Accurate Thermal Analysis Considering Nonlinear Thermal ... · Accurate Thermal Analysis...
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Accurate Thermal Analysis Considering NonlinearThermal Conductivity
Anand Ramalingam1 Frank Liu2 Sani R. Nassif2David Z. Pan1
1Department of Electrical and Computer Engineering,The University of Texas, Austin, TX 78712
2Austin Research Laboratory, IBM Research Division, Austin, TX 78758
ISQED 2006
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 1 / 18
Outline
1 Introduction & Motivation
2 Background
3 Steady State v RMS
4 Nonlinear thermal conductivity
5 Conclusion
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 2 / 18
Introduction & Motivation
Introduction & Motivation
Why thermal analysis?Temperature affects performanceTemperature and power are tightly coupledReliability issues
Is Electrothermal analysis any different in digital domaincompared to analog domain?
Essentially involves solving A(x)x = b
The size of A is the biggest challenge
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 3 / 18
Introduction & Motivation
Difference in temperature on ignoring thermalconductivity
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∆T [C]
x[mm] y[mm]
∆T [C]
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 4 / 18
Background
Heat conduction equation
A partial differential equation of the form,
ρCp
∂T(x, y, z, t)
∂t= κ(T(x, y, z, t)) 52 T(x, y, z, t) + h(x, y, z, t)
whereκ(T(x, y, z, t)) is the thermal conductivityh(x, y, z, t) is a heat source at (x, y, z, t).ρCp is the heat capacity
Since we are interested in steady state, the above equationreduces to,
κ(T(x, y, z)) 52 T(x, y, z) + h(x, y, z) = 0
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 5 / 18
Background
Finite difference in 1-dimension
Let us consider the problem in one dimension and in one layer(silicon) to simplify things The heat conduction equation reducesto,
κ ��� (T)∂2T(x)
∂x2+ h(x) = 0
Applying finite difference to the heat conduction equation,
κ ��� (Ti)
(
Ti+1−Ti
∆x
)
−(
Ti−Ti−1
∆x
)
∆x
+ h(x) = 0
κ ��� (Ti)
(∆x)2(2Ti − Ti−1 − Ti+1) = h(x)
Can be written in matrix form as �( � ) � = � .
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 6 / 18
Background
Electrical Interpretation in 2-dimension
(i, j)
κ �� (Ti,j)∆x
∆y
(i, j + 1)
κ �� (Ti,j)∆x
∆y
(i, j − 1)
κ �� (Ti,j)∆y
∆x
(i − 1, j)
κ �� (Ti,j)∆y
∆x
(i + 1, j)
(∆x∆y) × h ��� ����
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 7 / 18
Background
Size of the Matrix
In literature, � ( � ) � = � is solved as � � = � [Cheng 1998, Wang2003]
Linear system of equationsStill difficult to solve due to sheer sizeConsider a 8000µm × 8000µm chip with grid size of∆x × ∆y = 10µm × 10µm
The size of the matrix K is R640,000×640,000
Fortunately, the matrix is sparseTo get an accurate solution need to solve the nonlinear system ofequations �
( � ) � = �Very hard since there are no black box methods to solve anonlinear system of equations
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 8 / 18
Steady State v RMS
Steady state thermal circuit
We study if steady state (SS) heatsources can be modeled as a DCsource having the RMS value of SSThe steady state problem is studiedin z direction, assume uniformity inx, y directionsPicture of chip layers from [Su,ISLPED 2003]
−+ T �������
R ������� R ����� R � ���2
α + βejωt
R � � R � + �
−+ T � �
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 9 / 18
Steady State v RMS
Steady State Response
The RMS analysis is done by setting the DC sources to zero andsetting the ac source to the RMS valueThe temperature rise at the substrate calculated using the RMSvalues for the ac sources is 2.65◦ CThe steady state response turns out to be zero!Thus the approximation is good
-8e-13
-6e-13
-4e-13
-2e-13
0
2e-13
4e-13
6e-13
8e-13
0 2 4 6 8 10
tem
per
atu
re[C
]
time [ns]
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 10 / 18
Steady State v RMS
Time constant of the substrate
1e-10
1e-08
1e-06
0.0001
0.01
1
100
10000
1e+06
1 100 10000 1e+06 1e+08 1e+10
A
ω [hz]
subSiO2
SiM + I
C4
The thermal time constant is dominated by the substrate which isaround !#"%$ and the operating frequency is &'"($
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 11 / 18
Nonlinear thermal conductivity
Nonlinear thermal conductivity
Thermal conductivity is a nonlinear function of temperatureThere is a 22 % change in thermal conductivity of Silicon over therange of [27, 80]◦C
Thermal gradients calculated assuming constant thermalconductivity is not very accurate
Challenge is designing efficient algorithms to solve system of m
nonlinear equations simultaneously
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 12 / 18
Nonlinear thermal conductivity
Algorithms to solve nonlinear equations
Newton-Raphson) (k) = ) (k−1) − [ * ( ) (k−1))]−1 + ( ) (k−1))
0x
f(x)
x0x1x2
Simple and fast convergenceNeed to factorize the Jacobian matrix during every iterationRamalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 13 / 18
Nonlinear thermal conductivity
Algorithms to solve nonlinear equations
Constant Jacobian) (k) = ) (k−1) − [ * ( ) (0))]−1 + ( ) (k−1))
0x
f(x)
x0x1x2
Only one factorizationBut slow convergenceRamalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 14 / 18
Nonlinear thermal conductivity
Sketch of Proposed Algorithm
Accelerate Constant Jacobian
0x
f(x)
x(0)cjx
(1)cjx
(2)cj
x(2)roj
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 15 / 18
Nonlinear thermal conductivity
Proposed Algorithm
Input: , ( ) ) , - ( ) ) ) − . : m nonlinear equations in m unknownsInput: p, the number of partitions - ( ) ) is divided into
1: k← 0
2: repeat3: // Use reduced Jacobian every qth iteration4: // after the first k = p + 1 iterations5: if ((k > p) and !(k / q)) then6: Use reduced Jacobian7: else8: Use constant Jacobian9: end if
10: k← k + 1
11: until Convergence
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 16 / 18
Nonlinear thermal conductivity
Difference in temperature on ignoring thermalconductivity
Difference in temperature in a silicon layer between having aconstant thermal conductivity (27◦ C) and incorporatingnonlinearityThe chip dimension is 8mm × 8mm and it dissipates 100 Wuniformly
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∆T [C]
x[mm] y[mm]
∆T [C]
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 17 / 18
Conclusion
Conclusion
RMS response is an upper bound on the steady state responsewhen thermal analysis is done at transistor levelAccurate thermal analysis needs to consider nonlinear thermalconductivityAn efficient algorithm to solve the system of nonlinear equationshas been proposed
Ramalingam, Liu, Nassif, Pan Thermal Analysis ISQED 2006 18 / 18