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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
2 p = 20 bar constant
3 p = 30 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
1 p = 10 bar constant
2 p = 20 bar constant
3 p = 30 bar constant
4 p = 50 bar constant
5 p = 100 bar constant
325 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
2 p = 20 bar constant
3 p = 30 bar constant4 p = 50 bar constant
5 p = 100 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
150 L/min nominal flow with a 10 bar valve pressure dif ferential
1 p = 10 bar constant
2 p = 20 bar constant
3 p = 30 bar constant4 p = 50 bar constant
5 p = 100 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...
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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
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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
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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
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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.
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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.
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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.
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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
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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
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for Plastic Injection Axes
Velocity and Pressure Control
Transducer and Control ler
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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
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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
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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
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D i ith T d
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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
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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
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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
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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
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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
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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
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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
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Increas
edAcceleratio
nFeedback
Effect of Parameters Case 1
Effect of Parameters Case 2
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Effect of Parameters Case 2
Increased P-Gain
IncreasedLa
gTimeT1
Effect of Parameters Case 3
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Effect of Parameters Case 3
Increased P-Gain
IncreasedDerivat
iveGainD
Performance Optimized with Parameters
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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
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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
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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
5 p = 100 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
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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
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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
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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
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Cylinder
PA PB
Valve Driver
I
V
Spool Opening
X
X
X
Valve
PS PT Command
Generator
Open Loop Response to a Force Step
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Cylinder
PA PB
Valve
PS PTValve Driver
I
V
Spool Opening
XX
F
Command
Generator
Elements of a Closed Loop System
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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
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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
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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
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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
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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
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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
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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
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Input
Time
Time
Time
Time
Time
Output
P Term
ITerm
D Term
Response of a PID Controller to a Ramp
Input
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Input
Time
Time
Time
Time
Time
Output
P Term
ITerm
D Term
Position Control - Improving Dynamics using
PD ControlP Control
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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
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Out
Out
Command
Feedback
Command
Feedback
PT1 Control
Posit
ion
Position
Time
Time
P Gain
P Gain + Lag Time
BoschRexroth Proportional Valves
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Which Proportional Valve should
I use?
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StartStart
herehere
Low Flow (Direct Operated)
Open Loop (No Cylinder Feedback)
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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)
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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)
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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)
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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
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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
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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
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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