JPS5843194A - Speed controller for motor - Google Patents

Abstract

PURPOSE:To smoothly start a motor in a short time by measuring the acceleration of the motor at the starting time, stopping the motor for a short time to measure the deceleration of the motor, and correcting the integrated value of the difference between the instructed value and the actual speed with the measured values. CONSTITUTION:A motor 2 is driven by a bridge circuit of transistors 20-23, diodes 24-27, the rotating signal N of an encoder 4, a speed instruction Nc, a normal or reverse command, and start command are inputted to a microcomputer, thereby controlling transistors 20-23 through a drive circuit 7. The proportional value of the difference between the speed instruction Nc and the actual speed N and the time integrated value are fed back to operate them, the acceleration obtained at the starting time and the deceleration obtained by stopping the motor for a short time are measured, thereby correcting the time integrated value with the measured values. Accordingly, an overshoot can be eliminated to attain to the constant speed in the minimum time to start the motor even if the load and the voltage vary.

Description

DETAILED DESCRIPTION OF THE INVENTION The present invention relates to a control device having a finger speed generator for measuring rotation of an operating shaft and having a switching element for driving an electric motor at a predetermined speed.

Conventionally, in order to accurately control the speed of a DC motor, a finger speed generator such as an encoder was attached to the motor or drive shaft.
A common method is proportional control, which feeds back this output to keep the speed constant.

However, in this method, since the deviation component between the speed command value and the actual speed detected by the encoder or finger speed generator is fed back together, the deviation component cannot be reduced to zero, resulting in an error between the command value and the motor speed.

A method to solve this problem is proportional control, which provides feedback proportional to the deviation between the speed command value and the motor speed.
There is also an integral control method in which the deviation amount obtained by integrating the above deviation component is fed back, and a control method in which # is added.

On the other hand, there are many information devices and the like as loads on motors, and the moment of inertia or load torque of the motor shaft to be used as an image also varies. In this case, in conventional equipment, the motor is directly connected to the equipment each time, and the control loop gain, constant of the integral term, etc. are all adjusted so that the speed rises as smoothly as possible without overshoot or undershoot at startup. I was doing it. Additionally, if the power supply voltage that drives the motor changes, this is equivalent to a change in the gain of the control loop, so either the gain, etc. must be readjusted each time, or a circuit that detects the voltage and automatically adjusts the gain must be added. There was a need.

The object of the present invention is to provide a speed control device that can be started and operated to a constant speed in the shortest possible time even if the load or the inertia of the load or the voltage applied to the motor changes, and that there is no overshoot or vibration during startup. Its purpose is to provide speed control.

Consider elements related to speed control. First, speed command value Nc
There is. This is a target value for controlling the speed of the Lh motor to a certain value. Check motor shaft speed in winter 61”: /1-15
There is a speed N at 1''1·rl''6''6°6゛-e. There are voltage components corresponding to the load torque and the back electromotive force of the motor, and let this component be Σ. The output voltage V @ ut of the circuit applied to the motor by these n is expressed by equation (12).

V −−t = G+ (Nc −N)+Σ,...
・・・・・・・・・(1), The first term on the right side of equation (1) is called a proportional term, and when the difference between the speed N and the command value Nc is large, that is, during acceleration, the motor output V o a t This term becomes zero when the speed N becomes equal to the command value Nc.

G is a constant of this term and corresponds to the gain. Therefore, at startup, this term becomes a dominantly large value and accelerates the motor. Including cases where digital control is performed, the second term is an integral term corresponding to the load torque and motor speed electromotive force, and the previous value is converted to Σ. When set to -1, the current Σ is given by the following equation.

To Σ,, = 1 to Σ. -8+ K a ・・・・・
(2) Ka is the value given by the command Nc and the speed N.

In such control, the first term on the right side of equation (1) acts by generating acceleration torque, which becomes zero in a steady state. Furthermore, the term ``term'' corresponding to the load and the back electromotive force of the motor is integrated during acceleration, and becomes 1 at a certain constant value Σ in a steady state. From this result, if the value of Σ at the end of acceleration is equal to the steady state value Σ, which is equal to 1, the speed becomes stable immediately.

In addition, the power generation constant 't- of the DC motor is (V/rp
m), the steady-state speed of the motor tNt (rp
The armature resistance of the mχ motor is R2O), and the load torque is Tt.
(Kt-m), torque constant t K T (Kf-m/
A), the voltage vt required for the motor in steady state is given by the following equation.

For simplicity, let Kdl be the constant value of Kat in equation (2). Assuming that the acceleration part is ttt, the integral term in equation (1) is expressed by the following equation at the end of acceleration.

Σk, e ” Ka I X fl ・・・・・・・・・
(4) If two accelerations are ideally performed, the condition is that equations (3) and (4) are equal. That is, if the acceleration is α, and Ka*' is determined from equation (6), equation (7) is obtained.

That is, returning to the beginning, the illusion regarding the integral term Σ in equation (1) is determined if the acceleration α1 and the load Tt are determined as shown in equation (7). The load inertia is measured by the acceleration α as a response to the load, and if n is calculated as Ka+ corresponding to n, it is possible to respond to changes in the load inertia. Load T compared to inertia
When t is small, it is sufficient to consider only the term for the back electromotive force, that is, the first term on the left side of equation (6), to calculate the steady state value.
1 is expressed as equation (8).

K-1= α, ・K-・・・・・・・・・・・・
... (8) The simplest way to incorporate load adaptation into speed control is to measure the acceleration α1 as shown in (8), then perform Ka'ft correction on the integral term of the equation (1). In this case, stable startup is completed in the shortest time, and the value Σ of the integral term at this time becomes equal to the value Σ for the back electromotive force in the steady state. ,become.

If the load torque is large compared to the acceleration torque for inertia, it is difficult to achieve stable startup in the shortest possible time even with the above method. Therefore, a method of measuring the load torque Tt and stably starting the engine in the shortest time will be described below.

This method completely decelerates the motor during a short period of time during startup.

If the drive of the motor is stopped while the motor is rotating at a certain speed, the motor is decelerated by the load torque Tt. Letting this deceleration be α2 and inertia tJ, the equation (9) is obtained.

- Similarly, if acceleration α1 is motor torque kTw (1
0) Expressed by the formula.

(9) formula and (10) formula, (11) military force! Lead or take.

If the motor torque TM is the motor current tI (12)
Substituting this relationship into equation (7), 7TM=
To? XI ・・・・・・・・・・・・・・・・・・
(12) and (13).

・・・・・・・・・・・・・・・・・・(13)
Even if the load inertia and load change, L can respond to the load change by measuring the acceleration α and deceleration α2 and calculating Kal corresponding to these values. In order to incorporate load adaptation into speed control and start up the motor to the set speed in the shortest time, acceleration/deceleration α1. α3. Stable startup is completed by correcting Ka' related to the integral term in equation (1) by performing all measurements, and the value of the integral term at this time Σ
Then, the value Σ for the back electromotive force in the steady state becomes equal to 1, and the motor speed N becomes ≦゛steady speed N1. ′
This is how it looks in Figures 1 and 6. This will be explained with a diagram. FIG. 1 shows the case of (8) only correction G, where the broken line shows the case without K at -correction regarding the integral term, and the solid line shows the case with correction. The example in FIG. 1 is an example in which the motor speed rises quickly when the load inertia and load are small. In the case of the broken line where Ka is not corrected, the motor speed N approaches the steady value and the proportional term G+ (Ns
Even if N) becomes small, the value of the integral term Σ remains the steady-state value Σ
Since the motor speed N does not reach 1, the motor speed N also does not reach a steady value.
The motor speed is not stable until time t, when the integral term reaches a steady value. On the other hand, Ka for the integral term shown by the solid line
In the method of correction using equation t(8), if the motor speed N is Nsl within a certain time t, the acceleration αt is N
It can be found by s1/1.1, and if time 'ut is chosen to be a sufficiently small value, the value KatKa related to the integral term at one time point is
Correct it to l and make it Σ, with a slope similar to the solid line.

In this way, when the motor speed N approaches the steady value N1, Σ also approaches the steady value X k ,l. Therefore, the start-up time t can be made much smaller than 1t.

Next, FIG. 2 shows a case where the correction of equation (13) is further applied to FIG. 1. The broken line is the corrected case using equation (8), and the solid line is the example of equation (13). In the example of equation (13), the weight is measured, and in 1, the motor speed N, tt is measured, and! ,2~t゛,3
The motor drive is cut off only during the time Δt until the motor speed α
t'f: Calculate the equation 13) to correct Kil regarding the integral term as shown by the solid line. In this way, when the motor speed N reaches the steady state value N1, it also reaches the steady state value Σ, and the proportional term Gt (Nc-N) becomes zero, so the motor speed smoothly shifts to the steady state. . Therefore, the motor starting time can be shorter at t3 than at t8 as indicated by the broken line.

Next, another fll--which starts up in the shortest time--will be explained. Substituting equations (11) and (12) into equation (3), we get (1
4) It becomes Eq.

These four equations are equations that provide the necessary voltage when the motor is in a steady state. By comparison, the value of 1 itself for the integral term Σ becomes (
14) When the motor speed tN reaches N1 at steady state, the value of the equation (14) is set to Σ, and when the motor speed tN reaches N1 at steady state, the value of the equation (14) is set to 1. It also saves the effort of integrating. This situation is shown in Figure 3. In Fig. 3, the acceleration α at startup is measured, and then the output Vat is turned off for a short time Δt, and the deceleration α at that time is
2, α1. Equation (14) is calculated using α2 to obtain the steady state voltage vt'fc. When the motor rotational speed N reaches the steady rotational speed N, the output V0ut'ii- is turned off, and the value obtained by equation (14) is entered into the integral term Σ. If this value is equal to Σ, the startup will be completed in the shortest possible time.

FIG. 4 shows an example of the configuration of the present invention, in which a power supply 1 is connected to an H-type bridge circuit of transistors 20-23 and a bridge circuit of diodes 24-27. Power supply 1
The collector of the transistor 20.22 and the cathode of the diode 24.26 are connected to the positive terminal side of the transistor 20.22, and the emitter of the transistor 21.23 and the diode are connected to the negative terminal side of the
25゜27,) A, -4, connect self 1'1. ing. 1
,7.

The emitter of 20 is connected to the collector of 21 registers,
The other end of the motor 2 is connected to the emitter of the transistor 22, the collector of the transistor 23, the anode of the diode 26, and the cathode of the diode 27.

The shaft of the motor 2 drives the load 3 and the encoder 4, and the rotation signal 10 which is the output of the encoder 4 is sent to the speed detection port W1.
15 is input. The speed signal 4N, which is the output of the speed detection circuit 5, is input to the microcomputer circuit 6.

Furthermore, the microcomputer circuit 6 receives from an external device,
Speed command Nc, forward/reverse command 16, start command 17
It takes away human power. The microcomputer circuit outputs a duty signal 12 that determines the duty when controlling the transistors 20 to 23 by chopping, and a forward/reverse signal 13 that generates forward rotation torque and reverse rotation torque in the motor.
It is input to the drive circuit 7. Drive time! The normal output 14, which is the output of transistor 2.7, is the output of transistor 2.7. ”'114
), 2'30 base, the reversal output 15 is connected to the pace of the tiger registers 21, 22.

The configuration is as described above, and the operation is as follows. When a speed command Nc, a reverse rotation command 16, and a start command 17 are given to the microcomputer circuit 6, the motor speed N is taken in from the speed detection circuit 5 and the command is issued. It is compared with the value Nc, and an error value (proportional term) is calculated according to the difference. In addition, the integral term is calculated based on the error value, the voltage to be applied to the motor is determined based on these, and the voltage to be applied to the motor is output as the duty signal 12' for chopping. Outputs signal 13. This causes the drive circuit 7 to operate, and in the case of normal rotation, the transistor 20.
23, the motor is turned on in response to the duty signal 12 to rotate the motor.

Since the motor speed N is low at the beginning of startup, the difference from the command value Nc is large, and the value Σ of the integral term is also sufficiently small, so the duty signal 12 for turning on the motor is Gt (N
Decided by c N)! It becomes a large duty signal and accelerates the motor. As the motor approaches the speed command value Nc, Gt (Nc N) becomes smaller and the integral term Σ
increases, the startup is completed, and at the 7'CEI point, the speed command value Nc and the motor speed N become equal. At exactly this point, the integral term is corrected so that the value Σ of the integral term becomes the steady value Σ, so the motor smoothly settles to the steady speed N.

Each block in FIG. 4 will be explained in more detail.

An example of the speed detection circuit 5 is constituted by a counter 51 and a latch 52 as shown in FIG. In FIG. 5, the rotation signal 10 from the encoder 4 is input to the clock input of the counter 51, the enable signal 53 for a certain period of time is input to the enable signal of the counter 51, and the reset signal 54 is input to the reset terminal. . Output S of counter 51.

~S11 is input to the latch 52, and the latch 52
The output is taken out as a speed signal N to the microcomputer. A strobe signal 55 is input to the latch 52. In this operation, as shown in the time chart of FIG. 6, the force counter 51 performs an opening operation when the enable signal 53 is received, counts the rotation signal 10, and outputs the signal to the outputs S0 to S1. Next, the strobe signal 55 of the latch 52 is applied.

The contents of ~S are latched by the latch 52, and the counter 51 is reset by the next instantaneous reset signal 54 to prepare for the next count. Therefore, the enable signal 5 for a certain period of time is
3, and a value proportional to the motor speed can be obtained as the speed signal N.

Next, the drive circuit 7 is composed of an inverter gate 71 and AND gates 72 and 73 as shown in FIG. The duty signal 12 is connected to one input of the AND gates 72 and 73, and the forward/reverse signal 13'i is connected to one input of the AND gate 72, and the other input of the AND gate 73 is connected through an inverter gate 71. With this configuration, as shown in FIG. 8, the output of the AND gate 72 is such that the duty signal 12 appears only when the forward/reverse signal 13 is at the level, and becomes the forward rotation output 14. Further, the output of the AND gate 73 becomes the reverse output 15 since the duty signal 12 is displayed only when the forward/reverse signal 13 is at the θ level.

A microcomputer has a central processing unit and RAM.

-::□:,: Consists of ROM, input/output, etc. n, recorded in ROM f'
Operates by Lfc program. FIG. 9 shows this operational block. Upon receiving a speed command Nc, a start command 17, and a forward/reverse command 16 from an external device, the motor speed N is
k and compare the speeds, and from the difference between the command value Nc and the motor speed N, the proportional term G1 (Nc
N) Calculate k, calculate the entire acceleration from the speed at a certain fixed time, calculate the integral term, calculate the duty, determine forward/reverse, and output the duty signal 12 and forward/reverse signal 13.

The operation of the present invention will be explained with reference to the flowchart in FIG.

When the start command 17 in FIG. 4 is input, the flowchart starts. First, all values of the reference time t for measuring the acceleration α and the integral term of equation (1) are set. Next, temperature command Nc, motor speed N1 forward/reverse command RW
Read in and calculate the proportional term in equation (1). Then, it is checked whether the difference between the speed command Nc and the motor speed N is zero or large, that is, whether the motor speed 1□1N has reached the steady speed N,
.

At the speed N1, the integral term Σ remains as before and outputs a chuty Dr and a forward/reverse signal RO commensurate with this.
Speed command again NC% Motor speed N1 Forward/reverse command Rw
? Load and repeat the same sound. If the motor speed N is not equal to the command value Nc, check whether the time from the start of startup is equal to the initially set t, 1, and if it is equal, calculate the acceleration α from the instantaneous speed Nm1 at that time. do. if,
If the time from the start of activation is outside the set value, the acceleration α1 is not calculated and kd is not corrected. Next, calculate Σ as shown in equation (2) to obtain the value of i tancho, store all of this value in memory, and use it for the next repeated calculation.

When the calculation of the acceleration α1 is completed, the duty Di' is set to zero in order to measure the deceleration α2. When this fixed time Δt has elapsed, the deceleration α2 during the fixed time Δt is calculated, − correction is made for the integral term, Σ is calculated, and the duty D is also calculated. When the startup is completed, the speed N is equal to the command speed Nc, the acceleration α1,
Calculation of deceleration α2 is not performed.

As described above, according to the present invention, by stopping the drive once at startup, stable startup is possible with the shortest time required to rise to steady state. In particular, even when the load or load inertia changes significantly, stable startup is always possible in the shortest possible time.

[Brief explanation of the drawing]

FIG. 1 shows an example of conventional starting characteristics. FIG. 2 is an example of the starting characteristics of the present invention. FIG. 3 shows another example of the starting characteristics of the present invention. FIG. 4 shows an example of the hardware configuration of the present invention. FIG. 5 shows an example of the speed detection configuration of the present invention. FIG. 6 is an explanatory diagram of the operation of the speed detection circuit of the present invention, and FIG. 7 is an example of the drive circuit of the present invention. FIG. 8 is an operational explanatory diagram of the drive circuit of the present invention, and FIG. 9 is an operational block diagram of the microcomputer of the present invention. FIG. 10 is a flowchart showing one example of the present invention. Nc...speed command value, Nc2...temporary start speed setting value, N...motor speed, NI...motor steady speed,
GI: constant determined by gain, etc., Kd: integrand value regarding integral term, 6 for Σ: value of integral term, 61 for Σ
...The value of the integral term at steady motor speed, Σ. ...Final value of the integral term for temporary startup, 1...Power source, 2...Motor, 3...Load inertia and load, 4...Encoder,
5... Speed detection circuit, 6... Microcomputer, 7... Drive circuit, 20-23... Transistor, 24-27... Diode, Dt... Chopping duty, RW... Fig. 1 Fig. 3 Otoko Fig. 5 Otoko Fig. 6 Fig. 53 - Torture Fig. 8 Fig. 9 Fig. 1 + 51'J- Fig. 10

Claims (1)

[Claims] 1. A motor having a finger speed generator for measuring speed on the operating axis is driven at an arbitrary speed by a speed command, and at least a proportional term due to the difference between the speed command value and the motor speed, and a speed command value In motor tube control using feedback that includes an integral term obtained by time-integrating the difference between measure,
1 depending on the acceleration and deceleration values. A motor speed control device characterized in that the integral term is corrected. 2. In the product described in claim 1 above, the motor output at the time of startup is set as the maximum value, and the value of the integral term in the steady state obtained from the acceleration/deceleration at the time of startup is used at the same time as the startup is completed. A motor speed control device characterized in that: