JPS6283503A - Digital valve closed loop control device - Google Patents

Abstract

PURPOSE:To enable the use of a digital valve for a closed loop control device, by reducing an amplification factor in case that a difference signal between an input indicating signal and a feedback signal is big and increasing the amplification factor in case that the difference signal is small. CONSTITUTION:An input indicating signal S and a feedback signal F are calculated on a computing element 10 to become a difference signal V, and it is inputted to a compression amplifier 11. And the output signal of the compression amplifier 11 is transmitted to a pulse motor control circuit 12, thereby, a digital valve 15 is driven and the control of an actuator 17 is carried out. As the amplification factor of the compression amplifier 11 is set for big in case that the difference signal is small and for small in case that the difference signal is big, the stability and the responsibility of a closed loop control system are improved, and the use of the digital valve becomes easy.

Description

[Detailed description of the invention] (Industrial application field) The present invention drives an actuator with a digital valve,
The present invention relates to a digital valve closed loop control device that controls the operation of the actuator using a feedback control method.

(Conventional Example) A conventional closed loop control device is shown in FIG. 4.

In the figure, reference numeral 1 denotes an arithmetic unit, which detects the difference between the value of an input instruction signal S for instructing the operation of the actuator to be controlled and the value of a feedback signal F, which will be described later, and generates a difference signal corresponding to the difference. ■Output.

A servo amplifier 2 has a constant amplification factor A, and outputs a drive signal AV obtained by multiplying the difference signal V by .

3 is an electric/hydraulic servo valve, 4 is a fluid pressure source, and 5 is an actuator. The servo valve 3 operates an internal torque motor in response to a drive signal AV, and uses a combination mechanism of a nozzle and a flapper to supply liquid. The hydraulic pressure signal is converted into a pressure signal, and the hydraulic pressure signal is used to move the spool to adjust the hydraulic pressure supplied from the fluid pressure source 4 to the actuator 5.

A sensor 6 detects the amount of operation of the actuator 5, and supplies this as a feedback signal F to the computing unit 1.

This feedback control method uses electrical analog signals for the input instruction signal S and the feedback signal F, and amplifies the difference signal V between them in an analog manner using a servo amplifier 2.
It supplies the torque motor of the servo valve 3, processes the input instruction signal S, feedback signal F, and difference signal
The signal is converted into an analog signal using a servo valve (nalog converter), etc., and then supplied to the torque motor of the servo valve 3.

Then, control is carried out so that the difference between the input instruction signal S supplied to cause the actuator 5 to perform the desired operation and the feedback signal F indicating the actual amount of operation of the actuator 5, that is, the value of the difference signal ■, is made zero. The input instruction signal S causes the actuator 5 to perform the operation instructed by the input instruction signal S. However, in a closed-loop control device using such a servo valve, the operation of the spool and nozzle flapper inside the servo valve is complicated by maintenance because even minute dust can cause sticks. Moreover, the servo valve has problems such as being large and expensive.

Therefore, in place of the servo valve shown in Fig. 4 above, a
Some attempts have been made to improve this problem by using digital valves that do not require special management of hydraulic fluid and are inexpensive.

(Problem to be Solved by the Invention) However, in a closed-loop control device using a digital valve, the response characteristics of the digital valve are defined by the maximum response pulse frequency F max of the pulse motor, so the response cannot be compared. If the amplification factor of the servo amplifier 2 is increased incorrectly, oscillation of the actuator output will occur, and if the amplification factor of the amplifier 2 is decreased, the input signal S and the actuator 5
The difference between the output and the steady-state deviation became large, and it was not possible to obtain sufficient performance.

That is, according to the conventional feedback control method, the fourth
A difference signal V from an arithmetic unit 1 in the figure is amplified by a servo amplifier 2, which is a so-called linear amplifier having a constant amplification factor, and a digital valve is opened and closed according to the indicated value. However, if this amplification factor is increased too much in order to reduce the steady-state deviation between the output F of the actuator 5 and the input instruction value S and further improve the output response time, as shown in FIG. In response to an input instruction signal S of There are problems in that an output higher than the value H is generated and has an adverse effect on the equipment, it takes time to settle down to the target value, and in the worst case, an oscillation state occurs.

On the other hand, in the case where the operation is stabilized and overshoot etc. are eliminated by lowering the amplification factor, the error between the output F of the actuator 5 and the input instruction value S, that is, the steady-state deviation δ becomes large, making it impossible to achieve satisfactory performance by 1q. Therefore, there was a problem that the sufficient effect of feedback control could not be obtained.

(Means for solving the problem) The present invention has been made in view of the above problems,
The response performance of a digital valve is determined by the maximum response frequency of the pulse motor, and the response time is usually slow when the width is wide, but the response time when the width is small is extremely fast. If the speed is 4000 pps, 50 m5 is required to open and close for 200 pulses, but it takes 2.5 mS to open and close for 10 pulses and 250 μs for 1 pulse, and the response speed changes extremely depending on the amount of opening and closing. Focusing on this point, we aimed to provide a digital valve closed loop control device with low steady-state deviation and high stability.To achieve this purpose, we adopted a configuration in which a digital valve drives an actuator, and a sensor that detects the amount and generates a fadeback signal, an arithmetic unit that detects the difference between the feedback signal and the input instruction signal that instructs the actuator to operate, and outputs a difference signal, and Equipped with a compression amplifier that amplifies with a compression amplification factor having /N-th power characteristics, etc., and a pulse motor control circuit that drives a pulse motor in a digital valve and instructs the valve opening based on a signal output from the compression amplifier. When the value of the difference signal is large, the amplification factor is suppressed to stabilize the operation of this closed-loop device, and when the difference signal is small, the amplification factor is increased to stabilize the response of the entire control system. The technical point is to construct a closed-loop control system with extremely small steady-state deviations while ensuring sufficient security.

(Embodiment) FIG. 1 is a block diagram showing an embodiment of a digital valve closed loop control device according to the present invention.

First, to explain the configuration, 10 is an arithmetic unit that detects the difference between the input instruction signal S supplied from the outside to instruct the actuator to perform a predetermined operation and the feedback signal F described in 1, and uses it as a difference signal ■. Output.

A compression amplifier 11 amplifies the difference signal V and outputs an opening instruction signal W.

Here, the input/output characteristics of the compression amplifier 11 are expressed as the difference signal v=S−
The following equations (1) and (2
) or (3), the larger the difference signal V is, the more the amplification factor is set to be suppressed.

w=K・Ivl"'=K・l5−Fl"'...(1
) W = -8・(l S−F l /S) ”' ・・
・・(2)w=L・log(l5−Fl/S)・
...(3) However, in the above formulas (1) and (2>,
If the constant S-F≧O, then K is a positive coefficient; if S-F<0, then K is a negative coefficient, n is a real constant, and in the above formula (3), L is a negative coefficient, and a is a real These are constants, and the characteristics can be set by appropriately changing these coefficients.

In order to realize this characteristic, the compression amplifier 11 is equipped with a device having an arithmetic function such as a microcomputer and performs an arithmetic operation based on any one of the above equations (1) to 3), or In order to obtain a result similar to that obtained by performing an operation based on one of the expressions (1) to 3), a ROM (read only memory) is used that stores data equal to the operation result of one of the above expressions in advance.
y), and the storage address of the ROM is designated according to the magnitude of the difference signal V to obtain the data.

A pulse motor control circuit 12 includes an excitation signal generation circuit 13 and a drive circuit 14, and controls the operation of the pulse motor PM of the digital valve 15 to adjust the opening degree of the variable throttle 15a.

The excitation signal generation circuit 13 determines the magnitude of the opening instruction signal W, and generates an excitation signal to rotate the pulse motor in the closing direction if the current pulse motor position, that is, the valve opening is larger than the instruction value, or in the opening direction if the opposite is true. Outputs φ1 to φ4.

FIG. 2 is a block diagram showing the configuration of the excitation signal generation circuit 13, including a comparator 13a, a discriminator 13b, and an oscillator 13C1.
Position counter 13d, pulse motor excitation phase conversion circuit 13
The opening instruction signal W is input to the comparator 13a, and the pulse motor excitation phase conversion circuit 13e generates excitation signals φ1 to φ4 for controlling each excitation coil of the pulse motor.

The details are described in "Patent Application No. 207207 of 1982" previously filed by the inventor of the present invention, but to give an overview here, the oscillator 13G generates a reference clock signal C consisting of a pulse train. , is set equal to the frequency at which the pulse motor PM can be driven at the maximum response speed.

The comparator 13a detects the magnitude of the current position signal A of the pulse motor PM and the opening instruction signal W supplied from the position counter 13d using an addition/subtraction counter, and outputs the current position signal @A.
When the current position signal Δ is larger than the opening instruction signal W, that is, when A>W, an output Q1 is generated, and when the current position signal Δ is smaller than the opening instruction signal W, that is, when A<W, an output Q2 is generated.

The discrimination circuit 13b specifies the addition/subtraction operation of the position counter 13d based on the outputs Ql and Q2.

When the output Q1 is input, the reference clock signal CL from the oscillator 13G is supplied to the position counter 13d as a subtraction pulse DOWN, and the position counter 13d realizes that A=
Counting is performed until W is reached, and the subtraction result l A−w 1
A difference correction signal @CW consisting of pulses is output to the pulse motor excitation phase conversion circuit 13e. The pulse motor excitation phase conversion circuit 13e converts the difference correction signal CW into an excitation signal of φ1 to φ4 in the case of a four-phase pulse motor, and rotationally drives the pulse motor in a direction to close the variable throttle 15a of the digital valve 15.

On the other hand, when output Q2 is input, position counter 1
3d, the reference clock signal C from the oscillator 13G is supplied as an addition pulse UP, and the position counter 13d
Counting is performed until = W, and the addition result is 1 w-A
A difference correction signal CCW consisting of one example of pulses is output to the pulse motor excitation phase conversion circuit 13e. The pulse motor excitation phase conversion circuit 13e outputs a difference correction signal C when a four-phase pulse motor is used.
Converts CW to φ1 to φ4 excitation signals and operates digital valve 1.
The pulse motor PM is rotationally driven in the direction of increasing the opening degree of the variable diaphragm 15a of No. 5.

Further, the discrimination circuit 13b receives the signal Ql from the comparator 13a.
, Q2 are not supplied, that is, when the current position signal A and the opening degree instruction signal W are equal, the supply of the reference pulse CL of the generator 13c to the position counter 13d is stopped.

Therefore, the position counter 13d does not perform addition/subtraction operation, and the excitation signals cw and c to the pulse motor excitation phase conversion circuit 13e are
CW is not supplied and the pulse motor PM is not driven.

Next, in FIG. 1, the drive circuit 14 uses drive signals φ1 to
The current of φ4 is amplified and supplied to each excitation coil of the pulse motor.

In this way, the pulse motor PM is controlled to be driven by the difference between the current position signal A and the opening instruction signal W.
It always follows the opening position indicated by the opening instruction signal W.

Accordingly, 16 is a fluid pressure source, and 17 is an actuator. Working fluid pressure corresponding to the opening degree of the variable throttle 15a is supplied from the fluid pressure source 16 to the actuator 17, and the actuator 17 is driven.

18 detects the amount of operation of the actuator 17, and supplies the detected value as a feedback signal F to the calculator 10.

The operation of the digital valve closed loop control device having such a configuration will be explained.

In the computing unit 10, an input instruction signal S that instructs the actuator 17 to perform a predetermined operation and an actual actuator 17 are input.
A feedback signal F from the sensor 18 indicating the amount of operation of
, and the difference signal V is amplified by the compression amplifier 11 based on any one of the above equations (1), (2), or (3), for example, equation (1), and The output signal W is output as a signal W.

The excitation signal generation circuit 13 compares the opening instruction signal W and the current position of the pulse motor PM, and generates excitation signals φ1 to φ4 for driving the pulse motor PM by the difference between the current position and the opening instruction signal W. is generated, the current is amplified by the drive circuit 14,
The signal is supplied to the pulse motor PM to cause the pulse motor PM to equally follow the opening instruction signal W.

The pulse motor PM rotates forward or reverse according to the excitation signals φ1 to φ4 to adjust the opening degree of the variable throttle 15a and adjust the hydraulic pressure supplied from the fluid pressure source 16 to the actuator 17.

FIG. 3 shows the steady-state deviation and response i of the actuator 17 to the input instruction signal S by this feedback control.
It shows that the quality of life has improved.

Here, when compared with FIG. 5, FIG. 5 shows the compression amplifier 11
Arithmetic unit 1 is calculated by using an amplifier with a constant amplification factor without using
The results are shown using a conventional feedback control means for amplifying the difference signal (■) from 0 and controlling the operation of the pulse motor PM, where the amplification factor G of the amplifier is 2000, the input instruction signal S is 0.4 m/sec, This is a case where the input instruction signal S is supplied in a stepwise manner.

As shown in FIG. 5, large overshoots and undershoots occur in the opening degree θ of the variable throttle 15a of the digital valve 15 in response to the input instruction signal S, and it takes a long time to stabilize at a constant value. . Furthermore, due to the instability of the opening θ, the operation η of the actuator 17 is also unstable and requires a long time to stabilize. The difference δ from the target value H specified by is extremely large.

Therefore, it can be said that the actuator 17 cannot perform the desired operation quickly, resulting in poor responsiveness.

On the other hand, FIG. 3 uses the digital valve 15, fluid pressure source 16, actuator 17, etc. under the same conditions as FIG. 4, and uses the compression amplifier 11 to which the above equation (2) is applied. In addition, in the above formula (2), n=3, K=400.
The input instruction signal S is Q, 4 m/sec, and is set so that even the lowest amplification factor is 400, the amplification factor G'' = W/V of the compression amplifier 11 at this time is shown in the following table as an example. It changes with respect to the difference signal as follows.

As shown in FIG. 3, the opening degree θ1 of the variable throttle 15a of the digital valve 15 in response to the input instruction signal S stabilizes at a constant value in a very short time, and since this opening degree θ is also stable, the actuator 17 The operation η also stabilizes in a short time.

Furthermore, the difference between the value when the actuator 17 is stable (settling state) and the target value H indicated by the input instruction signal S, that is, the steady-state deviation δ is extremely small.

In this way, since the amplification factor can be increased overall compared to conventional amplifiers, it is possible to obtain the effect that the difference δ" between the target (1fH and the set value of the actuator) can be made extremely small, and Even if the amplification factor is increased, an unstable state such as oscillation does not occur, so the actuator 17 can quickly perform the desired operation, and responsiveness is improved.

In the compression amplifier 11 of this embodiment, the above formulas (1) to 3 are used; however, the amplification factor is not limited to this, and as the value of the input difference signal becomes smaller, the amplification factor increases. Any device can be applied as long as it has a characteristic that the amplification factor decreases as the signal value increases.

(Effects of the Invention) As explained above, according to the present invention, the difference signal between the input instruction signal and the feedback signal is amplified by the compression amplifier and the actuator is used as a feedback side. Since the rate can be increased overall, the actuator can be driven and controlled extremely stably and accurately to a predetermined value indicated by the input instruction signal, and responsiveness can be improved.

[Brief explanation of drawings]

FIG. 1 is a block diagram showing an embodiment of a digital valve closed loop control device according to the present invention, FIG. 2 is a block diagram showing the configuration of a pulse motor control circuit in the embodiment of FIG. 1, and FIG. FIG. 4 is a block diagram showing an example of a conventional closed-loop control device, and FIG. 5 is an explanatory diagram illustrating the problems of the conventional closed-loop control device. be. 10: Arithmetic unit 11: Compression amplifier 12: Pulse motor control circuit 13: Excitation signal generation circuit 14: Drive circuit 15: Dijoital valve 16 Two-fluid pressure source 17: Actuator 18: Sensor Fig. 8 Fig. 5

Claims (1)

[Claims] A digital valve that controls a working liquid from a fluid pressure source driven by a pulse motor, and detects the amount of operation of an actuator driven by the working liquid controlled by the digital valve. A digital valve closed loop control device that performs feedback control of the actuator includes a sensor that detects the amount of operation of the actuator and generates a feedback signal, and a sensor that detects the difference between an input instruction signal that instructs the operation of the actuator and the feedback signal. an arithmetic unit that detects and generates a difference signal; and an amplification factor that increases the amplification factor as the value of the difference signal becomes smaller and decreases the amplification factor as the value of the difference signal becomes larger. A digital valve closed-loop control comprising: a compression amplifier for amplification; and a pulse motor control circuit for driving a pulse motor in a digital valve and setting the valve opening based on a signal output from the compression amplifier. Device.