microelectromechanical systems (mems) electrical fundamentals · mems electrical fundamentals page...

© March 15, 2008 Dr. Lynn Fuller, Motorola Professor

Rochester Institute of TechnologyMicroelectronic Engineering

MEMS Electrical Fundamentals

ROCHESTER INSTITUTE OF TEHNOLOGYMICROELECTRONIC ENGINEERING

3-15-2008 mem_elec.ppt

Microelectromechanical Systems (MEMs)Electrical Fundamentals

Dr. Lynn FullerWebpage: http://people.rit.edu/lffeee

Microelectronic EngineeringRochester Institute of Technology

82 Lomb Memorial DriveRochester, NY 14623-5604

Tel (585) 475-2035Fax (585) 475-5041

Email: [email protected] webpage: http://www.microe.rit.edu




OUTLINE

ResistorsHeatersTemperature SensorsCapacitorsElectrostatic ForceGap Comb DrivesMagnetic ActuatorsPiezoelectric DevicesDiode Temperature SensorsPhotodiodeSeeback EffectPeltier EffectTunnel Sensors




INTRODUCTION

Resistors are used as heaters, temperature sensors, piezoreistorsensors and photoconductors. Heaters are used in many MEMS applications including ink jet print heads, actuators, bio-mems, chemical detectors and gas flow sensors. Diodes, Capacitors, electrostatic comb drives, magnetic devices, and many other devices are used in many applications. This module will discuss these devices as they apply to MEMS.




SURFACE MICROMACHINED GAS FLOW SENSOR

Upstream Polysilicon Resistor

Polysilicon heater

Downstream Polysilicon Resistor

GAS





Vee Chee Hwang, 2004





L of heater & resistor = 1mm

W (heater) = 50um

W (resistors) = 20um

Gap = 10um

V applied = 27V to 30.5V

Temp ~600 °C at 26 volts

Lifetime > 10 min at 27 volts (possibly longer, did not test)

Vee Chee Hwang, 2004




BULK MICROMACHINED GAS FLOW SENSOR

Measured Resistance, V/I=1.2KohmsTheoretical Resistance, L*Rhos/W= 400µm*60/20µm = 1.2Kohms

Raunak Mann, 2004




THERMIONIC GAS DETECTOR

§ Polysilicon Micro-filament heater

ResistiveHeater

Ioniccurrent

BiasingPlate

CollectionPlate

AmplifiedCurrent

Make hotThermionic emission occurs causing ionizationForce ions to a collection plateMeasure resulting current or voltage

+

-

Robert Manley, 2004





Si

Polysilicon

Silicon Nitride

Un-etchedSacox

Robert Manley, 2004





Cold Hot

Robert Manley, 2004




OHMS LAW

I

V

Resistor a two terminal device that exhibits a linear I-V characteristic that goes through the origin. The inverse slope is the value of the resistance.

R = V/I = 1/slope

I

V

-

+




THE SEMICONDUCTOR RESISTOR

Resistance = R = ρ L/Area = ρs L/w ohms

Resistivity = Rho = ρ = 1/( qµnn + qµpp) ohm-cm

Sheet Resistance = Rhos = ρs = 1/ ( q µ(N) N(x) dx) ~ 1/( qµ Dose) ohms/square

L Area

R

wt

ρs = ρ / t

Note: sheet resistance is convenient to use when the resistors are made of thin sheet of material, like in integrated circuits.

R = Rho L / Area= Rhos L/W

Rho is the bulk resistivity of the material (ohm-cm)Rhos is the sheet resistance (ohm/sq) = Rho / t




MOBILITY

Total Impurity Concentration (cm-3)

0200400600800

1000120014001600

1013

1014

1015

1016

1017

1018

1019

1020

ArsenicBoronPhosphorus

Mob

ility

(cm

2 / V

sec

)

electrons

holes

Parameter Arsenic Phosphorous Boronµmin 52.2 68.5 44.9µmax 1417 1414 470.5Nref 9.68X10^16 9.20X10^16 2.23X10^17α 0.680 0.711 0.719

µ(N) = µ mi+ (µmax-µmin)

1 + (N/Nref)α

Electron and hole mobilitiesin silicon at 300 K as functions of the total dopantconcentration (N). The values plotted are the results of the curve fitting measurements from several sources. The mobility curves can be generated using the equation below with the parameters shown:

From Muller and Kamins, 3rd Ed., pg 33




TEMPERATURE EFFECTS ON MOBILITY

Derived empirically for silicon for T in K between 250 and 500 °K and for N (total dopant concentration) up to 1 E20 cm-3

µn (T,N) =

µp (T,N) =

88 Tn-0.57

54.3 Tn-0.57

Where Tn = T/300From Muller and Kamins, 3rd Ed., pg 33

1250 Tn-2.33

407 Tn-2.33

1 + [ N / (1.26E17 Tn 2.4)] ^0.88 Tn -0.146

1 + [ N / (2.35E17 Tn 2.4)]^ 0.88 Tn -0.146

+

+




EXCELL WORKSHEET TO CALCULATE MOBILITY

MICROELECTRONIC ENGINEERING 3/13/2005

CALCULATION OF MOBILITY Dr. Lynn Fuller

To use this spreadsheed change the values in the white boxes. The rest of the sheet isprotected and should not be changed unless you are sure of the consequences. Thecalculated results are shown in the purple boxes.

CONSTANTS VARIABLES CHOICESTn = T/300 = 1.22 1=yes, 0=no

Temp= 365 °K n-type 1N total 1.00E+18 cm-3 p-type 0

<100>

Kamins, Muller and Chan; 3rd Ed., 2003, pg 33mobility= 163 cm2/(V-sec)




EXCELL WORKSHEET TO CALCULATE RESISTANCE

MICROELECTRONIC ENGINEERING 7/23/2007Rochester Institute of Technology Dr. Lynn Fuller

CALCULATION OF RESISTANCE FROM LENGTH, WIDTH, THICKNESS AND IMPLANT DOSE

To use this spreadsheed change the values in the white boxes. The rest of the sheet isprotected and should not be changed unless you are sure of the consequences. Thecalculated results are shown in the purple boxes.

Calculation of Mobility

CONSTANTS VARIABLES CHOICESTn = T/300 =1.00 1=yes, 0=no

Temp= 300 °K n-type 0N total 2.00E+16 cm-3 p-type 1

<100>

Kamins, Muller and Chan; 3rd Ed., 2003, pg 33mobility = µ = 433 cm2/(V-sec)

Calculation of ResistanceLength, L = 6000 µm

R = Rho L / W / t Width, W = 20 µmR = Rhos L / W Thickness, t = 0.5 µm

Implant Dose = 1.00E+12 ions/cm2Rho = bulk resistivityRhos = sheet resistance =1/(q µ Dose) N total = Average Doping = 2.00E+16 atoms/cm3

q = 1.6e-19 coul/ion Mobility, µ = 433 cm2/v-secResistance = 1.73E+09 ohms




TEMPERATURE COEFFICIENT OF RESISTANCE

∆R/∆T for semiconductor resistors

R = Rhos L/W = Rho/t L/W

assume W, L, t do not change with T

Rho = 1/(qµn + qµp) where µ is the mobility which is a function of temperature, n and p are the carrier concentrations which can be a function of temperature (in lightly doped semiconductors)

as T increases, µ decreases, n or p may increase and the result is that R usually increases unless the decrease in µ is cancelled by the increase in n or p




RESISTANCE AS A FUNCTION OF TEMPERATURE

R(T,N) = 1

qµn(T,N) n + q µp (T,N) pL

Wt

L, W, t, are physical length width and thickness and do not change with TConstant q = 1.6E-19 coulµn (T,N) = see expression on previous pageµp (T,N) = see expression on previous pagen = Nd and p = 0 if doped n-typen=0 and p = Na if doped p-typen = p = ni(T) if undoped, see exact calculation on next page

ni(T) = A T 3/2 e - (Eg)/2KT = 1.45E10 @ 300 °K




EXACT CALCULATION OF n AND p

CONSTANTS VARIABLESK 1.38E-23 J/Kq 1.60E-19 CoulEgo 1.16 eVa 7.02E-04B 1.11E+03h 6.63E-34 Jsec Nd = 3.00E+16 cm-3 Donor Concentrationεo 8.85E-14 F/cm Ed= 0.049 eV below Ecεr 11.7 Na = 8.00E+15 cm-3 Acceptor Concentrationni 1.45E+10 cm-3 Ea= 0.045 eV above EvNc/T^3/2 5.43E+15Nv/T^3/2 2.02E+15 Temp= 300 °K

Donor and Acceptor Levels (eV above or below Ev or Ec)Boron 0.044

Phosphorous 0.045Arsenic 0.049

CALCULATIONS: (this program makes a guess at the value of the fermi level and trys to minimizethe charge balance)

KT/q 0.026 VoltsEg=Ego-(aT^2/(T+B)) 1.115 eVNc 2.82E+19 cm-3Nv 1.34E+01 cm-3Fermi Level, Ef 0.9295 eV above Evfree electrons, n = Nc exp(-q(Ec-Ef)KT) 2.17E+16 cm-3Ionized donors, Nd+ = Nd*(1+2*exp(q(Ef-Ed)/KT))^(-1) 2.97E+16 cm-3holes, p = Nv exp(-q(Ef-Ev)KT) 3.43E-15 cm-3Ionized acceptors, Na- = Na*(1+2*exp(q(Ea-Ef)/KT))^(-1) 8.00E+15 cm-3Charge Balance = p + Nd+ - n - Na- 3.22E+12 cm-3

Click on Button to do Calculation

ButtonButton




DIFFUSED RESISTOR

Aluminum contacts

The n-type wafer is always biased positive with respect to the p-type diffused region. This ensures that the pn junction that is formed is in reverse bias, and there is no current leaking to the substrate. Current will flow through the diffused resistor from one contact to the other. The I-V characteristic follows Ohm’s Law: I = V/R

n-wafer Diffused p-type region

Silicon dioxide

Sheet Resistance = ρs ~ 1/( qµ Dose) ohms/square




RESISTOR NETWORK

500 Ω 400 250 Ω

Desired resistor network

Layout if Rhos = 100 ohms/square

L

W




R AND C IN AN INTEGRATED CIRCUIT

Estimate the sheet resistance of the 4000 ohm resistor shown.

RC

741 OpAmp




DIFFUSION FROM A CONSTANT SOURCE

N(x,t) = No erfc (x/2 Dptp)

SolidSolubilityLimit, No

xinto wafer

Wafer Background Concentration, NBC

N(x,t)

Xj

p-type

n-type

erfc function




DIFFUSION FROM A LIMITED SOURCE

for erfc predepositQ’A (tp) = QA(tp)/Area = 2 No (Dptp) / π = Dose

N(x,t) = Q’A(tp) Exp (- x2/4Dt)π Dt

Where D is the diffusion constant at the drive in temperature and t is the drive in diffusion time, Dp is the diffusion constant at the predeposit temperature and tp is the predeposit time

Gaussian function




DIFFUSION CONSTANTS AND SOLID SOLUBILITY

DIFFUSION CONSTANTSBORON PHOSPHOROUS BORON PHOSPHOROUS

TEMP PRE or DRIVE-IN PRE DRIVE-IN SOLID SOLIDSOLUBILITY SOLUBILITY

NOB NOP900 °C 1.07E-15 cm2/s 2.09e-14 cm2/s 7.49E-16 cm2/s 4.75E20 cm-3 6.75E20 cm-3950 4.32E-15 6.11E-14 3.29E-15 4.65E20 7.97E201000 1.57E-14 1.65E-13 1.28E-14 4.825E20 9.200E201050 5.15E-14 4.11E-13 4.52E-14 5.000E20 1.043E211100 1.55E-13 9.61E-13 1.46E-13 5.175E20 1.165E211150 4.34E-13 2.12E-12 4.31E-13 5.350E20 1.288E211200 1.13E-12 4.42E-12 1.19E-12 5.525E20 1.410E211250 2.76E-12 8.78E-12 3.65E-12 5.700E20 1.533E21

DpDp or D D




ION IMPLANTED RESISTOR

Like the diffused resistor but more accurate control over the sheet resistance. The dose is a machine parameter that is set by the user.

Sheet Resistance = ρs ~ 1/( qµ Dose) ohms/square

Also the dose can be lower than in a diffused resistor resulting in higher sheet resistance than possible with the diffused resistor.

Aluminum contacts

n-wafer Ion Implanted p-type region

Silicon dioxide




THIN FILM RESISTOR

Aluminum contacts

n-wafer Thin Layer of Poly SiliconOr Metal

Silicon dioxide

For polysilicon thin films the Dose = film thickness ,t, x Solid Solubility No if doped by diffusion, or Dose = ion implanter dose setting if implanted

The Sheet Resistance Rhos = ~ 1/( qµ Dose) ohms/square

For metal the Sheet Resistance is ~ the given (table value) of bulk resistivity, Rho, divided by the film thickness ,t.

Rhos = Rho / t ohms/square




HEATERS

P= IV = I2R watts

Final steady state temperaturedepends on power density inwatts/cm2

and

the thermal resistancefrom heater to ambient

I

V

-

+

R = ρs L/W




THERMAL CONDUCTIVITY

Tempambient

ThermalResistance, Rth

to ambient

Temp aboveambient

Powerinput

Rth = 1/C L/Area

whereC=thermal conductivityL= thickness of layer between heater and ambientArea = cross sectional areaof the path to ambient




THERMAL PROPERTIES OF SOME MATERIALSMP Coefficient Thermal Specific°C of Thermal Conductivity Heat

Expansionppm/°C w/cmK cal/gm°C

Diamond 1.0 20Single Crystal Silicon 1412 2.33 1.5Poly Silicon 1412 2.33 1.5Silicon Dioxide 1700 0.55 0.014Silicon Nitride 1900 0.8 0.185Aluminum 660 22 2.36 0.215Nickel 1453 13.5 0.90 0.107Chrome 1890 5.1 0.90 0.03Copper 1357 16.1 3.98 0.092Gold 1062 14.2 3.19Tungsten 3370 4.5 1.78Titanium 1660 8.9 0.17Tantalum 2996 6.5 0.54Air 0.00026 0.24Water 0 0.0061 1.00

1 watt = 0.239 cal/sec




HEATER EXAMPLE

Example: Poly heater 100x100µm has sheet resistance of 25 ohms/sq and 9 volts is applied. What temperature will it reach if built on 1 µm thick oxide?

Power = V2/R = 81/25 = 3.24 watt

Rthermal = 1/C L/Area = (1/0.014 watt/cm °C)(1e-4cm/(100e-4cm x100e-4cm))= 71.4 °C/watt

Temperature = Tambient + (3.24) (71.4) = Tambient + 231 °C




TEMPERATURE SENSOR EXAMPLE

Example: A diffused heater is used to heat a sample. The temperature is measured with a poly silicon resistor. For the dimensions given what will the resistance be at 90°C and 65°C

R(T,N) = 1

q µn (T,N) nL

Wt

10µm

50µm

t=1µmNd=n=1E18cm-3

q µn (T=365,N=1e18) n = 1.6e-19 (163) 1e18 = 26.1

q µn (T=390,N=1e18) n = 1.6e-19 (141) 1e18 = 22.6

50,000

R=2212 ohms

R=1916 ohms




THERMAL FLOW SENSORS

SacOx

Silicon Substrate

Si3N4

Polysilicon Aluminum

Spring 2003EMCR 890 Class Project

Dr. Lynn Fuller

Flow

Downstream Temp Sensor

Heater

Upstream Temp Sensor




GAS FLOW SENSORS

Constant heat (power in watts) input and two temperature measurement devices, one upstream, one downstream. At zero flow both sensors will be at the same temperature. Flow will cause the upstream sensor to be at a lower temperature than the down stream sensor.

Upstream Polysilicon Resistor

Polysilicon heater

Downstream Polysilicon Resistor

GAS




FLOW SENSOR ELECTRONICS

Gnd

+6 Volts

-6 Volts

Vout+-

Constant Power Circuit

Vout near Zero so thatit can be amplified

R1

R2

AD534

10 OHM

AD534_b.pdf




SINGLE WIRE ANEMOMETER

A single heater/sensor element is placed in the flow. The amount of power supplied to keep the temperature constant is proportional to flow. At zero flow a given amount of power Po will heat the resistor to temperature To. With non zero flow more power Pf is needed to keep the resistor at To.

Flow

Heater/Sensor




CONSTANT TEMPERATURE CIRCUIT

AnalogDividerUsingAD534

Gnd

+9 Volts

I

Setpoint

+-

Heater

10 Ω

-+

+-

R=V/I

I

V

1000

AD534_b.pdf




PHYSICAL FUNDAMENTALS - PIEZORESISTANCE

Piezoresistance is defined as the change in electrical resistance of a solid when subjected to stress. The piezorestivity coefficient is Π and a typical value may be 1e−10 cm2/dyne.

The fractional change in resistance ∆R/R is given by:

∆R/R = Π σ

where σ is the stress. Other references use the gage factor GF to describe the piezoresistive effect where

∆R/R = (∆L/L) / GF




EXAMPLE: PIEZORESISTANCE

Example: Find the maximum stress in a simple polysilicon cantilever with the following parameters. Ymax = 1 µm, b=4 µm, h=2µm, L=100 µm

σx=0 = 5.6e7 newton/m2 = 5.6e8 dyne/cm2

From example in mem_mech.ppt

Continue Example: What is the change in resistance given Π = 1e-10 cm2/dyne

∆ R/R = Π σ = (1e-10 cm2/dyne)/(5.6e8 dyne/cm2)= 5.8%




CALCULATION OF EXPECTED OUTPUT VOLTAGE FROM A PIEZORESISTIVE PRESSURE SENSOR

R1R3

R2R4

Gnd

+5 Volts Vo2

Vo1

The equation for stress at the center edge of a square diaphragm (S.K. Clark and K.Wise, 1979)

Stress = 0.3 P(L/H)2 where P is pressure, L is length of diaphragm edge, H is diaphragm thickness

For a 3000µm opening on the back of the wafer the diaphragm edge length L is 3000 – 2 (500/Tan 54°) = 2246 µm




CALCULATION OF EXPECTED OUTPUT VOLTAGE

Stress = 0.3 P (L/H)2

If we apply vacuum to the back of the wafer that is equivalent to and applied pressure of 14.7 psi or 103 N/m2

P = 103 N/m2

L= 2246 µmH= 25 µm

Stress = 2.49E8 N/m2

Hooke’s Law: Stress = E Strain where E is Young’s Modulusσ = E ε

Young’s Modulus ofr silicon is 1.9E11 N/m2

Thus the strain = 1.31E-3 or .131%





The sheet resistance (Rhos) from 4 point probe is 61 ohms/sqThe resistance is R = Rhos L/WFor a resistor R3 of L=350 µm and W=50 µm we find:

R3 = 61 (350/50) = 427.0 ohms

R3 and R2 decrease as W increases due to the strainassume L is does not change, W’ becomes 50+50x0.131%W’ = 50.0655 µmR3’ = Rhos L/W’ = 61 (350/50.0655) = 426.4 ohms

R1 and R4 increase as L increases due to the strainassume W does not change, L’ becomes 350 + 350x0.131%R1’ = Rhos L’/W = 61 (350.459/50) = 427.6 ohms





Gnd

5 Volts

R1=427 R3=427

R2=427 R4=427

Vo2=2.5vVo1=2.5v

Gnd

5 Volts

R1=427.6 R3=426.4

R2=426.4 R4=427.6

Vo2=2.5035vVo1=2.4965v

No stressVo2-Vo1 = 0

With stressVo2-Vo1 = 0.007v

=7 mV




CAPACITORS

I = C (dV/dt)

I

C V+

-

C = εo εr Area / d

Capacitors are used to store charge in memory cells, build filters, oscillators, integrators and many other applications in electronics.

Symbol:

d

Area

εo = 8.85E-14 F/cmεr air = 1εr SiO2 = 3.9




CAPACITOR SENSOR

C = εo εr Area / d

d

Move one plate relative to the other results in a change

in capacitance

Pressure

Aluminum diaphragm pressure sensor




CAPACITIVE ELECTROSTATIC FORCE

F

++++++++

- - - - - - -

The energy stored in a capacitorcan be equated to the force timesdistance between the plates

W = Fd or F = W/d

2d2εo ε r AV2

F =

d

area A

Energy stored in a parallel plate capacitor Wwith area A and space between plates of d

W = QV = CV2

since Q = CV

dε oε rAC =

ε o = permitivitty of free space = 8.85e-12 Farads/mε r = relative permitivitty (for air ε r = 1)




ELECTROSTATIC FORCE EXAMPLE

Example: 100 µm by 100 µm parallel platesspace = 1 µm, voltage = 10 V

Find the force of attraction between the two platesε oε r AV2

F =2d2

(8.85e-12)(1)(100e-6)(100e-6)(10)2

F =2(1e-6)2

F = 4.42e-6 newtons




MEMS CANTILEVER WITH ELECTROSTATIC ACTUATOR

Poly 2

Poly 1

Metal

y

V0 10 20 30

1

2

3um

3µm gap

As the voltage is increased the electrostatic force starts to pull down the cantilever. The spring constant opposes the force but at the same time the gap is increased and the force increases. The electrostatic force increases with 1/d2 so eventually a point is reached where it is larger than the spring and the cantilever snaps down all the way. The voltage has to be reduced almost to zero to release the cantilever.




MEMS CANTILEVER WITH ELECTROSTATIC ACTUATOR

Example: Plot the displacement versus current assuming poly sheet resistance of 200 ohms/sq, a piezoresistance coefficient of 1e-10 cm/dyne, d=1.5 µm, h=2µm and other appropriate assumptions.

1st layer poly

2nd layer polyI

100 x100 µmpads

400 µm

40 µm

100 x100 µmpads




SOLUTION FOR y VERSUS I

y

I

I

V

V = I(R+πσ)

R=2000 Ω

R=2000 Ω

ε oε r AV2

F =2d2

Ymax 3 E bh3

F =12L3

12 F Lσx=0 =

2b h2

Ymax 3 E bh3

F =12L3




GAP COMB DRIVE MICROACTUATORS

Electrostatic movement parallel to wafer surface

AnchorAnchor

Anchor

From Jay Zhao




CALCULATION OF DISPLACEMENT VS VOLTAGE

t

L

d

movement

C = εr εo t L/d

F = εr εo t V2 / 2 d




COMBINED SPRING ELECTROSTATIC DRIVE AND CAPACITIVE OR PIEZORESISTIVE READ OUT

R

Anchor

C1

C2

C1




MAGNETIC TORSIONAL MIRROR

W

Contact

L

Supporting arm

z

Bcoil

Bm

Figure 2: Cross sectional view labels with variables.

( ) ( )( )

Nz

BLWRz

IR

zmm

Fo

m

o

coilm

µ

ππ

µ

µ

22/322

20 2

4

+==

Metal contact

Topside Hole

Diaphragm

Figure 1: Top down CAD design of single axis Torsional Mirror

Paper by Eric Harvey




MAGNETIC DEVICES

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfie.html

Magnetic flux density B in the center of a coil

( )

+= 2/322

20 2

4 Rz

IRBloopπ

πµ

loopcoil BNB *= where

The magnetic pole strength is m (webers) = B A where A is the pole area

Magnetic flux density of a permanent magnet B is given by the manufacturer in units of weber/sq.meter or Tesla. (some of the magnets we use in MEMS are 2mm in diameter and have B=0.5 Tesla)

The force between two poles is Force = µo z2

m1 m2

µo = 4πx10-7 w2/Nm2




MAGNETIC DEVICES

Force on a straight conductor in a uniform magnetic field.

F = I L X B

( ) ( )( )

Nz

BLWRz

IR

zmm

Fo

m

o

coilm

µ

ππ

µ

µ

22/322

20 2

4

+==

Force on a coil with current I in a uniform magnetic field




MAGNETIC SOLENOID

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

at center of a solenoidMagnetic Flux Density (Tesla)

B = µoµr NI/L

where µo=4π e-7 H/mµr=600 for Ni

Energy = Wm = B2/2µoµror

Wm = (µoµr/2) (NI/L)2

L




MAGNETIC SOLENOID

The energy stored in a small slice (dx) in the core of area (A) is

Wm/dx = (µoµr/2) (NI/L)2A

the energy difference to remove the core by dx is dWm

dWm = Wm/dx - Wm’/dx

the change in energy equals force times change in distance

Fdx = dWm

F=dWm/dx = (µo/2)(µr-1) (NI/L)2A




MAGNETIC SOLENOID

Example: Given a nickel core solenoid of length 200 µm with 25 turns and 0.1 amp of current. The cross-sectional area is 40 µm by 2µm. Calculate the force needed to move the core.

F = (µo/2)(µr-1) (NI/L)2A

F = (4π e-7/2)(600-1)[(25)(0.1)/200e-6)]2(40e-6)(2e-6)

= 4.7e-6 Newton




MAGNETIC - SOLENOID

2nd layer poly

I

100 x100 µmpads

40 µm

2nd layer poly

I

100 x100 µmpads

40 µm




EXAMPLE - SOLENOID




MAGNETIC FIELD SENSORS

Lateral MagnetotransistorMOS Magnetotransistor

F=q(v x B)

See UGIM 1999Add Picture




PIEZOELECTRIC DRIVES

A piezoelectric material will exhibit a change in length in response to an applied voltage. The reverse is also possible where an applied force causes the generation of a voltage. Single crystal quartz has been used for piezeoelectric devices such as gas grill igniters and piezoelectric linear motors. Thin films of various materials (organic and inorganic) exhibit piezoelectric properties. ZnO films 0.2 µm thick are sputtered and annealed 25 min, 950C giving piezoelectric properties. Many piezoelectric materials also exhibit pyroelectricproperties (voltage - heat).

more




SEEBECK EFFECT

When two dissimilar conductors are connected together a voltage may be generated if the junction is at a temperature different from the temperature at the other end of the conductors (cold junction) This is the principal behind the thermocouple and is called the Seebeck effect.

∆V

Material 2Material 1

Hot

Cold

Nadim Maluf, Kirt Williams, An Introduction to Microelectromechanical Systems Engineering, 2nd Ed. 2004

∆V = α1(Tcold-Thot) + α2 (Thot-Tcold)=(α1-α2)(Thot-Tcold)

Where α1 and α2 are the Seebeck coefficients for materials 1 and 2




PELTIER EFFECT

Heat pump device that works on the gain in electron energy for materials with low work function and the loss in energy for materials with higher work function. Electrons are at higher energy (lower work function) in n-type silicon.

I

heat

heat

n nn pp

Cu

electronsCu

Cu

CuCu Cu

Cu




PELTIER HEAT PUMP

Ferrotec America Corp1050 Perimeter Rd, #202Manchester, NH 03103(603) 626-0700




TUNNELING SENSOR

I = IO e (-β ΦZ)

where Io = scaling factorβ = conversion factor (~10.25 eV-0.5/nm)φ = tunnel barrier height in eV (~0.5eV)z = tip to surface separation (100 Å)

Show Picture




INTEGRATED DIODES

p-wafer

n+ p+n-well

p+ means heavily doped p-typen+ means heavily doped n-typen-well is an n-region at slightly higher

doping than the p-wafer

Note: there are actually two pn junctions, the well-wafer pnjunction should always be reverse biased




DIODE TEMPERATURE DEPENDENCE

Id = IS [EXP (q VD/KT) -1]Neglect the –1 in forward bias, Solve for VD

VD = (KT/q) ln (Id/IS) = (KT/q) (ln(Id) – ln(Is))

Take dVD/dT: note Id is not a function of T but Is is

dVD/dT = (KT/q) (dln(Id)/dT – dln(Is)/dT) + K/q (ln(Id) – ln(Is))zero VD/T from eq 1

RewrittendVD/dT = VD/T - (KT/q) ((1/Is) dIs/dT )

Now evaluate the second term, recall

Is = qA (Dp/(LpNd) +Dn/(LnNa))ni2

eq 1

eq 2

Note: Dn and Dp are proportional to 1/T




DIODE TEMPERATURE DEPENDENCE

and ni2(T) = A T 3 e - qEg/KT

This gives the temperature dependence of IsIs = C T 2 e - qEg/KT

Now take the natural logln Is = ln (C T 2 e - qEg/KT)

Take derivative with respect to T(1/Is) d (Is)/dT = d [ln (C T 2 e -qEg/KT)]/dT = (1/Is) d (CT2e-qEg/KT)dT

eq 3

= (1/Is) [CT2 e-qEg/KT(qEg/KT2) + (Ce-qEg/KT)2T]

= (1/Is) [Is(qEg/KT2) + (2Is/T)]Back to eq 2

dVD/dT = VD/T - (KT/q) [(qEg/KT2) + (2/T)])

dVD/dT = VD/T - Eg/T - 2K/q)




EXAMPLE: DIODE TEMPERATURE DEPENDENCE

dVD/dT = VD/T - Eg/T - 2K/q

Silicon with Eg ~ 1.2 eV, VD = 0.6 volts, T=300 °K

dVD/dT = .6/300 – 1.2/300) - (2(1.38E-23)/1.6E-19= -2.2 mV/°

I

V

T1T2

T1<T2




PHYSICAL FUNDAMENTALS - PHOTOELECTRIC

B -

P+ Ionized Immobile Phosphrous donor atomIonized Immobile Boron acceptor atom

Phosphrous donor atom and electronP+-

B-+ Boron acceptor atom and hole

n-type

p-typeB - P+

B -B -

B -B -

P+ P+P+P+

P+

P+

P+-

B-+

εB -B -

P+-

P+-

P+-

B-+

-+

-+

I

electronand hole

pair

-+

-+

space charge layer




PHOTODIODE

I

V

n

p

IV

I+V-

More Light No Light

Most Light




CHARGE COLLECTION IN MOS STRUCTURES

ε

-+

-+

electronand hole

pair

-+

-+

depletion region

-+

-- -

p-type

+ V

B -

B -

B -

B -

λ1 λ3 λ4λ2

thin poly gate

E = hν = hc / λ

h = 6.625 e-34 j/s = (6.625 e-34/1.6e-19) eV/s

E = 1.55 eV (red)E = 2.50 eV (green)E = 4.14 eV (blue)




LIGHT EMITTING DIODES (LEDs)

P-side N-side

SpacechargeLayer

LightLight

Hole concentration vs distnace

xx

Electron concentration vs distance

In the forward biased diode current flows and as holes recombine on the n-side or electrons recombine on the p-side, energy is given off as light, with wavelength appropriate for the energy gap for that material. λ = h c / E

h = Plank’s constantc = speed of light




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. “Etch Rates for Micromachining Processing”, Journal of Microelectromechanical Systems, Vol.5, No.4, December 1996.

4. “Design, Fabrication, and Operation of Submicron Gap Comb-Drive Microactuators”, Hirano, et.al. , Journal of Microelectromechanical Systems, Vol.1, No.1, March 1992, p52.

5. “Piezoelectric Cantilever Microphone and Microspeaker”, Lee, Ried, White, Journal of Microelectromechanical Systems, Vol.5, No.4, December 1996.

6. “Crystalline Semiconductor Micromachine”, Seidel, Proceedings of the 4th Int. Conf. on Solid State Sensors and Actuators 1987, p 104

7. Fundamentals of Microfabrication, M. Madou, CRC Press, New York, 19978. Element properties: http://web.mit.edu/3.091/www/pt/




HOMEWORK – MEMS ELECTRICAL

1. Calculate the voltage needed to pull down a polysilicon cantiliverby electrostatic force. Make appropriate assumptions for dimensions.

2. Calculate the number of fingers needed in an electrostatic comb drive to create a force of 10 micro newtons. Make appropriate assumptions for dimensions.

3. What coil current is needed to create a force of 10 micro newtonsin a magnetic field of 0.5 Tesla. Make appropriate assumptions for dimensions.

4. In a thermocouple made of aluminum on n+ poly what voltage will be generated for a temperature difference of 70 °C?

5. A diode is used as a temperature sensor and is forward biased with a 1.5 Volt battery in series with a 10Kohm resistor. If the device is used to measure body temperature (nominal 98.6 °F) how much change in voltage across the diode if someone had a temperature of 102.6°F.

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