Foundations of MEMS 2rd Edition Chang Liu Solution Manual Full Doawnload: https://solutionsmanualbank.com/d https://solutio nsmanualbank.com/download/solutionownload/solutionmanual-for-foundations-of-mems-2-e-chang-liu/
Instructor Manual Foundations of MEMS
Chang Liu Northwestern University Chapter 3 Visit http://www.memscentral.com Visit http://www.memscentral.com , a compani companion on webs webs ite of the the book for book for addition additional al teach teachin ingg mater i a s ls .
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Probl Problems Problem 1: Review
Solution:
Silicon assumes diamond lattice. In each unit lattice (bound by a box with each side being 5.43 Å, there are 8 atoms per cell. The volume concentration is 8 22 3 atoms / cm = 5 × 10 −8 3 (5.43× 10 ) The density of silicon is 5 × 1022 (atoms / cm 3 ) × 28.1( g / mole) 3 = 2.33 g / cm 23 6.02 ×10 (atoms / mole) Problem 2: Design
Solution:
The resistivity of gold at room temperature is 2.271x10-8Ωm according to CRC Handbook of Chemistry and Physics, 75th edition. Since 50 = 2.271×10 −8
l 0.1 ×10 −12
=
227100l
we find the value of length being l =220 =220 µm Problem 3: Design
Solution:
The semiconductor is doped n type. type. The concentration of minority carrier is 3 2 p = ni / n = 2250cm The resistivity is −
ρ =
=
=
1
σ
=
1
q ( µ n n + µ p p) 1 −19
1.6 ×10 0.046
×
17
(1350 × 10
V ⋅ s ⋅ cm C
=
0.046
+
480 × 2250)
V ⋅ cm A
=
0.046Ω ⋅ cm
Problem 4: Design
Solution:
2
The piece of semiconductor material material is doped p type. The concentration of minority carrier is 2 3 n = n / p = 2250cm The resistivity is −
ρ
=
1
σ
1
=
q ( µ n n + µ p p) 1
=
−19
1.6 ×10
×
(1350 × 2250
+
17
480 × 10 )
0.13Ω ⋅ cm
=
Problem 5: Design
Solution:
The semiconductor is doped n type. type. The concentration of minority carrier is 9 2 3 p = ni / n = 2.25 ×10 cm The resistivity is −
1
1
ρ
=
=
1 11 9 1.6 ×10−19 × (1350 × 10 + 480 × 2.25 ×10 )
=
4.6 ×10
σ
=
q ( µ n n + µ p p)
4
Ω ⋅ cm
Problem 6: Design
Answer :
The volume doping concentration is simply
10 14atoms / cm 2 −4
10 cm
=
1018 atoms / cm3 .
Problem 7: Design
Solution :
The resistor consists consists of 20 squares. The total resistance is 20x50=1k Ω. Given the junction depth (d), the resistivity is ρ
=
ρ s d
=
50 × 0.3 ×10−4
=
1.5 ×10
−3
Ωcm
Assuming the majority carrier concentration dominates, we have ρ =
=
=
1
σ
1
=
q ( µ p p) 1 −19
1.6 ×10
−3
1.5 ×10
×
(480 × p )
Ω ⋅ cm
Hence the concentration p is 8.69 × 1018 atoms / cm3 .
3
Problem 10: Design
Solution:
Two vectors, <111> and <100>, are perpendicular to (111) and (100) planes. The inner product of two vectors a and b is given by a ⋅ b = a b cos θ If a=<111> and b=<100>, the magnitude of the th e inner product is 1 + 0 + 0 = 1 = 3 ⋅1 ⋅ cos θ Hence 1 θ = cos 1 = 54.7356 3 −
Problem 11: Design
Solution:
For a 10oC temperature change, the relative thermal expansion and longitudinal strain is 2.6 × 10
−6
×
10
=
2.6 × 10
−5
The equivalent longitudinal stress, assuming the Young’s modulus is 120 GPa, is 6 − 2.6 × 10 5 × 120 × 109 = 3.12 × 10 N / m 2 The longitudinal strain causes the cross-section of the bar to change. Each dimension of the cross-section change by an amount − 2.6 × 10
−5
γ
where γ is the Poisson’s ratio. The relative change of resistance, which is related to the length and cross-section by R = ρ dR =
∂ R
∂l
dl +
∂ R
∂ A
dA =
ρ A
dl −
ρl A2
dA
Problem 12: Review
Solution:
The stiffness matrix i s 0.64 0.64 0 0 0 0 0 0.64 1.66 0.64 0 0 0 0.64 0.64 1.66 0 11 0 0 0.8 0 0 0 10 Pa 0 0 0 0 0.8 0 0 0 0 0 0 0.8
1.66
C S i ,
100>
<
=
4
l
A
, is
The compliance matrix is therefore 0.767 − 0.213 − 0.213 0 0 0 0 − 0.213
S S i ,
100>
<
=
− 0.213
0.767 0 0
0 0 0 0 0.8 0 0 0.8
0 0 − 10 11 Pa 0 0
−1
Problem 13: Design
An axial force with magnitude of U is applied at the end of a bar attached to a wall. Find the reactive force and torque (if any) at the anchored end of the bar and at section A.
Answer :
Both reaction forces, at the anchor and at section A, are in the longitudinal direction of the rod and equal to U. Problem 14: Design
Two independent axial forces with magnitude of U and F are applied at the end of a bar attached to a wall. Find the reactive force and torque (if (if any) at the anchored end of the bar and at section A. Assume the overall reaction under a combined force is the linear sum of reactions under two forces acting individually.
Solution:
At the anchored end, there are two reactive forces – one one equals to U and acts along the longitudinal direction, another equals to F and acts in the transverse direction. There is one reactive torque, with the magnitude being FL.
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At section A, there are two reactive forces – one one equals to U and acts along the longitudinal direction, while another equals to F and acts in the transverse direction. There is one reactive torque with magnitude being 0.5FL. The torque at the intersection is acting in the counterclockwise direction. Problem 15: Design
An axial torque with magnitude of M is applied at the end of a bar attached to a wall. Find the reactive force and torque (if any) at the anchored end of the bar and at section A.
Solution:
At the anchored end, there is no reactive force but one o ne reactive torque – the the magnitude is M and the torque is clock-wise with axis along the longitudinal direction of the rod. At the section A, there is no reactive force but one reactive torque. torque. The magnitude and direction of the torque is exactly exactl y equal as the one at the anchor. Problem 16: Design
A slender silicon beam is under a longitudinal tensile stress. stress. The force is 1 mN, and the crosssectional area is 20 µm by 1 µm. The Young’s modulus in the longitudinal direction is 120 GPa. Find the relative elongation of the beam (percentage). (percentage). What is the force necessary to fracture the beam if the fracture strain of silicon is 0.3%? Solution:
The longitudinal stress is approximately 0.001 N 2 7 5 σ = 2 = ×10 N / m −12 20 × 10
m
The corresponding longitudinal strain is therefore ε
=
σ E
=
5 × 10 7 120 × 10
9 =
0.04167%
The force necessary to fracture the beam is 0.3 0.001 N × = 0.0072 N = 7.2mN 0.04167
Problem 17: Design
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Solution :
Compressive normal normal stress is introduced under this this loading condition. The magnitude of the stress is 0.001 N 2 2 = 10 N / m −4 1×10 m
Problem 18: Design
Solution:
The maximum allowable force is 140 ×109 × 0.002 × 5 × 10−12
=
0.0014 N
Problem 19: Design
Solution :
For case (a), I For case (b), I
=
=
wt 3
12 tw
3
=
3
12
=
20× (40) ×10
−24
12 40× (20)
3 ×
12
10
19
4
=
1.067 × 10− N
=
0.267 × 10− N 4
−24
19
Problem 20: Design
Find the force constant of the beam in Problem if a force is applied in the the longitudinal direction of the cantilever. The beam is 800 µm long in this case. Solution :
7
For a given force, the degree of elongation of the free end is F d
=
wt × l E
The force constant is k
=
F wtE = d l
=
20× 10
−6
×
40 ×10 −6×120 × 10 −6 800 ×10
9 =
20000 N / m
This meant the beam acts as a relatively stiff force constant. Problem 21: Design
Solution:
(10 −6) 3 4 − 25 (1) The moment of inertia is I = m . = 4.167 × 10 12 (2) The silicon bar is doped primarily primarily with n-type carriers. carriers. The electrical resistance associated with the silicon bar is 6 100 ×10 l 1 l 6 9.24 ×10 Ω ρ ≈ = 12 = −19 16 A q ( µ n n) A 1.6 ×10 ×1350 ×10 × 5 ×10 5× 10
−6
×
−
−
Problem 22: Design
Answer:
The moment of inertia scales with L4. For small devices, the moment of inertia is reduced dramatically with length. The reduced I can translate into into small force force constant of mechanical elements (such as as cantilevers). For sensors, MEMS devices have soft springs and therefore high sensitivity. Problem 23: Design
Solution: 3
For case d, the moment of inertia is
w1 t1
12
. The spring spring constant is k
=
. The spring spring constant is k
=
3
For case e, the moment of inertia is
t1 w 1
12
3 EI Ew1t13 =
4l 3 l 3 3 EI Et 1w31 = 4l 3 l 3
.
3
For case f, the moment of inertia is
t1 w 1
12
3l . The force force constant is approximately k
, but the effective length of the beam is approximately =
3 EI (3l ) 3
8
3
Ew 1t 1 = 4(3l ) 3
3
Ew1t 1 . = 108l 3
Problem 24: Design
Solution:
According to formula in Appendix B, the maximum displacement in the center of a fixed-fixed beam under a transverse force F is Fl 3 d= 192 EI
If we consider a fixed-fixed beam with a central force loading as two fixed-guided beams with end loading, then the amount of force on each end is F/2. The maximum displacement, displacement, according to formula in appendix B, is F (l / 2)3 d = 2 12 EI
Fl 3 = 192 EI
Since the displacement under F is identical, the force constant is identical. Problem 25: Design
A torsional bar is anchored on two ends, with a lever attached in the middle of the bar. A force, F=0.01 µ N, is applied to the end of the lever. Determine the degree of angular bending due to the rotation of the torsional bars. Do not consider the bending of the flexural lever lever segment. The values of L, w, and t are 1000, 10, and 10 µm, respectively. respectively. The beam is made of polycrystalline silicon. (Hint: find Young’s modulus and Poisson’s r atio atio from the literature and handbook, and cite the source of information (other than this textbook))
Solution:
We will use a Poisson’s ratio of 0.22 for polysilicon. polysilicon. There are many ways ways the value of the Young’s modulus can be found or guessed. Based on Ref. [20], we will use 158GPa as the value of E. The torsional moment of inertia of the torsion beams is J
=
w 2.25( ) 4
=
0.1406 w 4 . The torque
2 experienced by the torsion beam is M = FL . Since there are two symmetric beams, the torque acting on each torsional beam with length L is M/2.
9
The total angular displacement is therefore M TL
M
L
2 Φ= = JG 0.1406w 4G
2
=
0.1406 w
L
4
E
=
0.0549rad = 3.15 o
2(1 2(1 + ν)
Problem 26: Design
Solution:
This is a typical typical design synthesis synthesis problem. We have the expression of the force constant as 3 F 3 EI Ewt k
=
d
=
l 3
=
4l 3
The resonant frequency of the first mode is k EIg 3.52 EIg = f n = n = 0.56 4 4 2π wl
wl
2π
Ewt 3 g g (l wt ) 4 ρ 12 l l
=
0.16
E t 2
ρl
4
=
0.16
E t
ρ l 2
Solving both equations simultaneously, for example by raising the left and right hand of the formula for the resonant frequency to the 3 rd power, f n3
=
0.004096
E
ρ
3 2
t3 6
l
one find l=0.00293 m. Answer (2) is correct.
Foundations of MEMS 2rd Edition Chang Liu Solution Manual Full Doawnload: https://solutionsmanualbank.com/download/solutionhttps://solutionsmanualbank.com/dow nload/solutionmanual-for-foundations-of-mems-2-e-chang-liu/
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