PIEZOELECTRIC MATERIALS
4.4 COMMON OPERATING MODES OF A PIEZOELECTRIC TRANSDUCER
4.4.4 Piezoelectric Stack Actuating a Linear Elastic Load
A common application of a piezoelectric stack is motion control in which a piezoelec- tric stack is applying a force against an object that can be modeled as an elastic load.
The simplest model of an elastic object is a linear spring, as shown in Figure 4.14a.
To facilitate the analysis, rewrite equation (4.79) in the form us= 1
ksE f +uov, (4.84)
whereksE=Y3EA/Ls is the short-circuit stiffness of the piezoelectric stack anduo
is the free displacement of the stack per unit voltage input. For the case shown in Figure 4.14,us=uand f = −klu, whereklis the stiffness of the load. Substituting these expressions into equation (4.84), we have
u= −kl
kEsu+uov. (4.85)
Solving for the displacement as a function of the applied voltage, we have u= uo
1+kl/ksEv= δo
1+kl/ksE. (4.86)
From equation (4.86) we see that if the stiffness of the load is much less than the stiffness of the piezoelectric,kl/ksE1 and the displacement is approximately equal to the free displacement of the stack. If the stiffness of the stack is much lower than the stiffness of the load,kl/kEs 1 and the displacement of the stack is much smaller than the free displacement. A plot of the normalized displacement as a function of
u
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0.2 0.4 0.6 0.8 1
kl/ks u/δo
(a) (b)
spring constant kl
E
Figure 4.14 Piezoelectric stack actuator driving an elastic load modeled as a spring.
kl/ksEis shown in Figure 4.14b. From the plot we see that the displacement becomes
1
2δo when the stiffness of the load is equal to the stiffness of the actuator. This is sometimes referred to as thestiffness match pointdue to the fact that stiffness of the load and actuator are equal.
The stiffness match point is also important in the analysis of the force and work transferred from the piezoelectric to the load. If we solve for the blocked force from equation (4.84), we see that
fbl=δoksE. (4.87)
The force applied by the actuator to the load is f = −klu; therefore, we can solve for the ratio of the force to the blocked force as
f fbl
= klδo/
1+kl/ksE
δoksE = kl/ksE
1+kl/kEs
. (4.88)
This function is plotted in Figure 4.15a. From the plot we see that the force output follows a trend that is opposite to the displacement. When the load is much softer than the actuator,kl/kEs 1 and the output force is much less than the blocked force.
In the opposite extreme we see that the force is nearly equivalent to the blocked force of the stack.
A final metric that is of importance is the output work of the stack. Recall that the work is defined as the produce of the force and displacement. For the piezoelectric stack we can define it asW = f u. The maximum work output of the device is equal to the product of the blocked force and the free displacement. If we normalize the output work with respect to the product of the blocked force and free displacement,
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0.2 0.4 0.6 0.8 1
kl/ksE kl/ksE
f/fbl
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0.2 0.4 0.6 0.8 1
fu/fblδo
(b) (a)
Figure 4.15 (a) Output force of a piezoelectric stack normalized with respect to the blocked force; (b) output work of a piezoelectric stack normalized with respect to the maximum possible work.
we can show that
f u fblδo
= kl/ksE
1+kl/ksE2. (4.89)
This function is plotted in Figure 4.15b and we see that the maximum output work is one-fourth of the product of the blocked force and the free displacement. Furthermore, the result demonstrates that this maximum occurs at the stiffness match between the actuator and the load. This is an important number to remember because it provides a straightforward method of computing the maximum energy transfer between the actuator and the elastic load if the blocked force and the free displacement are known.
Often, the specifications from a manufacturer will list the blocked force and free displacement of the transducer. The maximum achievable energy transfer is then estimated as one-fourth of the product of fblandδo.
These results emphasize that there are three operating regimes when using the actuator to produce force and displacement on an elastic load. In the regime in which the actuator stiffness is much lower than the load stiffness,kl/kEs 1, and we see that very little load transfer occurs between the transducer and the load. This is often desirable when using the material as a sensor because it indicates that very little force is transferred to the load from the transducer. Thus, the load cannot “feel”
the presence of the smart material, and its motion is not going to be affected by the presence of the transducer. At the opposite extreme we see that the motion of the transducer is not affected by the presence of the load; therefore, it is often desirable for ksEkl for applications in motion control where the objective is to achieve maximum displacement in the piezoelectric actuator and the load. The regime in whichkl ≈kEs is typically desirable when the application requires maximizing energy transfer between the transducer and the load. This is often the case when one is trying to design a system that dissipates energy in the load using a piezoelectric transducer.
Example 4.6 An application in motion control requires that a piezoelectric actu- ator produce 90 µm of displacement in a structural element that has a stiffness of 3 N/µm. The applications engineer has chosen a piezoelectric stack that produces a free displacement of 100µm. Determine (a) the stiffness required to achieve 90µm of displacement in the load, and (b) the amount of force produced in the load for this stiffness value.
Solution (a) From equation (4.86) we can solve for the stiffness ratio as a function of the free displacement,
kl
ksE = δo
u −1,
and substitute in the values given in the problem:
kl
ksE = 100µm
90µm −1=0.11.
Solving forksEyields
ksE=3 N/µm
0.11 =27.3 N/µm.
(b) To compute the blocked force, we see from equation (4.87) that the blocked force is simply equal to the product of the free displacement and the stiffness; there- fore,
fbl=(100µm) (27.3 N/µm)=2730 N.
The amount of force applied to the load can be computed from equation (4.88) as f =(2730 N) 0.11
1+0.11 =270.5 N.
This design can be visualized by drawing the relationship between force and displacement for the stack and placing the load line that represents the stiffness of the load. This is illustrated in Figure 4.16, where the solid line represents the relationship between force and displacement for the actuator. It is a linear relationship with intercepts at 2730 N and 100µm. The stiffness of the load is represented by the dashed line, which intercepts the force–displacement curve at a point defined by a displacement of 90µm and 270 N. This is represented by a square in the figure.
750 1500 2250 3000
25 50 75 100
displacement (µm)
force (N)
load line
Figure 4.16 Load line for a piezoelectric actuator design.
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10–3 10–2 10–1 100 101 102
ω/ωnE
|u/δ V|o
k
l/k
sE= 0 k
l/k
sE= 10
k
l/k
sE= 100
(b) (a)
k
lu(t)
m
Figure 4.17 Piezoelectric stack actuating a mass–spring system.
The previous analysis concentrated on the static response of a piezoelectric stack actuator exciting a system modeled as an elastic spring. If we generalize the analysis such that the load is the mass–spring system shown in Figure 4.17a, the force applied to the piezoelectric stack is equal to
f(t)= −mu¨(t)−klu(t), (4.90) and equation (4.84) can be rewritten
u(t)= 1
kEs[−mu(t)¨ −klu(t)]+uov(t). (4.91) Rewriting equation (4.91) as a mass–spring system, we have
mu(t¨ )+ kl+ksE
u(t)=uokEsv(t). (4.92) Taking the Laplace transform of equation (4.92) (assuming zero initial conditions) and rewriting as the transfer functionu(s)/v(s) yields
u(s)
uov(s)= kEs/m s2+
kl+ksE
/m. (4.93)
Denoting the ratiokEs/mas the square of the short-circuit natural frequencyωnE2, we can write the transfer function in the frequency domain as
u(ω)
uov(ω) = 1
1+kl/kEs − ω/ωEn
2. (4.94)
The low-frequency asymptote of the frequency response as ω/ωEn →0 matches the analysis previously performed for an elastic spring. This is expected be- cause the analysis for an elastic spring is equivalent to the present analysis with m=0.
A plot of the magnitude of the frequency response for three values of stiffness ra- tio is shown in Figure 4.17b. The low-frequency response is the quasistatic response which matches that of the analysis for an elastic spring. We note that the quasistatic response is flat, which indicates that the harmonic response at frequencies well be- low resonance has a peak amplitude equal to the static response. The sharp rise in the magnitude is associated with the resonant response of the system. The resonant frequency increases with increasing stiffness ratio, due to the additional stiffness of the load spring.