550
Modeling MEAM and Design of MEMS
Spring 2004
Solution to homework #6 Problem 1
p2L V
Flexure
L p3L
Narrow beam Connector
T0 p2L p1L
p2L
p2L
(1-p1)L Ks
Wide beam
Out-of-plane thickness
= ptL
Fig. 1 Basic electro-thermal microactuator Electrical analysis The four segments (viz. narrow beam, connector, wide beam, and flexure) in the actuator can be treated as electrical resistors in series. The resistance of an axial conductor of this type is given by the electrical resistivity times the length divided by the cross-section area. Thus, we have
Ri =
e Li
Ai
for i = 1,2,3,4
(1)
Noting the relative proportions shown in Fig. 1, we get the combined series-resistance as
R = R1 + R2 + R3 + R4 =
e 1
p 1 (1 p1 ) + + + 1 = e e L pt p 2 pt p t p3 pt p 2 L
(2)
The current J and the electrical input power Pe are given by J= V LV = R e e LV 2 (3) (4)
Pe = J 2 R =
e e
& The Joule heating per unit volume per unit time in each of the four segments, i.e., Qei (i = 1L 4) ,
which acts as the heat source for the thermal analysis, can be written as follows using Eqs. (1) through (3). J 2 Ri L2V 2 & Qei = = 2 2 for i = 1,2,3,4 Ai Li e Ai e i.e, (5)
MEAM 550
Modeling and Design of MEMS
Spring 2004
& Qe1 = & Qe2 =
V2 2 e2 pt2 p2 L2 e V2 2 e2 pt2 p 2 L2 e
V2 & Qe3 = 2 2 2 2 e p t p3 L e & Qe4 = V2 2 e2 pt2 p 2 L2 e
(6)
Thermal analysis One-dimensional thermal conduction modeling is appropriate for the slender segments of this microactuator. Since the connector segment is very short, the temperature distribution in it is neglected but the heat generated in it is taken into account in balancing the flux across its two interfaces. The temperature Ti (s) in each of the three remaining segments is governed by the diffusion equation shown below.
& d 2Ti ( s ) Qei + = 0 for i = 1,3,4 kt ds 2
(7)
where s runs from zero through the length of the respective segment. With convection and radiation neglected, the only boundary conditions at either end of each segment are either due to the temperature being equal to the ambient or the continuity of heat flux across the interface between two segments. The differential equation in Eq. (7) can be readily solved as
Ti ( s ) =
& Qei 2k t
s 2 + ai s + bi
(8)
where the constants ai and bi ( i = 1,3,4 ) are solved using the following six boundary conditions:
1. Temperature at the left end is at the ambient temperature: T1 ( s = 0) = T0 , i.e., b1 = T0 (9a)
2. Temperature at the right end of the narrow beam is equal to that at the right end of the wide beam: T1 ( s = L1 ) = T3 ( s = 0) , i.e.,
& Qe1
2k t
L2 + a1 L + T0 = b3 which can be simplified as
t V 2 1 a1 L b3 = 2 T0 = c1 with t = 2 e e k t 2 pt2 p 2
1
1
(9b)
3. Continuity of heat flux across the connector along with the heat generated in it:
MEAM 550
Modeling and Design of MEMS
Spring 2004
k t A1
dT1 ds
k t A3
x = L1
dT3 ds
x =0
& + Qe2 A2 L2 = 0 which can be simplified as
& Qe 3 & k t pt p 2 L2 1 L + a1 k t pt p3 L2 a3 + Qe2 pt p 2 L3 = 0 kt
p 2 La1 + p3 La3 =
t V 2 1 = c 2 with t = 2 pt e e k t
2 2
1 1 p2
(9c)
4. Temperature at the left end of the wide beam is equal to that at the right end of the flexure: T3 ( s = L3 ) = T4 ( s = 0) , i.e.,
(1 p1 ) La3 + b3 b4 =
& Qe3
2k t
3
(1 p1 ) 2 L2 + a3 (1 p1 ) L + b3 = b4 which can be simplified as
t V 2 (1 p1 ) 2 = c3 with t = 2 e2 e k t 2 pt2 p3
3
(9d)
5. Heat flux continuity across the interface between the wide beam and the flexure:
k t A3 dT3 ds
2
k t A4
x = L3
dT4 ds
= 0 which can be simplified as
x =0
& Qe3 k t pt p3 L (1 p1 ) L + a3 k t pt p 2 L2 a 4 = 0 kt
p3 La3 p 2 La 4 =
t V 2 (1 p ) = c4 with t = 2 1 2 pt p 3 e e k t
4
4
(9e)
6. Temperature at the left end of the flexure is at the ambient temperature: T4 ( s = L4 ) = T0 , i.e.,
p1 La 4 + b4 =
& Qe4
2k t
p12 L2 + a 4 p1 L + b4 = T0 which can be simplified as
t V 2 p12 + T0 = c5 with t = 2 2 pt2 p2 e2 e k t
5 5
(9f)
The linear equations in Eqs. (9b) through (9f) can be solved for {a1 , a3 , b3 , a 4 , b4 } . Notice that, for consistency, bi s should have units of temperature, and ai s temperature unit per length. Thus, in view of Eq. (8), the temperature profile depends on the material properties and relative proportions but not on the size-factor, L .
MEAM 550
Modeling and Design of MEMS
Spring 2004
Problem 2
The circular conductor can be modeled as a 1-D conductor for electrical and thermal analysis.
L T=0 V T=0
The electrical resistance at the ambient temperature is given by L (2r ) (1) R0 = = A wt Due to TCR effect, the resistance at any other temperature (denoted as temperature raise from the ambient, which is taken as zero) is R = R0 (1 + T ) (2) The current can then be calculated as V Vwt I= = (3) R 2 r (1 + T ) The Joule heating induced power is equal to
MEAM 550
Modeling and Design of MEMS
Spring 2004
V2 V 2 wt & Q = I 2R = = (4) R 2 r (1 + T ) The above expression gives the total power in the loop. The power generated per unit volume is given by V2 V 2 wt 1 V2 & q= = = R(volume) 2 r (1 + T ) 2rwt 4 2 r 2 (1 + T ) For 1-D thermal analysis, let us consider a differential element and sum up the heat entering, leaving it, and being generated within it. hTAconv = hTwds
Ac / s dT dT = wpt L ds ds
ds dT d 2T wpt L ds + ds 2 ds
& & qdV = qwpt Lds
By balancing heat we get d 2T & wpt L 2 ds hTwds + qwpt Lds = 0 ds d 2T hT & 2 +q =0 (5) pt L ds In view of Eq. (4), Eq. (5) becomes d 2T hT V2 + =0 (6) 2 pt L 4 2 r 2 (1 + T ) ds which is an inhomogeneous second order differential equation that is difficult to solve analytically (at least based on inspection). The boundary conditions are: Ts =0 = Ts = 2r = 0 . So, it is a two-point boundary value problem (BVP). We will use bvp4c function in Matlab to solve Eq. (6). For this purpose, we define two functions as follows. y1 ( s ) = T ( s ) (7) dT ( s ) y 2 (s) = ds From Eqs. (7) and (6), we can write dy1 y2 ds (8) =0 = 2 V dy 2 1 hy1 w 4 2 r 2 (1 + y1 ) ds
MEAM 550
Modeling and Design of MEMS
Spring 2004
The above notation helps understand the Matlab script below. Save as hw6p2.m
% Solution to problem 2 in homework #6 in MEAM 550 in Spring 2004 % The script is below is adapted from Constantin Hatzis's code % submitted as a solution. % % Solution of the two-point boundary value problem (bvp) concerning % a Joule-heated circular loop with convection from top considering % TCR effect and zero temperature at the two ends. clear all clc % Global variables global t kappa alpha V h rho r % Data r = 150E-6; w = 3e-6; t = 1e-6; kappa = 146.4; alpha = 2000E-6; rho = 4...