6.777J/2.372J Design and Fabrication of Microelectromechanical Devices
Spring Term 2007
Massachusetts Institute of Technology
PROBLEM SET 3 (13 pts)
By Brian Taff (with acknowledgement to Feras Eid and Xue'en Yang)
Due: Lecture 9
Problem 9.14 (2 pts): Bending of an AFM Cantilever
For LPCVD stoichiometric silicon nitride, E = 270 GPa.
From the lecture on structures, for a cantilever with the following configuration and loading
F
2
d w
the differential equation of beam bending is:
dx
2
=−
x
M
L
EI
z
where M is the internal moment at distance x:
M = -F (L-x)
and I is the moment of inertia of the cross section at the distance x. For a rectangular cross section:
I = bh3/12
For our case, h = 0.5 μm and b is a function of x due to the triangular shape of the cantilever.
b
For our geometry:
L−x
=
a
L
a = 50µm
b
x
a ( L − x ) h3
I=
12 L
Therefore:
2
and
d w
dx
Integrating twice:
2
=
12 F ( L − x ) L
Ea ( L − x ) h
w=
3
=
L= 250µm
12 FL
Eah3
6FL 2
x + Ax + B
Eah3
Applying the boundary conditions: w(x=0) = dw/dx (x=0) =0, we get: A = B = 0.
Therefore:
w( x ) =
6 FL 2
x
Eah3
Cite as: Carol Livermore and Joel Voldman, course materials for 6.777J Design and Fabrication of
Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/),
Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
6 FL3
= 55.55 F [w in m, F in N ]
Eah3
F
Eah3
=
=
= 0.018 N / m
6 L3
w( L )
w( L ) =
At the tip (x = L), the deflection is:
And the equivalent spring constant is:
keq
For a tip deflection of 2 μm:
F = keq ⋅ w( L ) = 0.018 ⋅ 2 x10 = 3.6x10 N
Mh F ( L − x )h12 L 6 FL
σ max =
=
= 2 = 4.32 MPa
2I
2ah3 ( L − x )
ah
−6
And the maximum stress at the support is:
-8
Problem 14.12 (2 pts): Circuit loading
Rs
a) Assuming we have an ideal op-amp
+
i+ = 0 Î V+ = Vs
i- = 0 Î V- = V1, i1 = i2
+
R2
and V = V Î Vs = V1
+
i2
-
-
Vs
+
-
V2
-
Also V1 =i1 R1 , VL = i1 ( R1 + R2 )
i1
+
R1
-
+
VL
RL
-
V1
Therefore VL / Vs = 1+ R2 / R1
b) Assuming again we have an ideal op-amp
i- = 0 Î
R1
is = iL
is
and V+ = V- = 0
Rs
R2
-
iL
+
Also V- - VL = iL R1 , Vs - V-= is ( Rs + R2 )
Vs
Therefore VL / Vs = - R1 / (Rs + R2 )
+
-
+
VL
-
The ratio is negative, hence the name "inverting".
c) The gain G is defined as the ratio of the output voltage to the input voltage, or VL / Vs.
For the non-inverting amplifier: Gideal = 1+ R2 / R1 = Gactual for all values of Rs
Gactual / Gideal = 1
For the inverting amplifier:
Gideal = - R1 / R2
Gactual / Gideal = R2 / (Rs + R2 )
Thus for the non-inverting amplifier, the ideal and actual gains are always equal, while for the inverting
amplifier, the actual gain approaches the ideal gain only when the sensor impedance approaches zero
(assuming both amplifiers are ideal). Clearly, the non-inverting amplifier is less sensitive to changes in Rs than
the inverting one.
Cite as: Carol Livermore and Joel Voldman, course materials for 6.777J Design and Fabrication of
Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/),
Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
RL
Problem 9.13 (3pts): Leveraged Bending with Electrostatic Actuation
F
a) The governing equation for the deflection of a doubly clamped beam
under a point load as shown in the figure is,
d 4w
EI 4 = − F < x − a > −1 ,
a
dx
−1
where < x − a > denotes a concentrated load at a. Integrating, we
have,
d 3w
EI 3 = − F < x − a > 0 +C1
dx
d 2w
EI 2 = − F < x − a >1 +C1 x + C2
dx
dw
F
1
EI
= − < x − a > 2 + C1 x 2 + C2 x + C3
dx
2
2
F
1
1
EIw = − < x − a > 3 + C1 x 3 + C2 x 2 + C3 x + C4
6
6
2
Apply the boundary conditions
w(0) = w( L ) = w '(0) = w '( L ) = 0 ,
we get,
F ( L − a ) 2 ( L + 2a )
C1 =
L3
− F ( L − a )2 a
C2 =
L2
C3 = C4 = 0
Hence, the center deflection is,
2
⎛ L ⎞ Fa (4a − 3L)
w⎜ ⎟ =
48EI
⎝2⎠
L
If we next examine the effect of the point load acting at the right hand side of the beam (acting alone as if
the load at “a” had been removed), we get a similar setup.
d 4w
= − F < x − ( L − b ) > −1
dx 4
d 3w
EI 3 = − F < x − ( L − b ) > 0 +C1
dx
d 2w
EI 2 = − F < x − ( L − b ) >1 +C1 x + C2
dx
dw
F
1
EI
= − < x − ( L − b ) > 2 + C1 x 2 + C2 x + C3
dx
2
2
1
1
F
EIw = − < x − ( L − b ) >3 + C1 x3 + C2 x 2 + C3 x + C4
6
6
2
EI
Applying the boundary conditions
w(0) = w( L ) = w '(0) = w '( L ) = 0 ,
we get,
Cite as: Carol Livermore and Joel Voldman, course materials for 6.777J Design and Fabrication of
Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/),
Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
C1 =
Fb 2 ( 3L − 2b )
L3
Fb 2 ( b − L )
C2 =
L2
C3 = C4 = 0
In this case, the center deflection is:
2
⎛ L ⎞ Fb (4b − 3L)
w⎜ ⎟ =
48EI
⎝2⎠
Applying superposition to the beam, we can obtain the center deflection due to both loads to be,
δ=
F ⎡⎣a 2 (4a − 3L) + b2 (4b − 3L) ⎤⎦
48EI
Problem 5.9 (2 pts): Circuit representation of a lumped mechanical system
F
k2
k1
k3
m1
m2
b1
x2
x1
In the e → V convention, the flows are the time derivatives of the displacements xi of the different bodies, and
effort is the external force. Since k1 and b1 share the same displacement, and hence, the same flow, they are in
series. The same holds for k2 and m2 and also for k3 and m1. By observation, the net flow through b1 is ( x 2 − x1 ) .
Hence, the equivalent circuit can be represented as shown in Figure.
m2
F
+
–
1/k2
m1
F
1/k1
x 2
m2s
1/k3
x1
+
–
k3/s
k1/s
x2s
s domain
b1
m1s
k2/s
x1s
b1
Apply KVL for the left loop, we have
2
m 2 s x 2 + k 2 x 2 + ( k1 / s + b1 )( x 2 s − x1 s ) = F
⇒ − (b1 s + k1 ) x1 + ( m 2 s 2 + b1 s + k1 + k 2 ) x 2 = F
Cite as: Carol Livermore and Joel Voldman, course materials for 6.777J Design and Fabrication of
Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/),
Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
(1)
Apply KVL for the right loop, we have
2
m1s x1 + k 3 x1 + ( k1 / s + b1 )( x1s − x2 s ) = 0
⇒ ( m1s 2 + b1s + k1 + k 3 ) x1 − (b1s + k1 ) x2 = 0
⇒ x2 =
m1s 2 + b1s + k1 + k 3
x1
b1s + k1
Substitute into (1), we have,
x1 ( s )
F (s)
=
b1s + k1
− (b1s + k1 ) + ( m1s + b1s + k1 + k 3 )( m2 s 2 + b1s + k1 + k 2 )
2
2
Cite as: Carol Livermore and Joel Voldman, course materials for 6.777J Design and Fabrication of
Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/),
Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].