JPH0338442B2 - - Google Patents

Description

DETAILED DESCRIPTION OF THE INVENTION [Field of Industrial Application] The present invention relates to a method of controlling a fluid actuator used in construction machinery and the like.

[Conventional technology]

FIG. 6 is a system diagram showing a conventional hydraulic switching valve operating device, in which a hydraulic switching valve 5 that hydraulically controls a hydraulic motor 6 has switching positions 5a, 5b, and 5c. The switching position is configured to be operated by the hydraulic actuator 1 or the lever 1c, and pressure oil is forced to be sent or removed from the displacement chamber 1a or 1b in the hydraulic actuator 1 by operating the switching valve 2 or 3. It is structured as follows.

The solenoids 2d and 3d for operating the switching valves 2 and 3 are manually turned on and off by operating a switch box 11 via electrical wiring.

The operation of the hydraulic switching valve operating device shown in FIG. 6 will be explained as follows.

The hydraulic motor 6 is driven by hydraulic power from the hydraulic switching valve 5, and the hydraulic switching valve 5 is in the switching position 5b.
, the flow of hydraulic power is closed and the operation of the hydraulic motor 6 is stopped.

On the other hand, when the hydraulic switching valve 5 is at the switching position 5a or 5c, hydraulic power is force-fed to one or the other of the hydraulic motors 6, driving the hydraulic motor 6 in one or the other direction.

Further, the hydraulic motor 6 drives a crane or the like (not shown).

The operation of the hydraulic switching valve 5 is performed by operating the hydraulic actuator 1 or manually operating the lever 1, and the operation of the hydraulic actuator 1 is as follows.

When the switching valves 2 and 3 are set to switching positions 2b and 3b, respectively, as shown,
Both displacement chambers 1a and 1b are each open to the reservoir 7, and the hydraulic actuator 1 is in a completely free state.

Therefore, in this state, when the driver operates the lever 1c from the driver's seat, the piston 1d in the hydraulic actuator 1 can be freely operated, and the operated piston 1d
operates the hydraulic switching valve 5.

In this way, by manually operating the lever 1c, the above switching position 5a, 5b or 5c can be arbitrarily selected, and the operation of the hydraulic actuator 1 is controlled by the driver from the driver's seat. The lever 1c can be operated while the user is in the room.

In addition, in the hydraulic actuator 1, the lever 1
A centering action is exerted on the piston 1d connected to the piston 1d by the springs 5e and 5f of the hydraulic switching valve 5.
Therefore, in a state where both the displacement chambers 1a and 1b are open to the reservoir 7 as described above,
When the driver releases the lever 1c, the hydraulic actuator 1 is in the illustrated position, and its switching position is set to 5b (neutral position).

For such manual operation, move the hydraulic actuator 1 away from the driver's seat, for example near the load of the hydraulic motor 6, while checking the movement of the load.
When remotely controlling the
The switch is manually operated to drive the solenoid 2d or 3d, and this drive operates the switching valve 2 or 3, and the operated switching valve 2 or 3 operates the hydraulic actuator 1.

Here, the operation of operating the hydraulic actuator 1 by operating the switching valve 2 or 3 is as follows.

When it is desired to fix the piston 1d in the hydraulic actuator 1 at an arbitrary position, the switching valves 2 and 3 are set to the switching position 2a and the switching position 3a, respectively.

When the switching positions of the switching valves 2 and 3 are set in this way, the pressure oil from the hydraulic source 4 is forced to be sent to both displacement chambers 1a and 1b, and the switching positions of the switching valves 2 and 3 to which the pressure oil is sent are Since check valves 2c and 3c are provided at the inlet portions, the movement of the piston 1d is fixed in the hydraulic actuator 1.

In this state, the displacement chambers 1a and 1b
Since oil pressure from the oil pressure source 4 is introduced to both of them, oil pressure is applied to the piston 1d from both sides of the piston 1d.

If only the switching valve 2 is set to the switching position 2b in such a state, the hydraulic pressure in the displacement chamber 1a is released to the reservoir 7 via the switching valve 2, so the piston 1d is moved to the hydraulic pressure in the displacement chamber 1b. Therefore, it is pressed downward in Fig. 6,
The pressure changes the switching position of the hydraulic switching valve 5 to 5.
Shift to side a.

Conversely, when the piston 1d is fixed and only the switching valve 3 is set to the switching position 3b, the hydraulic pressure in the displacement chamber 1b changes to the switching valve 3.
The hydraulic pressure in the displacement chamber 1a presses the piston 1d upward in FIG. is moved to the switching position 5c.

As mentioned above, the operation of the hydraulic switching valve operating device is as follows:
The operator operates the switching valve 2 or 3 using the manual switch, and determines whether the operation is appropriate or not while observing the movement of the load on the hydraulic motor 6.

However, since the remote control of the switching valve 2 or 3 shown in FIG. 6 is based on on/off operations using a manual switch, it is difficult to perform delicate operations.

For this reason, in such remote operation, a method is desired in which the position of the piston 1d is commanded for the operation, and the operating position of the hydraulic switching valve 5 is feedback-controlled in response to the command.

Furthermore, it is desired that the feedback control method be inexpensive and durable, and the following methods are examples of such durable and inexpensive control methods.

When a deviation occurs between the command value and the feedback value, the hydraulic actuator is controlled in a direction to correct the deviation. The steps up to this point are normal feedback control.

In the control for correcting this deviation, the signal that controls the hydraulic actuator 1 is to force-feed pressure oil from the solenoid valve to the displacement chamber of the hydraulic actuator 1 as a single pulse, and the pulse application time duration of the single pulse is as follows: The values are based on experimental data that correspond to the fact that the deviation can be just corrected by the single pulse alone.

This means that the deviation is corrected each time it occurs by sending a single pulse of pressure oil to the hydraulic actuator 1 or sending a single pulse signal to its solenoid valve. .

In other words, from the transmission of a command signal instructing the position of the piston 1d, a feedback signal indicating the current position of the piston 1d is obtained, the difference between them and the resulting deviation are calculated, and the above-mentioned single pulse is calculated based on the deviation. Until the signal is sent, normal feedback control is in effect. Subsequently, a single pulse signal is transmitted to the solenoid valve based on this deviation result, and until the control of the hydraulic actuator 1 is completed for the time when the signal is transmitted, open-loop control is performed. .

If there is still a deviation in the control result of that time, the same single pulse control is performed based on the new deviation from the control result to converge the position of the piston 1d to the position specified by the command signal. let

Therefore, this method is inexpensive in that an inexpensive on/off solenoid valve can be used without using an expensive electro-hydraulic servo valve for proportional control.
Moreover, the on/off operation of the solenoid valve is different from normal solenoid valve control in which the solenoid valve is always turned on and off while changing the duty ratio depending on the magnitude of the deviation. That is, this control method uses an inexpensive solenoid valve, and the number of times the solenoid valve is turned on and off is extremely small, resulting in improved durability of the solenoid valve.

Furthermore, the application time of the single pulse signal in this control method is the sum of a fixed basic application time component that only starts the no-load hydraulic actuator 1 and an application time component that is proportional to the deviation. There is.

[Problem to be solved by the invention]

In the above control method, the fact that the basic application time is a fixed constant time means that no matter in which direction the hydraulic actuator operates or in which region the hydraulic actuator operates, the piston 1 d This is satisfied only when the conditions required for activation are the same.

However, as can be understood from the above explanation in FIG. Piston 1d at
The driving force that starts to start is different. And also,
Depending on the initial settings of the centering springs 5e and 5f or differences in their spring constants, the driving force at which the piston 1d starts to actuate in each operating region on the spring 5e or 5f side differs.

An object of the present invention is to provide a method for controlling a fluid actuator in which a single digital signal is used to control a fluid actuator whose activation force differs depending on the operating direction and each operating region of the piston 1d as described above. It's about doing.

[Means to solve the problem]

In the present invention, the stroke of the piston is in the first region and in the second region.
The target area is the control of a hydraulic actuator that generates different load forces when the piston moves to the forward side and when the piston moves to the return side. .

Its control is turned on and off by switching the switching valve.
The pressure oil supply generated off-line controls the position of the hydraulic actuator.

The on/off switching of the switching valve is driven by a single pulse signal Vn with a finite application time width tn, which is the basic application time width to and a correction application time width te, and the correction application time width is The value is proportional to the deviation Er between a detection signal detecting the position of the piston in the axial direction and a command signal instructing the position of the piston in the axial direction,
And for the deviation, only the value immediately before the transmission is used for each single pulse signal transmission. In the above configuration, the basic application time width is set for each region as follows: a: when the piston moves in the forward direction; b: When the piston is controlled in the return direction, the other basic application time width for each region is It is characterized by consisting of the above operations, which are the values of the time span.

[Effect]

As mentioned above, in the hydraulic actuator to be controlled used in the present invention, the piston stroke is the first stroke.
A configuration in which the hydraulic actuator is divided into a region and a second region, and in each region, different magnitudes of load force are generated when the piston in the hydraulic actuator moves to the forward side and when it moves to the return side. It is becoming.

In this case, from when the switching valve opens to when it closes, pressure oil is being forced into or discharged from the displacement chamber of the hydraulic actuator due to the opening, and the opening time is As time passes, the differential pressure in the displacement chamber increases.

This means that the increase in the differential pressure in the displacement chamber increases as the switching valve opens for a longer period of time.

Moreover, this corresponds to the length of the opening time of the switching valve and the magnitude of the load force on the hydraulic actuator.

For this reason, in the present invention, the application time of a single pulse signal for controlling the transition position of the hydraulic actuator and commanding the opening of the switching valve is the basic application time of the total application time. The width component is set to have a different value for each region of the piston when going forward and returning.

In other words, in each region, if the load resistance of the piston increases, the time width of the basic application time width component is increased for that region.
As the length increases, the differential pressure in the displacement chamber in the hydraulic actuator increases, and it is possible to cope with the increased load resistance.

Conversely, in the case of return where the load resistance of the piston decreases in each region, if the basic application time width component is shortened for each region, the displacement in the hydraulic actuator will be reduced by the shorter time. As the differential pressure within the room decreases, it is possible to cope with the decrease in load resistance in each area.

This also means that when a deviation occurs in control, it is possible to correct the deviation to almost zero using only a single pulse signal corresponding to the deviation. There is.

〔Example〕

The present invention will be explained below using examples.

FIG. 1 shows an embodiment of a hydraulic switching valve control device for carrying out the fluid actuator control method of the present invention, and the illustration is shown in the form of a system diagram.

The portion of the hydraulic switching valve operating device Y in FIG. 1 is the same as the conventional one in FIG. 6, as described above, and therefore a description thereof will be omitted.

In FIG. 1, the signal generator 10a is the lever 1
By operating 0i, a voltage signal is transmitted to the wiring 10b, and the voltage signal is a command signal that instructs the position of the piston 1d in the hydraulic actuator 1, and the command signal is sent to the electronic control device. 10 is entered.

Output wirings 10g and 10h in the electronic control device 10 are connected to solenoids 3d and 2, respectively.
d.

The movement of the piston 1d, that is, the movement of the hydraulic switching valve 5, is detected by a potentiometer 5d,
The output wiring 10e for transmitting the detected signal serves as a feedback signal wiring in the electronic control device 10.

The circuit configuration of the electronic control device 10 is shown in FIG. 4, and is illustrated by a block diagram.

In FIG. 4, 10A is a detection storage means;
B is a deviation calculation means, 10C is a drive judgment means, 10
E represents a tolerance setting means, 10F a basic time change storage means, 10J a pulse width calculation means, 10K a basic time selection means, 10L an FB reference value setting means, and 10M an output transistor.

The operation of the configuration of the present invention shown in FIGS. 1 and 4 will be described below.

First, in the hydraulic actuator 1 shown in FIG. 1, the driving force at the beginning of activation will be considered.

FIG. 2 shows a skeleton diagram of the hydraulic actuator 1 in FIG. 1. but,
In FIG. 2, the spring 5 in FIG.
e and 5f are adapted to apply their biasing forces to the piston 1d within the hydraulic actuator 1.

In the case of FIG. 1, the springs 5e and 5f exert a biasing force on the piston 1d in the hydraulic switching valve 5, but this biasing force is the same even if it acts on the piston 1d in the hydraulic actuator 1. That's true.

Therefore, in FIG. 2, for convenience of explanation, the springs 5e and 5f are enclosed within the hydraulic actuator 1.

That is, in FIG. 2, the cylinder 1e has:
A piston 1d is fitted to enable sliding movement in the axial direction.
The biasing forces of springs 5e and 5f act in opposing directions. The action of these two biasing forces is such that when no external force is acting on the piston 1d, the piston 1d is set at the neutral position n.

In the configuration shown in FIG. 2 as described above, when the piston 1d moves a distance x to the right from the neutral position n as shown, an axial force of approximately k1·xi is applied to the piston 1d to the left. This means that it is working.

In addition, k1 is the spring constant of the spring 5f,
xi is the total compression amount of the spring 5f, and the relationship between x and xi is xi=xo+x, where xo is the initial compression amount set from the free length of the spring 5f to the neutral position n.

Therefore, the starting force for moving the piston 1d further to the right from this position of the piston 1d requires approximately a force of k1·xi.

On the contrary, the starting force for moving the piston 1d further leftward from this position is the value driven by k1·xi.

For the above case, the piston 1d is at the neutral position n
When the piston 1d is located at a distance x to the left from that position, the force required to activate the piston 1d is approximately k2·xi when the piston 1d moves further to the left from that position.

Here, k2 is the spring constant of the spring 5e, and xi is used as the total compression amount of the spring 5e, similar to the above spring 5f. Therefore, the relationship between x and xi in this case is also the same as in the case of the spring 5f.

Moreover, the starting force when moving the piston 1d to the right from the position of the piston 1d is
The value is driven by k2・xi.

That is, the types of starting force required for the piston 1d are summarized as follows.

When the piston 1d is on the right side from the neutral position n, 1) When the piston 1d tries to move in the direction of arrow a1, and 2) When the piston 1d tries to move in the direction of arrow a2, 2 of In contrast, when the piston 1d is to the left from the neutral position n, 1) the piston 1d attempts to move in the direction of arrow a3, and 2) the piston 1d moves in the direction of arrow a4 When attempting to move in the direction of , there are two types of cases: .

In the end, there are four basic types of force required to start the piston 1d in FIG.

Next, the above-mentioned starting force required for the piston 1d,
The relationship between the hydraulic pressure of the displacement chamber 1a or 1b necessary for its activation will be examined.

Taking as an example the case where the piston 1d is activated in the direction of the arrow a1 from the position shown in FIG. 2, the force required for activation is k1·xi, as described above.

On the other hand, in order to activate the piston 1d by hydraulic pressure, it must be activated by the differential pressure dp between the hydraulic pressure generated in the displacement chamber 1a and the hydraulic pressure generated in the displacement chamber 1b. Become.

Such a pressure difference is caused by the above-mentioned switching valve 2 or 3.
When the switching valve 2 or 3 is driven by a single pulse signal with a finite time duration, the phenomenon of the differential pressure occurring is as follows.

As mentioned above, in order for the piston 1d to be fixed at the position shown in FIG. 2, the switching valves 2 and 3 in FIG. 1 are set to the switching position 2a and the switching position 3a, respectively. .

When moving the piston 1d from the fixed position of the piston 1d in the direction of the arrow a1,
Only the switching valve 3 is set to the switching position 3b, and the switching valve 2 is set to the switching position 3b.
The switching position in is left as 2a. In addition, the method of setting the switching valve 3 to the switching position 3b is the third position of the pattern in which the solenoid 3d is turned off.
A single pulse signal V1 as shown in the figure is sent to the solenoid 3d.

In FIG. 3, the horizontal axis t indicates elapsed time, the vertical axis V indicates the output voltage of the pulse signal V1, V2, or V3, and the vertical axis p indicates the hydraulic pressure characteristics pa1,
It shows the hydraulic pressure of pa2 or pa3.

In this case, at the moment when the solenoid 3d is turned off, the working oil pressures p in the displacement chambers 1a and 1b are both approximately the same pm.
The value is .

From this state (t=0 in FIG. 3), time t
As time elapses, the working oil pressure p in the displacement chamber 1b changes as the time t elapses, as shown in the working oil pressure characteristic pa1 in FIG. At the same time, the pressure decreases toward p=0.

However, since the voltage of the pulse signal V1 is turned on again at point a, the switching position of the switching valve 3 becomes 3a, and the pressure oil in the hydraulic source 4 is again forced into the displacement chamber 1b. As shown in the hydraulic pressure characteristic pa1, the pressure then increases toward p=pm.

Therefore, the pulse signal V1 is applied to the solenoid 3d.
When is transmitted, the pressure difference in the working oil pressure between the displacement chamber 1b and the displacement chamber 1a remains at dp1.

In this case, the value of pressure difference dp1 is the above-mentioned starting force k1・
When the value is smaller than the value corresponding to the value of xi, the piston 1d cannot be moved from the x position in FIG. 2 in the direction of the arrow a1.

In this way, if the negative application time during which the pulse signal is turned off is short, the piston 1d cannot be driven, and when the application time is set to the appropriate application time to shown in Fig. 3, the hydraulic pressure characteristics are as follows. pa2, the pressure difference becomes dp2, and the pressure difference dp2 becomes a value corresponding to the value of k1·xi that activates the piston 1d.

This value to becomes the basic application time for controlling the hydraulic actuator 1, and as explained in FIG.
must be selected in each case.

However, the pressure difference dp2 caused by this basic application time to is the minimum value necessary to start moving the piston 1d.

Therefore, in order to actually move the piston 1d, it is necessary to further use an application time obtained by adding the corrected application time te to the basic application time to, and the resulting pressure difference must be set to dp2+dpa.

That is, out of dp2+dpa, the pressure difference dpa is a component necessary to accelerate the piston 1d, and the correction application time te in this case is a value proportional to the above-mentioned deviation.

In addition, the working hydraulic characteristics resulting in the pressure difference dp2 + dpa are:
It is indicated by pa3 in FIG.

In addition, each drive direction a1, a in the above-mentioned FIG.
Strictly speaking, when the piston 1d is driven in directions 2, a3, or a4, the spring biasing force applied to the piston 1d differs depending on the position of the spring 5e or 5f within each driving direction.

Therefore, if we strictly consider the starting force for each driving position of the piston 1d, the value of the above basic application time to will be
Even if it is 3 or a4, it must change depending on the position.

However, without considering this strictly, in controlling within each drive direction a1, a2, a3, or a4, the basic application time to is set to an average constant value for each drive direction for the following reason. You can also use it as

As described above, when no external force is applied to the piston 1d, the piston 1d returns to the neutral position n. This means that when springs 5e and 5f are set in cylinder 1e, initial compression is applied to both springs 5e and 5f, and both springs 5e and 5f are given initial compression due to the initial compression of both springs 5e and 5f.
A neutral position n of the piston 1d is determined at a position where the applied forces of are balanced.

Here, for convenience of explanation, the spring 5f side will be explained as a representative.

In the explanation of FIG. 2 above, if the above-mentioned initial compression amount in the spring 5f is xo, when the piston 1d is at the position of the transition amount x from the neutral position n in FIG. 2, the spring 5f moves the piston 1d to the left. The urging force Fx is Fx=k1(xo+x) (1).

Therefore, at x=0, the piston 1d is a
The additional force Fo received from the spring 5f when attempting to shift in one direction is Fo=k1·xo (2).

When the auxiliary force Fx at this arbitrary position x is divided by the initial auxiliary force Fo, the auxiliary force ratio ζ becomes ζ=1+(x/xo) (3) from equations (1) and (2).

In the case of FIG. 2, the piston 1d does not use the entire stroke of the cylinder 1e, and the maximum value xm of the transfer amount x is considerably smaller than the initial compression amount xo.

For this reason, the value of (xm/xo) where x=xm in equation (3) is also much smaller than 1.0.

On the other hand, when the piston 1d is at an arbitrary position x as shown in FIG. 2, the difference between whether it is driven from that position in the forward direction of a1 or in the return direction of a2 is ) As can be understood from the equation, in driving in the a1 direction, the auxiliary force
A driving force that overcomes the force of Fx = k1 (xo + x) is required, and in the case of drive in the a2 direction, a small drive in which the driving force is assisted by the same auxiliary force Fx = k1 (xo + x) as the force that overcomes the force. It becomes power.

In other words, compared to the large difference between the driving force of the piston 1d in the a1 direction and the driving force in the a2 direction,
In driving only one side of the piston 1d in each driving direction a1, a2, a3 or a4, when the piston 1d is driven between x and xm, the change in force at each transition position of the spring 5f or spring 5e is as follows. , not that big.

For this reason, in the explanation of the operation in the following embodiments, each drive direction a of the piston 1d is
No consideration is given to the fact that the actuation force of the piston 1d changes depending on each transition position within 1, a2, a3, or a4.

In FIG. 1, when the operator operates the lever 10i, an analog voltage signal is generated on the wiring 10b, and this signal is input to the detection storage means 10A in FIG. 4.

Further, a signal from the potentiometer 5d that detects the position of the piston 1d is input to the detection storage means 10A as an analog signal via a wiring 10e.

The calculations in FIG. 4 after inputting to the detection storage means 10A in this way are as follows.

Calculation in the detection storage means 10A: The command signal in the wiring 10b and the feedback signal in the wiring 10e are calculated in the detection storage means 10A.
The converted command signal value is sequentially AD converted (converted from analog value to digital value) by switching in the multiplexer contained within the
cs and feedback signal value cf are temporarily
Stored in RAM.

Calculation in the deviation calculation means 10B: Calculate the deviation Er between the command signal value c _S and the feedback signal value cf, Er=c _S −cf.

Calculation in the drive determining means 10C: The deviation Er is compared with the allowable deviation Ero, and it is determined whether a negative drive pulse signal should be output to the wiring 10g or 10h.

To specifically explain the judgment in this case,
It looks like below.

When a deviation Er occurs because the feedback signal value cf (the actual position of the piston 1d) is ahead of the command signal value c _S , and the deviation Er is larger than the allowable deviation Ero, the position of the piston 1d When the deviation Er occurs because the feedback signal value cf lags the command signal value _cS , and the deviation Er is larger than the allowable deviation Ero, the position of the piston 1d is advanced. If the above deviation Er is smaller than the allowable deviation Ero,
Assuming that the control is satisfied, piston 1d
Fix it in the same position.

Note that the above tolerance Ero is the tolerance deviation setting means 1.
This is the value stored in 0E.

Calculation in the basic time change storage means 10F: This means includes each basic application time determined depending on which direction of the arrows a1, a2, a3, or a4 the piston 1d starts to move as shown in FIG. to is memorized.

Each of the stored basic application times to is modified as follows.

In other words, if any required basic application time to is used and the control is underdriven or overdriven, this basic application time to
is corrected by the deviation Er mentioned above.

That is, the new basic application time to is obtained by adding or subtracting a time proportional to the absolute value of the deviation Er to the previous basic application time to.

Calculation in basic time selection means 10K: This calculation is performed as shown in the flowchart in FIG.

The flowchart in FIG. 5 will be explained below.

Calculation A1: Start calculation in the basic time selection means 10K.

Operation A2: Determine whether the feedback signal value cf is greater than or equal to the feedback reference value cfo.

Here, the value of the feedback reference value cfo corresponds to the feedback signal value cf when the piston 1d is located at the neutral position n in FIG.
This is a value stored in the reference value setting means 10L.

Therefore, the determination in calculation A2 is to determine whether the piston 1d is present on the right side in FIG. 2, including the neutral position n.

Operation A3: When the determination in operation A2 is Yes, it is determined whether the command signal value c _S is greater than the feedback signal value cf.

That is, it is determined whether the feedback signal value cf is behind or ahead of the command signal value _cS .

Operation A4: When the determination in operation A3 is Yes, the basic application time to is set to1.

The basic application time to1 in this case corresponds to the basic application time to in the case of arrow a1 in FIG.

Operation A5: When the determination in operation A3 is No, the basic application time to is set to2.

The basic application time to2 in this case corresponds to the basic application time to in the case of arrow a2 in FIG.

Operation A6: When the determination in operation A2 is No, it is determined whether the command signal value c _S is greater than the feedback signal value cf.

Operation A7: When the determination in operation A6 is Yes, the basic application time to is set to4.

The basic application time to4 in this case corresponds to the basic application time to in the case of arrow a4 in FIG.

Operation A8: When the determination in operation A6 is No, the basic application time to is set to3.

The basic application time to3 in this case corresponds to the basic application time to in the case of arrow a3 in FIG.

The basic application times of to1, to2, to3, and to4 are stored in the basic time change storage means 10F, as described above.

In addition, in calculation A2 in FIG. 5 above, the feedback signal value cf is equal to the feedback reference value cfo.
It also determines whether it is equal to or not. That is, it is also determined whether the piston 1d is located at the neutral position n in FIG.

In this determination, the piston 1d is at the neutral position n
, the following piston 1
d is driven in the a1 direction or a from the configuration shown in FIG.
It can only be driven in one of three directions. Therefore, in the calculations after the judgment of calculation A2 becomes yes, it is judged whether cf=cfo or not, and the command signal value c _S is changed to the feedback reference value cfo (neutral position n) based on the judgment result. In which direction,
In other words, it is necessary to determine whether the vehicle is facing the a1 direction or the a3 direction.

As a result, if the judgment result is in the a1 direction, the basic application time to is set to1, and if the judgment result is in the a2 direction, the basic application time should be
A program with to as to2 must be added to Figure 5.

However, in FIG. 5, this part of the program is omitted because the program explanation is complicated.

In the program of FIG. 5 omitted in this way, when the piston 1d is at the neutral position n, as can be understood from the above explanation, the calculation is
It will proceed from A3 to operation A4 or A5.

As a result, when cf = cfo in calculation A3 and the command signal value c _S outputs a command to control in the a1 direction, the basic application time to in the a1 direction is changed to to1 for the a1 direction. do. Therefore, there is no problem in this case.

However, when cf = cfo in calculation A3, and the command signal value c _S outputs a command to control in the a3 direction, the basic application time to in the a3 direction is changed to to2 for the a2 direction. It will end up being put away.

In this case, the value of the basic application time to2 corresponds to the case where the piston 1d is about to be pushed back to the neutral position n side by the spring 5f in FIG. In order to control in the a3 direction, the basic application time to2 used in that case is smaller than the basic application time to3, which is the ideal value in this case.

For this reason, driving the piston 1d by a single pulse signal from the initial pulse width calculation means 10J, which controls the piston 1d from the neutral position n to the a3 side using the program shown in FIG.
This results in a deterioration in control responsiveness.

Here, the corrected application time is added to the shortened basic application time as described above. For this reason, in control from the neutral position, the drive is performed by the added correction application time component. However, since the basic application time uses a shorter value for the a2 direction, the response is degraded.

However, the deterioration of its responsiveness is due to its neutral position n
It is only the control of the first single pulse transmission from .

Then, based on the control result in which the piston 1d moves in the a3 direction by the first single pulse transmission from the neutral position, for a new deviation result that naturally occurs, from the program shown in Fig. 5, Basic application time to1, to2, to3 or to4 adapted for each direction of a1, a2, a3 or a4, respectively
is used, and subsequent control is in an ideal state.

After the value of the basic application time to is determined in this way, the calculation moves to calculation 9.

Calculation A9: Return the calculation to that of the pulse width calculation means 10J, and in this case, calculate the basic application time to corresponding to the current situation calculated in any of calculations A4, A5, A7, or A8 to the pulse width calculation means 10J.
Output to.

Calculation in the pulse width calculation means 10J: In the drive determination means 10C, as described above,
When it is determined that it is necessary to drive the hydraulic actuator 1, the following calculation in this means is started.

Using the deviation Er calculated by the deviation calculating means 10B described above, the corrected application time te=b・Er and the application time tn=to+te are calculated, and during this application time tn, V3 in FIG. 3 is applied from the output transistor 10M. Outputs a single pulse signal corresponding to .

Here, b is the proportionality constant and to is the operation described above.
This is the basic application time to output from A9.

Operation in the output transistor 10M: The output transistor 10M includes transistors Tr1 and Tr2 that turn on and off the voltages in the wirings 10g and 10h, respectively.

Among these, when the transistor Tr1 transmits a negative pulse signal (corresponding to the pulse signal V3 in FIG. 3) only to the wiring 10g and the voltage remains on to the wiring 10h, the hydraulic switching valve operating device Y
As explained in the explanation of the operation, the piston 1
d upwards in FIG. Conversely, when the transistor Tr2 sends a negative pulse signal only to the wiring 10h and the voltage on the wiring 10g remains on, the piston 1d is moved downward in FIG. 1.

In such a configuration, the output transistor 1
0M performs the following calculation.

1) When the content of the deviation Er in the deviation calculating means 10B is such that the piston 1d should be further moved upward in FIG. 1, only the transistor Tr1 is connected to the wiring 10g.
Sends a negative pulse signal to

2) When the content of the deviation Er in the deviation calculating means 10B is such that the piston 1d should be further moved downward in FIG. 1, only the transistor Tr2 is connected to the wiring 10h.
The above negative pulse signal is transmitted.

In this case, the application time tn for transmitting the negative pulse signal is the application time tn output from the pulse width calculation means 10J.

3) When the deviation calculating means 10B determines that the deviation Er is smaller than the allowable deviation Ero and the control is satisfied, the transistor Tr1
and Tr2 both keep the voltages on wirings 10g and 10h on.

As a result, the piston 1d remains fixed in that position.

Although the embodiment of the present invention is constructed and operates as described above, FIG. 1 may be as shown below.

In FIG. 1, the hydraulic actuator 1 has a centering effect using springs 5e and 5f, whose center is set as a neutral position n, and the value of the basic application time to is determined based on the neutral position n. ing.

However, this configuration is different from that shown in FIG. 1 when the hydraulic actuator is used alone.
The entire stroke of the piston 1d is divided into one region and the other region, and when operating in one region, only one spring gives a reaction force to it, and the piston 1d
When d operates in the other region, in addition to this one spring, the other spring also acts on the piston 1.
It can also be applied to a hydraulic actuator that applies a reaction force to d.

It is indicated by arrows a1 to a in FIG. 2, using the boundary between one area and the other area as a reference position
This is because it is possible to select the basic application time to divided into 4 sections.

In addition, the switching valve 2 or 3 is operated by a solenoid 2d.
Alternatively, the electrical pulse signal given to 3d is converted into a hydraulic signal, and the hydraulic actuator 1 is controlled by this hydraulic pressure.

However, this hydraulic actuator 1 may be directly controlled by a hydraulic pulse signal.

Furthermore, it will be easily understood that the hydraulic actuator 1 may be replaced with an electric actuator, and the output signal from the electronic control device 10 may directly control this electric actuator.

〔Effect of the invention〕

As is clear from the above description, the method for controlling a fluid actuator according to the present invention
Divide the process of d into a first region and a second region,
If the piston 1d has different loads in its forward and return directions for each region, the basic application time to of the single pulse signal application time transmitted to drive the hydraulic actuator 1 is set for each region. This is a control method that deals with and changes the load in each direction.

Therefore, even if the hydraulic actuator 1 used in the present invention is configured to generate a different starting resistance force for each control area and each control direction, it is possible to control the starting resistance in different areas and directions. Each time, a single pulse signal corresponding to those starting resistances will be emitted to correct the deviation.

This makes it possible to quickly control the piston 1d to the target position by transmitting the single pulse signal once.

In the end, control for each transmitted pulse becomes less wasteful control, and has the effect of easily increasing control responsiveness.

[Brief explanation of the drawing]

FIG. 1 shows a system diagram of a hydraulic switching valve control device for implementing the fluid actuator control method of the present invention, and the hydraulic switching valve operating device Y in FIG. 1 shows a conventional configuration,
Figure 2 shows the hydraulic actuator 1 in Figure 1.
is shown in a skeleton diagram, and FIG. 3 shows the hydraulic pressure characteristics pa1, pa2, or pa3 of the displacement chamber 1a or 1b in FIG. 1 and the pulse signals V1, V2, or V3 transmitted from the electronic control device 10, respectively. Figure 4 shows the characteristics of the first
5 shows a flowchart of the basic time selection means 10K in FIG. 4, and FIG. 6 is a system diagram of a conventional hydraulic switching valve control device. This is shown by. The symbols used in the examples are as follows.
1: hydraulic actuator, 1a and 1b: displacement chamber, 1d: piston, n: neutral position, x: distance, 2 and 3: switching valve, 2d and 3d: solenoid, 10: electronic control unit.

Claims (1)

[Scope of Claims] 1 A piston 1d is fitted into the cylinder 1e so as to be able to slide in the axial direction, and a switching valve 2 is provided on one side and the other side of the piston in the axial direction. One displacement chamber 1a and the other displacement chamber 1b are provided in which pressurized fluid is selectively pumped or removed through the switching valve. It is driven by a single pulse signal Vn having a time width tn, and the correction time width includes a detection signal detecting the position of the piston in the axial direction and a command signal instructing the position of the piston in the axial direction. The value is proportional to the deviation Er between the two pulses, and for each single pulse signal transmission, only the value immediately before the transmission is used, and the stroke of the piston is divided into the first region and the second region. The above structure is divided into sections, and in each region, load forces of different magnitudes are generated on the piston when the piston moves to the forward side and when the piston moves to the return side in the axial direction. In each of the regions, the basic application time width is: a: when the piston is controlled in the forward direction, it becomes the value of one basic application time width for that region, b: the piston When is controlled in the return direction, the value of each other basic application time duration for that region is set.