MEMS and Microsystems Design, Manufacture, and Nanoscale

55 Pages·2009·0.41 MB·English

MEMS and Microsystems
Design, Manufacture, and Nanoscale
nd
Engineering (2 Edition)*

SOLUTION MANUAL

Tai-Ran Hsu, Professor**
Department of Mechanical and Aerospace Engineering
San Jose State University
San Jose, CA 95192-0087
USA

August 15, 2008

_____________________________________
* John Wiley & Sons, Inc., Hoboken, New Jersey, USA
©2008, ISBN 978-0-470-08301-7
** Telephone: [phone]; Fax: [phone]
E-mail: [email protected]
1
Contents

Chapter 1 Overview of MEMS and Microsystems 3

Chapter 2 Working Principles of Microsystems 3

Chapter 3 Engineering Science for Microsystems Design
and Fabrication 9

Chapter 4 Engineering Mechanics for Microsystems Design 12

Chapter 5 Thermofluid Engineering and Microsystems Design 27

Chapter 6 Scaling Laws in Miniaturization 36

Chapter 7 Materials for MEMS and Microsystems 36

Chapter 8 Microsystems Fabrication Processes 41

Chapter 9 Overview of Micromanufacturing 50

Chapter 10 Microsystems Design 51

Chapter 11 Assembly, Packaging, and Testing of Microsystems 52

Chapter 12 Introduction to Nanoscale Engineering 55
2
Chapter 1
Overview of MEMS and Microsystems
(P. 32)

1. (b); 2. (a); 3. (b); 4. (c); 5. (a); 6. (c); 7. (c); 8. (c); 9. (b); 10. (c)
11. (a); 12. (a); 13. (b); 14. (a); 15. (b); 16. (c); 17. (c); 18. (a); 19. (a); 20. (c)

Chapter 2
Working Principles of Microsystems
(P. 77)

Part 1. Multiple Choice

1.(a); 2. (c); 3. (a); 4. (b); 5. (b); 6. (c); 7. (a); 8. (a); 9. (a); 10. (b); 11. (c); 12. (a); 13. (c);
14. (c); 15. (b); 16. (b); 17. (a); 18. (c); 19. (a); 20. (c); 21. (b); 22. (b); 23. (b); 24. (c);
25. (c); 26. (a); 27. (a); 28. (c); 29. (b); 30. (a); 31. (b); 32. (a); 33. (b); 34. (b); 35. (a)

Part 2. Description Problems

Problem 2:

Transducers Advantages Disadvantages
Piezoresistors High sensitivity. Sensitive to temperature.
Small sizes. Produced by doping foreign
substances to silicon substrates.
Capacitors Simple in structure, hence less Exhibit nonlinear input/output
expensive to produce. relationship-require careful calibration
Not sensitive to temperature- prior to applications.
suitable for operations at elevated Much bulkier than piezoresistors-takes
temperatures. up precious space in micro devices.

Problem 3: The three principal signal transduction methods for micro pressure sensors are:

(a) Piezoresistors.
(b) Capacitors.
(c) Resonant vibrating beams.

Advantages of (a) and (b) have been presented in Problem 2. Advantage of (c) is high
resolution and sensitivity, especially for high temperature applications. Principal
disadvantages of this method are the high cost involved in manufacturing and the bulky size.

3
Problem 5:

The assembly of minute overlapped electrodes (known as “comb drives”) can produce
electrostatic forces. The scaling laws in Chapter 6 will prove that electrostatic force actuation
scale down two orders of magnitude better than electromagnetic force for actuation. A major
drawback of electrostatic forces is their low magnitudes, which make them impractical for
actuation in macroscale.

Problem 6:

The natural frequency of a device is related to its geometry, which governs the stiffness of the
device, and its mass. Varying the stress state in the device made of an elastic solid, such as the
sensing element of a micropressure sensor will result in the change of its geometry, and thus the
shifting of its natural frequency.

Problem 7:

These holes in the back plate can mitigate the change of gap between the thin diaphragm and the
back plate. Such gap change can produce unwanted output in capacitance change, and thus
malfunctioning of the microphone

Problem 8:.

We may compute and tabulate the ratios of the output voltage, Vo to the input voltage, Vi vs. the
corresponding gaps between a pair of parallel electrodes and follow the procedure as outlined in
Example 2.2 on P. 47:

Gap, d 2 1.75 1.50 1.00 0.75 0.50
Vo/Vi 0 0.033 0.071 0.167 0.227 0.300

We may plot the relation of the gap, d versus Vo/Vi using the above data in the table. The curve
in Vo/Vi vs. the gap d is close to be a straight line. We realize that Vo/Vi → ∞ when d → 0.

Problem 9:

The output voltage from a thermopile with 3 thermocouple pairs can be obtained from Eq. (2.4)
as:
∆V = Nβ∆T

with N = 3, and ∆T = (120 – 20) + 273 = 373 K, the Seebeck coefficient, β = 38.74x10-6 V/oC
for copper/Constantan from Table 2.3.

Thus, the output voltage is: ∆V = 3x38.74x10−6x373= 0.04335volt or 43.35mv

4
Problem 10:

Actuation techniques Advantages Disadvantages
Thermal force Simple in structure. Response may not be instant due to
thermal inertia of the material.
Shape-memory alloys Actuation is more precise. Same problem as in the thermal actuation
case. It is functional only with a thermal
source.
Piezoelectric Simple and it is less costly Cannot maintain the actuated movement
to produce. Usually for sustained period of time due to
provides precise actuation. overheating.
Electrostatic force Takes up the least amount Low in magnitudes.
of space. Actuation is
instant.

Problem 11:

We assume that there is no friction between the electrodes and the dielectric Pyrex glass. By
following the geometry and the dimensions given in Example 2.1 on P. 45 with:

L = W = 800x10-6 m; ε = 8.85x10-9 F/m; ε = 4.7 (Table 2.2); V = 70 v; and d = 2x10-6 m
o r

From Equation (2.10), we may compute the electrostatic force in the width-direction: F =
w
0.0815 N.
From Equation (2.11), for the force in the length-direction: F = 0.0815 N
L

Problem 12:

We will model the comb drive actuator from a simplified model as illustrated below:

V

Spring constant
Spring constant
k
k

Moving electrodes Moving electrodes

Fixed electrodes

5
The required traveling distance of the moving electrodes is δ = 10x10-6 m, which corresponds
to the spring force with a spring constant, k = 0.05 N/m:

F = kδ = 0.05x10x10-6 = 0.5x10-6 N

There are five pairs of electrodes by each of the two moving electrodes. The force needs to
be generated by each pair of electrodes is thus equal to:

f = F/10 = 0.05x10-6 N
From Eq. (2.11),
1εεW
F = r o V2
L 2 d
with F = f = 0.05x10-6 N; ε = 1.0; ε = 8.85x10-12 C/N-m2; W = 5x10-6 m; d = 2x10-6 m:
L r o

11x8.85x10−12x5x10−6
0.05x10−6 = V2
2 2x10−6

We may solve for the required voltage to be V = 21.26 volts

Problem 13:

The geometry and dimensions of the microgripper is shown in Figure 2.45 below.

““AA””

FFlleexxiibbllee ““DDrriivvee AArrmm””

RReeqq’’ddttiipp
mmoovveemmeenntt::
““AA”” 55 µµmm

RRiiggiiddllyy hheelldd
““ CClloossuurree AArrmm”” WWiiddtthh ooff eelleeccttrrooddeess,, WW == 55µµ mm
RReeqq’’ddttiipp
mmoovveemmeenntt::
55 µµmm
GGaapp,, dd == 22 µµmm 1100 µµmm
115500 µµmm 88 µµmm 55 µµmm
330000 µµmm 1100 µµmm
VViieeww ““AA--AA””

We will first find the necessary voltage supply to the electrodes on both drive arms to provide a 5
µm movements at the free end of each of these two arms. We will treat the Drive arms as two
elastic cantilever beams and the generated electrostatic forces by the electrodes as concentrated
forces acting at the distance that equals to a distance b = 150 + 0.5x8 = 154 µm away from the
support-end as illustrated below:

6
bb == 115544 µµmm FF

δδ == 55 µµmm
mmaaxx

LL == 330000 µµmm

Since the expression for the maximum deflection at the free-end of the cantilever with a load, P
applied at a distant, b from the support (see the illustration above) is:
Fb2
δ = (3L−b)
max 6EI

with the Young’s modulus, E = 1.9x1011 Pa from Table 7.3 for silicon, and the area moment of
inertia, I = 4.17x10-22 m4 (for the cross-section of the beam shown in View “A-A”in the sketch of
the gripper), we will have the following relationship for the equivalent force, P:

F(154x10−6)2(3x300x10−6 −154x10−6)
5x10−6 =
( )( )
6x1.9x1011 4.17x10−22

Solve for the equivalent applied force, F = 0.1343x10-3 N

We are now ready to estimate the voltage supply to the electrodes to generate the above actuation
force.

There are 5 pairs of electrodes for each arm. From Equation (2.11), the electrostatic force is:
1εεW
F = r o V2
L
2 d

with ε = 1.0; ε = 8.85x10-12 C/N-m2; W = 5x10-6 m; and d = 2x10-6 m
r o

Since the electrostatic force in Equation (2.11) is for a single pair of electrodes, the total
electrostatic force generated by n-pair of electrodes can be expressed to be:

⎛1ε ε W ⎞
F = n⎜ r o ⎟V2
L ⎝2 d ⎠

We thus have:
1 1x8.85x10−12x5x10−6
0.1343x10−3 = 5 V2
2 2x10−6

7
We may solve for the supply voltage to be V = 1558.2 volts, which is an unusually high voltage
for a microgripper. The reduction of required voltage supply to the microgripper can be achieved
by a combination of increase the number of pairs of electrodes, as illustrated in Figure 2.29 for
Example 2.4, and the geometry and dimensions of the microgripper. Reduction in the length, or
the depth of the drive arm would result in the reduction of the required voltage for actuation too.
However, with the current geometry and dimensions of the microgrupper in Figure 2.45, it is not
realistic to drop the required actuation voltage to 40 volts.

Problem 14

Let us first show Equation (2.13) as F = 2mVxΩ, in which F is the induced Coriolis force, V
c c
is the velocity vector, and Ω is the angular displacement of the object.

Expressing Equation (2.13) in a full-length form, we have the following:

i j k
F i+F j+F k = 2m V V V
cx cy cz x y z
Ω Ω Ω
x y z

where i, j, and k = unit vector along x-, y- and z-coordinate respectively in a Cartesian
coordinate system. V , V and V = velocity component along x-, y- and z-coordinate
x y z
respectively, and Ω , Ω , and Ω = angular rotation component about x-, y- and z-coordinate
x y z
respectively.

Expansion of the above expression will lead to the following relations:

F i+F j+F k =
cx cy cz

[( ) ( ) ( ) ]
2m V Ω −V Ω i+ V Ω −V Ω j+ V Ω −V Ω k
y z z y x z z x x y y x

We observe from the setup illustrated in Figure 2.39 with the following zero quantities:

V = V = 0 and Ω = Ω = 0
y z x y

We thus from the above equality, the only non-zero Coriolis force component to be:

F = - 2m V Ω in the y-direction
cy x z

The numerical value of the Coriolis force can be obtained with the substitution of the mass m = 1
mg = 10-6 kg and V = 2 (maximum amplitude of vibration)/period of vibration.
x

We get V = 2 x (100 x 10-6) m/0.001 s = 0.2 m/s
x

The corresponding Coriolis force with an angular displacement Ω = + 0.01 rad in
z
counterclockwise direction is:
8
F = -2x10-6 x 0.2 x 0,01 = -4x10-9 N
cy

Problem 15

With a given equivalent spring constant k = 100 N/m, we have the displacement of the proof
mass in positive y-direction as:

δ = F /k = 4x10-9/100 = 4 x 10-11 m
y cy

where the value of F is obtained from Problem 2.14.
cy

Chapter 3
Engineering Science for Microsystems Design and Fabrication
(P. 105)

Part 1: Multiple Choice:

1.(b); 2. (b); 3. (a); 4. (a); 5. (a); 6. (a); 7. (b); 8. (c); 9. (b); 10(c); 11. (c); 12. (a); 13. (b); 14.(a);
15. (c); 16. (a); 17. (a); 18. (c); 19. (b); 20. (a); 21. (a); 22. (c); 23. (c); 24. (a); 25. (b); 26. (a);
27. (b); 28. (a); 29. (b); 30. (a); 31. (b); 32. (c); 33. (b); 34. (a); 35. (c); 36. (a); 37. (c); 38. (b);
39. (c); 40. (a); 41. (b); 42. (b); 43. (b); 44. (a); 45. (b).

Part 2: Descriptive Problems:

Problem 1:

We have learned from this chapter that the mass of a proton in an atom is 1.67x10-27 kg, which is
1800 times greater than the mass of an electron. We may thus assume that the total mass of
protons in an atom to be the mass of the same atom. We are also aware of the fact that a neutron
in the nucleus of an atom has the same mass as that of a proton.

Since each hydrogen atom has one proton and one electron, and each silicon atom has 14 each
protons and neutrons, we may thus obtain the mass of a single hydrogen atom to be 1.67x10-27
kg, whereas (14+14)x1.67x10-27 = 46.76x10-27 kg to be the mass of a silicon atom.

The radii of hydrogen and silicon atoms are available in Table 8.7, from which we may obtain
radii at 0.046 nm and 0.117 nm for hydrogen and silicon atoms respectively.

Problem 3:

A reasonable resistivity of a conductor is 10-5 Ω-cm, the same as that of platinum as indicated in
Table 3.3.

9
Problem 4:

The negative signs in these equations mean that the concentration of the diffused substance
decreases as the distance of diffusion into the base substance increases.

Problem 5:

Doping process allows engineers to humanly manipulate the electric resistivity of
semiconductors by creating localized positive or negative junction in the bulk material. With
such arrangements, engineers can control the way how electric current flow in the material,
which is the basic function of transistors in miniaturization.

Problem 6:

Advantages Disadvantage
Ion implantation A faster process at room Hard to control (see Figure 8.4)
temperature.
Diffusion Easier to control the diffusion A slow process at high temperature
zone (see Figure 8.6)

Problem 7:

By following what is shown in Figure 3.11, the optimum temperatures for As, P, and B are the
temperatures at which the maximum solubility of diffusion take place. Thus, the corresponding
optimum diffusion temperatures are ≈ 1220oC, ≈1200oC and ≈1330oC for As, P and B
respectively. The corresponding solid solubility of these materials are: 12x1020 for As, 5.5x1020
for P and 7.5x1020 for B with unit of atoms/cm3.

Problem 8:

Equation (3.5) is used for the solution of this problem:
⎛ x ⎞
( )
C x,t = C erfc⎜ ⎟
s ⎜ ⎟
⎝2 Dt ⎠
The coefficient C in the above equation is maximum possible input concentration. Inthis case,
s
we have the solubility of phosphorus at the given diffusion temperature of 1260oC at 5.45x1020
atoms/cm3 as obtained from Figure 3.11.

The concentration of phosphorus at the depth x = 0, 0.2, 0.4,……2.0 µm at selected time of t =
0.5, 2, and 3 hrs can be computed from the above equation with (D)1/2 = 1.05 µm/(h)1/2 from
Figure 3.12. The equation that we will use to compute the distribution of phosphorus
concentration at the above 3 selected time instants will thus take the form:

⎛ x ⎞ ⎛0.4762x⎞
C(x,t)=5.45x1020erfc⎜ ⎟=5.45x1020erfc⎜ ⎟
⎜ ⎟ ⎜ ⎟
⎝2x1.05 t ⎠ ⎝ t ⎠
10