# Analysis of static and dynamic characteristics of robot electro-hydraulic pressure servo valve

### Motor dynamics equations

The electromagnetic driving torque endured by the motor rotor is^{14}:

$${T}_{em}={k}_{t}{i}_{0}-{k}_{m}{\alpha }^{2}$$

(1)

Where \({i}_{0}(\mathrm{A})\) is the input current of the motor; \(\alpha (\mathrm{rad})\) is the rotation angle of the motor rotor; \({k}_{t}\) is the current torque coefficient; \({k}_{m}\) is the angular torque coefficient.

The equation of motion of the motor rotor is^{15}:

$${T}_{em}={J}_{r}\ddot{\alpha }+{B}_{r}\dot{\alpha }+{T}_{f}$$

(2)

Where \({T}_{f}(\mathrm{N}\cdot \mathrm{m})\) is the load torque; \( J_{r} ({\text{kg}} \cdot {\text{m}}^{2} ) \) is the moment of inertia of the motor rotor; \({B}_{r}\) is the damping coefficient of the motor rotor.

### Slide valve dynamics equation

During the opening process, the slide valve has two directions of movement: translation along the slide valve axis and rotation around the slide valve central axis. By decomposing the motion in two directions, we have:

$${F}_{x}={m}_{v}{\ddot{x}}_{v}+{B}_{v}{\dot{x}}_{v}+{k }_{v}\left({x}_{v}+{x}_{v0}\right)+{F}_{s}+{F}_{f}$$

(3)

$${F}_{s}=2{C}_{d}\pi {D}_{v}\mathrm{cos}\varphi \left[{x}_{v}\left({P}_{s}-{P}_{c}\right)-\left(U-{x}_{v}\right)\left({P}_{c}-{P}_{0}\right)\right]$$

(4)

$${T}_{\beta v}={J}_{\beta v}{\ddot{\beta }}_{v}+{B}_{\beta v}{\dot{\beta } }_{v}$$

(5)

Where \({F}_{x}(\mathrm{N})\) is the driving force for axial translation of the slide valve; \({m}_{v}(\mathrm{kg})\) is the spool quality; \({x}_{v}(\mathrm{m})\) is the displacement of the slide valve; \({B}_{v}\) is the axial translation damping coefficient of the slide valve; \({k}_{v}(\mathrm{kN}/\mathrm{m})\) is the stiffness of the bias spring; \({x}_{v0}(\mathrm{m})\) is the pre-compression amount of the bias spring; \({F}_{s}(\mathrm{N})\) is the hydraulic pressure; \({F}_{f}(\mathrm{N})\) is friction; \({CD}\) is the flow coefficient at the slide valve port; \({D}_{v}(\mathrm{m})\) is the diameter of the end of the spool; \(\varphi (\mathrm{rad})\) is the jet angle of the slide valve orifice; \(U(\mathrm{m})\) is the pre-opening amount of the slide valve; \({P}_{s}(\mathrm{MPa})\) is the oil supply pressure; \({P}_{c}(\mathrm{MPa})\) is the brake pressure; \({P}_{0}(\mathrm{MPa})\) is the oil return port pressure;\({T}_{\beta v}(\mathrm{N}\cdot \mathrm{m})\) is the driving torque for slide valve rotation; \({J}_{\beta v}(\mathrm{kg}\cdot {\mathrm{m}}^{2})\) is the moment of inertia of the slide valve relative to its central axis; \({\beta }_{v}(\mathrm{rad})\) is the rotation angle of the valve core around the valve core axis; \({B}_{\beta v}\) is the rotational damping coefficient of the slide valve.

In the slide valve, the valve core and the valve sleeve have a clearance fit. Due to the pressure difference at both ends of the valve core, the liquid in the gap will form an asymmetric pressure distribution from the high-pressure end to the low-pressure end, which can be called the slide valve side pressure. According to the slit flow theory of fluid mechanics, the pressure on the slide valve side changes with the size of the gap. If the clearance changes symmetrically along the longitudinal axis of the valve plug, the resultant pressure on the spool side is zero. However, when the gap changes asymmetrically along the longitudinal axis of the valve core, the resultant pressure on the side of the side slide valve is not zero, and side pressure friction is formed as the valve core moves.

Therefore, due to various reasons such as machining accuracy or assembly, the valve core and sleeve are not concentric or the gap is asymmetrical, which will cause asymmetric pressure on the side of the slide valve, thereby causing side pressure friction. If the pressure on the spool side further acts on the spool surface and eccentrically causes it, the clearance will become more asymmetrical and the pressure on the spool side will increase. In severe cases, the oil film between the valve core and the valve sleeve will be destroyed, causing dry friction, which will significantly increase the friction between the valve core and the valve sleeve, causing “hydraulic clamping” and causing equipment failure.The friction force can be expressed as^{6}:

$${F}_{f}=\frac{\pi }{4}LD\Delta P$$

(6)

Where \(L(\mathrm{m})\) is the spool shoulder width, and \(D(\mathrm{m})\) is the spool diameter, \(\Delta p(\mathrm{MPa})\) It is the pressure difference between the two ends of the valve core.

Steady-state hydraulic pressure is the hydraulic force acting on the valve core due to changes in flow and direction when the fluid flows into the valve chamber and passes through the control window of the valve, and is always in the window of the directional spool valve trying to close the control. According to the momentum theorem, the steady-state hydraulic pressure is:

$${F}_{s}=0.43W{x}_{v}({P}_{h}-{P}_{s})$$

(7)

Where \(W(\mathrm{m})\) is the width of the slide valve window, \({x}_{v}(\mathrm{m})\) is the spool displacement, \({P}_{h}(\mathrm{MPa})\) is the pressure at the entrance of the slide valve cavity, \({P}_{s}(\mathrm{MPa})\) It’s the brake pressure.

Transient hydraulic pressure is generated by the acceleration of the fluid.Assuming that the oil cannot be compressed and the oil quality in the valve cavity remains unchanged, the change rate of the oil speed in the valve cavity \(\mathrm{d}v/\mathrm{d}t\)the oil acceleration force is^{16}:

Suppose the axial distance between the inlet and outlet of the slide valve is \(L(\mathrm{m})\)the oil density is \(\rho ({\mathrm{kg}/\mathrm{m}}^{3})\)and \(Q(\mathrm{L}/\mathrm{s})\) is the flow rate in the valve cavity. Then, Eq. (7) can be modified as:

$$F=\rho L\frac{dQ}{dt}$$

(9)

According to the continuity equation, the flow rate in the valve cavity is the flow rate at the valve port, and the throttling formula can be obtained:

$$Q={C}_{v}W{x}_{v}\sqrt{\frac{2\Delta P}{\rho }}$$

(10)

Where \({CV}\) is the flow coefficient; \(\Delta p(\mathrm{MPa})\) is the pressure difference between the inlet and outlet of the valve chamber.

According to the above formula, we can get:

$$F={C}_{v}WL\sqrt{2\rho \Delta P}\frac{\mathrm{d}{x}_{v}}{\mathrm{d}t}+\frac{ L{C}_{v}W{x}_{v}}{2\sqrt{\Delta P/\rho }}\frac{\mathrm{d}(\Delta P)}{\mathrm{d} t}$$

(11)

Overall, the impact \(\mathrm{d}(\Delta p)/\mathrm{d}t\) The transient hydraulic pressure is very small and can be ignored.Transient hydraulic pressure is always related to \（F\）the transient hydraulic pressure is:

$${F}_{t}=-{C}_{v}WL\sqrt{2\rho \Delta P}\frac{\mathrm{d}{x}_{v}}{\mathrm{d }t}$$

(12)

### Kinematics and mechanical equations of eccentric mechanisms

The rotation of the motor drives the eccentric mechanism, which pushes the valve core to move linearly. The eccentric ball generates eccentric rotation and also drives the valve core to rotate around its axis. Therefore, the spool motion can be decomposed into linear motion along the X-axis and rotational motion around the X-axis.As shown in Figure 3, the displacement of linear motion is *X*_{v} \((\mathrm{m})\)then the rotation angle of the valve core around the valve core axis is \({\beta }_{v}(\mathrm{rad})\).Kinematics between spool motion and motor angle \(\α\) It can be expressed as^{number 17}:

$${x}_{v}=e\,\mathrm{sin}\,\alpha $$

(13)

$$\mathrm{tan}{\beta}_{v}=\frac{e\left(1-\mathrm{cos}\,\alpha\right)}{h}$$

(14)

Where *H* is the distance from the center of the ball to the plane, *e* is the eccentricity between the eccentric shaft and the motor rotation center,

According to Figure 3, the spool force on the ball is \({F}_{xb}(\mathrm{N})\) and \({F}_{\beta vb}(\mathrm{N})\),We have

$${F}_{xb}={F}_{x}$$

(15)

$${T}_{\beta v}={F}_{\beta vb}\frac{h}{\mathrm{cos}\,{\beta }_{v}}$$

(16)

$${T}_{f}={F}_{xb}e\,\mathrm{cos}\,\alpha +{F}_{\beta vb}e\,\mathrm{sin}\,\ α\, \mathrm{cos}\,{\beta }_{v}$$

(17)

### Controller Mathematical Model

The main functions implemented by the controller include signal conditioning, closed-loop pressure control, motor angle feedback control and PWM amplification.Among them, signal conditioning is to convert the input control current signal into \({i}_{i}\) converted into voltage signal \({u}_{i}\), and perform noise reduction filtering processing.Closed loop pressure control uses PID control to convert hydraulic pressure \({personal computer}\) Converted into the corresponding voltage, the voltage signal is linearly amplified by the amplifier as a pressure feedback signal \({up}\).According to the difference between the input signal and the pressure feedback signal, proportional, integral, and differential operations are performed to obtain the output signal. \({u}_{m}\) pressure control.An angular displacement sensor is installed in the motor angle feedback control to convert the motor angle into \(\α\) converted into the corresponding voltage.The voltage signal is linearly amplified by the amplifier and used as an angle feedback signal. \({u}_{\alpha }\) motor. The proportional coefficient of this process is the motor angle electrical feedback coefficient.Difference in output signal \({u}_{m}\) Pressure feedback control and feedback signal \({u}_{\alpha }\) Obtain the motor angle and output the final motor control signal.The control signal of the motor requires power amplification, and PWM power amplification is used to amplify the signal. \({K}_{PWM}\).To sum up, the mathematical model of the controller part is^{18}:

$${u}_{i}=k{i}_{i}$$

(18)

$${u}_{m}=\left({u}_{i}-{P}_{c}{k}_{f2}\right)\left({K}_{p}+{ K}_{I}\frac{1}{s}+{K}_{D}s\right)$$

(19)

$${u}_{0}=\left({u}_{m}-\alpha {k}_{f1}\right){K}_{PWM}$$

(20)

### Local linearization of RDPPV mathematical model

Under actual operating conditions of RDPPV, the valve core displacement is very small, about 0.1×10^{−3} rice. According to Eq. (12) Then the rotation angle of the motor is 4.78°, so equations (12) (12) and (13) can be linearized as

$${x}_{v}=e\alpha $$

(twenty one)

$${\beta }_{v}=\frac{e{\alpha }^{2}}{2h}={k}_{\beta v}\alpha $$

(twenty two)

To sum up, the motion equation of the motor rotor at the working point can be obtained as:

$${T}_{em}=\left({J}_{r}+{m}_{v}{e}^{2}+{J}_{\beta v}eh\alpha {k }_{\beta v}\right)\ddot{\alpha }+\left({B}_{r}+{B}_{v}{e}^{2}+{B}_{\beta v}eh\alpha {k}_{\beta v}\right)\dot{\alpha }+\left({k}_{v}+{k}_{s}\right){e}^{ 2}\alpha +{k}_{v}e{x}_{v0}$$

(twenty three)