The theory of harmonically driven, clamped silicon micro-beams
Henrik Bruus
Mikroelektronik Centret, bldg. 345, Technical University of Denmark, DK-2800 Lyngby
(Dated: June 10, 2002)
In this note the theory of harmonically driven, clamped silicon micro-beams is presented. The
specific example of a beam of dimensions 1 µm × 10 µm× 100 µm is studied in detail. The beam
is placed in a perpendicular magnetic field B, and it is forced to oscillate through the Lorentz force
by passing an ac-current J cos(ωt) through it. The amplitude of the oscillations at resonance are of
the order 0.1 µm and the amplitude of the associated, induced electromotive force E is of the order
1 µV. The whole setup is at room temperature in air.
PACS numbers: PACS numbers:
I.
INTRODUCTION: THE MICRO-BEAM
As a part of a DTU-MIC three weeks course we have
set out to design, fabricate, and test a MEMS-oscillator
in the form of a clamped silicon micro-beam etched out
from a very pure single-crystal. This constitutes the first
step at MIC to fabricate such devices. The micro-beam is
forced to oscillate by utilizing the magnetic force per unit
length, K(t) = BJ cos(ωt), that arises when the beam is
placed in a homogeneous magnetic field perpendicular to
the length axis of the beam, and an ac-current J cos(ωt)
is passed through the beam.
The oscillations of the beam are detected by measuring
the induced electromotive force, E(t) = −B ∂A/∂t, that
arises when the oscillating beam sweeps through the area
A(t) in the direction perpendicular to the homogeneous
magnetic field B.
The experiment is inspired by the one performed by
Cleland and Roukes,1 but instead of vacuum and cryogenic temperatures we plan to carry out our experiment
at room temperature and in air.
To be able to make a good design of the micro-beam
some theory is called for. The purpose of this note is to
provide the necessary theoretical understanding of the
experiment. The theory presented in the following is
based on Landau and Lifshitz.2
II.
In this note the x axis is placed along the length axis of
the beam. The magnetic field is applied along the y axis,
and consequently the beam is deflected in the z direction
when an ac-current J(t) is passed through the beam. The
beam has the length L, and a rectangular cross section
h × w, where h is the height in the z direction, and w the
width in the y dirction. In the specific example presented
in Sec. V we shall use
w = 10 µm,
L = 100 µm.
(1)
Furthermore, silicon has the following Young’s modulus
E and mass density ρ,
E = 1.6 × 1011 Pa,
ρ = 2.33 × 103 kg m−3 .
z 0 (0) = 0,
z(0) = 0,
z(L) = 0,
z 0 (L) = 0.
(3)
In this section we study the bent beam in equilibirium,
and find the basic differential equation for the deflection
in terms of an applied force per unit length, K(x). The
argument involves the internal forces F(x) acting on a
cross section S(x) in the yz plane at a given position x
and the associated internal force moment M(x) around
the center of mass of the same cross section.
Consider first a beam bent slightly downwards. In an
infinitessimal neighborhood of x the shape is approximated by a circle having a radius of curvature R. On
the top side the beam is stretched, on the bottom side
it is compressed. In the center, denoted z = 0, it keeps
its original length, dl. At the position z the length is
stretched from dl to dl0 . By the simple geometry of
concentric circle arches we find dl0 = dl (R + z)/R =
(1 + z/R) dl, whence the strain in the x direction is given
by ux (z) = z/R. By Hooke’s law the xx component of
the stress tensor σ then becomes
σxx (x, y, z) = E ux (z) =
E
z.
R(x)
(4)
From this we can easily calculate the corresponding y
component My of the internal moment M,
THE BENT BEAM IN EQUILIBRIUM
h = 1 µm,
The central goal of this note is to calculate the form of
the deflected beam, z(x, t). Throughout the calculation
we assume the clamped-beam boundary conditions,
(2)
Z
My (x) =
dydz z σxx =
S
EIy
,
R(x)
Iy =
wh3
.
12
(5)
Here we have introduced the y component Iy of the geometrical moment of inertia of a square cross section h×w,
R
R w/2
R h/2
Iy = S dy dz z 2 = −w/2 dy −h/2 dz z 2 . For small deviations the curvature is equal to the second derivative of
the deflection, 1/R(x) = z 00 (x), and we arrive at
z 00 (x) =
My (x)
.
EIy
(6)
While this relation constitutes a fundamental relation between curvature and the internal force momentum, we
2
however still need to study how the deflection is affected
by variations in M(x) induced by the externally applied
force per unit length K(x).
First we consider the forces acting on the infinitessimal
element of the beam between x and x + dx. This element
is influenced by three forces. At the surface
R S(x) we have
the internal force −F, given by (F)i = S(x) dydz σix . At
the surface S(x + dx) we have the internal force F +
dF. And finally we have the external force K(x) dx. In
equilibrium we must require force balance, whence
(F + dF) − F + K(x) dx = 0
dF
= −K.
dx
⇒
(7)
Next we consider the force moments acting on the element. Again three contributions must be considered.
In equilibrium they must add to zero. We study the
moments acting on the center of mass C of the surface
S(x + dx). The moment around C arising from the internal forces acting on the surface S(x + dx) is denoted
M + dM. The moment around C due to the internal
forces acting on the surface S(x) is found from the parallel theorem to be −M−dx×F. Finally, the moment from
the external force is neglected since it is proportional to
K(x) dx2 , i.e. of second order in dx. In equilibrium we
must require force moment balance,
(M+dM)−(M+dx×F) = 0
⇒
dM
= −ex ×F. (8)
dx
Now by combining Eqs. (5), (7) and (8) we arrive at the
fourth-order differential equation, which describes the deflection z(x) of the beam due to the z component Kz (x)
of the external force per unit length,
EIy z 0000 (x) = Kz (x).
(9)
For the static case with a constant external force Kz
the deflection is given by a simple fourth-order polynomium in x. We will not pause to examine the statics
further, but push on to study the dynamical eigenmodes.
The general solution to this homogeneous fourth-order
differential equation is the sum of four linearly independent solutions,
z(x) = a cos(kx) + b sin(kx) + c cosh(kx) + d sinh(kx).
(13)
By imposing the boundary conditions z(0) = 0 and
z 0 (0) = 0 we find immediately
z(x) = a[cos(kx)−cosh(kx)]+b[sin(kx)−sinh(kx)]. (14)
The boundary conditions z(L) = 0 and z 0 (L) = 0 lead to
a standard 2 × 2 secular equation,
µ
¶µ ¶ µ ¶
cos(kL)−cosh(kL) sin(kL)−sinh(kL)
a
0
=
.
−sin(kL)−sinh(kL) cos(kL)−cosh(kL)
b
0
(15)
Non-trivial solutions exist only when the determinant of
the coefficient matrix is zero. The condition for this to
happen is readily shown to be
cos(kL) =
(16)
Generally, the wavenumber k of the eigenmodes can
only be found by numerical methods. However, since
1/ cosh(kL) tends to zero exponentially fast for large kL,
good approximate solutions can easily be found:
2n + 1 π
, n = 1, 2, 3, . . . .
2
L
(17)
The exact equation Eq. (16) have no solutions corresponding to n = 0 in Eq. (17). The eigenfrequencies
become
s
s
2
EIy 2
E h
2 (2n + 1)
√
ωn =
k
≈ π
. (18)
ρS n
ρ L2
8 3
cos(kL) ≈ 0
⇒
kn ≈
Using the vaules from Eqs. (1) and (2) the exact solution
of the lowest eigenfrequency is
ω1 = 5.35 MHz
III.
1
.
cosh(kL)
⇒
f1 = 0.85 MHz.
(19)
THE EIGENMODES OF THE BEAM
The eigenmodes are found by extending the static
equation Eq. (9) to the dynamic case. When allowing
for motion it follows from Newton’s second law that the
external force is changed by the addition of the inertial
force per unit length, −ρS z̈(x, t),
EIy z 0000 (x, t) = Kz (x, t) − ρS z̈(x, t).
(10)
Looking for eigenmodes we put Kz (x) = 0 and make the
ansatz z(x, t) = z(x) cos(ωt). This results in
z 0000 (x) = k 4 z(x),
(11)
where we have introduced the wavenumber k given by
k4 ≡
ρS 2
ω .
EIy
(12)
IV.
THE HARMONICALLY DRIVEN BEAM
Having identified the eigenmodes we now turn to the
problem of the harmonically driven beam. As mentioned we use the magnetic force per unit length, K(t) =
BJ cos(ωt) as the driving force Kz (t). In this case the
driving force does not depend on position. To make the
calculation more realistic we add the damping force per
unit length −(γ/L)ż due to the friction between the air
and the beam. The damping coefficient γ is proportional
to the viscosity η of air and, since the beam oscillates
in the z direction,
to the square root of the area in the
√
xy plane, i.e. wL. I have not have time to look up the
numerical prefactor, but simply assume
√
(20)
γ = wL η.
3
The differential equation Eq. (10) for z(x, t) with these
assumptions becomes
γ
ż(x, t).
L
(21)
Due to the appearence of both z̈ and ż it is useful to
use complex functions. Let ζ(x, t) = ζ(x)eiωt be the solution to the equation with a complex drive K(t) = Keiωt .
By complex conjugation we find that ζ ∗ (x, t) therefore
is a solution with the complex drive K ∗ (t). Now, since
the differential equation Eq. (21) is linear we find that
Re[ζ(x)eiωt ] = [ζ(x, t) + ζ ∗ (x, t)]/2 is a solution with the
drive [K(t)+K ∗ (t)]/2 = K cos(ωt), which is the physical
drive. Hence our solution is
h
i
z(x, t) ≡ Re ζ(x) eiωt .
(22)
We therefore insert ζ(x) eiωt and K(t) = Keiωt into
Eq. (21) and obtain after dividing by eiωt
EIy ζ(x)0000 (x) = K + ω 2 ρS ζ(x) − iω
γ
ζ(x).
L
(23)
In analogy with Eqs. (11) and (12) we write this as
ζ 0000 (x) = q 4 ζ(x) +
K
,
EIy
(24)
where we have introduced the complex wavenumber q
given by
q4 ≡
ρS 2
γ
ω −i
ω.
EIy
EIy L
(25)
The differential equation Eq. (24) is inhomogeneous.
The complete set of solutions is generated by adding one
particular solution ζ0 (x) of the inhomogeneous equation
to the general solution Eq. (13), with k = q, of the corresponding homogeneous equation, i.e. where K = 0. By
insertion it is easily seen that the constant
ζ0 = −
1 K
q 4 EIy
0.3
z (um)
EIy z 0000 (x, t) = K(t) − ρS z̈(x, t) −
0.4
0.2
0.1
0
0
ζ(x) = ζ0 [1+a cos(qx)+b sin(qx)+c cosh(qx)+d sinh(qx)].
(27)
Note how the amplitude ζ(x) of the driven beam is fixed
by the driving force through the value of ζ0 . This is in
contrast to the case of the freely swinging beam Eq. (13),
where no scale factor is provided. By imposing the
boundary conditions z(0) = 0 and z 0 (0) = 0 to Eq. (27)
we find
n
ζ(x) = ζ0 a[cos(qx) − cosh(qx)]
(28)
o
+b[sin(qx) − sinh(qx)] + [1 − cosh(qx)] .
0.4
0.6
0.8
1
x/L
FIG. 1: The shape z(x) of the 1 µm × 10 µm × 100 µm Si
micro-beam at time t = T1 /4 when driven at the fundamental
resonance frequency ω = ω1 with an ac-current of amplitude
J = 1 mA in a magnetic field B = 10 mT. The damping is
given by the viscosity of air, which at room temperature and
standard pressure is η = 1.8 × 10−5 Pa s.
The boundary conditions z(L) = 0 and z 0 (L) = 0 lead to
a simple invertable 2 × 2 matrix equation for the coefficients a and b,
µ
¶µ ¶
cos(qL)−cosh(qL) sin(qL)−sinh(qL)
a
−sin(qL)−sinh(qL) cos(qL)−cosh(qL)
b
µ
¶
cosh(qL) − 1
=
. (29)
sinh(qL)
In the following section we study this solution.
V.
A SPECIFIC EXAMPLE: THE BEAM SHAPE
Once the clamped silicon beam has been specified by
Eqs. (1), (2) and (3) the resonance frequencies are fixed.
However, the amplitude ζ0 is controlled by the ac-current
amplitude J and the strength B of the magnetic field.
ζ0 = −
(26)
is a particular solution to Eq. (24), and we thus arrive at
0.2
K
BJ
=− 4 .
4
q
q
(30)
Moreover, the damping coefficient and thus the Q-factor
of the oscillating beam is controlled through the viscosity
η of the surrounding air. In this example we choose the
following vaules of the external control parameters, which
for η means room temperature and standard pressure
J = 1 mA,
B = 10 mT,
η = 1.8 × 10−5 Pa s. (31)
In Fig. 1 is shown the shape of the beam at maximal
deflection when driven at the first resonance frequency
ω1 . This occurs at t = T1 /4, where T1 ≡ 2π/ω1 is the
period of the oscillation. Thus the oscillation of the beam
is π/2 radians out of phase with the driving ac-current.
In Fig. 2 is shown the shape of the beam for t ≈ T1 /2
when it suffers almost no deflection.
4
0.001
0.04
Ar (um^2)
z (um)
0.0005
0
-0.0005
-0.001
0
-0.02
-0.04
-0.0015
0.2
0.4
0.6
x/L
0.8
1
0
0.2
0.4
0.6
0.8
1
0.8
1
0.8
1
t/T1
FIG. 2: The shape z(x) of the beam driven at the first resonance frequency as in Fig. 1, however, here at t ≈ T1 /2
where the beam happens to suffer a very small deflection.
Eleven different shapes are shown for times ranging from
t = T1 /2 − 2×10−6 T1 to t = T1 /2 + 2×10−6 T1 .
0.03
20
10
Ar (um^2)
0
0
-10
0.02
z (um)
0.02
0.01
-20
0
0
0.2
0.4
0.6
t/T1
-0.01
0.04
-0.03
0
0.2
0.4
0.6
x/L
0.8
1
FIG. 3: The shape z(x) of the beam as in Fig. 1, however,
here driven at the third resonance, ω = ω3 , and at time t =
T3 /4 = 2π/(4ω3 ).
Ar (um^2)
-0.02
0.02
0
-0.02
-0.04
VI.
0
THE INDUCED EMF E
dΦ
dA(t, ω)
=B
,
dt
dt
0.4
0.6
t/T1
To detect the resonance frequency and Q-factor of the
oscillating beam we plan to measure the electromotive
force E(t) induced by the current carrying beam moving
in the applied constant magnetic field B. According to
Fraday’s law of induction we have
−E(t, ω) =
0.2
(32)
FIG. 4: The top panel shows the time dependence of the area
amplitude A(t, 0.99 ω1 ), i.e. at a driving frequency slightly
below the first resonance frequency. The middle panel shows
the time dependence of the area amplitude A(t, 1.00 ω1 ), i.e.
when driving the beam at the first resonance frequency. The
bottom panel shows the time dependence of the area amplitude A(t, 1.01 ω1 ), i.e. at a driving frequency slightly above
the first resonance frequency. Notice the changes in amplitude
and phase of the three panels.
where A(t, ω) is the area of the closed current loop which
the beam is a part of. Let A0 be the area of the loop when
the beam is at its equilibrium position. Then
A(t, ω) = A0 + A(ω) ei(ωt+φ) ,
(33)
with the frequency dependent area-amplitude A given by
hZ
A(ω) = maxt
0
L
i
dx z(x, t) .
(34)
The amplitude E of the induced emf is therefore
E(ω) = ω B A(ω).
(35)
Once the maximal area is found from calculations as
the ones depicted in Fig. 4, we can find the frequency depedence of the induced emf E(ω) from Eqs. (34) and (35).
5
From this follows ζm directly:
1.2
ζm (ω) =
emf (uV)
1
0.8
BJ
− ω 2 ) − iω
γ
L
,
(39)
and consequently
0.6
zm (ω) = |ζm | = p
0.2
0.9999
1
w/w1
1.0001
FIG. 5: The amplitude E(ω) of the induced emf of the
1 µm × 10 µm × 100 µm Si micro-beam. The resonance frequency is at ω1 = 5.35 MHz or f1 = ω1 /2π = 0.85 MHz.
The amplitude of the driving ac-current is J = 1 mA and
the magnetic field is B = 10 mT. The damping is given by
the viscosity of air, which at room temperature and standard
pressure is η = 1.8 × 10−5 Pa s.
In Fig. 5 is shown the resulting resonance curve of E(ω)
near the first resonance frequency ω1 . It is seen that the
maximal emf-amplitude is 1.21 µV. This is a fairly large
signal that ought to be easy to measure.
From the graph we can also determine the Q-factor
Q1 for the first resonance. It is given by
ω1
5.35 × 106 Hz
Q1 ≡
=
= 1.26 × 104 ,
(36)
∆ω1
4.23 × 102 Hz
where ∆ω is the full width at half maximum of the resonance curve. This is a rather high value, so we can expect to observe well-defined resonance peaks in the emfspectrum.
VII.
(ρS)2 (ω12
in good agreement with the 0.43 µm of Fig. 1. The
estimate Qm for the Q-factor is even better. By setting
zm (ω1 − ∆ω/2) = zm (ω1 )/2 we find
h
i √ ω γ
1
1
(ρS) ω12 − (ω1 − ∆ω)2 = 3
,
(42)
2
L
and from this the perfect estimate
ω1
ω1 ρLwh
Qm =
= √
= 1.26 × 104 .
∆ω
3γ
(43)
The area A(ω) is approximated with that of a triangle
with base line L and height zm (ω1 ):
A(ω1 ) =
1
BJL2
zm (ω1 )L =
= 16.4 µm2 ,
2
2ω1 γ
(44)
which compares favorably with the 22.7 µm2 of Fig. 4.
Finally, we estimate the amplitude of the electromotive force E by inserting Eq. (44) into Eq. (35)
E(ω1 ) = ω1 B A(ω1 ) =
B 2 JL2
= 0.88 µV.
2γ
(45)
This is fairly close to the amplitude 1.21 µV of Fig. 5.
SIMPLE ESTIMATES
The calculation in the previous section are straightforward but not easy to understand directly. We can,
however, get a good grasp of the solution by making a
simple approximation valid for the tiny damping forces
present in the problem. According to Eqs. (11) and (12),
when damping is neglected, the shape of the oscillating
beam at resonance ω = ω1 is given by the equation
EIy ζ 0000 (x) = ρSω12 ζ(x).
BJ
. (40)
− ω 2 )2 + ω 2 (γ/L)2
The simple estimate for the amplitude zm (ω1 ) at
resonance ω = ω1 is found form this equation. The
result
BJL
zm (ω1 ) =
= 0.33 µm,
(41)
ω1 γ
0.4
(37)
If we let the entire beam be represented by the mid-point
ζm , we can derive an approximate equation for this point
by inserting Eq. (37) into Eq. (23)
γ
(38)
ω12 ρS ζm = BJ + ω 2 ρS ζm + iω ζm .
L
1
ρS(ω12
A. N. Cleland and M. L. Roukes, Appl. Phys. Lett., 69
2653 (1996)
VIII.
CONCLUSION
The calculation presented here shows that if we succeed
in fabricating a 1 µm × 10 µm × 100 µm Si micro-beam,
then fairly sharp emf-resonance curves are expected with
Q-factors around 104 and peak amplitudes of 1 µV at the
first resonance frequency of 0.85 MHz. This is remarkable
since the parameters in the calculation corresponds to a
situation where the beam is oscillating in air at room
temperature and standard pressure, while exposed to a
moderate magnetic field of strength 10 mT and driven
by an ac-current of 1 mA.
The series of simple estimates of the system values
presented in Sec. VII provides us with a good tool for
obtaining a good intuitive feel for the system.
2
Landau and Lifshitz, Theory of Elasticity (Pergamon Press
3rd edition, New York, 1986).