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EEL5225: Principles of MEMS Transducers (Fall 2003) Instructor: Dr. Hui-Kai Xie Lumped Modeling in Thermal Domain ; Last lecture z z z Self-heating resistor Other dissipation mechanisms Coupled flows z Today: z Heat flow – DC steady state L z Finite difference method z Eigenfunction method T(0)=0 T(L)=0 Reading: Senturia, Chapter 12, p.299-314 11/10/2003 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= 11/10/2003 κ C . EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 2 Heat Flow Equation--DC Steady-State z 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 z • Requires S(r) and boundary conditions for X(r) and dX(r)/dn, Dirichlet and Neumann B.C. To solve Poisson equation z z Superposition: Impulse response Æ Convolution integral Lumped-element equivalent z z 11/10/2003 Finite-difference method Eigenfunction method EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 3 DC Steady-State: Finite-Difference Method z 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) ⇒ 11/10/2003 hκ [ X ( x + h) + X ( x − h) − 2 X ( x) ] = − h3 S ( x) = −iS ,n EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 4 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 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 5 DC Steady-State: Finite-Difference Method z Matrix representation = Ref. Senturia, Microsystem Design, p. 304. z z 11/10/2003 MATLAB SPICE EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 6 DC Steady-State: Eigenfunction Method z 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 z 2 Eigenfunction solution z z z 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 11/10/2003 = λi xi eigenvector EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 7 DC Steady-State: Eigenfunction Method (∇ ) x 2 i eigenvalue = λi xi eigenvector 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 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 8 DC Steady-State: Eigenfunction Method z 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 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 9 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 11/10/2003 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 10 DC Steady-State: Eigenfunction Method L 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! 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 11 Heat Flow – Transient z 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 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 12 Heat Flow – Transient z 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 ). 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 13 Heat Flow – Transient z 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. 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 14 Heat Flow – Transient z 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 11/10/2003 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 15 Heat Flow – Transient z 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 ) = 11/10/2003 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 16 Heat Flow – Transient z Impulse Response z z z z 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 11/10/2003 4 nπ x −α nt sin ∑ e L n odd nπ EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 17 Heat Flow – Transient z Equivalent circuit for single mode z 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 ( 11/10/2003 ) EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 18 Heat Flow – Transient z 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 19 Heat Flow – Transient z 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 11/10/2003 4 VC nπ 8 S0 = VS0 2 2 C nπ EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 20 Heat Flow – Transient z 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 21 Heat Flow – Transient Example z 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 ∞ 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 22 Heat Flow – Transient Example z 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 11/10/2003 EEL5225: Principles of MEMS Transducers (Fall 2003) Dr. Xie 23