30794082 Semiconductor Thermal Design
Transcript of 30794082 Semiconductor Thermal Design
Semiconductor Device Thermal Characterization and System DesignRoger Stout, PE Research Scientist Corporate R&D: Technology DevelopmentRPS 2010 April
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Course outline Part I: Characterization (75 minutes) Why Everything you Thought You Knew (about device thermal characteristics) is Wrong Characterization Techniques Miscellaneous Measurement Techniques
Part II: Linear Superposition (120 minutes) Basic Theory The Reciprocity Theorem A Detailed Example and its Implications Building a System Model Time Varying Heat Sources
Part III: Thermal Runaway (45 minutes) Theory Datasheet example High-Temperature Reverse Bias qual example1 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Part I Characterization
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Can this device handle 2 W?
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Why is ONs SOT-23 thermal number so much worse than the other guys? ON SOT-23 package 60x60 die Solder D/A Copper leadframe Min-pad board Still air
some other guy SOT-23 package 20x20 die Epoxy D/A Alloy 42 leadframe 1 x 2oz spreader Big fan
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Fundamental ideas Heat flows from higher temperature to lower temperature The bigger the temperature difference, the more heat that flows Three modes of heat transfer Conduction (solids, fluids with no motion) Convection (fluids in motion) Radiation (it just happens)
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Thermal-electrical analogytemperature power temp/power voltage current resistance
energy/degree
capacitance
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Junction temperature?Historically, for discrete devices, the junction was literally the essential pn junction of the device. This is still true for basic rectifiers, bipolar transistors, and many other devices. More generally, however, by junction these days we mean the hottest place in the device, which will be somewhere on the silicon (2nd Law of Thermodynamics).
This gets to be somewhat tricky to identify as we move to complex devices where different parts of the silicon do different jobs at different times.7 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Whats wrong with theta-JA?
2
JA
TJ Ta PdJA
TJJtab
Ttab Pd
TJ8
Pd
Ta
TJCorporate R&D : Packaging Technology
Jtab
Pd
Ttab
Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
Theta-JA vs. copper area
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Typical thermal test board types
Min-pad board
1-inch-pad board
Minimum metal area to attach device (plus traces to get signals and power in and out)
Device at center of 1x1 metal area (typically 1-oz Cu); divided into sections based on lead count
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Theta ( ) vs.. Psi ( ) JEDEC terminology Z JX , R JA older terms ref JESD23-3, 23-4 JA ref JESD 51, 51-1 JMA ref JESD 51-6 JT, TA ref JESD 51-2 JB, BA ref JESD 51-6, 51-8 R JB ref JESD 51-8 Great overview, all terms: JESD 51-12
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Theta (Greek letterWe know actual heat flowing along path of interest
Txxy
Ty
qpath
TyTx
true thermal resistance12 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Psi (Greek letterWe dont know actual heat flowing along path of interest
Txxy
Ty
qtotal
??Tx all we know is total heat input13 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
Ty
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Facts and fallacies Basic idea: thermal resistance is an intrinsic property of a package Flaws in idea: there is no isothermal surface, so you cant define a case temperature Plastic body (especially) has big gradients
different leads are at different temperatures multiple, parallel thermal paths out of package
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An example of a device with two different Max Power ratings Suppose a datasheet says: Tjmax = 150C But it also says: JA = 100C/W JL = 25C/W Pd = 1.25W (Tamb=25C) Pd = 3.0W (TL=75C)
25 100 *1.25 25 125 150
75 25 * 3 75 75 150
Wheres the inconsistency?15 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Wheres the inconsistency?TJ =150C25C/W ( JL) 100C/W ( JA)Whats TL? Not 75C !!
(try about 119C)
TA =25C
of the way from ambient to Tj
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Back in the good old days ...metal can -fair approximation of isothermal surface axial leaded device -only two leads, heat path fairly well defined
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Which lead? Where on case?
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Archetypal package10%
convection case
wire/clip
silicondie attach
flag/leadframe10%
20%
60%
circuit boardCorporate R&D : Packaging Technology
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Then we change things add an external heatsink optional heatsink
flip the die over optional heatsink
40%
60%
mold compound/ case wire/clip silicondie attach
optional case
die attach pads/ balls optional underfill
silicon
flag/leadframe
20%
40%application board
20%
application board
20%
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A bare flip chip10%silicon pads/ balls underfill application board
90%
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Same ref, different valuesJ tab
1.2 C/WJ tab
Pd Tc
50 W 25 C
0.8 C/W
Pd Tc
1.5W 25 C
1 GPM of H 2 O
still air
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Even when its constant, its not!TjR1 (path down to board) constant at 20package environment
R3 (path through case top) constant at 80
TL
TCR4 (case to air path resistance) constant at 500, or 20x R2
1000 900 800 700 600 500 400 300 200 100 0 1
thetaJA - var brd only thetaJA - var airflow
JA
TJ Tamb Q total
1 1 R1 R 2board
1 R3 R 41000
R2 (board resistance) vary from 1 to 1000
10 rstnc. [C/W] 100
theta-JA
Tamb25 psi-JL - var brd only psi-JL - var airflow 60 50 40 15 20
psi-JTpsi-JC - var brd only psi-JC - var airflow
psi-JLJL
T J TL Q total
R1 10 R1 R 2 1 5 R3 R 40 1 10board rstnc. [C/W]
30 20 10 0 100 1000 1
JT
TJ TC Q total
10
R3 R3 R 4 1 board R1 R 2 rstnc. [C/W] 100
1000
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TjJA
Ta
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Fallacies recap: Package resistance isnt fixed: multiple heat paths exiting package boundary conditions dictate heat flow Heat sinks Neighboring devices/power dissipation Single vs.. double-sided boards Local convection vs.. board-edge cooling Multiple layers/power/ground planes
Therefore, different application environments will see different package resistanceSemiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Characterization TechniquesTypical TSP behaviorcalibrate forward voltage at controlled, small (say 1mA) sense current
Characterization Techniques125C sense current
T
Vf
25C
0.5 V
Vf
0.7 V
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How to measure Tjtrue const. current supply(1 mA typical)
approximate const. current supply
10K DUT
OR
10.7V DUT If V f-0.7V, then I-1mA
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How to heatsample current is off while heating current on sample current is always on
10K heating power supply
10K heating power supply
10.7VDUT
10.7V
OR
DUT
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The importance of 4-wire measurements+(1.00 V) Power Output = 1 W supply -(0 V) 0.18 V 0.64 W 0.70 W 0.05 V29 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
1A
0.82 V 0.90 W 0.85 V
0.15 V
0.95 V
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Which raises an interesting question:+(1.00 V) Power Output = 3 W supply -(0 V) 0.45 V 1.3 W 0.02 V 0.55 V
3A
1.3 W 0.3 W 0.98 V
Is this a fair characterization of a low-Rds-on device?30 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Bipolar transistor TSP is Vce at designated constant current Heating is through Vce Choose a base current that permits adequate heatingbias resistor TSP supply
10K
switch
TSP=Vcebias supply heating supply
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Schottky diode TSP is forward voltage at low current Voltages are typically very small (especially as temperature goes up) Highly non-linear, though maybe better as TSP current increases; because voltage is low, higher TSP current may be acceptable Heating current will be large
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MOSFET / TMOS Typically, use reverse bias back body diode for both TSP and for heating May need to tie gate to source (or drain) for reliable TSP characteristicTSP supply
10K
switch
TSP=Vsd
heating supply +
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MOSFET / TMOS method 2 If you have fast switches and stable supplies Forward bias everything and use two different gate voltages
TSP supply
10Kclose switch to heat
close switch to heat
close switch to measure
+ V-gate for heating -
+ V-gate for measure -
TSP=Vds
+ heating supply -
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RF MOS They exist to amplify high frequencies (i.e. noise)! Feedback resistors may keep them in DCTSP supply
10Kclose switch to heat
TSP supply + + heating supply -
close switch to heat
close switch to measure
+ V-gate for heating -
TSP = body diode
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IGBT Drain-source channel used for both TSP and heating Find a gate voltage which turns on the drain-source channel enough for heating purposes Use same gate voltage, but typically low TSP current for temperature measurementTSP supply
10K
switch
TSP=Vdsgate voltage heating supply
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Thyristor Anode--to-cathode voltage path TSP supply used both for TSP and for heating 10K typical TSP current probably lower switch than holding current, so gate must be turned on for TSP anode gate readings; try tying it to the anode (even so, we used 20mA to test SCR2146) TSP=Vac Hopefully, with anode tied to gate, cathode enough power can be dissipated to heat device without exceeding gate voltage limit37 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
heating supply
Logic and analog Find any TSP you can ESD diodes on inputs or outputs Body diodes somewhere Heat wherever you can High voltage limits on Vcc, Vee, whatever Body diodes or output drivers Live loads on outputs (be very careful how you measure power!)
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Heating curve method vs.. cooling curve method
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Quick review: Basic Tj measurementfirst we heat then we measure
10K heating power supply
10K heating power supply
10.7V DUT
10.7V DUT
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Question What happens when you switch from heat to measure?
Answer: stuff changes More specifically, the junction starts to cool down
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Basic heating curve transient methodVfcalibrate forward voltage @ 1mA sense current convert cooling volts to temperature.5V
measurementsvoltage current
1 ma power-off cooling power-off cooling power-off cooling power-off coolingsteady state reached
highcurrent heating
highcurrent heating
highcurrent heating
highcurrent heating
T25C
Temperature
125C
measured temperatures Time
Vf
.7V
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Heating curve method #2voltage
1 ma
highcurrent heating
highcurrent heating
highcurrent heating
highcurrent heating
Temperature
power-off cooling
power-off cooling
power-off cooling
power-off cooling
Time
measured temperatures43 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Time
current
measurementsvoltagecurrent
Basic cooling curve transient methodVfcalibrate forward voltage @ 1mA sense current Temperature.5V
1 ma
highcurrent heating
power-off cooling convert cooling volts to temperature
125C
T25C
steady state reached
Temperature
Vf
.7V
Time (from start of cooling)
Time transient cooling period (data taken)
heating period
subtract cooling curve from peak temperature to obtain heating curve equivalent
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Whoa! that last step there Heating vs.. cooling Physics is symmetric, as long as the material and system properties are independent of temperature
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Heating vs.. cooling symmetry
Start of constant power input (step heating)
Start of (constant) power offjunction flag
lead
(all the same curves, flipped vertically)
back of board edge of board
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For a theoretically valid cooling curve, you must begin at true thermal equilibrium (not uniform temperature, but steady state) So whatever your JA, max power is limited to:power47 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
T j max
TambientJA
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By the way Steady-state vs.. transient ? Since you must have the device at steady state in order to make a full transient cooling-curve measurement, steady-state JA is a freebie. (given that you account for the slight cooling which took place before your first good measurement occurred)
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Effect of power on heating curve24x steadystate power 6x steadystate power 3x steadystate power 2x steady-state power
Tj-maxjunction temperature
steady-state max power
< steady-state max power
Tamb49 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
timeCorporate R&D : Packaging Technology
Some initial uncertaintya few initial points may be uncertain
high-current heating Temperature steady state reached
but once were past the uncertain range, all the rest of the points are goodpower-off coolingtransient cooling period (data taken)
heating period Time
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Heating vs.. cooling tradeoffsHEATING starting temperature ambient limited by tester closer to ambient all points similar errorCorporate R&D : Packaging Technology
COOLING ? limited to steady-state closer to Tj-max error limited to first few points
heating powertemperature of fastest data error control51 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
Still air vs.. moving air Varying the air speed is mainly varying the heat loss from the test board surface area, not from the package itself You just keep re-measuring your boards characteristics
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100
total system thermal resistance90
8070 60 50 40 30 20
little package
package resistance
theta-JA [C/W]
mediumsized package
board resistance10 0 0.1
big package
1
10
air speed [m/s]
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Different boards Min-pad board 1 heat spreader board Youre mainly characterizing how copper area affects every package and board, not how a particular package depends on copper area
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1" pad vs min-pad350
Roger Stout 5/25/2000
300
Source: Un-derated thermal data from old PPD database
SOD-323
250TSOP-6
SOT-23
1" pad thetaJA (C/W)
SOT-23
200SOD-123 SOD-123 TSOP-6(AL42) SOT-23
150Micro 8
TSOP-6
SO-8
SOD-123 SMA & Pow ermite
100SMB Dpak D2pak & TO220 Top Can Top Can SMC SOT-223 SO-8
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overall linear fit is: 1" value = [0.51*(min-pad value) - 7]
0 0 100 200 300 400 500 600 700 min pad thetaJA (C/W)
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Standard coldplate testing infinite heatsink (that really isnt) for measuring thetaJC on high-power devices If both power and coldplate temperature are independently controlled, two parameter compact models may be created
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Standard coldplate testing Detailed design and placement of case TC can have significant effect on measured value
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2-parameter data reductionQheat up, Q1
Q1 Q2
T1
QR1
1 T j T1 R1
(
)
1 T j T2 R2
(
)
This has the form of a two-variable linear equation: heat in, Q Tj R2
y
m1 x1 m2 x2 b
where:
m1T2 heat down, Q2Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
m2
1 R1 1 R2
x1 x2
(T
j
T1 )j
(T
T2 )
b
0
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A single coldplate testambient Ta Rja80.0 0 120. 00 100. 00
Tj
Ta
Tjc ( C)
Tc
coldplate
Tj Rjc Tc coldplate
60.0 0 40.0 0 20.0 0 Incr easing Power, Chuck H eld Cold No P ower, Chuck T emper ature Incr eased 20.0 0 40.0 0 60.0 0 80.0 0 100. 00
0.00 -20. 00 0.00 -20. 00
Tja (C)
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ambient Ta Rba Tb Rjb Tj Rjc Tc coldplateTjc ( C)
A single coldplate test, package down120. 00 100. 00 80.0 0 60.0 0 40.0 0 20.0 0 0.00 -10. 00 0.00 -20. 00 Incr easing Power, Chuck H eld Cold No P ower, Chuck T emper ature Incr eased 10.0 0 20.0 0 30.0 0 Tjb (C) 40.0 0 50.0 0 60.0 0
Tj
Ta
Tb
Tc coldplate
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Miscellaneous Measurement Topics
Thermocouple Theory 101 Infrared Theory 202
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Thermocouple Theory 101
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Thermocouple Theoryregion B - 20cm of wire length between the junction and the door of the temperature chamber region C 10cm of TC wire length passing from inside of chamber door to outside of chamber door
TC junction B Aperfectly uniform temperature chamber at 100C
C D
TC scanner
perfectly uniform ambient temperature of 25C
region A - a 1cc box around the junction
region D 100cm of TC wire length from outside of chamber to TC measuring instrument
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Thermocouple Theory Question #1: where is the temperature actually being sensed? or to put it another way, which region of the four defined above (A, B, C, D), is the one that matters, as far as generating the EMF being measured by the TC scanner? Why?The emf is generated where there is a temperature gradientB C D
TC junction Aperfectly uniform temperature chamber at 100C
perfectly uniform ambient at 25C
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Thermocouple Theory Question #2: if the purpose of the TC scanner is to measure an EMF, theoretically how much current has to flow in the wire to make the correct measurement?
None. Ideally, a galvanometer circuit balances the EMF at the scanner until no current flows.
Corollary: wire size, and therefore resistance, may affect how long it takes the galvanometer circuit to stabilize.65 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Thermocouple Theory Question #3: what do the two wires in the TC actually do, and why do they have to be made of different materials?100C V=0.0025V
100C
Different materials have different Seebeck coefficients, i.e., V/C. So if you have two different materials, the same temperature gradient appearing in along both wires will result in two different EMFs.V=0.0037V
25C
25C
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Thermocouple Theory Question #4: what does the junction in a TC actually do?25C
V=0.0025V
100C
Provides the common electrical point that relates the EMFs in the two wires to each other, hence the net difference appears at the open end of the circuit.
Net V=0.0012V
V=0.0037V
The junction DOES NOT measure the temperature; the wires do!25C
Corollary: if the junction itself is isothermal, it can be made of any material whatsoever!67 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Thermocouple Theory Question #5: if you really wanted to calibrate this thermocouple, where would you need to apply the test conditions?
TC junction B A C
Everywhere along its length!D
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Infrared Theory 202
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Infrared TheoryA opaque surface radiates thermal energy in proportion to its absolute temperature (to the fourth power), and some other things
q70 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
AT
4
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Infrared TheoryIts also absorbing infrared energy from everything else around it (since everything else has a temperature and is therefore emitting)
It turns out that the absorptivity equals the emissivity.71 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Infrared TheoryThere is thus a net energy transfer which, in the simplest situations, may be described by:
q
obj
Aobj (Tobj
4
Tencl
4
)
(typically applies to a small, isothermal object in a big, isothermal enclosure)72 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Infrared TheoryBut it quickly gets more complicated when there are multiple objects (or temperatures) hanging around. Each object has a view factor of every other object Each object has its own emissivity (usually NOT unity)
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Infrared TheoryThe emissivity, , is also related to the reflectivity, , as follows:
1So the worse something emits, the better it reflects, and vice-versa.
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Infrared Theory
In other words, unless youre looking at a perfect emitter (aka a blackbody), you dont simply see less radiation than you would for a blackbody, you see some of the radiation of whats behind you!
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Infrared TheoryIn even the simplest real-life situation, you usually have four objects, and their associated temperatures, to consider: The test specimen (130C) The surrounding room (25C) You (33C) The detector (-195C)
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Infrared Theory
(the room) The target(the detector itself)
(you)
(the room)Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Infrared TheorySo if youre looking at a shiny, low emissivity specimen, youll see a combination of these four temperatures (that is, the radiation representing these four temperatures) all bouncing into the detector! Depending on the angles involved, therefore, it is quite possible to see vastly higher temperatures than are really there, or vastly lower temperatures are really there.78 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Hot water loop
Footprints
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Direct view
In mirror
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Reflection
Emissivity
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View at 10 am
View at 10 pm
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Infrared TheoryMoral:
For accurate temperature measurements, you must account for the emissivity of the target, and you must control the background!
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Infrared TheoryHeres one way to approach it:isothermal enclosure, high emissivity, opaque to infrared
IR camera
DUT
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Infrared TheoryGoverning equationTwo reference scans
I
I back a T 41
I1I2
I back a T14I back a T2 4
Solve for Temperature
Solve for the unknowns
T
I
I back a
4
a
I2 T24
I1 T14
I back
T2 4 I1 T14 I 2 T2 4 T14
In terms of the reference scan quantities:
T
T2 (I4
I1 ) T1 (I I 2 I14
I2 )
1
4
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Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Part II Linear Superposition
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Linear superposition what is it? The total response of a point within the system, to excitations at all points of the system, is the sum of the individual responses to each excitation taken independently.Tsource 2
Tcomposite
Tsource1
Tsource n
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Linear superposition when does it apply? The system must be linear in brief, all individual responses must be proportional to all individual excitations.
Tnet ATA89 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
TA
B
TA
C
TA
D
2 qB
3 qCCorporate R&D : Packaging Technology
1.2 qD
Linear superposition doesnt apply if the system isnt linear.
T
a(T , q1 ) q1 b(T , q2 ) q2
T
a q1
n1
b q2
n2
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Linear superposition how do you use it?Tamb
Tref5Tj6 Tj5
Tj3Tref3
Tj2
Tref1 Tj4 TB91 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Tj1
Linear superposition when would you use it?
When you have multiple heat sources (that is, all the time!)
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junction temperature vector
theta matrix assembled from simplified subsystems
power input vector
T j1 Tj2 T jn93
J 1A 21
12 J 2A
1n 2n
n1 n2
JnA
q1 q2 qn
Ta
self-heating termsSemiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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junction temperature vector
theta matrix assembled from simplified subsystems
power input vector
T j1 Tj2 T jn
JB1 21
BA1 JB 2
12 BA2
1n 2n
n1 n2
JBn BAn
q1 q2 qn
Ta
device resistance94 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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visualizing theta and psiheat in here measurements here are s
J1A J1B(idle heat source x)
measurements here are s (idle heat source y)
xA
BA
yA
thermal ground95 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
theta matrix doesnt have to be squarejunction temperature vector one column for each heat sourceJA1 12 x1 L1 1 B1 12 JA 2 x2 L1 2 B2 13 23 x3 L1 3 B3
power input vector
T j1 Tj2 TxA TL1 A TBA
q1 q2 q3(why is this and not ?)
one row for each heat source
one row for each temperature location of interest96 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Electrical reciprocity
+ 5V -
+ -
+ ?V -
+
0.3 V V-
I 0.3A 2 V
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Thermal reciprocityheat input here
response here
same response here
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Another thermal reciprocity example
(r)response here
heat input here same response here
(s)
(s)
(r)
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When does reciprocity NOT Apply? Upwind and downwind in forced-convection dominated applications
B
Cairflow A D Heat in at A will raise temperature of C more than heat in at C will raise temperature of A B and D may still be roughly reciprocal
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(square part of) matrix is symmetriccolumns are the heat sources J1 J2 J3 J4 rows are the response locations J5 J6 75 65 55 60 22 10 73 65 71 60 55 25 11 65 55 60 65 61 21 15 55 60 55 61 73 18 11 59 22 25 21 18 125 14 22 10 11 15 11 14 180 10
R1R3 R5 BSemiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
5520 65
6024 63
6314 62
6119 63
2195 21
1515 12
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Superposition exampleTamb=25 Tref5=47.0 Tj6=36.0 Tj5=49.2
Tj3=85.5 Tref3=85.5
Tj2=96.5
Device 1 heated, 1.1 W
Tref1=105.3 Tj4=91.0 Tj1=107.5 TB=96.5
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Reduce the dataTj1 Tambj1Aj1A
75 65 55
q1Tj2 Tamb q1
107.5 25 1.196.5 25 1.1
75
j2A j3A
j2 A
65
j4A j5A
6022 10 73 55 20 65
TBBA
j6A r1A
Tamb q1
96.5 25 1 .1
65
r3A r5A BA
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Device 2 heated, 1.2 Wj1A j2A
65 71
Tamb=25 Tref5=53.8 Tj6=38.2 Tj5=55.0
j3Aj4A j5A j6A r1A
6055 25 11
6560 24 63
Tj3=97.0 Tref3=97.0
Tj2=110.2r3A r5A
Tref1=103.0 Tj4=91.0 Tj1=103.0 TB=100.6104 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
BA
Device 3 heated, 1.3 Wj1A
55 60
Tamb=25j2A
Tref5=43.2 Tj6=44.5
j3A
6561 21 15
Tj5=52.3
j4A j5A j6A
Tj3=109.5 Tref3=106.9
Tj2=103.0r1A r3A
5563 14 62
Tref1=96.5
Tj4=104.3
Tj1=96.5TB=105.6
r5A BA
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Device 4 heated, 1.1 Wj1A
60 55
Tamb=25j2A
Tref5=45.9 Tj6=37.1
j3A
6173 18 11
Tj5=44.8
j4A j5A j6A
Tj3=92.1 Tref3=92.1
Tj2=85.5r1A r3A
5961 19 63
Tref1=89.9
Tj4=105.3
Tj1=91.0TB=94.3
r5A BA
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Device 5 heated, 0.7 Wj1A
22 25
Tamb=25j2A
Tref5=91.5 Tj6=34.8
j3A
2118 125 14
Tj5=112.5
j4A j5A j6A
Tj3=39.7 Tref3=39.7
Tj2=42.5r1A r3A
2221 95 21
Tref1=40.4
Tj4=37.6
Tj1=40.4TB=39.7
r5A BA
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Device 6 heated, 0.5 Wj1A
10 11
Tamb=25j2A
Tref5=32.5 Tj6=115.0
j3A
1511 14 180
Tj5=32.0
j4A j5A j6A
Tj3=32.5 Tref3=32.5
Tj2=30.5r1A r3A
1015 15 12
Tref1=30.0
Tj4=30.5
Tj1=30.0TB=31.0
r5A BA
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Collect the /J1 J2 J3 75 65 55 60 22 10 65 71 60 55 25 11 55 60 65 61 21 15
values60 55 61 73 18 11 22 25 21 18 125 14 10 11 15 11 14 180
columns are the heat sources
J4rows are the response locations J5 J6 R1 R3 R5 BSemiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
7355 20 65
6560 24 63
5563 14 62
5961 19 63
2221 95 21
1015 15 12
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Now apply actual power
Tamb=25Tref5=106.3
Actual power in applicationTj5=124.7
Tj6=139.1
Qj1
.4
Q j2Tj3=134.9 Tj2=140.1
.4.4 .4 .5 .2
Q j3 Q j4 Q j5 Q j6
Tref3=134.1Tref1=138.8 Tj4=135.8 Tj1=140.0 TB=139.1110 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Compute some effective /
values
Take Tj1, for instance. Remember when it was heated all alone, we calculated its self-heating theta-JA like this:
Tj1 Tambj1A
q1
107.5 25 1.1
75
Now lets see:
Tj1 Tambj1A
q1
140 25 0.4
288
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And thats not just a single aberration!Junction to ReferenceSelf heatingj1A j2A j3A j4A j5A j6A j1-R1
3.0 2.0 36.8
vs. 1.5x vs. 1.0x vs. 1.2x
2.0 2.0 30.0
288288 274 277 199 309
vs. 3.8x vs. 4.1x vs. 4.2x vs. 3.8x
7571 65 73
j3-R3 j5-R5
Junction to Boardj1-B
2.2
vs. 0.2x
10.0
vs. 125 1.6x vs. 180 1.7x
j2-Bj3-B j4-B
2.5-10.5 -8.3
vs. 0.3xvs. -3.5x vs. -0.8x
8.03.0 10.0
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Is the moral clear? You simply cannot use published theta-JA values for devices in your real system, even if those values are perfectly accurate and correct as reported on the datasheet and you know the exact specifications of the test conditions. Not unless your actual application is identical to the manufacturers test board and uses just that one device all by itself.
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So is it really this bad?Only sort-of. Lets revisit the math for one device
Tj1 Tj2 Tjn
J1A 12
12 J2 A
12 q2n
1n 2n
1nTj1Tj1
JnA1n qn
2nJ1A q1
q1 q2 qnTaTa
Ta
1nqn
J1A q1 2
effective ambient
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A graphical viewIsolated deviceTj1J1A q1
power, q
1
TaTa
J1A
junction temperature , TJ1
Device in a systemn
shift in effective ambient
Tj1
J1A q1 2 J1A q1
1n qn
TaTa
1J1A
still the same slope
Ta
Ta
junction temperature , TJ1
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What about that system theta we saw earlier that was so different?power q1 the system theta-JA 1J1A J1A
device #1 power/temperature perturbations will fall on this line
NOT this one
1n
Ta116
1n qn 2
Ta
the isolated-device theta-JA TJ1
junction temperature
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How does effective ambient relate to board temperature?system slope for isolated deviceif any of these are non-zero, Ta will be higher than Tan
Tj1
j1a
Q1i 2B1a
(
i1
Qi ) Ta
(
j1B
)QB1a
1
effective Ta ambient
j1B
Q1
Q1TB1a
TaTa
Tj1Btemperature rise, board to J1117 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
when Q1 is not zero, both zero, both of of these be these willwill be non-zero zero
temperature rise, ambient to boardCorporate R&D : Packaging Technology
Predicting the temperature of high power components The device and system are equally important to get right
Predicting the temperature of low power components The system is probably more important than the deviceSemiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Using the previous board example theta arrayJ1 J2 J3
75 65 55 60 22
65 71 60 55 25
55 60 65 61 21
60 55 61 73 18
22 25 21 18 125
10 11 15 11 14 Qj1
power vector0.5
J4J5 J6 R1 R3 R5 B
Q j2 0.5Q j3 0.5 Q j4 0.5 Q j5 0.2 Q j6 0.02
1073 55 20
1165 60 24
1555 63 14
1159 61 19
1422 21 95
18010 15 15
65
63
62
63
21
12Corporate R&D : Packaging Technology
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Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
Observe the relative contributionsFor junction 1 (a high power component) we have: the device itself the other devices = (75 x 0.5) + (65 x 0.5) + (55 x 0.5) + (60 x 0.5) + (22 x 0.2) + (10 x 0.02) + 25
=
37.5
+ 32.5 + 27.5 + 30 + 4.4 + 0.2
+
25
= 37.5 +120 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
94.6
+
25
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Graphically, a high-power device looks like this:power q1=0.5 W note the embedded theta-JA looks like 264 C/W decreasing power
increasing power
1264 C/W
175 C/W
=94.6 Cn
25 C
1n qn 2
=37.5 C (J1Aq1)
157 C TJ1
junction temperature
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Relative contributions to TJ6the other devices = (10 x 0.5) + (11 x 0.5) + (15 x 0.5) + (11 x 0.5) + (14 x 0.2) + (180 x 0.02) + 25 the device itself = 5.0 + 5.5 + 7.5 + 5.5 + 2.8 + 3.6 = 26.3 + 3.6 + 25 + 25
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Graphically, low-power device #6 looks like this:
power and just in case you were wondering, the embedded theta-JA looks like 1495 C/W !1 =3.6 C
q6=0.02 W=26.3 C
180 C/W
25 C
54 C TJ6
junction temperature
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Controlling the matrix
How to harness this math in Excel
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3x3 theta matrix, 3x1 power vector Excel mathMatrix MULTiply obtained by using multi-cell placement Ctrl-Shift-Enter rather {=array formula notation} than ordinary Enter of array formula
theta matrix
power vector
array reference to theta matrix
array reference to power vector
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7x3 theta matrix, 3x1 power vector Excel mathdont forget to use theta matrix is no longer square Ctrl-Shift-Enter # of columns still must equal # of rows of power vector to invoke array formula notation array formula now occupies 7 cells
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7x3 theta matrix, 3x2 power vector Excel mathpower vector is now a 3x2 array each column is a different power scenario, yet both are still processed using a single array (MMULT) formula the single MMULT array formula now occupies 7 rows and 2 columns (one column for each independent power scenario result)
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Temperature direct contributions and totals120
120
80
temperature rise [C] sum of sourcesJ1 J2 J3 J4 J5 J6 R1 R3 R5 B
temperature rise [C], each source
100
100
80
60
60
40
40
20
20
0 result location
0 J1 J2 J3 J4 J5 J6 R1 R3 R5 J3 at 0.4 W J6 at 0.2 W B result location J3 at 0.4 W J6 at 0.2 W J1 at 0.4 W J4 at 0.4 W J2 at 0.4 W J5 at 0.5 W
J1 at 0.4 W J4 at 0.4 W
J2 at 0.4 W J5 at 0.5 W
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Normalized responses at each location due to each source200
normalized response [C/W], each source
180 160 140 120 100 80 60 40 20 0 J1 J2 J3 J4 J5 J6 R1 response location R3 R5 B J1 at 1 W J2 at 1 W J3 at 1 W J4 at 1 W J5 at 1 W J6 at 1 W
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Filling in the theta-matrix
Handy formulas for quick estimates Utilizing symmetry
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Conduction resistancebasic heat transfer relationship for 1-D conductionq dT k A dx k A T L
if we defineR T q
thenR L k ACorporate R&D : Packaging Technology
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Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
Convection resistancebasic heat transfer relationship for surface coolingq h A T
if we defineR T q
thenR 1 hACorporate R&D : Packaging Technology
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Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
Radiation resistancebasic heat transfer relationship for surface radiationq F A (T 4 Ta4 ) F A (T 2 Ta2 )(T Ta )(T Ta ) F A (T 2 Ta2 )(T Ta ) T
if we defineR T q
thenR 1 FA (T 2 Ta2 )(T Ta )
temperatures must be expressed in degrees absolute!
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Thermal capacitance and time constantCapacitance is ability to store energy specific heat is energy storage/mass Based on simple RC concept, relate rate of storage to rate of flux, result is
Cso if
cp Vand if
= RC
Rthen
L and C k Ac pL2 k L2
c p (L A )
Rthen
1 and C h Ac pL h
c p (L A )
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Some useful formulas conduction resistance.. R convection resistance... RL k A 1 h A c pV c pL2 k c pL T h Q A 2 cp k t
thermal capacitance..... C characteristic time... (dominated by 1-D conduction) characteristic time... (dominated by 1-D convection) short-time 1-D transient response...
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Terms used in preceding formulas L - thermal path length A - thermal path cross-sectional area k - thermal conductivity k - density cp cp - heat capacity - thermal diffusivity cpk - thermal effusivity h - convection heat-transfer film coefficient) T - junction temperature rise Q - heating power t - time since heat was first applied
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When do these effects enter?hundreds of seconds tens of seconds a second or so
junction temperature
mainly environmental convection and radiation effects
mainly local application board conduction effects
typical heating curve for device on FR-4 board in still-airtime
mainly package
materials/conduction effects137 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
if
R
then
and
2R
4R138 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Cylindrical and spherical conduction (through radial thickness) resistance formulasro ln ri k L[solid angle]
Half-cylinder
R
R
1 ri 2 1 ri 4
1 ro k 1 ro k
Hemisphere
[included angle]
Full cylinder
R
r ln o ri 2 k L
R
Full sphere
where
L cylinder length ri inner radius ro outer radius
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Package-shrink gotchaRemember how much of theta-JA depends on what isnt the package?
Well, what if your cooling depends significantly on convection from the board surface (whether free or forced air)?
q
h A
T
A
q h T
So never mind the package resistance, the board can only transfer a certain amount of heat to the air:*Special thanks to Dave Billings for the idea behind this slide140 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Package-shrink gotcha1000 A 2 mm
1.0q WW 1E-5 mm2 C
h
100C TSo if 4 SOT23s each have 0.25 W, then T=0.25*200=50C PROBLEM: No problem! 4 SOT23s on a 1000 mm2 board, effective JA=400C/W
1 SOT23 on a 250 mm2 spreader, JA=200C/W
Fortunately, 0.25 W each x 400C/W100C*Special thanks to Dave Billings for the idea behind this slide141 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Whew!
Package-shrink gotchaNOW your favorite supplier comes out with the next generation version (much smaller) of the same component.SOT23 4 SOT723s on a 1000 mm2 SOT23s board, JA=400C/W, T=100C, power = 1 W total, power = 1 W total, 0.25 W each each
Decrease size (but not power and reduce power dissipation dissipation) (RDSON or other electrical performance)
How many SOT723s can you put in that same 1000 mm2? ANSWER: 4 88 SOT723s on a 1000 mm2 board, JA=800C/W, T=100C, power = 1 W total, 0.125 W each
SOT723*Special thanks to Dave Billings for the idea behind this slide142 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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PERIODIC and NON-PERIODIC POWER INPUT
Duty-cycle curves Normalized and non-normalized Linear superposition applied to time domain
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Square-wave duty cycles (aka ratios) What the heck are they? show peak junction temperature as a function of duty cycle and pulse width
Where do they come from? heating curve is equivalent to the limiting case of a 0% duty-cycle, where once the pulse width is over, theres never another cycle linear superposition allows you to generate the whole family from the heating curve
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Representative duty cycle response chart2.5 R(t), Thermal Resistance [C/W]
2
1.5
d=0.5
1
0.2 0.10.05 Single pulse
0.5
0 0.00001
0.0001
0.001
0.01 t, time (s)
0.1
1
10
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It begins with one of these:100
transient thermal response for axial lead rectifier Junction-to-Lead, 1/4" lead length
10
R(t)JL [C/W]
1
0.1 0.0001
0.001
0.01
0.1
1
time [s]
10
100
1000
A single-pulse response is the same as a 0% duty cycle - if you think in terms of a 0% duty cycle meaning you have only one pulse (however long it lasts), but once you turn if off, you never turn it back on again. So any finite on time is zero percent of infinitely long off.146 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
published formulasNon-dimensionalized versionsr(t , d ) dr(t , d ) d
dimensionalized versionsR(t , d ) d R
(1(1
d ) * r(t )d ) * r(t p) r(t ) r( p)
(1
d ) R(t )
r (t , d ) d
(1
d)*r t
t d
r (t ) r
t d
R(t , d ) d R
(1
d) R t
t d
R(t ) R
t d
lim R(t , d ) d Rt 0
lim R(t , d )t
R
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A normalized curve:1JL=27.4C/W
r(t)JL [normalized]
0.1
normalized transient thermal response for axial lead rectifier Junction-to-Lead, 1/4" lead length
0.01 0.0001
0.001
0.01
0.1
1
time [s]
10
100
1000
Is this really such a bright idea?
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Consider this family of non-normalized curves:100
Transient thermal response for axial lead rectifier Junction-to-Lead, varying lead length
10
R(t)JL [C/W]
R(t)-1" R(t)-3/8" 1 R(t)-1/32"
0.1 0.0001
0.001
0.01
0.1
time [s]
1
10
100
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The danger!100
Transient thermal response for axial lead rectifier Junction-to-Lead, varying lead length
10
r(t)JL [normalized]
See what happens to the shorttime response when you normalize!
R(t)JL [C/W]
R(t)-1" R(t)-3/8" 1 R(t)-1/32"
0.1 0.0001 1
0.001
0.01
0.1
time [s]
1
10
100
JL-1/32"=20.0C/W JL-3/8"=34.9C/W
0.1
JL-1"=54.9C/W
0.01
r(t)-1/32" r(t)-3/8" r(t)-1"
normalized transient thermal response for axial lead rectifier, Junction-to-Lead varying lead length
0.001 0.0001
0.001
0.01
0.1
time [s]
1
10
100
150
Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Compare the families of curves:100
R(t)JL [C/W]
Different steady states
square-wave duty cycles for 1/32" lead length
10
single pulse 1% duty cycle 2% duty cycle 5% duty cycle
1
10% duty cycle 20% duty cycle 50% duty cycle
Same shorttime 0% (single-pulse)
0.1 0.0001
0.001
0.01
0.1
time [s]
1
10
100
100
square-wave duty cycles for 1" lead length
R(t)JL [C/W]
Different everything between !!
10
single pulse 1% duty cycle 2% duty cycle 5% duty cycle
1
10% duty cycle 20% duty cycle 50% duty cycle
0.1 0.0001
0.001
0.01
0.1
time [s]
1
10
100
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What to do if youre given a normalized curve Find out what reference value it was normalized from Undo it
If you cant find out Throw the curve away
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Linear Superposition Applied to Time Domain
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Basic heating curve - a single pulseR(t)
t
Q
t
power input corresponding to single pulse heating curve154 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Single pulse (actually turned off)Finite pulse, decomposed into two infinite stepsQ
equalsa t
Q
plust
Q a t
Temperature response of a finite pulse (constructed from superposition of two single pulse responses)R(t) R(t) R(t) a t
plust
equalsa t
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Two differing pulsesTwo finite pulses decomposed into infinite stepsQ1 a Q2 b c Q1 t a -Q1 Q2 b c -Q2 t
is made up of
Temperature response constructed for two finite pulses (constructed from superposition of four single pulse responses)R(t)R(t)
results in this
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An arbitrary pulse train Temperature at end of nth pulsen
Tni 1
Pi R(t2 n
[
1
t 2i
2
)
R(t2n
1
t 2i
1
)]
Notes: times and pulses must be in strictly increasing chronological orderP1 Pn P2
When there is no off period between two consecutive pulses, set t 2i 2 t 2i 3 (i.e. do not combine them into a single value)t2n-1
Powert0 t1 t2
t3 time
t2n-2
For temperature at the beginning of the nth pulse, set t 2n 1 t 2n 2
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An exampleItem P1 P2 P3 t1 t3 t3-1 t3-2 t5 t5-1 t5-2 t5-3 t5-4t0 t6 t4
Junction Temperarure
Value 80 40 70 0.0001 0.0013 0.0012 0.0010 0.0035 0.0034 0.0032 0.0022 0.0002Item
unit W W W s s s s s s s s sunit
P(PK) Peak Power (Watts)
100 80 60 40 20 0 t0 t1 t2 1 t3 2 t, Time (ms) 3 t4 t5 4 P1 P2 P3
T1T2
P R(t1 ) 1P R(t3 ) R(t3 t1 ) 1 P2 R(t3 t 2 )
[
]
T3 T1 T2
Value
T33 t4 t5 4
P R(t5 ) R(t5 t1 ) 1
[
] ]
0.0000 s 0.0003 s 0.0012 s
t0 t1 t2
1 t3 2 t, Time (ms)
P2 R(t5 t 2 ) R(t5 t3 ) P3 R(t5 t 4 )
[
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How to read the curve For small time differences, and on a low resolution log-log plot, what can you do?100
Transient thermal response for axial lead rectifier Junction-to-Lead, varying lead length
10
1
0.1 0.0001
0.001
Item P1 P2 P3 t1 t3 t3-1 t3-2 t5 t5-1 0.01 t5-2 t5-3 t5-4
Value 80 40 70 0.0001 0.0013 0.0012 0.0010 0.0035 0.0034 0.0032 0.0022 0.0002
unit W W W s s s s s s 0.1 s s s
R(t)JL [C/W]
R(t)-1" R(t)-3/8" R(t)-1/32"
time [s]
1
10
100
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Assume power law (straight line on log-log plot)R(t ) a tn
t2
t1
t1 ta R(t1 ) t1n
R(t 2 ) t2n
R(t
) R(t )R(t
t t
n
n
R(t ) 1
log n
R(t 2 ) log t2
t
R(t1 ) t1
)
R(t )
R(t )n
t
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How to read the curve #2 For small time differences, and on a low resolution log-log plot, what can you do?100
anchors R(.0001) 0.22 R(.02) 3.3
Transient thermal response for axial lead rectifier Junction-to-Lead, varying lead length
10
1
0.1 0.0001
0.001
0.01
0.1
time [s]
1
interpolate R(t1) 0.22 R(t3) 0.82 R(t3-1) 0.79 R(t3-2) 0.72 R(t5) 1.36 R(t5-1) 1.34 10 R(t5-2) 1.30 R(t5-3) 1.08 R(t5-4) 0.32
R(t)JL [C/W]
R(t)-1" R(t)-3/8" R(t)-1/32"
100
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ResultsP(PK) Peak Power (Watts)100 80 60 40 20 0 t0 t1 t2 1 t3 2 t, Time (ms) 3 t4 t5 4 P1 P2 P3
T1 T2
P R(t1 ) 80 0.22 17 .6 C 1 P [ R(t3 ) R(t3 t1 )] P2 R(t3 t 2 ) 1 80 (0.82 0.79 ) 40 0.72 2.4 28 .8 31 .2 C
Junction Temperarure
T3 T1 T2
T3
P [ R(t5 ) R(t5 t1 )] 1 P2[ R(t5 t 2 ) R(t5 t3 )] P3 R(t5 t 4 ) 80 (1.36 1.34 ) 40 (1.30 1.08) 70 0.32 1.6 8.8 22 .4 32 .8 C
t0 t1 t2
1 t3 2 t, Time (ms)
3 t4
t5
4
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Another transient analysis examplepower
1825.2W
23
Heres our power input scenario:power 20 pulses 25.2W
repeat 20 times, then stop for 6800 microsecondstime
20 pulses
time 802 last of 20 pulses ends 7600
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equivalence of non-zero average response with (constant average) + (zero-average disturbance)Actual duty cycle and responsepeak Tavg valley Tavg due to Qavg
Superposition of average and disturbance
Power shifted to average zero
Q0Qavg= d Q
(1-d) Q
0-dQ
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Another example, contHow do we analyze it? Over three time scales:power
1825.2W
instantaneous excursions based on local duty cycle
23power
average Tj based on local avg power
average excursions based on global duty cycle
repeat 20 times
time
25.2W
burst
timepower 25.2W
7600
15200
average Tj based on global avg power20 pulses time 802 last of 20 pulses ends 7600
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Endless Possibilities
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A rampA finite ramp decomposed into several infinite stepsPt a
P
is made up of
a
t
Temperature response constructed for finite ramp (constructed from superposition of many single pulse responses)Q t a Q
results in thisa
t
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An arbitrary pulseAn arbitrary pulse decomposed into several infinite stepsQ t a Q
is made up of
Temperature response constructed for this arbitrary pulseQ Q
t
results in thist
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Multiple Time-Varying Heat Sources
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Basically the same theta-matrix idea, only now everything is a function of time as welljunction temperature vectorT j1 (t ) T j 2 (t ) TxA (t ) TL1 A (t ) TBA (t )
one column for each heat source(t ) 12 (t ) x1 (t ) L1 1 (t ) B1 (t ) (t ) JA 2 (t ) x 2 (t ) L1 2 (t ) B 2 (t )12
power input vector
JA1
(t ) 23 (t ) x 3 (t ) L1 3 (t ) B 3 (t )13
q1 (t ) q2 (t ) q3 (t )
one row for each heat source
one row for each temperature location of interest170 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Each heat source needs to be independently heated.
Each temperature needs to be measured whether heating itself or being heated by another.
Time (sec) thermal system boundary
Measurement cycle(s)Time (sec)
*Special thanks to Dave Billings for this slide171 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Temperature (C)
Power input (W)
but first, some words about
Thermal RC networks Grounded (Cauer) vs.. non-grounded (Foster) Orders of magnitude, rungs Pulses and periodic waveforms Short-time behavior, limits, and limitations
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Thermal capacitance (grounded vs.. non-grounded) In electrical circuit, voltage on a capacitor is difference between its two terminals, and you have physical access to both A lumped parameter thermal model is predicated on energy storage being determined entirely by one temperature Therefore, in a thermal circuit, you only have access to one terminal of each capacitor, namely the one where you identify the temperature. The other terminal is ground.173 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Models: grounded vs.. non-groundedCauer ladders Foster ladders
Physical significance: if thermal capacitors are grounded, they bear some relationship to a physical system; not so for non-grounded Cs Mathematical convenience: certain non-grounded networks are mathematically trivial Interchangeability: single-input thermal systems can be represented as either grounded or non-grounded; multiple-input thermal systems are easy to model as grounded-C networks; it can be done, though with some intricate bookkeeping, using non-grounded-C networks174 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Actual thermal RC networksNon-grounded CFoster ladder1.7004 C/W 2.6627 3.9740 10.0255 11.7747 35.3008 33.1212 Tj
vs..
grounded C comparisonCauer ladderTj
7.02E-04 W-sec/C 2.3207 C/W 4.41E-03 5.89E-02 1.55E-01 6.03E-01 1.46E+00 4.16E+002.7397 8.3551 14.6949 21.0538 39.637 9.7581 Ta
5.96E-04 W-s ec/C 4.20E-03 3.67E-02 1.01E-01 4.13E-01 9.20E-01 1.14E+01
Ta175 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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2-rung Foster model
R1
C1
R1
1 sC1
1 sC1 1 R1
R1 R1C1 s 1
R2 R2 C2 s 1
R2
C2
R2
1 sC2
R2 R2 C2 s 1
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2-rung Cauer modelC1 R11 sC2
R1
1 sC1
R1R2 R2 C2 s 1
1 sC1
R2
C2
R2
R1
R2 R2 C2 s 1
1 sC1
1 sC1 1 R1 R2 R2 C2 s 1
R1 R2 C2 s R1 R2 R1C1 R2 C2 s R1C1 R2 C1 R2 C2 s 12
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2-rung models compared Time constants are roots of denominators "tau" is not RC product when capacitors are grounded !
non-grounded capacitors:(Foster)R1 R1C1 s 1 1 1 C1 s 112
grounded-capacitors:R1 R2 C2 s R1 R2 R1C1 R2 C2 s R1C1 R2 C1 R2 C2 s 1where time constants are 2 R1 C11
(Cauer)s s R1 R2 R1 R2 C 2 11
can be written
1 C1
s
12
R2 R2 C2 s 1 1 1 C2 s 12
can be written
1
c r
c
1
c r
2
c
c 4 r
r 100 10 3 1 1
c 0.01 0.1 1 /3 0.1 1
1
2
R1C1
R2C2
2 R2 C22
where time constants are1
r 1 c
rr
r 1 cR2 R1 c
2
rC1 C2
4
r c
R1C1
2
R2 C2
and defining
0.99 0.91 0.73 0.90 0.38
1.01 1.10 1.36 1.11 2.62
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Interesting (and important) implicationsFoster ladderRungs can be in any order and Tj has identical behavior! So where do you split it?Tj
Cauer ladderOrder matters, so a split can make good physical sense. Tj2.3207 C/W 5.96E-04 W-s ec/C 4.20E-03 3.67E-02 1.01E-01 4.13E-01 9.20E-01 1.14E+01 Ta
1.7004 C/W 2.6627 3.9740 10.0255 11.7747 35.3008 33.1212
7.02E-04 W-sec/C 4.41E-03 5.89E-02 1.55E-01 6.03E-01 1.46E+00 4.16E+00
package
2.7397 8.3551
??
14.6949 21.0538 39.637 9.7581
environment
Ta
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Orders of magnitude and rungs It is a rare transient curve that cannot be followed very accurately with time constants no closer than about 1 order of magnitude apart. This means that you need only about one rung per decade of transient response.
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For what its worth ...Every Cauer (grounded-capacitor) ladder has a mathematical solution. By definition, its mathematical solution represents a corresponding Foster (non-grounded-capacitor) ladder. Every Foster (non-grounded-capacitor) ladder having distinct (i.e. non-repeated) time constants, has a corresponding Cauer (grounded-capacitor) equivalent. (However, there is no a priori guarantee that all Rs and Cs will be positive!)Ref: L. Weinberg, Network Analysis and Synthesis, McGraw Hill Book Company, Inc., 1962
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RC network mathCauer ladder continued fraction:Z sC1 R1 sC 2 1 1 1 1 R2 ... 1 ...
with some algebra, both may be written:Z Nn s Dn 1 s
Foster ladder, sum of simple fractions:Z R1 R1C1s 1 R2 ... R2 C 2 s 1 Ri ... Ri Ci s 1
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Arbitrary repetitive pulse
Q
t p 2p
p period of waveform
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Generalized periodic square waveSo consider a general rectangular pulse positioned arbitrarily within a
repetitive cycle of period p. We can derive expressions for theinfinite sum of responses in three regions :Q t arbitrary time of interestm t a
R(t )i 1
Ri 1 e Fi (t )j 1
i
t a jp
Ri 1 em
i
t
R(t )i 1
Ri 1 e
i
p b pulse turns off a pulse turns on
2p
t measured from edge of power stepm t b
R(t )i 1
Ri 1 et b jp
i
0 t184
a
a t
b
b t
p
Fi (t )j 1
Ri 1 e
i
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Infinite summationUtilizing the identitiesj 0
zj
1 1 z
andj 1
z
j
1 z 1
and doing some algebra, we can show these results:
0 tb t p
aa t p
a ti
bb t p a ti
b tepi i
pb t a t
Fi (t )
Ri
e
i
epi
Fi (t )
Ri 1
e
Fi (t )
Ri
e
i
epi
i
1 e
1 e
1 em
In each domain, we then sum over all rungs of the Foster model:
F (t )i 1
Fi (t )
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RCEnd result for arbitrary periodic power Model for Multiple Square Waves in One Cycle
An important point of this example is that for complex power inputs, absolute peak temperatures do not always correspond to the ends of the highest power pulse!Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Short-time limits What do RC models do at shortest time? roll off linearly with timet t
R(t )
R1 1 e
1
R2 1 e t1
2
t2
R1 1 R1 t1
1
t2 2 21
R2 1 R1 t1
1
t2 2 2 2
t2 2 21
t2 2 21
R(t )
R11
R22
t
R12 1
R22 2
t2 2
So for small t: R(t ) (const) t
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RC models and Surface Heating It turns out that if Rs and Cs grow at a uniform ratio, a sqrt(t) behavior ( b t ) is approached in the limit If ratio is 1:3 (increasing with distance from junction node), rung-to-rung time constant ratio will be about
one order of magnitude, and the network will follow theoretical sqrt(t) within a few percent
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Rungs vs.. sqrt(t) behavior1.0E+0
Exact sqrt(t) model1.0E-1
1.0E-2
1 rung 1.0E-3 2 rungs @ 3.0:1 3 rungs @ 3.0:1 4 rungs @ 3.0:1 1.0E-4 1.0E-8 5 rungs @ 3.0:1 1.0E-7 1.0E-6 1.0E-5 1.0E-4 1.0E-3 1.0E-2 1.0E-1 1.0E+0 1.0E+1
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Where does that leave us with respect to the applicability of RC networks? We need to pay attention to silicon thickness vs.. device technology if surface heating is a good model, be sure your RC network has time constants short enough to land within the silicon within the silicon timescaleL2
, if surface heating is a good
model, you have to extend the RC network into that region on the other hand, if surface heating is not a good model (implying that volumetric heating is better), the fastest rung of your RC network can be set to the actual R&C corresponding to the volume being heated190 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Multiple heat-source RC networks You need data (experiment or simulation) Fit an RC network (or networks) to the data Foster networks, well see an Excel-based approach Cauer networks, beyond the scope of this tutorial
Using SPICE: directly input your RC networks Foster networks can be complicated Cauer networks are straightforward
Using Excel: Foster networks, well see an Excel-based approach Cauer networks are not practical
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Typical 2-input heating/response curves250q1 self avg q2 self avg interact
200
q1R(t) [C/W]150
q1 RC-fit q2 RC-fit q1q2 fit
100
q2
50
0 1E-4
1E-3
1E-2
1E-1
1E+0
1E+1
1E+2
1E+3
heating time [s]Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Fitting Foster ladders in Excel
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Using RMSE as fitting parameterA 1 B C D E F G H I J K L
RMSE test pow er [W] Time
1.9 1.00
1.3 1.00
0.5 1.00 TABLE 1: Foster model taus 1.0E-4 1.0E-3 1.0E-2 1.0E-1 1.0E+0 1.0E+1 1.0E+2 q1 R's 11.2 7.6 12.5 76.6 70.0 2.3 41.0 q2 R's 0.8 2.9 2.3 45.6 67.5 2.2 37.7 interact R's -0.1 0.3 -0.7 -2.3 38.4 4.0 34.1
23
q1 self q2 self q1 RC- q2 RC- q1q2 fit 8.48 9.68 10.59 11.82 14.14 14.64 15.37 15.87 16.37 16.98 18.45 18.58 20.29 1.16 1.32 1.50 1.67 1.84 2.02 2.21 2.42 2.65 2.89 3.14 3.17 3.76 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 9.18 10.51 11.67 12.63 13.45 14.14 14.80 15.46 16.16 16.86 17.60 18.39 19.21 1.01 1.20 1.40 1.59 1.79 1.98 2.19 2.41 2.67 2.92 3.20 3.50 3.80 -0.03 -0.03 -0.03 -0.03 -0.02 -0.01 0.00 0.01 0.02 0.03 0.04 0.04 0.05
4 5 6 7 8 9
1.26E-4 1.64E-4 2.09E-4 2.60E-4 3.18E-4 3.82E-4 4.58E-4 5.48E-4 6.57E-4 7.78E-4 9.19E-4 1.09E-3 1.28E-3
1011 12 13
{=SQRT(SUMSQ(E4:E100TABLE 5: results at arbitrary times B4:B100)/COUNT(E4:E100))} 120.5 100.70.00 25.0 {=$G$2*SUM(interact_Rs 25.0 *(1-EXP(-A4/taus)))} 30.0 0.01 59.6 0.05 90.0 39.1
1415 16
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Excels SolverUse Excels solver to minimize the error between the input data and the RC model.1000 0.6 0.4 100 0.2 0
-0.2 10 -0.4 -0.6 1 Simlation data R-C model Fit Error -0.8 -1 -1.2 100 1000
0.1 1E06
1E05
1E- 0.001 0.01 04
Simulation data0.1
1
10
Time (sec)
After optimization
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Fit error (C/W)
R(t) (C/W)
Using SPICE to exercise the models
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Using Foster RC models in SPICETself1 (volts) Tself2 (volts)
Q1 (amps)+ + Tj1 Tpos1 Tj2 -
Tpos2 time
Tneg2
Tneg1
Q2 (amps)
time
different networks of self heating elementsSemiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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identical networks of negative interaction elements
197
1/4th of a 4-input Foster RC SPICE modelPiecewise linear current source for power input from each source generating heat.Summing tool to add voltages from the separate interaction networks with the self heating network Thermal equivalent Foster RC networks (note that all these Rs are positive) The output port (OUT1) will be where you want to monitor the temperature response Each heat source will require a similar block in order to simulate the temperature response of the self heating effect as well as the interactions. Thermal ground by adding a voltage potential to the ground point ambient temperature can be added.
*Special thanks to Dave Billings for this slide198 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
A 2-input Cauer network modelTj1 C11 R11 R12 C12 C22 R22 C23 R23 C24 time R21 Tj2 Apply q1 to Tj1 node
C21
q1 (watts)
C13
R33 R13 C14 R14 C5 R44
R24R5 C6 R6 q2 (watts) Apply q2 to Tj2 node
C7 R7 time
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Using Excel to exercise Foster models
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Setting up a 2-input Excel modelI1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
J
K
L
M
Nambient
O25
P
Q
R
S
T
U
V
W
X
Y
Z
TABLE 1: Foster model taus 1.0E-4 1.0E-3 1.0E-2 1.0E-1 1.0E+0 1.0E+1 1.0E+2 q1 R's 11.2 7.6 12.5 76.6 70.0 2.3 41.0 q2 R's 0.8 2.9 2.3 45.6 67.5 2.2 37.7 interact R's -0.1 0.3 -0.7 -2.3 38.4 4.0 34.1
TABLE 2: pow er input vs time t0 change_at q1_pow er q2_pow er delta q1 delta q2 time 0.5 0.00 1.0 0.5 1.0 0.5 t1 0.10 1.0 0.0 0.0 -0.5 t2 0.25 0.0 1.3 -1.0 1.3 t3 0.40 0.8 0.9 0.8 -0.4 t4 1.00 0.8 0.0 0.0 -0.9 t5 1.10 0.0 0.0 -0.8 0.0 t6 1.50 0.0 1.0 0.0 1.0 t7 2.00 0.0 0.0 0.0 -1.0 t8 2.60 0.0 0.4 0.0 0.4 t9 3.00 0.6 0.4 0.6 0.0 t10 0.6 0.7 0.0 0.3 t11 0.0 0.7 -0.6 0.0
5.00 10.00
TABLE 3: unit step relative to chosen time 0.50 141.5 51.7 0.40 0.25 0.10 68.4 -15.1 0.00 0.0 0.0 0.00 0.0 0.0 0.00 0.0 0.0 0.00 0.0 0.0 0.00 0.0 0.0 0.00 0.0 0.0 0.00 0.0 0.0 1.4 0.00 0.0 0.0
TABLE 5: results at arbitrary times 120.5 0.00 0.01 0.05 0.10 0.25 0.26 0.26 0.27 0.29 25.0 59.6 90.0 112.2 143.6 116.5 110.0 101.0 89.2 100.7 25.0 30.0 39.1 47.4 37.1 46.3 50.3 56.9 67.4
TABLE 4: sum of terms T1_terms T2_terms -5.1 -109.3 -36.6 75.7
Tj1 =SUM(T2_terms)+ambient 180 Tj2 160 140 120 100 q1_pow er q2_pow er
{=IF(time>change_at,time-change_at,0)} {=SUM((S7*q1_Rs+S8*interact_Rs)* (1-EXP(-time/taus)))} 1.21
{=SUM((R7*interact_Rs+R8*q2_Rs)*(1-EXP(-time/taus)))}{=TABLE(,time)}0.8Corporate R&D : Packaging Technology
=SUM(T1_terms)+ambientperature [C]
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Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
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Results from a 2-input Excel Foster model180.0 160.0 140.0 1.00 120.0 Tj1 Tj2 q1_pow er q2_pow er 1.20 1.40
temperature [C]
100.0 80.0 60.0
0.80
0.60
0.40 40.0 20.0 0.0 0 1 2 3 tim e [s] 4 5 6 0.20
0.00
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Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
Corporate R&D : Packaging Technology
power [W]
Organizing the sheet for transient solutionA cell for keeping track of the overall time progression of ALL blocks.
=IF(Master_Time>Row_Time,Master_time-Row_time,0)
Self heating column (each cell is a separate array formula) {=dP-D#*SUM(R1:R10*(1-EXP(-dtime/Tau1:Tau10)))}
Interaction heated columns (each cell is a separate array formula) {=dP-D#*SUM(R5:R10*(1-EXP(-dtime/Tau5:Tau10)))}
A section for power input to the heat sources*Special thanks to Dave Billings for this slide203 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010
A section for Time changes
A section for power changes
A section for Temperature response calculation
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Table for plotting temperature output=SUM(D1:D1_by_D4) +T_ambient
Note!
Time in this column can be independent of the time values in the power input sectionNext, Select this whole region Apply a Data > Table option
*Special thanks to Dave Billings for this slide204 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Results from a 4-input Excel Foster modelTemperature (C)1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0 0.05 0.1 0.15 Time (Sec) 0.2 0.05 0.1 0.15 Time (Sec) 0.2 0.05 0.05 0.1 0.15 Time (Sec) 0.2 D13 2.5 2 1.5 1 0.5 0 0 0.05 0.1 Time (Sec) 0.15 T_D1 T_D2 T_D3 T_D4
Power (W)
0.25 D2
0.2
0.25
0.3
Last power input0.1 0.15 Time (Sec) 0.2
0.25 D3
Temperature (C)
3 2.5 2 1.5 1 0.5 0 0 0.05 0.1 Time (Sec) 0.15 0.2 0.25 0.3
Power (W)
0.25
Temperature (C)
3 2.5 2 1.5 1 0.5 0 0 0.05 0.1 Time (Sec) 0.15 0.2 0.25 0.3
Power (W)
0.25
Temperature (C)
D4
3 2.5 2 1.5 1 0.5 0 0 0.05 0.1
Power (W)
Time (Sec) 0.15
0.2
0.25
0.3
*Special thanks to Dave Billings for this slide205 Semiconductor Device Thermal Characterization and System Analysis (RPS) April 2010 Corporate R&D : Packaging Technology
Recap With the right tools, a thermal RC n