rochester institute of tehnology …people.rit.edu/lffeee/capacitor_sensors.pdf · eo = 8.85e-14...

© March 25, 2016 Dr. Lynn Fuller, Professor

Rochester Institute of Technology

Microelectronic Engineering

MEMS Capacitor Sensors

ROCHESTER INSTITUTE OF TEHNOLOGY MICROELECTRONIC ENGINEERING

3-25-16 capacitor_sensors.ppt


and Signal Conditioning

Dr. Lynn Fuller Dr. FullersWebpage: http://people.rit.edu/lffeee

Electrical and Microelectronic Engineering


82 Lomb Memorial Drive

Rochester, NY 14623-5604

Email: [email protected]

ProgramWwebpage: http://www.microe.rit.edu

http://people.rit.edu/lffeee

mailto:[email protected]

http://www.microe.rit.edu/





OUTLINE

Capacitors

Capacitors as Sensors

Chemicapacitor

Diaphragm Pressure Sensor

Condenser Microphone

Capacitors as Electrostatic Actuators

Signal Conditioning

References

Homework





CAPACITORS

Capacitor - a two terminal device whose current is proportional to the time rate of change of the applied voltage; I = C dV/dt a capacitor C is constructed of any two conductors separated by an insulator. The capacitance of such a structure is: C = eo er Area/d where eo is the permitivitty of free space er is the relative permitivitty Area is the overlap area of the two conductor separated by distance d

eo = 8.85E-14 F/cm

I

C V

+

-

Area

d er air = 1 er SiO2 = 3.9





OTHER CAPACITOR CONFIGURATIONS

C2

C1

C Total = C1C2/(C1+C2)

Two Dielectric Materials between Parallel Plates

Example: A condenser microphone is made from a polysilicon plate 1000 µm

square with 1000Å silicon nitride on it and a second plate of aluminum with a

1µm air gap. Calculate C and C’ if the aluminum plate moves 0.1 µm.

C1= eo 1 (1000E-4)1000E-4/1E-4 = 8.85 pF

C1at 1.1um = 8.05pF or DC1= 0.80 pF

C1at 0.9um = 9.83pF or DC1= 0.98 pF

C2 = eo 7.5(1000E-4)1000E-4/0.1E-4 = 664pF

Ctotal = 8.73pF

DCtotal = 8.73 + (-7.94 or +9.70) pF = -0.79 or +0.97 pF

eo = 8.85E-14 F/cm

er air = 1 er Si3N4 = 7.5

air





DIELECTRIC CONSTANT OF SELECTED MATERIALS

Vacuum 1

Air 1.00059

Acetone 20

Barium strontium titanate

500

Benzene 2.284

Conjugated Polymers

6 to 100,000

Ethanol 24.3

Glycerin 42.5

Glass 5-10

Methanol 30

Photoresist 3

Plexiglass 3.4

Polyimide 2.8

Rubber 3

Silicon 11.7

Silicon dioxide 3.9

Silicon Nitride 7.5

Teflon 2.1

Water 80-88

http://www.asiinstruments.com/technical/Dielectric%20Constants.htm

Note: Water has er of 80-88

http://www.asiinstruments.com/technical/Dielectric Constants.htm





CALCULATIONS






Interdigitated Fingers with Thickness > Space between Fingers

h = height of fingers s = space between fingers N = number of fingers L = length of finger overlap

C = (N-1) eoer L h /s

Example: 200 fingers, h = 2um, s=1um, air

C = 0.35 pF






Interdigitated Fingers with Thickness << Space between Fingers

C = LN 4 eoer p

n=1

0 0 1

2n-1 Jo2 (2n-1)ps

2(s+w)

Jo = zero order Bessel function w = width of fingers s = space between fingers N = number of fingers L = length of finger overlap

Reference:

Lvovich, Liu and Smiechowski,





MODELING OF INTERDIGITATED CAPACITOR

N = number of fingers er1

er2

er = er1 + er2

d1

d2

d1 and d2 (interrogation depth)





PLANAR INTERDIGITATED SENSOR

dt

tktkK

ws

wk

kK

kKLNC rr

=

=

=

1

0222

2

1

2

021

11

1)(

)(2cos

)(2

])1[()()1(

p

eee

V.F.Lvovich, C.C.Liu, M.F.Smiechowski

Jo is zero order Bessel function

er1

er2

d1

d2

er = er1 + er2





er = eoxide

FINGERS ON SILICON SUBSTRATE

C = C fingers + C to substrate / 2

(if silicon substrate is treated as a conductor)

Silicon

SiO2

C fingers

C to Substrate C to Substrate

er1

er = er1

C fingers C substrate

C = eo er area / t

t





CALCULATIONS

Note: er = er1 + er2 is for the material above/below the fingers if both are insulators a few times thicker than the space between fingers






Two Long Parallel Wires Surrounded by Dielectric Material

C/L = 12.1 er / (log [(h/r) + ((h/r)2-1)1/2]

h = half center to center space r = conductor radius (same units as h)

Capacitance per unit length C/L

Reference: Kraus and Carver

Example: Calculate the capacitance of a meter long connection of parallel wires.

Solution: let, h = 1 mm, r = 0.5mm, plastic er = 3 the equation above gives

C/L = 63.5 pF/m

C = 63.5 pF






Coaxial Cable

C/L = 2 p eo er / ln(b/a)

b = inside radius of outside conductor a = radius of inside conductor

Capacitance per unit length C/L

Reference: Kraus and Carver

Example: Calculate the capacitance of a meter long coaxial cable.

Solution: let b = 5 mm, a = 0.2mm, plastic er = 3 the equation above gives

C/L = 51.8 pF/m

C = 51.8 pF





CAPACITORS AS POSITION SENSORS

C d

C1

C2

C

d

C1

d

C2

One plate moves relative to other changing gap (d)

One plate moves relative to other

changing overlap area (A)

Center plate moves relative to the two

fixed plates

Top plate moves changing C1 and C2 differentially





CAPACITORS AS SENSORS

Change in Space Between Plates (d) Change in Area (A)

Change in Dielectric

Constant (er)

microphone gyroscope

Oil level sensor or

position sensor





CHEMICAL SENSOR

Two conductors separated by a material that changes its dielectric constant as it selectively absorbs one or more chemicals. Some humidity sensors are made using a polymer layer as a dielectric material.

Change in Dielectric Constant (er)

chemical sensor





DIAPHRAGM PRESSURE SENSOR

Diaphragm:

Displacement

Uniform Pressure (P)

Radius (R)

diaphragm

thickness ()

Displacement (y)

E = Young’s Modulus, = Poisson’s Ratio

for Aluminum =0.35

Equation for deflection at center of diaphragm

y = 3PR4[(1/)2-1]

16E(1/)23

= (249.979)PR4[(1/)2-1]

E(1/)23

*The second equation corrects all units

assuming that pressure is mmHg,

radius and diaphragm is m, Young’s

Modulus is dynes/cm2, and the

calculated displacement found is m.

Mechanics of Materials, by Ferdinand P. Beer





DIAPHRGM WITH CAPTURED VOLUME

F1 = force on diaphragm = external pressure times area of diaphragm F2 = force due to captured volume of air under the diaphragm F3 = force to mechanically deform the diaphragm

h d

rs

rd

PV = nRT

F1= F2 + F3 F1 = P x Ad F2 = nRT Ad / (Vd + Vs)

where Vd = Ad (d-y) and Vs = G1 Pi (rs2 – rd2)(d) where

G1 is the % of spacer that is not oxide

F3 = (16 E (1/ )2 h3 y)/(3 rd4[(1/ )2-1])

F1

F2 + F3 Ad y

h

d





DIAPHRAGM WITH CAPTURED VOLUME

y (µm) P (N/m2)

0 0

0.1 5103.44

0.2 10748.09

0.3 17025.03

0.4 24047

0.5 31955.23

0.6 40929.06

0.7 51199.69

0.8 63070.47

0.9 76947.31

1 93386.12

1.1 113169.1

1.2 137433.1

1.3 167895.7

1.4 207281.3

1.5 260184.8

1.6 335009.7

1.7 448945.4

1.8 643513.4

1.9 1051199

2 2446196

P (N/m2) vs displacement y (µm)

0

500000

1000000

1500000

2000000

2500000

3000000

0 0.5 1 1.5 2 2.5





CONDENSER MICROPHONE

1 µm Aluminum

2.0 µm Gap

ALUMINUM DIAPHRAGM





ALUMINUM DIAPHRAGM PRESSURE SENSOR

Diaphragm

Drain

Source Gate

Bottom

Plate

Si Wafer

Insulator Contact

Air Gap (~1 atm)

Kerstin Babbitt - University of Rochester

Stephanie Bennett - Clarkson University

Sheila Kahwati - Syracuse University

An Pham - Rochester Institute of Technology





SIGNAL CONDITIONING FOR CAPACITOR SENSORS

Delta Capacitance to AC Voltage

Static Capacitance to DC Voltage

Capacitance to Current

Ring Oscillator Capacitance to Frequency

RC Oscillator Capacitance to Frequency

Frequency to Digital

Capacitance to Analog Voltage to Digital

Other

Wireless





CHANGING CAPACITANCE TO AC VOLTAGE

+

9V

-9 V

TL081

i

V

R

C Vo

i

i = V Cm 2 p f cos (2pft) Co = Average value of C Cm = amplitude of C change C = Co +Cm sin (2pft) V is constant across C

Vo = - i R

i = d (CV)/dt

Vo = - 2pf V R Cm cos (2pft)

amplitude of Vo





EXAMPLE CALCULATIONS

Vo = - i R = - 2pf V R Cm cos (2pft)

Let f = 5 Khz, V=5, Cm= 100fF, R=1MEG

Vo = - 0.0157 cos (2pft) volts

(15.7 mV amplitude sinusoid)

0

5

10

15

20

25

30

0 2000 4000 6000 8000 10000

Vo

ut

(mV

)

Frequency (Hz)

Predicted Frequency Response

Amplitude of Vo = - 2p f V R Cm





SPICE CHANGING CAPACITANCE TO AC VOLTAGE





SENSOR CAPACITANCE TO FREQUENCY

-

+ Vo

Csensor

+V

R2 R1

R

VT

+V R3

Vo

t

t1

+V

0

Let R1 = 100K, R2=R3=100K

and +V = 3.3

Then VTH = 2.2 when Vo = 3.3

VTL = 1.1 when Vo = 0

VTH

VTL





CAPACITANCE TO FREQUENCY

C1 is the sensor

C2 represents Scope Probe





RING OSCILLATOR CAPACITANCE TO FREQUENCY

T = 2 N td

td = gate delay

N = number of stages

T = period of oscillation

Seven stage ring oscillator,

N=7 with two output buffers

T

Vout

Buffer Vout

Csensor

Note: Csensor looks bigger

due to Miller effect





RING OSCILLATOR CAPACITANCE TO FREQUENCY

3 stage ring oscillator – no buffers





CAPACITANCE TO VOLTAGE

+

i

R

C

i

Vo = - i R

Vo = 2pf Vm R C cos (2pft)

amplitude of Vo

i = C d(Vm sin (2pft))/dt

Vin = Vm sin (2pft)

i = dQ/dt = d (CV)/dt = C dV/dt

Example: R = 2MEG, C = 2pF, Vm = .5 V, f = 100 Khz

Vo = 0.4 volts peak-to-peak





CAPACITANCE TO VOLTAGE – BLOCK DIAGRAM

+

i

R

C

i

Vin

Vout -

+

AC Vout p-p depends

on Value of C Buffer Peak Detector





CAPACITANCE TO VOLTAGE

Buffer

Peak Detector





STATIC CAPACITANCE TO VOLTAGE





CAPACITANCE TO DC VOLTAGE

Vin

Cf

Cx

-

+ Vo

1

1 2

Vo = - Vin Cx/Cf

Q = CV





SPICE FOR CAPACITANCE TO DC VOLTAGE

Q = CV





WIRELESS REMOTE SENSING OF L OR C

C

L

R

I

f

I

Capacitive Pressure Sensor

~ 13MHz

Exte

rnal

Ele

ctro

nic

s

Antenna





BLOCK DIAGRAM FOR REMOTE SENSING ELECTRONICS

Digital Controlled Oscillator

Filter

RS 232 To

Display

Microchip PIC18F452

µP

R

I

Antenna

Gnd

-

+

Power Amp

A to D

Amp

Memory

C

L

Remote Resonant

LC





SIMILAR IDEA FOR WIRELESS TEMPERATURE

Network Analyzer

I vs. Frequency





TEMPERATURE SENSING WITH COIL AND DIODES





PICKUP COIL CURRENT

Due to nearby LC

No Resonant Circuit Present Resonant LC Circuit Present





ZOOM IN ON RESONANCE DUE TO LC

Frequency at Which this Dip Occurs

Changes with Temperature

Resonant Frequency

wo = 1/(LC)0.5

fo = wo/2p





COMMERCIAL CHIPS

MS3110 Universal Capacitive Read-out IC (left)

and its block diagram (right). capable of sensing

capacitance changes down to 4.0 aF/rtHz, typical





SIGNAL CONDITIONING BUILDING BLOCKS

Operational Amplifier

Analog Switches

Non-Overlapping Clock





CMOS OPERATIONAL AMPLIFIER





ANALOG SWITCHES

For current flowing to the right (ie V1>V2) the PMOS transistor will be on if V1 is greater than the

threshold voltage, the NMOS transistor will be on if V2 is <4 volts. If we are charging up a capacitor

load at node 2 to 5 volts, initially current will flow through NMOS and PMOS but once V2 gets

above 4 volts the NMOS will be off. If we are trying to charge up V2 to V1 = +1 volt the PMOS will

never be on. A complementary situation occurs for current flow to the left. Single transistor switches

can be used if we are sure the Vgs will be more than the threshold voltage for the specific circuit

application. (or use larger voltages on the gates)

NMOS

Vt=+1

PMOS

Vt= -1

D S

I

zero

V1 V2

+5

S D





ANALOG SWITCH

CMOS Inverter

Analog Switch





TWO PHASE NON OVERLAPPING CLOCK

Switch capacitor circuits require some switches to be open before

others are closed. Thus a need for non-overlapping clocks. The

time delays t1, t2, t3 enable the non-overlapping feature.

1

2

CLK 1

2 t3

t2

t1

S

R

Q





TWO PHASE NON OVERLAPPING CLK WITH BUFFERS

CLOCK 1

2 t3

t2

t1

S

R

Buffers





TWO PHASE NON OVERLAPPING CLK WITH BUFFERS





REFERENCES

1. Mechanics of Materials, by Ferdinand P. Beer, E. Russell Johnston, Jr., McGraw-Hill Book Co.1981, ISBN 0-07-004284-5

2. Electromagnetics, by John D Kraus, Keith R. Carver, McGraw-Hill Book Co.1981, ISBN 0-07-035396-4

3. Fundamentals of Microfabrication, M. Madou, CRC Press, New York, 1997

4. Switched Capacitor Circuits, Phillip E. Allen and Edgar Sanchez-Sinencio, Van Nostrand Reinhold Publishers, 1984.

5. “Optimization and fabrication of planar interdigitated impedance sensors for highly resistive non-aqueous industrial fluids”, Lvovich, liu and Smiechowski, Sensors and Actuators B:Chemical, Volume 119, Issue 2, 7 Dec. 2006, pgs 490-496.





HOMEWORK – MEMS CAPACITOR SENSORS

1. Calculate the capacitance for a round plate of 100µm diameter

with an air gap space of 2.0 µm.

2. If the capacitor in 1 above has the air gap replaced by water

what will the capacitance be?

3. Design an apparatus that can be used to illustrate the attractive

force between two parallel plates when a voltage is applied.

4. Design an electronic circuit that can measure the capacitance of

two metal plates (the size of a quarter) separated by a thin foam

insulator for various applied forces.

rochester institute of tehnology …people.rit.edu/lffeee/capacitor_sensors.pdf · eo = 8.85e-14...

Documents