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

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

2 Statistics Mean and standard deviation calculations µ = N N x ( ) i i= 1 x x S = n i= 1 i n 1 95% confidence interval (assuming normal distribution): x ( ) x ± 2S X 2

3 Statistics (cont.) Modified Thomson t technique Determine mean x and std. dev. S X Find largest deviation from mean, δ i = If δ then data point i > τ S X reject xi If data point rejected, recompute mean and standard deviation Repeat process with new mean and standard deviation ( x) x i

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

5 Normal Distributions f(x) x Let z f = 2 2 ( x µ ) / ( ) ( 2σ x = e ) f σ x µ σ 1 2 π 1 z / 2 () z = e 2 2 π a µ b µ P ( a x b)= P( z1 z z 2 ), z 1 = z 2 = σ σ Transform your data to zero-mean, σ=1, and evaluate probabilities in that domain!

6 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).

7 Student s t Distribution Data with n 30. n P( t x )= α / 2 tα / 2 1 α Result we re looking for: α/2 -t α/2 t α/2 α/2 S µ = x ± t α / 2 n w/ confidence:1 α Based on calculating the area of the shaded portions. Total area = α. How do we get t α/2 /2?

8 Student s t Distribution

9 Uncertainty Analysis #4 To estimate the uncertainty of quantities computed from equations: Note the assumptions and restrictions given on p. 182! (Independence of variables, identical confidence levels of parameters) = W u W 2 z 2 y 2 x u Z W u Y W u X W W Z 2 Y 2 X W u Z W W u Y W W u X W + + =

10 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?

11 Exercise #1a Experimental gain from an op-amp circuit is found from the formula G = E E out in = Compute the uncertainty in gain, u G, if both E in and E out have uncertainty: E in = 2.65 ± + 1 E out E E out = 6.27 ± in volts 0.11 volts

12 Exercise #1c Equation: G E out +1 1 = = EoutEin Ein u G G = u + 1 E E out out 2 u + 1 E E in in 2

13 Exercise #1d E out Answers: G = = E in volt 2.65volt = 2.37 u G G = volt 0.08volt volt 2.65volt 2 = u G = ± 3.8% ( 0.038) ( 2.37) u = u G = G 09 ± 0.

14 Foil Element Strain Gage F = the ratio between strain and resistance is the gage, factor F R R L L = R R ε R R = Fε for foil gages, F~2 for semiconductor gages, F~ ~ 25 to 50

15 One Gage - Uniform Member in Pure Tension Strain Gage P t if we assume uniform loading across the width of the beam, P σ = = A P wt w

16 Quarter Bridge Analysis recall that the change in the gage resistance is very small, 2 R 2 R << R R R 1 R R V out = Vin 2 R 4 R 4 + R V in quarter bridge

17 Two Gages - Cantilever Beam x P t If mounted correctly, the 2 gages see the same strain magnitude, where one gage in tension (R + R) and one gage in compression (R - R) w

18 Half Bridge Analysis substituting resistance values, V R = V, V 2 = R + R 1 in V out = V 1 V2 V out = = 2 R 2R R R R R V ( R + R) + ( R R) in R R 2R V in 1 Note the factor 1/2 for V in half bridge

19 Four Gages - Cantilever Beam x P t two gages on top in tension (R+ R) and two gages on bottom in compression (R- R) w

20 Full Bridge - Pure Bending gages of opposite strain in adjacent legs of bridge, V V out in = 4 4 R R Find σ, ε, R, and V out, R = R Fε if out if P = 50 lb, F = 2.0, R = 350Ω w = 1.0 inch, t = 0.25 inch, x = 6.0 inch material is aluminum (E AL ~ 10.5x10 6 psi ) V in = 12.1 volts

21 Load Cell w/differential Amplifier + V in - RED BLACK R- R R+ R WHITE R+ R R- R R i GREEN R i R f R f V 0 -

22 Potentiometers (pp of text) potentiometers ( pots ) are electrical resistance elements made in both linear & rotary form a mechanical motion of the wiper changes the output voltage in proportion to the wiper displacement + X max V in - X = X X max + V out - V V out in wiper

23 LVDT (pp of text) LVDT Linear Variable Differential Transformer AC AC 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 )

24 LVDT Construction - Fig. 8.11

25 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 R2 R1 Vout LED Phototransistor (What s a transistor?)

26 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). B A

27 How it works Fraden, Jacob, Handbook of Modern Sensors, AIP Press, Woodbury, New York, 1997.

28 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" 8 3-bits = patterns. 270 ο 315 ο 225 ο 0 ο 45 ο 135 ο 90 ο (How many phototransistors do you need??) (Are you limited to three lines?)

29 Average Velocity Timer Method Count events per fixed time interval the fixed time interval (1 sec) starts/stops counting lobe counting sensor ω = N lobes * 1sec 1rev 8lobes 8 lobes on rotating wheel

30 Average Velocity Timer Method Clock at 1000 Hz Timing counter T 1 T Lobe counter N lobes 1rev ω = * ω = 421 lobes * 1sec 8 lobes 1sec 1rev 8 lobes * 60sec 1min ~ 3160 RPM

31 Instantaneous Velocity Timer Method Count known clock between events the external event starts/stops counting Fix clock at 100 khz Count number of clock cycles, k, from one lobe to the next 8 lobes on rotating wheel ω = 1/ 8rev k clocks * 100,000clocks sec

32 Instantaneous Velocity Timer Method Clock at 100,000 Hz Timing counter Lobe sensor output ω = start 1 / 8rev 100,000 clocks * k clocks sec stop ω = 1 / 8rev * 235 clocks 100,000 clocks sec 53.2 rev sec ~ 3190 RPM

33 Velocity Measurement +5V +5V ~330Ω ~10kΩ E o LED Phototransistor

34 Sketch scope output for 1 rev 16 slots/revolution ω = 750 RPM

35 Typical Frequency Response as shown in the Wilcoxon (p. 105) handout, Typical resonance Curves for Various Sensitivities db (re 1 V) Use in these regions High Sensitivity, 1 V/g Med. Sensitivity, 100mV/g Low Sensitivity, 10mV/g Frequency, khz

36 Circuit Model for Permanent Magnet DC Motor i a + + V a R a V b - V a = applied armature voltage R a = armature resistance - V b = back EMF i a = armature current

37 PMDC Motor Steady-State Equations i a 1 R a ( V V ) = from circuit a b V b = k b ω from dv= B v dl and v = rω τ = k i a a from df = i a dl X B and τ = rf

38 PMDC Motor Equation Part τstall ω = 0 #3 V = R a a i a,stall τ stall = k a V R a a Torque, τ constant V a Speed, ω At any point on load curve, R a = V a i a k b ω ω NL = no-load speed (i a =0) V a = k b ω NL

39 2nd In-Class Exercise A small DC motor has these parameter constants V a =? volts K a = 3.60 oz-in/amp K b = 2.67 volt/krpm R a = 50 ohms On a single graph, we will plot the torque vs. speed relationship for different input voltages - 24, 18, 12, 6 VDC

40 Number Assignments - Exercise #2 Both 1000 and 3000 RPM Group V a, volts Group #1 24 VDC #5 #2 18 VDC #6 #3 12 VDC #7 #4 6 VDC #8 Both 5000 and 7000 RPM

41 In-Class Exercise - Solution volts 12 volts 18 volts 24 volts Torque, oz-in Speed, RPM

42 DC Brushless Motor The magnetic field in the rotor is provided by permanent magnets on the rotor Hall effect sensors (or resolver output) are used to signal a motor driver when to switch the current stator s in the windings Motor driver depends on the controller to set desired torque output

43 DC Brushless Motor Permanent Magnet Rotor Wound Wire Stator

44 DC Brushless Motor Advantages No appreciable heat is generated in the rotor and 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 indefinitely as long as the winding temperature does not exceed the limit. No brushes to wear out or contaminate the surroundings

45 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

46 Pulse-Width Modulation + V max V a + - -

47 PWM - 80% of Vmax Armature Voltage, Va Average V a is 80% of V max sec sec

48 PWM - 60% of Vmax Armature Voltage, Va Average V a is 60% of V max sec sec

49 PWM - 40% of Vmax Armature Voltage, Va Average V a is 40% of V max sec sec

50 PWM - 20% of Vmax Armature Voltage, Va Average V a is 20% of V max sec sec

51 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!

52 Half-Wave Rectifier V AC V a + - 0

53 Full-Wave Rectifier V a + - V AC

54 Silicon Controlled Rectifier V AC V a + Gate - 0 Delay time is adjustable by gate signal

55 Silicon-Controlled Rectifier Drive Gate Gate V a + - V AC

56 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)

57 Series Wound DC Motor i f = i a + + V in R f R a V b - - V in = input voltage R a = armature resistance R f = field resistance V b = back EMF i a = armature current i f = field current

58 Shunt Wound DC Motor i f + i a i a + + V in R f R a V b - i f -

59 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.

60 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.

61 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 120 f n s = p n s = synchronous speed (in RPM), f = frequency (in Hz), and p = the number of poles

62 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 = 100 n s n s n a s = slip (in %), n s = synchronous speed n a = actual speed

63 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

64 Split-Phase AC Motor Advantages 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

65 Permanent Split Capacitor (PSC) Advantages Quieter, smoother than split phase Reduced starting current Longer life Higher reliability Capable of frequent start/stop cycles Disadvantages More expensive for same HP Lower performance when starting Need to always use manufacturer's desired capacitor value

66 Shaded Pole AC Motor 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 Disadvantages Low starting and running torque Low efficiency Available in subfractional to ~ 1/4 hp sizes

67 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

68 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 n s = 100 s n s n a 120 f n s = p

69 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

70 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

71 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

72 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

73 Parallel Shaft Gearboxes Pinion gears Output shaft speed, ω out Gear reductions ratio typically given as ω R = in > 1 ω out Input motor speed, ω in Spur gears Each gear pair reduces the overall efficiency of gearmotor τ =ηrτ out in

74 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

75 Figure Simple 12 step/rev hybrid motor South poles on this end of rotor North poles on other end of rotor

76 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

77 Wave Drive (full steps) Start one set of stator windings is energized, then the other 60 rotation 30 rotation 90 rotation

78 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

79 Figure Half Stepping Start 15 rotation 30 rotation Phase 1 - ON Phase 2 - OFF Phase 1 - ON Phase 2 - ON Phase 1 - OFF Phase 2 - ON

80 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?

81 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 required See Parker-Compumotor notes for additional details

82 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

83 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

84 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

85 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 )

86 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

87 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

PVA Sensor Specifications

Position, Velocity, and Acceleration Sensors 24.1 Sections 8.2-8.5 Position, Velocity, and Acceleration (PVA) Sensors PVA Sensor Specifications Good website to start your search for sensor specifications: