ee c247b –me c218 introduction to mems design …ee247b/sp17/modules/...ee c245: introduction to...

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1 EE 247B/ME 218: Introduction to MEMS Module 11: Equivalent Circuits I CTN 3/16/17 Copyright © 2017 Regents of the University of California EE C245 : Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 1 EE C247B – ME C218 Introduction to MEMS Design Spring 2017 Prof. Clark T.-C. Nguyen Dept. of Electrical Engineering & Computer Sciences University of California at Berkeley Berkeley, CA 94720 Lecture Module 11 : Equivalent Circuits I EE C245 : Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 2 Lecture Outline Reading: Senturia, Chpt. 5 Lecture Topics: Lumped Mass Lumped Stiffness Lumped Damping Lumped Mechanical Equivalent Circuits Electromechanical Analogies

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Page 1: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

1

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 1

EE C247B – ME C218Introduction to MEMS Design

Spring 2017

Prof. Clark T.-C. Nguyen

Dept. of Electrical Engineering & Computer SciencesUniversity of California at Berkeley

Berkeley, CA 94720

Lecture Module 11: Equivalent Circuits I

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 2

Lecture Outline

• Reading: Senturia, Chpt. 5

• Lecture Topics:Lumped MassLumped StiffnessLumped DampingLumped Mechanical Equivalent CircuitsElectromechanical Analogies

Page 2: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

2

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 3

Lumped Parameter Mechanical Equivalent Circuit

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 4

Equivalent Dynamic Mass

•Once the mode shape is known, the lumped parameter equivalent circuit can then be specified

• Determine the equivalent mass at a specific location x using knowledge of kinetic energy and velocity

Maximum Kinetic Energy

Maximum Velocity @ location x

Equivalent Mass =

Location x

Density

Maximum Velocity Function

z

Page 3: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

3

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 5

Equivalent Dynamic Mass

• For the folded-beam structure, we’ve already determined the maximum kinetic energy

• And in our resonance frequency analysis, we’ve already determined expressions for velocity

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 6

Equivalent Dynamic Stiffness & Damping

• Stiffness then follows directly from knowledge of mass and resonance frequency

• And damping also follows readily from knowledge of Q or other loss measurands

•With mass, stiffness, and damping lumped parameter equivalent circuit

Page 4: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

4

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 7

Get Potential Energy & Frequency

Folded-beam suspension

Anchor

Shuttle w/ mass Ms

h = thickness

Folding truss w/

mass Mt\2

Meq(shuttle) = 2.16x10-11 kg

Keq(shuttle) = 4.8 N/m

Ceq(shuttle) = 1.02x10-10 kg/s

Meq(truss) = 8.64x10-11 kg

Keq(truss) = 19.2 N/m

Ceq(truss) = 4.08x10-10 kg/s

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 8

Electromechanical Analogies

meq

keq

ceq

lx rxcx

Page 5: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

5

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 9

Electromechanical Analogies (cont)

•Mechanical-to-electrical correspondence in the current analogy:

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 10

Bandpass Biquad Transfer Function

meq

keq

ceq

Page 6: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

6

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 11

-60

-50

-40

-30

-20

-10

0

8.7 8.9 9.1 9.3Frequency [MHz]

Tra

ns

mis

sio

n [

dB

]

Pin=-20dBm

In Out

VP

Sharper roll-off

Sharper roll-off

Loss PoleLoss Pole

Performance:fo=9MHz, BW=20kHz, PBW=0.2%

I.L.=2.79dB, Stop. Rej.=51dB20dB S.F.=1.95, 40dB S.F.=6.45

Performance:fo=9MHz, BW=20kHz, PBW=0.2%

I.L.=2.79dB, Stop. Rej.=51dB20dB S.F.=1.95, 40dB S.F.=6.45

Design:Lr=40m

Wr=6.5m hr=2m

Lc=3.5mLb=1.6m VP=10.47VP=-5dBm

RQi=RQo=12k

[S.-S. Li, Nguyen, FCS’05]

3CC 3/4 Bridged Mechanical Filter

[Li, et al., UFFCS’04]

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 12

Micromechanical Filter Circuit

1/krmr cr 1/krmr cr 1/krmr cr-1/ks -1/ks

1/ks

-1/ks -1/ks

1/ks

1/kb 1/kb

-1/kb

Co Co

1:e e:1

1:c 1:cc:1 c:1

1:b b:1

/4

/4

/4Input

Outputvi

RQ

RQ

vo

VP

Bridging Beam

Coupling Beam

Resonator

vovi

Page 7: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

7

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 13

Micromechanical Filter Circuit

1/krmr cr 1/krmr cr 1/krmr cr-1/ks -1/ks

1/ks

-1/ks -1/ks

1/ks

1/kb 1/kb

-1/kb

Co Co

1:e e:1

1:c 1:cc:1 c:1

1:b b:1

/4

/4

/4Input

Outputvi

RQ

RQ

vo

VP

Bridging Beam

Coupling Beam

Resonator

vovi

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 14

Micromechanical Filter Circuit

1/krmr cr 1/krmr cr 1/krmr cr-1/ks -1/ks

1/ks

-1/ks -1/ks

1/ks

1/kb 1/kb

-1/kb

Co Co

1:e e:1

1:c 1:cc:1 c:1

1:b b:1

/4

/4

/4Input

Outputvi

RQ

RQ

vo

VP

Bridging Beam

Coupling Beam

Resonator

vovi

Page 8: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

8

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 15

vovi

Micromechanical Filter Circuit

1/krmr cr 1/krmr cr 1/krmr cr-1/ks -1/ks

1/ks

-1/ks -1/ks

1/ks

1/kb 1/kb

-1/kb

Co Co

1:e e:1

1:c 1:cc:1 c:1

1:b b:1

/4

/4

/4Input

Outputvi

RQ

RQ

vo

VP

Bridging Beam

Coupling Beam

Resonator

All circuit element values determined

by CAD layout

All circuit element values determined

by CAD layout

Amenable to automated circuit

generation

Amenable to automated circuit

generation

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 16

-60

-50

-40

-30

-20

-10

0

8.7 8.9 9.1 9.3Frequency [MHz]

Tra

ns

mis

sio

n [

dB

]

Pin=-20dBm

In Out

VP

Sharper roll-off

Sharper roll-off

Loss PoleLoss Pole

Performance:fo=9MHz, BW=20kHz, PBW=0.2%

I.L.=2.79dB, Stop. Rej.=51dB20dB S.F.=1.95, 40dB S.F.=6.45

Performance:fo=9MHz, BW=20kHz, PBW=0.2%

I.L.=2.79dB, Stop. Rej.=51dB20dB S.F.=1.95, 40dB S.F.=6.45

Design:Lr=40m

Wr=6.5m hr=2m

Lc=3.5mLb=1.6m VP=10.47VP=-5dBm

RQi=RQo=12k

[S.-S. Li, Nguyen, FCS’05]

3CC 3/4 Bridged Mechanical Filter

[Li, et al., UFFCS’04]

Page 9: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

9

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 17

Beam Resonator Equivalent Circuits(Pretty Much the Same Stuff)

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 18

Equivalent Dynamic Mass

•Once the mode shape is known, the lumped parameter equivalent circuit can then be specified

• Determine the equivalent mass at a specific location x using knowledge of kinetic energy and velocity

h

W

Maximum Kinetic Energy

Maximum Velocity @ location x

Equivalent Mass =

Location x

Density

Maximum Velocity Function

z

Page 10: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

10

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 19

Equivalent Dynamic Mass

•We know the mode shape, so we can write expressions for displacement and velocity at resonance

h

W

Location xz

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 20

Equivalent Dynamic Stiffness & Damping

• Stiffness then follows directly from knowledge of mass and resonance frequency

• And damping also follows readily

h

W

Location xz

Page 11: EE C247B –ME C218 Introduction to MEMS Design …ee247b/sp17/modules/...EE C245: Introduction to MEMS Design LecM11 C. Nguyen 11/6/08 5 Equivalent Dynamic Mass •For the folded-beam

11

EE 247B/ME 218: Introduction to MEMSModule 11: Equivalent Circuits I

CTN 3/16/17

Copyright © 2017 Regents of the University of California

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 21

Equivalent Lumped Mechanical Circuit

h

W

Location x

Meq(x)

Keq(x)

Ceq(x)

2

2

)]([

)]([)(

xu

xdxuAxM

l

oeq

)()( 2 xMxK eqoeq

Q

xMxC

eqoeq

)()(

z

EE C245: Introduction to MEMS Design LecM 11 C. Nguyen 11/6/08 22

Equivalent Lumped Mechanical Circuit

h

W

Meq(0) = 1.03x10-13 kg

Keq(0) = 19,927 N/m

Ceq(0) = 5.66x10-9 kg/s

Meq(l/2) = 2.78x10-13 kg

Keq(l/2) = 53,938 N/m

Ceq(l/2) = 1.53x10-8 kg/s

Meq(node) = ∞

Keq(node) = ∞

Ceq(node) = ∞

Example: Polysilicon w/ l=14.9m, W=6m, h=2m 70 MHz

z