Hydraulic Proportional Control Bosch Rexroth[1]

8/11/2019 Hydraulic Proportional Control Bosch Rexroth[1]

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Design & Control ofProportional and Servo Systems

Bosch Rexroth Industrial HydraulicsRob Decker & Dave Saaski

April 2008

Drive for TechnologyCMA/Flodyne/Hydradyne, Inc.


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History of Hydraulic Control

Mechanical No Electric

Flow Controls, Limit Switches

and Relay Logic


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Using Closed-loop Controllers

Examples of Closed Loop Control lers

Analog Lower Cost

Computer not required for set-up or adjustment

Examples:

VT-MACAS (AVPC-V or AVPC-mA) - Position or Velocity Control Card p/Q Cards Open Loop Flow, Closed Loop Pressure Control Card

Digital

HACD Hydraulic Axis Controller: Digital

HNC Hydraulic Numerical Controller


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VT-MACAS (AVPC) Analog Velocity Posit ion Control / V or mA

Material Number forVoltage Command

= 0811405139

Material Number formilliAmp Command

= 0811405140


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VT-MACAS (AVPC) Analog Velocity Posit ion Control / V or mA


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VT-MACAS (AVPC) Schematic


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AVPC / Applications

Position Control

Velocity Control


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P/Q Cards with Valve Amplifier


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P/Q Cards for Valves with OBE

FrontP

late


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P/Q Card Applications


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Controller for the Pressure Difference (Force Control)


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HACD

HACD (Hydraulic Axis

Controller - Digital)

Multi-loop 32-bit Digital

Controller

DeviceNet -

Available CANOpen - In

development

PROFIBUS - In

development New Generation

Upgrade of the DMX


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HACD Technology: European + American Product Experience

1997 to 2003

Joint Development


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6 analog Inputs, Voltage or Current

(selection via software) 2 analog Outputs, 1x voltage or current

(selection via software)

Digital Feedback

SSI or Incremental

8 digital inputs

(configuration via software)

7 digital outputs

(configuration via software)

Enable Input and OK Output

Display and Keys

Serial Interface RS 232

CAN Bus - DeviceNet and CANOpen

protocol

Bus Controllers - HACD


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One Configuration Software for

all Applications

Editor for the Configuration of

the Control Structure

using predefined functions

(programming knowledge not

required)

Clearly arranged Settings

for commands, controller

parameters, analog and digital

I/O setup

Oscilloscope Functionalso suitable as a

data recorder

Language Options

English / Deutsch

HACD - Setup Program


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Bus Controllers - HNC

HNC

Hydraulic Numerical Controller

32 Bit Multi-Axis Controller

User Programmed using NC G

codes

Bus Capability

PROFIBUS Available

CANOpen Available

SERCOS - Available

DeviceNet - In development


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HNC Technology: Connectivity + Drive Control Options


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HNC - Bus Applications


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HNC with SERCOS interface

SERCOS interface

Noise immune fiber optic ring connection

Distributed drive architecture

Mode: Closed-loop position control

(SERCOS cycle time 2 ms)

Preferred and freely configurable messages

Special control algorithms for interpolating with

electro-hydraulic drives

HNC100 closes the control loop

Digital interfaces for measuring systemEnDat absolute, incremental, SSI

Command value feedforward via Sercos

HNC100 looks and acts like an electric servo to

the CNC

Robust SERCOS

fiber optic interface


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MTXCNC control

MTC 200CNC control

Control andPowerElectronics

Ecodrive IDIAX Electrohydraulic actuator controlled by the HNC

additional axes can be added

HRS(HNC 100)

SERCOS(command valueand actual value)

Electromagnetic motors Electrohydraulic actuator

Control andPowerElectronics

Ecodrive IDIAX Electrohydraulic actuator control led by the HNC

additional axes can be added

HRS(HNC 100)

SERCOS(command valueand actual value)

Electromagnetic motors Electrohydraulic actuator

HNC with SERCOS Interface

CNC control system solutions by Bosch Rexroth


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HNC Technology: Demonstration and Development Tools

The Test Box comes with three Prj.-Files:

Demo_ana (analog Feedback)

Demo_ink (incremental Feedback)

Demo_abs (absolute SSI Feedback)

VT-HNC100DEMO

Refer to RD/RE 30133


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History of Hydraulic Control

Closed Loop Position Control with

Servo Solenoid Valve and Linear

Displacement Transducer


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Open Loop Systems vs. Closed Loop Systems


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Closed Loop System Components

Setpoint

Potentiometer

Set Point Card

PLC Analog

Output

Controller

AVPC

p/Q card

DMX

HACD

HNC

Amplifier

Amp Card

Amp Cube

OBE

Valve

p/Q

Overlap (W) Zero Lap (V)

Actuator

Motor

Cylinder

Measuring Device

Pressure Transducer

Ultrasonic

LVDT

Encoder

Potentiometer


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The Three Most Common Types of Control

Position Control

Velocity Control

Pressure Control


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Physics of Hydraulics

Hydraulic Stiffness

Example: Apply a 1 ton load to a cylinder.

The no-load position is 100 cm

Compare the cylinder

when filled with :

1. Oil

2. Water3. Air

4. Steel

100 cm

A = 10 cm2

PL= 0 bar

No Load


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Stiffness of Various Materials

Oil

X = 0.7 cm

PL = 100 bar

1T

99.3 cm

Water

X = 0.4 cm

PL = 100 bar

1T

99.6 cm

Air

X = 99 cm

PL = 100 bar

1T

1 cm

Steel

X = 0.002 cm

PL = 100 bar

1T

99.998 cm


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Solving Stiffness Limitations

In order to maintain a fixed cylinder position independent of load changes,

the following must be used:1. Mechanical stops (metal to metal)

or

2. Closed loop control

The same condit ions are true for a constant velocity drive. Again the

following solutions must be used:

1. Electric servo drive (Indramat)

or

2. Closed loop hydraulic drive using load sense, loadcompensator or electro-hydraulic closed loop


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Example of Posit ion Control

Command

F

Feedback

ErrorSU

Controller

ControlValve

Position Transducer

1. Mechanical-hydraulic

Closed loop

2. Electro-hydraulic

Closed loop

F


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Improving the System Stiffness

A four-way valve controls both sides of an actuator.

As a result, the load is held between two springs . With this configuration,

the spring-constant is higher !

Note: See Rexroth Hydraulic Trainer Vol. 2 or Using

Industrial Hydraulics for more details.


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Posit ioning with Higher Stiffness

1. Mechanical-hydraulic

2. Electro-hydraulic


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What Impacts Machine Design ?

How does a real hydraulic drive differ from an ideal drive?

Response

The ideal or linear drive converts all input (command) signals

into output signals without delays or distortions.

Examples of input signals can be:

On / Off, Stop / Go

Analog voltages

PLC program I/O

Example: A rod or a lever converts inputs directly into

corresponding outputs.

In Out


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Spring Mass Systems Respond Differently

m

T

As we will see later, hydraulic drives are spring-mass systems.

Spring-mass systems have two observable properties:

1. Natural frequency2. Damping

Example:

1. The number of oscillations per second is the natural frequency

fo

2. After time, the oscillation decays due to damping.

T =1

fo


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Predicting the Natural Frequency

The natural frequency is determined solely by:

Spring constant C of the drive

Mass M coupled to the drive

Why should the natural frequency be high as possible ?

A simple experiment will show:

Every machine has mass and is not completely rigid.

Consequently all machines are spring-mass systems.

Take a machine axis with a given fo, and oscillate it

between two defined positions, 0 and 10.

fo =2

C

M


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Machine Limits Imposed by Natural Frequency

If we move the axis slowly", the machine will respond ideally, i.e. it

moves from 0 to 10, as commanded.

If we increase the number of cycles per second, the machine output will

increase in stroke! This can be dangerous and destructive to the

machine!

Every machine has a limit of operation. If we exceed this limit, the

natural frequency of the machine is approached, and the machine

starts to resonate at that natural frequency.

This MUST be avoided; BAD things will happen.

http://timber.ce.wsu.edu/supplements/seismic/frequency.htm


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Benefits of Damping

Increasing damping or,

How to Control Resonant Response

If damping is increased in a spring-mass system, amplitude increase or

overshoot can be reduced.

The graph that follows shows how our experiment varies if we increase

damping.

To understand the effect, visualize the previous experiment with the

moving components immersed in:

Water (d 0.5)

Honey (d 2) Hot tar (d 20)


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Effect of Increased Damping

A damping ratio of d 0.7 results

in the best overall response.

Unfortunately, by increasing

damping, another problem occurs:

A time delay between the inputand output, so-called phase-lag,

increases.


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How to increase damping:

Dissipate kinetic energy by converting it into heat, known aspassive damping (used in car shock absorbers) or;

Actively counter the kinetic energy using a closed loop feedback to

cancel oscillations, called active damping. Both methods are

used in systems today.

Improving Damping

Note: A hydraulic machine

drive will normally have a

damping ratio between 0.05 ~0.4, and will respond

accordingly.

Proportional valves and servo

valves are designed with

damping ratios of 0.75. They

exhibit no overshoot.


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What Creates Damping

Note: It would appear that increasing friction is a simple way to increase

passive damping. This is a trap! Mechanical friction can be high at low speeds

(breakaway friction), and lower at higher speeds (running friction). High

breakaway friction deteriorates a systems performance and must be

minimized! Use of low friction PTFE cylinder seals, lubrication, hydrostatic

bearings, etc. is normally recommended for this reason.

Passive damping is present due to internal leakage

Laminar FlowQ

FL1p

p=R

LAMQ


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Passive Damping

Leakage caused by cylinder seal leakage, bypass

valves, and valve null-flow, resulting from under-lapped spools, improves damping.

Disadvantages:

Loss of energy (efficiency) Decreased static accuracy

Turbulent Flow

Uo

Sharp-edge

Orifice

Bypass

Valve

ValveUnder-

lapped

Spool

Q

FL1p

p=R

turb

Q2


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Example of Active Damping using Pressure Feedback

The (simplified) circuit shows an aircraft actuator in closed loop position control.A step-load can cause oscillation, dependent on the system gain.

( lever ratio in this example)

B

A

P T

Active Damping

B

A

Command


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P T

Orifice

Volume

Modulator to Provide Active Damping

A so-called modulator is added to the system. Pressure increases caused by

step-load changes allow the lever support to move. Response time and amplitude

of the modulator piston is determined by an orifice and a small accumulator. A high

damping is created with this type of control (P-DT1 control).

This idea is 80 years old!

In some applications today, an electronic

equivalent is used, also known as activedamping, or state variable control.

Modulator


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Stiffness of Actuators

Therefore:

F = loadX = deflection

The result is the deflection under load, or the stiffness.

If a load is suddenly applied to the drive, or the load is accelerated

dynamically, it will respond like a spring - mass system, and the dynamic

properties will be dominant.

Selecting a cylinder should be based on both the static and dynamic

requirements:

C = F

X

X = F

C


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Defining Drive Requirements

1. Static:

- Balance of forces, i.e. the cylinder must have enough force to hold

and move the load

F = p A

- Rod buckling must be considered. The cylinder attachment design

or rod diameter should be changed if required.

2. Dynamic:

- The drive should have sufficient dynamic response to accommodate

load changes and accelerations.As seen earlier in a bode-plot, a drive will appear to be rigid and

have minimal adverse dynamic effects when operated below its

limit frequency.


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Determining Performance Requirements

When designing a hydraulic drive system, it is desirable to have the

natural frequency as high as possible, while staying within the allocatedbudget.

Rule of thumb for good design:

A hydraulic designer can only improve the natural frequency of the drive

by varying the cylinder size, and minimizing the pipe length between

cylinder and the valve.

NOTE: M (mass) is not measured in pounds in English units. Rather

M(mass)= F (lbs)/A (in./sec2)

= (lbs.)/(32.16 (ft./ sec2) x 12 (in./ft))

= (lbs.) / 386 (in./ sec2)

C = E A2

Vofo = ,

2

CM


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Design Guidelines

P T

A B

F

A quick approach for finding an optimal cylinder:

1. Calculate F = p A

2. Check for rod buckling

Typical Arrangement

Single rod cylinder with pipe

connections to the valve fo calculation is difficult

Simplified Approximation

Equal area cylinder(added rod results inlower fo)

Eliminate pipe volume bymounting valve directly on

the cylinder (improve fo)

P T

A B

F


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Rule of Thumb Selection

Assuming the two changes in the above approximation cancel each other,

the value of Vo (cylinder centered) is:

Since:Result:

Vo = AA S/2AA = Annulus Area

S = Stroke

fo =2

C

M

C = C1 + C2

C1 = C2 =E A2

Vo

Vo = AA S/2

fo =

2 E AAS/2 M

2


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Typical Machine Requirements

Therefore:

This formula provides an annulus area of the cylinder required for a

desirable natural frequency.

The following are experience based guidelines:

fo < 4 Hz Poor dynamics, good static performance only

fo 15 Hz Good frequency for general machine designsfo 30 Hz Needed for machines requiring high dynamics

AA =fo2 2 S M

E


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Hydraulic Motor Estimation

The same rules apply to hydraulic motor drives:

I = Rotating inertia

Note: If a difference of opinion arises, such as the machine designer

insisting on using a small cylinder operating at high pressure, butcalculations dictate a larger cylinder and lower pressures, you can

compare the alternatives with a simulation program, such as HYVOS by

BoschRexroth

q = fo2 4 4 IE


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Machine Considerations

ATTENTION!

In many designs,

the mass M is not

directly driven by

the actuator.

In the examples shown here, the effective

mass or inertia must be calculated. The

effective value seen by the actuator

(reflected inertia) is the (lever-ratio)2

or (gear-ratio)2 !Therefore:

Valve

m

M

x

y

M

x

y

M

r

MEFF = M ( )2xy

JEFF

JEFF = M (r)2

N in

N out

N = rpm

n =N out

N in

MEFF = M ( )2x + y

y

JEFF = ( )2M

2P

P = pitch =in

rev

JEFF = J (n)2

fo =2

CMEFF


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Machine Stiffness

Important Note:

If the driven machine structure is flexible (typically the case), the result

is a complex pair of 2nd order spring - mass systems (4th order and

higher in most situations). Manual calculation of these systems is very

tedious. Machine designers should always try to make the driven

structure as rigid as possible.

A goal is to have the natural frequency of the driven structure be 3

times higher than the hydraulic drives natural frequency. With this

ratio, the machines dynamic response will have a minimal influence on

the overall drive performance. When the ratio is 10:1, the machine can

be considered a rigid structure, and not be a factor in the drives design.


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Valve Sizing

1. Pressure Drop

Pressure drop (p) is defined as the difference between the system

(pump) pressure and the load pressure.

Load pressure can only be measured directly when a meter- in

throttle circuit is used.

Example 1:

pL =

F

Ap

pV = pP pL

F

pL

Q

PV

pP = 100%

AP


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Pressure Drop with 2 Orif ices

It is obvious that the pressure drop is not selected by the designer. It

is a function of the system pressure and the load applied to the actuator.

Example 2:

F

pA

Qp1

pP = 100%

p2

pB

pL = pA pB = FAA

p1 = p2

(Also applies to Hydraulic Motors)

pv = 2 p1


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pP ApFAA

Meter Out and Double Throttles

Example 3:

Example 4:

pV = pB =F

QP1

pP

= 100%

pB

AP = Piston area

AA = Annulus area

F

pA

Q1p1

pP = 100%

p2

pB

pv = p1+ p2 for Q1

pP ApF

Ap +p1 = AA

2 =

ApAA


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How to Find the Proper Valve

Flow ratings of proportional valves are the rated flow at a NOMINAL

pressure drop, and normally have nothing to do with actually sizing aproportional valve.

First, we need to calculate the ACTUAL pV for the specific system.

After calculating the pV, refer to a valve data sheet.

Example 4: Flow Q = 450 l/min

Pump-pressure pP = 210 bar

Force F = 3600 daN

Cylinder Ap = 20cm2AA = 10 cm

2

= 2


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Which Proportional Valve do we choose?

p1 =

pP Ap F

Ap +AA

2

p1 =(3000 psi 3.1 in2) 7920 LB

3.1 in2 + 00001.6 in2

22

p1 = 394 psi

For a 2:1 cylinder ( = 2), we choose a

2:1 valve (QA

:QB

= 2), Then, p1

= p2

and:

pV = 2 pV1 = 788 psi (= 54 bar)

RA 29 115

Page 13-14

4WRZ valve

WRZ 16 is too small

WRZ 25 E 220 is 92% open

WRZ 25 E 325 is 78 % open


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Spool Flow Characteristics

Characteristic curves (measured with spools E, W6-, EA, W6A at v = 46 mm2/s and v = 40 C) Size 16

150 L/min nominal flow with a 10 bar valve pressure differential

1 p = 10 bar constant



4 p = 50 bar constant5 p = 100 bar constant(80)

(320)

(460) 5

4

3

2

1

(400)

(240)

(160)

1008070605040302015 90

21.1

84.5

121.5

0

105.7

63.4

42.3

PA / BT

orPB / AT

Command value in %

788 PSI


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Spool Flow Characteristics

Characteristic curves (measured with spools E, W6-, EA, W6A at v = 46 mm2/s and v = 40 C) Size 25

220 L/min nominal flow with a 10 bar valve pressure dif ferential







(200)

(100)

(300)

(400)

(500)

5

4

3

2

1

(800)

(700)(600)

1008070605040302015 90

52.8

26.4

79.3

106.0

132.0

0

211.0

185.0159.0

Command value in %

PA / BTorPB / AT 788 PSI

(200)

(100)

(300)

(400)

(500)

(870) 5

4

3

2

1

(800)

(700)

(600)

1008070605040302015 90

52.8

26.4

79.3

106.0

132.0

0

211.0

185.0

159.0

230.0

Command value in %

PA / BTor

PB / AT

788 PSI1 p = 10 bar constant


3 p = 30 bar constant4 p = 50 bar constant


p = Valve pressure differential to DIN 24311 (input pressure pP minus load pressure

pL minus return line pressure pT)


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Data Sheet Ratings

As shown, the available pressure drop across

the valve is a result of pump pressure and load

pressure. Data sheets typically give a nominal

flow at a nominal pressure drop. There are two

de facto standards used in the hydraulics industry

to define the flow through a control valve:

Servo & High-Response Proportional Valves

A pressure drop Pv = 1000 PSI (70 bar) is

used. When the valve is open 100%, the

measured flow is the nominal flow at

1000 PSI total pressure drop.

Proportional Valves A pressure drop Pv = 145 PSI (10 bar) is

used. The valve is opened from zero to

maximum. The typically non-linear flow

characteristic is recorded.

pP = 70 bar

QNom

Pressure Drop Test Circuit


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Valve Pressure Drop Rating

We use the sharp-edged orifice equation and the valve flow rating

to determine the flow at the pressure drop of our system.

Q2 = Q1 x (p2/p1)

Example: If a valve is rated for 50 LPM @ 1000 PSI,

the flow at 2000 PSI =

Q2 = 50 lpm x (2000/1000) = 71 LPM


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Valve Pressure Drop Selection

Since proportional valves often not linearized, a graph is generatedfor flow vs. spool-stroke. This also eliminates the need to calculateflows at various pressure drops. Instead, many data sheets will give a

family of flows for various pressure drops:pv = 20 bar (290 PSI)

30 bar (435 PSI)50 bar (725 PSI)

100 bar (1,450 PSI)

Example: 4 WRZ 16 E 150 valve

(80)

(320)

(460) 5

4

3

2

1

(400)

(240)

(160)

1008070605040302015 90

21.1

84.5

121.5

0

105.7

63.4

42.3

PA / BT

orPB / AT

Command value in %

QNOM




3 p = 30 bar constant4 p = 50 bar constant


p = Valve pressure differential to DIN 24 311 (input pressure pP minus load pressure pL minus return line pressure pT)


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Sizing Proport ional Valves for Acceleration and Deceleration

For transfer systems, the highest pressures are normally experienced

during acceleration of the mass. The natural frequency of a hydraulic system normally determines the

minimal allowable acceleration and deceleration.

tmin

(sec) = (3 x 6)/ o fo = o / (2 x )

tmin = minimum time of acceleration (sec)

o = natural frequency, undamped (radians/sec)

=(C/M)

C = spring constant

M = mass (NOT POUNDS)


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Simulation Software

Above, we used an ideal 1:1 double-rod cylinder in order to simplify

calculations and estimate natural frequency and an ideal annulus area:

Simulation software such as HYVOS allows extremely accurate

simulation of the performance of hydraulic systems without the above

simplification. But this software can be difficult to access.

A simple spreadsheet, though, can quickly calculate many of the most

important values necessary for design. Simplifications such as above

then become generally unnecessary.

AA =fo2 2 S M

EC1 = C2 =

E A2

Vo fo =

2 E AAS/2 M

2


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Poor Mans Simulation Software

26,947Cmin=

1293.21034.6775.9dPt=

681.0544.8408.6dP2 =35.54d-crit (in)=

612.2489.8367.3dP1 =

700.00F (friction) (lb)=37.68V-pipe 2 (ci)=

387.8510.2632.7P2 (PSI)=610.06F (accel) =14.72V-pipe 1 (ci)=

681.0544.8408.6P3 (psi)=48.00L-pipe 2 (in)=

Decel.Const. Vel.Accel.0.15t(min. accel.) (sec)=1.00d-pipe 2 (in)=

48.00L-pipe 1 (in)=

1000.0P-system (psi)=0.02t (time cont)(sec)=0.63d-pipe 1 (in)=

42.00stroke (in)=

700.0F-friction (lb) =40.65w(damp)(rad/sec)=

610.1F-acceler (lb)=121.96w(theor)(rad/sec)=1.66A2 (si)=

1.38d-rod2 (in)=

336.8A-max (in/ss)=26,947Ct (lb/in)=2.00d-bore2 (in)=

49.7V-max (in/sec)=

1.00C (friction)=3.14A1 (si)=

0.6t (stroke) (sec)=1.81Mass (lb*ss/in)=0.00d-rod1 (in)=

40.5Qmax=20.0Stroke (in)=700.00Weight (lb) =2.00d-bore1 (in)=

Valve Pressure Drops Required200000

Bulk Mod

(lb/in)=

Natural Frequency, 2:1 Cylinder, 1=a side, 2=b side, 2:1 Valve


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Proportional Acceleration Systems

Remember:

Total pressure drop can be greater than system pressure,especially if the valve ratio does not match the cylinder ratio

A negative pressure in one or the other side of the cylinder means

that cavitation will occur. Again, this normally occurs if the ratios

do not match.

Infinite acceleration is not possible, so actual velocity and flow will

be greater than average velocity and flow. Size valve accordingly.

Reducing volume between the valve and cylinder will increase

spring-constant C, and allow for faster acceleration.

Larger bore cylinders also increase spring-constant C.

If natural frequency is too low, theres always state control. But

figure this out before start up!


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Oversizing and Undersizing of Valves

Design Example:

Two engineers select the valve and the pump (system) pressure settingfor the same application.

Engineer #1: calculates PL = 66 bar for load pressure

Engineer #2: calculates same as above

Engineer #1 sets the pump pressure to 100 bar.

Therefore the valve pressure drop is Pv = 33 bar.

A valve is selected from the catalog that passes the

required flow at a pressure drop of 33 bar, when the

valve is fully opened.


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Valve Selection Example

Engineer #2 selects a valve which has a flow rate

twice that of Engineer #1s choice (for example, 200 liter

spool instead of 100 liter spool). It will have the same

flow with 1/4 of the pressure drop across the valve.

Engineer #2 sets his pump pressure to Pp = 74 bar.

The valve pressure drop is only 8 bar rather than 33 bar!

(see graph)

Design Note:

Engineer #2 will have a more efficient system, suitable for

non-dynamic systems. If dynamic acceleration and

decelerations are required, a higher pressure drop willresult in better control. The electrical command to the

valve will be more accurately followed when a higher

pressure drop is used.


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Pressure Drop Relationships

pL = 66 bar (2/3)

pv = 33 bar (1/3)pP = 100 bar (3/3)

pL = 66 bar (11/12)

pv = 8.5 bar (1/12)

pP = 74.5 bar (12/12)200%

50 2/3 100%11/12

8%

92%

50 2/3 100%0

50

100

2/3

33%

66%

Engineer #1

Engineer #2

p

p

Q, v

pL

P

F

Q, v

pL

P

F

P = F v

P = pL Q

Q,

V,pL,

p,P

Q = Flow

F = Force

v = Velocity

P = Power

pL

= Load pressure

p = Valve drop


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Closing the Loop

An undesirable characteristic of most hydraulic systems is that therelationship between the command signal and the output of a hydraulicdrive is not a linear (proportional) function.

- Valve response to a stepped electrical signal is:

(A servo valve would respond similarly but up to 10 times faster)

0 40 80 120 160 200 0 40 80 120 160

0-25

10

20

30

40

50

60

70

80

90

100 0-100

0-75

0-50

Time in ms

Signal change in %Size 6

Transfer function: Stepped electrical input signal

Proportional valve

type: 4WRA6...


73/126


Cylinder Response

Cylinder response to a stepped input (flow Q) is:

Two characteristics are observed:

1) A cylinder with an attached mass responds as a

spring - mass system

2) An integration occurs between the input and output.This results in a phase-lag of -90

(X)vi X

Servo

Valve

Cylinder

Q


74/126


Example of Phase-Lag

If the cylinder stroke versus the valve command signal(valve spool stroke) is plotted, the phase-lag can be seen:

In this example, the valve and cylinder are in open loop.

Note: As the valve is operated at increasing frequencies, the phase-lag increases due to inertia and spring compliance Important: If -180total phase-lag occurs in the system, the output will be inverted ascompared to the command signal!

0

90

180

270

360


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Closing the Loop Mechanically

When a feedback (lever mechanism in this example) is added, the

system responds in a proportional manner rather than as an integrator.

Response to a step is similar to the response of a valve spool,

proportional with some time lag (known as PT1).

Feedback Command

F


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Gain of the Closed Loop

If the pivot on the feedback lever is moved to the left, the amount of valve

spool opening is increased for the same position error. The system

corrects for errors faster, and the positional accuracy improves. If we

move the pivot more and more to the left (more gain), the system

response will exhibit increasing overshoot, and eventually become

unstable.

Example:X X X

t t t

Low Gain High Gain Too High Gain


77/126


Closing the Loop Electrically

This example is essentially the same as the mechanical feedbackexample shown before.

The ultimate system performance (response time and accuracy) isdetermined by the properties of all components in the system.

SU

Feedback Command

F

Controller

Amplifier

Control Valve

Transducer


78/126


Properties of the System Components

1. Cylinder

An ideal cylinder would have zero breakaway friction and no internalleakage. In actuality, cylinder friction results in positioning errors

(cylinders exhibit stick-slip friction at low speed). Additionally internal

cylinder leakage will result in a compensating flow (spool shift) in the

valve.

These limitations cause position errors.

2. Feedback Transducer

An ideal transducer would measure infinitesimal position errors for the

controller to correct. An actual transducer has limited resolution, whichresults in a position measurement error. The response time of some

transducers can also cause signal delays, which can limit system

response.


79/126


Properties of System Components

3. Valve

An ideal valve would have zero response time and react to any

input signal. Real valves have step response times as seen, andrequire some minimum signal to respond, known as threshold.

Additionally, a true zero-overlap spool, as shown in the examples, is

only possible with poppet type valves. Since spool valves are

normally used, a zero-overlap spool has null flow leakage in the

center position.

A hydraulic designer must therefore select the best available

components, within his budget, to achieve the highest static accuracy

possible.


80/126


Component selection for Overall System Performance

Ultimately the system performance is determined by the quality of all of

the components used.

What system improvements and tricks can be

used to get the most out of a system?

What can be done to optimize the system:

The goal is to use the highest gain possible

to achieve the highest accuracy, and the

fastest response time.


81/126


Cylinder Selection Guidelines

1. Select a cylinder with:

- Lowest friction seal material (PTFE composite, step seals, metal

rings, etc); or for highest performance applications, a servo cylinderwith hydrostatic bearings.

- A cylinder size selected for high natural frequency, as well as for

force and buckling. Except for the piping length between the

cylinder and valve (best design is the valve mounted directly on the

cylinder), the piston area is the only variable for increasing thenatural frequency!

C = E A2Vo

fo =2

C

M


82/126


Valve Selection Guidelines

2. Select a valve with:

- Minimum threshold and hysteresis (highest

performance valves use electrical spool feedback)- A pressure drop as high as possible under nominal

load conditions (at least 30% of the supply pressure)

- Utilizing as much spool stroke as possible, with some

reserve margin- Spool center position zero-lapped for position and under-lapped

for pressure control

- Overlapped proportional valves may only be used with

appropriate control compensation features. (DMX, HNC)

- Overshoot-free step response

- Dynamic response determined by the application

(highest available response is not always the best choice)

- Special spool, if required

Example of a Special Spool


83/126


for Plastic Injection Axes

Velocity and Pressure Control

Transducer and Control ler


84/126


Selection Guidelines

3. Select a transducer with:

- Resolution five to ten times better than the

required accuracy

- Good linearity

- Minimal time lag (high dynamic response)

4. Select a controller with:- Highest possible resolution

- Fastest scan time, if digital

- Control algorithms optimized for hydraulic drives:

direction-dependent gain, switched integrator forpositioning, spool linearization and overlap

compensation, following error compensation,

P-I-DT1 controller, etc


85/126


Performance and Stability

5. Improve the System Damping

Stability criteria defines the maximum system gain (goal for

performance) as:

Kvmax< 2 D 2 fo

Kvmax = Gain [ ]

D = Damping Ratiofo = Natural frequency

What are some ways to increase the system gain ?

Increase the Damping Ratio D by using:1. Passive damping

2. Active damping

m/sm


86/126


Increasing Damping

If velocity feedback is added to the position control, the closed loop frequencyresponse can be increased. However, this results in reduced damping. Thiscan cause reduced performance in some hydraulic closed loop systems.

If acceleration feedback is added, the system damping will be increased. Loadpressure feedback provides similar results when applied correctly. Using these

feedbacks to improve performance is known as active damping or state-

variable control.

m a = P A

a ~ P

The modulator shown earlier is an excellent example of improving damping by

using pressure feedback !

m = massa = accelerationP = pressure

A = cylinder area


87/126

D i ith T d


88/126


2) Closed loop state variable control using position,

velocity and acceleration transducers

Damping with Transducers

X X

XKX

KX

Command

Controller Valve

GainfoD

GainfoD

F

Cylinder

Acceleration

Velocity

D i ith Ob M d l V l


89/126


Damping with Observer Model Values

3) Closed loop using

calculated values for

velocity and acceleration

X X

XM

XM

XM

X - XM

KX

KX

CommandController Valve

GainfoD

Gainfo

D

F

Cylinder

Acceleration

Velocity

ObserverModel

ModelCorrector

Using Integrators


90/126


Using Integrators

In theory, using an integrator in the controller would result in perfect accuracy

(any deviation from the commanded position would result in an increasing

opening of the valve spool). However, due to friction and threshold errors, a

continuous integration limit cycle oscillation can occur.

An integrator also results in an additional -90 of phase lag. This can

cause a hydraulic positioning axis to becoming a high power phase

shift oscillator

In practical use, an integrator is enabled only below a preset minimum velocity,

and when within a position error "window" (switched integrator).

Along with electronic spool overlap compensation, high accuracy

results can be achieved using simple, low cost proportional valves, in

closed loop.

Supply Pressure Considerations


91/126


Supply Pressure Considerations

1. Supply Pressure

The system pressure should be constant. When high flow is required (high

system dynamics), a drop in supply pressure can occur due to acceleration ofthe oil mass in the lines, and due to the response time of the pumps pressure

control.

Recommendation: An accumulator should be used in the pressure

supply, mounted near the control valve.

If the system has a long return line, similar problems may occur.

Acceleration of the return oil mass can cause return pressure spikes, which

reduces the available working pressure. This will reduce the system

performance.

Recommendation: Use return line accumulators near the control

valve, as needed.

Supply Pressure Optimization


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Supply Pressure Optimization

Using accumulators

to improve dynamicresponse.

Note: The mass of the oil in the return line can have a very significant

effect !

PT

P

S

M

Valve Dynamic Selection


93/126


Valve Dynamic Selection

There are three general configurations of valve-cylinder dynamics:

Case 1: The valve is high response andthe cylinder is low response (fo )

Case 2: The valve and the cylinder aresimilar in response

Case 3: The valve is low response andthe cylinder is high response (fo )

focyl = natural frequency of the cylinder

fov = natural frequency of the valve

Relationship of Valve and

Cylinder Dynamics


94/126


Cylinder Dynamics

Case 1: 0.3 - Best results are achieved with pressure

or acceleration feedback

Case 2: 1 - A simple PT1 controller offers best results

Case 3: 3 - A PD controller will provide best results

focylfov

focylfovfocyl

fov

State-variable

Controller

PT1 Controller

P Controller

PDT1Controller

0.3 0.6 1.0 1.7 3.00.3

Case 2 Case 3Case 1

Effect of Controller Parameters


95/126


Effect of Controller Parameters

The following graphs show step response for all three cases.

The parameters are varied per:

1. Increase P Gain, left to right

2. Increase, bottom to top :

Case 1: Acceleration feedback

Case 2: Lag time T1

Case 3: Derivative gain

Effect of Parameters Case 1


96/126

Alle Rechte bei Bosch Rexroth AG, auch fr den Fall von Schutzrechtsanmeldungen. Jede Verfgungsbefugnis, wie Kopier- und Weitergaberecht, bei uns.Increased P-Gain

Increas

edAcceleratio

nFeedback




97/126



Increased P-Gain

IncreasedLa

gTimeT1



98/126



Increased P-Gain

IncreasedDerivat

iveGainD

Performance Optimized with Parameters


99/126


p

In the examples shown, it can be seen that a higher gain can be

achieved when other compensating parameters are used. By using

higher gain, increased accuracy and improved response is achieved,

and stability is maintained.

Other Considerations on System Dynamics


100/126


y y

Note:

Case 1: Fast valve - slow cylinder, is an example of a classicservo system, using a servo valve. Optimally, a double-rod, equal areacylinder should be used.

Case 3: Slow valve - fast cylinder, is a typical application using lowcost proportional valves, operating in closed loop. The control

performance is not perfect, but can usually meet the designrequirements. Due to the proportional valves spool overlap and non-linear flow characteristics, as well as the fact that differential cylindersare most often used, the controller must compensate for theseproperties. Controllers such as the HACD and HNC-100 utilize these

compensations.Proportional valves can provide a low cost solution with good accuracy,long life, and simple maintenance.

Closed Loop Application Using a

Proportional Valve and a Differential Cylinder


101/126


p y

(20)

(40)

(60)

(80) 5

4

3

2

1

1008070605040302010 90

5.3

10.6

15.9

21.1

Command value in %

PA / BT

orPB / AT

1 p = 10 bar constant2 p = 20 bar constant3 p = 30 bar constant4 p = 50 bar constant


Encoder Feedback

CylinderProportional

Valve Amplifier CNC

Feedback

Command

Incremental Signal

Drive Gain Electrohydraulic Gain CNC Gain

m/min

v

m

s in mm

D/A Control

Q

AL/min

UCNC

10V

ControlOutputstage

s

Q UCNCs

+Q in L/min + UCNC in V + s in mm

vinm/min

vin

m/min

Q

inL/min

Q

inL/min

UCNC

inV

U

CNC

inV

KvKv

Kv

v Q UCNC

U

Kv = inv

Q

m/min

L/min Kv = in

Q

UCNC

L/min

V Kv = in

UCNC

s

V

mm

Loop GainKv = Kv CNC Kv El-Hy Kv Drive Kv FB in

m

min mm

Electronic Compensation of Non-linear

Flow Characteristics and Valve Spool Overlap


102/126


Standard Linearization Compensation Linearization and Overlap

Output signal

Resultingsignal

Controllersignal

Positiveoverlap

Positiveoverlap

Positiveoverlap

Output signal Output signal

Valve

Amplifier

I

+U

I

-U

I

+U

I

-U

I

+U

I

-U

+U-U +U-U +U-U

+Q +Q +Q

-Q -Q -Q

Overall System Characteristics


103/126


The following pages show examples of system response, in open and

closed loop. System response is shown with step and ramp

command inputs, step input opposing force, and the effects of various

controller elements.

Elements of an Open Loop System


104/126


X = Position

X = VelocityX = Acceleration

Cylinder

PA PB

PSP

T

Spool Opening

Valve Driver

Command

Generator

I

V

Valve

State Variables

Open Loop Response to a Step Input


105/126


Cylinder

PA PB

Valve Driver

I

V

Spool Opening

X

X

X

Valve

PS PT Command

Generator

Open Loop Response to a Force Step


106/126


Cylinder

PA PB

Valve

PS PTValve Driver

I

V

Spool Opening

XX

F

Command

Generator

Elements of a Closed Loop System


107/126


PTPS

Valve

Cylinder

Valve Driver

I

V

Controller

Spool Opening

X

Command

Generator

Transduce

r Interface

Transduce

r

PA PB

Closed Loop Response to a Step Input

Transducer


108/126


PTPS

Valve

Cylinder

Valve Driver

I

V

Controller

Spool Opening

X

X

Transducer

Transducer

Interface

Command

Generator

PA PB

Valve

Signal

Command

Feedback

Effect of P Gain on Position Control


109/126


CylinderTransducer

Transducer

Interface

ControllerValve Driver

Valve

PA PB

Spool Opening

PTPS

Time

Time

Position

Velocity

GainIV

Command

Gain 3Gain 2Gain 1

Gain 1 > Gain 2 > Gain

3

Closed Loop Response to a Ramp Input

Transducer


110/126


CommandFeedback

PTPS

Valve

Cylinder

ValveDriver

I

V

Controller

Spool Opening

X

XPA PB

TransducerInterface

Command

Generator

Transducer

Valve

Signal

Dynamics and Errors in Closed Loop

P iti E


111/126


Time

Position

Position

CommandFeedback

Position Error

Following

Error

Phase Lag

Amplitude Loss

Response to a

Ramped

Position

Command

Response to a

Sinusoidal

Command

Closed Loop Response to a Force Step

Transducer


112/126


PTPS

Valve

Cylinder

Valve Driver

I

V

Controller

Spool Opening

X X

PA PB

Transducer

Interface

Command

Generator

F

Command

Feedback

Position Control - Correction for Opposing

Force using PI Control

Force Position Error


113/126


Position

Position

Time

Time

CommandFeedback

CommandFeedback

Force

Force

Position Error

Command

Command

Feedback

Feedback

P Control

PI Control

P Gain

P Gain + I Rate

Out

Out

Response of a PID Controller to a Step Input


114/126


Input

Time

Time

Time

Time

Time

Output

P Term

ITerm

D Term

Response of a PID Controller to a Ramp

Input


115/126


Input

Time

Time

Time

Time

Time

Output

P Term

ITerm

D Term

Position Control - Improving Dynamics using

PD ControlP Control


116/126


Command

Feedback

Command

Feedback

PD Control

Gain 2

Gain 1

Gain 1 > Gain 2

Gain 2

Gain 1

Gain 1 > Gain 2

Position

Position

Time

Time

Out

Out

P Gain

P Gain + D Lead Time

Position Control - Improving Dynamics using PT1

ControlP Control


117/126


Out

Out

Command

Feedback

Command

Feedback

PT1 Control

Posit

ion

Position

Time

Time

P Gain

P Gain + Lag Time

BoschRexroth Proportional Valves


118/126


Which Proportional Valve should

I use?


119/126


StartStart

herehere

Low Flow (Direct Operated)

Open Loop (No Cylinder Feedback)


120/126


4WRAB Flows to 25 Lpm,

minimum cost, limited performance

(similar to KDG4V-3 S)

4WRA Greater flow range to 60 Lpm,

better performance, more features

(Ramp)

4WRAB6

4WRAE10

4WRAE6

Low Flow (Direct Operated); Open Loop with Better

Performance (No Cylinder Feedback)

4WRE(E)


121/126


4WRE(E)

Designed for good

performance without high cost

Very repeatable

Greater flow capacity

High value

4WRP(E)

Proportional with LVDT

5 bar/ land or 10 bar delta-p,

E and W spools in housing (no

sleeve)

Overlap compensation

High reliability

Most robust OBE available

CE approved

4WREE10

4WRPE10 E

Proportional

High Flow (Pilot Operated)

Open Loop (No Cylinder Feedback)


122/126


4WRZ Wide flow range, Cost effective,

Accurate enough for most open loop needs

4WRL (HPP) with overlap E and W spools,OBE has overlap comp., High dynamics, Most

robust OBE, CE approved

4WRLE

(HPP)

HighPerformance

Proportional

4WRZE

Low Flow (Direct Operated)

Closed Loop (Cylinder or system Feedback)


123/126


4WRP H - Servo Solenoid Best

Choice, Excellent dynamics for

closed loop, Failsafe defined,Outperforms many old servos, Very

Robust OBE, High reliability, Wide

range of nominal flow 2 to 100 Lpm

at 70 bar (Higher delta-P for

Spool/sleeve valve)

4WRE Wider flow range, spool in

housing (normally V), proportional

flow rating, lower cost, Performance

not as high

4WRPEH 10

4WREE10

High Flow (Pilot Operated)

Closed Loop - System Feedback


124/126


4WRLE Servo Solenoid with V-spool,

High Flow range, Main spool in housing,

Flow rated at 10 bar delta-p,Very Robust OBE

4WRVE (HRV 2-Stage) Higher

dynamics, 10 bar drop, Very Robust

OBE, 12-pin Connector

4WRLE

(Servo Solenoid

2 stage)

4WRVE

(HRV - 2 stage)

Injection Valves are Preferred


125/126


4WRLE..Q4 Best choice for

Injection valve for Plastic Injection

Molding Machines Can replace Q2-spool

High reliability

Better Performance

Same advantages of ServoSolenoid valves

4WRLE..Q4.-3X/

pQ Valves are Preferred

For Closed Loop Pressure and Throttle Function


126/126


5WRPE (pQ Valve) Can combine

double port throttle and 3-way

for close loop pressure control,Requires transducer and PID card,

140 Lpm at 11 bar

Note: Use P1+B to P2+A to balance

flow forces, pmax is 210 bar

0 811 402 107

Documents

Hydraulic Proportional Control Bosch Rexroth[1]