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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

01

23

45

67 1

23

45

67

8-12-10-8-6-4-20

∆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

01

23

45

67 1

23

45

67

8-12-10-8-6-4-20

∆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