Exam 2 Topics• Statistics• Uncertainty• Strain gages• Load cells• PVA sensors• DC motors• AC motors• Stepper motors• Electrical control components

Statistics

• Mean and standard deviation calculations

=µN

xN

1ii∑

= ( )1

2−∑

=

xxn

ii=S

1−n

∈x ( )XS2x ±

• 95% confidence interval (assuming normal distribution):

Statistics (cont.)

• Modified Thomson t technique• Determine mean and std. dev. SX

• Find ______ deviation from mean,• If then _____ data point• If data point ________________ mean

and standard deviation• ______ process with new mean and

standard deviation

x( )xx ii −=δ

Xi S⋅> τδ ixlargestlargest

rejectrejectrejected, rejected, recomputerecompute

RepeatRepeat

Statistics (cont.)

(I will provide these if they are needed for the exam.)

Normal Distributions

-5 0 50

0.1

0.2

0.3

0.4

x

f(x) ( ) ( ) ( )22 2/xe

21xf σµ

πσ−−

⋅=

=σ

µ−xLet z

( ) 2/z2

e21zf −

⋅=

π

( ) =≤≤⇒ bxaP ( ),zzzP 21 ≤≤ =1z =2zσ

µ−aσ

µ−b

⇒Transform your data to zero-mean, σ=1, and evaluate probabilities in that domain!

Normal Distribution• Standard table available describing the area under the curve from “0 to

z” for a normal distribution. (Table 6.3 from Wheeler and Ganji.) So, if you want ±X%, look for (0→X/2).

Student’s t DistributionData with nData with n≤≤30.30.

Based on calculating the Based on calculating the area of the shaded portions.area of the shaded portions.

Total area = Total area = α.α.

tα/2-tα/2

α/2α/2α/2α/2

( ) =≤≤− 2/2/ txtP αα α−1

Result we’re looking for:Result we’re looking for:

=µn

Stx 2/ ⋅± α

α−1w/ confidence:w/ confidence:

How do we get How do we get ttαα/2/2??

Student’s t Distribution

Uncertainty Analysis #4• To estimate the uncertainty of quantities computed

from equations:

• Note the assumptions and restrictions given on p. 182! (Independence of variables, identicalconfidence levels of parameters)

=Wu W

2

z

2

y

2

x uZWu

YWu

XW

W1

+

+

⋅

∂∂

∂∂

∂∂

2Z

2Y

2X

Wu

ZW

Wu

YW

Wu

XW

+

+

=

∂∂

∂∂

∂∂

Uncertainty Analysis #14• Which of the three measurements X, Y, or

Z, contribute the most to the uncertainty in W?

• If you wanted to reduce your uncertainty in the measured W, what should you do first?

Exercise #1a

• Experimental gain from an op-amp circuit is found from the formula

• Compute the uncertainty in gain, uG, if both Ein and Eout have uncertainty:

==in

out

EE

G

volts08.065.2E in ±=volts11.027.6E out ±−=

1in

1out EE −+

Exercise #1c

=Gu G

11 −+== inoutin

out EEEEG

2

in

E2

out

E

Eu

1Eu

1 inout

−+

+

• Equation:

Exercise #1d

• Answers: ==in

out

EE

G

=Gu G

=+−

volt65.2volt27.6 37.2−

=→= Gu038.0

( ) ( ) =→−⋅ Gu37.2038.0

%8.3±

=Gu 09.0±

22

volt65.2volt08.01

volt27.6volt11.01

−+

−

Foil Element Strain Gage

• the ratio between strain and resistance is the ____ _____, F

=F

gage factorgage factor

=∆

∆

LL

RR

⇒∆

εR

RεF=

∆RR

for foil gages, for foil gages, FF~2~2for semiconductor gages, for semiconductor gages, FF~ 25 to 50~ 25 to 50

One Gage - Uniform Member in Pure Tension

• if we assume uniform loading across the width of the beam,

w

Strain GageP

t

=σ =AP

wtP

Quarter Bridge Analysis• recall that the change in the gage

resistance is very small,

⇒<<∆ 42R

R

inVRR

∆

41

≈∆

+R

R24 4

⇒ ≈

∆

+

∆

= inout V

RR

RR

V24

“quarter” bridge

Two Gages - Cantilever Beam

• If mounted correctly, the 2 gages “see” the same strain magnitude, where– one gage in tension _______ and– one gage in compression _______

Px

t

w

(R + (R + ∆∆R)R)(R (R -- ∆∆R)R)

Half Bridge Analysis• substituting resistance values,

,1 inVRR

RV+

= =2V ( ) ( ) inVRRRR

RR∆−+∆+

∆−

inVR

RRR

R

∆−

−22

=−= 21 VVVout

inVRR∆

21 Note the factor 1/2 for

“half” bridge=outV

Four Gages - Cantilever BeamP

x

t

w

• two gages on top in tension (R+∆R) and• two gages on bottom in compression (R-∆R)

“Full” Bridge - Pure Bending• gages of opposite strain in adjacent legs

of bridge,

=in

out

VV

Find Find σσ, , εε, , ∆∆R, and R, and VVoutout ifif–– P = 50 lb, F = 2.0, R = 350P = 50 lb, F = 2.0, R = 350ΩΩ–– w = 1.0 inch, t = 0.25 inch, x = 6.0 inchw = 1.0 inch, t = 0.25 inch, x = 6.0 inch–– material is aluminum (Ematerial is aluminum (EALAL ~ 10.5x10~ 10.5x1066 psipsi ))–– VVinin = 12.1 volts= 12.1 volts

,44

RR∆ εF

RR

=∆

GREEN

Load Cell w/Differential Amplifier

R+R+∆∆RR

R+R+∆∆RRRR--∆∆RR

RR--∆∆RR

VVinin

++

--

--

++RRii

RRff

RRff

RED

BLACK

WHITE

RRii

++

VV00

--

Potentiometers (pp. 232-233 of text)

in

out

max VV

XX

=

Xmax

• potentiometers (“pots”) are electrical resistance elements made in both _____ & ______ form

• a mechanical motion of the wiper changes the output voltage in proportion to the wiper displacement

linearlinear rotaryrotary

+

Vin

-Vout

-

+X

wiperwiper

LVDT (pp. 233-236 of text)

• LVDT – Linear Variable Differential Transformer– External ___ voltage applied to a primary coil– ___ voltages of the same frequency are

induced in two secondary coils– The difference in the two secondary voltages

is proportional to the position of a ferromagnetic core (“armature”)

ACACACAC

LVDT Construction - Fig. 8.11

Optical Encoders– Optical sensing of

encoder position is used– A light source (LED or

light-emitting diode) is placed on one side of the encoder disk

– A light detector (phototransistor) is on the other side

+V +V

Phototransistor

LED

Vout

R1 R2

(What’s a transistor?)(What’s a transistor?)

Incremental Encoders• Two sensors

(usually optical) are mounted such that one is halfway blocked by the "solid" area (Channel A) while the other is in the middle of the "clear" area (Channel B).

AB

How it works

Fraden, Jacob, Handbook of Modern Sensors, AIP Press, Woodbury, New York, 1997.

Absolute Encoder – gives a finite number of

unique patterns spread uniformly over 1 revolution.

– 3 output lines (or bits) and each line can be either "solid" or "clear"

– 3-bits = __ patterns.

0ο

45ο

90ο

135ο225ο

270ο

315ο

(How many phototransistors do you need??)(How many phototransistors do you need??)

88

(Are you limited to three lines?)(Are you limited to three lines?)

Average Velocity Timer Method• Count events per fixed time interval

– the fixed time interval (1 sec) starts/stops counting

=ω“lobe”countingsensor

*sec1

N lobeslobesrev

81

8 “lobes”on rotating wheel

Average Velocity Timer Method“Clock” at 1000 Hz

0 1 2 3 4 5 6 7 8 ….. 997 998 999 1000

Timing counterT1 T2

1 2 3 ….. 420 421 “Lobe” counter

→=lobesrevlobes

81*

sec1Nω =ω

sec1214 lobes *

81lobesrev ~

min1sec60

RPM3160*

“Instantaneous” Velocity Timer Method

• Count known clock between events– the external event starts/stops counting

8 “lobes”on rotating wheel

• Fix clock at 100 kHz• Count number of _____

_______ from one lobe to the next

=ω

clockclockcycles, k,cycles, k,

sec000,100 clocks*

k8/1

clocksrev

Instantaneous Velocity Timer Method

“Clock” at 100,000 Hz

0 1 2 3 4 5 6 7 8 ….. 234 235 …….. Timing counter

“Lobe” sensor output

start stop

sec000,100*

k8/1 clocks

clocksrev

=ω

=→ ω *352

8/1clocks

rev≈

sec000,100 clocks ~

sec2.53 rev RPM3190

Velocity Measurement

+5V +5V~10kΩ~330Ω

Photo-transistor

Eo

LED

Sketch “scope” output for 1 rev 16 slots/revolution

ω = 750 RPM

Typical Frequency Responseas shown in the Wilcoxon (p. 105) handout,

Typical resonance Curves for Various Sensitivities

-50

-40

-30

-20

-10

0

10

20

30

0 10 20 30 40 50 60

Frequency, kHz

dB (r

e 1

V)

High Sensitivity, 1 V/g

Med. Sensitivity, 100mV/g

Low Sensitivity, 10mV/g

Use in theseregions

Circuit Model for Permanent Magnet DC Motor

Va

+

-

ia

Vb

+Ra

-

Va= applied armature voltage Vb= back EMF

Ra= armature resistance ia= armature current

PMDC Motor Steady-State Equations

( )baa

a VVR1i −= from circuit

ωbb kV = from dV= B v dL and v = rω

aa ik=τ from df = ia dL X B and τ = rf

PMDC Motor Equation Part #3

stallτ 0=ω

aVconstant

→stalla,aa iRV =a

aastall R

Vk=τ

Torque, τ

Speed, ω

ωNL=“no-load” speed (ia=0)At any point on load curve,

a

baa i

kVR ω−=

NLωba kV =

2nd In-Class Exercise

Va = ? voltsKa = 3.60 oz-in/ampKb = 2.67 volt/KRPMRa = 50 ohms

On a single graph, we will plot the torque vs. speed relationship for different input voltages -

24, 18, 12, 6 VDC

A small DC motor has these parameter constants

Number Assignments - Exercise #2

Group Va, volts Group

#1 24 VDC #5

#2 18 VDC #6

#3 12 VDC #7

#4 6 VDC #8

Both1000and3000RPM

Both5000and7000RPM

In-Class Exercise - Solution

0.6

0.8

1.0

Torq

ue, o

z-in

6 volts12 volts18 volts24 volts

2.0

1.8

1.6

1.4

1.2

0.4

0.2

0.00 1000 2000 3000 4000 5000 6000 7000

Speed, RPM

DC Brushless MotorDC Brushless Motor

• The magnetic field in the rotor is provided by permanent magnets on the _____

• Hall effect sensors (or resolver output) are used to signal a motor driver when to switch the current in the ______________

• Motor driver depends on the controller to set desired torque output

rotorrotor

stator’s windingsstator’s windings

DC Brushless Motor

Wound Wire Stator

Permanent Magnet Rotor

DC Brushless Motor Advantages

DC Brushless Motor Advantages

• No appreciable heat is generated in the rotorand hence the heat conducted to the shaft is minimized.

• Due to the lack of brushes, motors can be operated at high torque and zero rpm indefinitelyas long as the winding temperature does not exceed the limit.

• No brushes to wear out or contaminate the surroundings

DC Brushless Motor Disadvantages

DC Brushless Motor Disadvantages

• Torque ripple is hard to minimize by design• Motor operation requires the purchase of an

electronic motor driver• Rotor magnets can become demagnetized in

high current or temperature environments• Most motor drivers brake DC brushless motors

by applying reverse current, in which almost as much power is expended to stop the motor as was required to start it moving

Pulse-Width ModulationPulse-Width Modulation

Vmax

+

-

Va

+

-

PWM - 80% of Vmax

0

20

40

60

80

100Ar

mat

ure

Volta

ge, V

a

Average Vais 80% of Vmax

0.00008 sec0.00010 sec

PWM - 60% of Vmax

0

20

40

60

80

100Ar

mat

ure

Volta

ge, V

a

Average Vais 60% of Vmax

0.00006 sec0.00010 sec

PWM - 40% of Vmax

0

20

40

60

80

100Ar

mat

ure

Volta

ge, V

a

Average Vais 40% of Vmax

0.00004 sec0.00010 sec

PWM - 20% of Vmax

0

20

40

60

80

100Ar

mat

ure

Volta

ge, V

a

Average Vais 20% of Vmax

0.00002 sec0.00010 sec

PWM Motor Drive

• If the time period T is short compared to the time constants of the system, the motor response will be the same– PWM switching frequencies in the 10 kHz (or

higher) range are frequently used.• Relatively high drive efficiency (up to 80%)

– inefficiency creates heat in the amplifier that must be dissipated!

Half-Wave Rectifier

VACVa

+

-

0

Full-Wave Rectifier

VAC

Va

+

-

Silicon Controlled Rectifier

VACVa

+

Gate-

0Delay time isadjustable bygate signal

Silicon-Controlled Rectifier Drive

VAC

Gate

Gate Va

+

-

DC Motor - Magnetic Field Generation

• Magnetic field on the stator can be generated two ways– with a permanent magnet (PM)– electro-magnetically with wound coils

• Wound DC motors• Series wound• Shunt wound• Compound wound ( series and shunt

windings)

Series Wound DC Motor

Vin

+

-

if = ia

Vb

+Rf Ra

-

Vin= input voltage Vb= back EMF

Ra= armature resistance ia= armature current

if = field currentRf = field resistance

Shunt Wound DC Motor

if + ia

Vin

+

-

Rf

ia

Vb

+Ra

if-

AC Induction Motors

• Simplest and most rugged electric motor • Consists of wound stator and rotor assembly• AC in the primary member (stator) induces

current in the secondary member (rotor)• Combined electromagnetic effects of the stator

and rotor currents produce the force (torque) to create rotation.

AC Induction Motors

• Rotors typically consist of a laminated, cylindrical iron core with slots for receiving the conductors.

• Common type of rotor has cast-aluminum conductors and short-circuiting end rings.

• This "squirrel cage" rotates when the moving magnetic field induces a current in the shorted conductors.

AC Motor Speed• The magnetic field rotates at the

synchronous speed of the motor• Determined by the number of poles in

the stator and the frequency of the AC power

ns = synchronous speed (in RPM), f = frequency (in Hz), and p = the number of poles

pfns

120=

AC Motor Speed• Synchronous speed is the absolute upper limit

of motor speed. • When running, the rotor always rotates slower

than the magnetic field (or no torque!)• The speed difference, or slip, is normally

referred to as a % of synchronous speed:s = slip (in %), ns = synchronous speedna = actual speed

s

asn

nns −= 100

Single-phase AC Motors

• Single phase AC motors require a "trick" to generate a 2nd "phase" to develop starting torque

• Three common methods:– split-phase (auxiliary winding is rotated 90°)– capacitor– shaded-pole

Split-Phase AC MotorAdvantages• Operate at ~ constant

speed, 4 pole, 60 Hz:– 1780 RPM (no load)– 1700/1725 RPM at full

load• Reversible at low

speed• Rapid acceleration• Relatively low cost

Disadvantages• Repeated start/stop

cycles heat the windings (high start resistance)

• Less useful for large inertial loads

• Requires large wiring to handle starting currents

Permanent Split Capacitor (PSC)

Disadvantages• More expensive for

same HP• Lower performance

when starting• Need to always use

manufacturer's desired capacitor value

Advantages• Quieter, smoother

than split phase• Reduced starting

current– Longer life– Higher reliability

• Capable of frequent start/stop cycles

Shaded Pole AC Motor

Disadvantages• Low starting and

running torque• Low efficiency• Available in sub-

fractional to ~ 1/4 hp sizes

Advantages• Simple in design and

construction• Suitable for low cost,

high volume app's• Relatively quiet and

free from vibration• "Fail safe" design -

starts in only 1 direction

AC Motor Efficiency

• Efficiency, η = Power Output / Power Input• Small universal motors have η ~ 30%• Large 3-phase motors have η ~ 95%• Depends on actual motor load vs. rated

load– efficiency best near rated load– efficiency drops rapidly for both under- and

over-load conditions

AC Induction Motor Speed Control

• So what can we do to control the speed of an AC induction motor?– Change the number of poles (in discrete

increments - inefficient & rarely done)– Change the frequency of the AC signal– Change the slip

pfns

120=

s

asn

nns −= 100

Change AC Frequency• Variable speed AC Motor adjustable speed

drives are known as – inverters, – variable frequency drives (VFD) , or– adjustable speed drives (ASD).

• Common ways to vary AC frequency:– Six-step inverter– Pulse-Width-Modulation– Vector Flux

Universal Motor

• Runs off AC or DC power• Commonly found in

household appliances• Wound like a DC series

motor– windings on both stator

and rotor– brushes like a DC motor

Universal Motor

• Nearly equivalent performance on DC or AC up to 60 Hz

• Highest horsepower-per-pound ratio of any AC motor – speeds many times higher than that of any

other 60-Hz motor

Gearmotors

• Motors are inherently high-speed, low torque devices

• Applications frequently require low-speed, high torque

• Manufacturers provide motors with integral gear sets - called “gearmotors”– both AC and DC versions– increased torque - lower speed available

Parallel Shaft Gearboxes

Spur gears

Output shaft speed, ωout

Pinion gears• Gear reductions

ratio typically given as

• Each gear pair reduces the overall efficiency of gearmotor

1>=out

inRωω

Input motor speed, ωin

inout Rτητ =

Permanent Magnet (or PM) Stepping Motors

• an electrical circuit alternately switches the polarity of the stator poles

• as the polarity of a stator pole changes, the rotor will move to approach an equilibrium position

• equilibrium positions where N/S rotor poles align with the S/N stator poles

Figure 1.4 - Simple 12 step/rev hybrid motor

South poles on thisend of rotor

North poles on otherend of rotor

Selection of Stepper Motors

• steps per revolution (or degrees/step)– actual output position assumed by the

motor depends greatly on the static friction in the system

• maximum stepping torque– cannot be exceeded or the motor will slip– causes serious problems in open-loop

control systems where the output is assumed to match the number of input pulses

Wave Drive (full steps)

30° rotation

60° rotation

90° rotation

Start

one set of stator windings is energized, then the other

Half Step Mode

• Twice the resolution (steps/rev) from the same motor

• Much better smoothness at low speeds• Less overshoot and ringing at end of each

step• Slight loss of torque

– can be improved with the "profiled current" method of Figure 1.10

Figure 1.7 - Half Stepping

15° rotation 30° rotationStart

Phase 1 - ONPhase 2 - OFF

Phase 1 - ONPhase 2 - ON

Phase 1 - OFFPhase 2 - ON

Electric Motor Selection

Two basic decisions to make:• What type of motor is needed?

– DC motor?– Stepper motor?– AC motor?

• Once type of motor is selected, what size motor is required?

Type Selection - DC Motor

DC motors are typically used when– low-cost, variable speed is advantageous

• but precise speed regulation not required– starting torque required up to 5-10 times

more than running torque • brief overloads OK, since motor has time to

cool– frequent start/stop cycles, reversing, or

closed-loop positioning requiredSee Parker-Compumotor notes for additional details

Type Selection - Stepper Motor

Stepper motors are typically used when– low-cost, open-loop positioning required

• no feedback sensors required to monitor position if max torque not exceeded

• noncumulative nature of positioning errors gives good accuracy over long motions

– reasonably high torques at low speeds• not able to handle large inertial loads due to

low acceleration requirements– energy efficiency not important

Type Selection - AC Motor

AC motors are typically used when– low-cost, constant speed is advantageous

• gearing required to deliver speeds that are significantly less than 1200 RPM

– starting torque less than twice running torque

• brief periods of high running torque frequently handled by flywheels

– available access to AC power

DC Motor Ratings

• DC motors are “rated” at a single speed and torque

• In most cases, the motor can operate at this point continuously– temperature rating will not be exceeded– DC motors rated with form factor of 1

• DC motors are “typically” used at – about 90% of rated speed– abour 10 to 40% of rated torque

AC Motor Ratings• AC motors are also “rated” at a single

speed and horsepower – 3450, 1725, 1140, 850 RPM are common

• AC motor can operate here continuously– temperature rating will not be exceeded

• AC motors are “typically” used at – about 90% of rated torque/power– much lower efficiency if motor is too large

(“over-rated”)

Brainstorming Scenario• Your group has been hired to select

motors for various applications and products

• Possibilities include:– DC motor (w/brushes)– Brushless DC motor– Stepper motor– Split phase AC motor– Permanent Split Capacitor (PSC) AC motor– Shaded pole AC motor– Universal motor

Question• Match one of the 7 different types of

motors to these applications & justify your selection:– “low cost” educational robot – electric knife for carving turkey– constant speed conveyor belt with

frequent start/stops– power window drive in auto– fan in indoor HVAC unit– industrial robot used in painting

applications