exam 2 topics - 2 topics_rev_c.pdf · exercise #1a • experimental ... _____ & _____ form •...
TRANSCRIPT
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Exam 2 Topics• Statistics• Uncertainty• Strain gages• Load cells• PVA sensors• DC motors• AC motors• Stepper motors• Electrical control components
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Statistics
• Mean and standard deviation calculations
=µN
xN
1ii∑
= ( )1
2−∑
=
xxn
ii=S
1−n
∈x ( )XS2x ±
• 95% confidence interval (assuming normal distribution):
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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
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Statistics (cont.)
(I will provide these if they are needed for the exam.)
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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!
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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).
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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??
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Student’s t Distribution
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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
+
+
=
∂∂
∂∂
∂∂
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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?
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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 −+
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Exercise #1c
=Gu G
11 −+== inoutin
out EEEEG
2
in
E2
out
E
Eu
1Eu
1 inout
−+
+
• Equation:
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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
−+
−
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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
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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
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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
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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)
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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
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Four Gages - Cantilever BeamP
x
t
w
• two gages on top in tension (R+∆R) and• two gages on bottom in compression (R-∆R)
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“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
=∆
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GREEN
Load Cell w/Differential Amplifier
R+R+∆∆RR
R+R+∆∆RRRR--∆∆RR
RR--∆∆RR
VVinin
++
--
--
++RRii
RRff
RRff
RED
BLACK
WHITE
RRii
++
VV00
--
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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
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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
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LVDT Construction - Fig. 8.11
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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?)
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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
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How it works
Fraden, Jacob, Handbook of Modern Sensors, AIP Press, Woodbury, New York, 1997.
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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?)
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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
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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*
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“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
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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
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Velocity Measurement
+5V +5V~10kΩ~330Ω
Photo-transistor
Eo
LED
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Sketch “scope” output for 1 rev 16 slots/revolution
ω = 750 RPM
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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
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Circuit Model for Permanent Magnet DC Motor
Va
+
-
ia
Vb
+Ra
-
Va= applied armature voltage Vb= back EMF
Ra= armature resistance ia= armature current
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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
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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 =
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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
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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
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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
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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
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DC Brushless Motor
Wound Wire Stator
Permanent Magnet Rotor
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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
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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
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Pulse-Width ModulationPulse-Width Modulation
Vmax
+
-
Va
+
-
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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
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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
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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
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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
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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!
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Half-Wave Rectifier
VACVa
+
-
0
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Full-Wave Rectifier
VAC
Va
+
-
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Silicon Controlled Rectifier
VACVa
+
Gate-
0Delay time isadjustable bygate signal
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Silicon-Controlled Rectifier Drive
VAC
Gate
Gate Va
+
-
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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)
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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
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Shunt Wound DC Motor
if + ia
Vin
+
-
Rf
ia
Vb
+Ra
if-
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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.
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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.
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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=
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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 =
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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
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Figure 1.4 - Simple 12 step/rev hybrid motor
South poles on thisend of rotor
North poles on otherend of rotor
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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
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Wave Drive (full steps)
30° rotation
60° rotation
90° rotation
Start
one set of stator windings is energized, then the other
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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
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Figure 1.7 - Half Stepping
15° rotation 30° rotationStart
Phase 1 - ONPhase 2 - OFF
Phase 1 - ONPhase 2 - ON
Phase 1 - OFFPhase 2 - ON
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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?
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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
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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
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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
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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
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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”)
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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
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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