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Passive Electronic Components
Lecture 4
Page 1 of 11
22-Mar-2017
Heat transfer in passive electronic components
Lecture Plan
1. Harmonic functions and related physical phenomena
2. Modes of heat transfer
1. Harmonic functions and related physical phenomena.

Consider some examples of stationary (not depending on the time) vector field A and conjugated
scalar field  related as
 
A   ,
(1)

where  - vector operator that is called nabla operator. In Cartesian coordinates x, y, z
  
 
 
  ix 
jy  kz .
x
y
z
In cylindrical coordinates r , , z where x  r cos , y  r sin  , z  z

iz

i

ir
   1  
 
  ir 
i  i z .
r
r 
z
In spherical coordinates r , , where x  r cos sin  , y  r sin  sin  , z  r cos
  
1
  1  
  ir 
i 
i .
r
r sin  
r 
Passive Electronic Components
Lecture 4
Page 2 of 11

ir

i

i
The examples of conjugated physical fields that are describable by (1) are:

(A) Electrostatic field E [V/m] (the force per unitary positive test charge) and electrostatic
potential  [V] in dielectric medium:


E   ,

(B) Electric current density j [A/m2] and electrical field potential  [V] in conductive medium (in
the case of the positive charge carriers):


j   ,
where  - electrical conductivity of the medium [-1m-1]. (Pay attention: the same letter  is used
below for designation of Stefan – Boltzmann constant).

(C) Heat flux q [W/m2] and temperature field T [K] in solid body (Fourier’s law):


q   KT ,
where K - thermal conductivity of the body material [W/(mK)],

It is important that in all these examples respective vector field A has zero total flux through any
closed surface S:
 
A
  ds  0.
(2)
S
Indeed:

 Electrostatic field E in dielectric medium has zero total flux through any closed
surface S supposing that there is no unbalanced charge inside S;
Passive Electronic Components
Lecture 4
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Page 3 of 11

Electric current density i in conductive medium has zero total flux through any
closed surface S supposing that there is no current source inside S;

Heat flux q has zero total flux through any closed surface S supposing that there is
no heat source inside S.
Let us apply the divergence theorem to the surface integral in the left side of equation (2):
 
 
A

d
s



  Adv.
S
V
(3)

 
The product   A is called the divergence of vector A .
Transform equation (2) using (1) and (3):
 
   dv  0;
V
 
    0;

 2  0;
  0,
(4)


where  or  2 is a scalar operator - Laplacian. We shall use  2 designation in order not to mix
up Laplacian designation  with designation  of increment (difference) of a variable.
In Cartesian coordinates x, y, z

2
2
2
2  2  2  2 .
x
y
z
In cylindrical coordinates r , , z
 2 1     1 2
2
.
 

r  
r r  r  r 2  2 z 2
In spherical coordinates r , ,
2 1   2  
1
2
1
 
 
 sin 
.
  2
 2
r
 2 2
2
 
r r  r  r sin  
r sin   
Any function that satisfies differential equation (4) is called harmonic function. Therefore, a
potential of stationary electric field in dielectric medium, an electrical potential in conductive
medium, and a stationary temperature field in a solid body are examples of harmonic functions of
spatial coordinates.
Harmonic function defined inside a closed surface attains its maximum (minimum) on the surface
(on the boundary) if it has no “sources” inside this surface.
Passive Electronic Components
Lecture 4
Page 4 of 11
2. Modes of heat transfer
2.1. Heat conduction is a mode of transfer of energy within and between bodies of matter, due to a
temperature gradient and with no visible motion.
Fundamental hypothesis of the Mathematical Theory of Heat Conduction (Fourier’s law) is that

heat flux q through elementary surface per unit time per unit area [W/m2] resulting from thermal
conduction is proportional to the magnitude of the temperature gradient and opposed to its sign:


q   KT ,
(5)
where
K - thermal conductivity of a body material [W/(mK)],
T - temperature [K] as function of the spatial coordinates.
We shall consider heat flux and temperature field in a solid body. Suppose that some volume
bounded by the surface S does not comprise heat sources and is kept in stationary condition (the
temperature field does not change in time). According to the law of energy conservation


 q  ds  0.
(6)
S

Let us substitute q by T in (6) using (5) and apply the divergence theorem. The result will be

 2T  0 .
That means that stationary distribution of temperature in homogeneous isotropic solid whose
thermal conductivity is independent of the temperature is a harmonic function.
Conception of thermal resistance. Consider some cylindrical solid body (Fig.1) that conducts a
heat power Q from one flat base having uniform temperature T 0  T1 to another flat base having
uniform temperature T l   T2 . It is important that there are two areas with constant temperature on
the body’s surface. They serve as the “thermal terminals” in the heat conduction process (similar to
ohmic resistor’s terminals in an electrical conduction process). Let us show that the values T2  T1
and Q are proportional. Proportionality coefficient Rt depends on the form and material of the body
and is called "thermal resistance" of this body.
Q
T1
T2
x
0
l
Fig.1
It may be easily shown that linear temperature distribution (7)
T T
T ( x)  T1  2 1 x .
l
(7)
Passive Electronic Components
Lecture 4
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2
Satisfies Laplace equation  T  0 and given boundary conditions: T 0  T1 ; T l   T2 .
Heat flow through cylindrical surface (in directions 0y, 0z perpendicular to axis x) is zero.
Heat flux in direction of axis 0x is




T  T1 
q   KT   K 2
 ix  0  j y  0  k z
l
T  T1
.
qx   K 2
l
Absolute value of entire power Q (heat rate) transferred through area S of each base will be
Q  q x S  KS
T2  T1
l
Rt 

1
T2  T1 .
Rt
T2  T1
.
Q
(8)
(9)
Parameter Rt is called "thermal resistance". It follows from (8) that for considered cylindrical solid
body
l
(9a)
Rt 
.
KS
Compare (9) with expressions for electrical resistance of the prismatic body that has the same shape
as shown in Fig.1 and is made of material with electrical conductivity  [-1m-1]:
R
U 2  U1
I
;R 
l
.
S
(10)
Identity of the expressions (9) and (10) stems from the fact that both temperature field and electrical
potentials are harmonic functions and satisfy the identical differential equations and boundary
conditions.
Calculation of thermal resistance. Suppose that arbitrary solid body conducts a heat power Q
from part a of its surface to part b of its surface. Suppose that both a and b are characterized by
uniformly distributed temperatures T1 and T2 respectively. The remaining part c of the body's
surface is supposed to be thermally isolated.
A. The temperature field inside the body is represented by harmonic function T that can be
found as solution of Laplace equation with the given boundary conditions:

 2T  0;
T
a
T
n
 T1 ; T b  T2
 0.
C
Passive Electronic Components
Lecture 4
Page 6 of 11

B. T determination.
C. Determination of thermal resistance between two surfaces having constant temperature
distributions:
Rt 
T2  T1
Q
T2  T1
  
 q  ds

S
T2  T1

 .
 KT  ds
(11)
S
Equation (11) means that thermal resistance may be expressed in terms of scalar temperature field

T . Alternatively thermal resistance may be expressed in terms of vector field of heat flux q by
substitution of (5) in (9):
a
Rt 
T2  T1
Q



 q  dl
T2  T1
b
  
  .
q

d
s
K
q

  ds
S
(12)
S
Example 1. Thermal resistance of a prism.
z
l
S
x
y
T1
T2
(A) Suppose that temperature distribution is given. It is
T  ax ,
(13)
where a is some constant. Check that T is harmonic function and complies to all boundary
conditions.



(B) Determination of T : T  aix .
(C) Determination of thermal resistance:
Passive Electronic Components
Lecture 4
Page 7 of 11
Rt 
ax 2  x1 
aK  ds
S

al
aK  ds

l
KS
S
Example 2. Thermal resistance of ring sector with thickness h.
T1
a
T2
1   T  1  2T  2T

 0;
r

r r  r  r 2  2 z 2

T  1 T  T 
T 
ir 
i 
iz .
r
r 
z
b
(A) Suppose that temperature distribution is given as linear function of  : T  c  d . It is
obvious that if   const then T  const . It may be found from the given boundary conditions
T r,1   T1 ; T r, 2   T2 and temperature distribution T  c  d that
c
T2  T1
T   T21
;d  1 2
.
 2  1
 2  1

Let us check that T is harmonic function and complies to boundary conditions. Indeed, 2T  0
because T is linear function. T r,1   T1 ;T r, 2   T2 ; qr a,   qr b,   0; q z r,   0 .


c
(B) Determination of T : T  i .
r
(C) Determination of thermal resistance:
1

c( 2  1 )

2
Rt 


.
b
chdr
b
b
hK ln   2hK ln  
K
r
a
a
a
2.2. Heat convection is the transmission of heat through a currents of a fluid (liquid or gas).
Newton’s law of cooling:
q  h(T  TS ) ,
(11)
where
q - heat transferred from the unit surface of a solid body in the unit of time [W/m2],
T - temperature of a solid body surface [K],
TS - temperature of oncoming fluid (gas) [K],
h - heat transfer coefficient [W/(m2K)].
Passive Electronic Components
Lecture 4
Page 8 of 11
Examples of heat transfer coefficient values [2, p.21]
Object
Air speed,
h,
T  TS ,
m/s
W/(m2K)
K
Vertical 0.3 m wall
0
30
4.3
1 m flat plate
30
70
80
The convection coefficient for natural convection in gas is generally in the range 1…20 W/(m2K).
Typical range for a liquid ambient is 100…1000 W/(m2K).
Let us extend the conception of thermal resistance to the case of convection. Multiplying (11) by S
gives:
Q  hS (T  TS ) .
Following the logic of equation (9) we come to the following definition:
T  TS T

Q
Q
1
Rt 
.
hS
Rt 
(12)
(  means difference, not Laplacian!)
2.3. Radiation is heat transfer by infrared rays.
Stefan – Boltzmann law: The power per unit area q radiated by a body is proportional to the fourth
power of its absolute temperature T and is given by equation:
q  T 4 ,
(13)
where
 = 5.67010-8 W/(m2K4) is Stefan – Boltzmann constant,
 - dimensionless emissivity coefficient (the fraction of the ideal blackbody spectrum energy which
a real body actually emits), T - absolute temperature of the body, K.
Surface material
Ideal blackbody
Black paint
Concrete
Polished aluminum foil

1.0
0.9
0.88
0.05
Each body radiates and absorbs a radiant energy simultaneously. In special case of a small object
with uniform surface temperature T placed in a relatively large enclosure having uniform
temperature T0 , the total heat transfer qnet from the object surface outwards is given by the
following equation:
q net   (T 4  T04 ) ,
Passive Electronic Components
Lecture 4
Page 9 of 11
Let us introduce new designation:
T  T  T0 .
(  means difference, not Laplacian!)
T
4
 T04
 4T
 (T0  T )

 T03 1 

4
 T04


 T04 1 

4
2
3
4



 T 
 T   T 
T 
T
4
4
  1   T0 1  4
  4
  
  1 
 6
T0 
T0



 T0 
 T0   T0 
2
3
3 T  T 
1  T  
  4T  T03 .





 


2 T0  T0 
4  T0  

qnet   (T 4  T04 )    4T  T03  hrad T ;
(14)
hrad  4T03 .
The smaller is T T0 ratio the more precise is equation (14).
T, K
Error, %
(T0=293K)
5
11
16
22
29
64
10
20
30
40
50
100
We can conclude that conception of thermal resistance that was introduced for heat conduction and
convection may be expanded to radiation of a body situated inside some enclosure:
Rt 
T
T

;
Q
q net S
q net  hrad T ;
Rt 
1
hrad S
(15)
.
Important conclusion results from universality of thermal resistance conception: in any conditions
of heat transfer a temperature rise (a temperature difference between heat generating body and
ambience) T is proportional to the power Q of heat generation. The proportionality coefficient
will be thermal resistance Rt .
When thermal resistance of complex system is evaluated the whole system be be split into the parts.
Then its heat analysis may be performed in the terms of thermal resistances of its parts. The
elementary thermal “resistors” may be combined in a network the same as electrical resistors in the
complex circuit.
Passive Electronic Components
Lecture 4
Page 10 of 11
Appendix 1
Example of Thermal Analysis of Axial Resistor
Consider axial resistor of CCF-2 style manufactured by Vishay. Resistor is mounted on PCB. The
contribution of each way of heat transfer outside from resistor body has to be evaluated.
Data for calculation:
Input Data
Dissipated power
Q
0.5 W
Ambient air and PCB temperature T 1
20 C
293 K
Resistor temperature
T2
74 C
347 K
Body length
l body
8.7 mm
8.70E-03 m
Body diameter
d body
3.7 mm
3.70E-03 m
Lead length
l lead
12 mm
1.20E-02 m
Lead diameter
d lead
0.81 mm
8.10E-04 m
Heat transfer coefficient
Emissivity coefficient
Stefan – Boltzmann constant
Copper heat conductivity
Thermal resistance of PCB
for heat flow from each lead
h conv
10 W/(m^2*K)
0.9
5.67E-08 W/(m^2*K^4)
400 W/(m*K)


K
R PCB
220 K/W
Body
Lead
PCB
Solution.
Rrad
Rconv
Heat dissipated
in ambient space
RPCB
Rlead
RPCB
Rlead
Heat released in
resistor body
Passive Electronic Components
Lecture 4
Page 11 of 11
Calculation results:
R lead
S body   d body l body  d body 2 
2
S lead   d lead
4
R conv  1 hconv S body
R r ad  1 4 T 2  T1  2 3 S b od y
Rlead  l lead KS lead
5.82E+01
K/W
R total
Rtotal  1 / 1 / R rad  1 / R conv  2 /( R lead  R PCB )  1.08E+02
K/W
Body surface
Lead cross section
R conv
R rad


1.23E-04
5.15E-07
8.16E+02
m^2
1.22E+03
K/W
m^2
K/W
Verification of solution
Resistor temperature
T   T1  R total Q
74.2
C
Contribution of each way of heat transfer is the following:
Convection
Radiation
Conduction
Total
13 %
9%
78 %
100 %
Example.
Maximum thermal resistance
between thick-film chip resistor surface at 125C and ambient space
(EN140401-802 European Standard)
Metric style
(Inch style)
RR 1005M (0402)
RR 1608M (0603)
RR 2012M (0805)
RR 3216M (1206)
Chip dimensions,
mm
1.00.50.35
1.60.80.50
2.01.20.50
3.21.60.56
Rt,
K/W
880
550
440
220
The actual thermal resistance of chip components is mainly influences by PCB design: soldering
pad dimensions, copper thickness, insulation material type (glass-polymer, ceramics).
1. Carslaw H.S., Jaeger J. C., Conduction of heat in solids. Oxford at the Clarendon Press,
1959.
2. J.H. Lienhard, A heat transfer textbook. Englewood Cliffs, N. J.: Prentice-Hall, 1981.