MEMS EMS and and Microsy Micro syst ste ems Design, Manufacture, and Nanoscale nd Edition)* Engineering (2 Edition)*
SOLUTION MANUAL Tai-Ra Tai-Ran n Hsu , Profess or** or ** 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:
[email protected]
1
Contents
Chapter 1
Overview of MEMS MEMS and Micro systems sys tems
3
Chapter Chapter 2
Working Principl es of Microsyst ems
3
Chapter Chapter 3
Engineering Engineering Science for Microsyst ems Design Design and Fabri Fabri catio n
9
Chapter Chapter 4
Engineering Mechanics Mechanics for Microsyst ems Design Design
12
Chapter Chapter 5
Thermof Thermof luid Engineering Engineering and Microsyst ems Design Design
27
Chapter 6
Scaling Laws in Miniatu rizatio n
36
Chapter 7
Materials Materials for MEMS MEMS and Micro syst ems
36
Chapter Chapter 8
Microsyst ems Fabrication Fabrication Processes Processes
41
Chapter Chapter 9
Overview Overview of Micromanufacturi ng
50
Chapter 10 Micro syst ems Desig Desig n
51
Chapter Chapter 11 Assembly, Packaging, Packaging, and Testing Testing of Microsyst ems
52
Chapter Chapter 12 Introduct ion to Nanoscale Nanoscale Engineering Engineering
55
2
Chapter 1 Overview of MEMS and Microsystems (P. 32) 1. (b); 2. (a); 3. (b); 4. (c); 5. (a); 6. (c); 7. (c); 8. (c); 9. (b); 10. (c) 11. (a); 12. (a); 13. (b); 14. (a); 15. (b); 16. (c); 17. (c); 18. (a); 19. (a); 20. (c)
Chapter 2 Working Principles of Microsystems (P. 77) Part 1. Multiple Choice 1.(a); 2. (c); 3. (a); 4. (b); 5. (b); 6. (c); 7. (a); 8. (a); 9. (a); 10. (b); 11. (c); 12. (a); 13. (c); 14. (c); 15. (b); 16. (b); 17. (a); 18. (c); 19. (a); 20. (c); 21. (b); 22. (b); 23. (b); 24. (c); 25. (c); 26. (a); 27. (a); 28. (c); 29. (b); 30. (a); 31. (b); 32. (a); 33. (b); 34. (b); 35. (a)
Part Part 2. Descrip Descrip tion Problems Problem 2:
Transducers Piezoresistors
Advantages High sensitivity. Small sizes.
Capacitors
Simple in structure, hence less expensive to produce. Not sensitive to temperaturesuitable for operations at elevated temperatures.
Disadvantages Sensitive to temperature. Produced by doping foreign substances to silicon substrates. 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, 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 will prove that electrostatic force actuation scale down two orders of magnitude better than electromagnetic electromagnetic force for actuation. 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 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 Vo/Vi
2 0
1.75 0.033
1.50 0.071
1.00 0.167
0.75 0.227
0.50 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 rea lize that Vo/Vi → ∞ when d → 0. Problem 9:
The output voltage from a thermopile with 3 thermocouple pairs can be obtained from Eq. (2.4) as: ∆V = N β ∆T with N = 3, and ∆T = (120 – 20) + 273 = 373 K, the Seebeck coefficient, β = 38.74x10 V/ C for copper/Constantan from Table 2.3. -6
Thus, the output voltage is:
∆V = 3 x38.74 x10 −6 x373 = 0.04335 volt or 43.35 mv
4
o
Problem 10:
Actuation techniques Thermal force Shape-memory alloys
Piezoelectric
Electrostatic force
Advantages Simple in structure.
Disadvantages Response may not be instant due to thermal inertia of the material. Actuation is more precise. Same problem as in the thermal actuation case. It is functional only with a thermal source. Simple and it is less costly Cannot maintain the actuated movement to produce. Usually for sustained period of time due to provides precise actuation. overheating. Takes up the least amount Low in magnitudes. of space. Actuation is instant.
Problem 11:
We assume that there is no friction between the electrodes and the dielectric Pyrex glass. By following the geometry and the dimensions given in Example 2.1 on P. 45 with: L = W = 800x10 m; εo = 8.85x10 F/m; εr = 4.7 (Table 2.2); V = 70 v; and d = 2x10 m -6
-9
-6
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
Spring constant k
Spring constant k
Moving electrodes
Moving electrodes
Fixed electrodes
5
The required traveling distance of the moving electrodes is δ = 10x10 m, which corresponds to the spring force with a spring constant, k = 0.05 N/m: -6
F = k δ = 0.05x10x10 = 0.5x10 N -6
-6
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: -6
f = F/10 = 0.05x10 N From Eq. (2.11),
F L =
1 ε r ε o W
2
V
2 d -12 2 -6 -6 with FL = f = 0.05x10 N; εr = 1.0; εo = 8.85x10 C/N-m ; W = 5x10 m; d = 2x10 m: -6
0.05 x10
−6
=
1 1 x8.85 x10 −12 x5 x10 −6 2
2 x10
−6
2
V
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. “A” Flexible “Drive Arm”
“A” Rigidly held “Closure Arm”
Req’d tip movement: 5 µm
Width of electrodes, W = 5µm
Gap, d = 2 µm 150 µm
10 µm
Req’d tip movement: 5 µm 5 µm
8 µm
10 µm
300 µm
View “A-A”
We will first find the necessary voltage supply to the electrodes on both drive arms to provide a 5 µm movements at the free end of each of these two arms. We will treat the Drive arms as two elastic cantilever beams and the generated electrostatic forces by the electrodes as concentrated forces acting at the distance that equals to a distance b = 150 + 0.5x8 = 154 µm away from the support-end as illustrated below:
6
F
b = 154 µm
δmax = 5 µm L = 300 µm
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:
δ = max
F b
2
6 EI
(3 L − b ) 11
with the Young’s modulus, E = 1.9x10 Pa from Table 7.3 for silicon, and the area moment of -22 4 inertia, I = 4.17x10 m (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:
5 x10
−6
=
(
) (3 x300 x10 − − 154 x10 − ) 6 x(1.9 x10 )(4.17 x10 − )
F 154 x10
−6
2
6
11
6
22
Solve for the equivalent applied force, F = 0.1343x10-3 N We are now ready to estimate the voltage supply to the electrodes to generate the above actuation force. There are 5 pairs of electrodes for each arm. From Equation (2.11), the electrostatic force is: 1 ε r ε o W 2
F L =
2
d
V
with εr = 1.0; εo = 8.85x10 C/N-m ; W = 5x10 m; and d = 2x10 m -12
2
-6
-6
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: F L
⎛ 1 ε ε W ⎞ = n⎜ r o ⎟V 2 ⎝ 2 d ⎠
We thus have: 0.1343 x10
−3
=5
1 1 x8.85 x10 −12 x5 x10 −6 2
2 x10 −6
7
2
V
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 Fc
=
2m V x Ω , 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: i
j
k
F cx i + F cy j + F cz k = 2m V x
V y
V z
Ω x Ω y Ω z 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: F cx i + F cy j + F cz k =
[
2m (V y Ω z
− V z Ω y )i + (V x Ω z −V z Ω x ) j + (V x Ω y −V y Ω x )k ]
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 -6 mg = 10 kg and Vx = 2 (maximum amplitude of vibration)/period of vibration. -6
We get Vx = 2 x (100 x 10 ) m/0.001 s = 0.2 m/s The corresponding Coriolis force with an angular displacement Ωz = + 0.01 rad in counterclockwise direction is:
8
-6
-9
Fcy = -2x10 x 0.2 x 0,01 = -4x10 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: -9
-11
δy = Fcy/k = 4x10 /100 = 4 x 10 m where the value of Fcy is obtained from Problem 2.14.
Chapter 3 Engineering Science for Micro syst ems Design and Fabri cation (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: -27
We have learned from this chapter that the mass of a proton in an atom is 1.67x10 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 -27 protons and neutrons, we may thus obtain the mass of a single hydrogen atom to be 1.67x10 -27 -27 kg, whereas (14+14)x1.67x10 = 46.76x10 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 Table 3.3.
-5
Ω-cm, the same as that of platinum as indicated in
9
Problem 4:
The negative signs in these equations mean that the concentration of the diffused substance decreases as the distance of diffusion into the base substance increases. Problem 5:
Doping process allows engineers to humanly manipulate the electric resistivity of semiconductors by creating localized positive or neg ative 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 A faster process at room temperature. Easier to control the diffusion zone (see Figure 8.6)
Ion implantation Diffusion
Disadvantage Hard to control (see Figure 8.4) 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 o o o optimum diffusion temperatures are ≈ 1220 C, ≈1200 C and ≈1330 C for As, P and B 20 20 respectively. The corresponding solid solubility of these materials are: 12x10 for As, 5.5x10 20 3 for P and 7.5x10 for B with unit of atoms/cm . Problem 8:
Equation (3.5) is used for the solution of this problem: ⎛ x ⎞ C ( x, t ) = C s erfc⎜⎜ ⎟⎟ ⎝ 2 Dt ⎠ The coefficient Cs in the above equation is maximum possible input concentration. Inthis case, o 20 we have the solubility of phosphorus at the given diffusion temperature of 1260 C at 5.45x10 3 atoms/cm 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 = 1/2 1/2 0.5, 2, and 3 hrs can be computed from the above equation with (D) = 1.05 µm/(h) from Figure 3.12. The equation that we will use to compute the distribution of phosphorus concentration at the above 3 selected time instants will thus take the form:
⎛ x ⎞ ⎛ 0.4762 x ⎞ ⎟⎟ = 5.45 x10 20 erfc⎜⎜ ⎟⎟ t ⎠ ⎝ 2 x1.05 t ⎠ ⎝
C ( x, t ) = 5.45 x10 erfc⎜⎜ 20
10
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) 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
t = 0.5 hr 5.45 4.69 3.82 2.86 2.45 1.91 1.42 0.95 0.68 0.53 0.33
t = 2 hr 5.45 5.07 4.69 4.2 3.82 3.49 3.11 2.73 2.45 2.18 1.96
t = 3 hr 5.45 5.18 4.77 4.52 4.09 3.87 3.49 3.16 2.89 2.73 2.4
Graphical representation of the distribution of phosphorus concentration in the silicon substrate at various times is presented below:
Diffusion o f Phos pho rus into Silicon 0.5 hr
2 hrs
3 hrs
6 ) m5 f / c o s n i m4 o t i a t ( 3
a r t s 3 n u r e o c n h 2 o p s C o h 1 P
0 0
0.5
1
1.5
2
2.5
Depth in Silicon (micrometers)
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.
11
Problem 9: o
We will first assume that diffusion process takes place at the same temperature of 1250 C as in 20 3 Example 3.1. The corresponding solubility for boron is Cs = 7x10 atoms/cm 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 -3 obtained the corresponding concentration of boron at resistivity of 10 Ω-cm from Figure 3.8 to 20 3 be C = 10 atoms/cm . Thus by using Equation (3.5), we have the following relation for tf :
⎛
(
C 2µ m, t f
⎞ ⎟ ⎟ ⎝ 2 D t f ⎠
) = C erfc⎜⎜ s
2
in which, tf has a unit of hr. 1/2
o
The diffusivity, (D) in the above expression for boron in silicon at 1250 C is 1.05 µm/(h) from Figure 3.12.
10
20
1/2
⎛ ⎞ ⎞ 1 ⎜ ⎟ = 7 x10 20 erfc⎛ ⎜ 0.4762 ⎟ = 7 x10 erfc⎜ ⎜ t f ⎟ ⎜ 2 x1.05 t f ⎟⎟ ⎝ ⎠ ⎝ ⎠ 20
⎛ 0.4762 ⎞ ⎟= From which we have erfc⎜ ⎜ t f ⎟ ⎝ ⎠
1 7
=
0. 1429
The corresponding value of the argument X in complementary function erfc(X) = 0.1429 from Figure 3.14 is X ≈ 1, which leads to: 0.4762 t f
≈ 1 , 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: -6
-6
6
2
We have d = 600x10 m, a = d/2 = 300x10 m, and P = 20x10 N/m . 12
11
2
The Young’s modulus, E = 0.7x10 N/m for aluminum from Table 7.3 on P. 257. The Poisson’s ratio, ν = 0.3 for aluminum from a metal handbook. -6
The maximum deflection of the circular diaphragm with a thickness, h = 13.887x10 m is obtained from Equation (4.7) as: wmax
=−
3W (m
2
− 1)a 2
16 π E m 2 h 3
in which W = (πa )P = 3.14(300x10 ) (20x10 ) = 5.652 N, m = 1/ν = 1/0.3 = 3.33. 2
-6 2
6
One may thus calculate the maximum deflection, wmax = -1.4745x10 m, or 147.45 µm. -4
Problem 2:
The geometry of the die as illustrated below: 3 mm
Applied Pressure, P=75 MPa
3 mm
h
H
a
∆L
L = 3000µm
∆L
Cross-section of the Die
Plan View
We will then designate the dimension of the thickness of the die h and the size of the foot print ∆L as shown below: h
H = 500 µm
∆L = 250 µm
o
7 4 5 4.
∆a
Edge Length of Diaphragm, a
L =3000 µm Detail Dimensions of Foot Print
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.
13
It is clear from the diagram that
∆a =
H − h o
tan(54.74 )
( H − h)
=
∆a
tan(54.74 o ) , in which H = 500 µm. We thus have:
= 353.6 − 0.707h
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: 0.308 p[(1792.78 + 1.414h) x10 −6 ] 2
h
=
or p
= 350 x10 6
2
350 x10 6 h2
0.308 x10
−12
(1792.78 + 1.414h)
2
=
1.1364 x10 21 h2 (1792.78 + 1.414h)
2
2
N / m or Pa
We may tabulate the results of the diaphragm thickness vs. applied pressure as follows: Diaphragm thickness (µm) Maximum pressure (MPa)
500 88.47
300 31.85
200 14.16
100 3.54
50 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: F
k eq =
where
δ
F = applied load to the beam δ = deflection of the beam under the load
Case 1 Simply-supported beams: F
L
From the strength of materials theory, we have the deflection of the beam under the concentrated force, F to be: δ
=
3
FL
48 EI
14
from which we may obtain the equivalent spring constant, k eq to be: F 48 EI
k eq =
δ
=
3
L
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:
L The deflection of the beam under the concentrated force, F is: F L3 δ = 192 EI Hence the equivalent spring constant is: F 192 EI
k eq =
δ
=
L
3
Case 3 Cantilever beams (Extra, not requested in Problem 3):
F
L
The deflection of the beam at the free-end is: F L3 δ = 3 EI which leads to the following expression for the equivalent spring constant:
k eq =
F L
=
3 EI
L
3
Problem 4: -3 The mass, m attached to the beam is 5 g, or 5x10 kg; The equivalent beam spring constant k eq in the arrangement shown below, and from Case 2 of Problem 3 is 18240 N/m
15
20x10-6 m
y (a) From Eq. (4.16), we have the equivalent natural frequency,
k eq
ω n =
m
=
18240 5 x10
−3
= 1910
rad / s
(b) The equivalent motion of the mass in the y-direction, according to Eq. (4.14) is: 2 y (t ) d + k eq y (t ) = 0 m 2 d t -6 with y(0) = 5x10 m, and y’(0) = 0. Substitute these values into the above equation:
d
2
y( t )
dt
2
+
3.648x10 6 y( t )
= 0
(a)
The solution of the differential equation is: y (t )
= C 1 cos1910t + C 2 sin 1910t
(b)
-6
-6
From the condition y(0) = 5x10 , we get C1 = 5x10 From y’(0) = 0, we have C2 = 0 Thus, the amplitude of vibration, y(t) is: y (t )
= 5 x10 −6 cos1910t
(c)
The maximum amplitude of vibration is the coefficient of the cosine function in the solution in -6 Eq. (c), or ymax = 5x10 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: d 2 y( t ) dt
2
+
ω y(t) = Fm cos ωt 2
o
o
16
-6
with the specified conditions: y(0) = 5x10 m and y’(0) = 0. The proper differential equation as derived from Problem 4 is: 2 d y (t ) + 3.648 x10 6 y (t ) = 5 cos1910t 2 3 d t 5 x10 −
in which the natural frequency of the beam spring system,
ω
o
= 3.648x10 6 = 1910 rad / s
The solution of the above differential equation is: t = c1 cos ωo t + c2 sin ωo t + Fo 2m
y( t )
ω
with
sin ωt
o
ω = ωo = 1910 rad/s at the resonant vibration situation and Fo = 5 N -6
-6
Use the first condition, i.e. y(0) = 5x10 m will result in c 1 = 5x10 . The other condition y’(0) will result in c2 = 0. Thus the solution for the amplitude of the vibration mass being:
= 5 x10 −6 cos1910t +
y (t )
0.2618 t sin 1910t -3
Now, if we let tf = the time at which the beam spring breaks at y(t f ) = 1 mm = 10 m, we will have: 10 −
3
= 5 x10 −6 cos1910t f +
0.2618 t f sin 1910 t f
We may solve for tf from the above equation, or by an approximate relationship of 10 ≈ 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. -3
Problem 6:
The beam is loaded as illustrated in Example 4.8 1 µm
Dynamic force
h µm
L=600µm Beam Cross-section
The area moment of inertia of the beam cross-section is: I
=
1 12
(10 )(10 −6
−6
h)
3
=
10 −24 12
h3
m4
17
in which h is in micrometers. The equivalent spring constant, k eq is as computed in Case 2 of Example 4.8 and 4.9 for fixedends as:
k eq =
192 EI
L
3
=
(
)(
)
192 1.9 x1011 10 −24 h3
(
12 600 x10
)
−6 3
=
0.014074 h3 N / m -11
The proof mass of the vibrating beam m = 16.1x10 kg as computed in Example 4.9. The corresponding circular frequency of the balanced force accelerometer is: 2 k eq
ω=
m
=
2 x 0.014074 h
3
16.1x10 −11
= 13220.54 h3
rad / s
From Example 4.9, the amplitude of vibration of the beam is: X(t )
= c1 cos ωt + c 2 sin ωt
in which the arbitrary constants c1 and c2 can be determined by the initial conditions: X (t ) t =0
=0
and
dX (t ) dt
= 50
km / h
= 13.8888
m/ s
t =0
We thus have: c1 = 0 and c2 = 13.8888/ω = 1.051x10 h -3
-3/2
Thus, we have the amplitude of vibration to be: X(t )
= 1.051x10 −3 h −1.5 sin 13220.54 h3 t -3
With the given condition X(tf ) = 5 mm = 5x10 m for the beam to break, mathematically as: 5x10 −3
or
1.5
= 1.051x 10 −3 h −1.5 sin 13220.54 h3 t f
h =
1.051 x 10 −3 5x10 −3
1.5
sin (13220.54 h
t) f
1.5
The approximate value of tf is when sin 13220.54 h leads to h = 0.21 µm.
18
t = 1.0 for a maximum value of h, which f
Problem 8:
We may illustrate the balanced force accelerometer system below: Beam Springs
5 µm
µ m 0 0 4
Beam Mass
10 µm
From Example 4.12 on P. 139, we get the damping coefficients for the balanced force accelerometer to be: -12
cair = 2.625x10 N-s/m for air as the damping fluid, and -10 csi = 1.036x10 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 3 3 3 g/cm or 2.3x10 kg/m 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 -6
-4
-4
-14
3
v = bBL b = (1x10 ) (10 ) (7x10 ) = 7x10 m 3
-14
-11
Consequently, the beam has a mass, m = (2.3x10 )(7x10 )= 16.1x10
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 = 2k eq 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 (λ - ω ) as described in Eq. 2 2 (4.20a), (4.20b) or (4.20c). We will thus need to compute both λ and ω first in order to select which of the above solutions for the case under consideration. 2
2
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:
k eq =
192 EI
L
3
11
2
with E = 1.9x10 N/m (Table 7.3) and I
=
1 12
(5 x10 − )(10 x10 − ) 6
19
6 3
=
4.167 x10 − 22 m4
Thus, k eq
=
)(
(400 x10 ) −6
2 k eq
ω=
and
(
192 1.9 x1011 4.167 x10 −22
2 x 237.52
=
m
16.1x10
−11
which leads to ω = 2.95x10 2
12
3
) = 237.52
= 1.7177 x10 6 2
N / m
rad / s
2
rad /s
The damping parameters:
λ λ
=c = air
air
−11
2m
2 x 16.1x10
= c =
1.036 x10 −10
si
si
2.625x10 −12
2m
= 8.152 x 10 −3
=
2 x16.1x10 −11
for air
0.32174 for silicone oil
from which we have:
λair − ω = (8.152 x10 −3 ) − 2.95x1012 < 0 2
2
2
λsi − ω = (0.32174)2 − 2.95x1012 < 2
2
0
The values of (λ - ω ) 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. 2
2
Problem 9:
The balanced-force accelerometer is illustrated in Fig. 4.25, and also as below: L = 700 µm
h = 5 µm Beam Mass, m
b = 1 µm Dimensions of the Beam Mass
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
20
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: Dimensions of the Beam Springs 600µm
5
µm
1 µm
We may compute the area moment of inertia of the beam springs to be:
( 10 − )(5 x 10 − ) I =
6 3
6
12
= 10.4167 x 10 − 24 m 4
The equivalent spring constant for beam springs with fixed-ends is k eq
=
192 EI 3
L
=
192(1.9 x 1011 )(10.4167 x 10 −24 )
(600 x10 )
3
−6
= 1.76 N / m as in Case 2 of Example 4.9.
&& = − 22.22 m / 2 from Since the maximum deceleration of the car in the present case is X s Example 4.14. By neglecting the mass of the beam springs, we may express the dynamic force associated with the moving beam mass as:
F (t )
&& (t ) = m X
The mass of the beam mass, m = 16.1 x 10-11 kg as computed in Example 4.9. The force acting 2 on both beam springs at the time of deceleration of –22.22 m/s is: F
=
&& mX
=
(16.1x10 −
11
kg )(22.22 m / s 2 ) = 35.77 x10 −10 kg − m / s 2
= 35.77 x10 −10
N
The induced deflection of the beam springs by the above dynamic force of the magnitude is: -10
-10
P = F/2 = 35,77x10 /2 = 17.885x10 N -6
-24
4
11
2
with L = 600x10 m, I = 10.42x10 m (from Example 4.9), E = 1.9x10 N/m (Table 7.3), and k eq = 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:
21
δ max
( 17.885 x 10 − )(600 x 10 − ) = = = 1.0166 x10 − − 192EI 192(1.9 x 10 )(10.4167 x 10 ) 6 3
10
PL3
15
11
24
m
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: SiO2 strip 5 µm
1000µm
5 µm
Silicon strip
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: 2h ρ = 3(α 2 − α 1)T 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: C ρ = T 2h
in which the constant C =
3(α 2 − α 1)
=
2 x10 x10 −6 3(2.33 − 0.5) x10 −6
= 3.643 -6 o
where α2 = coefficient of thermal expansion of silicon = 2.33 x 10 / C, and α1 = coefficient of -6 o thermal expansion of SiO2 = 0.5 x 10 / C, as obtained from Table 7.3. From Example 4.17, we have the movement of the free-end, δ to be: δ
≈ ρ (1 − cosθ )
where
θ
360 L 2πρ
-6
with L = 1000x10 m
360 x1000 x10 −
6
5.7325 x10 −
2
and the movement of the free-end of the bi-layer = 6.28 ρ ρ beam can be obtained from the following expression: Hence θ
=
=
22
o
δ = 3.643(1 – cosθ)/T in which T is temperature in C. We may tabulate the values of the temperatures vs. the movement of the free-end of the beam actuator as follows: o
T ( C) 10 20 30 40 50
ρ = C/T (m)
θ(o)
δ = ρ(1 - cosθ) (µm)
0.3643 0.1822 0.1214 0.0911 0.0729
0.1574 0.3147 0.4720 0.6294 0.7880
1.373 2.747 4.120 5.496 6.890
The plot of the above tabulated data is shown below. Movement of Free-end of a Contilever Beam
8
7
6
) r e t e 5 m o r c i m4 ( t n e m e 3 v o M 2
1
0 0
10
20
30
40
50
60
o
Temperature ( C)
Problem 11:
The beam has the following geometry and dimensions: y b = 5 µm 2h = 10 µm
Silicon
x
0
Beam
L = 1000 µm
The temperature variation in the beam is: 6
T(z) = 2x10 z + 30 -6
o
C -6
o
At the top face, i.e. z = 5x10 m, we have T(5x10 ) = 40 C, and
23
H = 10 µm
-6
-6
o
at the bottom face at z = -5x10 m, the temperature is T(-5x10 ) = 20 C. Material properties of the silicon beam are given in Example 4.19 on P. 161: Mass density, ρ = 2.3 g/cm ; Specific heats, c = 0.7 J/g- C; Thermal conductivity, k = 1.57 J/cmo -6 o 11 2 C-s; Coefficient of thermal expansion, α = 2.33x10 / C; Young’s modulus, E = 1.9x10 N/m ; Poisson’s ratio, ν = 0.25. 3
o
We will first compute the thermal force, NT and the thermal moment, MT from the respective Equations (4.55a) and (4.55b) as:
N T = (2.33 x10
−6
5 x106
)(1.9 x10 )∫− 11
11 −6 M T = (2.33 x10 )(1.9 x10 )∫
(
)
(
)
2 x10 6 z + 30 dz = 132.81 N 6 − 5 x10
5 x106
2 x10 6 z + 30 zdz = 73.78 x10 −6 N − m 6 − −5 x10 -11
2
-22
4
From Example 4.19, we have A = 5x10 m and I = 4.167x10 m . From Equation (4.56), we have the thermal stress along the x-direction to be: b N T z (b M T ) + σ xx ( x, z ) = − α ET ( z ) + A I -6 with σxx,max occurs at z = 5x10 m. Thus, σxx,max = σ(x,5x10 ) = -2600 Pa -6
We will compute the associate thermal strains from Equations (4.57a) and (4.57b) with -6 maximum values occurring at z = 5x10 m: 1 ⎡ b N T z ⎤ + (b M T )⎥ ε xx ( x, z ) = ⎢ E ⎣ A I ⎦ which leads to εxx,max = εxx(x,5x10 ) = 0.00932% ν ⎡ b N T z 1 + ν ⎞ ⎤ + (b M T )⎥ + ⎛ ⎜ ⎟α T ( z ) ε zz ( x, z ) = − ⎢ E ⎣ A I E ⎝ ⎠ ⎦ -6
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: x ⎡ b N T z ⎤ + u ( x, z ) = b ( ) M T ⎥ E ⎢⎣ A I ⎦ -6
-6
with umax at x = 500x10 m and z = 5x10 m: -6
-6
umax = u(500x10 , 5,10 ) = 0.0466
µm
24
The deflection of the beam in the z-direction, w(x,z) is obtained from Equation (4.58b): w( x, z ) = −
b M T
2 EI
x − 2
ν ⎡ b N T
⎢
⎤ ⎛ 1 + ν ⎞ zT ( z )dz z + z (b M T )⎥ + α ⎜ ⎟ ∫0 2
2 I
E ⎣ A
⎝
⎦
-6
E
⎠
-6
with wmax occurs at x = 500x10 m and z = 5x10 m, we have: wmax = w(500x10 , 5x10 ) = -0.582 µm -6
-6
Problem 12: -6
The width of the beam has been increased to 100x10 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. 6
T(z) = 2.1x10 z + 28.8
in degree C -6
We realize that the thermal force, NT = 127.5 N and the thermal moment, MT = 77.4725x10 Nm 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: 5 6 6 12 σ xx = σ yy = − 5.9027 x10 (2.1 x10 z + 28.8) + 17 x10 + 1.2396 x10 z
The associated thermal strains are obtained from Equation (4.53a): −5 ε xx = ε yy = 6.711 x10 + 4.893 z 6 12 6 −11 −6 ε zz = − 0.3509 x10 (12.75 x10 + 0.9297 x10 z ) + 3.8833 x10 (2.1 x10 z + 28.8)
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): u ( z )
=
v( z )
=
x 11
(12.75 x10
11
(12.75 x10
1.9 x10 y 1.9 x10
and w( x, y, z )
6
+ 0.9297 x1012 z )
6
+ 0.9297 x1012 z )
= − 2.4465 x2 + y 2 + 0.7018 x10 −11 (96.84 − 6.375 x10 6 z − 0.2324 x1012 z 2 ) -6
-6
The maximum values of stress, strains and displacements occur at: x = 500x10 m, y = 50x10 -6 m and z = 5x10 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% 25
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 bea m 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 = -2 -3 -4 -4 10 m; b = 5 mm = 5x10 m; and the width, B = 24x10 m, and c = 100 µm = 10 m. Pcr = 40 MN
B = 2400 µm b = 5000 µm
s = 10000 µm 6
2
The critical load, Pcr that breaks the specimen is 40x10 N/m . 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: 2
3
⎛ c ⎞ ⎛ c ⎞ ⎛ c ⎞ ⎛ c ⎞ ⎛ c ⎞ F ⎜ ⎟ = 1.09 −1.735⎜ ⎟ + 8.2⎜ ⎟ − 14.18⎜ ⎟ + 14.57⎜ ⎟ ⎝ b ⎠ ⎝ b ⎠ ⎝ b ⎠ ⎝ b ⎠ ⎝ b ⎠
4
with c/b = 0.02 Hence F(c/b) = 1.0586 Equation (4.64) is used to compute fracture toughness: ⎛ c ⎞ K c = σ c π c F ⎜ ⎟ ⎝ b ⎠ The σc in the above expression is obtained from the bending stress in a “solid” beam subjected to three-point bending as follows:
σ c = where c
M c I
= b/2 =
2.5 x10
The bending moment, M
−3
m
and I =
s ⎞ = Pcr ⎛ ⎜ ⎟= ⎝ 2 ⎠
Bb
3
12
=
250 x10 −13 m 4
2 x10 N − m 5
Thus, we have the critical stress corresponding to Pcr to be:
26
( 2 x10 − )(2.5 x10 − ) = = 2000 5
σ c
3
Pa
250 x10 −13
which leads to the fracture toughness, K c to be: K c
=
2000 3.14 x10 −4 (1.0586 )
= 37.52
Pa m
Problem 15:
For the width of the specimen, B to be increased to 100x240 µm, we will have I = 2.5x10 m . This new value will change the critical stress according to the following expression: -9
( 2 x10 − )(2.5 x10 − ) = = 20 5
σ c
4
3
2.5 x10 −9
Pa
K c = 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 K c for design purpose must be independent of the specimen geometry and size. A great deal of research effort is needed in the measurements of K c 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); 2 3. (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: -6
-6
We have d1 = 500x10 m and d2 = 50x10 m -6
3
-14
3
The flow rate is Q = 1x10 cm /min = 1.67x10 m /s
27
A1 =
π(500x10-6)2/4
= 19.64x10
m
A2 =
π(50x10-6)2/4 = 19.64x10-10
m
V 1
V 2
=
=
Q A1 Q A2
1.67 x10
=
2
2
−14
19.64 x10 −8
=
-8
1.67 x10 −14 19.64 x10 −10
= 0.085 µ m / s
= 8.503 µ m / s
Problem 3: m µ 0 3 0 = W , h t d i W
The opening of the valve may be illustrated as follows: Valve Plate Thickness: 4 µm
0 0 µ m L = 4
, V e o c i t y l e v E x i t
o 1 5
Valve Opening, H
Fluid Flow o
o
-6
o
o
-6
(a) The opening of the valve is H = (L sin15 )cos15 = 400x10 x sin15 cos15 = 100x10 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:
M x1 V e = ρ Ae in which
ρ = mass density of the H2 gas = 0.0826 kg/m3 (from Example 5.2) -6
Mx1 = the mass flow rate in the direction of Ve = 15.3x10 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 -6 300x10 m. -6
-6
-8
2
Hence Ae = (100x10 )(300x10 ) = 3x10 m . Thus, the exit velocity is:
V e =
15.3 x10 −6 0.0826 x3 x10 −8
= 6174.33
m/ s
NOTE: This exit velocity is unrealistically high for a microvalve. This high value on the -8 2 velocity is a result of extremely small opening at the exit (A e = 3x10 m ), and large mass flow -6 rate (Mx1 = 15.3x10 kg/s)
28
(c) The volumetric flow at the exit can be computed as follows: 15.3 x10 −6 M x1 Q = = = 185.23 x10 −6 m3 / s ρ
0.0826
3
or Q = 11,113.8 cm /min, which is significantly smaller than the intended design capacity of 3 30,000 cm /min. Problem 4:
The uniformly distributed load that is required to bend a cantilever beam (plate) such as the -6 closure plate with a free-end displacement of H = 100x10 m in Problem 3 (see illustration below) can be obtained by the following expression: L = 400 µ m
δ = max
w L
4
8 EI
W N/m
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: 1 6 6 3 = 16 x10 −22 m4 I = 300 x10 − 4 x10 − 12
(
)(
)
with a Young’s modulus, E = 1.9x10 N/m from Table 7.3 for silicon, and δmax = H = 100x10 m as shown in the figure in Problem 3, we may determine the required load, W from the following relation: 11
(100 x10 ) = −6
(
w 400 x10
8 x (1.9 x10
11
−6
2
-6
)
4
)(16 x10 − ) 22
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 N. However, there is a fluid-induced force acting on the plate too. This force is Fy = -8 40x10 N as computed from Example 5.2. The net required electrostatic force is thus equal to -6 -8 the difference of the above two forces, or Fd = 3800x10 – 40x10 ≈ 3.8 mN. -6
The corresponding required voltage to generate such force can be obtained by using Eq. (2.8) to give:
29
V = 2
2 d 2 F d
ε r ε o WL
=
(
) (3.8 x10 − ) = 29.443 x10 − − )(300 x10 )(350 x10 )
2 x 6 x10 −
(
1 x 8.85 x10
−12
6 2
3
6
4
6
which leads to V = 542 v – a rather high voltage.
Problem 5: m 0. 1 L =
The problem is illustrated below:
V 2
m/ s 0 0 µ 6 =
m 0 µ = 5
V 1
d 2 µ m 0 0 1 =
d 1
d a v e
Assume that the average velocity of the fluid is computed at the cross-sectional area of the o conduit at its mid-section is used. The 30 inclination is neglected. From Example 5.3 on P. 194, -3
V2 = 2.4x10 m/s. We may calculate the Vave in the mid-cross section to be: -3
Vave = 0.5 (V1 + V2) = 1.5x10 m/s -6
-6
Let dave = 0.5 (d1 + d2) = 75x10 m, which leads to the radius at the mid-section, aave = 37.5x10 m. The pressure drop, ∆P in the conduit using the Hagen-Poiseuille equation in Equation (5.17) is: 8 LQ ∆P = π a 4 where µ = dynamic viscosity of the fluid = 1199.87x10 N-s/m (Table 4.3 on P. 138 for alcohol) -6
2
L = length of the conduit = 0.1 m, and Q
= AaveV ave
from Eq. (5.6 )
= (π a2ave )(V ave ) = 3.14(37.5 x10 −6 ) (1.5 x10 −3 ) = 66.2343 x10 −13 m3 / s 2
We will have the approximate pressure drop, ∆P to be:
30
∆P =
8 x1199.87 x10 −6 x 0.1 x 66.2343 x10 −13
(
3.14 37.5 x10
−6
)
4
= 1024
Pa
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: 1000 µm
The radius of the capillary tube is -6 -5 a = d/2 = (20x10 )/2 = 10 m
20 µ m dia.
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: ρ DV Re = = 1000 µ with the density of water, ρ = 1000 kg/m , and the inside diameter of the tube, D = 20x10 m, and -6 2 the dynamic viscosity, µ = 1001.65x10 N-s/m from Table 4.3, we will have, from the above expression, the velocities of flow to be: 3
-6
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: π 6 2 10 3 Q = AV = 20 x10 − x5 = 15.708 x10 − m /s 4 The corresponding pressure drop, ∆P by the Hagen-Poiseuilli’s equation is:
(
∆P =
)
8 x(1001.65 x10 −6 )(10 −3 )(15.708 x10 −10 )
⎛ 20 x10 ⎝ 2
3.14⎜⎜
−6
⎞ ⎟⎟ ⎠
4
=
400,000 Pa
31
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 = -16 3 31.4x10 m /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:
∆P =
(
)(
)(
8 x 1001.65 x10 − 10 − 31.4 x10 − 6
3
⎛ 20 x10
3.14 x⎜⎜
⎝
16
−6 ⎞ 4
2
) = 0.8013
Pa
⎟⎟ ⎠
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: 3γ ∆Pst = a in which γ = surface tension coefficient of water = 0.07 3 N/m as in Eq. (5.23) and a = the radius of the tube. We will thus have the required pressure:
∆ Pst =
3 x 0.073 10 −5
= 21,900 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 -5 tube has a diameter of 20 µm, which gives a radius, a = 10 m. 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: h
=
2γ cosθ
=
2 x0.073 x cos(0)
wa 9810 x10 −5 the capillary tube in only 1 mm.
= 1.488 m, which is unrealistically high, as the total length of
32
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: v
h r ∂T (r , t ) + T (r , t ) r =r = r ∂n r =r k r
r
s
r r
s
h k
T f
which leads to the following special cases: (A) When h = 0: Equation (5.45) becomes: r
∂T (r , t ) = r ∂n r r =r sr
0
which is equivalent to have qin and qout = 0 in Eq. (5.44a) and (5.44b) respectively. This means r r that no heat is allowed to flow across the boundary at r = r s . (B) When h → ∞: By dividing each term in Eq. (5.45) by h and then letting h → ∞, we will have: r
T (r , t ) r r =r r
s
= T f
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 SiO 2/Si bilayer strip, but does not impose any mechanical constraint on the overall structure. One end of o the strip is maintained at 20 C 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
33
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 2 0oC. 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.
Heat supply
2 µm thick copper film
Si O2 C o 0 2
4 µm
silicon
x
20 µm
1000 µm
y
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 d istributions, T1(x,y,t) and T2(x,y,t) with the following initial and boundary c onditions: The initial conditions: o T 1 ( x, y, t ) t =0 = T 2 [ x, y, t ] t =0 = 20 C
The boundary conditions for the temperature distribution in the SiO2 layer are: o T 1 ( x, y, t ) x =0 = 20 C at the left end, and
∂ T 1 ( x, y , t ) = 0 thermally insulated at the right end, and ∂ x x =1000 µ m T 1 ( x, y, t ) y =4 µ m = T 2 ( x, y, t ) y =4 µ m for compatibility at SiO2 and silicon interface, and 2 ∂ T 1 ( x, y, t ) R i =− ∂ y k 1 y =0
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 k 1 = o thermal conductivity of SiO2 (W/m- C) The boundary conditions for temperature distribution in the silicon strip are:
34
o T 2 ( x, y, t ) x =0 = 20 C at the left end, and
∂ T 2 ( x, y , t ) = 0 thermally insulated at the right end, and ∂ x x =1000 µ m
∂ T 2 ( x, y, t ) ∂ ( x, y, t ) = T 1 for compatibility at the silicon and SiO2 interface, and ∂ y ∂ y y = 4 µ m y = 4 µ m ∂ T 2 ( x, y, t ) = ∂ y y = 24 µ m
0
for thermally insulated at the bottom surface.
Problem 11:
The respective heat conduction equations for SiO2 and silicon stripes are:
∂2 T 1 ( x, y, t ) + ∂2 T 1 ( x, y, t ) = ∂ x2
∂ y 2
∂ T 1 ( x, y, t ) ∂t α 1 1
for 0 ≤ x ≤ 1000 µm and 0 ≤y ≤ 4 µm; t > 0 and
∂ 2 T 2 ( x, y, t ) ∂ 2 T 2 ( x, y, t ) 1 ∂ T 2 ( x, y, t ) + = ∂t ∂ x2 ∂ y 2 α 2 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: o T 1 ( x, y, t ) x =0 = 20 C
35
∂ T 1 ( x, y , t ) h h + T 1 ( x, y , t ) x =1000 µ m = (20 ) ∂ x k 1 k 1 x =1000 µ m 2 ∂ T 1 ( x, y, t ) R i =− ∂ y k 1 y =0
T 1 ( x, y, t ) y =4 µ m = T 2 ( x, y, t ) y =4 µ m
∂ T 1 ( x, y, t ) ∂ ( x, y, t ) = T 2 ∂ y ∂y y = 4 µ m y = 4 µ m (B) In silicon strip: o T 2 ( x, y, t ) x =0 = 20 C
∂ T 2 ( x, y , t ) h h + T 2 ( x, y, t ) = (20 ) ∂ x k 2 k 2 x =1000 µ m x =1000 µ m ∂ T 2 ( x, y , t ) h h + T 2 ( x, y , t ) = (20 ) ∂ y k 2 k 2 y = 24 µ m y = 24 µ m where k 1 and k 2 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 f or MEMS and Microsys tems (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.
36
(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: A
=
π 4
2 d in which d = the diameter of the wafer.
The ratio of plane areas of wafers with 300 mm and 200 mm diameters is: 2
A2 = Area of 300 mm wafer = ⎛ 300 ⎞ = 2.25 ⎜ ⎟ A1 Area of 200 mm wafer ⎝ 200 ⎠ 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:
⎛ V ⎞ ⎛ 0.001 ⎞ N = ⎜ ⎟ n = ⎜ ⎟ ⎝ v ⎠ ⎝ 0.543 x10 −9 ⎠
3 20 3 x18 = 1.12 x10 atoms / mm
in which v = the volume of a single silicon crystal. Likewise, the number of atoms per cubic micrometer of silicon is:
⎛ 10 −6 ⎞ ⎟ N = ⎜⎜ −9 ⎟ ⎝ 0.543 x10 ⎠
3
x18
= 1.12 x1011
atoms / µ m
3
Problem 4:
A piezoresistor has the following geometry and dimension:
max
2 µm
µ m 1 0
max =
4 µm
37
235.36x106 Pa
The area on which the maximum normal stress exists is: A = 2 x 10 = 20 µm = 20x10 2
-12
2
m.
From Equation (7.8), we have:
∆ R R
= π L σ L + π T σ T
Since the piezoresistor is attached to the cantilever beam as illustrated in Figure 7.18, we will 6 have: σL = σmax = 235.36x10 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 -11 -1 case. We will have the coefficient πL = 0.02π44, with π44 = 138.1x10 Pa from Table 7.8. -11 -1 We will thus have the piezoresistive coefficient πL = 2.762x10 Pa . The corresponding rate of the change of electric resistance by the piezoresistor is:
∆ R R
= π L σ L =
6
(2.762 x10 −11 ) x( 235.36 x10 )
=
6.5 x10 −3
But since the resistance of a material is defined as:
=
ρ L
in which ρ is the resistivity of the material, which is a p-type piezoresistor. We find A -3 4.5 the values of ρ vary from 10 to 10 Ω-cm from Table 7.1. We will adopt a value of ρ = 2 7.8 Ω-cm = 7.8x10 Ω-m from Table 7.8. R
-6
-12
2
Thus, with L = 4x10 m and A = 20x10 m , the resistance R in the piezoresistor is:
( 7.8 x10 )(4 x10 − ) = 1.56 x10 R = 2
20 x10
6
8
−12
Ω
or 156 M Ω
The net change of resistance in the piezoresistor at 235.36 MPa stress is:
∆ R =
(6.5 x10 − )(1.56 x10 ) = 10.14 x10 3
8
5
Ω
or 1.014 M Ω
Problem 5: -12
The piezoelectric coefficient, d, for PVDF polymer films can be found to be 18x10 m/v from -5 Table 7.14. Consequently, the induced voltage by the induced strain of 123.87x10 m/m from Example 7.4 is:
38
V
ε
=
d
123.87 x10 −5 ε max = = = 6.88 x10 7 v / m −12 d
18 x10
with the piezoelectric film being 4 µm long as shown in Figure 7.18, the output voltage is: 7
-6
v = Vl = (6.88x10 )(4x10 ) = 275.3 v Problem 6:
If the length of the imaginary lattice is (a) in the (100) plane, then both diagonal (110) and inclined (111) planes in Figure 7.8.
2a
= 1.414a is the lattice for
Problem 7:
The lattices for the three planes in a single silicon crystal are: a
a
0.707a
0.707a
a
L
L a
L
L a
a
0.707a
(a) The (100) Plane: L =
1 2
L
a 7 0 7 . 0
a
0.707a
0.707a
(b) The (110) Plane:
2a = 0.707a
La =
3 4
a 7 0 7 . 0
0.707a
(c) The (111) Plane
a = 0.433a
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 <100> <110> <111>
G , GPa 79.0 61.7 57.5
ν 0.25 0.25 0.25 39
Eth, GPa 197.50 154.25 143.75
Etable , GPa 129.5 168.0 186.5
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 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: 4 3
π r 3
⎞ = ⎛ ⎜ D 2 ⎟ t in which t is the thickness of ink dot on the paper. ⎝ 4 ⎠ π
Again we will use t = 1 µm as in Example 7.5. This assumption will lead to r = 6.954x10 m -6
The volume of the ink dot is computed by using the right-hand-side of the above expression -18 3 to be Vdot = 1408x10 m . The corresponding expansion of the piezoelectric cover for the ejection of ink volume, Vdot is:
W
=
4V dot π ∆
2
=
4 x1408 x10 −18
(
3.14 2000 x10 −6
)
2
=
448 x10 −12 m
The corresponding strain in the piezoelectric cover is ε =
W L
=
448 x10 −12 10 x10 −6
= 44.8 x10 −6
m/m
The required voltage for 1 m thick cover is: 44.8 x10 −6 ε V = = = 9.3418 x10 4 v/m −12 d 480 x10
40
The required voltage for the present case for a 10 µm thick cover is thus: -6
4
v = LV = (10x10 )(9.3418x10 ) = 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: -9 -9 -7 Rp = 42x10 m, and ∆Rp = 19.5x10 m = 19.5x10 cm 18
3
We further have the maximum concentration, Nmax = 30x10 atom/cm as given in Example 8.1. -9 It is at x = Rp = 42x10 m = 0.042 µm. (a) The supplied dose is: Q
= 2π ∆ Rp N max = 6.28 x(19.5 x10 −7 )(30 x1018 ) = 1.466 x1014
atoms / cm
2
(b) We will use the following relationship to find the concentration at x = 0.15 µm:
N (0.15 µ m)
= 30 x10
18
⎡ (0.15−0.042)2 ⎤ exp ⎢− = 6.57 x1012 2 ⎥ ⎣ 2 x(0.0195) ⎦
atoms / cm
3
(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:
N ( xo)
= 0.0001 x(30 x10
18
⎡ ( xo − 0.042) 2 ⎤ )= exp ⎢− ⎥ 2 x3.14 x19.5 x10 −7 ⎢⎣ 2 x(0.0195) 2 ⎥⎦ 1.466 x1014
41
Solve for xo = 0.1257 µm Problem 2:
We have energy level for the ion implantation to be 100 KeV and maximum concentration Nmax 18 3 = 30x10 atom/cm . 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) 307 135 67.8
B P As
∆Rp (nm) 69 53.5 26.1
By following the procedure outlined in Example 8.1, we have: 1) The dose of ion beams by Q
= 2π ∆ Rp / N max
We thus have: Q p
=
2π (53.5 x10 −7 )(30 x1018 ) = 4.022 x1014 atom/cm for phosphorus
Q A
=
2π (26.1 x10 −7 )(30 x1018 ) = 1.962 x1014 atom/cm for arsenic
2
2
2) Concentration of P and As at x = 0.15 µm beneath the surface of silicon substrate by using Equation (8.1):
⎡ (0.15 − 0.135)2 ⎤ 3 18 N p (0.15 µ m ) = (30 x10 ) exp ⎢− ⎥ = 28.84 x10 atom/cm for phosphorus 2 ⎣ 2 (0.0535) ⎦ ⎡ (0.15 − 0.0678)2 ⎤ 3 18 18 N A (0.15 µ m ) = (30 x10 ) exp ⎢− ⎥ = 0.21 x10 atom/cm for arsenic 2 2 (0.0261) ⎣ ⎦ 18
16
3
3) Location xo at which N(xo) = 0.01 Nmax = 3 x 10 atom/cm : We may derive the following relationship by using Equation (8.1): 14
2
For the case of phosphorus with Q p = 4.022x10 atom/cm :
42
⎡ ( xop − 135 x10 −7 )2 ⎤ ⎥ N p ( x op ) = exp ⎢− 2 x 3,14 (53.5 x10 − 7 ) ⎢⎣ 2(53.5 x10 −7 )2 ⎥⎦ ⎡ ( xop − 135 x10 −7 )2 ⎤ ⎥ = 3 x1016 = 29.999 x1018 exp ⎢− −11 ⎢⎣ 5.7245 x10 ⎥⎦ 4.022 x 1014
from which we solve for xop = 0.334 µm for phosphorus. 14
2
For the case of arsenic with QA = 1.962 x 10 atom/cm :
⎡ ( xoA − 67.8 x10 −7 )2 ⎤ exp ⎢− N A ( x oA ) = ⎥ 2 2 x 3.14 (26.1 x10 −7 ) ⎢⎣ 2 (26.1 x10 −7 ) ⎥⎦ ⎡ ( xoA − 67.8 x10 −7 )2 ⎤ 18 = 30 x10 exp ⎢− ⎥ = 3 x1016 −14 ⎢⎣ 1362.42 x10 ⎥⎦ 1.962 x1014
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) 14 2 ( 10 atom/cm )
Boron (P) Phosphorus (P) Arsenic (As)
5.2 4.022 1.962
Concentration at xo = 0.15 µm beneath surface 18 3 (10 atom/cm ) 2.27 28.84 0.21
Depth at which N(xo) = 1% of Nmax (µm) 0.5635 0.334 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), ln( D )
=
aT ' + b with T’ = 1000/T = 0.8525, in which T = 900 + 273.
= − 19.982 x0.8525 + 13.1109 = − 3.924 → µm /h, or D = 0.10844x10-6 µm2/s Hence ln D
D
= 0. 01976 or D = 0.0003904
2
(b) Equation (8.4) can be used to obtain the concentration function, N(x,t) follows:
43
⎡ ⎡ x ⎤ 11 10 erfc = ⎢ ⎥ ⎣ 2 Dt ⎦ ⎢⎣ 2
N ( x, t ) = 10 erfc ⎢ 11
⎤ ⎛ 1518 x ⎞ = 1011 erfc⎜⎜ ⎟⎟ ⎥ −6 ⎥ t ⎝ ⎠ 0.10844 x10 t ⎦ x
or N ( x, t )
⎡ ⎛ 1518 x ⎞⎤ = 1011 ⎢1 − erf ⎜⎜ ⎟⎟⎥ ⎝ t ⎠⎦ ⎣
For x = 0.1 µm and t = 1 h = 3600 s: N (0.1µ m,1h )
⎡ 151.8 ⎞⎤ = 1011 ⎢1 − erf ⎛ ⎜ ⎟⎥ = 1011 [1 − erf (2.53)] = 1011 (1 − 0.99999) ≈ 0 ⎝ 60 ⎠⎦ ⎣
(c) For diffusion to take place at 800 C, we have T ' =
1000
o
800 + 273
= 0. 932 .
This will lead to the
computation of the diffusivity, D as follows: ln D
= − 19.982 x0.932 + 13.1109 = − 5.51165 →
D = 0.000163 µ m / h 2
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. o
(d) Let us raise the diffusion temperature to 1100 C: We will have T’ = 1000/(1100 + 273) = 0.72833, which leads to D = 0.05584 µm /h, -6 2 or 15.5x10 µm /s. This diffusivity will result in a concentration at x = 0.1 µm and t = 1 h to be: 2
N (0.1 µ m,1 h )
⎡ ⎛ ⎞⎤ 0.1 ⎟⎥ = 1011 [1 − erf (0.21167 )] = = 1011 ⎢1 − erf ⎜⎜ ⎟ ⎝ 2 15.5 x10 −6 x3600 ⎠⎦⎥ ⎣⎢ 1011 (1 − 0.2352 ) = 7.648 x1010 atoms / cm 3
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: o
Temperature, C 2 Diffusivity, D, µm /h Concentration, N(0.1 µm, 1h), 3 atoms/cm
800 0.000163 0
900 0.0003904 ≈0
1000 0.005676 10 3.482x10
1100 0.05584 10 7.648x10
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.
44
Problem 4: 10
3
The time required reaching the same concentration of dopant of 3.482x10 atoms/cm as in o Example 8.2 at 0.2 µm beneath the surface at 1000 C can be obtained by solving the following equation: 3.482 x1010
⎡ ⎛ 398.21 x 0.2 ⎞⎤ = 1011 ⎢1− erf ⎜⎜ ⎟⎟⎥ t ⎝ ⎠⎦ ⎣
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 0.04532 0.006516
B/A, µm/h B, µm/h
Wet oxidation 0.6786 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:
Equation (8.9) for small time Equation (8.10) for larger time
Time for dry oxidation, h 22.065 153.47
Time for wet oxidation, h 1.474 4.836
Problem 6:
The dilution of the hydrogen gas is η = 1%, and the deposition takes place at 800 C. 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. o
(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 o 800 C to be:
⎛ ⎞
⎛ 20 + 273 ⎞
T 1 ⎟ x 44.643 = 12.1905 mol / m 3 d 2 = ⎜⎜ ⎟⎟ d 1 = ⎜ ⎝ 800 + 273 ⎠ ⎝ T 2 ⎠ The concentration of H2 per cubic meter is:
45
23
23
3
NG = (6.022x10 )x12.1905 = 73.4111x10 molecules/m (b) The density (ρ) of the carrier gas, H2 is:
ρ = (2 g/mol)(12.1905 mol/m3) = 24.38
3
g/m
(c) The Reynolds number (Re): Re
=
ρ DV µ
with the gas density, ρ = 24.38 g/m , 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 o g/m-s from Table 8.6 for H2 gas at 825 C. 3
We may compute the Reynolds number to be Re = 11.39 (d) The thickness of the boundary layer (δ): L 0.2 δ = = = 0.0593 m Re 11.39 (e) The diffusivity of the carrier gas (D): We may use the same equation presented in Example 8.5 as shown below: v δ N D = η ( N G − N s ) r
24
2
with N = 10 molecules/m -s (given) 23 3 NG = 73.4111x10 molecules/m (in Part (a)) Ns = 0 and η = 1% = 0.01 We thus have the diffusivity, D to be:
D
=
0.0593 x10 24
(
0.01 73.4111 x10
23
− 0)
= 0.8078
2
m /s
(f) The surface reaction rate (k s): Following the expression used in Example 8.5, this rate can be computed from the expression: v
k s =
D N
v
D N G − δ N
=
0.8078 x10 24 0.8078 x73.4111 x10 23 − 0.0593 x10 24
46
= 0.1376
m/ s
(g) The deposition rate (r): We first compute δk s = 0.0593x0.1376 = 0.00816 << D = 0.8078
N G k s
This will justify us using the expression for estimating the rate of deposition, r = η
γ From Table 8.7 and Example 8.5, we have the number of SiO2 molecules per unit film 28 volume to be: γ = 4.3074x10 , which leads to the rate of deposition, r to be: 0.01 x73.4111 x10 23 x0.1376
r =
4.3074 x10 28
= 0.2345 µ m / s
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: o
The following changes in computing the rate of deposition will take place with a 490 C process temperature: (a) The number of molecules in one cubic meter volume of gas mixture (NG):
⎛ T 1 ⎞ ⎛ 293 ⎞ x 44.643 = 17.1433 = ⎟ d 2 ⎜⎜ ⎟⎟ d 1 = ⎜ 763 ⎝ ⎠ ⎝ T 2 ⎠
mol / m
3
The corresponding concentration is 23 23 3 N G = (6.022 x10 )(17.1433) = 103.24 x10 molecules / m
(b) The density:
=
ρ
2 x17.1433
= 34.2866
g/m
3
(c) The Reynolds number: Re
=
ρ DV µ
=
34.2866 x0.2 x 0.05 0.0167
= 20.53
(d) The thickness of the boundary layer: δ
=
L
Re
=
0.15 20.53
=
0.033 m
47
(e) The diffusivity of the carrier gas: r
=
D
δ N η ( N G − N s )
=
0.033 x10 24 0.01(103.24 x10
23
− 0)
= 0.3196
2
m /s
(f) The surface reaction rate: v
0.3196 x10
D N
k = D − δ N r = N s
G
0.3196 x103.24 x10
23
24
= 0.0978
− 0.033 x10 24
m/ s
(g) The deposition rate: 0.01 x103.24 x10 23 x0.0978 N G k s = = 0.2344 µ m / s r = η 28 γ
4.3074 x10
Problem 9: o
We have 1% H2 carrying gas at 800 C. The standard conditions have: P1 = 101,325 Pa, T1 = -3 3 273+20 = 293 K and V1 = 22.4 x 10 m /mol. The process conditions are: P2 = 1 Torr = 133.322 Pa, T2 = 800 + 273 = 1073 K. From the ideal gas law, we get: P1 V 1 P2 V 2
=
T 1 T 2
or
101325 (22.4 x10 −3 ) 133322 x V 2
=
293 1073
3
, from which we solve for V 2 = 62,344.04 m /mol
(a) The molar density d2: -5
3
d2 = 1/V2 = 1/62344.04 = 1.604 x 10 mol/m – a very low molar density! The concentration of H2 gas in the carrier gas is: 23
-5
18
3
NG = Avogardro’s number x d2 = (6.022 x 10 )(1.604 x 10 ) = 9.66 x 10 molecules/m (b) The corresponding density of H2 gas is: -5
3
-5
3
ρ = (2 g/mol)(1.604 x 10 mol/m ) = 3.208 x 10 g/m (c) The Reynolds number is: Re
=
ρ D V
= 1.5 x10 −5
µ with D = 0.2 m, V = 50 mm/s = 0.05 m/s, µ = 0.0214 g/m-s
48
(d) The thickness of the boundary layer is: δ
=
L
=
Re
0.2
= 51.64
1.5 x10 −5
m - 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: r δ N D = η ( N G − N s ) r
with N = 10 24 molecules/m -s (given), η= 1%, NG = 9.66 x 10 molecules/m , and Ns = 0, we have: D
=
2
0.2 x10 24 0.01(9.66 x10
18
)
=
18
3
2
2.07 x10 6 m /s (very high!)
(f) The surface reaction rate: r ( D N 2.07 x10 6 )(10 24 ) r = k s = (2.07 x10 6 )(9.66 x1018 ) − 0.2(10 24 ) D N G − δ N
= 1.045 x10 5
m/s (very large!)
(f) Rate of deposition: 5
5
6
Check: δk s = 0.2 x 1.045 x 10 = 0.21 x 10 < D = 2.07 x 10 Hence the deposition rate is: r = η
N G k s
γ
(9.66 x10 )(1.045 x10 ) = 0.2343 = 0.01 18
5
4.307 x10 28
µm/s
28
3
The value of SiO2 molecules per unit volume γ = 4.307 x 10 molecules/m 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.
49
Problem 10: o
We have the temperature T = 800 C = 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: 23
3
(a) NG = 73.4111 x 10 molecules/m (same as in Problem 6) 3 (b) ρ = 24.38 g/m (same as in Problem 6) (c) Re with reduced velocity becomes: Re
(24.38)(0.2 )(0.025 )
=
0.0214
= 5.6963
in which we used D = 0.2 m and µ = 0.0214 g/m-s (d) The boundary layer thickness: δ
=
L
Re
=
0.2 5.6963
=
0.0838 m
(e) Diffusivity of the carrier gas: r
D
=
δ N η ( N G
− N s )
=
(0.0838)(10 24 )
0.01(73.4111 x10
23
= 1.1415 m2/s
− 0)
(f) The surface reaction rate: r D N (1.1415) 10 24 r = k s = (1.1415) 73.4111 x10 23 − (0.0838) 10 24 D N G − δ N
(
(
)
)
(
)
=
0.1376 m/s
(g) The deposition rate: Check δ k s = 0.0838 x 0.1376 = 0,01153 << D = 1.1415 N k Hence r = η G s γ
=
0.01
(73.4111 x10 23 )(0.1376 ) 4.3074 x 10 28
= 0.2345
µm/s
Chapter 9 Overview of Micromanufacturing (P.344) Part 1 Multiple Choice:
50
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 b ecomes 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. A a a’
h
Figure 10.35 Micro Pressure Die with Double Height (Legend in Figure 10.16)
H Θ=54.74
2H
o
Cavity c
Constraint Base
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
51
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 c ourse, 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: L
a 2a
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 d p = 30 µm. The resistance to water flow in this conduit may be obtained by summing the resistance from a complete circular a nd that from a rectangular cross-sections using the equations provided in Section 10.7.
Chapter 11 As sembly, Packaging, and Testi ng of Mi cr osys tem s (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).
52
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 u p a cylinder with a mass of 2 mg, i.e., half of the -6 2 normal force associated with the friction force, or N = (2x10 kg)(9.81 m/s )/(µ = 0.94) = -6 20.08x10 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 -6 is: F = 20.08x10 N. By referring to the microgripper illustrated in Figure 2.45 o n page 80, each pair of electrode would generate electrostatic force f that is equa l to what is given in Equation (2.11), or:
f L
=
1 ε r ε o W V 2
2
d -12
2
in which εr = 8.85x10 C/N-m 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: nf L
=
n
1 ε r ε oW V 2
d
2
2
=
F
=
20.08 x 10 −6 , from which we arrive at the following expression:
6
nV = 1.81514x10
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:
V
=
1.81514 x10 6
14 for the actuation.
= 189.76 volts
53
Problem 3:
The diameter of the cylindrical object in Problem 2 can be obtained by the following expression: The mass of the cylinder m
π ⎞ = ρ ⎛ ⎜ d 2 ⎟( L ) ⎝ 4 ⎠ 3
3
3
where ρ = mass density of silicon = 2.3 g/cm (Table 7.3) = 2.3 x 10 kg/m , L = the length of the cylinder = 200 µm. -6
-3
Thus with m = 2 mg = 2x10 kg, we can computer the diameter d = 2.3528 x 10 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:
F e
=
q
2
4πε d 2
(1.6 x10 − ) 4(3.14 )(8.85 x10 − )(2.3528 x10 − ) 6 2
=
3 2
12
=
4160 N
We have computed from Problem 2 that the normal force gripping force for the cylindrical object -6 was 20.08 x 10 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: 11
2
For quartz:
Yong’s modulus E1 = 0.865x10 N/m (an average value from Table 7.3) -6 o Coefficient of thermal expansion α1 = 7.1x10 / C (Table 7.3)
For silicon:
Young’s modulus E2 = 1.9x10 N/m (Table 7.3) -6 o Coefficient of thermal expansion α2 = 2.33x10 / C (Table 7.3)
11
2
The ratio of Young’s modulus in Equation (4.49) is n = E 1/E2 = 0.4553 -3
We have the thickness of quartz layer to be t1 = 1 mm = 10 m and the thickness of the silicon -4 layer to be t2 = 300 µm = 3x10 m, from which we have m = t1/t2 = 3.3333. We further have the -3 thickness of the bi-layer beam to be h = t1 + t2 = 1.3x10 m. o The temperature drop ∆T in Equation (4.49) is: ∆T = 20 – 500 = -480 C By substituting the above material properties and physical parameters into Equation (4.49), we have the following:
54
