Page 106 - The Mechatronics Handbook

0066_Frame_C07 Page 16 Wednesday, January 9, 2002 3:39 PM

FIGURE 7.14 Coupled lumped parameter electromechanical system with single degree of freedom mechanical
motion x(t).
circuits by defining the charge on the capacitor, Q, as another generalized coordinate along with x, i.e.,
in Lagrange’s formulation, q 1 = x, q 2 = Q. Then we add to the kinetic energy function a magnetic energy
function W m (Q ˙ , x), and add to the potential energy an electric field energy function W e (Q, x). The
equations of both the mass and the circuit can then be derived from

[
d∂ T +[ W m ] ∂ T + W m ] ∂ W + W e ]
[
------ ----------------------- – -------------------------- + --------------------------- = Q k (7.44)
dt ∂q ˙ k ∂q k ∂q k
The generalized force must also be modified to account for the energy dissipation in the resistor and the
–
–
energy input of the applied voltage V(t), i.e., Q 1 = cx ˙ , Q 2 = RQ ˙ + V(t). In this example the magnetic
energy is proportional to the inductance L(x), and the electric energy function is inversely proportional
to the capacitance C(x). Applying Lagrange’s equations automatically results in expressions for the
magnetic and electric forces as derivatives of the magnetic and electric energy functions, respectively, i.e.,

1
W m = 1 ˙ 2 1 2 W e = ---------------Q 2 (7.45)
--Lx()Q =
--LI ,
2 2 2Cx()
(
∂W m x,Q) 1 2dL x() ∂W e x,Q) 1 2 d 1
(
˙
F m = -------------------------- = --I --------------, F e = – ------------------------- = −--Q ------ ----------- (7.46)
∂x 2 dx ∂x 2 dx Cx()
These remarkable formulii are very useful in that one can calculate the electromagnetic forces by just
knowing the dependence of the inductance and capacitance on the displacement x. These functions can
often be found from electrical measurements of L and C.
Example: Electric Force on a Comb-Drive MEMS Actuator
Consider the motion of an elastically constrained plate between two grounded fixed plates as in a MEMS
comb-drive actuator in Fig. 7.15. When the moveable plate has a voltage V applied, there is stored electric
field energy in the two gaps given by

∗
W e V, x) = 1 2 d 0 (7.47)
(
--e 0 V A---------------
2
2 d 0 – x 2
In this expression the electric energy function is written in terms of the voltage V instead of the
charge on the plates Q as in Eqs. (7.45) and (7.46). Also the initial gap is d 0 , and the area of the plate is A.

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