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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 1

Heat flux Temperature gradientHeat flux, (2.1)

or heat rate, (2.2)

For significant temperature ranges, kconstant.

2. Fouriers Law of Heat Conduction

In some situations e.g. heat conduction across a slab, radial

conduction across a cylindrical or spherical shell/layer, the

conduction can be regarded as one-dimensional and relatively

simple heat transfer calculations can be made.

x

TkAqx

xTkqx

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Conduction in Slab or Insulated Bar

1-D Steady Heat Conduction in plane slab or insulatedbar oflength L, cross-sectional area,A:

(2.3) 1 2T TT

Q kALL

kA

Thermal resistanceQ

R

T1 T2 Analogy with electrical resistance.

T1 T2

A qT2

T1

qL

L

q

q

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Multilayer Composite Wall

Assumptions:

1. Uniform T across vertical interfaces.

2. No heat transfer between C-D andD-E interfaces.

q

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1 2 3

L1

4

q1 q2 q3

W

5

L2 L3 L4

Thermal Resistance Network

Assume steady-state, only conduction heat transfer, no heat loss

through sides of printed circuit board (PCB).

q5

T1 T5T2 T3T4

q1 q2 q3

q4 R4R3R2R1

Above thermal resistance

network can be constructed to

model the heat conduction from

chips to edges of PCB.

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Analysis of Thermal Resistance Network

1 2 3

L1

4

q1 q2 q3

5

L2 L3 L4

q5

T1 T5T2 T3T4

q1 q2 q3

q4 R4R3R2R1

Considering energy balance at each

node:

0

0

0

0

0

5

4

53

4

1

41

3

4

35

3

32

2

3

23

2

21

1

2

12

1

14

qR

TT

qR

TT

qR

TT

R

TT

qR

TT

R

TT

qR

TT

R

TT

Applying boundary conditions:

e.g. T4 = 35C, T5= 38C

Unknown T1, T2, T3, can be calculated.

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Typical Multilayer Construction of PCB

Copper

Dielectric Material

(e.g. FR-4, G10,

polyimide)

Material W/m.K

Copper layer 386

Dielectric layer:

G10 0.30

FR-4 0.35

Polyimide 0.52

Thermal conductivity of

basic PCB materials

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The two equivalent thermal conductivities of a printed circuit board

(PCB) are the planar (lateral) thermal conductivity kxyand the normalthermal conductivity kz(in the zdirection).

Planar Thermal Conductivity kxyThe laminated lavers are considered as parallel paths in calculation

of the equivalent thermal conductivity.

For a parallel network, the total resistance is

1 2 3

1 1 1 1...

xyR R R R

Equivalent thermal conductivity of PCB

(2.5)

where Ri= thermal resistance for the ithlayer, i= 1, 2, 3, ...

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Li= length of ithlayer

Ai= cross-sectional area of ithlayer (normal to heat flow)

=

ki= thermal conductivity of ith, layer

ci= fraction of total coverage for ithlayer(e.g. if the cutout for a given layer (usually copper) is

80%, ci= 0.2 )

ii

i i i

LR

k A c

Planar thermal conductivity of PCB

(2.6)

where

LiWi=W

ti

Wi ti

tPCBq

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where the subscripts in and Cu represent the dielectric insulation

material and the copper, respectively.Ais the total cross-sectional

area over the thickness of a PCB; i.e.

xy i i i i i i

i iin Cu

k A c k A c k A

L L L

Planar thermal conductivity of PCB

Eqn (2.5) can be rewritten as: (2.7)

(2.8)SinceLi= L and ci,in= 1:

A=WtPCB

Cu

iii

in

iixy

A

Akc

A

Akk

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For heat conduction normal to

each layer, it may be assumed thatthe resistances of the layers are in

series:

PCB

z

. . i i

z i i i i i iin Cu

t t ti ek A c k A c k A

Normal thermal conductivity of PCB

LiWi=W

ti

tPCB

q

1 2 3 ...zR R R R (2.10)

(2.12)

(2.11)

PCB

z

i i i i i iin Cu

tk

t c k t c k

SinceAi= Az,

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Simple Thermal Model of IC Package

Paths of heat transfer from component

Rjc

Rca

qd

Ta

Tj

Tc

Case temperature Tc assumed

constant.

Heat transfer from junction to

case characterized by oneresistance,Rjc, and that

from case to ambient byRca

.

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For IC chips, manufacturers usually provide

values forRjaandRjc, the junction-to-ambientand junction-to-case thermal resistances,

respectively:

qd= (T

c- T

a)/ R

ca (2.20)

qd= (T

j- T

c)/ R

jc(2.21)

qd= (T

j- T

a)/ R

ja(2.22)

where qd = heat dissipation in chip. UsuallyTj < 125C.

Rjc

Rca

qd

Ta

Tj

Tc

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Rja

= Rjc

+ Rca

(2.23)

Internal Resistance

(depends onconstitution of chip

package)

External Resistance

(depends on methodof cooling)

Rca is usually based on natural convection (worst case).


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Under 135oCUnder 115oC

(oC/W)aPackage

-1.18032-38PLCC 84

-0.96041-47PLCC 68-1.07038-42PLCC 52

-0.85044-53PLCC 44

0.8900.62059-73PLCC 28

0.8100.56070-80PLCC 20

0.9300.64070SO 28

0.8100.56070-80SO 24

0.7200.50080-90SO 20

-0.45090-110SOL 16

-0.375110-120SO 16

-0.340110-130SO 14

Power (W) for junction

temperature

aThese typical thermal resistance

values are based on copper lead

frames with components mounted on

sockets. Actual thermal resistance wilvary with die size, component

mounting and molding compound.

Maximum power rating for

surface mount packages for

junction temperatures of

115oC and 135oC and an

ambient temperature of 70oC

(M. Kastner, and P. Melville,

Ed. Signetics SMD Thermal

Considerations. Signetics

Publication 98-9800-010,

1986.)

Typical Thermal Resistance of IC Packages

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Improved Thermal Model of IC Package

Limitations of Rjc:

- assumes uniform case temperature- assumes all surfaces uniformly cooled

- dependent on actual boundary conditions

due to existence of multiple heat flow paths

Improved model:

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For thin, small outline and flip chip packages, Rside

and heat conduction can be assumed to occur only through

the top and bottom surfaces.

d

sidetopbo t

side

topbo t

bo t

sidetop

topsidebo t

j

side

sidej

bo t

bo tj

top

topj

d

qR

RRRT

R

RRT

R

RRT

R

RRT

R

TT

R

TT

R

TT

q

sidetopsidebo tbo ttop RRRRRRR where

(2.24)

(2.25)

(2.26)

Improved Thermal Model of IC Package

M ti C t

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Mounting Components

on Printed Circuit Boards

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Using metal plates to enhance conduction cooling

copper strips over

PCB and under

components

aluminum plate bonded to

a thin PCB

clearance holes in

aluminium plate with

PCB on back side

(Source: D.S. Steinberg, 1991)

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Conduction Strip

PCB with chip packages

If packages are closely placed on strip, a uniform distributed heat

load may be assumed.

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Analysis of Conduction Strip

Assuming all heat transferred by conduction and no losses:

10, 0 : 0dT

x cdx

From energy balance,

Integrating,

From Fouriers heat conduction eqn,

From symmetry, at

(2.28)

(2.29)

(2.30)

(2.27)

x=L0

To Te

x dx

x dx

q'dx

qx+ dxqx

qdx

dq

dx

qq

dxqqq xxdxx

dxxdxx

0lim.

.1cxqqx

.1cxqdx

dTkAqx

q'

A l i f C d ti St i

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Let T=Teatx=L :

2

2

2

1cxqkAT

kA

LqTT eo

2

2

x=L0

To Te

(2.31)

Integrating (2.29),

Analysis of Conduction Strip

x dx

q'dx

qx+ dxqx

The maximum temperature is Toat x=0:

ekATLqc 2

2

2

1

222

1xLq

kATT e

(2.32)

q'

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A row of 6 IC packages each dissipating 100 mW are mounted onto a

conduction strips on a PCB of thickness 0.071 mm (2 oz Cu per sq ft).The heat must be conducted to the edges of the PCB, where it flows

into a heat sink. Determine the temperature difference between the

centre and edge of the copper strip to see if the design will be

satisfactory.

Example Problem

q'= 0.6/0.152 = 3.947 W/m

152 mm 50.071 mm

q'

Assuming that the heat is spread uniformly over the length of the strip:

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k = 287 W/mK, (copper alloy)

A = 0.005 x 0.000071 = 3.55 10-7 m2

From Eqn (2.32) ,

Example Problem

With 4 ounce copper (thickness 0.14mm), T = 56C,giving a cooler package temperature.

Rule of thumb:Design is satisfactory if package surface

temperature 100C.

7

22

1055.32872

076.0947.3

2

kA

LqTTT eo

Th l R i f I f

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q Unless surfaces are very well

bonded together, a thermal

interface resistance will exist.

Interface Resistance,

(2.35)

(2.33)

where hi is the unit (area) interface or contact conductance.

Interface Conductance,

(2.34)

Thermal Resistance of Interfaces

q

TRi

ii

i

AhRT

q

1

TAhq ii

Th l R i t f I t f

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Interface thermal resistance depends on :a. Effective contact area (contact pressure)

b. Interface medium

c. Surface roughness of contact surfaces

d. Flatness of surfacese. Hardness of materials in contact

Heat flow across interface


Table 3 5 Interface Conductance for Various Materials with

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Table 3.5 Interface Conductance for Various Materials with

an Interface Pressure of 10 psi (71 kPa)

6780452012008008070Bronze AMS 4846

113007330200013006060

7620226013504008570

9040452016008009015

10175180017165051 Aluminium

12435220031SAE 4141 steel

OilaDryOilaDry

Rms

(in)bSurface 2

Rms

(in)bSurface

1Material

W /m2KBtu /(hr ft2oF)

Interface Conductance

akoil= 0.073 Btu/hr ft F = 0.126 W/m K.b 1 in = 0.025m


Th l R i t f I t f

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Fig. 4 Interface conductance for various interface

pressures at sea level.



I t f C d t t Hi h Altit d

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Interface Conductance at High Altitudes

At high altitudes air is depleted andhidrops drastically (see Fig. below).

To improve heat transfer, thermal greases or epoxy glue should beemployed. The use of some intervening material may also help.

Fig. 5 Interface

conductance with

contact pressureof 2 psi as a function

of altitude.


Effect of altitude on thermal interface resistance

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Effect of altitude on thermal interface resistance

PCB Edge Guides

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PCB Edge Guides

PCB Edge Guides

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PCB

Guide

Turning screw of wedge

clamp causes wedges to

exert firm pressure

against PCB resulting insecure installation and

low thermal resistance.

PCB Edge Guides

Fig. 7 Board edge guides with typical unit thermal resistance. (a) G guide, 0.3 K.m/W;(b) B guide, 0.2 K.m/W; (c) U guide, 0.15 K.m/W; (d) wedge clamp, 0.05 K.m/W.

Thermal Resistance of PCB Edge Guides

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Thermal Resistance of PCB Edge Guides

Tg

Te

Ac

q

( )

where contact conductance

contact area

c c e g

c

c

Q h A T T

h

A total

( ) 1 1

(2 )

e g g

g c c c g g

T T RR

Q h A h wL L

L g

w

where = unit thermal resistance of edge guide.g

R

q

q

(2.36)

Example Problem

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TEMPERATURE RISE ACROSS A PCB EDGE GUIDE

Determine the temperature rise across a G-guide if it is 125

mm long and the total power dissipation of the PCB is 10 W,

uniformly distributed. What would be this temperature rise

when the equipment is operated at an altitude of 100,000 feet?

Example Problem

Example Problem

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Since there are two edge guides, half of the total power will be

conducted through each guide. The temperature rise at sea level

conditions can be determined from

Given: = 0.3 K.m/W(unit thermal resistance)

q = 10/2 = 5 watts (on half of the PCB), L = 0.125 m.

At an altitude of 100,000 ft, the resistance across the edge guide will

increase about 30%. The temperature rise at this altitude will then be

T= 1.30 x 12K = 15.6 K

0.3 512 K

0.125T

Example Problem

gR

L

RqqRT

g

E l P bl

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Example Problem

Cross section through a chasis with chip packages on PCBs.

qp= 200 mW

E l P bl

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Example Problem

Each IC package of 9 mm x 9 mm and dissipating

qp= 200 mW is attached onto a 0.20 mm thick PCB lamina (k= 1.5W/mK) with an air gap of 0.14 mm. The edge guide has a unit

thermal resistance of 0.305 K.m/W. The interface conductance

between the base of the box and the cold plate is 1130 W/m2K at

sea level.

Assume conduction heat transfer only. Maximum allowable IC

case temperature is 100 C. Is above design satisfactory for

operation at sea level?

Solution to Example Problem

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Solution to Example Problem

The hottest IC package is located at the centre of the top row of

packages. This maximum temperature can be determined bycalculating the temperature drops along individual segments of

the heat flow path from the hottest component to the liquid cooled

cold plate heat sink, namely

T1

= T from IC case to centre of aluminium plate.

T2= T from centre of aluminium plate to the edge of the plate.

T3= T across board edge guide to chassis side wall.

T4= T down chassis side wall to the base of the chassis.

T5= T across the bolted interface to the cold plate.

T1 from IC case to centre of aluminium plate

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T1

: 0.2 mm thick PCB lamina plus air gap of 0.14 mm.

2 2

0.0002 0.00014

1.5 0.009 0.03 0.009

1.7 57.6 59.3 /

a al

a al

L LLR

k A k A k A

K W

qp= 0.2 W

T1= 0.2 x 59.3 = 11.9C

T1from IC case to centre of aluminium plate

T2 from centre of aluminum plate to edge of plate.

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T2: Assuming uniform heat distribution over the PCB, eqn (2.32)

can be used with L = 0.075 m,

T2from centre of aluminum plate to edge of plate.

= 10.4C

q'= 40 x 0.2/0.15 =53.33 W/m.

001.0102.06.1432

075.033.53

2

22

2

kA

LqT

T3 across board edge guide to chassis side wall.

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T3: Unit thermal resistance of edge guide at sea level is

= 0.305K.m/W

thermal resistance, R3= 0.305/0.102

= 3 K/W

T3= 4 x 3 = 12C

T3across board edge guide to chassis side wall.

gR

T4down the chassis side wall to the base of chassis

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T4:

4

A

Side view

T4down the chassis side wall to the base of chassis

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Along AB, heat input is uniform at 4/0.102 W/m and edge at A is

insulated as heat loss is negligible here. Hence AB represents half a

standard conduction strip and eqn. (2.32) can be used with q = 4/0.102= 39.22 W/m, L = 0.102 m

T4= 24.7+12.1+2.3 = 39.1 C

C.

...

.

kA

qLT

C....

.

kA

qLT

C....

..

kA

qLT

CD

BC

AB

32006002506143

012504

1120023002506143

02504

72400230025061432

10202239

2

22

4

T5across the bolted interface to the cold plate.

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T5: Since the average interface conductance at sea level is

1130 W/m2K,

Hence total T = 11.9+10.4+12+39.1+5.7

= 79.1 C

C...hA

qT os 75

025002501130

4

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Maximum component surface temp. = 27+79.1

= 106.1 C

Hence design is not satisfactory.

IC case temperature can be reduced by various measures such

as increasing the thickness of the supporting bracket, cementingIC to PCB, etc.

me4225 2 conduction 2014

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