ETC 18448

CPU Thermal Management
Advanced
Micro
Devices
Application Note
This application note describes CPU thermal management practices using a heat sink/fan combination. The heat
sink/fan assembly helps to guard against processor overheating in systems where the airflow may be restricted by
the addition of add-in cards or modules that can block airflow necessary for proper CPU cooling.
OVERVIEW
Heat is generated by all semiconductors while operating. Most microprocessors to date have been able to
dissipate the heat directly to the ambient air without heat
sinks or fans. With faster processors that dissipate more
heat than the slower processors, it is no longer possible
to ignore thermal management. The objective is to ensure the generated heat is dissipated into the ambient
air while a safe operating temperature is maintained.
There are several methods for keeping the processor
cool. All of these methods include a combination of heat
sink and airflow. In general, the trade-off is heat sink
versus airflow. A smaller and less costly heat sink requires more airflow. Analogously, larger heat sinks require less airflow to maintain a safe case temperature.
There are several choices of motherboards and computer cases that manufacturers can use in their assembly. After receiving the system, the end customer can
populate the system with a myriad of add-on cards and
peripherals; hence, it is extremely difficult to guarantee
that the processor will be adequately cooled in all the
different combinations of systems.
AMD has researched several products that aid in thermal management design. The product that effectively
provides thermal management at a reasonable cost is
the heat sink and fan combination. This product consists
of a small fan mounted on a heat sink. The fan is powered by the standard power supply and connects via a
cable. The heat sink is clipped on or glued to the microprocessor. The heat sink adds between 11/16" to 3/4"
of height to the processor. The space above the fan and
to the sides of the processor should be cleared to allow
for proper airflow. A typical heat sink and fan assembly
is shown in Figure 1.
AMD has tested various fan/heat sink devices and found
the Thermalloy 2321B-TCM cooling module to provide
reliable operation.
1 3/4"
Fan
1/4"
Heat sink
11/16"
1 3/4"
Figure 1. Heat Sink and Fan Assembly
HEAT SINKS AVAILABLE FOR Am486 AND
Am5X86 CPUs
Thermalloy’s Heat Sinks
Omnidirectional
Models: 2321B, 2332B, 2333B, 2342B
■ 20% greater performance than extruded
equivalents
■ Heat Sinks may be bonded to the PGA with epoxy
or with the PGA E-Z Mount frame (p/n 8317) and
spring (p/n PF17)
This document contains information on a product under development at Advanced Micro Devices. The information is
intended to help you evaluate this product. AMD reserves the right to change or discontinue work on this proposed
product without notice.
Publication#18448 Rev: D Amendment/0
Issue Date: August 1995
AMD
Other Available Heat Sinks
AMP Low Insertion Force PGA Sockets
Models: SCA17-1, SCA17-2
(Heat sink with tabs for spring clips)
■ Spring clip (SCA17-x) attaches a Thermalloy pin fin
heat sink (23xx series) to PGA in an AMP LIF PGA
socket
■ Clip easily snaps over the edges of the PGA socket
and requires no special tools
AAVID’s Heat Sinks
SINK-to-SOCKET Clip Heat Sinks
Models: 3333, 3334, 3335, 3336, 3337
Figure 2. Omnidirectional Heat Sink
Thermalloy Cooling Modules (TCM)
(A clip that attaches the heat sink to an AMP Socket
with the CPU in between)
■ Removable heat sink and clip with built-in quick
release/load latch
SINK-to-PROCESSOR Clip Heat Sinks
Models: 2321B-TCM, 2333B-TCM
(Heat sink with fan attached)
■ High performance relative to its low cost
Models: 3600, 3331, 3601, 3602, 3603, 3329
■ Fans available in 5 or 12 volts, 12 V recommended
(A clip that attaches the heat sink directly to the CPU)
■ TCM assembly may be attached to PGA with Ther-
■ No keep-clear areas required
malloy’s innovative PGA E-Z Mount frame and
spring or epoxy.
■ May be used in conjunction with PGA sockets, such
as AMP 382624-1 and -2
■ Functions on socket or direct mount CPUs
PGA KLIPS Heat Sinks
(Uses PGA Klips for easy installation)
Bidirectional
Models: 3300, 3301, 3302, 340011
■ Low cost heat sink
■ Ideal for directional and high airflow patterns
Omnidirectional
Models: 3305, 3306, 3307, 340021
■ Utilizes airflow from any direction
■ Ideal for impingement airflow patterns
Fan-Sink Heat Sinks
Model 351055
Figure 3. Thermalloy Cooling Module
Notes:
1. Thermal paste is recommended in order to provide
the best heat transfer.
2. When applying thermal paste, it should be applied
in a thin, smooth even layer across the entire CPU
package.
3. In no circumstance should an air gap exist between
the CPU package and the heatsink. If a gap exits,
the heatsink will provide little or not heat dissipation
and therefore is useless.
2
(Heat sink that uses a fan)
■ Low profile design
■ Shrouded design maximizes cooling capacity
For Further Information Contact:
AAVID Engineering, Inc.U.S.A.: (603) 528-3400
Thermalloy, Inc.
CPU Thermal Management
U.S.A.: (214) 243-4321
U.K.: 0793537861
Hong Kong: 852-4647312
AMD
APPENDIX−Background Information
Thermal Resistance
Thermal characteristics of integrated circuits (IC) have
long been a major concern for both electronic product
manufacturers and designers. This is because an increase in junction temperature can have an adverse
effect on the long term performance and operating life
of an IC. With the 486 CPU, for example, squeezing 1.2
million transistors on board and running at faster
speeds, more heat is generated which can not be easily
vented out of the computer with the usual fans. Unvented, the heat builds up and destroys the transistors. Heat
sinks are finding their way into 486 systems but they
may not be good enough for future generations of
CPUs.
The maximum case temperature of some Am486 CPUs
is specified to be 65°C. The cooling module must dissipate the heat into the ambient air, which must be below
65°C. How much lower the ambient temperature must
be is given by the thermal resistance times the power.
Therefore, to calculate the maximum ambient temperature that the processor with cooling module can operate, the following formula is used:
T Max – Ambient = 65 – ( P Max • θ CA )
The maximum power consumption (PMax) of the
Am486 processor is given as:
P Max = 5.35 [ V ] • 1200 [ mA ] = 6.3 Watts
With unit 1 using the thermal grease, the maximum ambient temperature for safe operation will be:
T Max – Ambient = 65 – ( 6.3 • 3.3 ) = 44.21°C
For comparison, by using the Thermalloy 2321B-TCM,
the maximum ambient temperature is:
T Max – Ambient = 65 – ( 6.3 • 1.4 ) = 56.18°C
When a transistor is turned on, power is dissipated
equal to the product of the voltage across the collector
junction and the current through it. As a result, the collector junction’s temperature begins to rise. Eventually,
a steady state is reached when the transistor dissipates
the same energy supplied to it. This energy is in the form
of heat and is given off through the case to the surrounding environment. The temperature depends upon the
power level and the thermal resistance of the device
package. Thermal resistance is the ability of the package to conduct heat away from the CPU and into the
surrounding environment. A low thermal resistance value means that for a given amount of power, the integrated circuit junction will operate at a lower
temperature, thereby providing a longer life time.
used. With the trend toward higher density circuits, increasing circuit complexity and increasing number of
pin outs, total power dissipation is increasing. Hence,
management of thermal characteristics remains a valid
concern.
Thermal resistance (theta jc (θjc)) is expressed as the
rise in the collector junction temperature (Tj) above the
case temperature (Tc) per unit of power dissipated (Pd)
in the device.
(1)
θjc = ( Tj – Tc ) ⁄ ( Pd )
Where θjc is expressed in °C/W.
Thermal resistance can also be calculated between
junction temperature and ambient temperature (Ta).
( 1a )
θja = ( Tj – Ta ) ⁄ ( Pd )
Figure 4 illustrates the path of heat flow through a device
with and without a heat sink, and Figure 5 shows a schematic representation of the thermal resistance paths between the junction and ambient temperatures.
The temperature of the junction (Tj) is related to the
power dissipation and the ambient temperature (Ta) by
the following equation:
Tj = ( Pd • θja ) + Ta
If a heat sink is applied, the heat passes from the case
to the sink before being emitted into the air. The purpose
of a heat sink is to increase the effective heat-dissipation
area and quickly remove heat from the device, permitting the device to work at higher power levels. The heat
sink provides an additional low-thermal resistance path
from case to ambient air.
Once a certain case temperature is reached, the maximum power rating drops off linearly as shown in Figure
6. This is called the derating curve. The derating factor
(Df) is a measure of how fast the curve drops off (i.e.,
the slope of the curve). Its units are in W/°C. Derating
factor (Df) is the reciprocal of θjc.
Several variables affect junction temperature. Some are
controlled by the IC vendor, while others are controlled
by the user and the environment in which the device is
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APPENDIX
P
Ta
Tc
DF = Pmax
Tcm– Tc0
Tj
= 1
= 12 mW/ ° C
θ jc
Pmax
(500 mW)
ID
200 mW
Tc
a. without heat sink
TC0
Tc
Tj
TC1
TcM
100°C
125°C
Figure 6. Derating Curve
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Heat Transfer
Insulator
Heat sink
Any discussion of heat transfer should begin with a brief
overview of how heat is transferred. There are three
mechanisms by which heat may be transferred: convection, radiation, and conduction.
θsa
b. with heat sink
Figure 4. Heat Flow Path
Tj
θ jc
Tc
θ ja
θ ca
θ cs
Ts
θ sa
Ta
Tj = Junction Temperature
Tc = Case Temperature
Ts = Sink Temperature
Ta = Air Temperature
Convection involves the transfer of heat by the mixing
of fluid. It is the primary process for heat transfer from
a solid to a liquid or gas in contact with it. The rate of
convection heat flow is mainly a function of surface area,
position of the solid in contact with the fluid, the fluids
velocity and properties, and gravity force.
Thermal radiation is heat transferred by electromagnetic radiation. It exists always, but is the only means of
heat transfer between entities separated by a vacuum.
Heat transfer by conduction involves the transfer of kinetic energy from one molecule to another. It is the primary mechanism for heat transfer between solids.
Conduction heat transfer is governed by Fourier’s law.
The thermal resistance equations, 1 and 1a, mentioned previously can be derived using Fourier’s law
stating that the rate of heat flow (P) through a material
is proportional to the cross sectional area (A) of the
material normal to the heat flow, the temperature gradient (T) along the thickness (x) of the material, and
the thermal conductivity (K), a constant and a basic
property of the material. The value of K is in units
of W/°C-cm mathematically:
Without heat sink
θja = θjc + θca
With heat sink
θja = θjc + θcs + θja
Figure 5. Thermal Resistance Paths
(Schematic Representation)
(2)
P = KA • ( dT ⁄ dx ) = ( kA ⁄ x ) • ( T2 – T1 )
expressed in units of W/cm2,
and hence:
(3)
T = θ • P where θ = x ⁄ KA
Now look at the definition of heat capacity or the time
rate of heat flow.
P = dQ ⁄ dt
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Equation 3 shows that:
θ = ( T2 – T1 ) ⁄ Pd
Equation 3 illustrates that thermal resistance is a function of the geometry and thermal conductivity of the device, varying inversely with cross sectional area.
Therefore, assuming that larger chip sizes are contained in larger packages, it can be concluded that the
larger the device package area, the lower the thermal
resistance.
This can also be shown by understanding the concept
of thermal spreading. Heat spreads both laterally and
vertically through the IC layers, primarily by conduction.
A cross sectional view of the die mounted on a substrate
package is shown in Figure 7. The spread angle varies
for each type of material.
In a small package with restricted thermal spreading,
more heat builds up within the package area (i.e., a
higher thermal constant). While in a larger package, increasing the area beyond full spreading does not affect
the thermal constant because the area normal to the
heat flow does not increase.
The graph in Figure 8a shows the relationship between
thermal resistance (θ) and the device package area (A).
Figure 8b shows the θjc versus the ratio of thickness (X)
to the area (A).
θ
Thermal Resistance (θjc)
Where Q is the quantity of heat in calories. Thus, P
equals the power dissipation in watts.
P = cal ⁄ sec = w atts
Thermal Resistance (θjc)
APPENDIX
60
50
40
30
20
10
0
0.128
0.13
0.185 0.211 0.372 0.438 0.537
Area (in. Square)
a. Thermal Resistance versus Area
130
110
90
70
50
30
10
–10
0
0.2
0.4
0.6
0.8
Thickness to Area Ratio (x/A)
b. Thermal Resistance versus Thickness
Figure 8. Thermal Resistance Curves
The challenge in computing the thermal resistance of
the layers of a packaged device is in finding the boundaries with which to define the area of heat. This is not
an easy task because the spread angle of heat varies
for each type of material, increasing with larger thermal
conductivity. Table 1 shows some spread angles of various materials.
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A1
Restricted Thermal Spreading
Table 1. Material Spread Angles
Material
Silicon
BeO
Al2O3
Kovar
Epoxy
Eutectic
Copper
Aluminum
θ
A
Full Thermal Spreading
θ
A2
Figure 7. Spread Angles (A1<A<A2)
Material Spread Angles
40°
60°
25°
25°
0°
0°
70°
65°
The model thus far has provided a means for predicting
thermal conditions for a constant power input. In the
case of a transient response to a pulse input of power
or a series of pulses, thermal capacitance is introduced.
When the die is subjected to a short pulse of power, the
layers below act as a thermal capacitor, absorbing and
storing the thermal energy. Upon termination of the
pulse, the die cools and the thermal energy is dissipated
CPU Thermal Management
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APPENDIX
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through the layers to the case of the device. Figure 9
shows the transient response to a single heat pulse and
a series of pulses to the die.
The junction temperature of a semiconductor device
is related to the temperatures of the various other
elements surrounding it by mathematical relationships similar to those that define the properties of an
electrical circuit that contains resistance and capacitance. It is convenient, therefore, to show the thermal
characteristics of a solid state device as analogous
to the electrical properties of a circuit.
The flow of heat is analogous to the flow of charge;
thermal resistance and capacitance are analogous to
electrical resistance and capacitance; and the potential
difference (voltage) between two points in an electrical
circuit is analogous to temperature difference. Table 2
shows these relationships.
170
Steady State Response
Temperature (°C)
160
Heating curve
150
140
Cooling curve
Table 2. Comparison of Electrical Parameters to
Analogous Thermal Parameters
Electrical
Resistance
R
(Ohms)
Capacitance
C
Amps-s/V
Voltage
V
Volts
Current
I
Amps
Conductivity
p
Ohms/cm2
Charge
q
Coulomb
V1 – V2
R2 = --------------------P
Thermal
Thermal resistance
q
°C/W
Thermal capacitance
C
W-s/°C
Temp. difference
T
°C
Power dissipation
P
Watts
Thermal conductivity
K
W/in-°C
Quantity of heat
Q
Calories
130
T1 – T2
θ2 = -------------------P
120
0
0.01
0.02
0.03
0.04
0.05
Transient
Power
Input
Time (seconds)
Electrical
a. 20-ms Heat Pulse
Temperature (°C)
200
Steady State Response
150
Table 3. Comparison of Electrical Parameters to
Analogous Thermal Parameter
Resistance
R
(Ohms)
Capacitance
C
Amps-s/V
Voltage
V
Volts
Current
I
Amps
Conductivity
p
Ohms/cm2
Charge
q
Coulomb
V1 – V2
R2 = --------------------P
100
Thermal
50
Thermal resistance
q
°C/W
Thermal capacitance
C
W-s/°C
Temp. difference
T
°C
Power dissipation
P
Watts
Thermal conductivity
K
W/in-°C
Quantity of heat
Q
Calories
0
0
0.05
0.1
0.15
0.2
Transient
Power
Input
Time (seconds)
0.25
b. 9-ms Square Pulse with a 12-ms period
Figure 9. Transient Responses
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CPU Thermal Management
T1 – T2
θ2 = -------------------P
AMD
APPENDIX
Table 4. Comparison of Electrical Parameters to
Analogous Thermal Parameters
P
Electrical
Resistance
R
(Ohms)
Capacitance
C
Amps-sec/V
Voltage
V
Volts
Current
I
Amps
Conductivity
p
Ohms/cm2
Charge
q
Coulomb
C1
R1
C2
R2
C3
R3
P
V1 – V2
R2 = --------------------P
Thermal
Thermal resistance
q
°C/W
Thermal capacitance
C
W-sec/°C
Temp. difference
T
°C
Power dissipation
P
Watts
Thermal conductivity
K
W/in-°C
Quantity of heat
Q
Calories
ZeQ
R = x/KA
C=p•c•v
p = density
c = heat capacity
v = volume
Figure 10. Electrothermal Circuit
for a Three-Layer Device
T1 – T2
θ2 = -------------------P
Thermal capacitance is equal to the product of the specific heat (H) of the material used in the sample and the
mass (M) of the sample; it is the quantity of heat absorbed by the sample when its temperature rises 1°C.
Therefore, if a sample absorbs a quantity of heat (Q)
when its temperature is increased from T1 to T2, the
thermal capacitance of the sample, expressed in wattseconds per °C, can be determined from the following
equation:
Q
C = -------------------T2 – T1
Figure 10 shows the electro-thermal circuit for a three
layer device. Thermal impedance, like electrical impedance, is a complex variable because of the time
dependence associated with the thermal capacitance.
In this circuit, the thermal resistances closest to the
heat source are large because the cross section of the
semiconductor is small (all the heat flows through a
small area). Thermal resistance varies inversely with
cross sectional area (as shown in equation 3). Thermal
resistance becomes progressively smaller as distance
from the semiconductor increases.
Since thermal capacitance varies directly with both
specific heat and mass, the small mass of the semiconductor causes the thermal capacitance to be smallest
at the heat source and to become pressively larger as
distance from the heat source increases. The final thermal capacitance in the series is considered an infinite
capacitance, which is electrically equivalent to a short
across the end of the line.
To simplify the equation, assume a steady state condition where capacitance does not play a role. Thus, for
a three-layer device (die, epoxy, and package), total
thermal resistance equals three resistances in a series.
R die + R epoxy + R package
Consider a square heat source of side dimension a in
contact with a package of thickness x. For simplicity,
assume that the thermal spread angle of the material is
45°. Thus, the area is the truncated pyramid and theta
is calculated by the following equation:
a
a
x
45°
θ = ( 1 ⁄ K ) ∫ dx ⁄ area
CPU Thermal Management
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7
APPENDIX
AMD
and therefore:
Therefore:
θ die + θ epoxy + θ pkg
θ = x ⁄ [ K • a ( a + 2x ) ]
For this example, the package material thermal conductivity equals 0.1524 W/°C-in and the thickness of 0.1
inches (100 mils). The heat source (i.e., the die) is 160
mils x 160 mils.
1
θpkg = -------------------------------------------------------------------------------- = 11.39°C ⁄ W
0.1524 [ 0.16 • ( 0.16 + 2 • ( 0.1 ) ) ]
Continuing to determine the total thermal resistance, the
resistance of the die and adhesive is calculated.
Silver-filled conductive epoxy of thickness 1.0 mil, having a thermal conductivity of 0.0063 W/°C-in is used.
Again, the heat source is the die of the side dimension
160 mils; however, the spread angle of the heat through
epoxy is 0°. Thus,
0.001
θepoxy = --------------------------------------------- = 6.20 °C ⁄ W
( 0.0063 ) • ( 0.16 ) 2
The thermal conductivity of silicon is 3.27 W/°C-in; die
thickness is 20 mils and the heat source is the sum of
the collector junction region totaling a dimension of approximately 10 mils.
= 12.23°C ⁄ W + 6.20°C ⁄ W + 11.39°C ⁄ W
θtotal = 29.82°C ⁄ W
A more accurate calculation of θjc would also include
the lead frame and wire bonding thermal resistances as
described in Figure 11.
The graphs in Figure 12 illustrate how thermal resistance varies with package area and die size. All values
were calculated as described above using the dimensions of LCC packages and associated die.
Thermal Properties of Materials
Several factors affect the thermal resistance of an IC
package. Package variables include leadframe material
and construction, case material and construction, and
die attach method and material.
Accurate information on thermal properties of materials
is not readily available. The thermal conductivity of ceramics, for example, is dependent on the purity of the
material, and that of silicon is dependent on temperature.
0.02
θdie = ---------------------------------------------------------------------------------- = 12.23°C ⁄ W
3.27 [ ( 0.01 ) • ( 0.01 + 2 • ( 0.02 ) ) ]
1
Bond wire
R die
7
Die
Silver Epoxy
2
R epoxy
R wire
R pkg 1
3
8
Package body
R lead frame 1
Lead frame finger
R lead frame 2
Nodes:
1. collector junction
2. within die under junction
3. lead frame surface
4. lead frame periphery
5. lead frame finger
6. ambient
7. bond pads
8. package bosy
4
5
R pkg 2
R socket
6
Figure 11. Total Thermal Resistance
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Lead frame
APPENDIX
AMD
60
50
40
30
20
10
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
Die Area (in. square)
a. for various die sizes
70
Thermal Resistance (°C/W)
Thermal Resistance (°C/W)
70
60
50
40
30
20
10
0
0
0.1
0.2
0.3
0.4
0.5
0.6
Package Area (in. square)
b. for various packages
Figure 12. Calculated Thermal Resistance
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APPENDIX
AMD
Table 5 lists several materials and their thermal
conductivity.
Table 5. Typical Thermal Properties of Materials
Silicon
K = 1.05 W/cm°C
@ T = 100°C
K = 0.976 W/cm°C
@ T = 120°C
K = 1.29 W/cm°C
@ T = 25°C
Kovar
K = 0.21 W/cm°C
@ T = 20–100°C
K = 0.134 W/cm°C
@ T = 25°C
Gold-Silicon Eutectic
K = 2.96 W/cm°C
@ T = 25°C
K = 2.16 W/cm°C
@ T = 25°C
Beryllia
K = 1.54 W/cm°C
@ T = 100°C
K = 2.05 W/cm°C
@ T = 20°C
K = 2.54 W/cm°C
@ T = 25°C
Conductive Silver Filled
Epoxy
K = 0.016 W/cm°C
@ T = 120°C
Alumina (90–92%)
K = 0.06 W/cm°C
@ T = 25°C
K = 0.07 W/cm°C
@ T = 100°C
Solder
K = 0.492 W/cm°C
@ T = 25°C
Sapphire
K = 0.033 W/cm°C
@ T = 38°C
K = 0.026 W/cm°C
@ T = 93°C
K = 0.023 W/cm°C
@ T = 149°C
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CPU Thermal Management