Page 632 - The Mechatronics Handbook

0066_Frame_C20.fm Page 102 Wednesday, January 9, 2002 1:44 PM

Applying Newton’s second law for translational motion, we have

2
d x
Ft() = m-------- + B v dx ( k s1 x + k s2 x ) + F e t()
------ +
2
dt 2 dt
where x denotes the microplunger displacement, m is the mass of a movable member (microplunger),
B v is the viscous friction coefficient, k s1 and k s2 are the spring constants, and F e (t) is the magnetic force

∂W c i, x( )
(
F e i, x) = ----------------------
∂x

2
It should be emphasized that the restoring/stretching force exerted by the spring is given by (k s1 x + k s2 x ).
1 2
Assuming that the magnetic system is linear, the coenergy is found to be W c (i, x) = L(x)i and the
--
2
electromagnetic force developed is given by
F e i, x) = 1 2 dL x()
(
--i --------------
2 dx
In this formula, the analytic expression for the term dL(x)/dx must be found. The inductance is

2
2
N
Lx() = ------------------ = -------------------------------------------------
N m f m 0 A f A g
ℜ f + ℜ g A g l f + 2A f m f x + 2d)
(
where ℜ f and ℜ g are the reluctances of the magnetic material and air gap; A f and A g are the cross-sectional
areas; l f and (x + 2d) are the lengths of the magnetic material and the air gap.
Thus

2 2 2
dL – --------------------------------------------------------
------ =
2N m f m 0 A f A g
(
dx [ A g l f + 2A f m f x + 2d)] 2
Using Kirchhoff’s law, the voltage equation for the electric circuit is

u a = ri + dy
-------
dt

where the flux linkage ψ is ψ = L(x)i.
Thus, one obtains

dL x()dx
di
u a = ri + L x()----- + i--------------------
dt dx dt
Therefore, the following nonlinear differential equation results:

2 2 2
di r 2N m f m 0 A f A g 1
----- = – -----------i + -------------------------------------------------------------------iv + -----------m a
[
(
dt Lx() Lx() A g l f + 2A f m f x + 2d)] 2 Lx()
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