Dr. S.Meenatchi Sundaram, Department of Instrumentation & Control Engineering, MIT, Manipal

ICE 4010: MICRO ELECTRO MECHANICAL SYSTEMS (MEMS)

Lecture #09

Scaling Laws

Dr. S. Meenatchi SundaramEmail: meenasundar@gmail.com

1

Effect of size reduction

2Dr. S.Meenatchi Sundaram, Department of Instrumentation & Control Engineering, MIT, Manipal

As you decrease the size

• Friction > inertia

• Heat dissipation > Heat storage

• Electrostatic force > Magnetic Force

2

Surface Area to Volume Ratio

10, 101010 1000

0.1 V0.1x0.1x0.10.001

0.1 10 0.01

Scale for change in length Scale for change in Volume

0. 001 1000 0.000001

From the above example it can be seen that 0.01

Conclusion: If a body changes size by a scale S, the volume changes by a scale .

The same can be shown for surface area, The surface area will change by scale

Characteristics of MEMS

Design Issues of Micro Sensors

3

What are the implications of this?Heat Storage ∝ Volume

Heat Dissipation ∝ Surface Area

Buoyant Forces

From the example above: 1000 !

= 0.001 !

!1000000 !" #.

• If both cubes were heated to the same temperature, the smaller cubewould contain 1000000 times less heat than the larger cube. On theother hand, surface area is only 10,000 times less than the larger cube.

• This will mean 100 times more heat dissipation on the smaller cube.Volume relates, for example, to both mechanical and thermal inertia.Thermal inertia is a measure on how fast we can heat or cool a solid. Itis an important parameter in the design of a thermally actuated devices.

Design Issues of Micro Sensors

4

Scaling effects on spring constant (k)

• Consider a beam, length L, width w, Thickness t, and Young's Modulus E.

Design Issues of Micro Sensors

5

• Step 1: Derive eqn for parameter of interest.

$ %&

4

• Step 2: Identify all scale related parameters.Let l be L, w be aL and t be bL

$ %()(

4(

• $ +,-.

/-0= (

+,

/Hence: $ ∝ (

• This implies : As L decreases, k decreases.

• Therefore, the smaller the beam, the smaller the spring constant k, or the more flexible it is.

Design Issues of Micro Sensors

6

Stress in a rod connected to a mass experiencing a constant acceleration

Step 2: Identify all parameters related to length

Step 3: Redefine length related parameters

1 &#2

3

( )( 4( 2

3(

Step 4: Re- write expression:

1 )42

3(

(

1 ∝ (

Step 1: Derive governing question for

tensile stress

1

5

1

3

Design Issues of Micro Sensors

7

Size related to massConsider a cantilever beam with L=500 um, W=50 um and t=5um. Let 10mg mass is attached at the free end with a size of l=w=50um.

m=rho x vol

10 mg = 2330 kg/m3 x vol

10 x 10^-6 kg = 2330 kg/m3 x vol

0.00429 x 10^12 um3 = 50 x 50 x t

um3

429x10^7/2500 um = t

1.71x10^6 um = t

Mass of the cantilever = 500 x 50 x 5 um^3 x rho

= 0.29 x 10^-3 mg

Design Issues of Micro Sensors

8

Resistance

• Given a conductor with a length L, cross-

section A, and resistivity ρ, the resistance is

R = ρ L / A

6 ρ(

)(

6 ∝ 7

As the resistor is scaled down by scale factor

S, its resistance increases as S-1.

Design Issues of Micro Sensors

9

Consider a small piece of piezoresistor embedded

on a silicon diaphragm with L= 20um, W=5 um and

t=1um. Resistivity(ρ) of p type silicon with 100

orientation =7.8Ω-cm.

R = ρL/A = 7.8 Ω(10-2m) * 20 (10-6m)/5(10-12m)

= 31.2 104 Ω = 312 kΩ

But, usually the resistance value for design is

around 2 kΩ and may vary up to 2.5kΩ.

Sheet Resistance needs to be used

Design Issues of Micro Sensors

10

Sheet Resistance

R = ρ L / A = ρ L / w t = ρ8

L:

= Rs

L:

Sheet resistance is a special case of resistivity for a uniform sheet

thickness. Commonly, resistivity (also known as bulk resistance,

specific electrical resistance, or volume resistivity) is in units of Ω·m,

which is more completely stated in units of Ω·m2/m (Ω·area/length).

When divided by the sheet thickness, 1/m, the units are

Ω·m·(m/m)·1/m = Ω. The term "(m/m)" cancels, but represents a

special "square" situation yielding an answer in ‘ohms’.

An alternative, common unit is "ohms per square" (denoted "Ω/sq" or

" Ω/ "), which is dimensionally equal to an ohm, but is exclusively

used for sheet resistance.

This is an advantage, because sheet resistance of 1 Ω could be

taken out of context and misinterpreted as bulk resistance of 1 ohm,

whereas sheet resistance of 1 Ω/sq cannot thus be misinterpreted.

Design Issues of Micro Sensors

11

Electromagnetism• Faraday’s law governs the induced force (or a motion)

in the wire under the influence of a magnetic field.

• The scaling of electromagnetic force follows: F ∝ S4.

• For electromagnets, as S decreases, these forces decrease because it is difficult to generate large magnetic fields with small coils of wire.

• However permanent magnets maintain their strength as they are scaled down in size, and it is often advantageous to design magnetic systems that use the interaction between an electromagnet and a permanent magnet.

Design Issues of Micro Sensors

12

Fluid MechanicsA: Volumetric Flow Q,

; 3/∆=

8?(

Hence:

; ∝ /

This implies that a reduction of

10 to the radius, will lead to a

10000 time reduction in

volumetric flow.B: Pressure Drop ∆P

@8?ABC(

∆= ∝ 7

A reduction of 10 times in conduit radius leads 1000 times increase in

pressure drop per unit length

Design Issues of Micro Sensors

13

Design Issues of Micro Sensors

14

Benefits Of Scaling

• Speed (Frequency increase, Thermal Time constraints reduce)

• Power Consumption (actuation energy reduce, heating power reduces)

• Robustness (g-force resilience increases)

• Economy (batch fabrication)