EE C245 – ME C218 Introduction to MEMS Design Fall 2007gamescrafters.berkeley.edu/~ee245/fa07/lectures/Lec25.SensingCkt… · Lecture 25: Sensing Circuits EE C245: Introduction

1

EE C245: Introduction to MEMS Design Lecture 25 C. Nguyen 11/19/07 1

EE C245 – ME C218Introduction to MEMS Design

Fall 2007

Prof. Clark T.-C. Nguyen

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

Berkeley, CA 94720

Lecture 25: Sensing Circuits


Lecture Outline

• Reading: Senturia Chpts. 6, 14 (op amp parts), 19• Lecture Topics:

Sensing CircuitsIdeal Op AmpsVelocity Sensing CircuitsPosition Sensing CircuitsNon-Ideal Op Amps

MEMS/Transistor Integration

2


Complete Electrical-Port Equiv. Circuit

V1

VP

Fd1

d1 d2

Elec

trod

e 2

Elec

trod

e 1

I1C1 C2

k

x

m

b

ηe2:1cxlx rx

Co1

1:ηe1

Co2

V2

I2

+

-V1

I1

+

-V2

I2

x

mlx =

kcx

1=

brx =

1

111 d

CVxCV o

PPe =∂∂

=η

2

222 d

CVx

CV oPPe =

∂∂

=η

Static electrode-to-mass overlap

capacitance


Condensed Equiv. Circuit (Symmetrical)

ηe:1cxlx rx

Co1

1:ηe

Co2

+

-V1

I1

+

-V2

I2

x

CxLx Rx

Co1 Co2

+

-V1

I1

+

-V2

I2

2e

xmLη

=

kC e

x

2η=

2e

xbRη

=

If ηe1 = ηe2, then …

where

Holds for the symmetrical case, where port 1 and

port 2 are identical

Holds for the symmetrical case, where port 1 and

port 2 are identical

3


Phasings of Signals

• Below: plots of resonance electrical and mechanical signals vs. time, showing the phasings between them

V1

VP

Fd1

d1 d2

Elec

trod

e 2

Elec

trod

e 1

I1C1 C2

k

x

m

b

V2

I2


Port 1 to 2 TransG Across the Circuit

ηe2:1cxlx rx

Co1

1:ηe1

Co2

+

-V1

I1

+

-V2

I2

x

•What is the transconductance from port 1 to port 2 with port 2 shorted to ground?

4


Port 1 to 2 vi-to-io Transfer Function


Sensing Circuits

5


Sense Electrodes

Tuning Electrodes

Sense Electrodes

Tuning Electrodes

Drive Electrode

zΩr

Drive

Sense

[Zaman, Ayazi, et al, MEMS’06]

Drive Voltage Signal

(-) Sense Output Current

(+) Sense Output Current

Drive Oscillation Sustaining Amplifier

Differential TransRSense

Amplifier

MEMS-Based Tuning Fork Gyroscope


Detecting Velocity Versus Position

• Transfer Function:

• Detect position when the output is varying slowly, i.e., at low frequencies

Position(i.e., displacement)

22

2

)(1

)()(

oo

o

d sQsksFsX

ωωω

++=

ω

)()(

sFsX

dLowpassBiquad

ωo

Velocity


• Detect velocity when the output is at resonance or when a bandpass response is required

22

2

)(1

)()(

oo

o

d sQss

ksFs

ωωωυ

++=

ω

)()(sFs

d

υ

ωo

BandpassBiquad

6


Detecting Velocity Versus Position


Position(i.e., displacement)

22

2

)(1

)()(

oo

o

d sQsksFsX

ωωω

++=

Velocity


22

2

)(1

)()(

oo

o

d sQss

ksFs

ωωωυ

++=


Output Current Measures Velocity

ηe2:1cxlx rx

Co1

1:ηe1

Co2

+

-V1

I1

+

-V2

I2

x

• Relationship between output current and velocity:

• To turn current into voltage (for a voltage output), send the current into a resistor RD

• To get position, must integrate → send the current into a capacitor CD

xi eo &2η= Output current is proportional to velocity, and thus, directly measures velocity

7


Velocity-to-Voltage Conversion

• To convert velocity to a voltage, use a resistive load


Position-to-Voltage Conversion

• To sense position (i.e., displacement), use a capacitive load

8


Ideal Operational Amplifiers


Ideal Op Amp

• Equivalent Circuit of an Ideal Op Amp:

• Properties of Ideal Op Amps:

01 =i1v

-

+02 =i

( )−+ − vvA

0R

( )−+ −= vvAv0

( )12 vvA −=

2v

inR

+-

+-+-+v

−v

Voltage-Controlled Voltage Source (VCVS)

Single-ended outputDifferential input

∞=inR00 =R∞=A

1.2.3.

0== −+ ii

finite assuming , 0 == −+ vvv

4.

5. Why?

9


Ideal Op Amp (cont)

∞=inR00 =R∞=A

1.2.3.

0== −+ ii

finite assuming , 0 == −+ vvv

−+−+ =→=−∴ vvvv 0

ground) (virtualcircuit short virtual 0 ⇒∞v

4.

5.

( ) finite 0 ==−∞ −+ vvv

Why? Because for

( )finite 0 =v

• Properties of Ideal Op Amps:

• Big assumption!• How can we assume this? We can assume this only when there is an appropriate negative feedback path!


Negative Feedback

0

0

SSSaSS

i βε

ε

−==

[ ] finite! 10 ==≈⇒∞→ββa

aSSa

i

finite! 10 ==∴ iSS

β

Where S could be a current, voltage, displacement, etc.,…

Negative feedback acts to oppose or subtract from input.

( )

( )β

β

β

aa

SSaSaS

SSaS

ii

i

+=→=+

−=

1 1 0

0

00

Overall transfer function.

(When there is negative FB around the amplifier.)

↑Δ βS

↑Δ iS↑Δ εS

εS+ a

β

↓Δ εS

↑Δ 0S0SiS

12

5

4

3

+-

10


Negative Feedback

• Comments:1. Negative FB can insure S0 = finite even with a = ∞.2. Overall gain dependent (or overall T.F.) dependent only

on external components. (e.g., β).3. Overall (Closed-loop) gain (So/Si) is independent of

amplifier gain a.→ Very important, since amplifiers using transistors

can be designed to have large gain, but it’s hard to get an exact gain. i.e., if you’re shooting for a = 50,000, you might get 47,000 or 60,000 instead.

→ Comment 3 makes this less consequential.


Positive Feedback

• Contrast with Positive Feedback:

• But for a bounded, controllable function, need negative FB around the op amp

↑Δ βS

↑Δ iS ↑↑Δ SS+ a

β

↑↑Δ 0S0SiS

+

+Output blows up!

(for aβ > 1)

Will be the case for a = ∞.

If β is a bandpassbiquad transfer function → get oscillation at the

resonance frequency ωωο

β

11


Inverting Amplifier

0viv-

++- 1i 0=−i

2i1R

2R

Virtual ground

OV

1. Verify that there is negative FB.2. node attached to (-) terminal is virtual

ground.3.

→=→=∴ −+ finite 0 vvv

21 0 iii =∴=−

22220

211

1

0

0

RiRiv

iRv

Rvi ii

−=−=

==−

=

1

20

1

22

10

RR

vvv

RRR

Rvv

ii

i −=∴−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=⇒

NOTE: Gain dependent only on R1 & R2 (external components), not on the op amp gain.

Benefit: Any shunt C at this node will be grounded out.


• Put the feedback around the (+) terminal and see what happens

• This is not (-) FBCannot analyze using the ideal op amp methodCircuit will “rail out”

Flip the Op Amp Around

2R

0viv +

-

1R

−+ ≠≠∴ vvv finite, 0

( ) −+ →−= Lv

( ) ++ →+= Lv

conditions initialon depending or 0−+= LLv

12


Transresistance Amplifier

1. Verify that there is neg. FB → yes, since same FB as inverting amplifier

2. Thus, vo = finite → v+ = v- → (-) terminal is virtual ground3. i- = 0 → i1 = i2

0vii-

+

1i

0=−i

2i

2R

Virtual ground

OV

2220 RiRiv i−=−= ⇒ 20 R

iv

i−=

An inverting amplifier is just a transresistanceamplifier with an R1 to

convert voltage to current!

• Take R1 away

Again, shunt C at this node will be grounded out.


Velocity Sensing Circuits

13


Problems With Purely Resistive Sensing

v1

VP

Fd1

d1 d2

Elec

trod

e 2

Elec

trod

e 1

io

i1 C1 C2

k

mCp

x b

vo

RD

Includes Co, line C, bond pad C, and next stage C


Problems With Purely Resistive Sensing

v1

VP

Fd1

d1 d2

Elec

trod

e 2

Elec

trod

e 1

io

i1 C1 C2

k

mCp

x b

vo

RD

• In general, the sensor output must be connected to the inputs of further signal conditioning circuits → input Riof these circuits can load RD

Ri

Dx

iD

i

o

ss

RRR

svv

ω+⋅Θ⋅=1

1)()(( ) piDD CRR

1=ω

Now:

,

These change w/ hook-up → not good.

14


The TransR Amplifier Advantage

0v-

+

2i

2R

• The virtual ground provided by the ideal op amp eliminates the parasitic capacitance Cp and Ri

v1

VP

Fd1

d1 d2

Elec

trod

e 2

Elec

trod

e 1

i2

i1 C1 C2

k

m Cp

x b

• The zero output resistance of the (ideal) op amp can drive virtually anything

Ω= 0oR


Position Sensing Circuits

15


Integrator

1. Verify that there is neg. FB → need a large resistor R2 to DC connect the output and (-) terminal

2. With R2, vo = finite → v+ = v- → (-) terminal = virtual gnd3. i- = 0 → i1 = i2

0vii-

+

1i

0=−i

2i

Virtual ground

OV

22

20 sC

isCiv i−=−= ⇒

2

0 1 sCi

v

i−=

• Replace R2 with C2

Again, shunt elements at this node will be grounded out.

C2


Problems With Pure-C Position Sensing

v1

VP

Fd1

d1 d2

Elec

trod

e 2

Elec

trod

e 1

io

i1 C1 C2

k

mCp

x b

vo

CD

16


The Op Amp Integrator Advantage

• The virtual ground provided by the ideal op amp eliminates the parasitic capacitance Cp

v1

VP

Fd1

d1 d2

Elec

trod

e 2

Elec

trod

e 1 io

i1 C1 C2

k

mCp

x b

0v-

+

C2

R2

(for biasing)2

21

sCR >>


Non-Inverting Amplifier

2R1R

+

- 0v

iv1i

0=−i

2i

iv

21 0 iii =∴=−

21

1 iRvi i ==

⇒

Again, gain depends only on R1 & R2(external or ratioedcomponents), not on the op amp gain.

1. Verify that there is negative FB → same feedback circuit as for inverting amplifier, so neg. FB

−+ =→=∴ vvv finite 02.3. Analysis:

( )211

22110 RRRvRiRiv i +=+= 1

20 1 RR

vv

i+=

17


Unity Gain Buffer

+

- 0v

iv

0=−iiv

1

20 1 RR

vv

i+=

2R1R

+

- 0v

iv1i

0=−i

2i

iv

1 0 =iv

v

R1=∞R2=0

Non-Inverting Amplifier

Unity Gain Buffer


Differential Position Sensing

18


Differential Position Sensing

Suspension Beam in Tension

Proof Mass

Sense Finger

C2

Fixed Electrodes

Tethers with fixed ends

C1

C2

C1Vo

VP

-VP

• Example: ADXL-50


Buffer-Bootstrapped Position Sensing

+VP

-VP

+

-0v

Unity Gain Buffer

Cp

Includes capacitance from interconnects, bond pads, and Cgs of the op amp

Cgd

0v

0v

Cgd = gate-to-drain capacitance of the input MOS transistor

• Bootstrap the ground lines around the interconnect and bond pads

No voltage across CpIt’s effectively not there!

1×

InterconnectGround Plane

Documents

EE C245 – ME C218 Introduction to MEMS Design Fall 2007gamescrafters.berkeley.edu/~ee245/fa07/lectures/Lec25.SensingCkt… · Lecture 25: Sensing Circuits EE C245: Introduction