Lumped-Element Modeling

Transcription

1 EE55: Principles o MEMS Trnsducers (Fll 003) Instructor: Dr. HuiKi Xie st lecture umpedelement Modeling Oneport elements Generlized Cpcitor Generlized Inertnce Kirchho s ws Exmple Tody: Singleport element exmples Twoport elements Trnsormer Gyrtor Reding: Senturi, pp. 03. EE55: Principles o MEMS Trnsducers (Fll 003) ecture 7 by H.K. Xie 0/3/003

2 SinglePort Elements Idel OnePort Elements: There re 5 oneport elements idel sources: eort nd low (i.e. voltge, current, motors, etc.) dissiptor: dissiptes energy, e R complince: stores P.E. due to displcement, e jω C inertnce: stores K.E. due to momentum, e jω I EE55: Principles o MEMS Trnsducers (Fll 003)

3 umped Element Determintion Question: Given the vrible distribution o the physicl phenomen, how do we generte the lumpedelement? Solve the sttic PDE (i.e. Poisson s/plce s Eqution) nlyticl, FDE, FEM, etc. Set reerence displcement sptil loction Compute the incrementl P.E. nd K.E. Set the integrted distributed nd lumped energies equl to solve or the storge elements 3 EE55: Principles o MEMS Trnsducers (Fll 003)

4 Exmple: Mechnicl Resontor A clmpedcirculr plte: C me M me p(t)/a distributed delection lumped, D delection EE55: Principles o MEMS Trnsducers (Fll 003)

5 Exmple: Mechnicl Resontor A clmpedcirculr plte: Assumptions: linerlyelstic, uniormlyloded, xisymmetric. The governing dierentil eqution or the trnsverse sttic delection wr ( ) is d d dw pr Eh r, where D, p is the trnsverse dr r dr dr D ( ν ) pressure loding, h is the plte thickness, E is the modulus o elsticity, nd ν is Poisson's rtio. 3 Boundry conditions: dw w ( ) 0 nd w(0) <. dr r D: Plte Flexurl Rigidity 5 EE55: Principles o MEMS Trnsducers (Fll 003)

6 Exmple: Mechnicl Resontor pr Ar The generl solution is wr ( ) Blog( r) C. The requirement 6D o inite solution requires B 0 nd pplying the zeroslope B.C., then the zerodelection B.C. gives the inl solution wr ( ) p 6D r. This is the sttic delection shpe or n pplied lod. Since there is distributed delection, the potentil nd kinetic energies re lso distributed. Next we will ind the P.E. or the system... 6 EE55: Principles o MEMS Trnsducers (Fll 003)

7 EE55: Principles o MEMS Trnsducers (Fll 003) 7 Eective Complince () [ ] [ ] 0 0 (0) (0) 6 (0) ), ( ), ( ), ( ) ( D w Dw D p D pdp rdr r D πprdrdw r W dp r D dp p p r w p r dw p r dw rdr p Fdx ed dw p w PE PE π π π π π [ ] me PE C w W ) 0 ( Thereore, ( ) Eh v D C me π π

8 Eective Mss * * Now we need to determine the WKE o the plte. Recll, WKE mu. For simple hrmonic motion, u( r) ωw( r) nd u(0) ωw(0) * ρ r dwke u(0) πrdr, where ρ is the mss / unit re. * W KE * W KE. * r πρ u(0) dwke πρ u(0) rdr or 0 5 M me u(0) πρ Thereore, M or /5 o the ctul mss. me 5 8 EE55: Principles o MEMS Trnsducers (Fll 003)

9 Mechnicl Resontor How good is the lumped pproximtion? Compre irst mode resonnt requency rom bending wve eqution versus tht predicted by the lumped model. ρ( ν ) w h E w 0 predicts Eh t ρ( ν ) 80 h E h E 0.75 π C M π 9 ρ( ν ) ρ( ν ) me me umped element model predicts irst resonnt requency to within %! 9 EE55: Principles o MEMS Trnsducers (Fll 003)

10 Exmple: Resistnce in Pipe minr Flow in Circulr Pipe: The pipe rdius is nd the length is. Assumptions: incompressible, stedy, ully developed, Newtonin, nd continuum. Re. F. M. White, Fluid Mechnics, p EE55: Principles o MEMS Trnsducers (Fll 003)

11 Exmple: Resistnce in Pipe minr Flow in Circulr Pipe: The governing dierentil eqution d du p or the xil velocity u( r) is r, where p is the rdr dr µ pressure drop over length o pipe nd µ is the dynmic viscosity o the luid. Boundry conditions: u ( ) 0 (noslip) nd pr The generl solution is ur ( ) Cln ( r) C µ u(0) <.. EE55: Principles o MEMS Trnsducers (Fll 003)

12 Generlized Dissiptor A inite solution requires tht C 0. Applying the noslip B.C gives the inl velocity distribution cross the pipe crosssection p r ur ( ). In order to compute the dissiption R µ, we must irst compute the volumetric low rte Q ( V n) da, or A π π Q u( r) rdrdθ p. Reclling tht p is the eort nd Q is the 0 0 8µ 8µ low in the luid domin, Q R p, we ind tht R. π EE55: Principles o MEMS Trnsducers (Fll 003)

13 Idel TwoPort Elements: TwoPort Elements A lossless wy to chnge energy domins (constnt P) There re twoport elements Trnsormer Gyrtor Sign convention is very importnt! TwoPort e Element e P 0 e e NET Thereore, e e. Note, i both lows re into the twoport element, e e. 3 EE55: Principles o MEMS Trnsducers (Fll 003)

14 Trnsormers n: :n Trnsormer lw:, so by reclling tht the power is constnt, the reltionship between e n e e e eort nd low becomes n 0 e e 0 n Exmples: levers, gers, electrodynmic trnsduction, etc. EE55: Principles o MEMS Trnsducers (Fll 003)

15 Gyrtors e r e Gyrtor lw: e r, so by reclling tht the power is constnt, the reltionship between eort nd low becomes 0 r e e 0 r Exmples: levers, gers, mgnetic trnsduction, etc. Note: Gyrtors re usully used when going rom Impednce to Admittnce nlogies. 5 EE55: Principles o MEMS Trnsducers (Fll 003)

16 Impednce Trnsormtions Z in (s) e :n e Z(s) Z in (s) e n e Z(s) Z( s) Z e e in ( S) ne / n Z( s) n n e Z( s) Z ( S) e e in n e / n n Z( s) n e Impednce Scling. Impednce Admittnce. Cpcitor Inductor 6 EE55: Principles o MEMS Trnsducers (Fll 003)

17 ever Exmple ever/trnsormer: F u F u u 0 : u Blncing the moments requires, F F u ω u lw requires n or i not blnced. So the trnsormer F F 7 EE55: Principles o MEMS Trnsducers (Fll 003)

18 ever Exmple ever/gyrtor: F u F u 0 F u F u The gyrtor lw requires. Mth check: e r or u u, ine. Note: r. n r 8 EE55: Principles o MEMS Trnsducers (Fll 003)