Dissipation and the Thermal Energy Domain Part I

Dissipation and The Thermal Energy Domain Part I

Joel Voldman* Massachusetts Institute of Technology *(with thanks to SDS)

Figure 1 on p. 55 in: Leonov, V. N., N. A. Perova, P. De Moor, B. Du Bois, resistance across the pixel C. Goessens, B. Grietens, A. Verbist, C. A. Van Hoof, and J. P. Vermeiren. "Micromachined Poly-SiGe Bolometer Arrays for Infrared Imaging and Spectroscopy." Proceedings of SPIE Int Soc Opt Eng 4945 (2003): 54-63.

Silicon nitride and vanadium oxide 50 µm IR 0.5 µm Better design… Radiation Y-metal 2.5 µm Image removed due to copyright restrictions.

B Figure 2 on p. 56 in: Leonov, V. N., N. A. Perova, P. De Moor, B. Du Bois, E C. Goessens, B. Grietens, A. Verbist, C. A. Van Hoof, and Monolithic bipolar J. P. Vermeiren. "Micromachined Poly-SiGe Bolometer X-metal transistor Arrays for Infrared Imaging and Spectroscopy.“ Image by MIT OpenCourseWare. Proceedings of SPIE Int Soc Opt Eng 4945 (2003): 54-63. Honeywell

Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. J. Voldman: 2.372J/6.777J Spring 2007, Lecture 12 - 5 Thermal flow sensing > A time-of-flight flowrate sensor > One resistor creates a heat pulse > A downstream resistor acts as a temperature sensor > Time for heat pulse to drift downstream is inversely related to Figure 1 on p. 686 in: Meng, E., and Y.-C. Tai. "A Parylene MEMS Flow Sensing Array.“ flowrate Technical Digest of Transducers'03: The 12th International Conference on Solid-State Sensors, Actuators, and Microsystems, Boston, June 9-12, 2003. Vol. 1. Piscataway, NJ: IEEE Electron Devices Society, 2003, pp. 686-689. ISBN: 9780780377318. © 2003 IEEE. > This example illustrates the important benefit of MEMS materials • Large range in thermal conductivities • From vacuum (~0) to metal (~100’s W/m-K) Figure 3 on p. 687 in: Meng, E., and Y.-C. Tai. "A Parylene MEMS Flow Sensing Array.“ Technical Digest of Transducers'03: The 12th International Conference on Solid-State Sensors, Actuators, and Microsystems, Boston, June 9-12, 2003. Vol. 1. Piscataway, NJ: IEEE Electron Devices Society, 2003, pp. 686-689. ISBN: 9780780377318. © 2003 IEEE.

Image removed due to copyright restrictions. OMRON flow sensor. > Convection alters temperature profile in a predictable fashion

Flow Calm

Thermopile B

Temperature distribution Heater

Image by MIT OpenCourseWare.

perfectly ) g / 0.177 V ( 0.176 y

The “extra” energy goes into the t

• i 0.175 v i t i

thermal domain (e.g., heat) s 0.174 n

e 0.173 S • We can never totally recover that 0.172 0.171 heat energy 0.170 -30 -20 -10 0 10 20 30 40 50 60 70 80 Temperature (oC) ADI ADXL320 Image by MIT OpenCourseWare.

I R 1 2 + Energy stored in capacitor: WCVCS= + 2 VS C VC Energy delivered by power supply: - - V 2 Pt()== IVS e−tRC/ Image by MIT OpenCourseWare. SsR VV=−()1 e−tRC/ ∞ CS WPdtCV==2 ss∫ S dV V IC==CS e−tRC/ 0 dt R

> It is easier to put ENERGY INTO than get WORK OUT of thermal domain

> All domains are linked to Elastic Electric Magnetic thermal domain via dissipation Chemical Fluids

> Thermal domain is linked to Thermal energy domain all domains because temperature affects Image by MIT OpenCourseWare. constitutive properties Adapted from Figure 11.2 in Senturia, Stephen D. Microsystem Design. Boston, MA: Kluwer Academic Publishers, 2001, p. 271. ISBN: 9780792372462.

> Heat engines convert heat into mechanical work, but not perfectly efficiently, just like the charging of a capacitor cannot be done perfectly efficiently > This is a statement of nd irreversibility: the 2 Law of Courtesy of Zyvex Corporation. Used with permission. Thermodynamics Electrical energy > Zyvex heatuator • One skinny leg and one fat leg Thermal energy • Run a current and skinny leg will heat up Lost to thermal • Structure will bend in response reservoir Strain energy and deflection

Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. J. Voldman: 2.372J/6.777J Spring 2007, Lecture 12 - 15 Governing Equations > Some introductory notation X ~ Q Thermal energy [J] = X ~ V > Be careful with units Q Thermal energy/volume [J/m3] and normalizations & = IQ Q Heat flow [W] Heat flux [W/m2] JQ ∂Q Heat capacity at constant volume (J/K) CV = ∂T Volume ∂Q Heat capacity at constant pressure (J/K) C = P ∂T Pressure Are the same for incompressible materials PV == CCC ~ 3 C = C Heat capacity/unit volume (J/K-m ) V ~ ~ C Heat capacity/unit mass (J/K-kg) AKA specific heat Cm = ρm Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. J. Voldman: 2.372J/6.777J Spring 2007, Lecture 12 - 16 Governing Equations

> Like many domains, ⎛ stuffnet ⎞ ⎛generation ⎞ d ⎛stuff in ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ = ⎜entering ⎟ + ⎜of stuff in ⎟ conservation of energy dt ⎝ volume ⎠ ⎜volume ⎟ ⎜ volume ⎟ leads to a continuity ⎝ ⎠ ⎝ ⎠ d bdV nF +⋅−= gdVdS equation for thermal dt ∫∫ ∫ energy Get point relation ∂b Divergence theorem and dV F +⋅∇−= gdVdV V bring derivative inside ∫∫∫∂t ∂b F +⋅−∇= g ∂t For heat b g ~ transfer Qd ~ Q =⋅∇+ PJ sources n dt F S Image by MIT OpenCourseWare.

• Often not important for MEMS, but JQc=−hT( 21 T) sometimes is… • Talk more about this later > Radiative heat transfer

• Between two bodies (at T1 and T2) • Stefan-Boltzmann Law 4 4 = σ SBQ ( 212 −TTFJ 1 ) • Can NEVER turn off

Q = −κ∇TJ ⇓

JQ ()κ∇⋅−∇=⋅∇ T ⇓ ~ ∂Q ~ ()κ +∇⋅∇= PT ∂t sources dQ% For homogeneous materials, with = C% dT

∂T κ 2 1 ~ ~ T +∇= ~ P ∂t C C sources

Cite as: Joel Voldman, course materials for 6.777J / 2.372J Design and Fabrication of Microelectromechanical Devices, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu/), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. J. Voldman: 2.372J/6.777J Spring 2007, Lecture 12 - 19 Thermodynamic Realities > The First law of thermodynamics implies that entropy is a generalized displacement, and temperature is a generalized effort; their product is energy > The Second Law states that entropy production is ≥0 for any process • In practice it always increases > Entropy is not a conserved quantity…. > Thus, it does not make for a good generalized variable > Therefore, we use a new convention, the thermal modeling convention, with temperature as effort and heat flow (power) as the flow. Note that the product of effort and flow is no longer power!!! • But heat energy (thermal displacement) is conserved • Just like charge (electrical displacement) is conserved

> Heat current source IQ is represented with a flow source IQ > Temperature difference source ΔT is represented with an effort source ΔT + -

Thermal Electrical conductivity conductivity κ∇−= TJ Relation between = σ = −σ ∇VEJ Q effort and flow eE e

∂T κ 2 ~ ~ +∇= PT ∂t C sources 2 2 T =∇ 0 Laplace’s Eqn. V =∇ 0

= TT Continuity of effort 12 = VV 12 No No heat charge ()()⋅=⋅ nJnJ Continuity of flow storage Q 1 Q 2 ()()E 1 E ⋅=⋅ nJnJ 2 storage

IQ RT 1 L R = [K/W] For bar of uniform T cross-section + ΔT - κ A > Plus, heat conduction and current flow obey the same differential equation > Thus, we can use exactly the same solutions for thermal resistors as for electrical resistors • Just change σ to κ

types of heat flow? T1 > Convection I > Use linear resistor Q

JhTTQc=−( 21) IhATThAT=−=Δ Qc()21 c IQ R 1 T, c on v RT, conv = hAc + ΔT -

Large-signal model

IQ IQ T T 2 - 1 T1 + T2

4 4 = σ SBQ ( 212 −TTFJ 1 ) 4 4 = σ SBQ ()212 −TTAFI 1

> Radiation 4 • Many ways to linearize IFATQSB≈ σ 12 2 • We show two approaches 3 δ IFATTQSB= ()4σδ12 2

T2>>T1

δIQ RT,rad

44 IQSB=−σ FAT12( 2 T 1 ) + δT -

T ≅T 2 1 4 4 IQSB≈+−σδFA12(() T 1 T T1 )

3 δσIFATTQSB= ()4 12 δ 1

IQ CT

+ ΔT -

> In electromechanical energy transduction, we introduced the two- port capacitor • Energy-storing element coupling the two domains • Capacitor because it stores potential energy > What will we use to couple electrical energy into thermal energy? • In the electrical domain, this is due to Joule dissipation, a loss mechanism » Therefore it looks like an electrical resistor • In the thermal domain it looks like a heat source » Therefore it looks like a thermal current source

+ 2 > We reverse convention on direction of IR + port variables in thermal domain V R ΔT • OK, because IQ·ΔT does not track power - • Reflects fact that heat current will - always be positive out of transducer > This is not energy-conserving • Dissipation is intrinsic to transducer > This is not reciprocal • Heat current does not cause a voltage > Thermal domain can couple back to the electrical domain • See next time…

Silicon nitride and vanadium oxide 50 µm IR 0.5 µm Radiation Y-metal 2.5 µm + B C E IQ T ΔT RT Monolithic bipolar X-metal transistor - Image by MIT OpenCourseWare. ⎛ ⎞ IR > Heat input is a current ⎜ 1 ⎟ QT =Δ ⎜ RIT TQ // ⎟ = ⎝ T sC ⎠ 1+ TT sCR > Heat loss is a resistor Δ = IRT > Heat storage in mass is a capacitor QTss τ = CR TT

> Increasing RT increases response • This means better thermal isolation • There is always some limiting value determined by radiation

> Given a fixed RT, decreasing CT improves response time • This means reducing the mass or volume of the system

Δ = IRT QTss

τ = CR TT

> What goes into RT?

> RT is the parallel combination of all loss terms • Conduction through the air and legs • Convection • Radiation > We can determine when different terms dominate

Material κ (W/m-K) Silicon 148 1 L R = Silicon Nitride 20 Tair, κ A ()2.5 μm Thermal Oxide 1.5 = ()10−5 W/m-K()() 50 μμ m 50 m Air (1 atm) 0.03 8 RTair, =10 K/W Air (1 mtorr) 10-5 Conduction through legs dominates

> Convection 1 R = • At low pressure, there Trad, 3 4σ SB F AT will be no air movement 12 1 • Convection will not exist 1 R ,radT = ()⋅ −8 W/m1067.54 42 ()()()()505.0K 50m μμ300m K 3 > Radiation R ⋅= 103.1 8 K/W • There is transfer ,legsT between plate and body Leg conduction dominates loss • Use case where bodies are close in temp • Radiation negligible % > Very fast time constant CCTmm= ρ V • ~ms is typical for = ()700 J/kg-K() 3000 kg/m3 ()()() 50 μm 50 μμm 0.5 m thermal MEMS =⋅310−9 J/K

RCTT=1.5 ms

> How can we make Silicon nitride and vanadium oxide responsivity higher? 50 µm IR 0.5 µm Radiation Y-metal > Change materials 2.5 µm

E > Decrease thickness or Monolithic bipolar X-metal transistor width of legs, or increase Image by MIT OpenCourseWare. length • This reduces mechanical rigidity