Lecture 09 scaling laws
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Dr. S. Meenatchi Sundaram discusses scaling laws relevant to micro electro mechanical systems (MEMS). As devices are scaled down in size: (1) Surface area to volume ratio increases, meaning heat dissipation increases relative to heat storage; (2) Spring constants and stresses decrease with decreasing length/size, making smaller structures more flexible; (3) Resistance increases with decreasing size due to the scaling of resistivity. Sheet resistance, which accounts for thickness, is often used instead for microscale devices. (4) Forces related to fluid flow and pressure drop scale differently with size, with implications for design of microfluidic systems. While electromagnetic forces decrease with size, permanent magnet
Lecture 09 scaling laws
- 1. 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 Sundaram Email: meenasundar@gmail.com 1
- 2. 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
- 3. 2 Surface Area to Volume Ratio ܸ݁݉ݑ݈ ൌ ܽଷ ܽ ݎܨ ൌ 10, ܸ ൌ 10 01 ݔ 01 ݔ ൌ 1000 ܽ ݎܨ ൌ 0.1 V ൌ 0.1 x 0.1 x 0.1 ൌ 0.001 ܵ ൌ 0.1 ൊ 10 ൌ 0.01 Scale for change in length ሺܵሻ Scale for change in Volume ሺܵሻ ܵ ൌ 0. 001 ൊ 1000 ൌ 0.000001 From the above example it can be seen that ܵ ൌ 0.01ଷ Conclusion: If a body changes size by a scale S, the volume changes by a scale ܵଷ. The same can be shown for surface area, The surface area will change by scale ܵଶ Characteristics of MEMS
- 4. 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: ܸଵ ൌ 1000 ܷ݊݅ݏݐଷ ܸଶ= 0.001 ܷ݊݅ݏݐଷ ܸଵ ܸ݅ ݄݊ܽݐ ݎ݁ݐܽ݁ݎ݃ ݏ݁݉݅ݐ 0000001 ݏଶ. • If both cubes were heated to the same temperature, the smaller cube would contain 1000000 times less heat than the larger cube. On the other 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. It is an important parameter in the design of a thermally actuated devices.
- 5. 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.
- 6. Design Issues of Micro Sensors 5 • Step 1: Derive eqn for parameter of interest. ݇ ൌ ݐݓܧଷ 4݈ଷ • Step 2: Identify all scale related parameters. Let l be L, w be aL and t be bL ݇ ൌ ܧሺܽܮሻሺܾܮሻଷ 4ܮଷ • ݇ ൌ ாర ସయ = ܮ ா ସ Hence: ݇ ∝ ܮ • This implies : As L decreases, k decreases. • Therefore, the smaller the beam, the smaller the spring constant k, or the more flexible it is.
- 7. 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 ߪ ൌ ݈ܽߩ݄ݓ ߨݎଶ ൌ ܮ ܾܮ ܿܮ ߩܽ ߨሺ݁ܮሻଶ Step 4: Re- write expression: ߪ ൌ ܾܿߩܽ ߨ݁ଶ ܮଷ ܮଶ ߪ ∝ ܮ Step 1: Derive governing question for tensile stress ߪ ൌ ܨ ܣ ߪ ൌ ݉ܽ ߨݎଶ
- 8. Design Issues of Micro Sensors 7 Size related to mass Consider 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
- 9. 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 ܴ ൌ ρ ܮ ሺܾܮሻଶ ܴ ∝ ܵିଵ As the resistor is scaled down by scale factor S, its resistance increases as S-1.
- 10. 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 =7.8 -cm. R = ρL/A = 7.8 (10-2m) * 20 (10-6m)/5(10-12m) = 31.2 104 = 312 k But, usually the resistance value for design is around 2 k and may vary up to 2.5k . Sheet Resistance needs to be used
- 11. Design Issues of Micro Sensors 10 Sheet Resistance R = ρ L / A = ρ L / w t = ρ ௧ L ௪ = Rs L ௪ 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 ·m, which is more completely stated in units of ·m2/m ( ·area/length). When divided by the sheet thickness, 1/m, the units are ·m·(m/m)·1/m = . 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 " /sq" or " /□ "), which is dimensionally equal to an ohm, but is exclusively used for sheet resistance. This is an advantage, because sheet resistance of 1 could be taken out of context and misinterpreted as bulk resistance of 1 ohm, whereas sheet resistance of 1 /sq cannot thus be misinterpreted.
- 12. 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.
- 13. Design Issues of Micro Sensors 12 Fluid Mechanics A: Volumetric Flow Q, ܳ ൌ ߨܽସ∆ܲ 8ߤܮ Hence: ܳ ∝ ܵସ This implies that a reduction of 10 to the radius, will lead to a 10000 time reduction in volumetric flow.B: Pressure Drop P ൌ െ 8ߤܸ௩ܮ ܽଶ ∆ܲ ∝ ܵିଷ A reduction of 10 times in conduit radius leads 1000 times increase in pressure drop per unit length
- 14. Design Issues of Micro Sensors 13
- 15. 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)