Transcript Lecture Notes Part 5a

ET 438B Sequential Control and Data Acquisition
et438b-5a.pptx
1
Angular and Linear Position Sensors
Simple linear displacement
transducers: Slide
Potentiometer
Simple angular transducers: Rotary
Potentiometers, multi-turn or single,
Wire-wound or carbon composite
Slider
Es
L
x
Eout
Eout 

Es
L
Eout
Es
x
Eout  Es
L
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Wire-wound potentiometers do not have infinite resolution
due to wire diameter
Resistance varies with
position
Cross
section
Number of turns per unit length determines resolution of
the potentiometer
Resolution (%) = 100/N
Where
N = number of turns in the pot.
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Es
Req
Rp = total potentiometer R
aRp = fraction of potentiometer R
from wiper to ground
Rp
+
aRp
RL
E0
-
RL is load on potentiometer. Must
include in calculation
R L || a  R p
Solve using voltage division
a


E0  
 E
 1  a  r  a2  r  s
R
r p
RL
Loading error LE defined as = aEs -Eout
 a2  r  (1  a) 
%LE  
  100
 1  a  r  (1  a) 
Rp
r
RL
R eq  (1 a )  R p  R L || a  R p
et438b-5a.pptx
Percent Loading Error
4
Problem Statement
A linear wire-wound potentiometer has a total
resistance of 50,000 ohms The voltage output of the
potentiometer will be converted to a digital value by a
12 bit ADC. Determine minimum number
of turns required so that the resolution of the
potentiometer does not exceed the quantization error of
the ADC.
What percent loading error can be expected if the input
resistance of the ADC is 100 kW and the range of
measurement is from 20% to 80% of the potentiometer
value?
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Find the resolution of the 12-bit ADC then convert it to a percentage.
Assume 1 volt range
Q.E.
VLSB
2
VLSB 
VFS 1 V
1


 2.441 10 4 V
n
12
2
2
4096
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Loading error depends on position of potentiometer
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Note: dots indicates
instantaneously positive
voltages in transformers
Secondary
Primary
Ls1
Vp
Vp must be AC typical
frequency 50k - 15 kHz
Vs1
Lp
Core Motion causes Vs1
and Vs2 to change
Ls2
Vs2
Lp and Ls1 coupled Vs1 Increasing
Movable
Core
Lp and Ls2 coupled Vs2 Increasing
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Coils connected so voltages subtract
V out +
Output V
0.06-
0
+
0.06
Core Position
-
+ core
position
V out opposite phase
core coupling
Lp and Ls1
et438b-5a.pptx
core neutral
core coupling
Lp and Ls2
9
Three parts to system 1.) Control Transmitter 2.) Control Transformer
3.) Control Differential
ac
source
Control
transmitter
Control
transformer
Vout
+
-
Vout
time
Sinusoidal output with maximum value from
this position
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ac
source
Control
transmitter
Control
transformer
a
c
-
+
Vout
Output voltage is zero when rotors in this position
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ac
source
Control
transmitter
Control
transformer
a
c
Vout
+
Output voltage 180 degrees out of phase from initial position
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Output voltage magnitude is a function of the angular
displacement between the control transmitter and the
control transformer
Vout  (Em cos())sin(t)
Where
Em = the maximum amplitude of excitation
 = the angular position difference between
the transmitter and the transformer rotor
 = frequency of excitation voltage
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Control differential transformer can be added to add
constant phase shift.
Vout  (Em cos(  d ))sin(t)
Sending and receiving devices will have a constant
angular difference given by the value of d.
System can be used to maintain a constant position
difference between two shafts. It can also be used to
keep two shafts synchronized.
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A syncho system operates at a frequency of 400 Hz. The maximum
amplitude of the transformer rotor voltage is 22.5 V. Determine the ac
error signal produced by each of the following pairs of angular
displacements
a.) = 90° d=0°
c.) = 135° d=-15°
At 400 Hz
b.) = 60° d=0
d.) = 100° d=-45°
  2  (400 Hz)  2570 rad/s
a.)
b.) V  (22.5  cos(60  0)) sin( 2570  t )
0
V0  (E m  cos(  d )) sin(   t )
V0  11.25 sin( 2570  t )
E m  22.5
V0  (22.5  cos(90  0)) sin( 2570  t )  0
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Velocity Measurements
Angular and Linear Velocity
Angular Velocity Measurement Methods
- dc tachometer
- ac tachometer
- optical tachometer
DC tachometer
N
Dc generator using
permanent magnets for
field flux
S
Magnetic Field
Field Poles
Brushes and
Commutators
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Tachometer produces a dc voltage that is proportional to
angular velocity
Proportionality constant = emf constant KE (V/rpm)
KE depends on the construction of tachometer
2  R  B  N  L emf
KE 
constant
60
30  K E   Induced
E  KE s 
emf

Where
E = tachometer output (V)
KE = emf constant
s = angular velocity (rpm)
 = angular velocity (rad/s)
R = average radius (m)
B = flux density (Wb/m2)
N = number of conductors
L = length of conductor in field (m)
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Example: A dc tachometer has the following parameters
R = 0.03 m
N = 220
L = 0.15 m
B = 0.2 Wb/m2
Find KE and the output voltage at the following speeds s = 1000,
2500, and 3250 rpm
2R  B  N  L 2(0.03 m)(0.2 Wb/m 2 )(220)(0.15 m)
KE 

60
60
K E  0.0207 V/rpm
For s = 1000 rpm
E  KE s
E  0.0207 V/rpm  1000 rpm   20.7 V
For s= 2500 rpm
And 3250 rpm
E2  0.0207 V / rpm(2500 rpm) = 51.8 V
E3  0.0207 V / rpm(3250 rpm) = 67.3 V
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Ac Tachometers Construction : a 3-phase alternator with a 3-phase
rectifier to convert the output to dc. Must have constant field excitation Permanent magnet field.
Non-linear at low speeds due to the forward drop of diodes Limited to lower
speed ranges due to this. Range 100-1
Ac Tachometers can also produce a variable frequency output that has a
constant voltage.
Use frequency-to-voltage conversionet438b-5a.pptx
to get proportional dc
20
Digital encoder
attached to shaft
produces a sequence
of pulses.
3 light
sources
3 light
sensors
Encoder tracks
Inner – locates home
Middle - gives direction info
Phase shift between outer
and middle tracks
Middle leads or lags outer
based on direction
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Optical Tachometers
Pulses counted over time interval. Counter is then reset.
The number of pulses counted in interval is proportional to
the angular velocity
Formulas
60  C
N  Tc
s  N  Tc
C
60
s
Where
s = shaft speed (rpm)
N = number of pulses per shaft
revolution
C = total count during time period
Tc = counter time interval (sec)
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Incremental encoder produces 2000 pulses/rev.
a.) determine count produced by shaft speed of 1200
rpm with a count interval of 5 mS
b.) determine the speed measured at a count of 224
for a timer interval of 5 mS
a.)
s  N  Ts 1200 rpm   2000   0.005 s 
C

 200
60
60
Number of counts for 1200 rpm speed
b.)
60  C
60  224 
s

 1344 rpm
N  Ts 2000  (0.005 s)
Speed for 224 counts
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Definition - acceleration is rate of change of velocity
dv
v
a
 lim it
dt
t
Sensing Methods
Newton’s Law f = Ma
Where: f = force acting on body
M = mass of body
a = acceleration
Can measure acceleration by measuring force required to accelerate
known mass.
For angular acceleration, differentiate the velocity
signal from velocity sensors
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Sensing Methods
For angular acceleration, differentiate the velocity signal
from velocity sensors
For sampled data systems, derivative is given by difference of
readings
v v t 1  v t
at 

t
Ts
Where
at = acceleration at time t
vt-1 = velocity sample at t-1
vt = velocity sample at t
Ts = sampling time
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Springs
General Structure
K/2
Mass
M
Accelerometer attached to device under
test and experiences same acceleration
as measured object. Motion of mass
damped by viscous fluid around mass
Theory of operation same for
Integrated devices.
K/2
Springs
Displacement
Sensor (LVDT)
et438b-5a.pptx
Applications
Gaming
Automotive
Appliance control
Disk Drive protection
Bearing Condition Monitoring
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Accelerometers
Accelerometer forms a second order mechanical system.
Accelerometer in action
For small deflections,
spring force offsets
force due to acceleration so....
Displacement
caused by
acceleration
Kx = Ma
Which gives
x = (M/K)a
so acceleration is
proportional to displacement
f = Ma
Direction of
acceleration
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Acceleration is dynamic measurement, interested in
a(t), acceleration as function of time.
Consider second order response
1 K
fo 
2 M
b2

4KM
Where fo = resonant frequency of accelerometer (Hz)
 = damping ratio
K = spring constant (N/m)
fa = max. frequency of a(t)
M = mass (Kg)
b = damping constant (N-s/m)
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Accuracy of response related to resonant frequency of
device
Must have sufficient range to measure changes
if fa << fo then measurements will be accurate
if fa >>fo then the mass does not have time to react
incorrect measurement
fa = fo then displacement is greatly exaggerated
incorrect measurement
Make fo at least 2.5 times greater than fa max for < 0.5% error
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The accelerometer shown below has the following specifications:
M = 0.0156 kg
K = 260 N/m
b = 2.4 N-s/m
xmax = +- 0.3 cm
K/2
b
K/2
M
Find the following
a.) maximum acceleration that can be
measured
b.) resonant frequency
c.) damping ratio
d.) maximum frequency that can be used
with 0.5% error or less
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Accelerometer Example
M = 0.0156 kg K = 260 N/m
b = 2.4 N-s/m xmax = +- 0.3 cm
K/2
b
k  x max
 a max
a.) k  x  M  a 
M
a max 
M
(260 N/m)  0.003 m 
 50 m/s 2
0.0156 kg
b.)
1
k
1
260 N/m
f0 



 20.6 Hz
2 M 2 0.0156 kg
K/2
c.)
b2
(2.4 N - s/m )2


 0.595
4k M
4(260 N/m)  (0.0156 kg)
d.) Maximum frequency of a(t) with 0.5% error is…..
f 0 / 2.5
f max  20.6 Hz/2.5  8.25 Hz
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All force measurement based on force balance
f = M∙a
Where : f = force (N) M=Mass (kg) a=acceleration (m/s2)
Null Balance – unknown force off set by know weight (beam balance)
When weights equal
scale balanced
Weight= f
M∙a
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Displacement – displacement of elastic material with unknown
balancing force determines measurement (Spring Scale) (Strain gage
load cell)
F
Elastic properties of beam determines
Displacement due to external force
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Strain gages turn changes in displacement into changes in electric
resistance. Fine wire bonded to plastic base.
Bonded strain gages measure strain at specific location on deformable
body. Cemented to material and change length when body deforms.
L
Change in length = L+L
Define strain:
L

L
Where L = change in length due to force
(m)
L = unforced length (m)
L
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Gage factor determines the sensitivity of the sensor
R
G R
L
L
Typical values of strain gage parameters:
G: 2 to 4
R: 50 to 5000 ohms
L: 0.5 to 4 cm (unstrained)
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Define Stress
f
A
Where
S
S  stress (N/m 2 )
f  force (N)
A  area (m 2 )
Stress and strain related through the modulus of elasticity for a given
material
S
E

Where E = modulus of elasticity (N/m2 )
S = stress (N/m2)
 = strain (m/m)
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Stress depends on the geometry of the object. Consider
the cantilever beam (diving board) below.
L
f
Strain Gage
6f L
S
b  h2
A
Where:
A
Section
AA
Combining previous equations
R  6  G  L 

f
2

R bh E
b
h
S = stress (N/m2)
L= distance from fixed end (m)
f = applied force (N)
b = width of beam (m)
h = height of beam (m)
Normalized gage resistance change proportional
to force
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L
f
Strain Gage
A
Length =L = 2 m
Width =b = 20 cm
Height=h = 6 cm
A
Section
AA
Find the applied force
The beam structure
shown has the following
parameters
b
h
et438b-5a.pptx
It is made from
Aluminum with a
modulus of elasticity of
E = 6.9x1010 N/m2
The strain gage has an
unstrained R=100 Ω
G=3
R =0.073 Ω
38
Define the variables
Restate the formula
2
L  2 m
E  6.9 10  N m
b  20 cm
R  100 W
R  6  G  L 

f
2

R bh E
h  6 cm
R  0.073 W
Solve for force
10
G 3
f 

2

R b  h  E
( R 6 G L)
Substitute values and simplify
0.073W
  0.20m
  ( 0.06m
 )  6.9 10  N m
2
f 
10
 2

100 W  6 3 2 m
f  1007N
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Survey of Types
Resistance Temperature Detector (RTD)-change in resistance of pure metals
relates to temperature. Key features: wide temperature range, high accuracy,
excellent repeatability, good linearity. Need constant current source and other
electronics to produce output signal
Thermistor – temperature-sensitive semiconductor. Resistance inversely
proportional to temperature. Increasing temperature causes decreasing
resistance. Key features: high sensitivity, small size, fast response., narrow
temperature range Not recommended in applications requiring high accuracy.
Thermocouple – junctions of two dissimilar metals produces small (mV) voltages
when placed at different temperatures. Magnitude of voltage depends on
temperature difference. Key features: small size , low cost, rugged, wide
measurement range. Limitations: noise pickup, low signal levels, high minimum
span (40° C.)
et438b-5a.pptx
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Survey of Types
Integrated Circuit Temperature Sensors –precision solid-state devices with
linear output to temperature. Directly calibrated to various temperature scales (
Celsius, Fahrenheit, Kelvin, Rankine). Key features: calibrated output voltages,
linear scaling, low voltage and current draws, good linearity. Typical devices
LM34, LM 35. Limitations: low temperature range typically (-55 C to 155 C)
et438b-5a.pptx
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Construction: Coil of Nickel or Platinum wire in protective tube. Need
to detect small changes in resistance.
Platinum –high accuracy, linearity, and cost
Nickel –moderate cost, higher output than Platinum
Typical Relationship
R  R o (1  a1  T  a 2  T 2 )
Where: R = resistance at given Temp, T C
Ro = resistance at 0 degrees C
T = temperature
a1, a2 = constants
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Finding constants from data
Typical values for Platinum
Use table values to find constants
Ro, a1 and a2. Substitute data into previous
equation.
T(°C)
Resistance (W)
100  R o 1  a 1 0  a 2 0 2
0
100.00
25
109.90
50
119.80
75
129.60
100
139.30


 R o  100 W


139.9  1001  a 100  a 100   1001  a 100  a 10000
109.9  100 1  a 1 25  a 2 252  1001  a1 25  a 2 635
2
1
2
1
2
Simplify equations and solve simultaneously
0.099  a1 25  a 2 625
0.399  a1100  a 210000
a1  0.00395
Mainly linear
a 2  4 107
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Must convert RTD resistance changes into usable voltage or current
signals
Direct Methods (2 and 4 wire) - Constant current supplied to RTD
and voltage measured across it. The 4 wire method removes lead-wire
error.
Bridge Methods (2 and 3 wire) –Use dc bridge to convert resistance
changes into voltage changes. Three wire method removes most leadwire resistance error.
et438b-5a.pptx
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Iin≈0
is
Rs
is
is
+
vs
-
Voltage
To
Current
Converter
4-20 mA
output
Design circuit using Platinum RTD
Zin very high
Iin approx. zero
1.) constant current source is = 1 mA
2.) Instrumentation amp. G=50
3.)Summing amp
4.) V-to-I converter 4 mA @ 0 C
20 mA @ 100 C
et438b-5a.pptx
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5.1 V regulated source available. Remember the V-to-I converter . Use LM324\
quad OP AMPs.
12V
R1
Vs-
R2
1k
-12V
U1A
LM324
RTD
Rs
100
Vs+
+
D1
1N751
12V
Is
Is  1 mA
Vin  5.1 V
Io 
Vin
V
5.1 V
 R1  in 
 5.1 kW
R1
Is 1 mA
Now design the Instrumentation
amplifier using the two stage
amplification circuit
et438b-5a.pptx
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Av1
12V
+
Vs+
G=50 . Divide gain between
the two stages.
U1B
LM324
Av2
-12V
R3
R8
R5
R4
R9
Av1 = 5
Av2 = 10
R6
-12V
+
-12V
U1D
LM324
12V
Vo
R7
Vs-
+
U1C
LM324
12V
Gain formula
 2R 3  R 6 
Vo  
 1
(Vs )  (Vs ) 
 R5
 R 8 
Where R3=R4 R8=R9 and R6=R7
 2R 3  R 6 
Vo  
 1
(Vs )  (Vs ) 
 R5
 R 8 
 2R 3 
A v1  
 1 Let R3  10 kW
R
5


20,000
 2(10,000) 
5
 1  R5 
 5 kW
R
5
4


Use 5.1 kW standard value
et438b-5a.pptx
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Stage 2 gain is 10
 R6 
A v1  
 Let R6  56 kW
R
8


56,000
 56 kW 
10  
 5.6 kW
  R8 
R
8
10


Now determine the span of the RTD output voltage
@ 0 C (0.001 A)(100 W)  0.1V
@ 100 C (0.001 A)(139.3 W)  0.1393V
Vo (min)  50(0.1 V)  5.0 V
After
Amp
Vo(max)  50(0.1393V)  6.965 V
Desired output span 16 mA – 4 mA
Input span -6.965 - -5.0
Size the R of the V-to-I converter based on span ratio

I o (max)  I o (min) 
20 mA  4 mA   16 mA
1


R (Vin (max)  Vin (min) ) (6.965 V  5.0 V) 1.965 V
R
et438b-5a.pptx
1.965 V
 123 W
16 mA
48
Now determine the offset voltage for the 0 degree input to the converter.
Set Io to minimum and compute Vin using the computed R.
12V
Zero Set
Vb
R14
10k 37.25%
Vo
R10
470k
R11
470k
Span Set
R12
470k
R15
500 24.3%
-12V
123 W
-12V
RL
100
Io
+
U2A
LM324
12V
Vin
+
U2B
LM324
12V
Vb  Vin (max)  Vin  (1)( 5.0  0.492 V)
Vin
 Vin  I o  R
R
Vin  (0.004 A)(123 W)  0.492 V @ 4 mA
Io 
U2A inverting V  4.508 V
b
summer
Find value of Vb
et438b-5a.pptx
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Thermister Characteristic Curve
Thermister Reistance (Ohms)
6000
Photoresistor
has similar
characteristic
4000
2000
0
0
20
40
60
80
100
Temperature (Degrees C)
120
140
160
Very Non-linear- High sensitivity in (0-40 C range)
et438b-5a.pptx
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Semi-log plot (y axis logarithmic) of thermister data
Thermister Characteristic Curve
1 10
4
Thermister Reistance (Ohms)
Approximately
linear over small
range 40-100 C
Log(R)=mT+b
1 10
3
100
10
0
20
40
100
80
60
Temperature (Degrees C)
120
140
160
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Linearizing Thermister Characterisitcs
Vcc
Voltage Divider Output Curve
10
Thermister
R=2000 ohms
RT
R
 R 
V cc
Vo  
 R  RT 
8
Output Voltage (Vo)
Vo
6
4
Changing value of R
changes output curve
shape
2
0
20
40
et438b-5a.pptx
60
80
100
Temperature (Degrees C)
120
140
160
52
Junctions of dissimilar metals produces voltage
proportional to temperature
Iron
V
Constantan
0˚C
Heat
Cold Junction
et438b-5a.pptx
ANSI Standard Type J
-190-800 ˚C
53