Positive displacement to the right.
Displacement is limited by hard stops -l ≤x ≤l
Mechanical force:
Capacitance of right, left comb drive for |x|≤l :
When the comb drive fingers disengage, capacitance is
significantly decreased and can be neglected, i.e.
For x>l:
For x< -l
Electrostatic force:
For |x|≤l:
For x> l :
For x< -l:
In equilibrium:
Introduce
For |x|≤l:
For x> l :
For x< -l:
a)V0=V
No voltage is applied to the right actuator, only the left actuator is ON.
Since Electrostatic force is attractive, displacement is always to the left,
X<0.
Force and displacement are quadratic in voltage until the hard stop is reached.
b)V0 =const:
The situation becomes much more interesting.
Depending on the position of the shuttle there are 5 cases – only left, only
right or both combs
may be engaged, and the shuttle may be in contact with either left or right
hard stop.
When both combs are engaged, the displacement is a linear -equations.html">LINEAR function of voltage V.
When only one comb is engaged, the displacement is a quadratic function of
voltage V.
In the quadratic regions there are two voltages V for a given displacement,
corresponding to
two branches on a parabola x (V).
When the voltage V is changed , whenever displacement reaches
, a transition occurs,
which may involve a discontinuous jump in displacement. This leads to
various hysteresis loops.
Except for the transition points, all branches are stable since
Several branch configu rations may be possible, depending on values of parameters
V0, β, l, L.
Some of them are sketched be low .
Specifically, the stable branches are
1.Both combs engaged, linear dependence
x=-4 βV0V for
-l <x < l and
2.Only right comb engaged, quadratic dependence
x=β(V0 - V)2 for l <x <l
This happen in two voltage region
and 
3.Only left comb engaged, quadratic
dependence
x=-β(V0 + V)2 for -l <x <-l
Also in two voltage regions:
and
4.Hard stop to the right
5.Hard stop to the left
Some possible configurations:
