FLOW CONTROL SERVOVALVE MODEL
C. ROMAN
Imagine, Roanne
1
CONTENTS
I
Introduction ................................................................................................................. 3
II
Description ................................................................................................................... 4
III
Modeling ................................................................................................................... 5
III.1 Armature, flapper, flexure tube, wire ...................................................................... 5
III.1.i
torque motor ................................................................................................... 8
III.1.ii hydraulic forces on flapper.............................................................................. 8
III.1.iii flexure tube action .......................................................................................... 8
III.1.iv feedback spring force...................................................................................... 8
III.1.v state equation.................................................................................................. 8
III.2
torque motor, electromagnetic model ................................................................... 12
III.3
Hydraulic amplifier .............................................................................................. 21
III.4
valve spool ........................................................................................................... 22
IV
Simulation............................................................................................................... 24
IV.1
parameters ........................................................................................................... 26
IV.2
hydraulic characteristics & static performances................................................... 26
IV.3
dynamic performance ........................................................................................... 32
ANNEX .............................................................................................................................. 35
2
I
Introduction
The aim of this document is to explain the model of a 2-stages servovalve, built with
AMESet and AMESim utilities. It is a flow control servovalve, with two stages, double nozzle
and mechanical feedback. This kind of system uses many libraries as it deals with electrics,
magnetism, hydraulics and mechanics.
Such servovalves are composed of a magnetic torque motor, an armature with flexure
tube and feedback spring, an hydraulic amplifier (with nozzles), a valve spool. Each part can
be modeled by a specific library. It’s a perfect example to use the new AMESim’s magnetic
library.
For specific parts such as armature, AMESet allows you to code your own model. We
will try, through this document, to describe the modeling, step by step, and the way to set your
configuration.
Then, we will give the characteristics of this servovalve obtained by simulation
(hydraulic characteristics, static and dynamic performance) that are usually obtained
experimentally. Those characteristics are really similar to the real ones.
3
II
•
•
•
•
•
•
•
•
•
Description
Electrical current in
torque motor coils
creates magnetic forces
on ends of armature.
Armature and flapper
assembly rotates about
flexure tube support.
Flapper closes off one
nozzle and diverts flow
to that end of spool.
Spool moves and opens
Ps (supplied pressure)
to one control port and
opens other control port
to R (tank pressure).
Spool pushes ball end
of
feedback spring
creating
restoring
torque
on
armature/flapper.
As feedback torque
becomes equal to torque
from magnetic forces,
armature/flapper moves
back
to
centered
position.
Spool stops at a
position
where
feedback spring torque
equals torque due to
input current.
Therefore
spool
position is proportional
to input current.
With
constant
pressures, flow to load
is proportional to spool
position.
a)
b)
MOOG Documents
Figure 1 a) and b): scheme of the 2-stages servovalve
4
III
Modeling
Servovalve model is built with the following elements:
• Electromagnetic part: electromagnetic library in industrialization (see technical
document in Annex 1).
• Hydraulic part: built with Hydraulic Component Design Library.
• Armature, flapper and flexure tube: specific model built with AMESet utility.
III.1 Armature, flapper, flexure tube, wire
θ
+
l
armature
d1
flexure tube
L
d2
center of
gravity
flapper
d3
Figure 2 : scheme of the
armature and flexure
With :
•
•
•
•
•
•
l: half armature length
L: flexure tube length
d1: distance between center of gravity and flexure tube start
d2: distance between center of gravity and nozzles
d3: distance between center of gravity and spool
θ: angle of rotation
To make a complete model of armature, we need few more parameters:
• m : mass of armature
• J: moment of inertia
• E: Young modulus of flexure tube material
• ∅e : exterior diameter of flexure tube
• ∅i : interior diameter of flexure tube
• k : stiffness of feedback spring
• br: damping ratio for rotation
• bt: damping ratio for translation
5
Those parameters are sufficient to fully describe the armature.
Hypothesis: rigid body, small displacements are considered.
At first, we need to know armature displacement in a few points. Armature has 2 degrees of
freedom: θ for rotation and xg, center of gravity displacement (z displacement of the gravity
center is neglected). We need to express all other displacements with these 2 degrees.
Here are the points we need to see:
z1
z3
1
xt
z2
3
5
2
6
z4
4
xg
8
7
xflapper
9
Figure 3 : principal displacements
xfictitious
This is a multiport system: there are 8 ports with the following variables:
• Port 1: force (f1) on armature (input), velocity (vz1) (output), displacement (z1)
(output)
• Port 2: force (f2) on armature (input), velocity (vz2) (output), displacement (z2)
(output)
• Port 3: force (f3) on armature (input), velocity (vz3) (output), displacement (z3)
(output)
• Port 4: force (f4) on armature (input), velocity (vz4) (output), displacement (z4)
(output)
• Port 5 : flexure tube actions (internal), velocity (vt), displacement (xt)
• Port 6 : center of gravity velocity (vg) and displacement (xg), rotation (θ) and rotary
speed (ω) (state variables).
• Port 7: right hydraulic force on flapper (fr)(input), flapper displacement and velocity
(output).
• Port 8: left hydraulic force on flapper (fl)(input), flapper displacement and velocity
(output).
• Port 9: wire displacement (input), fictitious displacement (internal), feedback spring
force (output).
6
We easily find the relations:
z1

 
z2

 
z3
=
d
z4
 
dt 
xt
 xflapper  
 xfictitiou s  
0
0
0
0
1
1
1
l 
−l 
−l  d  xg 
l ⋅
d1  dt  θ 
− d2
−d3
The relations are identical for their displacements (small displacements).
Now, we need to list the different forces which act on this armature: There is:
• the torque motor
• hydraulics forces on the flapper
• flexure tube action
• feedback spring force
7
III.1.i
torque motor
Torque motor creates magnetic forces on ends of armature. There are 4 forces f1, f2, f3, f4
which corresponds with the displacements z1, z2, z3, z4.
The torque motor is:
CTorqueMotor = (f2+f3-f4-f1) . l
III.1.ii hydraulic forces on flapper
We can separate hydraulic action on flapper by 2 components that we will call fl (left force)
and fr (right force). Those forces acts on the flapper.
Fnozzle-flapper= fl-fr
III.1.iii flexure tube action
Litterature gives classical relations between warping variables and torque and force (ref: [8]
Karnopp, Margolis, Rosenberg).
3
2
F = E⋅I ⋅ 12 L −6 L ⋅ xt = K11 K12 ⋅ xt = K 11 K 12 ⋅ 1 d1⋅ xg = K11 K12 + d1⋅K11 ⋅ xg
2
τ
−6 L 4 L θ K 21K 22 θ K 21K 22 0 1 θ K 21K 22 + d1⋅K12 θ
with:
E= Young modulus of flexure tube material
I= Moment of inertia of flexure tube.
L= Flexure tube length
I = π ⋅(De4 − Di 4 )
64
III.1.iv feedback spring force
This is a simple stiffness:
Fw= - kw . ∆x
= - kw . ( xfictitious – xw)
= - kw . ( (xg – d3 . θ) - xw)
xw corresponds to spool position.
III.1.v state equation
We already have detailed the different actions on armature. It is now easy to find:
J ⋅θ=(f2+ f3− f1− f4)⋅l −br⋅θ−τ tube − fw⋅d3+(fr − fl)⋅d2
m⋅xg =−bt⋅xg + fl − fr + fw− Ftube
8
Replacing some expressions, we have:
J ⋅θ=CTM −br⋅θ−(K 21⋅xg +(K 22 +d1⋅K12 )⋅θ )+kw((xg −d3⋅θ )− xw)⋅d3+( fr − fl )⋅d2
m⋅xg =−bt⋅xg + fl − fr −kw⋅((xg −d3⋅θ )− xw)−(K11⋅xg +(K12 +d1⋅K11 )⋅θ )
In a vectorial writing:
−br 0 −(K 22 +d1⋅K12 +kw⋅d3 )− K12 +kw⋅d3
J
J
J
CTM −kw⋅xw⋅d3+( fr − fl )⋅d2
ω
ω
−
bt
−
K
12 −d1⋅K11 + kw⋅d3
−
K
11 −kw
J
vg 0 m
vg +
kw⋅xw−( fr − fl )
m
m
=
⋅
θ
θ
m
0
0
xg
xg 1 0
0
0
0 1
0
0
2
This is a state equation of the classical form:
x = Ax + Bu
A is called state matrix, B is called control matrix.
The code of the armature model is based on this state equation.
We have tested this model by 2 ways. First, we made a system composed only of this model
with no motor torque nor hydraulic components in order to obtain its eigenvalues. Then, we
tested it with the motor torque to ensure the gain was good and to see the influence of the
motor torque on the previous eigenvalues.
First, we needed to set parameters for the armature. We didn’ t have any exact values but, we
managed to find a coherent set.
File: test_armature.ame
9
m = 5.5 g
J = 5.8 10-7 Kg.m²
l = 14.5 mm (half armature length)
L = 4.8 mm (flexure tube length )
∅e = 4.46 mm
∅i = 4.36 mm
E = 1.2 106 bars (Young modulus)
d1 = 3 mm
d2 = 8.6 mm
d3 = 22 mm
kw = 3400 N/m
br = 0.001 N/(rad/s) (damper rating for
rotation)
bt = 10 N/(m/s) (damper rating for translation)
l
θ
+
d1
Table 1 : armature parameters
L
d2
d3
Figure 4 : scheme of armature
After linearization, we found 2 pairs of conjugate eigenvalues, which corresponds to 2
frequencies and modal shapes.
Those two modes do not effect on rotation and translation in the same way. The second mode
(4034Hz) is much more effective on translation than the first one.
To be sure of it, when can have a look to the modal shapes.
10
The first mode doesn’ t
effect
much
on
translation as we can see
on figure 5.
Scale factor : 1
Scale factor : 10
Scale factor : 1
Scale factor : 10
Figure 5 : modal shape
for first mode
The second mode shows
that translation and
rotation
are
both
influenced
by
this
frequency. We can note
that translation is much
more important for this
frequency than for the
first mode.
Figure 6 : modal shape
for second mode
11
III.2 torque motor, electromagnetic model
The torque motor consists in an armature (the one we studied before), 2 permanent magnets, 2
pole pieces, 4 air gaps.
air gap
Upper pole piece
N
N
armature
S
S
Motor coil
Lower pole piece
magnet
magnet
Figure 7 : scheme of torque motor
Armature is to rotate, which change air gap lengths. Equilibrium position results from the
simultaneous actions of an electromagnetic torque, due to 2 serial coils set on the armature,
and a mechanical return torque. This return torque is null when all air gaps are equal.
To simplify the problem, we will only consider a half of the problem, and moreover, we will
consider that the only reluctance important enough is the air gap ones.
Then, to simplify again, we will work with an equivalent air gap instead of 2 air gaps.
Figure 8 : simplified problem
12
An air gap is characterized by his pole area, and his length. The equivalent air gap of 2 serial
air gaps is an air gap of the same pole area but with length 2 times bigger.
We shall take into account the influence of the motor coils, which produce demagnetizing
ampere-turns.
The aim of the problem is to find the operating point of the system.
First, we must evaluate the permeance of the equivalent circuit. For an air gap the permeance
is given by the following relation:
P = µ 0⋅ S
e
with:
µ0= permeability of free space
S= pole area
e = air gap
You must remember that equivalent air gap is 2 times the normal air gap.
Then, you calculate the equivalent permeability of the circuit with the dimensions of the
magnet (effective section, lenght).
µ =P⋅ la
Sa
This is the main parameter because it represents the linear characteristic of the system with
the relation:
∆: B= µ . H
But, now we need to evaluate the demagnetizing ampere-turns (Atd), to know exactly in
which range the circuit will function.
Atd= N . I
Then, we have the equation of a new linear characteristic of the circuit:
(
∆’ : B= µ ⋅ H − Atd
la
)
The operating point is located at the intersection of the demagnetizing curve and the load line
defined by the relation above. This line intersects the H-axis at point H= Atd/la.
This point is defined by its coordinates, Bmin and Hmin, by which passes the recoil line. The
recoil line is the real characteristic of the system, it means that the operating points will be
situated on this line.
13
∆
∆’
Remanent flux
density (Br)
Recoil line
Bmin
Coercitive
field (Hc)
Hmin
Atd/la
Figure 9 : Recoil line determination
In the torque motor supercomponent, you must define your recoil line by specifying the
coordinates of the operating point (Hmin, Bmin) and the remanent flux density of the recoil
line. It’ s really important for your model to have a good values for your recoil line and the
best way is to apply the method above.
1
3
Air gap
?
Corresponding
ports on
armature
model
Permanent
magnet
Electrical ports
2
4
Figure 10 : sketch of torque motor
14
For example, we want to locate your operating point for a system composed of:
Magnet:
Hc = -30 000 A/m
Br = 0.2 T
la = 25 mm
Sa = 60 mm²
Air gap:
e = 310 µm
S= 15 mm²
Serial coil: N = 4200 turns
R = 400 Ω
Command intensity : 10 mA
P=4⋅π ⋅10− 7 ⋅
15.10−6
=3.04⋅10−8
2⋅310.10−6
25⋅10−3
=1.27⋅10−5
60⋅10−6
Atd = 4200⋅0.01=1680
la
25⋅10−3
µ =3.04⋅10−8 ⋅
With these values, we can find some good values for your operating point coordinates as
(Hmin=16 000, Bmin=0.13). Supposing you know the recoil permeability of your permanent
magnet (for example 3.1 . 10-6 for Alnico 1500), you easily find Brrecoil = Bmin -µr . Hmin =
0.18 (see Annex 2: Alnico 1500 demagnetizing curve).
If you don’ t know the recoil permeability, you can see that using Br instead of Brrecoil is not so
important.
In any case, you would better put values for Bmin and Hmin less than the real ones but not
too far, else your model could have problems.
Now that AMESim can calculate your torque motor, let’ s see how it looks like. Literature
gives a good approximation of that torque with the following relation (Annex 1, ref: [7] M.
Lebrun):
()
 2x
l⋅µ0⋅S
 x 2 + N ⋅i ⋅ x  

2 ⋅ E0 ⋅ + N ⋅i⋅E0 1+
E0 e  
 e
 x2   e
2
e ⋅1− 2 
e 

with x, the displacement of the armature in air gap (note that θ=x/l), E0, magnetomotive force
of magnet.
CTM =
We want to linearize this equation, so we transform it in the first order equation:
CTM= K1.i + Km.θ
With:
15
K1=
l 2⋅µ0 ⋅S ⋅E02
l⋅µ0 ⋅S ⋅N ⋅E0
,
Km
=
e2
e3
It’ s interesting to see what happens in the state equation when we replace CTM by the linear
approximationK1.i+Km.θ.
We find:
2
−br 0 −(K 22 + d1⋅K12 +kw⋅d3 )+ Km − K12 +kw⋅d3
J
J
J
K 1⋅i −kw⋅xw⋅d3+( fr − fl )⋅d2
ω
ω
12 −d1⋅K11 + kw⋅d3
11 −kw
−
bt
−
K
−
K
J
vg 0 m
vg +
kw⋅xw−( fr − fl )
m
m
=
⋅
θ
θ
m
0
0
xg
xg 1 0
0
0
0 1
0
0
When you connect the armature submodel and the torque motor, the eigenvalues will be
modified as you changed the state matrix. Magnetic stiffness is opposed to others stiffness so
that it will reduce eigenvalues.
It’ s maybe the most difficult part of the model for setting parameters as we could have seen it
earlier in the notice. To test the model, we found some parameters in different catalogues and
here are the values we chose:
Electrical characteristics:
Series coils
R = 200 Ω for each coil (total resistance 400 Ω for 2 coils)
iR = 10 mA
N = 2100 turns per coil ( 4200 turns for 2 series coils)
Elements characteristics:
Air gap: 310 µm
Pole area: 15 mm²
Armature length: 29 mm
Cross section: 12 mm²
The most important is to set parameters for the magnet as we said above. Notices always give
the torque motor gain (ref: [4], Moog), which is its main characteristic. We want to set our
parameters in order to find a standard value for this gain: 0.03 N.m for i = 10 mA.
Nota: CTM = K1 . i + Km . θ (linear model)
To examine the motor torque, we built a simple system in which you can change only the
intensity and observe the torque ( it means K1), or only the angle ( it means Km).
To control intensity and rotation angle, and to observe the resulting torque, we used signal
elements. (file: test_torquemotor.ame)
16
Figure 11: Benchmark for motor torque
results:
Good values for magnets are:
Hmin = -20 000 A/m
Brrecoil = 0.15 T
Bmin = 0.1 T
AMESim find an operating point defined by:
H = -9900 A/m
B = 0.12 T
Which corresponds to a magnetomotive force E0 = 248 A.
Analytic calculus (as in paragraph III) gives:
K1 = - 2.96 N.m/A
Km = 8.18 N.m/rad
We obtained the following characteristics:
17
a)
b)
Figure 12 a) and b) : torque motor constants
The simulation finally gives:
Km = 7.86 N.m/rad (figure 12 a) )
K1 = -2.73 N.m/A
(figure 12 b) )
It means we have a difference of 4% for Km and 8% for K1 with our linear approximation.
As we said earlier, motor torque as an influence on armature’ s eigenvalues because torque
motor is function of rotation angle (CTM = K1 . i + Km . θ). This test consists in linking torque
motor and armature and observing new eigenvalues. (file: test_armature+torque.ame)
Figure 13 : sketch to test armature and torque motor
18
As we can see in the state equation, the more the magnetic constant Km grows, the more it
will reduce the frequency of eigenvalues. For a certain value of Km, it will annul this
frequency, and over this value the system will get unstable. The problem is that the value used
for the complete servovalve makes this sytem unstable. But, when we link this system with
hydraulic amplifier, it won’ t be unstable anymore, because of hydraulics stabilizing effects.
We will test this model with 3 sets of parameters, in order to see the influence on eigenvalues.
We want to modify Km, it comes to change the magnets’ magnetomotive force. The best way
to illustrate this phenomen, is to translate our recoil line.
B
∆’
Recoil line 3
Recoil line 2
Recoil line 1
H . la
E3 E2 E1
Figure 14: Recoil line translation
We will choose 3 sets for Bmin and Br, and see the influence on eigenvalues frequencies and
damping.
•
Bmin= 0 T, Br = 0.05 T
E1= 96.7 A
Table 3 : influence of torque motor on armature’ s eigenvalues
19
•
Bmin= 0.06T, Br = 0.11T
E2= 198.8 A
Table 4
•
Bmin= 0.08 T, Br = 0.142 T
E1= 255A
Table 5
As we can see, the rotation mode is to disappear when permanent magnets get sufficient
induction.
20
III.3 Hydraulic amplifier
Flexure tube
Flapper
nozzle
orifice
Ps
Ps
Inlet orifice
C1
T
Drain orifice
C2
Figure 15 : Hydraulic amplifier
Let’ s see what happens in the hydraulic amplifier:
• Armature and flapper are rigidly joined and supported by thin-wall flexure tube.
• Fluid continuously flows from pressure Ps, through both inlet orifices, through another
orifice, past nozzles into flapper chamber, through drain orifice to return R (tank).
• Rocking motion of armature/flapper throttles flow through one nozzle or the other.
• This diverts flow to A or B (and builds up pressure if A and B are blocked).
This system can be easily built with AMESim’ s Hydraulic Component Design library.
Sketch:
Armature submodel
nozzle
Inlet orifice
Ps
orifice
Tank ( T )
Drain orifice
C1
C2
Figure 16 : Sketch for hydraulic amplifier
21
nota: There is a model of elastic contact behind the nozzles. Its aim is to absorb an impact
between nozzle and flapper. In case of excessive rotation of the armature, it will stop it.
III.4 valve spool
The spool is an essential part of the servovalve because it assumes the distribution of pressure
between orifices, and by the way, the flow control.
•
•
•
•
Spool slides in bushing (sleeve) as pressure in C1 and C2 varies.
Bushing contains rectangular holes (slots) or annular grooves that connect to supply
pressure Ps and return T.
At null position, spool is centered in bushing; spool lobes (lands) just cover Ps and T
openings.
Spool motion to either side of null allows fluid to flow from Ps to one control port,
and from other control port to T.
Feedback spring
spool
Ps
T
T
Ps
C1
C2
A
Ps
B
Bushing
a)
T
T
Ps
C2
C2
A
B
b)
Figure 17 a) and b) : scheme of valve spool
22
This time again, you can build the valve spool with an assembly of spools with annular
grooves, pistons for the chambers A and B, and a mass with end-stops to take in account the
inertia.
Here’ s what it looks like:
C1
To feedback spring
port
Piston
C2
spool
Ps
T
A
B
Figure 18: sketch of spool valve
23
IV
Simulation
Complete model:
(file: servovalve.ame)
sketch:
Figure 19 : Sketch of complete servovalve
24
The system is now complete. First, we are going to check the motor gain. To obtain it, we just
have to plot (f2+f3-f4-f1).l for a sinusoidal current in input (amplitude: rated current=10 mA).
Torque motor (Nm)
As you can see, torque motor is well adjusted with the parameters we set in the magnets. This
is surely the most difficult to configurate in the complete model.
Time (s)
Figure 20 : Torque motor for sinusoidal rated current
25
IV.1 parameters
Here is the list of parameters we used for the complete model. We kept the same value for
motor torque and armature. We will only detail hydraulic amplifier and spool configuration.
Supply pressure:
Nozzle diameter :
Nozzle-flapper opening :
Inlet orifice :
Drain orifice :
Nozzle orifice :
Spool mass:
Spool diameter:
Rod diameter:
Spool/bushing radial clearance :
Spool stroke:
Spool rounded corner radius:
Underlap:
Spool viscous friction:
210 bars
0.4 mm
0.06 mm
0.2 mm
0.3 mm
1 mm
10 g
10 mm
5.5 mm
2 µm
500 µm
5 µm
10 µm
50 N/(m/s)
IV.2 hydraulic characteristics & static performances
rated flow:
simulation conditions:
i = rated current (10mA)
Ps= 50 bars → 275 bars
During 20s.
Figure 21
26
Flow-load function:
This is the servovalve control flow when there is load pressure drop, expressed with input
current as parameter.
Nominal flow to load is given by:
QV = K ⋅i⋅ ∆ PV
Qv = valve flow to load
K = servovalve characteristic constant
i = input current
∆Pv = valve pressure drop
note: ∆Pv = Ps - R - ∆PL
Ps= supply pressure
R = return pressure
∆PL = load pressure drop
QV = K ⋅i⋅ Ps − R − ∆ PL
Control flow % rated
where
Load pressure drop % of supply pressure
Figure 22
27
flow curve:
This graph represents
the flow gain, and
show good linearity
and
low
hysteresis(only due to
system’ s
dynamic).
Rated flow is 84 l/min
for Ps = 210 bars.
Maximum valve flow
can reach 135 % rated
flow with oversignal.
28
Spool lap:
In sliding spool valve, the relative axial position relationship between the fixed and movable
flow metering at null. The actual configuration of our servovalve is set with an amount of
underlap of 10 µm. But, you can specify an amount of underlap or overlap in order to:
Underlap (10 Pm):
• increase flow gain at null.
• reduce valve pressure gain at null.
• increase valve null leakage.
Overlap (10 Pm):
• reduce flow gain at null
• reduce null leakage flow
• degrade pressure gain (into load)
Zero lap
Flow control
underlap
overlap
Input current
Figure 24: influence of spool lap on flow rate at null
29
Static performances with blocked load
Internal leakage:
This characteristic is obtained with blocked load.
Internal leakage includes first stage hydraulic amplifier flow, spool null leakage flow, and
bushing laminar leakage flow.
• Spool leakage is essentially zero when spool is off-null.
• Servovalve internal leakage excluding spool null leakage is called Tare flow.
Hydraulic amplifier flow largely determines servovalve frequency response
• Lower flow degrades response.
Spool null leakage is related to maximum valve flow (slot width) and null cut.
null leakage
Tare flow
null
Figure 25 : internal leakage at null
30
Pressure gain:
The blocked load
differential pressure
will change rapidly
from one limit to the
other as input current
causes the valve spool
to traverse the null
region.
Here,
the
pressure gain at null
exceeds 100 % of
supply pressure for
1% of rated current.
Figure 26 : Pressure gain
31
IV.3 dynamic performance
frequency response(file: bode-servovalve.ame):
Frequency response will depend upon signal amplitude, supply pressure…Natural frequency
ranges from 45 to 80 Hz. To make a bode diagram, we used a transferometer. You can’ t
realize the standard linearization because of numeric problems due to non-linearities.
Figure 27
Figure 28
32
Step response:
Step of current:
We observe that time
response is depending
of
signal
input
amplitude.
From 50 % of rated
current, spool velocity
is constant.
This phenomenon is
due
to
flapper
displacement saturation.
We can observe flapper
displacement on figure
30.
100 %
75 %
50 %
25 %
Figure 29
Figure 30
33
REFERENCES
[1]
« Hydrauliques et électrohydraulique » - J. Faisandier – Dunod
[2]
« Analyse et conception des systèmes hydrauliques » - Imagine
[3]
« Systèmes dynamiques et automatique linéaire » - Imagine
[4]
Moog documents
[5]
« Specification standards for electrohydraulic flow control servovalve » - Moog
[6]
« Manuel technique des aimants permanents » – Chambre syndicale des producteurs
d’ aciers fins et spéciaux
[7]
« Modélisation et simulation d’ asservissements electrohydrauliques» - M. Lebrun
[8]
« System Dynamics, Modeling & simulation of mechatronic systems » - Karnopp,
Margolis, Rosenberg
HTML reports for the different model:
test_armature.html
test_torquemotor.html
test_armature+torque.html
bode-servovalve.html
servovalve.html
34
ANNEX
ANNEX 1 : Torque motor expression.
ANNEX 2 : Permanent magnet (Alnico 1500) demagnetizing curve.
ANNEX 3: Bond graph of armature.
ANNEX 4: Bond graph of torque motor.
ANNEX 5: Bond graph of complete servovalve.
35
ANNEX 1
Torque motor:
Here is the demonstration of the analytic formula for torque motor.
R1
Φ1
Φ2
R2
Φ1
R1
N.i
R2
Φ2
E0 = magnetomotive force of the permanent magnet
R0 = air gap reluctance for neutral reluctance
We will consider that the only reluctance important enough is air gap reluctance, as a
numerical application could confirm.
x = armature displacement in air gap.
e = air gap
R1= e+ x = R0⋅1+ x
Air gap reluctances are:
µ 0⋅S
e
e
−
x
R2=
=R0⋅1− x
µ 0⋅S
e
( )
( )
− N ⋅i+ R2⋅φ2 = R1⋅φ1
E0 = R2⋅φ2 + R1⋅φ1
Magnetomotive force evaluation gives:
We obtain the following flow expressions:
φ1= E0− N ⋅i = E0− N ⋅i
2⋅R1 2⋅R0⋅1+ x
e
( )
φ2 = E0 + N ⋅i = E0 + N ⋅i
2⋅R2 2⋅R0 ⋅ 1− x
e
( )
Torque exerted by the motor is:
C = l (φ22 −φ12 )
µ0⋅S
ie:
CTM =
()
 2x
l⋅µ0⋅S
 x 2 + N ⋅i ⋅ x  

2 ⋅ E0 ⋅ + N ⋅i⋅E0 1+
 e E0 e  
 x2   e
2
e ⋅1− 2 
e 

36
ANNEX 2
37
ANNEX 3
Magnétique
1
Magnétique
z1
TF : l
1
Magnétique
z 2
1
TF : -l
TF : -l
1
TF: z4
TF: z3
z3
Magnétique
1
TF : l
I:J
θ
TF: -z2
z 4
C
K11 K12 + d1⋅K11
K 21K 22 + d1⋅K12
1
0
1
0
0
x flapper
x fictitious
Fw
0
C : Kw
1
x G
I:m
1
x w
Bond graph of the armature
38
−1
ANNEX 4
C
0
1
1
0
C
C
1
C
1
Air gap
1
1
C
0
C
1
1
Permanent
magnet
0
Permanent
magnet
coil
1
C
F
x
C
1
C
1
GY
1
C
0
1
SE
C
0
1 port capacitor with
derivative causality
Bond graph of torque motor
39
1
ANNEX 5
I:J
I:m
C : -1/Km
SF
R :br
K11 K12 + d1⋅K11
K 21K 22 + d1⋅K12
C
K1
i
GY
R :bt
1
θ
−1
1
x G
TF : -d3
TF : -d2
0
0
x flapper
1
TF
R
R
TF
R
C
R
C
C
1
0
Pc1
Pt
1
0
R
SE
Ps
1
1
R
x fictitous
0
I : mspool
1
TF
Pc2
SE
Ps
TF
x spool
0
Bond graph of complete servovalve
C : Kw
40
1