Sammakia 4-06Thermal Management of Flexible Electronic Systems(what are the unique thermal/mechanical design features of a flexible electronic system)Sammakia 4-06Modes of Heat Transfer in Electronic SystemsSammakia 4-06•Conduction (relatively simple analysis, will introduce today)•Convection (mostly empirical and numerical)•Radiation (often negligible at low operating temperatures)•Multimode (combinations of the above modes)Modes of heat transferSammakia 4-06Classification of levels of packaging…why ???Thermal Mechanical Electrical Reliability FunctionalSammakia 4-06Thermal aspectsFirst level….conduction heat transferSecond level….convection and conductionSystem level….thermodynamic considerations, overall system balance, applicationSammakia 4-06•The design of a system should not be a set of individual requirements such as thermal, mechanical, electrical and functional. There should be one design that is optimal and meets all of the requirements. •Case in point thermal and mechanical requirements.Sammakia 4-06•Thermal and mechanical requirements are often conflicting.•For example a large Cu heat sink is likely to adversely impact interconnect reliability (fatigue, shock, vibration)Sammakia 4-06•Interconnects are vulnerable failure points in all systems•Chip level interconnects are the most vulnerable (size and materials) •This has a direct impact on thermal management and the mechanical design of the system•Flexible electronics offer an advantageSammakia 4-06•Conduction (relatively simple analysis, will introduce today)•Convection (mostly empirical and numerical)•Radiation (often negligible at low operating temperatures)•Multimode (combinations of the above modes)Modes of heat transferSammakia 4-06ConductionHeat transfer mode in solids and stationary fluids. Heat is transferred via random molecular interactions. In gases: random molecular collisions (translational + vibration + rotational components) In solids: combination of free electron transport and latticevibrations In liquids: similar to gases but closer bonding between moleculesSammakia 4-06ConductionRate of heat transfer: Fourier’s “Law” (isotropic material):dxdTkqTkq −=∇−="";~~1-D conductionTemperature predictions: Heat diffusion equation(Fourier Law + energy balance)tTkckqzTyTxT∂∂=+∂∂+∂∂+∂∂ρ&222222Rectangular coordinates;k is constant• Need 6 boundary conditions + 1 initial conditionSammakia 4-06Concept of Thermal Resistance0=q&022=dxTd()112TLxTTT +−=LkTTq )(21"−=TLkAAqq ∆⎟⎠⎞⎜⎝⎛=="Voltage (V)Analogous to V= IRororL••T1T2q′′x• Steadystate:Current (I)kALqTRth/=∆=Slab:⎟⎟⎠⎞⎜⎜⎝⎛=12ln21rrLkRthπ⎟⎟⎠⎞⎜⎜⎝⎛−=211141rrkRthπCylindrical shell:Spherical shell:Assuming:• 1-D• No internal heatgeneration:Sammakia 4-06Concept of Thermal ResistanceIn electronic packaging, a single chip package contains manymaterials. An overall resistance is defined:qTTRcjjc−=: Internal thermal resistanceJunction temperatureCase temperatureTypical values:80 K/W: plastic package, no spreader12 -20 K/W: plastic package with spreader5 - 10 K/W: ceramic packageSammakia 4-06•Conduction (relatively simple analysis, will introduce today)•Convection (mostly empirical and numerical)•Radiation (often negligible at low operating temperatures)•Multimode (combinations of the above modes)Modes of heat transferSammakia 4-06Thermal management in a second level packageGoverning Equations for fluids in motion:Assuming the flow to be steady with constant properties except for the thermal conductivity in general and the density in the buoyancy term specifically, and neglecting viscous dissipation, the governing equations are,:Continuity (mass conservation):∇. V= 0Momentum conservation:ρ (V.∇)V= - ∇p + µ∇2V+ ρg β(T-T∞)iEnergy conservation:ρcp(V. ∇)T = ∇.(k∇T)Sammakia 4-06Sammakia 4-06•Navier Stokes equations fairly complex to solve•Most practical applications require a numerical solution•Several commercial codes are available, provide good design tools•It is possible to solve conjugate conduction/convection./radiation problemsThermal management in a second level package…cont.FLOTHERM (TM)Sammakia 4-06diffuser ductscomputer equipment racksreturn vents for hot exhaust air(on parallel walls)cold aisles(chilled air supply)13.42 m long6.05 m widediffuser ductscomputer equipment racksreturn vents for hot exhaust air(on parallel walls)cold aisles(chilled air supply)13.42 m long6.05 m wideSammakia 4-06z=0.8mz=6.5mz=12mz=16myxzz=20mMost Racks ~19 kWz=0.8mz=6.5mz=12mz=16myxzyxzz=20mMost Racks ~19 kWSammakia 4-06On the other hand there are very interesting small scale problems at the micro level; Example of micro-channel coolingSammakia 4-06Channels in Perpendicular DirectionTemperature DistributionFluid flow directionFluid flow directionSammakia 4-06Temperature at various cross-sections along the lengthSammakia 4-06Variation of Maximum Temperature with Velocity(Single Channel with Adiabatic BCs)0.0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0020406080100120140160Chi^2/DoF = 2.77172R^2 = 0.99924 y0 31.89508 ±1.50026A1 224.53191 ±16.14905t1 0.03871 ±0.00338A2 41.0406 ±3.37751t2 0.62888 ±0.12918 y0 + A1e^(-x/t1) + A2e^(-x/t2) Simulated Data Second order exponential decayTemperature (oC)Velocity (m/s)Sammakia 4-06Variation of Pressure Drop with Velocity(Single Channel with Adiabatic BCs)0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.00400080001200016000200002400028000Pressure (Pa)Velocity (m/s) B Polynomial Fit of Data2_BY = A + B1*X + B2*X^2Parameter Value Error------------------------------------------------------------A -30.44094 18.76507B1 4372.04107 37.73392B2 630.1282 9.51187Sammakia 4-06•Conduction (relatively simple analysis, will introduce today)•Convection (mostly empirical and numerical)•Radiation (often negligible at low operating temperatures)•Multimode (combinations of the above modes)Modes of heat transferSammakia 4-06Example using thermal grease as the interface materialSammakia 4-06Thermal grease in a ceramic single chip packageWithout greaseRint = 10 C/WWith grease Rint= 3.5 C/WWhy not use adhesive ?Sammakia 4-06AirAirAirRcap-airRcard-air1Rcard-air2Sammakia 4-06Resistance Equation Value oC/WSpreading in Si Rj-c = ln(ro-ri)/2 π kt 0.6 Through a solder joint Rc4 = l/kA (Assume all solder joints conduct in parallel) 1.1 Spreading in the substrate Rsubstrate= ln(ro-ri)/2 π kt 6.4 Through