coefficient r _r /l _r I ₀
make. This data I ₂ r _r /l _r I ₀ is integrated by an integrator 15 to obtain data σ ^* (t). Integrator 1
As a specific method for calculating the output data σ ^* (t) of step 5, when the output data of the integrator 15 at the previous sampling time is σ ^* (t-T), the output data of the integrator 15 at the current sampling time is calculated as follows: Data σ ^*
(t) may be calculated using the following formula. σ ^* (t) = σ ^* (t-T) + TI ₂ r _r /l _r I ₀ (where T is the sampling time of 1/1000 sec) Thus, data σ ^* (t) is filled every T seconds.
Since TI ₂ r _r /l _r I ₀ is cumulatively added, we have calculated the integral value of the second term on the right side of equation (27): σ ^* (t) = r _r /l _r I ₀ ∫I ₂ dt Become. Next, when the program control unit 11 executes the program in step 3, it reads out the data 8192/2π·θ temporarily stored in step 1 from the buffer register 12, and uses this data in the coefficient multiplier 13.
is multiplied by the coefficient 2π/8192 to create data θ corresponding to the first term on the right side of equation (27). This data θ and the value σ ^* (t) obtained in step 2 are added by an adder 23 to obtain ^* (t). This ^* (t) corresponds to the magnetic flux angle in equation (27). Next, the program in step 4 is executed, and the trigonometric function generator 28 uses the magnetic flux angle created in step 3 to generate trigonometric function values cos, sin, cos (-
2π/3) and sin(−2π/3). Next, the program in step 5 is executed, and the magnetic flux angle ^* (t) obtained in step 3 is differentiated by the differentiator 16 to obtain data 〓 ^* (t). To calculate the output data of the differentiator 16 = ^* (t), the magnetic flux angle at the previous sampling time is
^* (t-T), and the magnetic flux angle at the current sampling time is ^* (t), then differential value data =
^* (t) can be calculated using the following formula. 〓 ^* (t) = 1/T [ ^* (t) - ^* (t - T) (T is the sampling time 1/1000sec) This differential value data 〓 ^* (t) is given a coefficient l _n I by the coefficient unit 17 Multiply by ₀ to create data l _n I ₀ 〓. Next, in the program of step 6, the output data I ₂ of the command torque supply means 9 is multiplied by a constant coefficient r _s in the coefficient unit 18 to generate data r _s I ₂ . This data r _s I ₂
and the data l _n I ₀ 〓 calculated in step 5 are subtracted by the adder 24, resulting in data -(l _n I ₀ 〓 + r _s I ₂ )
This corresponds to the amplitude value of the sine function of the second term on the right side of equation (26). Next, the program in step 7 is executed, and the cos, sin, cos (-
2π/3), sin (-2π/3), and -(l _n I ₀ 〓 + r _s I ₂ ) obtained in step 6, V _a and V _b of the voltage control equation of equation (26) are calculated using the coefficient multiplier 19. and 20, multiplier 21
and 22, and the adders 25 and 26 perform the following calculation process. V _a and V _b thus obtained are subtracted by an adder 27 to obtain V _c =-(V _a +V _b ). This is because the following equation V _a +V _b +V _e =0 (28) always holds true from equation (26). Next, the program in step 8 is executed, and the coefficient multiplier 3 is applied to V _a , V _b , and V _c obtained in step 7.
2, 33, and 34 are each multiplied by a constant coefficient 2048/E to create data V ^* _a , V ^* _b , and V ^* _c , and these data are converted to PWM (Pulse Width
Modulation) Three sets of binary codes 12 in the circuit 38
Each bit is written to a hold register. Here, the above coefficient E is calculated as V which can be considered in this example.
Assuming the maximum voltages of _a , _Vb , and _Vc , V ^* _a , V ^* _b ,
Since V ^* _c falls within the range of -2048 to +2047, the 12-bit hold register will not overflow. As above, according to the calculator 10, the stator voltage command value V
I explained the process of calculating ^* _a , V ^* _b , and V ^* _c , but
Next, the operation of the PWM circuit 38 will be explained.
The detailed circuit inside the PWM ^circuit 38 is as ^shown in _{FIG .}
^* _c is stored in the hold register 39A, hold register 39B, and hold register 39C, respectively. Note that these three sets of hold registers 39A
-39C are binary coded 12-bit registers that can store data in the range of -2048 to +2047. The contents of hold register 39A are then input to coincidence circuit 52A via a 12-bit data line. In addition, the oscillator 40 generates a clock pulse CLK of a constant frequency (5MHz), and the reversible counter 41
is a binary value with counting data in the range of -2048 to +2047
It is a 12-bit reversible counter, and when the UP input is “1” (hereinafter, when the logic level is High level, it is “1” and when it is Low level, it is “0”).The contents of the reversible counter 41 are clock pulses.
It increases every time CLK comes, and the count value is the maximum value +
When the number reaches 2047, the MAX output of the counter 41 becomes "1". This MAX output sets the flip-flop (hereinafter simply referred to as FF) 42, and
Set the "1" side output signal DN to "1" and the "0" side output signal UP to "0". Here, the DN signal puts the reversible counter 41 in the subtraction mode, and the contents of the reversible counter 41 decrease every time the clock pulse CLK comes, and finally, when it reaches the minimum value -2048, the MIN output of the counter 41 becomes "1". . This MIN output resets the FF 42, sets its output UP to "1", and sets the DN signal to "0". In this way, the count contents of the reversible counter 41 are -2048~ as shown in Figure 8a.
The range of +2047 continues to rise and fall at a constant slope over time, and the period is 4096 x 2/5 MHz =
It becomes 1.64ms. Note that the DN signal of the "1" side output of the FF 42 is as shown in FIG. 8b. Further, the contents of reversible counter 41 are inputted to matching circuits 52A-52C via 12-bit data lines. Therefore, the coincidence circuits 52A to 52C match the contents of the hold registers 39A to 39C and the reversible counter 4.
Only when the contents of 1 match, the output is set to “1”. J-K flip-flop FF47A~47
AND circuits 45A to 45C on the J side input of C
When the AND condition is met, FFs 47A to 47C are set and their "1" side outputs are set to "1". In addition, the AND circuits 46A to 46C of the K side input
When the AND condition is satisfied, FFs 47A to 47C are reset and their "0" side outputs are set to "1". Here, data V ^* _a output from the computer 10
is +500 as shown by the straight line 54 in FIG. 8a, the contents of the hold register 39A and the reversible counter 41 match at time A and B in FIG. 8a, and the output of the matching circuit 52A is " It becomes 1”. Since the DN signal is “1” at time A, FF47
A is set, and since the UP signal is "1" at time B, FF47A~ are reset. Thus, the "1" side output of FF47A ~ is as shown in Fig. 8c,
The "0" side outputs change as shown in FIG. 8d. On the other hand, delay circuits 48A to 48C are FF47A
When the "1" side output of ~47C changes to "1", an output signal a1 as shown in FIG. 8e delayed by time D is generated. Moreover, the delay circuits 49A to 49C are
When the "0" side outputs of the FFs 47A to 47C change to "1", an output signal a2 as shown in FIG. 8f delayed by time D is generated. These output signals a1 and a2 are transmitted to a transistor power amplifier 3, which will be explained next.
9 to 41 are driven. As can be seen from FIG. 8, the output signals a1 and a
2 is the content of the hold registers 39A to 39C pulse width modulated, and when V ^* _a is -2048, the output signal a1
is always "0", and as V ^* _a increases from -2048, the period during which the signal a1 is "1" increases, and V ^* _a
When is +2047, the output signal a1 is always "1". Note that when the output signal a1 is "1", the signal a2
is always "0", and when signal a2 is "1", signal a1 is "0". Here, the period (D) in which the signals a1 and a2 are both "0" is determined by the delay circuit 48.
The purpose of making transistor power amplifiers 39 to 4 by A to 48C and 49A to 49C is as follows.
This is to protect transistor No. 1. Therefore, the output signals b1 ^and _b2 of the B action in FIG. It is exactly the same as a1 and 2 are made. This concludes the explanation of the operation of the PWM circuit 38, and next the transistor power amplifiers 39-41 will be explained. As shown in FIG. 6, the phase output a1 of the PWM circuit 38
and a2 are input to a power amplifier 39 having a transistor configuration, and the output of this amplifier 39 becomes the stator voltage V _a of the induction machine 5. Similarly, the B-phase outputs b1 and b2 are input to a power amplifier 40, whose output supplies the stator voltage V _b , and the C-phase outputs c1 and c2 are input to a power amplifier 41, whose output supplies the stator voltage V _c. . Here, a detailed circuit diagram of the power amplifiers 39 to 41 is shown in FIG. 9, but since the power amplifiers 39 to 41 have the same configuration, the power amplifier 3
I will explain only 9. That is, the power amplifier 39 has power transistors 55 and 56 connected in series, both ends of which are connected to the +E ^(V) and -E ^(V) terminals of the DC power supply 59, so that the transistors 55 and 56 are turned on and off alternately. Therefore, the stator voltage V _a of the induction machine 5 is set to +E ^(V) or -E ^(V) . The base drive circuit 57 of the transistor 55 is
When the output signal a1 of the PWM circuit 38 is "1", the transistor 55 is turned on and the voltage V _a becomes +E ^(V) , and when the output signal a1 is "0", the transistor 55 is turned on and the voltage V a is +E (V).
Turn off 5. Similarly, the base drive circuit 58 of the transistor 56 turns on the transistor 56 to set the voltage V _a ^to -EV when the output signal a2 of the PWM 38 is "1", and turns off the transistor 56 when the output signal a2 is "0". Make it. Here, when the signal a1 changes to "0" and the transistor 55 turns from on to off, the transistor 55 does not turn off immediately due to the switching delay of the transistor 55. Therefore signal a1
When changes from “1” to “0”, the signal a immediately
When 2 is changed from "0" to "1", the DC power supply 59 is short-circuited, a large current flows through the transistors 55 and 56, and the transistors are destroyed. Therefore, the signal a1 becomes "0" and the transistor 55
Delay circuits 48A to 4 are installed in the PWM circuit 38 so that the signal a2 becomes "1" after the
It contains 8C and 49A to 49C. In addition, an AC voltage is always applied to the primary side (not shown) of the isolation transformer 67 inside the base drive circuit 57, and the DC power supply voltage floated by the bridge rectifier 66 and the capacitor 68 is applied to both ends of the capacitor 68. occurs in The negative terminal of this power supply is connected to the emitter of the transistor 55, and since the potential V _a of the emitter of the transistor 55 alternately changes to ±E ^(V) , such a floating DC power supply is required. Furthermore, when the output signal a1 of the PWM circuit 38 input to the photocoupler 60 inside the base drive circuit 57 becomes "1", the output terminals 61 and 62 of the photocoupler 60 are turned on, turning off the transistor 65, and the resistor 64 A base current flows from the floating DC power source to the power transistor 55 through the power transistor 55, turning on this transistor and setting the potential of V _a to +E. On the other hand, when the signal a1 becomes "0", the output terminals 61 and 62 of the photocoupler 60 are turned off, and the base current of the transistor 65 flows through the resistor 63, turning on the transistor 65 and turning off the power transistor 55. . Note that the base drive circuit 58 also performs the same operation. Now, the data V ^* _a , V output from the computer 10
^* _b , V ^* _c change with time _t as shown in FIG _. , V _c takes two values of +E and -E as shown in Fig. 10 b, c, and d, and V ^* _a , V ^* _b , and V ^* _c are the triangular waves 53 in Fig. 10 a.
It becomes a pulse width modulated square wave. Here, the average values (DC components) V〓 _a , V〓 _b , V〓 _c of the rectangular wave voltages V _a , V _b , V _c from which the triangular wave carrier frequency harmonic components are removed are V〓 _a = E/ 2048・V
^* _a , V〓 _b E/2048・V ^* _b , V〓 _c = E/2048・V ^* _c
Therefore, the average values, V〓 _a , V〓 _b , V〓 _c are completely the voltage command values V ^* _a , V ^*
_b ,
Proportional to V ^* _c . Therefore, Fig. 10 b, c, d
When a PWM square wave like this is applied to the stator winding of the induction machine 5, the stator current becomes a smooth current with the harmonic components of the square wave voltage removed due to the reactance of the induction machine 5, which is generated inside the induction machine 5. Magnetic flux is proportional to this smoothed current. From this, Figure 10b,
When PWM rectangular wave voltages c and d are applied to the induction machine 5, if the average value of this rectangular wave voltage is proportional to the voltage command value, it will be equivalent to applying the voltage of equation (26), ( The induction motor is driven by the voltage control equations given by Equations 26) and (27). From the above description, it is clear that even with the configuration shown in FIG. 6, a torque proportional to the data I ₂ given by the command torque supply means 9 can be obtained in the induction motor 5. Embodiment 2 (Maximum Voltage Increase Control) In the above-described Embodiment 1, V _a , V _b , and V _c are controlled using the equation (26) above. Therefore, this equation (26) can be rewritten as follows. However, tan δ=l _n I ₀ + _rs I ₂ / _rs I ₀ Also, from FIG. 9, the maximum voltage of V _a , V _b , and V _c cannot be larger than the DC power supply voltage ±E ^(V) . Therefore, the amplitude value of equation (29)√( _{s 0} ) ² +( _{n 0} + _s
I ₂ ) ² is never greater than E. Furthermore, in applications where it is necessary to increase the amplitude value √( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ² , it is necessary to increase the DC power supply voltage ±E ^(V) , but this requires the power transistor 55 and 56 with high breakdown voltage are required, which may be difficult to realize. On the other hand, the line voltage applied to the induction motor 5 is V _ab =V _a −V _b , V _bc =
V _b -V _c , V _ca =V _c -V _a are given by the following equation from equation (29). Here, √( _{s 0} ) ² + (
Since _n I ₀ + r _s I ₂ ) ² will never be greater than E, the line voltage will never be greater than √3E. That is, in the first embodiment, the maximum value of the line voltage applied to the induction machine 5 is suppressed to √3E. On the other hand, since the power amplifiers 39 to 41 in FIG. 9 can output a maximum line voltage of up to 2E, the first embodiment is
This means that you are losing a ratio of 2 at the maximum voltage.
On the other hand, in this second embodiment, the maximum line voltage is 2E.
It is possible to produce up to In this second embodiment, the voltage control equation is as follows. By the way, this equation (31) is equivalent to adding the same variable V _N to the right side of the above equation (26). When controlling using equation (31), the line voltages V _ab , V _bc , and V _ca applied to the induction machine 5 are the same as equation (30), so they are controlled using the principle of the induction machine described above. . In addition, in this Example 2, the amplitude √ of equation (31)
( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ² is the highest

[Formula] is allowed. Therefore, the variable V _N of the voltage control equation of equation (31) is determined as follows. In other words, the first term on the right side of equation (31) √( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ² co
s
(+δ), √( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ² co
s(
+δ−2π/3), √( _{s 0} ) ² +( _{n 0} + _s
2 ) If both of ² cos (+δ-4π/3) are within the range of -E to +E, set V _N =0. Then, when any of the first terms on the right side of equation (31) becomes less than -E, V _N is determined so that the entire right side of that phase becomes -E, and when it becomes more than +E, V N is determined so that the entire right side of that phase becomes -E. Decide V _N so that the entire right side of the phase becomes +E. Here, Fig. 11 shows the amplitude value of equation (31) √
( _rs I ₀ ) ² + (l _n I ₀ + r _s I ₂ ) ² is

This is a diagram illustrating how V _N changes when [Equation] is satisfied, and the dotted line in a of the figure indicates the first term on the right side of Equation (31). +
When δ=0 to π/6, the first term on the right side of V _a in equation (31) √( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ² cos(+δ)
Ha+
Since it is greater than E, V _N is determined so that V _a = +E, and when +δ = π/6 to π/2, the first term on the right side of V _c is √( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ² cos(+
δ−
4π/3) is less than -E, so V _N is determined so that V _c = -E. Thereafter, V _N is similarly determined as shown in FIG. 11b, and V _a , V _b , and V _c in equation (31) change together to +δ as shown by the solid line in FIG. 11 a. From Figure 11a, the line voltages V _ab , V _bc ,
It can be seen that the maximum value of V _ca is allowed up to 2E. Next, the specific configuration of this second embodiment will be explained with reference to FIG. 12. This second embodiment has the same structure as the first embodiment except for block 10 in FIG. In this second embodiment, the computer processes a block 69 surrounded by a dotted line. Block 10 in block 69 is completely similar to the processing of block 10 in the first embodiment of FIG. Thus, the computer 69 allows the program control unit 50 to execute the processes from step 1 to step 7 in exactly the same way as in the first embodiment.
Calculate V _a −V _N , V _b −V _N , and V _c −V _N in equation (31). Here, the program control unit 5
0 executes the program from step 1 to step 7, then executes the program from step 8,
V _a -V _N , V _b -V _N , V _c - obtained in step 7
A _constant coefficient 2048/
Multiply each by E to create data V ^* _a , V ^* _b , and V ^* _c .
By the way, in Example 1, V _a determined by equation (26),
The permissible range of values of V _b and V _c was -E to +E, but in Example 2, V _a -V _N and V _b - of equation (31)
V _N , V _c −V _N is

Since the range of [Formula] is allowed, V ^* _a , V ^* _b , V ^* _c in Fig. 12 are

It will take a value within the range of [Formula]. Next, the program control unit 50 executes the program in step 9, and the minimum value selector 7
0 is the minimum value of data V ^* _a , V ^* _b , V ^* _c
This minimum value MIN is subtracted from the constant value -2048 by the subtracter 72, and data (-2048-MIN) is input to the line function generator 74. Here, the linear function generator 74 is a function generator whose output V _N1 is "0" when the input is negative, and whose output is equal to the input when the input is positive. Therefore, when the minimum value MIN is less than -2048, V _N1 = -2048-
MIN, when the minimum value MIN is greater than -2048, V _N1 =
It becomes 0. Next, the program in step 10 is executed, and the maximum value selector 71 outputs the maximum value among the data V ^* _a , V ^* _b , and V ^* _c as MAX. This maximum value MAX
is subtracted from the constant value +2047 by the subtracter 73 (+
2047−MAX) is generated by the line function generator 75.
is input. Here, the linear function generator 75 is a function generator whose output V _N2 is "0" when the input is positive, and whose output V _N2 is equal to the input when the input is negative. Therefore, when the maximum value MAX is +2047 or more, V _N2 = 2047 - MAX, and when it is less than +2047, V _N2 =
It becomes 0. By the way, V _a − calculated by equation (31)
V _N , V _b -V _N , V _c -V _N

As long as it is within the range of [formula], the minimum value MIN is less than -2048,
Furthermore, the maximum value MAX cannot exceed +2047. Therefore, one of the data V _N1 and V _N2 is always 0. Next, the program in step 11 is executed, and the adder 76 adds the data V _N1 and V _N2 to obtain data V ^* _N . This data V ^* _N is added to the data V ^* _a , V ^* _b , and V ^* _c obtained by the program in step 8 by adders 77, 78, and 79, respectively, to create data V _a
', V _b ', V _c ', and these data are input to the computer 6.
9 to three sets of 12-bit hold registers in the PWM circuit 38. Now, V ^* _a ,
When either of V ^* _b or V ^* _c is smaller than -2048 or larger than +2047, the data V ^* _N is added to that data.
Since the data V ^* _N is determined so that the value added is -2048 or +2047, the data V _a ′, V _b ′, V _c
′ is always in the range from −2048 to +2047,
The three sets of hold registers A, B, and C in the PWM circuit 38 can be 12 bits. Therefore, the same PWM circuit 38 and power amplifiers 39 to 41 as the components of the first embodiment described above can be used in this second embodiment as well. Example 3 (Magnetic flux change torque control) In the maximum voltage increase control of Example 2 described above, the voltage amplitude value √ ( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ² is the maximum

It has already been mentioned that the voltage E is allowed up to [Equation] and there is a limit to the voltage E due to the withstand voltage of the power transistor in the power amplifier shown in FIG. On the other hand, in applications where the motor rotation speed θ〓 must be high, the magnetic flux angular velocity〓 also becomes large, and the amplitude value √( _{s 0} ) ² + ( _{n 0} + _s
I ₂ ) ² is

[Formula] There are cases where the above is necessary. When 〓 is large like this, in Example 2, the above amplitude value is

[Formula] In order to achieve the following, the magnetic flux current I ₀ can be made smaller, but in this case, as can be seen from equation (17), the output torque T _e decreases for the same rotor current I ₂ , causing the inconvenience. . Therefore, in this third embodiment, in order to eliminate the above-mentioned drawbacks, when the motor rotation speed θ is a constant value (base speed) θ _{or less} , the exciting current I ₀ is set to a constant value (rated value).
I _0B , and when θ〓 _B or more, the excitation current I ₀ is decreased and the amplitude value is always √( _{s 0} ) ² + ( _{n 0} + _{s 2} ) ^2.

[Formula] A method is provided in which the rating of the induction motor can be fully utilized by making the following equation. In this embodiment 3, the excitation current I ₀ changes, so
The following voltage equation is used by adding the variable U _N to equation (25). Also, equation (32) can be rewritten as follows. however,

[Formula] As mentioned above, the amplitude value of the first term on the right side of equation (34) is due to power amplifier limitations.

It cannot be more than [expression]. That is, [r _s I ₀ + (l _n +l _r r _s /r _r )d
I ₀ /dt] ² + [l _n I ₀ 〓+ _rs I ₂ ] ² 4/3E ² must hold true. Now, if the sampling period T = 1/1000 (sec) and the value of current I ₀ one sampling time before is I ^* ₀ , then [r _s I ₀ + (l _n + l _r r _s /r _r )1/T (I ₀ −I ^* ₀
] ² + [l _n I ₀ 〓+ _rs I ₂ ] ² 4/3E ² ...(A) must hold. (33) 〓=θ〓+
Substituting r _r /l _r I ₂ /I ₀ into this equation (A), [ _rs I ₀ + (l _n + l _r r _s / r _r )1/T (I ₀ - I ^* ₀
] ² + [l _n I ₀ θ〓 + (l _n /l _r r _r + r _s ) I ₂ ] ² 4/
3E ² ...(35) is found. Therefore, in the third embodiment, in order to satisfy this equation (35), the absolute value of the motor rotation speed θ〓 |
When θ〓| is greater than or equal to the base speed θ〓 _B , the excitation current I ₀ is controlled so that the following equation holds. ( _rs I ₀ ) ² + [l _n I ₀ | θ〓 | + (l _n /l _r r _r + r _s ) I _2n ] ² = 4/3E ² −α ² ... (36) This (36) In the formula, I ₀ is always the correct value, |θ〓
| is the absolute value of θ〓, and I _2n is the maximum value that the rotor current I ₂ can take (|I ₂ |I _2n ). Furthermore, α ² is a constant value, and for any absolute value of excitation current change |I ₀ −I ^* ₀ | in the range of |θ〓|>θ〓 _B , it is the minimum value among the values that satisfy the following formula. Let them choose the value of the value. [r _s I ₀ + (l _n +l _{r rs} /r _r )1/T | I ₀ −I ^* ₀
|] ² −r _s I ₀ ) ² <α ² (37) If formulas (36) and (37) are satisfied, the condition of formula (35) is satisfied. This is because by substituting equation (36) into equation (37), [r _s I ₀ + (l _n +l _r r _s / r _r )1/T | I ₀ − I ^* ₀
|] ² + [l _n I ₀ | θ〓 | + (l _n /l _r r _r + r _s ) I _2n ]
² <4/3E ² ...(38) Therefore, the following condition always holds true. [r _s I ₀ + (l _n +l _{r rs} /r _r )1/T(I ₀ −I ^* ₀
)] ² [ _rs I ₀ + (l _n +l _r r _s /r _r )1/T | I ₀ −I
^* ₀ |] ² … …(39) [l _n I ₀ θ〓+(l _n /l _r r _r +r _s )I _2n ] ² [l _n I ₀ |θ〓|+(l _n /l _r r _r + r _s ) I _2n ]
² ... (40) From these equations (38), (39), and (40), the above equation (35) is established. Also, the value of α ² in equation (37) is
From the formula, the smaller ^α2 is, the more the voltage is at the limit of the power amplifier.

It is efficient because [formula] can be added to the induction motor. To that end, from equation (37) |I ₀
It can be seen that the smaller the value of -I ^* ₀ |, the better. On the other hand, since the current I ₀ is always controlled to satisfy equation (36), |I ₀ −I ^* ₀ | becomes a function of the rate of change in motor speed d|θ|/dt, and the function is It can be found in this way. That is, if we differentiate both sides of equation (36) with respect to time t, we get So, in this formula

If you substitute [formula] becomes. Therefore, the absolute value of d|θ|/dt is It is. Now, as an example, let's take a 3.7 ^(kW) -200 ^(V) -14.6 ^(A) two-pole three-phase standard induction motor and enter the constant of equation (42) below to limit the acceleration |d|θ|/dt| Let's calculate the value. Furthermore, in this example, let's choose α=10 ^(V) as α. Therefore, by substituting I ₀ = I _0B = 7.4 ^(A) into equation (36), |θ〓|=337 (rad/sec) (=3220 ^r . ^p . ^m
is obtained, so the base speed θ〓 _B 〕=337(rad/
sec). Here, if the following formula holds true in any case, the above-mentioned formula (37) always holds true. When the right side of this equation (B) is constant α=10 ^(V) , the larger the positive value of the exciting current I ₀ is, the smaller it becomes.
Therefore, the minimum value on the right side is when I ₀ = I _0B , Then, equation (37) always holds true. Also, the coefficient on the right side of equation (41) is given by equation (36). , and the first term on the right side of equation (C) is

[Formula] is the minimum value when I ₀ = 0 The second term on the right side |θ|/I ₀ is |θ〓|=θ〓 _B in the range of |θ〓|>θ〓 _B , and when I ₀ = I _0B , the minimum position is

Take each [formula]. Therefore, the following equation (44) always holds true. Now, if we set the acceleration |d|θ|/dt| in equation (41) to |d|θ|/dt|=45.555×70.64=3218 (rad/sec ² ) (=30729r.pm/sec), we get ( 41) From equation is obtained, and from this equation (D) and equation (44), |I ₀ -I〓|/
Since T<70.64 (A/sec), equation (43) always holds true. In other words, the acceleration |d|θ|/dt| is 3218 (rad/sec
² ) If the following, then in any case |I ₀ - I〓|/T is 70
.64 (A/sec) or less, equation (43) holds, and therefore equation (37) also holds, and α in equation (36) is set to α=10 ^(V).
It can be done. In general applications, the acceleration
It is almost impossible for the constant to exceed 3218 (rad/sec) (=30729 r.pm/sec), so when the constant of the induction machine is expressed by equation (42), this Example 3 is sufficient to satisfy the rating of the induction machine. can be fully demonstrated. Now, when an appropriate α is determined as described above, in the third embodiment, the motor rotation speed |θ〓| is first detected, and when the |θ〓| is less than the base speed θ〓 _B , the exciting current I ₀ is set to a constant value _I0B , and when |θ〓| is greater than _θ〓B , the current _I0 is determined according to equation (36). When the above equation (42) is a constant, the first term ( _rs I ₀ ) ² on the left side of equation (36) becomes maximum when I ₀ = I _0B = 7.4A, and this maximum value is (0.368×7.4) ² = 7.4, but this value is the right side of equation (36) 4/3E ² −α ² = 4/3×141 ² −10 ²
= 26408, so when it is a constant as described above, the first term (r _s I ₀ ) ² on the left side of equation (36) can be ignored and equation (36) can be simplified as follows. However, applications where r _s becomes large (for example, applications where a series resistor is inserted between the primary side of an induction machine and a power amplifier to smoothly control the excitation current I ₀ near 〓 = 0 by voltage) ), the current I ₀ must be determined accurately from equation (36). Here, if we derive a formula for calculating the current I ₀ for the rotation speed |θ〓| from formula (36), we get (r _s ² +l _n ² |θ〓| ² )I ₀ ² +2l _n (l _n /l _r r _r +r _s ) I _2n | θ〓 | I ₀ + (l _n /l _r r _r + r _s ) ² I ₂
Since _n ² + α ² −4/3E ² = 0 ……(46)′ becomes. Therefore, in this example, in equation (46),

Since E, I _2n and α are selected so that [Equation] is positive, the current I ₀ obtained by Equation (46) is always obtained as a real number. Note that the curve 80 in FIG. 13 is the relationship between the motor rotational speed |θ〓| and the exciting current I ₀ obtained from the equation (46) when the above-mentioned equation (42) is a constant. This curve 8
It can be seen that 0 is almost equal to the inversely proportional relationship of equation (45). In this third embodiment as well, the program control unit 100 sequentially executes the program from step 1 to step 14, which will be described below, every time the sample pulse SP, which occurs every sampling time T=1/1000 seconds, is input. That is, when the program in step 1 is executed, the computer 81 calculates the contents of the reversible counter 7 (8192/2π
θ) and apply the coefficient multiplier 107 to this data 8192/2πθ.
is multiplied by a coefficient 2π/8192 to form data θ. Next, the program in step 2 is executed, and the above data θ is differentiated by the differentiator 108, and the data θ
is obtained. Next, the program in step 3 is executed, and the absolute value |θ| of θ is obtained by the absolute value unit 109, and I ₀ is calculated from this |θ| by the I ₀ calculator 110. This I ₀ calculator 11 calculates I ₀ from |θ〓| according to the above-mentioned equation (46), and if this I ₀ is less than a constant value I _0B , I ₀ calculated by equation (46) is directly used as I _{0 The} output data I ₀ of the calculator 110 is set as I 0 , and when I ₀ is greater than or equal to the constant value I _0B , the constant value I _0B is set as the output data I ₀ of the calculator 110 . Next, the program in step 4 is executed, and the command current data supply means 51 receives digital data.
I ₂ ' is read and the limiter 106 limits the data I ₂ ' to a limit value ±I _2n to form rotor current data I ₂ . Here, the Smit value I _2n corresponds to the constant I _2n in equation (36). Next, the program in step 5 is executed, and the divider 111 divides the current I ₂ obtained in step 4 by the current I ₀ obtained in step 3 to obtain data I ₂ /I ₀ . The coefficient unit 112 converts this data I ₂ /I ₀ into a coefficient r _r /l _r
to form data r _r /l _r I ₂ /I ₀ . Next, the program in step 6 is executed, and the data r _r /l _r I ₂ /I ₀ obtained in step 5 are integrated by the integrator 113 to obtain data -θ. That is, −θ=∫r _r /l _r I ₂ /I ₀ dt in the above-mentioned equation (20) is calculated. This data -.theta. is added by an adder 85 to the data .theta. obtained in step 1 to form data. This corresponds to the magnetic flux angle in the voltage equation (32). Next, the program in step 7 is executed, and the trigonometric function generator 99 calculates trigonometric function values cos, sin, cos (-2π/3), and sin (-2π/3) from the magnetic flux angle determined in step 6. Next, the program in step 8 is executed, and the data I ₀ obtained in step 3 is differentiated by the differentiator 114 to obtain data dI ₀ /dt. By the way, differentiator 1
In order to calculate the output data dI ₀ /dt of 14, if the value of I ₀ one sampling time before is I ^* ₀ , it is sufficient to calculate dI ₀ /dt=I ₀ -I〓/T. This data dI ₀ /
dt is multiplied by a coefficient (l _n +l _r r _s /r _r ) in a coefficient unit 117 to form data (l _n +l _r r _s /r _r )dI ₀ /dt. Next, the program in step 9 is executed, and the coefficient r of the coefficient unit 118 is added to the data I ₀ obtained in step 3.
Multiply by _s to create data r _s I ₀ and combine this data with the data obtained in step 8 (l _n +l _r r _s /r _r )dI ₀ /
dt is added by the adder 87 and the data r _s I ₀ +(l _n +l _r r _s /r _s
)dI ₀ /dt. This value is the first value on the right side of voltage formula (32).
corresponds to the amplitude value of the cos function of the term. Next, the program in step 10 is executed, and the θ〓 obtained in step 2 and the data r _r /l obtained in step 5 are
_r I ₂ /I ₀ are added by an adder 86 to obtain data 〓.
That is, 〓=θ〓+r _r /l _r I ₂ in the above equation (33)
/I ₀ is calculated. This 〓 and the data I ₀ obtained in step 3 are multiplied by a multiplier 101 to produce data I ₀ 〓. this
I ₀ 〓 is multiplied by the coefficient l _n in the coefficient unit 115 to obtain the data l _n
I ₀ 〓 is obtained. Next, the program in step 11 is executed, and the coefficient r _s is added to the data I ₂ obtained in step 4 using the coefficient unit 116.
is multiplied to create data r _s I ₂ , and this data and the data l _n I ₀ 〓 obtained in step 10 are added to create data l _n I ₀ 〓 + r _s I ₂ . This value is the amplitude value of the sine function of the second term on the right side of voltage equation (32). Next, the program in step 12 is executed, and cos, sin, cos (-2π/3), sin (-2π/3) obtained in step 7 and r _s I ₀ + (l _n +l _s r _s /r _r )dI ₀ /dt and step 11
From l _n I ₀ 〓 + r _s I ₂ found in equation (32), V _a −V _N , V _b −V _N
is obtained by the following calculation process using multipliers 102, 103, 104, 105 and subtracters 80, 90. Next, the program in step 13 is executed, and the coefficient multiplier 1 is applied to V _a -V _N and V _b -V _N obtained in step 12.
19 and 120 by a constant coefficient 2048/E, respectively, to create data V ^* _a and V ^* _b . Further, the data V ^* _a and V ^* _b are subtracted together by the adder 91, and V ^* _c =-(V
^* _a , V ^* _b ) are found. In this Example 3, the amplitudes of V _a -V _N , V _b -V _N , and V _c -V _N in equation (32) is always

[Formula] These data V ^* _a , V ^* _b , V ^* _c are

[Formula] Takes a value within the range of . Next, the program in step 14 is executed and processed in exactly the same manner as described in the second embodiment to calculate the output data V _a ', V _b ', and V _c ' of the computer 81. As explained in Example 2, these data V
_a ′, V _b ′, and V _c ′ always fall within the range of −2048 to +2047, and the
Set of 12-bit hold registers 39A, 39
B and 39C, respectively. The process by which the calculator 81 of FIG. 14 calculates the voltage value of the voltage equation of equation (32) has been described above. Example 4 (Speed control) In the above-mentioned Examples 1 and 2, the excitation current I ₀
Since the output torque T _e of the induction motor is kept constant,
As is clear from equation (17) above, is completely proportional to the command torque data _I2 . Thus, when the torque control devices of Embodiments 1 and 2 are applied to a speed control device, the control system follows the linear automatic control theory, and a speed control system with good responsiveness as intended by the designer can be realized. FIG. 15 is a block diagram of an embodiment of the speed control device. In this fourth embodiment, the configuration is the same as that shown in FIG. 6 except for the computer processing portion of block 128 and the command speed data supply means 129, so a description thereof will be omitted. The sample pulse SP that occurs every sample time T = 1/1000 seconds is calculated by the computer 128.
Each time the program comes, the program control unit 141 sequentially executes the program from step 1 to step 3 described below. First, when you run the program in step 1,
The calculator 128 calculates the contents of the reversible counter 7 [8192/2π
θ]. This data is differentiated by a differentiator 132 to obtain velocity data 8192/2πθ〓. Differentiator 1
32 output data 8192/2πθ〓, the contents of the reversible counter 7 before the sampling time are calculated as [8192/2πθ] ^*
and the current sampling time is 8192/2πθ
Then, basically 8192/2πθ〓=1/T{[8192/2πθ]−
[8192/2πθ] ^* }...(48) However, since the reversible counter 7 has a capacity of only 13 binary bits, its contents [8192/2πθ] only have a value between 0 and 8191. That is, as the motor rotates forward and θ increases from 0, the contents of the counter 7 increase from 0, and when θ is slightly smaller than 2π ^(rad) (one rotation of the motor), the contents of the counter become 8191. When θ=2π ^(rad) (one rotation of the motor), the contents of the counter become 0 again, and when θ increases from θ=2π ^(rad) , the contents of the counter increase from 0 again. That is, the contents of this counter 7 are expressed in digital quantities within one rotation of the motor rotation angle θ, but for two or more rotations, the contents are the same as the first rotation. Therefore, in the calculation of equation (47), there may be cases where accurate calculation results cannot be obtained if the calculation is performed as is. For example, if the position of θ one sampling time before was 2π/8192 (8000) rad and the position of θ at the current sampling time was 2π/8192 (8200) rad, the previous counter contents were [8192/8000] rad. 2π
θ] ^* = 8000, the contents of the counter this time are [8192/2π
θ〕=8200−8192=8. If we put this into equation (47) and calculate it, it becomes 8192/2πθ = 1/T [8-
8000], and even though θ is increasing in the positive direction,
The speed 8192/2πθ〓 gives rise to the disadvantage that a negative result is obtained. Now, the maximum change in θ for T=1/1000 seconds is ±π ^{(r
ad )} , then in the calculation of equation (47), the calculated value in { } is [8192/2πθ] − [8192/
2πθ] ^* is in the range of −4096 to +4095, then it is the correct answer,
Otherwise you are wrong. Therefore, in order to eliminate the above-mentioned disadvantage, an accurate value can be obtained by calculating the speed data 8192/2πθ〓 using the following equation. Correct speed data 8192/2πθ〓 is obtained from equation (49) by the differentiator 132, and data proportional to the actual speed θ〓 (rad/sec) of the motor 5 is obtained. Next, the program in step 2 is executed, and the speed command digital amount R is read from the command speed data supply means 129 of the remote control device.
The correct feedback data 8192/2πθ〓 obtained in step 1 is subtracted by the subtracter 136 to obtain the speed error data VE. This velocity error data VE is multiplied by a constant coefficient K in a coefficient multiplier 131 to obtain data I ₂
is created (in this case, the data I ^* ₀ input to the other terminals of the coefficient multiplier 131 is irrelevant). This data _I2 is directed to block 130 within computer 128. Here, the block 130 has exactly the same configuration as the torque control computer processing section 10 in FIG. 6 of the first embodiment, or has exactly the same configuration as the torque control computer processing section 69 in FIG. 12 of the second embodiment. shall be taken as a thing. That is, the data of block 130 in FIG.
I ₂ corresponds to torque data I ₂ in FIG. 6 or torque data I ₂ in FIG. 12, and corresponds to block 130 in FIG. 15.
The data [8192/2πθ] input to the input terminal 135 corresponds to the reversible count data 8192/2πθ of FIG. 6 or the reversible counter data 8192/2π of FIG. 12. Next, when the program control unit 141 executes the program in step 3, the block 130 performs the same calculation as explained in the first or second embodiment, and the block 130 calculates the torque data I ₂ and the reversible count data [8192/2πθ ] to calculate motor control voltage data and output it to the PWM circuit 38. Thus, as explained in Embodiment 1 or Embodiment 2, the induction machine 5 generates a torque T _e that is completely proportional to the data I ₂ applied to the block 130 in the configuration shown in FIG. With the configuration shown in FIG. 15, now the command speed data R
When is larger than the actual motor speed data 8192/2πθ〓, the speed error VE becomes positive and a positive torque T _e proportional to this VE is generated in the induction motor 5, and the motor accelerates so that the speed error VE becomes zero. A speed control system based on a completely linear automatic control theory is realized by feedback to the motor speed, and the speed is controlled so that the feedback amount 8192/2πθ which is proportional to the motor speed matches the command speed data R. On the other hand, the torque control of the third embodiment, which controls the excitation current I ₀ to be weakened above the base speed θ〓 _B, can also be used for the speed control shown in FIG. 15. In this case, the block 130 shown in FIG. 15 has exactly the same configuration as the torque control computer processing section 81 of the third embodiment shown in FIG. That is, the _I2 data of block 130 in FIG.
In the I ₂ ' data of block 81 in Figure 4, [8192/2π
θ] data is 8192/2πθ data, output is V in Figure 14
Corresponds to _a ', _Vb ', and _Vc ' data. Now, in the case of Embodiment 3, the output torque T _e of the induction machine 5 in FIG. 15 is expressed as T _e =K _T as shown in equation (17).
l _n ² /l _r I ₀ I ₂ . Now, the coefficient unit 131 in FIG.
When the coefficient of is set to a constant value K, I ₂ =KV _E , and the output torque of the induction machine T _e =K _T l _n2 /l _r I ₀ KV _E. On the other hand, the open loop gain G of the speed control system is determined by T _e /V _E , so if I ₀ decreases above the base speed θ〓 _B , the gain

[Formula] also decreases, causing the inconvenience that the response of the automatic control system becomes worse at base speed θ〓 _B or higher. Therefore, the coefficient K of the coefficient multiplier 131 in FIG. 15 is changed according to the data I ^* ₀ as follows. That is, K=K ₀ /I ^* ₀ , K ₀ is a constant coefficient, and I ^* ₀ is the calculated value of I ₀ at the sample time of the I ₀ calculator 110 of the third embodiment. This I ^* ₀ enters the input terminal 144 of the coefficient multiplier 131 from the block 130 in FIG.
A calculation of =K ₀ /I ^* ₀ is performed by the coefficient unit 131. Thus, the open loop gain

[Formula] is a constant value. This is because I ^* ₀ one sampling ago and I ₀ at the time of the current sampling are almost equal. In this way, the gain remains constant no matter what the motor speed θ, and the response will not deteriorate. In the fourth embodiment, it has been explained that the torque control devices of the first, second, and third embodiments can be used for speed control.
The present invention is not limited to the above-mentioned speed control, but can be similarly applied to position control, and therefore an embodiment relating to position control will be described next. Embodiment 5 (Position Control Device) FIG. 16 is a block diagram of an embodiment of the position control device, and block 145 is a part processed by a computer. Therefore, the command position data supply means 146
supplies the command position data of the position control system to the computer 145, and when this embodiment is applied to a numerical control device etc., the command position data supply means 146 is supplied to the computer 145.
The position data supply means 146 may be included in the computer 145 and generate a function generator for straight lines and circular arcs necessary for the numerical control device using the computer 145. In the configuration diagram of FIG. 16, the components other than the computer 145 and the command position data supply means 146 are the same as the speed control configuration of the fourth embodiment, so the explanation will be omitted. Each time a sample pulse SP, which is generated every 1/1000 seconds, comes to the computer 145, the program control unit 151 sequentially executes the following programs from step 1 to step 6. First, when you run the program in step 1,
The calculator 145 calculates the contents of the reversible counter 7 [8192/2π
θ] is read and temporarily stored in the buffer register 159. However, the contents of this counter 7 [8192/
2π θ] is expressed as a digital amount within one rotation of the motor rotation angle θ, as explained in the fourth embodiment, but for two or more rotations, it has the same content as the first rotation. Therefore, in order to ensure that the position data 8192/2πθ that completely represents the motor rotation angle θ even when the motor rotation angle is one rotation or more appears in the output of the accumulation register 156, the following processing is performed. When the program control unit 151 executes the program in step 2, the subtracter 157 transfers the contents of the buffer register 159 stored in step 1 to the register 158 which stores the contents [8192/2πθ] ^* .
The following equation is calculated based on the contents of , and output data A of the subtracter 157 is obtained. Next, the program in step 3 is executed, and the data A obtained in step 2 is added to the content 8192/2πθ of the previous sampling time in the accumulation register 156, resulting in the new content 8192/2πθ of the accumulation register at the current sampling time. . This value indicates the cumulative value of the motor position, and becomes data indicating the actual rotational position of the motor 5. The accumulation register 156 is cleared to zero when the power is turned on, and its data storage capacity is large enough to completely cover the maximum position of motor position θ, so that for all motor positions θ, the register is 1: It is possible to have position data corresponding to 1. Next, the program in step 4 is executed, and the contents of buffer register 159 stored in step 1 are transferred to register 158. The contents of this register 158 are used as the previous contents of the reversible counter 7 [8192/2πθ] ^* at the next sampling time. Next, the program in step 5 is executed, and the command position data supply means 146 of the position control device
The position command digital quantity PSN is read into the computer 145, and this command value PSN is converted to the feedback data 8192/2πθ obtained in step 3 by the subtracter 155.
is subtracted to obtain position error data PE. Data R is created by multiplying this position error PE by a constant coefficient KP in a coefficient multiplier 152. This data R is calculated by computer 1
45 to block 150. The block 150 has exactly the same configuration as the speed control computer processing section 128 in the fourth embodiment. Next, when the program control unit executes the program in step 6, block 150 performs calculations similar to those described in the speed control of the fourth embodiment, and calculates the R data and the reversible counter data [8192/2πθ
], the motor control voltage data is calculated, and this data is output from block 150 to the PWM circuit 38.
Thus, as explained in the fourth embodiment, the velocity data 8192/2πθ is controlled so as to match the data R added to the block 150 in the configuration shown in FIG. With the configuration shown in FIG. 16, now the command position data,
When PSN is greater than the actual motor position data 8192/2πθ, the position error PE becomes positive, the motor rotates at a positive speed proportional to this PE, the motor position data 8192/2πθ increases in the positive direction, and the position Feedback control is performed to make the error PE zero, and position control is performed so that the motor position data 8192/2πθ matches the command position data PSN.

[Brief explanation of the drawing]

FIG. 1 is a sectional view of a two-phase two-pole induction motor to which the present invention can be applied, FIG. 2 is a diagram for explaining its operation, and FIG. FIG. 4 is a cross-sectional view of the polar induction motor for explaining the principle of the invention in correspondence with FIG. 1, FIG. 5 is a cross-sectional view for explaining the principle of operation thereof, and FIG. is a circuit diagram showing a first embodiment of the present invention, FIG. 7 is a circuit diagram showing a specific example of the configuration of the PWM circuit 38, and FIGS.
A time chart showing an example of the operation shown in the figure.
10 a to d and FIGS. 11 a and b are diagrams for explaining their operations, and FIG. 12 is a circuit diagram showing a specific example of the configuration of the power amplifiers 39 to 41 shown in the figure. A circuit configuration diagram showing an example, FIG. 13 is a characteristic diagram for explaining the example, and FIG. 14 is a third diagram of the present invention.
FIG. 15 is a circuit diagram showing a fourth embodiment of the present invention, and FIG. 16 is a circuit diagram showing a fifth embodiment of the present invention. 1-1', 2-2'...Stator winding, 3-
3', 4-4'...Rotor winding, 5...Induction motor, 7...Reversible counter, 8...Sample pulse generator, 9...Command torque data supply means, 1
0,69,81,128,145...calculator, 1
1, 50, 100, 141, 151...Program control unit, 28...Trigonometric function generator, 38
...PWM circuit, 39-41 ...power amplifier.

Claims (1)

[Scope of Claims] 1. In an induction motor with a number of poles P and a stator winding with a number of phases m, the phase voltage V _os of the nth phase given by the following formula is applied to the stator of the induction motor in order to generate torque T _e . A torque control device for an induction motor, comprising a stator voltage generating means applied to a winding. V _os = {r _s I ₀ + (l _n +l _r r _s /r _r )dI ₀ /dt} cos (
-2(n-1)/mπ) -(l _n I ₀ d/dt+ _rs I ₂ ) sin(-2(n-1
)/mπ) However, =P/2θ+r _r /l _r ∫I ₂ /I ₀ dt I ₀ = I _0B (When I _0B I _0N ) I ₀ = I _0N (When I _0B > I _0N ) I _0B : Base excitation current (constant value) [A] r _s , r _r , l _r , l _n : Constant determined by the induction motor θ : Rotor rotation angle of the induction motor [rad] I _2n : Maximum rotor current value [A] E: Maximum allowable voltage of stator voltage generation means [V] α: Constant value [V] K _T : Constant value 2 In order to generate torque T _e in an induction motor with a number of poles P and a stator winding with a number of phases m,
The first term on the right side of the following equation (2) is provided with a stator voltage generating means for applying the n-th phase voltage V _os given by the following equation (1) to the stator winding of the induction motor, and the first term on the right side of the following equation (2) is characterized by controlling the second term V ₀ on the right side of the following equation (2) so that when the phase voltage V _os exceeds the maximum allowable voltage value of the voltage generating means, the phase voltage V os becomes equal to the maximum allowable voltage value. Torque control device for induction motors. However, =P/2θ+r _r /l _r ∫I ₂ /I ₀ dt [rad
] I ₀ : Excitation current (change) [A] is I ₀ = I _0B (when I _0B I _0N ) I ₀ = I _0N (when I _0B > I _0N ) I _0B : Base excitation current (constant value) [A] r _s , r _r , l _r , l _n : Constant determined by the induction motor θ : Rotor rotation angle of the induction motor [rad] I _2n : Maximum rotor current value [A] E: Maximum allowable voltage of stator voltage generation means [V] α: Constant value [V] K _T : constant value