1

MEMS and Microsystems Design, Manufacture, and Nanoscale

Engineering (2nd 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: (408)924-3905; Fax: (408)924-3995 E-mail: tairan@email.sjsu.edu

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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

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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.

Small sizes. Sensitive to temperature. Produced by doping foreign substances to silicon substrates.

Capacitors Simple in structure, hence less expensive to produce. Not sensitive to temperature-suitable for operations at elevated temperatures.

Exhibit nonlinear input/output relationship-require careful calibration prior to applications. Much bulkier than piezoresistors-takes 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.

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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:

TNV ∆=∆ β 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: mvorvoltxxxV 35.4304335.03731074.383 6 ==∆ −

5

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 to produce. Usually provides precise actuation.

Cannot maintain the actuated movement for sustained period of time due to overheating.

Electrostatic force Takes up the least amount of space. Actuation is instant.

Low in magnitudes.

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; εo = 8.85x10-9 F/m; εr = 4.7 (Table 2.2); V = 70 v; and d = 2x10-6 m From Equation (2.10), we may compute the electrostatic force in the width-direction: Fw = 0.0815 N. From Equation (2.11), for the force in the length-direction: FL = 0.0815 N Problem 12: We will model the comb drive actuator from a simplified model as illustrated below:

V

Moving electrodes Moving electrodes

Fixed electrodes

Spring constantk

Spring constantk

6

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),

VF dWor

L2

21 εε=

with FL = f = 0.05x10-6 N; εr = 1.0; εo = 8.85x10-12 C/N-m2; W = 5x10-6 m; d = 2x10-6 m:

Vx

xxxxx 26

6126

1021051085.81

211005.0

−

−−− =

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.

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:

Flexible “Drive Arm”

Rigidly held“Closure Arm”

Req’d tipmovement:5 µm

Req’d tipmovement:5 µm

Gap, d = 2 µm 10 µm

Width of electrodes, W = 5 µm

150 µm 8 µm

300 µm

“A”

“A”

5 µm

10 µm

View “A-A”

Flexible “Drive Arm”

Rigidly held“Closure Arm”

Req’d tipmovement:5 µm

Req’d tipmovement:5 µm

Gap, d = 2 µm 10 µm

Width of electrodes, W = 5 µm

150 µm 8 µm

300 µm

“A”

“A”

5 µm

10 µm

View “A-A”

7

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:

( )bLEI

F b −= 36

2

maxδ

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:

( ) ( )

( )( )2211

66266

1017.4109.161015410300310154105 −

−−−− −

=xxx

xxxxFx

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:

VF dWor

L2

21 εε=

with εr = 1.0; εo = 8.85x10-12 C/N-m2; W = 5x10-6 m; and d = 2x10-6 m 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:

2

21 V

dW

nF orL ⎟

⎠⎞

⎜⎝⎛=

εε

We thus have:

Vxxxxxx 2

6

6123

1021051085.81

215101343.0 −

−−− =

b = 154 µm F

L = 300 µmδmax = 5 µm

b = 154 µm F

L = 300 µmδmax = 5 µm

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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 ΩVFc xm2= , in which Fc is the induced Coriolis force, V 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:

zyx

zyxczcycx VVVmFFFΩΩΩ

=++kji

kji 2

where i, j, and k = unit vector along x-, y- and z-coordinate respectively in a Cartesian coordinate system. Vx, Vy and Vz = velocity component along x-, y- and z-coordinate respectively, and Ωx, Ωy, and Ωz = angular rotation component about x-, y- and z-coordinate respectively.

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

( ) ( ) ( )[ ]kji

kji

xyyxxzzxyzzy

czcycx

VVVVVVm

FFF

Ω−Ω+Ω−Ω+Ω−Ω

=++

2

We observe from the setup illustrated in Figure 2.39 with the following zero quantities: Vy = Vz = 0 and Ωx = Ωy = 0 We thus from the above equality, the only non-zero Coriolis force component to be: Fcy = - 2m Vx Ωz in the y-direction The numerical value of the Coriolis force can be obtained with the substitution of the mass m = 1 mg = 10-6 kg and Vx = 2 (maximum amplitude of vibration)/period of vibration. We get Vx = 2 x (100 x 10-6) m/0.001 s = 0.2 m/s The corresponding Coriolis force with an angular displacement Ωz = + 0.01 rad in counterclockwise direction is:

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Fcy = -2x10-6 x 0.2 x 0,01 = -4x10-9 N 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: δy = Fcy/k = 4x10-9/100 = 4 x 10-11 m where the value of Fcy is obtained from Problem 2.14.

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.

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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

temperature. Hard to control (see Figure 8.4)

Diffusion Easier to control the diffusion zone (see Figure 8.6)

A slow process at high temperature

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:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=

DtxerfctxC Cs 2

,

The coefficient Cs in the above equation is maximum possible input concentration. Inthis case, 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:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

txerfcx

txxerfcxtxC 4762.01045.505.12

1045.5, 2020

11

in which x is in µm and t is in hr. The value of complementary error function erfc(X) in the above expression may be obtained by using the curve shown in Figure 3.14. We may summarize the computed results in the flowing table:

x-depth (µm) t = 0.5 hr t = 2 hr t = 3 hr 0 5.45 5.45 5.45

0.2 4.69 5.07 5.18 0.4 3.82 4.69 4.77 0.6 2.86 4.2 4.52 0.8 2.45 3.82 4.09 1 1.91 3.49 3.87

1.2 1.42 3.11 3.49 1.4 0.95 2.73 3.16 1.6 0.68 2.45 2.89 1.8 0.53 2.18 2.73 2 0.33 1.96 2.4

Graphical representation of the distribution of phosphorus concentration in the silicon substrate at various times is presented below:

Diffusion of Phosphorus into Silicon

0

1

2

3

4

5

6

0 0.5 1 1.5 2 2.5

Depth in Silicon (micrometers)

Con

cent

ratio

n of

Ph

osph

orus

(at

ims/

cm3 )

0.5 hr 2 hrs 3 hrs

The curves in the above figure are not “smooth” due to approximated values of the erfc(X) value obtained by visual means from Figure 3.14. The trend of more even distributions of the phosphorus in the silicon substrate at larger times into the diffusion follows what has been depicted in Figure 3.10.

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Problem 9: We will first assume that diffusion process takes place at the same temperature of 1250oC as in Example 3.1. The corresponding solubility for boron is Cs = 7x1020 atoms/cm3 as given in Figure 3.11. Let the time required to dope boron into silicon substrate at a depth of 2 µm to be tf. We obtained the corresponding concentration of boron at resistivity of 10-3 Ω-cm from Figure 3.8 to be C = 1020 atoms/cm3. Thus by using Equation (3.5), we have the following relation for tf:

( ) ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

tCt

fsf DerfcmC

22,2µ

in which, tf has a unit of hr. The diffusivity, (D)1/2 in the above expression for boron in silicon at 1250oC is 1.05 µm/(h)1/2 from Figure 3.12.

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

fft

erfcxx

erfcxt

4762.010705.12110710 202020

From which we have 1429.0714762.0==⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛

fterfc

The corresponding value of the argument X in complementary function erfc(X) = 0.1429 from Figure 3.14 is X ≈ 1, which leads to:

14762.0≈

ft , and thus tf ≈ 0.2268 hr

Chapter 4 Engineering Mechanics for Microsystem Design

(P. 178) Part 1. Multiple Choice 1. (b); 2. (c); 3. (a); 4. (a); 5. (b); 6. (c); 7. (c); 8. (a); 9. (b); 10. (a) 11. (c); 12. (a); 13. (c); 14. (a); 15. (c); 16. (a); 17. (b); 18. (c); 19. (a); 20. (a) 21. (b); 22. (c); 23. (a); 24. (c); 25. (b); 26. (c); 27. (a); 28. (c); 29. (b); 30. (c) Part 2. Computation Problems Problem 1: We have d = 600x10-6 m, a = d/2 = 300x10-6 m, and P = 20x106 N/m2.

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The Young’s modulus, E = 0.7x1011 N/m2 for aluminum from Table 7.3 on P. 257. The Poisson’s ratio, ν = 0.3 for aluminum from a metal handbook.

The maximum deflection of the circular diaphragm with a thickness, h = 13.887x10-6 m is obtained from Equation (4.7) as:

( )

32

22

max 1613hmEamWw

π−

−=

in which W = (πa2)P = 3.14(300x10-6)2(20x106) = 5.652 N, m = 1/ν = 1/0.3 = 3.33.

One may thus calculate the maximum deflection, wmax = -1.4745x10-4 m, or 147.45 µm.

Problem 2: The geometry of the die as illustrated below:

We will then designate the dimension of the thickness of the die h and the size of the foot print ∆L as shown below:

The assigned die thickness, H = 500 µm is the standard thickness of 100 mm diameter wafer as indicated in Section 7.4.2 on P. 249, whereas the footprint ∆L = 250 µm is given.

In order to make use of Eq. (4.10) for the required thickness of the diaphragm, we need to determine the edge length of the diaphragm, a, first. Referring to the diagram of the footprint as illustrated above can do this.

h

H = 500 µm

∆L = 250 µm ∆a

54.74o

L =3000 µm

Detail Dimensions of Foot Print

Edge Length of Diaphragm, a

3 mm

3 mm

Plan View

Applied Pressure, P=75 MPa

h H

L = 3000 µm ∆L∆L

Cross-section of the Die

a

14

It is clear from the diagram that )74.54tan()( o

ahH

=∆− , in which H = 500 µm. We thus have:

hhHa o 707.06.353)74.54tan(

−=−

=∆

Consequently, the edge length of the square diaphragm, a is:

a = L – 2∆L - 2∆a = 1792.786 + 1.414h µm

From Eq. (4.10) with σmax = 350 MPa, we will have:

62

2610350]10)414.178.1792[(308.0 xxhp

h=

+ −

or PaorNh

xhx

xp mhh 22

221

212

26/

)414.178.1792(101364.1

)414.178.1792(10308.010350

+=

+=

−

We may tabulate the results of the diaphragm thickness vs. applied pressure as follows:

Diaphragm thickness (µm) 500 300 200 100 50 Maximum pressure (MPa) 88.47 31.85 14.16 3.54 0.88

Any combination of maximum applied pressure and the diaphragm thickness will produce a maximum stress of 350 MPa at the mid-span of the edges of the square diaphragm.

Problem 3:

The equivalent spring constant of elastic beams can be obtained from the following expression:

δF

keq =

where F = applied load to the beam δ = deflection of the beam under the load

Case 1 Simply-supported beams: From the strength of materials theory, we have the deflection of the beam under the concentrated force, F to be:

EIFL

48

3=δ

L

F

15

from which we may obtain the equivalent spring constant, keq to be:

348

LEIF

keq ==δ

where E = Young’s modulus of the beam material I = Area moment of inertia of the beam cross-section Case 2 Beams with fixed-ends: The deflection of the beam under the concentrated force, F is:

EIF L

192

3=δ

Hence the equivalent spring constant is:

Lk

EIFeq 3

192==

δ

Case 3 Cantilever beams (Extra, not requested in Problem 3): The deflection of the beam at the free-end is:

EIF L3

3=δ

which leads to the following expression for the equivalent spring constant:

Lk

EILF

eq 33

==

Problem 4: The mass, m attached to the beam is 5 g, or 5x10-3 kg; The equivalent beam spring constant keq in the arrangement shown below, and from Case 2 of Problem 3 is 18240 N/m

L

F

L

16

(a) From Eq. (4.16), we have the equivalent natural frequency,

sradxm

keqn /1910

10518240

3 ===−ω

(b) The equivalent motion of the mass in the y-direction, according to Eq. (4.14) is:

0)()(2

2=+ ty

dtym k

td

eq

with y(0) = 5x10-6 m, and y’(0) = 0. Substitute these values into the above equation:

0)t(y10x648.3d

)t(y 62

2

td =+ (a)

The solution of the differential equation is:

ttty CC 1910sin1910cos)( 21 += (b)

From the condition y(0) = 5x10-6, we get C1 = 5x10-6 From y’(0) = 0, we have C2 = 0

Thus, the amplitude of vibration, y(t) is:

txty 1910cos105)( 6−= (c)

The maximum amplitude of vibration is the coefficient of the cosine function in the solution in Eq. (c), or ymax = 5x10-6 m, or 5 µm. Problem 5: By referring to the forced vibration analysis in Section 4.3.2, we have the following differential equation to solve for the amplitude of the vibrating mass:

tcosm

)t(yd

)t(yd Ft

o2

o2

2

ω=+ ω

20x10-6 m

y

17

with the specified conditions: y(0) = 5x10-6 m and y’(0) = 0.

The proper differential equation as derived from Problem 4 is:

36

2

2

1051910cos5)(10648.3)(

−=+

xttyx

dty

td

in which the natural frequency of the beam spring system, s/rad191010x648.3 6o

==ω The solution of the above differential equation is:

tsinm2

ttsintcos)t(y

o

oo2o1

Fcc ω+ω+=ωω

with ω = ωo = 1910 rad/s at the resonant vibration situation and Fo = 5 N Use the first condition, i.e. y(0) = 5x10-6 m will result in c1 = 5x10-6. The other condition y’(0) will result in c2 = 0. Thus the solution for the amplitude of the vibration mass being:

tttxty 1910sin2618.01910cos105)( 6 += − Now, if we let tf = the time at which the beam spring breaks at y(tf) = 1 mm = 10-3 m, we will have:

ttt fffx 1910sin2618.01910cos10510 63 += −−

We may solve for tf from the above equation, or by an approximate relationship of 10-3 ≈ 0.2618tf from the above expression. This approximation is justified by letting sin1910tf = 1.0 and cos1910tf = 0. This approximation leads to tf = 3.82 ms, which is the time the strip spring will reach a breaking amplitude of 1 mm. Problem 6: The beam is loaded as illustrated in Example 4.8 The area moment of inertia of the beam cross-section is:

( )( ) 4324

366 mh12

10h1010121I

−−− ==

L=600µm

Dynamic force 1 µm

h µm

Beam Cross-section

18

in which h is in micrometers. The equivalent spring constant, keq is as computed in Case 2 of Example 4.8 and 4.9 for fixed-ends as:

( )( )( )

mNx

xEIhh

Lkeq /014074.0

1060012

10109.1192192 336

32411

3 ===−

−

The proof mass of the vibrating beam m = 16.1x10-11 kg as computed in Example 4.9. The corresponding circular frequency of the balanced force accelerometer is:

s/rad54.1322010x1.16

014074.0x2m

2hhk 3

11

3eq ===ω

−

From Example 4.9, the amplitude of vibration of the beam is:

tsinctcosc)t(X 21 ω+ω= in which the arbitrary constants c1 and c2 can be determined by the initial conditions:

smhkmdt

tdXandtXt

t /8888.13/50)(0)(0

0 ====

=

We thus have: c1 = 0 and c2 = 13.8888/ω = 1.051x10-3 h-3/2

Thus, we have the amplitude of vibration to be:

t54.13220sin10x051.1)t(X hh 35.13 −−= With the given condition X(tf) = 5 mm = 5x10-3 m for the beam to break, mathematically as:

thh f

35.133 54.13220sin10x051.110x5 −−− =

or )54.13220(sin10x5

10x051.1 thh f

5.13

35.1

−

−

=

The approximate value of tf is when 0.154.13220sin th f

5.1 = for a maximum value of h, which leads to h = 0.21 µm.

19

Problem 8: We may illustrate the balanced force accelerometer system below: From Example 4.12 on P. 139, we get the damping coefficients for the balanced force accelerometer to be:

cair = 2.625x10-12 N-s/m for air as the damping fluid, and csi = 1.036x10-10 N-s/m for silicone oil as the damping fluid.

We further have the mass of the silicon beam to be: m = ρv, in which the mass density, ρ = 2.3 g/cm3 or 2.3x103 kg/m3 from Table 7.3 on P. 257, and the volume of the beam = v. By referring to the geometry and dimensions of the beam in Example 4.11, we have

v = bBLb = (1x10-6) (10-4) (7x10-4) = 7x10-14 m3

Consequently, the beam has a mass, m = (2.3x103)(7x10-14)= 16.1x10-11 kg We will use the model illustrated in Fig. 4.7(b) on P. 119 to assess the motion of the beam mass, and Eq. (4.19) with the spring constant k = 2keq in Example 4.9 is used to compute the displacement of the beam mass, X(t) in the equation. The solution of Eq. (4.19) depends on the cases with the values of (λ2 - ω2) as described in Eq. (4.20a), (4.20b) or (4.20c). We will thus need to compute both λ2 and ω2 first in order to select which of the above solutions for the case under consideration. Let us assume that both beam springs have fixed ends, and the equivalent spring constants can be computed from the following expressions as presented in Case 2 on P. 131:

Lk

EIeq 3

192=

with E = 1.9x1011 N/m2 (Table 7.3) and ( )( ) mxxxI 422366 10167.41010105121 −−− ==

Beam Mass

Beam Springs

5 µm

10 µm

400 µm

20

Thus, ( )( )( )

mNx

xxkeq /52.237

10400

10167.4109.119236

2211==

−

−

and s/rad10x7177.110x1.16

52.237x2m

26

11eqk ===ω −

which leads to ω2 = 2.95x1012 rad2/s2

The damping parameters:

airfor10x152.810x1.16x2

10x625.2m2

311

12air

airc −

−

−

===λ

oilsiliconefor32174.010x1.16x2

10x036.1m2 11

10si

sic === −

−

λ

from which we have:

( ) 010x95.210x152.8 122322

air<−=− −ωλ

( ) 010x95.232174.0 12222

si<−=−ωλ

The values of (λ2 - ω2) shown above for the two distinct damping media of air and silicone oil will lead to the use of Equation (4.20c) for the movement of the beam mass. The movement of the proof mass will be of an undesirable oscillatory nature as illustrated in Figure 4.11. Problem 9: The balanced-force accelerometer is illustrated in Fig. 4.25, and also as below: The dimensions of the two beam springs are not given in the problem. We may either assume the unspecified dimensions are identical to those given in Example 4.8 and 4.9, or by using an open

Beam Mass, m

L = 700 µ m

h = 5 µm

b = 1 µm

Dimensions of the Beam Mass

21

size of the beam springs that will withstand the specified conditions as described in Example 4.12. We will assume the dimensions of the beam springs as shown below: We may compute the area moment of inertia of the beam springs to be:

( )( ) 424366

m10x4167.1012

10x510I −−−

==

The equivalent spring constant for beam springs with fixed-ends is

( )( )( )

m/N76.110x600

10x4167.1010x9.1192L

EI192k 36

2411

3eq ===−

−

as in Case 2 of Example 4.9.

Since the maximum deceleration of the car in the present case is smX 2/22.22−=&& from Example 4.14. By neglecting the mass of the beam springs, we may express the dynamic force associated with the moving beam mass as:

( )tXmtF &&=)( The mass of the beam mass, m = 16.1 x 10-11 kg as computed in Example 4.9. The force acting on both beam springs at the time of deceleration of –22.22 m/s2 is: ( )( ) N10x77.35s/mkg10x77.35s/m22.22kg10x1.16XmF 10210211 −−− =−=== && The induced deflection of the beam springs by the above dynamic force of the magnitude is: P = F/2 = 35,77x10-10/2 = 17.885x10-10 N with L = 600x10-6 m, I = 10.42x10-24 m4 (from Example 4.9), E = 1.9x1011 N/m2 (Table 7.3), and keq = 1.76 N/m, we will calculate the maximum movement of the proof beam mass from a simple beam with both ends rigidly fixed and subject to an equivalent concentrate force P in the middle span. Mathematical expression for the maximum deflection under the load is available in handbooks such as (Roark 1965) as:

600 µ m

5 µ m 1 µm

Dimensions of the Beam Springs

22

( )( )( )( ) m10x0166.1

10x4167.1010x9.119210x60010x885.17

EI192PL 15

2411

36103

max−

−

−−

===δ

which is a too small a movement to be detectable. Problem 10: This bi-layer strip is subjected to a uniform temperature rise, T as illustrated below: The width of the bi-layer strip b = 5 µm and the overall depth h = 10 µm. The radius of curvature, ρ from Equation (4.51) is:

( )Thαα

ρ123

2−

=

where α1 and α2 are coefficients of thermal expansion of SiO2 and silicon strips respectively (available in Table 7.3), and h is the thickness of the individual strips. Let us express the radius of curvature of the bi-layer strip in a different form from the above expression:

TC

=ρ

in which the constant ( ) ( )643.3

105.033.2310102

32

6

6

12=

−=

−=

−

−

xxxhC

αα

where α2 = coefficient of thermal expansion of silicon = 2.33 x 10-6/oC, and α1 = coefficient of thermal expansion of SiO2 = 0.5 x 10-6/oC, as obtained from Table 7.3. From Example 4.17, we have the movement of the free-end, δ to be:

( )θρδ cos1−≈

where πρ

θ2

360L= with L = 1000x10-6 m

Hence ρρ

θ26 107325.5

28.6101000360 −−

==xxx and the movement of the free-end of the bi-layer

beam can be obtained from the following expression:

5 µm

5 µm

1000 µm

SiO2 strip

Silicon strip

23

δ = 3.643(1 – cosθ)/T in which T is temperature in oC. We may tabulate the values of the temperatures vs. the movement of the free-end of the beam actuator as follows:

T (oC) ρ = C/T (m) θ(o) δ = ρ(1 - cosθ) (µm) 10 0.3643 0.1574 1.373 20 0.1822 0.3147 2.747 30 0.1214 0.4720 4.120 40 0.0911 0.6294 5.496 50 0.0729 0.7880 6.890

The plot of the above tabulated data is shown below. Problem 11: The beam has the following geometry and dimensions: The temperature variation in the beam is: T(z) = 2x106z + 30 oC At the top face, i.e. z = 5x10-6 m, we have T(5x10-6) = 40 oC, and

b = 5 µm

H = 10 µmx

y

2h = 10 µm

L = 1000 µm

0SiliconBeam

Movement of Free-end of a Contilever Beam

0

1

2

3

4

5

6

7

8

0 10 20 30 40 50 60

Temperature (oC)

Mov

emen

t (m

icro

met

er)

24

at the bottom face at z = -5x10-6 m, the temperature is T(-5x10-6) = 20 oC. Material properties of the silicon beam are given in Example 4.19 on P. 161: Mass density, ρ = 2.3 g/cm3; Specific heats, c = 0.7 J/g-oC; Thermal conductivity, k = 1.57 J/cm-oC-s; Coefficient of thermal expansion, α = 2.33x10-6/oC; Young’s modulus, E = 1.9x1011 N/m2; Poisson’s ratio, ν = 0.25. We will first compute the thermal force, NT and the thermal moment, MT from the respective Equations (4.55a) and (4.55b) as:

( )( ) ( ) Ndzzxxx x

xTN 81.13230102109.11033.2

66

105

105

6116 =+= ∫ −−

−

( )( ) ( ) mNxzdzzxxx x

xTM −=+= −

−

− ∫ −6105

105

6116 1078.7330102109.11033.26

6

From Example 4.19, we have A = 5x10-11 m2 and I = 4.167x10-22 m4. From Equation (4.56), we have the thermal stress along the x-direction to be:

( ) ( )I

bzA

bzETzx MN TT

xx)(

, ++−= ασ

with σxx,max occurs at z = 5x10-6 m. Thus, σxx,max = σ(x,5x10-6) = -2600 Pa We will compute the associate thermal strains from Equations (4.57a) and (4.57b) with maximum values occurring at z = 5x10-6 m:

( ) ( )⎥⎦⎤

⎢⎣

⎡ += MNT

Txx b

Iz

Ab

Ezx 1,ε

which leads to εxx,max = εxx(x,5x10-6) = 0.00932%

( ) ( ) )(1, zTE

bIz

Ab

Ezx MN

TT

zz αννε ⎟

⎠⎞

⎜⎝⎛ +

+⎥⎦

⎤⎢⎣

⎡ +−=

results in εzz,max = εzz(x,5x10-6) = -0.0023% The deflection of the beam in the x-direction = u (x,z) can be computed from Equation (4.58a) as:

( ) ( )⎥⎦⎤

⎢⎣

⎡ += MNT

T bIz

Ab

Exzxu ,

with umax at x = 500x10-6 m and z = 5x10-6 m: umax = u(500x10-6, 5,10-6) = 0.0466 µm

25

The deflection of the beam in the z-direction, w(x,z) is obtained from Equation (4.58b):

( ) ( ) ∫⎟⎠⎞

⎜⎝⎛ +

+⎥⎦

⎤⎢⎣

⎡+−−= z

TTT dzzT

Eb

Iz

Ab

EEIb

zxw MzNxM0

22 )(1

22, ναν

with wmax occurs at x = 500x10-6 m and z = 5x10-6 m, we have: wmax = w(500x10-6, 5x10-6) = -0.582 µm Problem 12: The width of the beam has been increased to 100x10-6 m. The “wide” beam now is effectively a “plate”. As such, the thermal stress formulation for thin plates will be used to solve this problem, with the temperature variation across the plate thickness, i.e.

T(z) = 2.1x106z + 28.8 in degree C We realize that the thermal force, NT = 127.5 N and the thermal moment, MT = 77.4725x10-6 N-m remain unchanged as in Example 4.19 on P. 163 and P. 164.

The thermal stresses in both x- and y-directions can be computed from Equation (4.52) as:

( ) zxxzxxyyxx

12665 102396.110178.28101.2109027.5 +++−==σσ

The associated thermal strains are obtained from Equation (4.53a):

zxyyxx 893.410711.6 5 +== −εε

( ) ( )8.28101.2108833.3109297.01075.12103509.0 6612611 +++−= −− zxxzxxxzzε with εxy = εyz = εzx = 0

The induced displacements of the plate in the x-direction, u(z) and that in the y-direction, v(z), and w(x,y,z) in the z-direction can be computed from Equations (4.54a,b and c):

( ) ( )zxxxxzu 126

11 109297.01075.12109.1

+=

( ) ( )zxxxyzv 126

11 109297.01075.12109.1

+=

and ( ) ( ) ( )zyx xzxxzyxw 21261122 102324.010375.684.96107018.04465.2,, −−++−= − The maximum values of stress, strains and displacements occur at: x = 500x10-6 m, y = 50x10-6 m and z = 5x10-6 m. Thus, we will have the following maximum stress, strains and displacements: σxx,max = σyy,max = 4000 Pa εxx,max = εyy,max = εzz,max = 0.00915%

26

umax = 0.046 µm; vmax = 0.0046 µm; wmax = -0.6173 µm We have realized that by extend the beam into a plate with a width of 100 µm has not produced significant difference in the results from those obtained from a beam with a width of 5 µm. Problem 14: We have the dimensions of the specimen as shown in the diagram below, in which s = 1 cm = 10-2 m; b = 5 mm = 5x10-3 m; and the width, B = 24x10-4 m, and c = 100 µm = 10-4 m. The critical load, Pcr that breaks the specimen is 40x106 N/m2.

We will use Equation (4.65a) for the function F(c/b) as s/b = 2 < 4, as indicated in Section 4.5.2 on P. 168:

432

57.1418.142.8735.109.1 ⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛

bc

bc

bc

bc

bcF

with c/b = 0.02 Hence F(c/b) = 1.0586 Equation (4.64) is used to compute fracture toughness:

⎟⎠⎞

⎜⎝⎛=bcFcccK πσ

The σc in the above expression is obtained from the bending stress in a “solid” beam subjected to three-point bending as follows:

IcM

c =σ

where 4133

3 1025012

105.22/ mxBbIandmxbc −− ====

The bending moment, mNxsM Pcr −=⎟⎠⎞

⎜⎝⎛= 5102

2

Thus, we have the critical stress corresponding to Pcr to be:

s = 10000 µm

Pcr = 40 MN

b = 5000 µm

B = 2400 µm

27

( )( ) Pax

xxc 2000

10250105.2102

13

35==

−

−−

σ

which leads to the fracture toughness, Kc to be:

( ) mPaxK c 52.370586.11014.32000 4 == − Problem 15: For the width of the specimen, B to be increased to 100x240 µm, we will have I = 2.5x10-9 m4. This new value will change the critical stress according to the following expression:

( )( ) Pax

xxc 20

105.2105.2102

9

35==

−

−−

σ

Kc = 0.375 Pa m , which is 100 times smaller than the case in Problem 14. This result, of course, is computed on the basis that the enlarged specimen breaks at the same critical load, Pcr, which is not quite a realistic hypothesis. We would expect a much greater value of Pcr for larger specimens. It nevertheless underlines the importance of the size effect on the measurement of the fracture toughness of specific materials. A credible Kc for design purpose must be independent of the specimen geometry and size. A great deal of research effort is needed in the measurements of Kc for microsystems materials in micro scale.

Chapter 5 Thermofluid Engineering and Microsystem Design

(P. 221) Part 1. Multiple Choice

1. (c); 2. (a); 3. (b); 4. (b); 5. (c); 6. (a); 7. (b); 8. (a); 9. (a); 10. (c); 11. (b); 12. (c); 13. (b); 14. (a); 15. (c); 16. (a); 17. (b); 18. (c); 19. (c); 20. (b); 21. (a); 22. (a); 23. (c); 24. (b); 25. (b); 26. (a); 27. (b); 28. (a); 29. (b); 30. (c); 31. (c); 32. (a); 33. (b); 34. (c); 35. (a); 36. (c); 37. (a); 38. (c); 39. (c); 40. (b). Part 2. Computational Problems Problem 2: We have d1 = 500x10-6 m and d2 = 50x10-6 m

The flow rate is Q = 1x10-6 cm3/min = 1.67x10-14 m3/s

28

A1 = π(500x10-6)2/4 = 19.64x10-8 m2 A2 = π(50x10-6)2/4 = 19.64x10-10 m2

smx

xAQ

V /085.01064.19

1067.18

14

11 µ=== −

−

smxx

AQ

V /503.81064.19

1067.110

14

22 µ=== −

−

Problem 3: The opening of the valve may be illustrated as follows: (a) The opening of the valve is H = (L sin15o)cos15o = 400x10-6x sin15ocos15o = 100x10-6 m (b) We will next estimate the velocity of the gas flow at the exit of the valve, i.e.Ve. Base on the

law of continuity, we have:

AMV

e

xe ρ

1=

in which ρ = mass density of the H2 gas = 0.0826 kg/m3 (from Example 5.2) Mx1 = the mass flow rate in the direction of Ve = 15.3x10-6 kg/s (from Example 5.2) The exit cross-sectional area, Ae = HW, in which W is the width of the plate valve = 300 µm, or 300x10-6 m. Hence Ae = (100x10-6)(300x10-6) = 3x10-8 m2. Thus, the exit velocity is:

smxx

xV e /33.6174

1030826.0103.15

8

6==

−

−

NOTE: This exit velocity is unrealistically high for a microvalve. This high value on the velocity is a result of extremely small opening at the exit (Ae = 3x10-8 m2), and large mass flow rate (Mx1 = 15.3x10-6 kg/s)

15o

L = 400 µm

Widt

h, W

= 300 µ

m

Exit velocity, Ve

Fluid FlowValve Opening, H

Valve Plate Thickness:4 µm

29

(c) The volumetric flow at the exit can be computed as follows:

sxxQ mM x /1023.1850826.0

103.15 366

1 −−

===ρ

or Q = 11,113.8 cm3/min, which is significantly smaller than the intended design capacity of 30,000 cm3/min. Problem 4: The uniformly distributed load that is required to bend a cantilever beam (plate) such as the closure plate with a free-end displacement of H = 100x10-6 m in Problem 3 (see illustration below) can be obtained by the following expression:

EI

wL8

4

max=δ

in which δmax is the maximum deflection of a cantilever beam at the free-end due to uniformly distributed load, w per unit length; E is the Young’s modulus of the beam material; I is the area moment of inertia of the beam cross-section. The cross-section of the plate is 300 µm wide x 4 µm thick, which leads to an area moment of inertia, I to be:

( )( ) mxxxI 422366 101610410300121 −−− ==

with a Young’s modulus, E = 1.9x1011 N/m2 from Table 7.3 for silicon, and δmax = H = 100x10-6 m as shown in the figure in Problem 3, we may determine the required load, W from the following relation:

( ) ( )( )( )2211

466

1016109.181040010100 −

−− =

xxxxwx

which leads to w = 9.5 N/m The force required to lift the plate of 400 µm long at the free end is wL, or 9.5x(400x10-6) = 3800x10-6 N. However, there is a fluid-induced force acting on the plate too. This force is Fy = 40x10-8 N as computed from Example 5.2. The net required electrostatic force is thus equal to the difference of the above two forces, or Fd = 3800x10-6 – 40x10-8 ≈ 3.8 mN. The corresponding required voltage to generate such force can be obtained by using Eq. (2.8) to give:

W N/m

L = 400 µm

30

( ) ( )( )( )( )

46612

32622 10443.29

10350103001085.81108.310622

xxxxx

xxxWLor

dFdV ===−−−

−−

εε

which leads to V = 542 v – a rather high voltage. Problem 5: The problem is illustrated below: Assume that the average velocity of the fluid is computed at the cross-sectional area of the conduit at its mid-section is used. The 30o inclination is neglected. From Example 5.3 on P. 194, V2 = 2.4x10-3 m/s. We may calculate the Vave in the mid-cross section to be: Vave = 0.5 (V1 + V2) = 1.5x10-3 m/s Let dave = 0.5 (d1 + d2) = 75x10-6 m, which leads to the radius at the mid-section, aave = 37.5x10-6 m. The pressure drop, ∆P in the conduit using the Hagen-Poiseuille equation in Equation (5.17) is:

aLQP 4

8πµ

=∆

where µ = dynamic viscosity of the fluid = 1199.87x10-6 N-s/m2 (Table 4.3 on P. 138 for alcohol) L = length of the conduit = 0.1 m, and

( ) ( )( ) ( ) ( ) sxxxEqfromQ mVaVA aveaveaveave /102343.66105.1105.3714.36.5. 3133262 −−− ==== π We will have the approximate pressure drop, ∆P to be:

V1 = 600 µm/s

V2

L = 0.1 m

d 1 = 100 µm

d 2 = 50 µm

d ave

31

( )Pa

x

xxxxxP 1024105.3714.3

102343.661.01087.1199846

136==∆

−

−−

The pressure drop, ∆P obtained from the Hagen-Poiseulle’s equation is about 2.5 larger than that by the Bernoulli’s equation in Example 5.3. This indicates that the scaled down effect of the conduits on the pressure drop of a fluid flowing in a small conduit is significant. Problem 6: The purpose of this problem is to compare the estimated pressure drop in a fluid (water in this case) flowing through a capillary tube computed by using the Hagen-Poiseuille’s equation and that induced by the surface tension in the minute water cylinder. The capillary tube section is illustrated below:

The radius of the capillary tube is a = d/2 = (20x10-6)/2 = 10-5 m

Since the volumetric flow rate of the water is not given in the problem, we will first work on an hypothesis that water flows in the capillary tube in a laminar flow pattern. This pattern of flow of water requires the Reynolds number, Re, be kept below 1000, which leads to the following relationship:

1000Re ==µ

ρDV

with the density of water, ρ = 1000 kg/m3, and the inside diameter of the tube, D = 20x10-6 m, and the dynamic viscosity, µ = 1001.65x10-6 N-s/m2 from Table 4.3, we will have, from the above expression, the velocities of flow to be: V = 50 m/s for Re = 1000, and V = 5 m/s for Re = 100. If we use the lower bond velocity, V = 5 m/s with Re = 100, the volumetric flow, Q would be:

( ) sxxxAVQ m /10708.15510204

31026 −− ===π

The corresponding pressure drop, ∆P by the Hagen-Poiseuilli’s equation is:

( )( )( ) Pax

xxxP 000,400

2102014.3

10708.15101065.1001846

1036

=

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

−

−−−

1000 µm

20 µm dia.

32

To maintain a water flow with this enormous pressure drop is beyond the capability of most volumetric pumping devices. Consequently, let us assume a typical velocity flow at V = 10 µm/s in the capillary tube. This water flow velocity of V = 10 µm/s, which leads to a volumetric flow rate, Q = AV = 31.4x10-16 m3/s, in which A is the cross-sectional area of the tube. Substituting the above values into the Hagen-Poiseuille’s equation, will lead to a pressure drop of:

( )( )( ) Paxx

xxxP 8013.0

2102014.3

104.31101065.1001846

1636=

⎟⎟⎠

⎞⎜⎜⎝

⎛=∆

−

−−−

Now, we will consider the pressure difference between the inside and outside of the water cylinder in that section of the capillary tube, namely the surface tension of the water. We will recognize that the pressure required to overcome the surface tension is the sum of that for the length of the two ends of the “water cylinder” in the tube, and the same surface tension between the circumferential surface and the tube wall. The surface tension of water in these areas can be found in Eqs. (5.24a) and (5.24b), or the required pressure is:

aPst

γ3=∆

in which γ = surface tension coefficient of water = 0.073 N/m as in Eq. (5.23) and a = the radius of the tube. We will thus have the required pressure:

900,2110

073.035 ==∆

−

xPst Pa

We realize that the pressure required to overcome the surface tension of the water in the capillary tube section is much greater than that for driving that tiny volumetric flow of water through the tube as predicted by the Hagen-Poiseulle’s equation. The above computations have underlined the dominance of the effect of surface tension in water (liquid) flowing in capillary tubes. Problem 7: The situation of a capillary tube inserted in the water is illustrated in Fig. 5.16 on P. 183. The tube has a diameter of 20 µm, which gives a radius, a = 10-5m. By following the same procedure for the solution in Example 5.6, we have the rise of water level in the capillary tube, h to be:

488.1109810

)0cos(073.02cos25 ===

−xxx

wah θγ m, which is unrealistically high, as the total length of

the capillary tube in only 1 mm.

33

Problem 8: Thermal diffusivity defined in Eq. (5.39) should be used as effective measure of materials’ response in thermal actuation. The listing should thus be constructed on this basis. Problem 9: Equation (5.45) provides:

( ) ( ) frrrr

TkhtrT

kh

ntrT

s

s

=+∂

∂=

=

rr

rr

rr

v,,

which leads to the following special cases: (A) When h = 0: Equation (5.45) becomes:

( ) 0,=

∂∂

= srrntrT

rrr

r

which is equivalent to have qin and qout = 0 in Eq. (5.44a) and (5.44b) respectively. This means that no heat is allowed to flow across the boundary at srr rr

= . (B) When h → ∞: By dividing each term in Eq. (5.45) by h and then letting h → ∞, we will have: ( ) T frr s

trT ==rr

r,

which is the prescribed surface temperature boundary condition as shown in Eq. (5.43). Problem 10: We assume that the copper film is so thin and ductile that it only generates heat to the SiO2/Si bilayer strip, but does not impose any mechanical constraint on the overall structure. One end of the strip is maintained at 20oC whereas the other end and the top and bottom surfaces are surrounded by stagnant air as illustrated in the figure below. We further postulate that heat flows in the strip predominantly in the y-direction with some dissipation through the left end at x = 0. This postulation on the heat flow is justifiable as the

34

dimension of the strip in the x-direction far exceeds that in the y-direction. Further, the assumption of thermal insulation of the right end at x = 1000 µm and the bottom surface (y =60 µm) is also reasonable. These surfaces are in contact with the surrounding stagnant air at 20oC. The surrounding air temperature is not expected to change significantly enough during the brief period of actuation to induce a natural convection that will dissipate heat from the strip through these surfaces. We will assign T1(x,y,t) and T2(x,y,t) to be the temperature distributions in the SiO2 and silicon strips respectively. Equation (5.48a) and (5.48b) are used to determine the respective temperature distributions, T1(x,y,t) and T2(x,y,t) with the following initial and boundary conditions: The initial conditions:

( ) [ ] Ctyxtyx ott TT 20,,,, 0201 ====

The boundary conditions for the temperature distribution in the SiO2 layer are:

( ) Ctyx oxT 20,, 01 ==

at the left end, and

( )0

,,

1000

1 =∂

∂

= mxxtyxT

µ

thermally insulated at the right end, and

( ) ( ) mymy tyxtyx TT µµ 4241 ,,,,

=== for compatibility at SiO2 and silicon interface, and

( )

kiT R

ytyx

y 1

2

0

1 ,,−=

∂∂

=

for heat input at the top surface of SiO2 layer, in which R = the

electric resistance of copper film, (Ω), and i = the passing electric current (amp), and k1 = thermal conductivity of SiO2 (W/m-oC) The boundary conditions for temperature distribution in the silicon strip are:

silicon

SiO2

1000 µm

x

y

20o C

20 µm

4 µm

2 µm thick copper filmHeat supply

Stagnant air at 20oC

35

( ) Ctyx oxT 20,, 02 ==

at the left end, and

( )0

,,

1000

2 =∂

∂

= mxxtyxT

µ

thermally insulated at the right end, and

( ) ( )

mymy ytyx

ytyx TT

µµ 4

1

4

2 ,,,,

== ∂∂

=∂

∂ for compatibility at the silicon and SiO2 interface, and

( )

0,,

24

2 =∂

∂

= myytyxT

µ

for thermally insulated at the bottom surface.

Problem 11: The respective heat conduction equations for SiO2 and silicon stripes are:

( ) ( ) ( )t

tyxtyxtyx Ty

Tx

T∂

∂=

∂+

∂∂∂ ,,1,,,, 1

12

12

21

2

α

for 0 ≤ x ≤ 1000 µm and 0 ≤y ≤ 4 µm; t > 0 and

( ) ( ) ( )t

tyxtyxtyx Ty

Tx

T∂

∂=

∂

∂+

∂∂ ,,1,,,, 2

22

22

22

2

α

for 0 ≤ x ≤ 1000 µm, 4 µm ≤ y ≤ 24 µm, and t > 0. The constant α1 and α2 in the above differential equations are the thermal diffusivities of SiO2 and silicon respectively. The appropriate initial and boundary conditions are presented in Problem 11. Problem 12: The differential equations are as shown in Problem 13, and the initial conditions are given in Problem 11. The following boundary conditions apply: (A) In SiO2 strip:

( ) Ctyx oxT 20,, 01 ==

36

( ) ( ) ( )20,,

,,

110001

11000

1

kT

kT htyxh

xtyx

mxmx

=+∂

∂=

=µ

µ

( )

kiT R

ytyx

y 1

2

0

1 ,,−=

∂∂

=

( ) ( ) mymy tyxtyx TT µµ 4241 ,,,,

== =

( ) ( )

mymy ytyx

ytyx TT

µµ 4

2

4

1 ,,,,

== ∂∂

=∂

∂

(B) In silicon strip:

( ) Ctyx oxT 20,, 02 ==

( ) ( ) ( )20,,,,

210002

21000

2

kT

kT htyxh

xtyx

mxmx=+

∂∂

== µµ

( ) ( ) ( )20,,

,,

2242

224

2

kT

kT htyxh

ytyx

mymy

=+∂

∂

== µµ

where k1 and k2 are thermal conductivity of SiO2 and silicon respectively, and h is the specified heat transfer coefficient.

Chapter 6 Scaling Laws in Miniaturization

(P.244)

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

Chapter 7

Materials for MEMS and Microsystems (P.281)

Part 1. Multiple Choice 1.(b); 2. (c); 3. (b); 4. (a); 5. (a); 6. (a); 7. (b); 8. (a); 9. (b); 10. (a); 11. (b); 12. (c); 13. (c); 14. (b); 15. (a); 16. (a); 17. (b); 18. (a); 19. (c); 20. (a); 21. (c); 22. (b); 23. (c); 24. (c); 25. (b); 26.

37

(c); 27. (a); 28. (c); 29. (b); 30. (c); 31. (a); 32. (b); 33. (a); 34. (c); 35. (a); 36. (c); 37. (3); 38. (a); 39. (a); 40. (a); 41. (c); 42. (a); 43. (c); 44. (a); 45. (b). Part 2. Computational Problems

Problem 2: The planar area of a circular wafer, A, can be computed by:

dA 2

4π

= in which d = the diameter of the wafer.

The ratio of plane areas of wafers with 300 mm and 200 mm diameters is:

25.2200300

200300 2

1

2 =⎟⎠⎞

⎜⎝⎛==

wafermmofAreawafermmofArea

AA

Hence a wafer with 300 mm diameter has 2.25 times greater area than that of a 200 mm wafer.

Problem 3: By following the same expression used in Example 7.1, the number of atoms per cubic mm of silicon is:

3203

9 /1012.11810543.0

001.0 mmatomsxxx

nvVN =⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=

−

in which v = the volume of a single silicon crystal.

Likewise, the number of atoms per cubic micrometer of silicon is:

3113

9

6/1012.118

10543.010 matomsxx

xN µ=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−

−

Problem 4: A piezoresistor has the following geometry and dimension:

10 µm

4 µm2 µm

σmax = 235.36x106 Paσmax

38

The area on which the maximum normal stress exists is: A = 2 x 10 = 20 µm2 = 20x10-12 m2. From Equation (7.8), we have:

σπσπ TTLLRR

+=∆

Since the piezoresistor is attached to the cantilever beam as illustrated in Figure 7.18, we will

have: σL = σmax = 235.36x106 Pa, and σT = 0 as in Example 7.4. Piezoresistive coefficients for several orientations of p-type silicon crystals is available in

Table 7.9. Let us assume that the piezoresistor of (100) plane in the <100> orientation is used in this

case. We will have the coefficient πL = 0.02π44, with π44 = 138.1x10-11 Pa-1 from Table 7.8. We will thus have the piezoresistive coefficient πL = 2.762x10-11 Pa-1.

The corresponding rate of the change of electric resistance by the piezoresistor is:

311 105.6)1036.235()10762.2(6 −− ===

∆ xxxxRR

LLσπ

But since the resistance of a material is defined as:

ALR ρ

= in which ρ is the resistivity of the material, which is a p-type piezoresistor. We find

the values of ρ vary from 10-3 to 104.5 Ω-cm from Table 7.1. We will adopt a value of ρ = 7.8 Ω-cm = 7.8x102 Ω-m from Table 7.8.

Thus, with L = 4x10-6 m and A = 20x10-12 m2, the resistance R in the piezoresistor is:

( )( )ΩΩ==

−

−

Morxx

xxR 1561056.11020

104108.7 812

62

The net change of resistance in the piezoresistor at 235.36 MPa stress is:

( )( ) ΩΩ==∆ − MorxxxR 014.11014.101056.1105.6 583 Problem 5: The piezoelectric coefficient, d, for PVDF polymer films can be found to be 18x10-12 m/v from Table 7.14. Consequently, the induced voltage by the induced strain of 123.87x10-5 m/m from Example 7.4 is:

39

mvxx

xdd

V /1088.61018

1087.123 712

5max ====

−

−εε

with the piezoelectric film being 4 µm long as shown in Figure 7.18, the output voltage is: v = Vl = (6.88x107)(4x10-6) = 275.3 v Problem 6: If the length of the imaginary lattice is (a) in the (100) plane, then aa 414.12 = is the lattice for both diagonal (110) and inclined (111) planes in Figure 7.8. Problem 7: The lattices for the three planes in a single silicon crystal are: (a) The (100) Plane: (b) The (110) Plane: (c) The (111) Plane

aaL 707.0221

== aaLa 433.043

==

Problem 8: The angle is 54.74 degree. Problem 9: We have been using ν = 0.25 as the Poisson’s ratio for silicon in our problems solving. By using this value for the Poisson’s ratio and the shear modulus of elasticity, G in Table 7.2, we will have the following values for the Young’s moduli, Eth, of silicon in the three orientations by using the relationship: Eth = 2(1 + ν)/G:

Orientations G , GPa ν Eth, GPa Etable , GPa <100> 79.0 0.25 197.50 129.5 <110> 61.7 0.25 154.25 168.0 <111> 57.5 0.25 143.75 186.5

a

a L a

0.707a 0.707a

0.707a 0.707a

L a LaL aL

a

0.707a 0.707a0.7

07a

0.707

a

40

We may make the following observations: (1) The Young’s moduli Etable in the above Table are the measured values as given in Table 7.2.

These values are lower than those calculated from linear theory of elasticity in the <100> orientation, but are higher in the other two orientations.

(2) The computed Young’s moduli, Eth in the <111> has the lowest value of the three. This is contrary to the measured values.

We thus conclude that the three elastic properties, E, G and ν of silicon do not follow the relationship established for isotropic elastic materials. Problem 10: We will use the geometry and the dimensions of the inkjet printer head as presented in Fig. 7.19 in Example 7.5.

For a printing resolution of 600 dots per inch (DPI), we should have the diameter of the dots to be D = 1 inch/600 = 25.4 mm/600 = 42.333 µm. The corresponding radius of the spherical ink dot (r) that is ejected by the printer head is:

tDr ⎟⎠⎞

⎜⎝⎛= 23

434 ππ in which t is the thickness of ink dot on the paper.

Again we will use t = 1 µm as in Example 7.5. This assumption will lead to r = 6.954x10-6 m The volume of the ink dot is computed by using the right-hand-side of the above expression to be Vdot = 1408x10-18 m3. The corresponding expansion of the piezoelectric cover for the ejection of ink volume, Vdot is:

( )mx

x

xxVW dot 12

26

18

2 1044810200014.3

10140844 −

−

−

==∆

=π

The corresponding strain in the piezoelectric cover is

mmxxx

LW /108.44

101010448 6

6

12−

−

−

===ε

The required voltage for 1 m thick cover is:

412

6

103418.910480108.44 x

xx

dV === −

−ε v/m

41

The required voltage for the present case for a 10 µm thick cover is thus: v = LV = (10x10-6)(9.3418x104) = 0.9342 v

Chapter 8 Microsystem Fabrication Processes

(P. 318) Part 1. Multiple Choice 1.(c); 2. (c); 3. (b); 4. (b); 5. (b); 6. (c); 7. (b); 8. (a); 9. (a); 10. (b); 11. (b); 12. (a); 13. (a); 14. (b); 15. (b); 16. (a); 17. (b); 18. (a); 19. (a); 20. (b); 21. (c); 22. (b); 23. (b); 24. (a); 25. (c); 26. (c); 27. (c); 28. (a); 29. (c); 30. (b); 31. (c); 32. (b); 33. (a); 34. (b); 35. (c); 36. (a); 37. (b); 38. (b); 39. (b); 40. (c); 41. (c); 42. (b); 43. (a); 44. (c); 45. (a); 46. (c); 47. (c); 48. (c); 49. (c); 50. (a) Part 2. Computational Problems Problem 1: We have phosphorous as the dopant and the doping is carried out with 30 KeV energy. From Table 8.2, we will have:

Rp = 42x10-9 m, and ∆Rp = 19.5x10-9 m = 19.5x10-7 cm

We further have the maximum concentration, Nmax = 30x1018 atom/cm3 as given in Example 8.1. It is at x = Rp = 42x10-9 m = 0.042 µm. (a) The supplied dose is:

214187max /10466.1)1030)(105.19(28.62 cmatomsxxxxNRpQ ==∆= −π

(b) We will use the following relationship to find the concentration at x = 0.15 µm:

( ) 3122

218 /1057.6

)0195.0(2042.015.0exp1030)15.0( cmatomsx

xxmN =⎥

⎦

⎤⎢⎣

⎡ −−=µ

(c) Let xo be the depth at which the dopant concentration is 0.1% of the maximum value. This

depth may be obtained by solving the following equation:

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−==

− 2

2

7

1418

)0195.0(2)042.0(

exp105.1914.32

10466.1)1030(0001.0)(xxxx

xxxN xx oo

42

Solve for xo = 0.1257 µm Problem 2: We have energy level for the ion implantation to be 100 KeV and maximum concentration Nmax = 30x1018 atom/cm3. We need to find the concentration of P and As at 0.15 µm beneath the surface. Let us first tabulate the projected range Rp and the scatter or straggle ∆Rp aat this energy level from Table 8.2:

Rp (nm) ∆Rp (nm) B 307 69 P 135 53.5

As 67.8 26.1 By following the procedure outlined in Example 8.1, we have: 1) The dose of ion beams by max/2 NRpQ ∆= π We thus have:

( )( ) 14187 10022.41030105.532 xxxQp == −π atom/cm2 for phosphorus

( )( ) 14187 10962.11030101.262 xxxQA == −π atom/cm2 for arsenic 2) Concentration of P and As at x = 0.15 µm beneath the surface of silicon substrate by using Equation (8.1):

( ) ( ) ( )( )

182

218 1084.28

0535.02135.015.0exp103015.0 xxmN p =⎥

⎦

⎤⎢⎣

⎡ −−=µ atom/cm3 for phosphorus

( ) ( ) ( )( )

182

218 1021.0

0261.020678.015.0exp103015.0 xxmN A =⎥

⎦

⎤⎢⎣

⎡ −−=µ atom/cm3 for arsenic

3) Location xo at which N(xo) = 0.01 Nmax = 3 x 1016 atom/cm3: We may derive the following relationship by using Equation (8.1): For the case of phosphorus with Qp = 4.022x1014 atom/cm2:

43

( ) ( )( )

( )( ) 16

11

2718

27

27

7

14

103107245.510135

exp10999.29

105.532

10135exp

105.5314,3210022.4

xxxx

x

x

xx

xxxxN

op

opopp

=⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−=

−

−

−

−

−

from which we solve for xop = 0.334 µm for phosphorus. For the case of arsenic with QA = 1.962 x 1014 atom/cm2:

( ) ( )( )

( )( ) 16

14

2718

27

27

7

14

1031042.1362108.67

exp1030

101.262

108.67exp

101.2614.3210962.1

xxxx

x

x

xxxx

xxN

oA

oAoAA

=⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−=

−

−

−

−

−

from which we solve for xoA = 0.1648 µm for arsenic. Observations on the results: We may summarize the computed results, combining that from Example 8.1 as follows:

Dopants Dose of ion beam (Q) ( 1014 atom/cm2)

Concentration at xo = 0.15 µm beneath surface

(1018 atom/cm3)

Depth at which N(xo) = 1% of Nmax

(µm) Boron (P) 5.2 2.27 0.5635 Phosphorus (P) 4.022 28.84 0.334 Arsenic (As) 1.962 0.21 0.1648 It appears phosphorus has the deepest penetration into the substrate of all three common dopants. Arsenic is definitely the worst of all. Problem 3: From Table 8.3, we have the constants required to evaluate the diffusion coefficient, D in Equation (8.6) to be: a = -19.982 and b = 13.1109. (a) From Equation (8.6), baTDn += ')(l with T’ = 1000/T = 0.8525, in which T = 900 + 273.

Hence ( ) 01976.0924.31109.138525.0982.19 =→−=+−= DxDnl or D = 0.0003904 µm2/h, or D = 0.10844x10-6 µm2/s

(b) Equation (8.4) can be used to obtain the concentration function, N(x,t) follows:

44

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎥⎥⎦

⎤

⎢⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

− txerfc

tx

xerfcDtxerfctxN 151810

1010844.0210

210, 11

61111

or

( ) ⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

txerftxN 1518110, 11

For x = 0.1 µm and t = 1 h = 3600 s:

( ) ( )[ ] ( ) 099999.011053.211060

8.1511101,1.0 111111 ≈−=−=⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−= erferfhmN µ

(c) For diffusion to take place at 800oC, we have 932.0273800

1000' =+

=T . This will lead to the

computation of the diffusivity, D as follows:

hmDxDn /000163.051165.51109.13932.0982.19 2µ=→−=+−=l

The diffusivity, D is so low in this case that it leads to a negligible concentration at x = 0.1 µm after 1 hour into the diffusion process. (d) Let us raise the diffusion temperature to 1100oC:

We will have T’ = 1000/(1100 + 273) = 0.72833, which leads to D = 0.05584 µm2/h, or 15.5x10-6 µm2/s. This diffusivity will result in a concentration at x = 0.1 µm and t = 1 h to be:

( ) ( )[ ]

( ) 31011

11

6

11

/10648.72352.0110

21167.01103600105.152

1.01101,1.0

cmatomsx

erfxx

erfhmN

=−

=−=⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

−µ

Tabulation of the results on the concentrations at x = 0.1 µm after 1 h into the diffusion process at various temperatures is given below:

Temperature, oC 800 900 1000 1100

Diffusivity, D, µm2/h 0.000163 0.0003904 0.005676 0.05584 Concentration, N(0.1 µm, 1h),

atoms/cm3 0 ≈0 3.482x1010 7.648x1010

It is clear from the above tabulation of results that the higher the diffusion temperature, the higher the diffusivity. Consequently, one can expect much higher concentration of the dopants beneath the surface of the substrate at higher temperatures.

45

Problem 4: The time required reaching the same concentration of dopant of 3.482x1010 atoms/cm3 as in Example 8.2 at 0.2 µm beneath the surface at 1000oC can be obtained by solving the following equation:

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−=

txerfx 2.021.39811010482.3 1110

By using Figure 3.14 and solve for the time, t = 13661.73 s, or 3.795 h Problem 5: Estimate the required time to achieve 1 µm thick SiO2 on a silicon substrate. The constants used in estimating the rate of oxidation in Equations (8.9) and (8.10) are available in the Table established in Example 8.3:

Dry oxidation Wet oxidation

B/A, µm/h 0.04532 0.6786 B, µm/h 0.006516 0.2068

Now if we let x = 1 µm in Equations (a) and (b) in Example 8.3, we will have the time required to oxidize the silicon substrate with 1 µm thick SiO2 obtained from Equations (8.9) and (8.10) to be:

Time for dry oxidation, h Time for wet oxidation, h Equation (8.9) for small time 22.065 1.474 Equation (8.10) for larger time 153.47 4.836 Problem 6: The dilution of the hydrogen gas is η = 1%, and the deposition takes place at 800oC. We assume that the process is used to deposit thin SiO2 film, and that other conditions for this CVD process are identical to those specified in Example 8.5.

(a) The number of molecules in one cubic meter volume of the gas mixture (NG):

We may follow the procedure in Example 8.4 and find the molar density of the gas mixture at 800oC to be:

31

2

12 /1905.12643.44

27380027320 mmolxd

TTd =⎟

⎠⎞

⎜⎝⎛

++

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

The concentration of H2 per cubic meter is:

46

NG = (6.022x1023)x12.1905 = 73.4111x1023 molecules/m3 (b) The density (ρ) of the carrier gas, H2 is:

ρ = (2 g/mol)(12.1905 mol/m3) = 24.38 g/m3

(c) The Reynolds number (Re):

µρDV

=Re

with the gas density, ρ = 24.38 g/m3, the diameter of the reactor, D = 20 cm = 0.2 m, the gas velocity, V = 50 mm/s (as given in Example 8.5), and the viscosity, µ = 214 µP = 0.0214 g/m-s from Table 8.6 for H2 gas at 825oC. We may compute the Reynolds number to be Re = 11.39 (d) The thickness of the boundary layer (δ):

mL 0593.039.11

2.0Re

===δ

(e) The diffusivity of the carrier gas (D):

We may use the same equation presented in Example 8.5 as shown below:

( )NN sG

ND−

=η

δv

with N

r = 1024 molecules/m2-s (given)

NG = 73.4111x1023 molecules/m3 (in Part (a)) Ns = 0 and η = 1% = 0.01 We thus have the diffusivity, D to be:

( ) smx

xD /8078.00104111.7301.0

100593.0 223

24=

−=

(f) The surface reaction rate (ks):

Following the expression used in Example 8.5, this rate can be computed from the expression:

smxxx

xND

NDN

kG

s /1376.0100593.0104111.738078.0

108078.02423

24=

−=

−= v

v

δ

47

(g) The deposition rate (r):

We first compute δks = 0.0593x0.1376 = 0.00816 << D = 0.8078

This will justify us using the expression for estimating the rate of deposition, γ

η kN sGr =

From Table 8.7 and Example 8.5, we have the number of SiO2 molecules per unit film volume to be: γ = 4.3074x1028, which leads to the rate of deposition, r to be:

smx

xxxr /2345.0103074.4

1376.0104111.7301.028

23µ==

Problem 7: Since the rate of the deposition is 0.2345 µm/s as calculated in Problem 6, a deposition of 0.5 µm thick film will take 0.5/0.2345 = 2.132 s. Problem 8: The following changes in computing the rate of deposition will take place with a 490oC process temperature: (a) The number of molecules in one cubic meter volume of gas mixture (NG):

31

2

12 /1433.17643.44

763293 mmolxd

TTd =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

The corresponding concentration is

( )( ) 32323 /1024.1031433.1710022.6 mmoleculesxxNG ==

(b) The density:

3/2866.341433.172 mgx ==ρ

(c) The Reynolds number:

53.200167.0

05.02.02866.34Re ===xxDV

µρ

(d) The thickness of the boundary layer:

mL 033.053.20

15.0Re

===δ

48

(e) The diffusivity of the carrier gas:

( ) ( ) smx

xNDNN sG

/3196.001024.10301.0

10033.0 223

24

=−

=−

=η

δr

(f) The surface reaction rate:

smxxx

xND

ND

NkG

s/0978.0

10033.01024.1033196.0103196.0

2423

24

=−

=−

= r

v

δ

(g) The deposition rate:

smx

xxxr kN sG /2344.0103074.4

0978.01024.10301.028

23µ

γη ===

Problem 9: We have 1% H2 carrying gas at 800oC. The standard conditions have: P1 = 101,325 Pa, T1 = 273+20 = 293 K and V1 = 22.4 x 10-3 m3/mol. The process conditions are: P2 = 1 Torr = 133.322 Pa, T2 = 800 + 273 = 1073 K. From the ideal gas law, we get:

2

1

22

11

TT

VPVP

= or ( )1073293

133322104.22101325

2

3

=−

Vxx , from which we solve for V2 = 62,344.04 m3/mol

(a) The molar density d2: d2 = 1/V2 = 1/62344.04 = 1.604 x 10-5 mol/m3 – a very low molar density! The concentration of H2 gas in the carrier gas is: NG = Avogardro’s number x d2 = (6.022 x 1023)(1.604 x 10-5) = 9.66 x 1018 molecules/m3 (b) The corresponding density of H2 gas is: ρ = (2 g/mol)(1.604 x 10-5 mol/m3) = 3.208 x 10-5 g/m3 (c) The Reynolds number is:

5105.1Re −== xVDµ

ρ

with D = 0.2 m, V = 50 mm/s = 0.05 m/s, µ = 0.0214 g/m-s

49

(d) The thickness of the boundary layer is:

mx

L 64.51105.12.0

Re 5===

−δ - an unrealistically thick boundary layer!!

The above computed large boundary layer thickness is due to the extremely low pressure in the process. At this point, we may choose a boundary layer thickness as thick as the available space in the chamber can accommodate, or use this unrealistically large number as computed. Since the chamber has a diameter of D = 0.2 m, we will use this number in the remaining calculation, i.e., δ = 0.2 m. (e) The diffusivity of the carrier gas to the silicon substrate:

( )sG NNND−

=η

δr

with 2410=N

rmolecules/m2-s (given), η= 1%, NG = 9.66 x 1018 molecules/m3, and Ns = 0,

we have:

( )6

18

24

1007.21066.901.0

102.0 xx

xD == m2/s (very high!)

(f) The surface reaction rate:

( )( )( )( ) ( )

524186

246

10045.1102.01066.91007.2

101007.2 xxx

xNND

NDkG

s =−

=−

= r

r

δ m/s (very large!)

(f) Rate of deposition: Check: δks = 0.2 x 1.045 x 10 5 = 0.21 x 105 < D = 2.07 x 106

Hence the deposition rate is:

( )( ) 2343.010307.4

10045.11066.901.0 28

518

===x

xxkNr sG

γη µm/s

The value of SiO2 molecules per unit volume γ = 4.307 x 1028 molecules/m3 given in Example 8.5 was used in the above calculation. The reader should not take the conditions stipulated in this problem to be realistic. Although the low pressure in the process is not unusual, the chamber size is unrealistically small to accommodate the thick boundary layer over the surface of the substrate. Nor is the low velocity that contributes to this unrealistically thick boundary layer.

50

Problem 10: We have the temperature T = 800oC = 1073 K. The velocity of the carrier gas, however, is reduced to V = 25 mm/s = 0.025 m/s We may follow the same procedure in computing the deposition rate as follows: (a) NG = 73.4111 x 1023 molecules/m3 (same as in Problem 6) (b) ρ = 24.38 g/m3 (same as in Problem 6) (c) Re with reduced velocity becomes:

( )( )( ) 6963.5

0214.0025.02.038.24Re ==

in which we used D = 0.2 m and µ = 0.0214 g/m-s

(d) The boundary layer thickness:

0838.06963.52.0

Re===

Lδ m

(e) Diffusivity of the carrier gas:

( )( )( )( ) 1415.1

0104111.7301.0100838.0

23

24

=−

=−

=xNN

NDsGη

δr

m2/s

(f) The surface reaction rate:

( )( )( )( ) ( )( ) 1376.0

100838.0104111.731415.1101415.1

2423

24

=−

=−

=xNND

NDkG

s r

r

δ m/s

(g) The deposition rate: Check δ ks = 0.0838 x 0.1376 = 0,01153 << D = 1.1415

Hence ( ) 2345.0103074.4

1376.0)104111.73(01.0 28

23

===x

xkNr sG

γη µm/s

Chapter 9 Overview of Micromanufacturing

(P.344) Part 1 Multiple Choice:

51

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

Chapter 10 Microsystem Design

(P.402) Part 1 Multiple Choice: 1. (a); 2. (b); 3. (c); 4. (a); 5. (c); 6. (b); 7. (b); 8. (b); 9. (a); 10. (b); 11. (c); 12. (a); 13. (c); 14. (b); 15. (a); 16. (c); 17. (b); 18. (c); 19. (b); 20. (c); 21. (a); 22. (b); 23. (a); 24. (b); 25. (c); 26. (a); 27. (c); 28. (b); 29. (a); 30. (b); 31. (c); 32. (c); 33. (a); 34. (b); 35. (a); 36. (b); 37. (b); 38. (a); 39. (b); 40. (c); 41. (b); 42. (a); 43. (a); 44. (c); 45. (b); 46. (a); 47. (a); 48. (b); 49. (c); 50. (b); 51. (a); 52. (c); 53. (a); 54. (b); 55. (c); 56. (c); 57. (c); 58. (c); 59. (a); 60. (c) Part 2 Descriptive Problems: Problem 3: There are times when flexibility of micro pressure sensor die becomes necessary due to significant difference of coefficients of thermal expansion between the silicon die, the die attach and constraint base, such as described in Section 11.20 in Chapter 11. One practical way of improving the flexibility of the die is to increase the height of the diaphragm to mitigate the parasite stresses induced by such difference of coefficients of thermal expansion of the attached parts. Intuitively one may increase the height of the silicon die by producing such these dies in fabrication as illustrated in dotted lines in Figure 10.35. Figure 10.35 Micro Pressure Die with

Double Height (Legend in Figure 10.16) One may show that the size of the diaphragm (i.e., the length of the square diaphragm a) is reduced by an amount ∆a = a – a’ = 1.414H, in which H = the original height of the die. This reduction of the diaphragm size will result in significant reduction in the maximum bending stress in the diaphragm, and thus the output of the sensor according to Equation (4.10), which is a situation that is hardly desirable. An alternative way to increase the flexibility of the die is by

Constraint Base

A

a a’h

Θ=54.74o H

2H

c

Cavity

52

inserting a spacer as illustrated in Figure 11.36 on Page 453. This arrangement will offer the required flexibility of the die, without reducing the maximum bending stress in the diaphragm for the output signal. Such insertion, of course, will add extra cost to the production of the pressure sensor. Problem 4: Hint: Since we are given the hydraulic diameter of capillary tubes of rhombus, obround and trapezoidal cross sections illustrated in Figure 10.19 on Page 383, the computation of the fluid resistance of each of the tubes of specific cross section can follow what are performed in Example 10.2 on Page 385. The conduit with rhombus cross section can be treated as two v-groove open channels; the tube with obround cross section, on the other hand needs to be handled in a different way. This geometry involves 2 semicircles with each connected to the two opposite sides of a rectangle as illustrated below: In the above figure, a is the radius of the circular sections. We may readily formulate the equation for the hydraulic diameter of the entire cross-section. One needs to optimize the relationship between the length of the rectangle L and the radius a of the semi-circular parts of the cross-section to satisfy the condition of dp = 30 µm. The resistance to water flow in this conduit may be obtained by summing the resistance from a complete circular and that from a rectangular cross-sections using the equations provided in Section 10.7.

Chapter 11 Assembly, Packaging, and Testing of Microsystems

(P. 458) Part 1 Multiple Choice 1. (c); 2. (c); 3. (a); 4. (a); 5. (a); 6. (a); 7. (b); 8. (b); 9. (a); 10. (a); 11. (b); 12. (a); 13. (b); 14. (a); 15. (a); 16. (c); 17. (a); 18. (c); 19. (b); 20. (a); 21. (a); 22. (b); 23. (c); 24. (b); 25. (a); 26. (b); 27. (c); 28. (a); 29. (c); 30. (c); 31. (a); 32. (b); 33. (a); 34. (c); 35. (b); 36. (a); 37. (a); 38. (b); 39. (a); 40. (b); 41. (b); 42. (a); 43. (b); 44. (c); 45. (b); 46. (b); 47. (b); 48. (a); 49. (a); 50. (b); 51. (c); 52. (b); 53. (a); 54. (b); 55. (c); 56. (b); 57. (c); 58. (b); 59. (c); 60. (c); 61. (c); 62. (b); 63. (c); 64. (b); 65. (c); 66. (b); 67. (b); 68. (c); 69. (a); 70. (c).

a

L

2a

53

Part 2 Descriptive Problems Problem 2 This is another typical open-ended design problem without having all the conditions given for getting straight solution from formula given from the textbook. What we need to do first is to assume a reasonable coefficient of static friction between the cylinder to be picked up by the gripper and the gripper arm. A number µ = 0.94 between dry glass to glass from mechanical handbook (Avallone and Baumeister 1996) appears to be a reasonable value to use. This will establish the gripping force required to pick up a cylinder with a mass of 2 mg, i.e., half of the normal force associated with the friction force, or N = (2x10-6 kg)(9.81 m/s2)/(µ = 0.94) = 20.08x10-6 N (Newton). This situation is depicted in a pick-n-place gripper illustrated in Figure 11.6 on page 418. We thus conceive that the electrostatic force that the gripper needs to generate is: F = 20.08x10-6 N. By referring to the microgripper illustrated in Figure 2.45 on page 80, each pair of electrode would generate electrostatic force f that is equal to what is given in Equation (2.11), or:

dVW

f orL

2

21 εε

=

in which εr = 8.85x10-12 C/N-m2 and εo = 1.0 (in air), W = the width = 5 µm, and the gap d = 2 µm. We already learned from Problem 2.13 that it will take huge number of pairs of electrodes to actuate this gripper with 25 volts. Let us assume the number of pairs required in this application to be n, which leads to the following relations:

62

1008.2021 −=== xF

dVW

nnf orL

εε, from which we arrive at the following expression:

nV2 = 1.81514x106

For the case with V = 25 v, we will need the number of pair of electrodes n = 2904.23, or 2905 pairs, which will take up 2905 x (10+2+10 = 22 µm) = 63920 µm >> 300 µm, the total length of the gripper arm. It is therefore not practical to expect a 25 v actuation voltage for the gripper. On the other hand, if we allow n = 300 µm/22 µm ≈ 14 pair of electrodes, then the actuation voltage will be:

76.18914

1081514.1 6

==xV volts

for the actuation.

54

Problem 3: The diameter of the cylindrical object in Problem 2 can be obtained by the following expression:

( )LdmcylindertheofmassThe ⎟⎠⎞

⎜⎝⎛= 2

4πρ

where ρ = mass density of silicon = 2.3 g/cm3 (Table 7.3) = 2.3 x 103 kg/m3, L = the length of the cylinder = 200 µm. Thus with m = 2 mg = 2x10-6 kg, we can computer the diameter d = 2.3528 x 10-3 m. Here, we need to assume that the attractive forces in microgripping as given in Equations (11.1) and (11.2) are also valid for cylindrical objects. The electrostatic attractive force in Equation (11.1) for the present case can thus be applied as:

( )( )( )( )

Nxx

xd

qFe 4160103528.21085.814.34

106.14 2312

26

2

2

===−−

−

πε

We have computed from Problem 2 that the normal force gripping force for the cylindrical object was 20.08 x 10-6 N, which is much less than the computed electrostatic attractive force Fe. We thus predict that the gripper will not release the cylindrical object upon the release of the gripper. Problem 4: The residual curvature of the anodically bonded quartz/silicon beam can be computed by using Equation (4.49) on page 153. Following material properties will be used in the computation: For quartz: Yong’s modulus E1 = 0.865x1011 N/m2 (an average value from Table 7.3) Coefficient of thermal expansion α1 = 7.1x10-6/oC (Table 7.3) For silicon: Young’s modulus E2 = 1.9x1011 N/m2 (Table 7.3) Coefficient of thermal expansion α2 = 2.33x10-6/oC (Table 7.3) The ratio of Young’s modulus in Equation (4.49) is n = E1/E2 = 0.4553 We have the thickness of quartz layer to be t1 = 1 mm = 10-3 m and the thickness of the silicon layer to be t2 = 300 µm = 3x10-4 m, from which we have m = t1/t2 = 3.3333. We further have the thickness of the bi-layer beam to be h = t1 + t2 = 1.3x10-3 m. The temperature drop ∆T in Equation (4.49) is: ∆T = 20 – 500 = -480oC By substituting the above material properties and physical parameters into Equation (4.49), we have the following:

55

( ) ( )( ) ( ) ( )[ ]

( ) ( )( )( ) ( ) ( ) ( )[ ] m

xxxxx

mnmmnmhTm

3083.24553.03333.3/13333.34553.03333.313333.313103.1

4801033.2101.73333.316

/1113161

223

662

2221

2

−=++++−−+

=

++++∆−+

=

−

−−

ααρ

We thus have the radius of curvature of the residual bent beam

m4332.03083.2

1−=−=ρ

The residual shape of the anodically bonded bi-layer beam is similar to that illustrated in the lower of the two bent shapes in Figure 4.37 on page 154 as illustrated below: The consequence of this residual shape may cause serious misalignment of the pressure sensor die in the silicon portion of the beam to the supporting base in the quartz portion.

Chapter 12

Introduction to Nanoscale Engineering (Page 503)

Part 1. Multiple Choice 1. (b); 2. (c); 3. (b); 4. (b); 5. (none); 6. (c); 7. (c); 8. (c); 9. (c); 10. (a); 11. (a); 12. (c); 13. (a); 14. (b); 15. (b); 16. (b); 17. (c); 18. (a); 19. (b); 20. (b); 21. (b); 22. (c); 23. (a); 24. (b); 25. (b); 26. (a); 27. (a); 28. (b); 29. (b); 30. (b); 31. (b); 32. (c); 33. (b or c); 34. (b); 35. (c); 36. (c); 37. (b); 38. (c); 39. (b); 40. (a); 41. (c); 42. (a); 43. (a); 44. (b); 45. (a); 46. (c); 47. (c); 48. (a); 49. (b); 50. (b); 51. (c); 52. (c); 53. (c); 54. (a); 55. (a); 56. (c); 57. (b); 58. (b); 59. (b); 60. (c); 61. (a); 62. (c); 63. (c); 64. (b); 65. (b); 66. (c); 67. (c); 68. (b); 69. (c); 70. (b).

ρ = 0.43 m

Quartz beam

Silicon beam