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8/12/2019 ME4225 2 Conduction 2014
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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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 9
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 10
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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Simple Thermal Model of IC Package
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).
Simple Thermal Model of IC Package
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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
dxqqq xxdxx
dxxdxx
0lim.
.1cxqqx
.1cxqdx
dTkAqx
q'
A l i f C d ti St i
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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 21
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 22
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 23
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 24
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 25
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
Thermal Resistance of Interfaces
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
(Source: D.S. Steinberg, 1991)
Th l R i t f I t f
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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 27
Fig. 4 Interface conductance for various interface
pressures at sea level.
Thermal Resistance of Interfaces
(Source: D.S. Steinberg, 1991)
I t f C d t t Hi h Altit d
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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 28
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.
(Source: D.S. Steinberg, 1991)
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 31
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 34
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 36
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 39
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 42
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 43
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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2013 Prof Andrew Tay ME4225 Industrial Heat Transfer 44
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.