Thermal Modeling© M. T. Thompson, 2009 1 Power Electronics Notes 29 Thermal Circuit Modeling and...

Thermal Modeling

© M. T. Thompson, 2009

1

Power Electronics Notes 29Thermal Circuit Modeling and Introduction

to Thermal System Design

Marc T. Thompson, Ph.D.

Thompson Consulting, Inc.

9 Jacob Gates Road

Harvard, MA 01451

Phone: (978) 456-7722

Email: [email protected]

Website: http://www.thompsonrd.com

Jeff W. Roblee, Ph.D.VP of EngineeringPrecitech, Inc.Keene, NHwww.ametek.com

Thermal Modeling


2

Summary

• Basics of heat flow, as applied to device sizing and heat sinking

• Use of thermal circuit analogies– Thermal resistance– Thermal capacitance

• Examples– Picture window examples– Magnetic brake– Plastic tube in sunlight

Thermal Modeling


• All components (capacitors, inductors and transformers, semiconductor devices) have maximum operating temperatures specified by manufacturer• High operating temperatures have undesirable effects on components:

Need for Component Temperature Control

Capacitors• Electrolyte evaporation rate increases significantly with temperature increases and thus shortens lifetime

Magnetic Components• Losses (at constant power input) increase above 100 °C• Winding insulation (lacquer or varnish) degrades above 100 °C

Semiconductors• Unequal power sharing in parallel or series devices• Reduction in breakdown voltage in some devices• Increase in leakage currents• Increase in switching times

3

Thermal Modeling


• Control voltages across and current through components via good design practices

• Snubbers may be required for semiconductor devices.• Free-wheeling diodes may be needed with magnetic

components• Maximize heat transfer via convection and radiation from component to ambient

• Short heat flow paths from interior to component surface and large component surface area.• Component user has responsibility to properly mount temperature-critical components on heat sinks.

• Apply recommended torque on mounting bolts and nuts and use thermal grease between component and heat sink.

• Properly design system layout and enclosure for adequate air flow

Temperature Control Methods

4

Thermal Modeling


Heat Transfer• Heat transfer (or heat exchange) is the flow of thermal

energy due to a temperature difference between two bodies

• Heat transfers from a hot body to a cold one, a result of the second law of thermodynamics

• Heat transfer is slowed when the difference in temperature between the two bodies reduces

5

Thermal Modeling


6

Intuitive Thinking about Thermal Modeling

• Heat (Watts) flows from an area of higher temperature to an area of lower temperature

• Heat flow is by 3 mechanisms– Conduction - transferring heat through a solid

body– Convection - heat is carried away by a moving

fluid• Free convection• Forced convection - uses fan or pump

– Radiation• Power is radiated away by electromagnetic

radiation• You can think of high- thermal conductivity

material such as copper and aluminum as an easy conduit for conductive power flow…. i.e. the power easily flows thru the material

Thermal Modeling


7

Thermal Circuit Analogy• Use Ohm’s law analogy to model thermal

circuits

• Thermal resistance

• k = thermal conductivity (W/(mK))• Thermal capacitance: analogy isn’t as

straightforward• cp = heat capacity of material (Joules/(kg-K))

TH

TH

CC

RR

PI

TV

kA

lR

A

lR TH

ELECTRICAL THERMAL Forcing variable Voltage (V) Temperature (K) Flow variable Current (A) Power (W) Resistance Resistance (V/A) Thermal resistance (K/W) Capacitance Capacitance (V/C) Thermal capacitance (J/K)

pTH McC

Thermal Modeling


8

Thermal Circuit Analogy

ELECTRICAL THERMAL

Forcing variable Voltage (V) Temperature (K)

Flow variable Current (A) Heat (W)

Resistance Resistance (V/A) Thermal resistance (K/W)

Capacitance Capacitance (V/C) Thermal capacitance (J/K)

• Heat transfer can be modeled by thermal circuits• Using Ohm’s law analogy:

Reference: M. T. Thompson, Intuitive Analog Circuit Design, Newnes, 2006.

thV R I T R P

Thermal Modeling


9

Thermal Circuit Analogy

• Elementary thermal network

Reference: M. T. Thompson, Intuitive Analog Circuit Design, Newnes, 2006.

Thermal Modeling


10

Thermal Resistance

• Thermal resistance quantifies the rate of heat transfer for a given temperature difference

• k = thermal coefficient (W/(mK))• A = cross section (m2)• l = length (m)

TH

l l KR ( ) R

A kA W

T 2T 1

Area A

l

P cond

Thermal Modeling


11

Thermal Capacitance

• Thermal capacitance is an indication of how well a material stores thermal energy

• It is used when transient phenomena are considered• Analogy isn’t as straightforward• M = mass (kg)• cp = heat capacity of material (Joules/(kg-K))

pTH McC TH

dTP C

dt

Thermal Modeling


12

Heat Flow Mechanisms

• Heat flows by 3 mechanisms; the driving force for heat transfer is the difference in temperature

1. Conduction

2. Convection • Free convection• Forced convection

3. Radiation

Reference: R. E. Sonntag and C. Borgnakke, Introduction to Engineering Thermodinamics, John Wiley, 2007

Thermal Modeling


13

Conduction• Heat is transferred through a solid from an area of

higher temperature to lower temperature

• To have good heat conduction, you need large area, short length and high thermal conductivity

• Example: aluminum plate, l = 10 cm, A=1 cm2, T2 = 25C (298K), T1 = 75C (348K), k = 230 W/(m-K)

condTH

cond R

TT

kA

lTT

P,

2121

WattCkA

lR o

condTH /35.4)10(230

)1.0(4,

WattsPcond 5.11

35.4

50

T 2T 1

Area A

l

P cond

Thermal Modeling


14

Thermal Conductivity of Selected Materials

Material Thermal conductivity @ 300K k, W/(m-K)

Copper 400 Aluminum 200

Iron 25-75 Steel (1008) 50

Stainless steel 12-35 Brick 0.4-0.5 Cork 0.04

Epoxies 0.16-1.4 Glass-epoxy 3

Window glass 0.78 Still air 0.027

References: 1. B. V. Karlekar and R. M. Desmond, Engineering Heat Transfer, pp. 8, West Publishing, 1977

2. Burr Brown, Inc., “Thermal and Electrical Properties of Selected Packaging Materials”

Thermal Modeling


Thermal Equivalent Circuits

• Heat flow through a structure composed of layers of different materials

• Thermal equivalent circuit simplifies calculation of temperatures in various parts of structure.

P

R sacsR

jcR

Junction Case Sink Ambient

jT cT sT aT++++

----

Chip T j

Case Tc

Isolation pad

Heat sink T s

Ambient Temperature T a

• Ti = Pd (Rjc+ Rcs + Rsa) + Ta

• If there parallel heat flow paths, then thermal resistances combine as do electrical resistors in parallel.

15

Thermal Modeling


16

Thermal Conductivity of Selected Materials

Reference: International Rectifier, Application note N-1057, “Heatsink Characteristics”

Thermal Modeling


17

Heat Capacity of Selected Materials

Reference: B. V. Karlekar and R. M. Desmond, Engineering Heat Transfer, West Publishing, 1977

• Heat capacity is an indication of how well a material stores thermal energy

Thermal Modeling


18

Heat Capacity of Alloys

Reference: http://www.engineeringtoolbox.com/specific-heat-metal-alloys-d_153.html

Thermal Modeling


19

Convection


• Convection can be free (without a fan) or forced (with a fan)

Thermal Modeling


20

Free Convection

Reference: http://www.freestudy.co.uk/heat%20transfer/convrad.pdf

Thermal Modeling


21

Heat Transfer Coefficient for Convection

• Heat is transferred via a moving fluid• Convection can be described by a heat transfer

coefficient h and Newton’s Law of Cooling:

• Heat transfer coefficient depends on properties of the fluid, flow rate of the fluid, and the shape and size of the surfaces involved, and is nonlinear

• Equivalent thermal resistance:

)( as TThAP

hAR convTH

1,

Test condition h, W/m2oC Air, free convection 5-15

Air, forced convection 15-50 Water, forced convection 300-12,000

Water, boiling 5,500-100,000

Reference: B. V. Karlekar and R. M. Desmond, Engineering Heat Transfer, pp. 14, West Publishing, 1977

Thermal Modeling


22

Free Convection• Heat is drawn away from a surface by a free gas

or fluid• Buoyancy of fluid creates movement• For vertical fin:

• A in m2, dvert in m• Example: square aluminum plate, A=1 cm2, Ta =

25C (298K), Ts = 75C (348K)

asvert

asconv TThA

d

TTAP

25.0

25.1

34.1T sT a

Area A

d vert

)/(2.11

56)01.0(

50)10(34.1

2

25.0

25.14

KmWh

mWP

eq

conv

Thermal Modeling


23

Free Convection Heat Transfer Coefficient (h)• For vertical fin:

• Area A in m2, fin vertical height dvert in m

T sT a

Area A

d vert

25.0

35.1

vert

As

d

TTh

Thermal Modeling


24

Forced Convection• With a fan


Thermal Modeling


25

Forced Convection

• In many cases, heat sinks can not dissipate sufficient power by natural convection and radiation

• In forced convection, heat is carried away by a forced fluid (moving air from a fan, or pumped water, etc.)

• Forced air cooling can provide typically 3-5 increase in heat transfer and 3-5 reduction in heat sink volume– In extreme cases you can do 10x better by using big

fans, convoluted heat sink fin patterns, etc.

Thermal Modeling


26

Thermal Performance Graphs for Heat Sinks

Reference: http://electronics-cooling.com/articles/1995/jun/jun95_01.php

• Curve #1: natural convection (P vs. Tsa)• Curve #2: forced convection curve (Rsa vs. airflow)

12

Thermal Modeling


27

Radiation

• Energy is transferred through electromagnetic radiation


Thermal Modeling


28

Radiation

• Energy is lost to the universe through electromagnetic radiation

• = emissivity (0 for ideal reflector, 1 for ideal

radiator “blackbody”); = Stefan-Boltzmann constant = 5.6810-8 W/(m2K4)

• Example: anodized aluminum plate, = 0.8, A=1 cm2, Ta = 25C (298K), Ts = 75C (348K)

P rad

T s

T a

Area A

Material Emissivity Aluminum

Polished Oxidized Anodized

0.04-0.06 0.2-0.33 0.7-0.9

Copper Polished Dull

0.02 0.15

Glass 0.8-0.95

44asrad TTAP

mWPrad 31)298348)(10)(107.5)(8.0( 4448

Thermal Modeling


29

Radiation

• Incident, reflected and emitted radiation; e.g. body in sunlight

Reference: http://www.energyideas.org/documents/factsheets/PTR/HeatTransfer.pdf

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Emissivity


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Emissivity


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32

Comments on Radiation• In multiple-fin heat sinks with modest temperature

rise, radiation usually isn’t an important effect– Ignoring radiation results in a more conservative

design• Effective heat transfer coefficient due to radiation for

ideal blackbody ( = 1) at with surface temperature 350K radiating to ambient at 300K is hrad = 6.1 W/(m2K), which is comparable to free convection heat transfer coefficient– However, radiation between heat sink fins is

usually negligible (generally they are very close in temperature)

Thermal Modeling


33

IC Mounted to Heat Sink

• Interfaces– Heat sink-ambient: convection (free or forced)– Heat sink-case of IC: conduction– Case – junction: conduction

Thermal Modeling


Multiple Fin Heat Sink

Reference: http://www.oldcrows.net/~patchell/AudioDIY/AudioDIY.html

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Thermal Modeling


35

IC Mounted to Heat Sink

Reference: International Rectifier, Application Note AN-997

Thermal Modeling


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IC Mounted to Heat Sink --- Close-up


• Thermal compound is often used to fill in the airgap voids

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IC Mounted to Heat Sink --- Contact Resistance vs. Torque (TO-247)


Thermal Modeling


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IC Mounted to Heat Sink --- Contact Resistance vs. Interface Material (TO-247)


Thermal Modeling


39

IC Mounted to Heat Sink --- Contact Resistance vs. Interface Material (TO-247)


• Dry vs. thermal compound vs. electrically-insulating pad

Thermal Modeling


40

Thermal Grease

Thermal Modeling


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Heat Sink Pad

Thermal Modeling


Transient Thermal Impedance

• Heat capacity per unit volume Cv = dQ/dT [Joules /oC] prevents short duration high power dissipation surges from raising component temperature beyond operating limits.

• Transient thermal equivalent circuit. Cs = CvV where V is the volume of the component.

P(t)R

jT (t)

aT

Cs

P(t)

t

RPo

tSlope = 0.5

log Z (t)

• Transient thermal impedance Z(t) = [Tj(t) - Ta]/P(t) •= π R Cs /4 =

thermal time constant

• Tj(t = ) = 0.833 Po R

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Thermal Modeling


Use of Transient Thermal Impedance• Response for a rectangular power dissipation pulse P(t) = Po {u(t) - u(t - t1)}.

P(t)

t

RPo

tt1 t1

Z (t)

1-Z (t - t )

net response

-R

• Tj(t) = Po { Z(t) - Z(t - t1) }

• Symbolic solution for half sine power dissipation pulse.

•P(t) = Po {u(t - T/8) - u(t - 3T/8)} ; area under two curves identical.

•Tj(t) = Po { Z(t - T/8) - Z (t - 3T/8) }T/8 3T/8

T/2

Po

t

P(t)

Equivalent rectangular pulse

Half sine pulse

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Thermal Modeling


Multilayer StructuresP(t)

Silicon

Copper mountHeat sink

Tj

CuT

cT

aT

• Multilayer geometry

• Transient thermal equivalent circuit

Tj CuTcT

aT

C (Si)s C (Cu)sC (sink)sP(t)

R (sink)R (Cu)R (Si)

R (Si)

R (Cu)R (Si)

+

R (sink)R (Cu)R (Si)

+ +

log(t)

log[Z (t)]

(Si) (sink) (Cu)

• Transient thermal impedance (asymptotic) of multilayer structure assuming widely separated thermal time constants.

44

Thermal Modeling


Heat Sinks• Aluminum heat sinks of various shapes and sizes widely available for cooling components.

• Often anodized with black oxide coating to reduce thermal resistance by up to 25%.

• Sinks cooled by natural convection have thermal time constants of 4 - 15 minutes.

• Forced-air cooled sinks have substantially smaller thermal time constants, typically less than one minute.

• Choice of heat sink depends on required thermal resistance, Rsa, which is determined by several factors.

• Maximum power, Pdiss, dissipated in the component mounted on the heat sink.

• Component's maximum internal temperature, Tj,max

• Component's junction-to-case thermal resistance, Rjc. • Maximum ambient temperature, Ta,max.

• Rsa = {Tj,max - Ta,max}Pdiss - Rjc • Pdiss and Ta,max determined by particular application.• Tj,max and Rjc set by component manufacturer.

45

Thermal Modeling


Heat Conduction Thermal Resistance

• Generic geometry of heat flow via conduction

Pcond

Temperature = T1Temperature = T2

d

h

b

T > T12

heat flow direction

• Heat flow Pcond [W/m2] =kA (T2 - T1) / d = (T2 - T1) / R cond

• Thermal resistance R cond = d / [k A]

• Cross-sectional area A = hb

• k = Thermal conductivity has units of W-m-1-oC-1 (kAl = 220 W-m-1-oC-1 ).• Units of thermal resistance are oC/W

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Thermal Modeling


Radiative Thermal Resistance

• Stefan-Boltzmann law describes radiative heat transfer.

• Prad = 5.7x10-8 EA [( Ts)4 -( Ta)4 ] ; [Prad] = [Watts]

• E = emissivity; black anodized aluminum E = 0.9 ; polished aluminum E = 0.05

• A = surface area [m2] through which heat radiation emerges.

• Ts = surface temperature [K] of component. Ta = ambient temperature [K].

• (Ts - Ta )/Prad = R ,rad = [Ts - Ta][5.7x10-8EA {( Ts/100)4 -( Ta/100)4 }]-1

• Example - black anodized cube of aluminum 10 cm on a side. Ts = 120 C and Ta = 20 C

• R,rad = [393 - 293][(5.7) (0.9)(6x10-2){(393/100)4 - (293/100)4 }]-1

• R,rad = 2.2 C/W

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Thermal Modeling


Convective Thermal Resistance

• Pconv = convective heat loss to surrounding air from a vertical surface at sea level having height dvert [in meters] less than one meter.

• Pconv = 1.34 A [Ts - Ta]1.25 dvert-0.25

• A = total surface area in [m2]

• Ts = surface temperature [K] of component. Ta = ambient temperature [K].

• [Ts - Ta ]/Pconv = R,conv = [Ts - Ta ] [dvert]0.25[1.34 A (Ts - Ta )1.25]-1

• R,conv = [dvert]0.25 {1.34 A [Ts - Ta]0.25}-1

• Example - black anodized cube of aluminum 10 cm on a side. Ts = 120C and Ta = 20 C.

• R,conv = [10-1]0.25([1.34] [6x10-2] [120 - 20]0.25)-1

• R,conv = 2.2 C/W

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Thermal Modeling


Combined Effects of Convection and Radiation

• Heat loss via convection and radiation occur in parallel.

• Steady-state thermal equivalent circuit

• R,sink = R,rad R,conv / [R,rad + R,conv]

• Example - black anodized aluminum cube 10 cm per side

• R,rad = 2.2 C/W and R,conv = 2.2 C/W

• R,sink = (2.2) (2.2) /(2.2 + 2.2) = 1.1 C/W

P R,conv,rad

R

sT

aT

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Thermal Modeling


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Cost for Various Heat Sink Systems

Reference: http://www.electronics-cooling.com/Resources/EC_Articles/JUN95/jun95_01.htm

• Note: heat pipe and liquid systems require eventual heat sink

Thermal Modeling


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Comparison of Heat Sinks

Reference: http://www.ednmag.com/reg/1995/101295/21df3.htm

STAMPED EXTRUDED “CONVOLUTED” FAN

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2N3904 Static Thermal Model

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Liquid Cooling

• Advantages– Best performance per unit volume– Typical thermal resistance 0.01-0.1 C/W

• Disadvantages– Need a pump– Heat exchanger– Possibility of leaks – Cost

Thermal Modeling


54

Heat Pipe

• Heat pipe consists of a sealed container whose inner surfaces have a capillary wicking material

• Boiling heat transfer moves heat from the input to the output end of the heat pipe

• Heat pipes have an effective thermal conductivity much higher than that of copper

Thermal Modeling


55

Thermoelectric (TE) Cooler

• “Cooler” is a misnomer; a TE cooler is a heat pump• Peltier effect: uses current flow to pump heat from

cold side to warm side• Pumping is typically 25% efficient; to pump 2 Watts

of waste heat takes 8 Watts or more of electrical power

• However, device cooled device can be at a lower temperature than ambient

• TE coolers can heat or cool, depending on current flow

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Thermoelectric (TE) Cooler

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Fan

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Example 1: Picture Window

• Consider picture window with A = 1 m2, 2.5 mm thick

• Ti = 70F (25C); Approximate To = 32F (0C) for 6 months (long winter !)

• What is total cost for heat loss at $0.10/kW-hr

T iT o

wR iw R w R ow

glass

P

T i T o

Thermal Modeling


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Example 1: Picture Window

• Assumptions:– Window glass k = 0.78 W/(m-K)– Inside and outside window, heat

transfer dominated by free convection; h = 10 W/(m2K)

• Riw = Row = 1/(hA) = 0.1 C/Watt• Rw = w/(kA) =0.0025/(0.78)(1) = 0.0032

C/Watt• Rtotal = 0.2032 C/Watt• P = DT/Rtotal = 25C/0.2032C/Watt =

123 Watts• E = 3 kW-hr/day or 539 kW-hr for winter• Cost = $53.9

T iT o

wR iw R w R ow

glass

P

T i T o

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Example 2: Picture Window with Double Pane

• Assumptions:– Still air in airgap k = 0.027

W/(m-K)– Ignore radiation

• 1 cm airgap: Rairgap = g/(kA) =0.01/(0.027)(1) = 0.37 C/Watt

• Rtotal = 0.58 C/Watt• P = DT/Rtotal = 25C/0.58C/Watt =

43 W• E = 1 kW-hr/day or 188 kW-hr for

winter• Cost = $18.80

– Cost will be lower if gap has vacuum

T i T airgap

wR iw R w R ow

P

T i T o

g

glass glass T o

R wR airgap

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Example 3: Temperature Rise in Magnetic Brake

Aluminum reaction rail (4 mm thick)

Steel back plate (11 mm thick)v

N

S

N

S

N

S

N

S

• Train mass M = 12,300 kg• Initial speed = 16 meters/second• Brake aluminum fin length 10 meters• Stopping time: a few seconds• Cycle time: 1200 seconds• What is temperature rise in aluminum fin and in

steel ?

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Example 3: Magnetic Brake Thermal Model

• Model for 1 meter long section of brake• Guesstimated dominant time constant = 4,500

seconds (0.5 x 9000 F) based on thermal model above

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Example 3: Magnetic Brake Temperature Profile

• PSPICE simulation

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Example 4: White Pipe in the Hot Sun

• How hot does the surface of a white pipe get? Assume R = 0.565 m, pipe length = 1m, sunlight = 1200 W/m2, h = 8 W/m2-K, = 0.9 and solar absorption coeff. solar = 0.26

• Assume no conductive heat transfer

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Example 4: Pipe in the Hot Sun

convradreflsun QQQQ

Qsun = 1356 WQrefl = (1-solar)Qsun = 1003 W

• Therefore, 353 Watts is absorbed by the pipe, then dissipated via radiation and convection

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Example 4: Pipe in the Hot Sun

• Given these assumptions, temperature rise above ambient (Ts – TA) 7 degrees C with Qconv = 195 W and Qrad = 158 W

For radiation:

)( 44Asrad TTAQ

with = 0.9, = 5.6810-8 W/(m2K4) and surface area A = 1.0 m2 .

For free convection:

Asconv TThA

Q

1

with free convection heat transfer coefficient estimated as h 8 W/(m2-K).

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Example 5: What Happens if Pipe is Black?

Qrefl goes way down (solar energy absorption goes up, as solar = 0.9)

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Other Important Thermal Design Issues• Contact resistance

– How to estimate it– How to reduce it

• Thermal pads, thermal grease, etc.• Proper torque for mounting screws

• Geometry effects– Vertical vs. horizontal fins– Fin efficiency (how close together can you put

heat sink fins ?)

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Some Heat Sinks

• TO-92 (small transistor package)

Reference: Aavid-Thermalloy

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Some Heat Sinks• TO-220


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Some Heat Sinks

• TO-247


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Some Heat Sinks

• Vicor power brick


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Some Heat Sinks

• Liquid cooled plate


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Extrusions


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Cooling Fins

References: J H. Lienhard IV and J H. Lienhard V, “A Heat Transfer Textbook,” 3 rd edition, Phlogiston Press, Cambridge, MA 2008

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Improving Conductive Heat Transfer

References: International Rectifier, Application note N-1057, “Heatsink Characteristics”

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Improving Forced Convection Heat Transfer


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Improving Forced Convection Heat Transfer


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Improving Radiation Heat Transfer


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Conversion Factors

References: J H. Lienhard IV and J H. Lienhard V, “A Heat Transfer Textbook,” 3 rd edition, Phlogiston Press, Cambridge, MA 2008

Thermal Modeling© M. T. Thompson, 2009 1 Power Electronics Notes 29 Thermal Circuit Modeling and...

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