Microelectromechanical Systems Assignment Help

Microelectromechanical Systems Assignment Help

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Problems Problem 9.14 A silicon nitride (LPCVD, stoichiometric) AFM cantilever has a thickness of 0.5 m and the triangular shape (top view) shown below. It is subjected to a point load F at the tip, pointing into the paper. Assuming small deformations, find the tip deflection in terms of F, and calculate an effective spring constant for this beam. For a load which produces a tip deflection of 2 m, calculate the maximum stress at the support.

a = 50 m

L = 250 m Problem 9.10 Wafer-scale fusion bonding is commonly used to create complex silicon structures. However, two wafers can be bonded together only if their bonding surfaces are about the same shape. If the shapes are too different, the energy required to deform the wafers exceeds the energy gained by bringing them into contact, and the wafers will not bond. Because stressed films deform substrates, film thicknesses are limited on wafers that will be wafer-bonded.

Consider a PECVD silicon dioxide film that is deposited on a 650 m thick silicon wafer. The wafer is then annealed at 1100 °C and cooled to room temperature. mechanicalengineeringassignmenthelp.com mechanicalengineeringassignmenthelp.com

Assume that the oxide is stress-free when it is above the oxide softening temperature, which we will take to be 950 °C. Also assume biaxial plane stress in the film, and approximate the substrate as isotropic. (a)Using constitutive relationships, calculate the thermal stress r in the film for the case where the film thickness is much less than the substrate thickness.

Experience shows that the wafer’s radius of curvature must be greater than about 100 m to ensure that the wafer can be bonded to a second, flat wafer. In the next two parts, we will determine the maximum thickness of the oxide film to ensure bondability. (b)The wafer curvature is determined by moment balance over the thickness of the wafer.

Find an expression for the net moment per unit width as a function of the geometry, materials properties, and thermal stress in the film. (c)Write an expression relating the film thickness to the radius of curvature. What is the maximum film thickness to ensure bondability for the case described above?

Problem 14.12 Circuit loading In this problem we will explore the differences among different op-amp topologies.

Assume we are trying to measure the signal Vs from a linear sensor and would like to amplify that before we measure it with a computer-based data acquisition card.

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Thevenin’s theorem states that any linear electrical sensor can be represented by a voltage source and a series impedance (in this case, a resistance RS). Your data acquisition board, like all circuits, has some input resistance denoted by RL.

Sensor R

Vs

+ -

s

R1

Computer Sensor R

+ -

R2

+

VL R1

-

RL

Vs

s

R2

+ -

+

Computer +

VL -

RL

One way to amplify the signal VS is to use a non-inverting amplifier (Figure, left). If we use this circuit, what is VL as a function of Vs and the four resistors, assuming that the opamp is ideal? (b) A second way to amplify the signal is to use the inverting amplifier (Figure, right).

Repeat part (a) for this circuit.

As RS changes, the gain of each circuit may change, leading to unpredictable behavior, which we’d like to avoid. What is the gain of each circuit when RS = 0? This is the ideal (c) gain. Determine analytically the (actual gain)/(ideal gain) for each circuit. Give the constraints on each circuit such that the actual gain approaches the ideal gain. Which circuit is less sensitive to changes in RS?

(a)

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Problem 9.13 Leveraged Bending with Electrostatic Actuation As we will learn in lecture 9, pull-in limits the useful analog travel of a voltage-controlled moveable capacitor plate to one third of the initial gap between the plates. One trick that MEMS designers sometimes use to get greater controllable travel at the center of the beam is leveraged bending (illustrated in the left hand figure below). In this scheme, the bottom capacitor plate is limited to the ends of the beam instead of extending to the center. Because no voltage difference is applied between the center of the beam and the substrate, the center can deflect more than one third of the initial gap before pull-in occurs.

L

F

F

a underlying electrodes

b L

Consider a fixed-fixed silicon beam of width w, length L, and height h that is actuated by leveraged bending. If the electrodes are narrow enough, we may approximate the load as two point forces applied at distances a and b from the ends of the beam, as shown in the right hand figure above. (We would typically design the system so that a = b, but fabrication nonuniformities may lead to some asymmetry.)

Assuming small deformations, calculate the deflection of the beam at the center. You are welcome to use any analytic method you like (explicit integration of the beam equation, use of tabulated beam bending solutions, etc.) mechanicalengineeringassignmenthelp.com mechanicalengineeringassignmenthelp.com

Problem 5.9 : Circuit representation of a lumped mechanical system A mechanically coupled micromechanical resonator can be lumped modeled as shown in the figure. Derive an equivalent circuit in the eV convention and find the transfer function x1(s)/F(s).

F

k2 m2

k1

k3 m1

x2 b 1

x1

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Solutions Problem 9.14 Bending of an AFM Cantilever For LPCVD stoichiometric silicon nitride, E = 270 GPa. From the lecture on structures, for a cantilever with the following configuration and loading d 2w the differential equation of beam bending is: d x L   x 2 M z EI where M is the internal moment at distance x: M = -F (L-x) and I is the moment of inertia of the cross section at the distance x. For a rectangular cross section: I = bh3/12 For our case, h = 0.5 m and b is a function of x due to the triangular shape of the cantilever. b a For our geometry: L  xL

a = 50µm

Therefore:

I 

a

 L  x h

3

12 L

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b

x

L= 250µm

Therefore: ( ) 3

and Integrating twice: 2

At the tip (x = L), the deflection is: And the equivalent spring constant is: For a tip deflection of 2 m: And the maximum stress at the support is: Problem 14.12 : Circuit loading a) Assuming we have an ideal op-amp Rs + -

Vs

+ -

i2

R2 i1

R1

+

V - 2 +

V - 1

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+

VL -

RL

b) Assuming again we have an ideal op-amp i- = 0  is = iL and V+ = V- = 0 Also V- - VL = iL R1 , Vs - V-= is ( Rs + R2 )L s 1 s 2 Therefore V / V = - R / (R + R )

R1 Rs

is

R2

+

Vs

+ -

iL +

VL -

RL

The ratio is negative, hence the name "inverting". c) The gain G is defined as the ratio of the output voltage to the input voltage, or VL / Vs. For the non-inverting amplifier: Gideal = 1+ R2 / R1 = Gactual for all values of Rs Gactual / Gideal = 1 For the inverting amplifier:

ideal = - R1 / R2 Gactual / Gideal (Rs + R2 ) G

2

= R

/

Thus for the non-inverting amplifier, the ideal and actual gains are always equal, while for the inverting amplifier, the actual gain approaches the ideal gain only when the sensor impedance approaches zero (assuming both amplifiers are ideal). Clearly, the noninverting amplifier is less sensitive to changes in Rs than the inverting one.

Problem 9.13 : Leveraged Bending with Electrostatic Actuation a) The governing equation for the deflection of a doubly clamped beam under a point load as shown in the figure is, mechanicalengineeringassignmenthelp.com mechanicalengineeringassignmenthelp.com

EI dx4

 F  x  a  ,

d4 w

F

a L

where  x  a 1 denotes a concentrated load at a. Integrating, we have,

Apply the boundary conditions w(0)  w( L)  w '(0)  w '(L)  0 , we get,

Hence, the center deflection is,

If we next examine the effect of the point load acting at the right hand side of the beam (acting alone as if the load at “a” had been removed), we get a similar setup. mechanicalengineeringassignmenthelp.com mechanicalengineeringassignmenthelp.com

Applying the boundary conditions w(0)  w( L)  w '(0)  w '(L)  0 , we get,

In this case, the center deflection is: Applying superposition to the beam, we can obtain the center deflection due to both loads to be,

Problem 5.9 : Circuit representation of a lumped mechanical system k2

F m2

k1

m1

x2

b1

k3

x1

In the e  V convention, the flows are the time derivatives of the displacements xi of the different bodies, and effort is the external force. Since k1 and b1 share the same displacement, and hence, the same flow, they are in series. The same holds for k2 and m2 and also for k3 and m1. By observation, the net flow through b1 is (x˙2  x˙1 ) . Hence, the equivalent circuit can be represented as shown in Figure. mechanicalengineeringassignmenthelp.com mechanicalengineeringassignmenthelp.com

m2 F

1/k2

m1

F

1/k1

+ –

x˙2

x˙1 b1

k2/s

m2s

1/k3 + –

m1s

k 3/ s

k1/s

x2 s

s domain

x 1s b1

Apply KVL for the left loop, we have (1)

Substitute into (1), we have,\

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