thermal management

School of Materials and Mineral Resources School of Materials and Mineral Resources EngineeringEngineering14 August 200714 August 2007

Thermal Thermal ManagementManagement

Assoc. Prof. Dr. Azizan AzizAssoc. Prof. Dr. Azizan Aziz

EBB 526 – Electronic PackagingEBB 526 – Electronic Packaging

Basic Thermal Analysis


• There are three mechanisms for heat transfer : conduction, convection, radiation

Heat Flow

• All active and passive devices are sources of heat and are always hotter than the average temperature of their immediate surrounding

• The second law of thermodynamics states that heat always flows from a hotter region to a cooler region


Conduction Convection Radiation

Heat Sink

Socket PCB

PPGA


Conduction

• Thermal conduction is a process in which heat flows through a solid,liquid, or gas or between two media that in intimate contact with each other.

• Dorminant mechanism for heat transfer within solids involving transfer of kinetic thermal energy from one electron to another.

• Conduction through dielectric solids is entirely due to lattice vibrations

• While through metallic solids has the added energy transport by free electron

T1

T2

q”

T1 > T2

Conduction


• Thermal transfer via electron is similar to that of electric charge – good electrical conductors such as copper and silver are good thermal conductors

• Liquid is also a good thermal conductor but to a lesser extent than solids

• Thermal conductivity of gases is quite low.

When liquid changes to a gas thereis a loosening of

molecular bonds compare to liquid. Gas molecules are free to move in any direction and the only constraint are randomly collisions which results in significantly lower

probability that contact areas will exist


Think of molecular activity - translational, rotational Think of molecular activity - translational, rotational and vibrational energy of the molecules.and vibrational energy of the molecules.

Collision between molecules occurs to transfer energy.Collision between molecules occurs to transfer energy.

No bulk motion of material.No bulk motion of material.

T1

T2

T

x

q”

T1

T2


Fourier’s Law

• Fourier’s law of heat conduction states that the rate of heat flow equals the product of the area normal to the heat path, the temperature gradient along the path, and the thermal conductivity of the medium

where k = thermal conductivity of medium in watts/m-K or watts/in-oC

A=cross-sectional area of medium normal to the heat flow in in2 or cm2

T = temperature of medium in oC

x = position along the medium in in or cm

t = time in seconds

q = heat generated per unit volume in joules/cm3

Q = heat flow in watts normal to the cross-sectional area of heat transfer

dx

dTkA

dt

dq (1)


dt

dq power in watts or calories per second

dx

dT Temperature gradient in oC/in or oC/cm

• The temperature gradient and cross-sctional area are defined at the same point x

• Heat flow considered positive when temperature is decreasing


• Equation (1) can be written as

dxkA

QdT k

(2)

• Intergrating both sides of Equation (2)

2

1

2

1

x

xk

k

T

T kA

dxQdT (3)

• Assuming that the thermal conductivity k does not vary over the length L=x1 – x2 , Eqn (3) reduces to

kA

LQTTT k 12 (4)

where ΔT = T2 – T1 is the temperature difference along the length L


• Thermal resistance θ can be defined as

kA

L (5)

kQ

T (6)


• As shown in Fig. 5.5 a heat source producing Qk watts is mounted on a block attached to a heat sink which is at a constant,uniform temp of Theat sink

• The temperature of the heat source in this configuration can be calculated as

kkheatheatsource QTT .sin (7)


Example

A copper rod 6 inches long, 0.5 in wide and 0.5 in thick as shown in Fig 5.6 has a 40oC temperature difference over its length. One end is fixed at 100oC. The themal conductivity of copper rod is 10 watt/oC-in.What is the power transferred.


The power transferred Qk can be calculated from eq (6)

kQ

T (6)

WCinininCWkA

L oo

/4.2)5.0)(5.0)(/10(

6

WWC

CTQ

o

o

k 67.16/4.2

40


Electrical analogies of Thermal Resistance,θ

• Thermal characteristic can be simplified by considering thermal properties analogous to electrical properties

• The temperature drop ΔT is similar to voltage drop

• When several materials are stacked in series such as a die attached with epoxy to a substrate wihich is soldered to a package base , the equivalent thermal resistance θequiv is

θequiv = θ1 + θ2 + θ3 + ………+ θN

• Temperature at particular interface may calculated as:

hsjjkkheatjj QTT :1,sin1, where Tj,j-1= temperature at interface of layers j and j-1

Σθj,j-1:hs = sum of thermal resistances from interface of layers j and j-1 to the heat sink


The Fig 5.7 illustrate the stack up of thermal resistances using the electrical circuit analogy

• Temperature at the interface of two layers cannot change discontinously

• Ex. Temp at the bottom of the die will be the same as the temperature at the top of the die attach material (in the Fig denoted as T1-2)


Example

• Refering to the above Fig, the heat dissipated in a junction is 10 watts. The thermal resistance of the copper is 0.5oC/watt. The case temperature is 100oC. Find the temperature of the top of the copper base.

hsjkkheatjj QTT sin1,

CxWxCT oo 1055.01010010100 55,4

• When there is more than one heat path from the dissipating element to ambient than the equivalent thermal resistance can be considered to be the parallel equivalent of the individual thermal resistance

Nequiv 1

.........1111

321

(8)


• In the above Fig. there are two heat paths from the die. The 1st path,θ1, is from the die thru the subsrtate,while θ2 is from the top of the die thru the lead frame.

21

111

equiv


Thermal Resistance - Series vs. ParallelThermal Resistance - Series vs. Parallel

We can treat one-dimensional conduction problem like a electrical We can treat one-dimensional conduction problem like a electrical circuit problem - resistances in series and parallel.circuit problem - resistances in series and parallel.Recall resistances in :Recall resistances in :

21 RRRseries R1 R2

R2

R1

21

111

RRRparallel

Series

Parallel


Convection

• Convection is the transfer of thermal energy between two surfaces as a consequence of a relative velocity between them

• It occurs only in fluid where the transfer mechanism is mixing of the fluids

• In practical , one is solid surface and the other is fluid surface

• Heat loss due to Newtonian cooling or convection cooling is proportional to the temperature difference, ΔT , between them.

TAhTTAhQ scAsscc (9)

Convection

Ts

Ta

q”


where Qc = heat transferred from a surface to ambient by convection in watts

As = surface area in cm2 or in2

Ts = surface temperature in oC

TA = ambient temperature in oC (temp to which the heat is being transferred)

hc = convection heat transfer coefficient in watts/cm3-oC or watts/in3-oC

• Eqn. (9) can rewritten as

csc

QAh

T1

(10)


• Convective surface thermal resistance

scs Ah

1 (11)

• Two types of convection cooling i) natural ii) forced

• Natural convection, is due entirely to differences in density within the fluids resulting from different temperatures

• Forced convection, thermal energy is transfered from solid to adjacent fluid particles as in natural convection but the subsequent fluid action occurs through artficially induced fluid motion generated by pumps or blowers.


• For natural or free convection, the convection heat transfer coefficient is given as

25.0

25.0

L

TDEh

(12)

where D = constant for air properties (see Fig 5.9)

E = constant for surface configuration (E=1.9 x 10-4 for flat surface)

L = characteristic length in cm or inches of dissipator surface with area factor

ΔT = temperature difference in oC between the dissipator and ambient air


• For forced convection, the convection heat transfer coefficient is given as

75.0

75.0

L

VBh (13)

where B = constant for air properties and surface configuration

V = linear velocity of air in cm/sec or in/sec

L = characteristic length of surface in direction of flow in cm or inches


Example

A flat plate with a characteristic length of 2.0 in as shown in the above Fig. can be considered a model for flat heat sink. The plate size is 2.0 x 2.0 in. The plat bottom is at 125oC while the ambient air is 25oC. The value of D for this configuration is 0.26. Calculate the natural convective transfer heat coefficient hc . (Ans: 1.31 x 10-2 )

In a forced convection cooling, air with a velocity of 500 feet/min is blown across a plate with characteristic length of 2.0 inches. The surface area is 40 in2 and the property/surface configuration constant is 1.0 x 10-3. Calculate the thermal resistance. (Ans: 0.518 oC/watt)


Radiation

• Radiation cooling is the transfer of heat by electromagnetic emission, primarily in the infrared wavelengths and may be considered a totally surface –related phenomena.

• Does not require a transport medium, and is maximized when there is no intervening material.

• The emissive power of a black body or surface is defined as Eb

• The emissivity of a body or surface,ε , is defined as the ratio of the radiated flux, E, emitted by a body to that of a black body at the same temperature

bE

E (14)

surface, T1

Radiation

surface, T2

q1”

q2”


• A black body or perfect emitter ε =1, while a perfect reflector ε =0

• Rate of emission of radiant energy from the surface of a body R can be expressed as

4TR (15)

• R is defined as (unit W/m2)

A

QR (16)


where ε = surface emissivity in joules/sec-cm2

σ = Stefan-Boltzmann constant (3.65 x 10-11 watts/in2-K4)

Q = heat transferred in watts

A = radiating surface area in m2

T = temperature of surface in K

• For non-black body surfaces, the heat transferred via radiation is

)( 22

41 TTSAQ (17)

Where T1 = temperature of hot body in K

T2 = temperature of cold body in K (air molecules or other absorbing body

S =shielding factor or view factor


• For the radiant heat transfer mode the thermal resistance is given as

)(

)(42

41

21

TTSA

TT

• The shielding factor S, whose values ranges from 0 to 1, is a measure of how well the emitter sees the absorber


Example

The bottom of a heat sink is at 150oC while the ambient air is at 25oC. The heat sink is nickel-plated with a surface area of 4.0 jn2. The shielding factor is 1.0. Calculate the amount of heat transferred due to radiation. (Ans: 0.388 watts)

thermal management

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