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EEL5225: Principles of MEMS Transducers (Fall 2003)
Instructor: Dr. Hui-Kai Xie
Lumped Modeling in Thermal Domain
;
Last lecture
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z
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Self-heating resistor
Other dissipation mechanisms
Coupled flows
z Today:
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Heat flow – DC steady state
L
z Finite difference method
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Eigenfunction method
T(0)=0
T(L)=0
Reading: Senturia, Chapter 12, p.299-314
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
Lecture 29 by
1
H.K. Xie 11/10/2003
Heat-Flow Equation
Consider a generalized continuity equation for variable X (r , t ),
∂X (r , t )
+ ∇i J X = S ( r , t )
C (r )
∂t
and a generalized flow equation,
J X = −κ (r )∇X
Combining the two equations gives:
⇒
C (r )
∂X (r , t )
− κ (r )∇ 2 X (r , t ) − ∇κ (r ) ⋅∇X (r , t ) = S (r , t )
∂t
For constant C and κ , we obtain the diffusion equation (applies
equally well for heat flow and diffusion):
⇒
∂X (r , t )
1
+ D∇ 2 X (r , t ) = S (r , t )
∂t
C
where D=diffusivity=
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κ
C
.
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow Equation--DC Steady-State
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In the DC steady state, the heat-flow equation
simplifies into the Poisson equation:
1
D∇ X (r ) = − S (r )
C
1
2
∇ X (r ) = − S (r )
2
κ
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• Requires S(r) and boundary
conditions for X(r) and dX(r)/dn,
Dirichlet and Neumann B.C.
To solve Poisson equation
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Superposition: Impulse response Æ Convolution integral
Lumped-element equivalent
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Finite-difference method
Eigenfunction method
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Finite-Difference Method
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Lumped element equivalent for finite-difference
solution to Poisson Equation
h
Mesh the solid into N smaller volumes equally spaced by h.
d2X
S ( x)
=
−
In one-dimension,
.
2
κ
dx
d 2 X X ( x + h) + X ( x − h) − 2 X ( x )
S ( x)
≅
=−
2
2
κ
dx
h
n-1
n n+1
To obtain heat current, multiply source (heat source/volume) by volume.
iS ,n = Q = h3 S ( x)
⇒
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hκ [ X ( x + h) + X ( x − h) − 2 X ( x) ] = − h3 S ( x) = −iS ,n
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Finite-Difference Method
⇒
hκ [ X n +1 + X n −1 − 2 X n ] = − h3 S ( x) = −iS ,n
1
1
Note that R=
or G=conductance= = hκ
hκ
R
X n +1 − X n X n − X n −1
⇒
−
= −iS ,n
Rn
Rn −1
Therefore, the finite-difference solution may be
represented as a resistor network. The equivalent
circuit for one-dimension is shown.
Ref. Senturia, Microsystem Design, p. 302
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Finite-Difference Method
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Matrix representation
=
Ref. Senturia, Microsystem Design, p. 304.
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MATLAB
SPICE
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Eigenfunction Method
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Heat-Flow Equation--DC Steady-State
∇ X (r , t ) = −
2
z
1
κ
S (r , t )
Finite-difference approach (divide and conquer)
X n +1 − X n X n −1 − X n
h
+
= −iS ,n where R n =
κ h ∇ X (r , t ) ≅
Rn
Rn −1
κA
3
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2
Eigenfunction solution
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z
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Set of global basis functions that satisfy B.C.s
‘Eigen’ is German for proper or nontrivial solution (i.e. nonzero)
What function gives itself when it is twice differentiated?
(∇ ) x
2
i
eigenvalue
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= λi xi
eigenvector
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Eigenfunction Method
(∇ ) x
2
i
eigenvalue
= λi xi
eigenvector
ƒ For one dimension, the possible
eignfunctions of the ∇2 operator
e
± jkx
∂2 X
1− d : ∇ X = 2
∂x
X = sin x
2
X ' = cos x
X " = − sin x ∝ X !!!
or cos( kx ) and sin( kx )
We wish to use a set of eigenfunctions and use superposition
to add the contributions of each to the solution.
If ψ i and ψ j are two eigenfunctions, we use some properties:
Orthogonality: ∫ψ i * ( x )ψ j ( x )dx = 0 for i ≠ j
Normalization: ∫ψ i* ( x)ψ i ( x) dx = 1 for i = j
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Eigenfunction Method
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Example:
L
Consider the problem of finding the temperature
distribution in an electrical resistor heated by
T(0)=0
Joule heating and connected on two sides to
heat sinks at T=0. We chose sin( kx) as the eigenfunction.
B.C. #1:
sin( kx) x =0 = 0
B.C. #2:
sin( kx) x = L = 0
T(L)=0
x
⇒ kL = nπ for n=1, 2, 3, ...
2
 nπ x 
Normalized eigenfunctions are: ψ n ( x ) =
sin 

L
 L 
General solution is a sum of 'proper' functions:
∞
T ( x) = ∑ Anψ n ( x)
n =1
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Eigenfunction Method
Substitute general solution into the D.C. heat
flow equation (Poisson equation) in 1-dimension:
∇ T (r ) = −
2
1
κ
S (r )
→
1
∂ 2T ( x )
=
−
S ( x)
2
∂x
κ
∞
L
T(0)=0
T(L)=0
where T ( x) = ∑ Anψ n ( x).
x
n =1
∞
Taking the derivative twice of T(x ) gives: ∑ k 2n Anψ n ( x ) =
n =1
L
∞
2
ψ
x
(
)
k
∫ m ∑ n Anψ n ( x)dx =
0
n =1
1
S ( x)
κ
.
L
ψ
∫
κ
m
( x)S ( x) dx
0
The left-hand-side is nonzero only for n=m. Assume S(x) is a constant S0 :
1
Am = 2
κ km
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S0
2
 mπ x 
sin 
S ( x) dx = 3
∫
L0
κ km
 L 
L
2
1 − (−1) m )
(
L
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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DC Steady-State: Eigenfunction Method
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Only the odd values of n contribute, and the final
solution of the temperature distribution is given by:
 nπ x 
sin


∞
4 S0 L2
L 

T ( x ) = ∑ Anψ n ( x) =
3 ∑
3
κπ
n
n =1
n odd
T(0)=0
Example: What is the temperature in the middle of the resistor, x=
T(L)=0
x
L
?
2
The maximum temperature is at x=L/2, and equal to
4 S0 L2 
1
1 
+
Tmax =
1−
...
3 
κπ  27 125 
Accurate results obtained with only a few terms!
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Finite difference method
Cn-1
Cn
Cn+1
∂T ( r , t )
1
− D∇ 2T (r , t ) = S (r , t )
∂t
C
∂T ( r , t ) 3
− h CD∇ 2T ( r , t ) = h3 S (r , t )
∂t
Note: Differentials in position and time.
⇒ h 3C
Tnp +1 − Tnp Tnp+1 − Tnp Tnp−1 − Tnp
⇒ −Cn
+
+
= −iS ,n
t p +1 − t p
Rn
Rn −1
Distributed Capacitor, Resistor, Heat current source Network
Solve using circuit simulator such as SPICE
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Eigenfunction method
∂T ( r , t )
1
2
− D∇ T ( r , t ) = S ( r , t )
∂t
C
Using separation of variables to write trial solution,
T ( r , t ) = Tˆ ( r )e −α t
Substitute into heat flow equation:
(
)
1
−α Tˆ ( r )e −α t − D ∇ 2Tˆ ( r ) e −α t = S (r , t )
C
1
−α t
2 ˆ
ˆ
⇒ α T ( r ) + D∇ T (r ) e = S ( r , t )
C
Our strategy will be to find the response for a time impulse source
(
)
in order to construct the overall solution. Let S(r,t) = S0 ( r )δ (t ).
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Eigenfunction method
Substituting S(r,t) = S0 ( r )δ (t ),
(
)
α Tˆ (r ) + D∇ 2Tˆ (r ) eα t =
1
S0 (r )δ (t )
C
For t>0+ , δ (t ) = 0, and the equation simplifies to:
2 ˆ
d
T ( x)
α ˆ
D∇ 2Tˆ ( r ) = −α Tˆ ( r ) ⇒
=
−
T ( x) in 1-dimension
2
dx
D
⇒−
α
is an eigenvalue of ∇ 2 .
D
Thus, the solution for the 2nd order ODE is:
αx
jk x x
jk i r
ˆ
ˆ
in 1-dimension where k x =
.
T (r ) = e
⇒ T ( x) = e
D
Note that e jk i r is a linear combination of sine and cosine terms.
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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One-Dimensional Example
Assume an spatially uniform, S0 (r)=constant, heat impulse at t=0.
The resistor is connected on two sides to heat sinks at T=0.
∴ sin(kx) is a suitable eigenfunction
B.C. #1:
sin( kx) x =0 = 0
B.C. #2:
sin( kx) x = L = 0
Since k=
αn
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D
⇒ kL = nπ for n=1, 2, 3, ...
, we know that
αn
D
=
nπ
.
L
Ref. Senturia, Microsystem Design, p. 258.
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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One-Dimensional Example
The overall solution is given by the sum of the contributions of
all of the trial functions to the response:
 nπ x  −α nt
= ∑ An sin 
e
 L 
n =1
n =1
The coefficients, A n , may be obtained similarly
∞
T ( r , t ) = ∑ Tˆ ( r )e
−α n t
∞
to the previous example by substituting into the
heat flow equation and integrating at t=0.
S 2
4 S0
 nπ x 
=
A n = 0 ∫ sin 
for n odd.
dx

π
C L0
L
n
C


The final solution is the impulse response:
L
T ( x, t ) =
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S0
C
4
 nπ x  −α nt
n
si
∑

e
 L 
n odd nπ
Ref. Senturia, Microsystem Design, p. 258.
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Impulse Response
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The total response is equal to the sum of the allowed
characteristic solutions (eigenfunctions or modes).
The contribution of each mode is given by its
amplitude, An.
Each mode decays with a characteristic decay time
constant, αn.
Overall response to a input, S(t), may be obtained by
convolution.
S0
T ( x, t ) =
C
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4
 nπ x  −α nt
sin 
∑
e
 L 
n odd nπ
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Equivalent circuit for single mode
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Eliminate position dependence by choosing a scalar
that depends only on time (not position). Choose
total heat current, Qconduction out of edges (for one mode)
I Q (t ) = Qconduction ( x = 0) − Qconduction ( x = L)
 W H ∂T

dydz 
dydz −  −κ ∫ ∫


x =0

 0 0 ∂x x = L
Substituting the expression for T ( x, t ),
W H
∂T
= −κ ∫ ∫
∂x
0 0
 8κWH
I Q (t ) = 
 L
∑e
n odd
-α n t
 S0
C

8κWH -α nt S0
e
Focus on one term: I Q ,n (t ) =
L
C
(
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)
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Equivalent circuit for single mode
8κWH -α nt S0
e
L
C
S0
8κWH
1
Laplace tranform gives: I Q ,n ( s ) =
αn L 1+ s / αn C
Focus on one term: I Q ,n (t ) =
(
)
It is a single-pole transfer function, which is equivalent to
a low pass RC filter with a time constant,
Tn
1
αn
= RnCn .
IQ,n(t)
Ref. Senturia, Microsystem Design, p.310.
11/10/2003
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Equivalent circuit for single mode
Cn : heat capacity for mode, n
 nπ x 
 2
Cn =C ∫ ∫ ∫ sin 
dxdydz = 
 L 
 nπ
0 0 0
WHL

 VC

Tn
IQ,n(t)
Dn 2π 2
 2 
=
Since RnCn =
and Cn = 
 VC ,
2
L
n
αn
π


1 L/2
⇒ Rn =
nπ κWH
Physically, we represent each mode as a lumped RC circuit driven by the modal temperature
at the center, Tn (x=L/2), referenced to the temperature at the two ends, Tn (x=0)=0 and Tn (x=L)=0.
1
The current source for each mode sets up the initial temperature distribution:
IS (t ) = Q0,nδ (t ) where Q0,n = CnTn (t = 0)
 2
or Q0,n = Cn An = 
 nπ
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  4
 VC 
  nπ
8
 S0
=
VS0

2 2
C nπ
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient
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Equivalent circuit for multiple modes
 ∞ 8

1
IQ ( s) =  ∑ 2 2
 VS0 ≅ ∑ first N modes
 n odd n π 1 + sRn Cn 
Approximate equivalent circuit
(3 modes)
IQ
Ref. Senturia, Microsystem Design, p. 312.
11/10/2003
EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient Example
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Example: sinusoidal steady-state
temporal heat source
Re
Electrical input current: ie = I 0 cos ωt
2
I
R
Heat source: S(t)=I02 Re cos 2 ωt = 0 e (1 + cos 2ωt )
2
Solution consists of DC and sinusoidal steady-state.
ie
• DC steady-state for self-heating obtained earlier.
• Sinusoidal steady-state obtained from circuit for s → j 2ω.

  8V  I 02 Re
1
⇒ I Q ( j 2ω ) = ∑ 
 2 2 
2
n odd  1 + j 2ω Rn Cn   n π 
∞
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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Heat Flow – Transient Example
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Example: sinusoidal steady-state temporal heat source

  8V  I 02 Re
1
I Q ( j 2ω ) = ∑ 
 2 2 
ω
j
R
C
1
2
2
+
n odd 
n n  n π 
Consider the transfer function where the response is the
heat current through the ends of the resistor and the input
is the total heat generated inside the resistor.
∞
 1

1
∑
 2

n
j
R
C
+
ω
1
2
n odd 
n n 
⇒ 3dB point of each mode is at ω Rn Cn = 1/ 2
I (2 jω ) 8
⇒ H ( s ) = X2
= 2
I 0 ReV / 2 π
∞
⇒ Roll-off accuracy dependent on number of modes.
• 1 mode: 1/ω roll-off (-20 dB/decade)
• ∞ modes (exact): 1/ ω roll-off (-10 dB/decade)
Ref. Senturia, Microsystem Design, p. 313
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EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie
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