MEMS技术 第四讲 机械零件原理(1).ppt

1、MEMS和微系统设计,课程内容,MEMS概述及MEMS设计的概述 工艺简要回顾 系统设计、工艺设计及版图设计 主要的机械、电子元件及其设计基础 多域耦合设计：以机电耦合为例子 器件性能的估计 简单的其他域的元件及其简要设计要点 设计实例,第4讲主要内容,1、弹簧设计原理及计算例子 2、薄膜设计原理及计算例子 3、电容设计原理及计算例子 4、电阻设计原理及计算例子 5、压电模型,弹簧设计原理及计算例子,1、基本知识 2、梁的静态分析 3、二阶系统 4、梁的动态分析与谐振,Yield Strength (cont.),Below: typical stress vs. strain curves

2、for brittle (e. g., Si) and ductile (e. g. steel) materials,Youngs Modulus and Useful Strength,Stored mechanical energy,弹簧的定义,F=KX,A Cantilever Beam,预备知识,What happen?,Define Strain is found to be proportional to stress ( for “small” stresses at “lower” temperatures):,possions Ratio,V=Possions Ratio,

3、Possion strains from off x-axis stress,Forces | to surfaces Sheer Stress,Note: compensating forces are applied to the vertical faces to avoid a net torque!,=sheer stress,G=sheer modulus,Youngs Modulus in (001) Plane,Volume Change for a Uniaxial Stress,Stresses acting on a differential volume element

4、,The resulting change in volume V,Isotropic Elasticity in 3D,Isotropic = same in all directions The complete stress-strain relations for an isotropic elastic solid in 3D: (i. e., a generalized Hookes Law),Basically, add in off-axis strains from Normal stresses in other directions,Important Case: pla

5、ne stress,Common case: very thin film coating a thin, relatively rigid substrate (e. g., a silicon wafer),At regions more than 3 thicknesses from edges, the top surface is stress-free Get two components of in-plane stress:,Important Case: plane stress (cont.),Symmetry in the xy-plane Thus, the in-pl

6、ane strain components are: where,where,Linear Thermal Expansion,As temperature increases, most solids expand in volume Definition: linear thermal expansion coefficient,Linear thermal expansion coefficient,Remarks: values tend to be in the 10-6 to 10-7 range Can capture the 10-6 by using dimensions o

7、f strian/k, where 10-6 K-1 = 1 strian/k In 3D, get volume thermal expansion coefficient For moderate temperature excursions, can be treated as a constant of the material, but in actuality, it is a function of temperature,T As a Function of Temperature,Madou, Fundamentals of Microfabrication, CRC pre

8、ss, 1998,Thin-Film Thermal Stress,Assume film is deposited stress-free at a temperature Tr, then the whole thing is cooled to room temperature Tr Substrate much thicker than thin film substrate dictates the amount of contraction for both it and the thin film,Stress Measurement Via Wafer Curvature,Co

9、mpressively stressed film bends a wafer into a convex shape Tensile stressed film bends a wafer into a concave shape Can optically measure the deflection of the wafer before and after the film is deposited Determine the radius of curvature R, then apply:,Linear Thermal Expansion,But the film is atta

10、ched to the substrate, so the actual stain is the film is the same as that in the substrate:,弹簧设计原理及计算例子,1、基本知识 2、梁的静态分析 3、二阶系统 4、梁的动态分析与谐振,Reaction Forces and Moments,Sign Conventions for Moments & Shear Forces,(+) moment leads to deformation with a (+) radius of curvature (i. e., upwards),(-) mome

11、nt leads to deformation with a (-) radius of curvature (i. e., downwards),(+) shear forces produce clockwise rotation,(-) shear forces produce counter-clockwise rotation,Beam Segment in Pure Bending,Small section of a beam bent in response to a tranverse load,Consider a segment bounded by the dashed

12、 lines defined by d: At z=0: (i.e., of the neutral axis): segment length=dx=Rd (1) At any z: segment length = dL =(R-z) d (2) Combining (1) & (2) :,Beam Segment in Pure Bending (cont.),Why a minus sign ? See Senturia ,pp.208-210,Radius of curvature geometric connection to strain,result from basic ca

13、lculus,Combining the curvature and moment results:,and,剪切力忽略的条件,Compliance is sum of bending and shear-stress contributions:,Solving for :,Transversal deflection of a beamwith a load at the end,with:,(187,1),(187,2),I := Area moment of inertia of the beam B := Strain at the surface of the beam F :=

14、Force acting at the end of the beam dB := Thickness of beam w := Deflection of beam bB := Width of beam LB := Length of beam EB := Youngs modulus of beam,187,* W. Beitz, K.-H. Grote, Dubbel, Taschenbuch fr den Maschinenbau”,*,(187,4),(187,5),Aria momentum of inertia,(188,1),(188,2),Rectangular:,(188

15、,3),I := Area moment of inertia of the beam EB := Youngs modulus of the beam FB := Elastic force of the beam at its end w := Deflection of the beam dB, bB, LB, RB := Thickness, width, length, and radius of the beam, respectively,188,More aria momentums of inertia are found in books like: W. Beitz, K

16、.-H. Grote, Dubbel, Taschenbuch fr den Maschinenbau” or R.D. Blevins, Formulas for Natural Frequency and Mode Shape“, Krieger, Malaba, FL (1987),bB,dB,(187,3),Circular:,RB,Trapezoid shaped:,dB,e,bB,1,bB,2,(188,4),with:,(188,5),Strain on the surface of a beam loaded at its end,(189,1),Rectangular bea

17、m:,(187,3) :,(189,2),189,* W. Beitz, K.-H. Grote, Dubbel, Taschenbuch fr den Maschinenbau”,*,I := Area moment of inertia of the beam EB := Youngs modulus of the beam F := Force acting at the end of the beam B := Strain at the surface of the beam w := Deflection of the beam dB, bB, LB : Thickness, wi

18、dth, and length of the beam, respectively,Rectangular beam clamped on one sideand loaded at the other end,x m,w m,(187,1), (187,3):,LB=800m bB=40m dB=20m EB=140GPa F=1mN,190,EB := Youngs modulus of the beam B := Strain at the surface of the beam F := Force acting at the end of the beam dB, bB, LB, w

19、 := Thickness, width, length, and deflection of the beam, respectively,Transversal deflection of a beamloaded at the end,The strain at the surface of a beam clamped at one end and loaded at the other end in transversal direction is largest at the fixed end.,Strain and deflection are not a functions

20、of an initial stress of the beam.,Rectangular beam:,(189,2):,Strain and deflection are proportional to the force (linear characteristic curve).,(187,1), (187,3):,191,EB := Youngs modulus of the beam B := Strain at the surface of the beam F := Force acting at the end of the beam dB, bB, LB, w := Thic

21、kness, width, length, and deflection of the beam, respectively,Transversal deflection of a beamloaded at the end,Rectangular beam :,Because of the transverse strain the beam gets narrower on the side with tensile stress and wider on the opposite side. (With the exception of the region next to the cl

22、amping),Cross-section of the beam:,Without load,With load,bB,bB (1 B B),dB,192,(189,2):,EB := Youngs modulus of the beam B := Poissons ratio of the beam B := Strain at the surface of the beam F := Force acting at the end of the beam dB, bB, LB, w: Thickness, width, length, and deflection of the beam

23、, respectively,Deflection of beams loaded by their weight,In the micro world the Golden Gate Bridge could be constructed as a simple beam.,198,Deflection of beams loaded by their weight,199,In the micro world the Golden Gate Bridge could be constructed as a simple beam.,x,z,F,(208,1),(208,4),208,*,*

24、 W. Beitz, K.-H. Grote, Dubbel, Taschenbuch fr den Maschinenbau”,B(z,x) := Strain of beam w := Deflection of beam F := Force acting at the end of the beam B, max := Maximum strain of beam EB := Youngs modulus of the beam I := Area momentum of inertia of the beam LB, dB, bB := Length, thickness, and

25、width of beam, respectively,(208,2) :,209,(208,3) :,x m,w0 m,B(z,x) := Strain of beam w := Deflection of beam F := Force acting at the end of the beam B, max := Maximum strain of beam EB := Youngs modulus of the beam I := Area momentum of inertia of the beam LB, dB, bB := Length, thickness, and widt

26、h of beam, respectively,n, D := Strain of neutral fiber and stress generated by it, respectively I := Area momentum of inertia of the beam w := Deflection of the beam w0 := Deflection in the center of the beam F := Force acting at the center of beam dB, bB, LB := Thickness, width, and length of the

27、beam EB := Youngs modulus of the beam,(208,2):,(212,1),The elastic strain n of the beam along the neutral fiber needs to be taken into account also. It is calculated from the extension in length. In Analogy to (90,2) it is assumed:,(212,2),212, 90,Effect of the stress,Analogous to the approach used

28、for the membranes, the equilibrium of forces at the ends of the beam is calculated:,x,z,F,w0,FR,z,FR,(213,1),An angular bending line is used as an approximation:,213, 90, 0, D := Stress, initial stress, and stress due to straining, respectively I := Area momentum of inertia of the beam w := Deflecti

29、on of the beam w0 := Deflection in the center of the beam F := Force acting at the center of beam dB, bB, LB := Thickness, width, and length of the beam EB := Youngs modulus of the beam,As an approximation the forces generated by bending moments and stress are simply added:,(214,1),(212,1) + (213,2)

30、:,Adding the forces is a rough approximation which shall show the interrelationships. More exact results are obtained by FE-calculations.,214, 0 := Stress and initial stress, respectively EB := Youngs modulus of the beam w := Deflection of the beam w0 := Deflection in the center of the beam dB, bB,

31、LB := Thickness, width, and length of the beam F := Force, 212,(214,1):,LB=800m dB=5m BB=50m EB=70GPa,0 = -400 MPa,0 = +400 MPa,w0 m,F mN,215, 0 := Stress and initial stress, respectively EB := Youngs modulus of the beam w := Deflection of the beam w0 := Deflection in the center of the beam dB, bB,

32、LB := Thickness, width, and length of the beam F := Force,Deflection of a rectangular beam clamped at both sides,(214,1):,If no force is acting on the beam (F = 0), there are three solutions for w0:,w0 m,F mN,(216,2),216,k, 0 := Critical stress and initial stress, respectively EB := Youngs modulus o

33、f the beam w := Deflection of the beam w0 := Deflection in the center of the beam dB, bB, LB := Thickness, width, and length of the beam F := Force acting at the center of beam,Deflection of a rectangular beam clamped at both sides as a function of the initial stress,If no force is acting on the bea

34、m (F = 0), there are three solutions for w0:,w0 = 0,0 k,w0 m,(216,1) :,(216,2) :,(214,1):,217,(216,3) :,with:,k, 0 := Critical stress and initial stress, respectively EB := Youngs modulus of the beam w0 := Deflection in the center of the beam F := Force acting at the center of beam dB, bB, LB := Thi

35、ckness, width, and length of the beam,Snapping over of a rectangular beam clamped at both sides,(218,1),(218,2),wU,wU,218,(216,2) :,(214,1):,k, 0 := Buckling stress and initial stress, respectively EB := Youngs modulus of the beam wU := Deflection of the beam at snapping over w0 := Deflection in the

36、 center of the beam dB, bB, LB := Thickness, width, and length of the beam F := Force acting at the center of beam, 217,Snapping over of a rectangular beam clamped at both sides,(218,2):,219,k, 0 := Buckling stress and initial stress, respectively EB := Youngs modulus of the beam wU := Deflection of

37、 the beam at snapping over w0 := Deflection in the center of the beam dB, bB, LB := Thickness, width, and length of the beam F, FU := Force and snapping force, resp.,Buckling stress,(216,3):,The approximation (216,3) is 20 % larger than the exact result (220,1).,220, 207,k, 0, EB:= Buckling stress,

38、initial stress, and Youngs modulus, respectively I := Area momentum of inertia of the beam F, Fk := Force and buckling load, respectively w0 := Deflection in the center of the beam dB, bB, LB := Thickness, width, and length of the beam,0, k, GB, EM, EB , M, B := Initial stress, critical stress, shea

39、r modulus, Youngs modulus, and Poissons ratio, respectively F, Mt := Force and Torque w0, := Deflection and angle of rotation dM, dB, LB, bB, RM := Thickness, length, width and radius wU := Deflection at snapping over,Calculations of deflection and force of circular membranes and rectangular beams,2

40、29,Force F,k,Membrane,Torque Mt,0,3,Beam,*The effect of initial stress is not taken into account,*,Integration of beams,Compliances add,弹簧设计原理及计算例子,1、基本知识 2、梁的静态分析 3、二阶系统 4、梁的动态分析与谐振,In frequency domain,Damping coefficient,transition function,信号与系统的相关知识,为什么要做傅立叶变换：周期函数与非周期函数的傅立叶变换 拉普拉斯变换 传递函数 常用的反变换

41、,频率响应,设阻尼器的阻尼系数为c ，则运动方程为,（4-19）,上式中质量块的瞬时位置X(t)有三种情况，取决于阻尼比=c/2m 的大小,图4-7,情况 1 ，过阻尼情况,图-9描述了系统的过阻尼情况,质量块的瞬时位置X(t)为:,（4-20a）,一、典型二阶系统的瞬态响应 下图所示为稳定的二阶系统的典型结构图。,开环传递函数为:,闭环传递函数为:,这是最常见的一种系统，很多高阶系统也可简化为二阶系统。,称为典型二阶系统的传递函数，称为阻尼系数， 称为无阻尼振荡圆频率或自然频率。,特征根为： ，注意：当 不同时，（极点）有不同的形式，其阶跃响应的形式也不同。它的阶跃响应有振荡和非振荡两种情况

42、。,特征方程为：, 当时 ，特征方程有一对共轭的虚根，称为零(无)阻尼系统，系统的阶跃响应为持续的等幅振荡。, 当时 ，特征方程有一对实部为负的共轭复根，称为欠阻尼系统，系统的阶跃响应为衰减的振荡过程。, 当 时，特征方程有一对相等的实根，称为临界阻尼系统，系统的阶跃响应为非振荡过程。, 当 时，特征方程有一对不等的实根，称为过阻尼系统，系统的阶跃响应为非振荡过程。,当输入为单位阶跃函数时， ，有：,分析：,此时输出将以频率 做等幅振荡，所以， 称为无阻尼振荡圆频率。,阶跃响应为：,极点的负实部 决定了指数衰减的快慢，虚部 是振荡频率。称 为阻尼振荡圆频率。,阶跃响应函数为：,即特征方程为,特

43、征方程还可为,因此过阻尼二阶系统可以看作两个时间常数不同的惯性环节的串联，其单位阶跃响应为,式中,上述四种情况分别称为二阶无阻尼、欠阻尼、临界阻尼和过阻尼系统。其阻尼系数、特征根、极点分布和单位阶跃响应如下表所示：,可以看出：随着 的增加，c(t)将从无衰减的周期运动变为有衰减的正弦运动，当 时c(t)呈现单调上升运动(无振荡)。可见 反映实际系统的阻尼情况，故称为阻尼系数。,二、典型二阶系统的性能指标及其与系统参数的关系,（一）衰减振荡瞬态过程 ：, 上升时间 ：根据定义，当 时， 。,解得：,从上图中，我们可以得出 在这种情况下，质量块的振动幅值迅速下降 过阻尼适应在易于过量振动的机械和器

44、件（包括微系统） 在这种微器件的设计中，设计工程师应该选择一个合适的阻尼器,情况 ，临界阻尼情况,图-10描述了系统的临界阻尼情况,质量块的瞬时位置X(t)为:,（4-20b）,情况 ，欠阻尼情况,图-11描述了系统的欠阻尼情况,质量块的瞬时位置X(t)为:,（4-20a）,Damping,Six kinds of damping Air damping,Energy Dissipation and Resonator Q,Thermoelastic Damping (TED),Occurs when heat moves from compressed parts to tensioned

45、parts heat flux = energy loss,TED Characteristic Frequency,Governed by Resonator dimensions Material properties Table 1. material properties,Q Vs. Temperature,Mechanism for Q increase with decreasing temperature thought to be linked to less hysteretic motion of material defects less energy loss per

46、cycle,Even aluminum achieves exceptional Qs at cryogenic temperatures,Disk Resonator Loss Mechanisms,air Damping,Many sources: ignore internal damping and acoustic radiation from the anchors focus on drag from the surrounding gas,Gap dimension can be On the order of mean-free Path continuum model do

47、esnt apply,Squeeze-Film Damping,Plates slide in y direction,Couette Damping,Drag force= Damping coefficient : b Effective viscosity at reduced pressure(p50 Torr):,Unfixed units,例题4-12,如图4.25所示，针对空气和硅油做为阻尼流体，估算力平衡微加速度计的阻尼系数。假定微加速度计工作在 20,图4-25 力平衡微加速度计的阻尼,解: 由上图知: 梁的质量块宽度，b=510-6m 梁的长度，L70010-6m 间隙宽度

48、，H=1010-6m 从表4.3知 : 在20时粘度是 air 18.7510-6Ns/m2 si 74010-6Ns/m2,How about the result in a squeeze damping situation?,Other Parameter in 2-order system,Resonant Frequency Quality factor,Physics meaning of f and Q,Energy must be conserved: Potential Energy +kinetic Energy = Total Energy Must be true at

49、 every point on the mechanical structure,Solving, we obtain for resonance frequency:,Example: ADXL-50,The proof mass of the ADXL-50 is many times larger than the effective mass of its suspension beams Can ignore the mass of the suspension beams (which greatly simplifies the analysis) Suspension Beam

50、: L=260m, h=2.3 m, W=2 m,Lumped spring-mass approximation,Mass is dominated by the proof mass 60% of mass from sense fingers Mass = M = 162 ng (nano-grams) Suspension: four tensioned beams Include both bending and stretching terms A. P. Pisano, BSAC Inertial Sensor Short Courses, 1995-1998,ADXL-50 S

51、uspension model,Bending contribution: Stretching contribution: Total spring constant: add bending to stretching,ADXL-50 Resonance Frequency,Using a lumped mass-spring approximation: On the ADXL-50 Data Sheet: f0=24kHz Why the 10% difference? Well, its approximate plus Above analysis does not include

52、 the frequency-pulling effect of the DC bias voltage across the plate sense fingers and stationary sense fingers something well cover later on ,To understand Q from different domain,Q=2.E/E=m.W0/b= ( .W0)/(W1/2.ln2) (参看参考资料中的谐振部分）,Quality Factor (or Q),Measure of the frequency selectivity of a tuned

53、 circuit Definition:,Example: series LCR circuit,Example: parallel LCR circuit,Selective Low-Loss Filters: Need Q,In resonator-based filters: high tank Q low insertion loss At right: a 0.1% bandwidth, 3-res filter 1GHz (simulated) Heavy insertion loss for resonator Q10000,Attaining High Q,Problem: I

54、Cs cannot achieve Qs in the thousands transistors consume too much power to get Q on-chip spiral inductors Qs no higher than 10 off-chip inductors Qs in the range of 100s Observation: vibrating mechanical resonances Q1000 Example: quartz crystal resonators (e.g., in wristwatches) Extremely high Qs 1

55、0000 or higher (Q106 possible) mechanically vibrates at a distinct frequency in a thickness-shear mode,弹簧设计原理及计算例子,1、基本知识 2、梁的静态分析 3、二阶系统 4、梁的动态分析与谐振,Vibrations of beams,The resonance frequency of a sensor may be the parameter to be measured.,Then it needs to be known, how the resonance frequency is

56、 affected by other parameters.,Besides this, the distance to other resonance frequencies needs to be as large as possible to obtain a large measurement range.,233,Useful frequency range,f1,Exciting frequency f,Amplitude of the beam,f2,f3,234,Actuators and sensors used at their resonance frequency,If

57、 actuators and sensors are driven at their resonance, its frequency needs to be known and how it is influenced, because:,235,Wave function of vibrating beams,LB := Length of the beam i := Phase angle of mode i fi := Resonance frequency of mode i w(x,t) := Wave function ui := Position dependent part

58、of a wave function of the mode i Ai := Amplitude of mode i,LB,w(x),(236,1),(236,2),The wave functions wi(x,t) of the modes are solutions of a differential equation.,236,Resonances of vibrating beams,LB := Length of the beam fi := Resonance frequency of mode i B := Density of beam AB := Cross-sectional area of the beam I := Area momentum of inertia of the beam EB := Youngs modulus of the beam i := Frequency parameter,(237,1),The resonance frequencies fi(i) of the modes are