EMA TUTORIAL 7 SOLUTIONS
Micro-Actuators
Question 1 Solution
a) The electrostatic energy for each pair of capacitor palates is
1
1 A
1  0 tV 2
We  CV 2   0 V 2  x 2
2 d 
2 d

The electrostatic force is given by
1  0 tV 2
Fe 
2 d

The total driving force acting on gripper for ( n capacitors  2 sides) is
F  2n  Fe  n

 0 tV 2 d For numerical values,
F n

 0tV 2 d  200 

8.55 10 12  50 10 6 140 2
 0.335 mN
5 10 6

b) From the motion equation,




F  m x  B x Kx


x

B K
F
x x  m m m 

By taking state variables x1  x and x2  x , and F as input variable, the state space equations are given by








x1  x  x2 x2  x  

B
K
F x2  x1  m m m The transfer function for the system when B is assumed to be zero:

X ( s)
1
1
K /m

  2
 2 n 2
2
F ( s) ms  K K s  K / m s   n
2

Thus, the resonant frequency can be calculated as follows,
n 

and f n 

1
2

K
,
m

K
1
 m 2

51
 2.48 kHz
2.110 7

Question 2 Solution

a)

When V  0 kx0  mg , where x0 is the initial stretching distance of the spring.

When V  0 , the spring stretches an additional displacement x  x0  x due to an additional electric energy storage

AV 2

. We have the following force

2x

balance equation at equilibrium: k (x  x0 )  mg 

AV 2
2x2

1 V2
 A 2  k ( x0  x)
2 x

for 0  x  x0

2kx2 ( x0  x)
V
.
A

Alternatively, this voltage can be obtained from the co-energy
W 

A
2x

V 2 and the net force:

F   F  k ( x0  x) 

W 
A
 k ( x0  x)   2 V 2  k ( x0  x) .
x
2x

At equilibrium F  0 which yields V 

2kx2 ( x0  x)
A

The plot V(x) is shown as following:

b) At x 

2 x0
2 x 2kx0
, we have a maximum Vm  0
. That is also the