MEMS and Microsystems: Design and Manufacture*

SOLUTION MANUAL

Tai-Ran Hsu, Professor**Microsystems Design and Packaging Laboratory

Department of Mechanical and Aerospace EngineeringSan Jose State UniversitySan Jose, CA 95192-0087

USA

June 3, 2002

_____________________________________* McGraw-Hill, Boston, 2002, ISBN 0-07-239391-2

** Telephone: (408)924-3905; Fax: (408)924-3995 E-mail: tairan@email.sjsu.edu

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 8

Chapter 4 Engineering Mechanics for Microsystems Design 10

Chapter 5 Thermofluid Engineering and Microsystems Design 24

Chapter 6 Scaling Laws in Miniaturization 35

Chapter 7 Materials for MEMS and Microsystems 36

Chapter 8 Microsystems Fabrication Processes 40

Chapter 9 Overview of Micromanufacturing 49

Chapter 10 Microsystems Design 49

Chapter 11 Microsystem Packaging 49

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Chapter 1Overview 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 2Working Principles of Microsystems

(P. 68)

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

Part 2. Description Problems

Problem 2:

Transducers Advantages DisadvantagesPiezoresistors 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.

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 macro scale.

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 micro pressure sensor will result in the change of its geometry, and thus the shifting of its natural frequency.

Problem 7:

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 0.50Vo/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 8:

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

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

Thus, the output voltage is:

4

Problem 9:

Actuation techniques Advantages DisadvantagesThermal 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 10:

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

From Eq. (2.10), we may compute the electrostatic force in the width-direction: Fw = 0.0815 N.From Eq. (2.11), for the force in the length-direction: FL = 0.0815 N

Problem 11:

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

5

V

Moving electrodes Moving electrodes

Fixed electrodes

Spring constantk

Spring constantk

From Example 2.4, we have the spring constant, k = 0.05 N/m.

The required traveling distance of the moving electrodes is = 10x10-6 m, which corresponds to the spring force:

F = k = 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 NFrom Eq. (2.11),

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

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

Problem 12:

We need to make a few assumptions before embarking the solution of the problem. The very first piece of information we need for the solution to this problem is the materials that make the gripper. For simplicity, we assume that both the “drive arm” and the “closure arm” are made of silicon with the surfaces of the “comb drive” “fingers” electrodes coated with thin electrically conducting copper films. The dielectric medium in the electrode gaps is air. Properties of all the materials are available in Table 7.3 on P. 247.

Next, we need to establish the dimensions of the gripper arms and the gaps between theelectrodes. We assume that the arrangements that are presented in Example 2.4 are applicable inthis Problem. We will consider the following geometry and dimensions:

6

Flexible “Drive Arm”

Rigidly held“Closure Arm”

Req’d tipmovement:5 m

Req’d tipmovement:5 m

Gap, d = 2 m 10 m

Width of electrodes, W = 5 m

150 m 38 m

300 m

“A”

“A”

5 m

10 m

View “A-A”

We will find the necessary voltage supply to the electrodes on both drive arms to provide a 5 m 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 + 0.5x38 = 169 m away from the support-end as illustrated below:

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:

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:

Solve for the equivalent applied force, P = 0.11385x10-3 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 Eq. (2.11), the electrostatic force is:

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

We thus have:

We may solve for the supply voltage to be V = 1434 volts, which is an unusually high voltage for a microgripper. The postulated geometry and dimensions thus require substantial modifications

7

b = 169 m P

L = 300 mmax = 5 m

in order to bring down the required voltage. Reduction in the length, or the depth of the drive arm would result in the reduction of the required voltage for actuation.

Chapter 3Engineering Science for Microsystems Design and Fabrication

(P. 93)

Part 1: Multiple Choice:

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

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 -cm, the same as that of platinum.

Problem 4:

The negative sign in the equation means that the concentration of the diffused substance decreases as the distance of diffusion increases.

Problem 5:

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 1400oC for As, P and B.

8

Problem 6:

Equation (3.5) is used for the solution of this problem:

The concentration in Example 3.1 at various instants into diffusion can be obtained by using the following parameters:x = 0.075; t = 1,1.5, 2, 2.5 h; Cs = 4.5x1020 atoms/cm3; and (D)1/2 = 0.085 m/h1/2, from which we have the following relationship:

Numerical solution of C(0.075, t) at various time, t can be tabulated as follows:Time, t (h) 1.0 1.5 2.0 2.5

Concentration, C(1020 atoms/cm3)

2.64 3.01 3.22 3.36

Problem 7:

Let the time required to dope boron into silicon substrate at a depth of 2 m to be tf. We obtained the corresponding concentration of boron at resistivity of 10-3 -cm from Fig. 3.8 to be C = 1020 atoms/cm3. Thus at a diffusion temperature at 1000oC as specified in Example 3.1, we will have the concentration of C at 2 m from the silicon substrate surface at time tf to be:

from which, we obtained the following expression for the solution of tf:

with (D)1/2 = 0.085 m/h1/2 as in Example 3.1.

One may readily solve tf = 3.72 h using Table 3.4 for the values of error functions, erf(x).

9

Chapter 4Engineering Mechanics for Microsystem Design

(P. 159)

Part 1. Multiple Choice

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

Part 2. Computation Problems

Problem 1:

We have d = 600x10-6 m, a = d/2 = 300x10-6 m, and P = 20x106 N/m2. The Young’s modulus, E = 0.7x1011 N/m2 for aluminum from Table 7.3 on P. 247. The Poisson’s ratio, = 0.3 for aluminum from metal handbook.

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

in which W = (a2)P = 3.14(300x10-6)2(20x106) = 5.652 N, m = 1/ = 1/0.3 = 3.33.

One may thus calculate the maximum deflection, Wmax = -1.4771x10-4 m, or 14.77 m.

Problem 2:

A few assumptions on the dimensions of the pressure sensor die need to be made before we can embark on the computation. First, let us assume the geometry of the die as illustrated below:

10

3 mm

3 m

m

Plan View

Applied Pressure, P

hH

L = 3000 m LL

Cross-section of the Die

a

We will then assign the dimension of the thickness of the die and the size of the foot print as shown below:

The assigned die thickness, H = 500 m is the standard thickness of 100 mm diameter wafer as indicated in Section 7.4.2 on P. 239, whereas the footprint L = 250 m is an assumed design parameter.

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.

It is clear from the diagram that , in which H = 500 m. We thus have:

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

a = L – 2L - 2a = 1792.786 + 1.414h

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

or

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

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

11

h

H = 500 m

L = 250 m a

54.74o

L =3000 m

Detail Dimensions of Foot Print

Edge Length of Diaphragm, a

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:

where F = applied load to the beam = 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:

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

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:

Hence the equivalent spring constant is:

12

L

F

L

Case 3 Cantilever beams:

The deflection of the beam at the free-end is:

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

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

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

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

13

F

L

20x10-6 m

y

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

(a)

The solution of the differential equation is:

(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:

(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 m.

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:

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

The proper differential equation becomes:

14

in which the natural frequency of the beam spring system,

The solution of the above differential equation is:

with = n = 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:

Now, if we let tf = the time at which the beam spring breaks at y(tf) = 1 mm = 10-3 m, we will have:

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.

Problem 6:

The beam is loaded as illustrated in Example 4.8

The area moment of inertia of the beam cross-section is:

with h in micrometers.

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

15

L=600m

Dynamic force1 m

h m

Beam Cross-section

The corresponding circular frequency of the balanced force accelerometer is:

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

in which the arbitrary constants C1 and C2 can be determined by the initial conditions:

We thus have:

C1 = 0 and C2 = 13.8888/ = 0.1434 h-3/2

Thus, we have the amplitude of vibration to be:

At X(tf) = 5 mm = 5x10-3 m for the failure of the beam:

or

The approximate value of tf is when for a maximum value of h, which

leads to h = 9.37 m.

Problem 8:

We may illustrate the balanced force accelerometer system below:

16Beam Mass

Beam Springs

5 m

10 m

400 m

From Example 4.11 on P. 125, we get the damping coefficients for the balanced force accelerometer to be:

cair = 1.3125x10-8 N-s/m for air as the damping fluid, andcsi = 51.8x10-8 N-s/m for silicone oil as the damping fluid.

We further have the mass of the silicon beam to be: m = v, in which the mass density, = 2.3 g/cm3 or 2.3x103 Kg/m3 from Table 7.3 on P. 247, and the volume of the beam = v.

By referring to the geometry and dimensions of the beam in Example 4.11, we have

v = bHL = (5x10-6) (10-5) (7x10-4) = 35x10-15 m3

Consequently, the beam has a mass, m = (2.3x103)(35x10-15)= 80.5x10-12 Kg

We will use the model illustrated in Fig. 4.7(b) on P. 107 to assess the motion of the beam mass, and Eq. (4.19) with the spring constant k = 2keq in Example 4.8 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 (2 - 2) as described in Eq. (4.20a), (4.20b) or (4.20c). We will thus need to compute both 2 and 2 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. 118:

with E = 1.9x1011 N/m2 (Table 7.3) and

Thus,

17

and

which leads to 2 = 5.9x1012 Rad2/s2

The damping parameters:

from which we have:

The above values of (2 - 2) for the two distinct damping media of air and silicone oil will lead to the use of Eq.(4.20c) for the movement of the beam mass. The movement will be of oscillatory nature.

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 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:

18

Beam Mass, m

L = 700 m

h = 10 m

b = 5 m

Dimensions of the Beam Mass

600 m

10 m5 m

Dimensions of the Beam Springs

The equivalent spring constant for beam springs with fixed-ends is keq = 1.76 N/m as in Case 2 of Example 4.8.

Since the maximum deceleration of the car in the present case is from

Example 4.12. By neglecting the mass of the beam springs, we may express the dynamic force associated with the moving beam mass as:

The mass of the beam mass, m = v, in which = 2.3 g/cm3 is the mass density of silicon from Table 7.3, and the volume of the beam mass, V is:

V = bhL = (5x10-6)(10x10-6)(700x10-6) = 35x10-15 m3

The mass of the beam mass is thus computed to be:

m = (2.3x103 Kg/m3)(35x10-15 m3) = 80.5x10-12 Kg

The force acting on both beam springs at the time of deceleration of –22.22 m/s2 is:

The induced deflection of the beam springs by the above dynamic force of the magnitude is:

P = F/2 = 1788.71x10-12/2 = 894.4x10-12 N is = P/keq.

with L = 600x10-6 m, I = 10.42x10-24 m4 (from Example 4.8), E = 1.9x1011 N/m2 (Table 7.3), and keq = 1.76 N/m, we will find that the maximum movement of the beam mass to be:

max = 508 m

Problem 10:

This bi-layer strip is subjected to a uniform temperature rise, T as illustrated below:

19

The radius of curvature, from Eq. (4.50) is:

where 1 and 2 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:

in which

From Example 4.14, we have the movement of the free-end, to be:

where with L = 1000x10-6 m

Hence

We may tabulate the values of the temperatures vs. the movement of the free-end of the beam actuator as follows:

T (oC) = C/T (m) (o) = (1 - cos) (m)10 0.3643 0.1574 1.37320 0.1822 0.3147 2.74730 0.1214 0.4720 4.12040 0.0911 0.6294 5.49650 0.0729 0.7880 6.890

Problem 11:

20

h = 10 m

h

1000 m

SiO2 strip

Silicon strip

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, andat 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 4.15 on P. 142:

Mass density, = 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, = 2.33x10-6/oC; Young’s modulus, E = 1.9x1011 N/m2; Poisson’s ratio, = 0.25.

We will first compute the thermal force, NT and the thermal moment, MT from the respective Eq. (4.54a) and (4.54b) as:

From Example 4.15, we have A = 5x10-11 m2 and I = 4.167x10-22 m4.

From Eq. (4.55), we have the thermal stress along the x-direction to be:

with xx,max occurs at z = 5x10-6 m.

Thus, xx,max = (x,5x10-6) = -2600 Pa

We will compute the associate thermal strains from Eq. (4.56a) and (4.56b) with maximum values occurring at z = 5x10-6 m:

21

b = 5 m

H = 10 mx

y

2h = 10 m

L = 1000 m

0SiliconBeam

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

results in zz,max = zz(x,5x10-6) = -0.0023%

The deflection of the beam in the x-direction = u (x,z) can be computed from Eq. (4.57a) as:

with umax at x = 500x10-6 m and z = 5x10-6 m:

umax = u(500x10-6, 5,10-6) = 0.0466 m

The deflection of the beam in the z-direction, w(x,z) is obtained from Eq. (4.57b):

with wmax occurs at x = 500x10-6 m and z = 5x10-6 m, we have:

wmax = w(500x10-6, 5x10-6) = -0.582 m

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 4.15 on P. 145.

The thermal stresses in both x- and y-directions can be computed from Eq. (4.51) as:

The associated thermal strains are obtained from Eq. (4.52a):

22

with xy = yz = zx = 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 Eq. (4.53a,b and c):

and

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:xx,max = yy,max = 4000 Pa xx,max = yy,max = zz,max = 0.00915%umax = 0.046 m; vmax = 0.0046 m; wmax = -0.6173 m

We have realized that by extend the beam into a plate with a width of 100 m has not produced significant difference in the results from those obtained from a beam with a width of 5 m.

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. We need to assume the crack length, c = 100 m = 10-4 m.

The critical load, Pcr that breaks the specimen is 40x106 N/m2.

We will use Eq. (4.64a) for the function F(c/b) as s/b = 2 < 4, as indicated in Section 4.5.2 on P. 149:

23

s = 10000 m

Pcr = 40 MN

b = 5000 m

B = 2400 m

with c/b = 0.02

Hence F(c/b) = 1.0586

Eq. (4.63) is used to compute fracture toughness:

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

where

The bending moment,

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

which leads to the fracture toughness, Kc to be:

Problem 15:

For the width of the specimen, B to be increased to 100x240 m, we will have I = 2.5x10-9 m4. This new value will change the critical stress according to the following expression:

Kc = 0.375 Pa , 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, P cr, 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.

24

Chapter 5Thermofluid Engineering and Microsystem Design

(P. 163)

Part 1. Multiple Choice

1.(3); 2. (1); 3. (2); 4. (2); 5. (3); 6. (1); 7. (2); 8. (1); 9. (1); 10. (3); 11. (2); 12. (3); 13. (2); 14. (1); 15. (3); 16. (1); 17. (2); 18. (3); 19. (3); 20. (2); 21. (1); 22. (1); 23. (3); 24. (2); 25. (1); 26. (3); 27. (1); 28. (1); 29. (2); 30. (1); 31. (3); 32. (1); 33. (1); 34. (3); 35. (2); 36. (2); 37. (1); 38. (2); 39. (1); 40. (2); 41. (3); 42. (3); 43. (1); 44. (2); 45. (3); 46. (1); 47. (3); 48. (1); 49. (1); 50. (2); 51. (3); 52. (3); 53. (3); 54. (1); 55. (1)

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

A1 = (500x10-6)2/4 = 19.64x10-8 m2

A2 = (50x10-6)2/4 = 19.64x10-10 m2

Problem 3:

The opening of the valve may be illustrated as follows:

25

15o

L = 400 m

Wid

th, W

= 3

00

m

Exit velocity, Ve

Fluid FlowValve Opening, H

Valve Plate Thickness:4 m

(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:

in which = 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 m, or 300x10-6 m.

Hence Ae = (100x10-6)(300x10-6) = 3x10-8 m2.

Thus, the exit velocity is:

NOTE: This exit velocity is unrealistically high for a micro valve. 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)

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

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:

26

W N/m

L = 400 m

in which max 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 m wide x 4 m thick, which leads to an area moment of inertia, I to be:

with a Young’s modulus, E = 1.9x1011 N/m2 from Table 7.3 for silicon, and max = H = 100x10-6

m as shown in the figure in Problem 3, we may determine the required load, W from the following relation:

which leads to W = 9.5 N/m

The force required to lift the plate of 400 m 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:

which leads to V = 542 v – a rather high voltage.

Problem 5:

The problem is illustrated below:

27V1

= 600 m/s

V2

L = 0.1 m

d 1 = 100 m

d 2 = 50 m

d ave

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. 175,

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 Eq. (5.17) is:

where = dynamic viscosity of the fluid = 1199.87x10-6 N-s/m2 (Table 4.3 on P. 124 for alcohol)

L = length of the conduit = 0.1 m, and

We will have the approximate pressure drop, P to be:

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.

28

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:

with the density of water, = 1000 Kg/m3, and the inside diameter of the tube, D = 20x10-6 m, and the dynamic viscosity, = 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, andV = 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:

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

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 m/s in the capillary tube.

This water flow velocity of V = 10 m/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:

29

1000 m

20 m dia.

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:

in which = 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:

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 m, 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:

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

the capillary tube in only 1 mm.

Problem 8:

The rarefied gas flows in a nano scaled tube section is illustrated below:

30

N2 gas

50x10-9 m

P = 0.5 Pa

30x10-9 m dia.

The use of the Hagen-Poiseuille’s equation in the form of:

will lead to the volumetric flow rate as:

with the dynamic viscosity of the N2 gas, = 17.48x10-6 N-s/m2 from Table 4.3 on

P. 124, the length of the tube section, L = 50x10-9 m, and the radius of the tube, a = 15x10-9 m. Thus with a pressure drop of P = 0.5 Pa, we will have the volumetric flow rate, Q to be:

and the mass flow rate = Q = 1.2506x1.1367x10-20 Kg/s = 1.422x10-20 Kg/s with the value = 1.2506 Kg/m3 for N2 gas. We notice that this mass flow rate is about two orders of magnitude less than = 3.116x10-18 Kg/s as in Example 5.7 for rarefied gas flow. The Hagen-Poiseuilli’s equation is thus inadequate for capillary and rarefied gas flow.

Problem 9:

We have the problem as illustrated below:

Following the same procedure as in Problem 8, we have the volumetric flow rate of air in the tube by the following expression:

with a = 5x10-6 m, P = 5 Pa, = 18.75x10-6 N-s/m2 from Table 4.3, and L = 10-2

m. We will thus have the numerical value of Q to be:

31

Air flow

1 cm = 10-2 m

P = 5 Pa

10x10-6 m dia.

The density of the air at 20oC is = 1.2929x10-3 g/cm3 = 1.2929 Kg/m3, which leads to the mass flow rate of the air to be:

We find that the mass flow rate of the air as shown above is about two orders of magnitude smaller than the value of 0.1352x10-12 Kg/s from Example 5.8. We thus conclude that the Hagen-Poiseuille’s equation is not suitable for assessing the rarefied gas flow.

Problem 10:

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 11:

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 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 m and the bottom surface (y =60 m) 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.

32

silicon

SiO2

1000 m

x

y

20o C

20 m

4 m

2 m thick copper filmHeat supply

Stagnant air at 20oC

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:

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

at the left end, and

thermally insulated at the right end, and

for compatibility at SiO2 and silicon interface, and

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

electric resistance of copper film, (), 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:

at the left end, and

thermally insulated at the right end, and

for compatibility at the silicon and SiO2 interface,

and

for thermally insulated at the bottom surface.

33

Problem 12:

Equation (5.45) provides:

which leads to the following special cases:

(A) When h = 0 :

Equation (5.45) becomes:

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 .

(B) When h :

By dividing each term in Eq. (5.45) by h and then letting h , we will have:

which is the prescribed surface temperature boundary condition as shown in Eq. (5.43).

Problem 13:

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

for 0 x 1000 m and 0 y 4 m; t > 0

and

34

for 0 x 1000 m, 4 m y 24 m, and t > 0.

The constant 1 and 2 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 14:

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:

(B) In silicon strip :

35

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

Problem 16:

The relaxation time, is defined in Eq. (5.54) as = /V, in which = average mean free path of phonons or electrons and V = average velocity of the heat carrier.

For metals such as gold, silver and copper, electrons are the principal energy (heat) carriers. Thus, the value of = 10-8 m are used for electrons and the velocity of the heat carrier, V = 106 m/s listed in Table 5.2 is used. These values lead to a “relaxation time”, to be:

= 10-8/106 = 10-14 seconds

Chapter 6Scaling Laws in Miniaturization

(P.234)

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

36

Chapter 7Materials for MEMS and Microsystems

(P.268)

Part 1. Multiple Choice

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

Part 2. Computational Problems

Problem 2:

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

in which d = the diameter of the wafer.

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

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:

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

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

37

Problem 4:

A piezoresistor has the following geometry and dimension:

The area on which the maximum normal stress exists is: A = 2 x 10 = 20 m2 = 20x10-12 m2.

From Eq. (7.8), we have:

Since the piezoresistor is attached to the cantilever beam as illustrated in Fig. 7.17, we will have: L = max = 235.36x106 Pa, and T = 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 L = 0.0244, with 44 = 138.1x10-11 Pa-1 from Table 7.8. We will thus have the piezoresistive coefficient L = 2.762x10-11 Pa-1.

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

But since the resistance of a material is defined as:

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

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

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

38

10 m

4 m2 m

max = 235.36x106 Pamax

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

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:

with the piezoelectric film being 4 m long as shown in Fig. 7.17, the output voltage is:

v = V = (6.88x107)(4x10-6) = 275.3 v

Problem 6:

If the length of the imaginary lattice is (a) in the (100) plane, then is the lattice for both diagonal (110) and inclined (111) planes in Fig. 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

39

a

a L a

0.707a 0.707a

0.707a 0.707a

L a LaL a

La

0.707a 0.707a

0.70

7a0.

707a

Problem 8:

The angle is 54.74 degree.

Problem 9:

We have been using = 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 + )/G:

Orientations G , GPa 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

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 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 m.

The corresponding radius of the spherical ink dot (r) that is ejected by the printer head is:

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

Again we will use t = 1 m as in Example 7.5. This assumption will lead to r = 6.954x10-6 m

40

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:

The corresponding strain in the piezoelectric cover is

The required voltage for 1 m thick cover is:

v/m

The required voltage for the present case for a 10 m thick cover is thus:

v = LV = (10x10-6)(9.3418x104) = 0.9342 v

Chapter 8Microsystem Fabrication Processes

(P. 305)

Part 1. Multiple Choice

1.(3); 2. (3); 3. (2); 4. (2); 5. (2); 6. (3); 7. (2); 8. (1); 9. (1); 10. (2); 11. (2); 12. (1); 13. (1); 14. (2); 15. (2); 16. (1); 17. (2); 18. (1); 19. (1); 20. (2); 21. (3); 22. (2); 23. (2); 24. (1); 25. (3); 26. (3); 27. (3); 28. (1); 29. (3); 30. (2); 31. (3); 32. (2); 33. (1); 34. (2); 35. (3); 36. (1); 37. (2); 38. (2); 39. (2); 40. (3); 41. (3); 42. (2); 43. (1); 44. (3); 45. (1); 46. (3); 47. (3); 48. (3); 49. (3); 50. (1)

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

41

We further have the maximum concentration, Nmax = 30x1018 atoms/cm3 as given in Example 8.1. It is at x = Rp = 42x10-9 m = 0.042 m.

(a) The supplied dose is:

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

(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:

Solve for xo = 0.1257 m

Problem 2:

The estimation of time required to dope a silicon substrate using ion implantation technique requires the understanding of the physics of this process. The movements of dopants in a substrate is extremely complicated as described in S.M. Sze’s book on “Semiconductor Devices”, John-Wiley & Sons, New York, 1985, pp.405-415. The method that we will propose here is a very “raw” approximation that is based on several somewhat not-too-realistic assumptions as presented below.

(i) As illustrated in the figure, we assume the mass of ions to be implanted is the concentration at a depth, i.e. N(x) atoms/cm2 at a velocity V(x).(ii) The implantation energy is E in Joules.(iii)The maximum penetration is 2Rp, which is the projected distance of maximum concentration f almost all dopants.(iv) The dopant ions penetrate into silicon substrate

without colliding with silicon atoms, so there is noscattering with changes of directions in their motion.

The silicon substrate however does offer resistance to ion movements. This resistance is

42

Ion massN atoms/cm2Implantation

Energy, E, Joules

Impact Velocity, Vo, m/s

x

Rp

Silicon Substrate

V(t)

responsible for the stopping of the moving ions after having traveled a distance 2Rp.

The deceleration of the dopant ions is estimated by the following derivation with abovehypotheses.

We have the input energy for the implantation, in Joule or N-m, in which M is the

mass of the ions in Kg and Vo is the initial impact velocity in m/s.

From the above expression, we will have the impact velocity to be:

(a)

in which with N(x) given in Eq. (8.1)

We assume that all ions stop at x = 2Rp and the following relationship holds:

in which aav is the average deceleration of the ions.

Since the final ion velocity Vf in the above expression is zero, so we have:

(b)

For the problem on hand, we have Nmax = 20x1020 atoms/cm3 at a depth of 0.2 m and Rp = 307x10-9 m and Rp = 69x10-9 m with E = 100 KeV as presented in Table 8.2.

By using Eq. (8.1), we establish the dose, Q:

Solve for Q = 1.151x1015 atoms/cm2

Now if we use the relations in Eqs. (a) and (b) for aav and Vo and apply them to the current case with:

43

(c)

in which x = 2 m = 2x10-6 m; E = 100 KeV = 1.6097x10-14 N-m.

The total mass of the ions, M with a concentration of 20x1020 atoms/cm3 is:

M = (20x1020)(10.81x1.66057x10-27) = 3.59x10-5 Kg

We need to work out the following:

and

By using Eq. (c), we may solve for the time required to have a concentration 20x1020 atoms/cm3

as:

2x10-6 = 2.995x10-5 t – 3.6515x10-4 t2

Solve the above equation for t = 0.27 s

Now, we are ready to find the depth of doping at which the concentration is asymptotically zero.

Let us first find the maximum concentration in the silicon substrate with a dose of Q = 1.151x1015 atoms/cm2. We may determine this value by following the expression presented in Step 1 in Example 8.1 as:

We assume that the concentration of the dopant is asymptotically zero at the depth xo at which the concentration N(xo) is 10-6 of Nmax, i.e. N(x0) = 6.6565x1013 atoms/cm3. We will then have the corresponding depth, xo from solving the following equation as in Eq. (8.1):

We may solve xo from the above equation to be xo = 669.7x10-7 cm or 0.67 m

44

Problem 3:

From Table 8.3, we have the constants required to evaluate the diffusion coefficient, D in Eq.(8.6) to be: a = -19.982 and b = 13.1109.

(a) From Eq. (8.6), with T’ = 1000/T = 0.8525, in which T = 900 + 273.

Hence or D = 0.0003904 m2/h, or D = 0.10844x10-6 m2/s

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

or

For x = 0.1 m and t = 1 h = 3600 s:

(c) For diffusion to take place at 800oC, we have . This will lead to the

computation of the diffusivity, D as follows:

The diffusivity, D is so low in this case that it leads to a negligible concentration at x = 0.1 mafter 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 m2/h, or 15.5x10-6 m2/s. This diffusivity will result in a concentration at x = 0.1 m and t = 1 h to be:

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

45

Temperature, oC 800 900 1000 1100Diffusivity, D, m2/h 0.000163 0.0003904 0.005676 0.05584

Concentration, N(0.1 m, 1h),atoms/cm3

0 0 3.482x1010 7.648x1010

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

Problem 4:

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

By using Table 3.4 on P. 83 and solve for the time, t = 13661.73 s, or 3.795 h

Problem 5:

Estimate the required time to achieve 1 m thick SiO2 on a silicon substrate.

The constants used in estimating the rate of oxidation in Eqs. (8.9) and (8.10) are available in the Table established in Example 8.3:

Dry oxidation Wet oxidationB/A, m/h 0.04532 0.6786B, m/h 0.006516 0.2068

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

Time for dry oxidation, h Time for wet oxidation, hEq. (8.9) for small time 22.065 1.474Eq. (8.10) for larger time 153.47 4.836

Problem 6:

46

The dilution of the hydrogen gas is = 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:

The concentration of H2 per cubic meter is:

NG = (6.022x1023)x12.1905 = 73.4111x1023 molecules/m3

(b) The density () of the carrier gas, H2 is:

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

(c) The Reynolds number (Re):

with the gas density, = 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, = 214 P = 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 ():

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

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

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with = 1024 molecules/m2-s (given) NG = 73.4111x1023 molecules/m3 (in Part (a))

Ns = 0 and = 1% = 0.01

We thus have the diffusivity, D to be:

(f) The surface reaction rate (ks):

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

(g) The deposition rate (r):

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

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

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

Problem 7:

Since the rate of the deposition is 0.2345 m/s as calculated in Problem 6, a deposition of 0.5 m 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):

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The corresponding concentration is

(b) The density:

(c) The Reynolds number:

(d) The thickness of the boundary layer:

(e) The diffusivity of the carrier gas:

(f) The surface reaction rate:

(g) The deposition rate:

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Chapter 9Overview of Micromanufacturing

(P.330)

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

Chapter 10Microsystem Design

(P.385)

1. (1); 2. (2); 3. (3); 4. (1); 5. (3); 6. (2); 7. (2); 8. (2); 9. (1); 10. (2); 11. (3); 12. (1); 13. (3); 14. (2); 15. (1); 16. (3); 17. (2); 18. (3); 19. (2); 20. (3); 21. (1); 22. (2); 23. (1); 24. (2); 25. (3); 26. (1); 27. (3); 28. (2); 29. (1); 30. (2); 31. (3); 32. (3); 33. (1); 34. (2); 35. (1); 36. (2); 37. (2); 38. (1); 39. (2); 40. (3); 41. (2); 42. (1); 43. (1); 44. (3); 45. (2); 46. (1); 47. (1); 48. (2); 49. (3); 50. (2); 51. (1); 52. (3); 53. (1); 54. (2); 55. (3); 56. (3); 57. (3); 58. (3); 59. (1); 60. (3)

Chapter 11Microsystem Packaging

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

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