Transcript Slide 1

1
ET 483b Sequential Control
and Data Acquisition
et438b-4.pptx
Static Characteristics
Error = (measured value) – (ideal value)
Ways of expressing instrument error
1.) In terms of measured variable
Example ( + 1 C, -2 C)
2.) Percent of span
Example (0.5% of span)
3.) Percent of actual output
Example (+- 1% of 100 C)
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The difference between the upper and lower
measurement limits of an instrument define the
device’s span
Span = (upper range limit) – (lower range limit)
Resolution is the smallest discernible increment of
output. Average resolution is given by:
Average Resolution (%) 
100
N
Where: N = total number of steps in span
100 = normalized span (%)
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Example: A tachogenerator (device used to measure
speed) gives an output that is proportional to speed. Its
ideal rating is 5 V/ 1000 rpm over a range of 0-5000 rpm
with an accuracy of 0.5% of full scale (span) Find the
ideal value of speed when the output is 21 V. Also find
the speed range that the measurement can be
expected to be in due to the measurement error.
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Determine the maximum output voltage
Vmax  n max  G
Where: Vmax = maximum output voltage
nmax = maximum speed
G = tachogenerator sensitivity (V/rpm)
Find Vmax
n max  5000rpm
G  5 V/1000 rpm
Vmax  (5000rpm) (5 V/1000 rpm) 25 V
Find ideal value of speed
Vout  21V
n ideal
V
21V
21V
 out 

 4200rpm
G
(5 V/1000 rpm) 0.005V/rpm
Accuracy +-0.5% of full scale
+-0.005(5000) = +-25 rpm
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Speed range
4200+25 = 4225 rpm
4200-25 = 4175 rpm
Ideal
speed
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A 1200 turn wire-wound potentiometer measures shaft position
over a range from -120 to +120 degrees. The output range is
0-20 volts. Find the span, the sensitivity in volts/degree, the
average resolution in volts and percent of span.
span  (120  (120 )  240
Vmax  Vmin 20  0
sensitivity 

 0.0833V/degree

span
240
100%
resolution(%) 
 0.0833% of span
1200
20 V
resolution(V) 
 0.01667V
1200
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Repeatability - measurement of dispersion of a
number of measurements (standard deviation)
Accuracy is not the same as repeatability
Example
+
+ +
+ ++
++ +
+
Not repeatable
Not accurate
+++
++
Repeatable
Not accurate
Ideal Value
+++
++
Repeatable
Accurate
Reproducibility - maximum difference between a number of
measurements taken with the same input over a time interval
Includes hysteresis, dead band, drift and repeatability
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Determining the accuracy of a measuring instrument is called
calibration. Measure output for full range of input variable. Input
could be increased then decreased to find hysteresis. Repeat input
to determine instrument repeatability.
0
1
Out put1i
0.06
I nput 2i
100
2
Out put2i
100
100.08
10
9.80
90
87. 24
20
19. 69
80
77. 26
30
29. 65
70
66. 22
40
39. 70
60
57. 12
50
49. 85
50
46. 80
60
60. 2
40
34. 70
70
70. 16
30
25. 73
80
80. 21
20
16. 75
90
90. 19
10
8.83
100
100.08
0
80
60
Output1
Output2
i
i
40
20
1
0
20
0.01
2
I nput 1i
0
20
40
60
80
Input1 , Input2
i
i
Increasing Input
Measurements
Decreasing Input
Measurements
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Plot the data
8
100
Hysteresis and Dead Band
Difference between upscale and downscale tests
called hysteresis and dead band
100
Hysteresis &
Dead Band
80
60
Output1
Output2
i
i
40
20
0
20
0
20
40
60
80
100
Input1 , Input2
i
i
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Linearity
Ideal instruments produce perfectly straight calibration
curves. Linearity is closeness of the actual calibration
curve to the ideal line.
Types of Linearity Measure
Terminalbased
line
% Output
Least-squares
minimizes the
distance
between
all data points
Least-squares
line
Zero-based
line
Average up
and down
scale values
et438b-4.pptx
% Input
10
First order instrument response
First order model
transfer function
Where :
C m (s)
G

C(s) 1  s
Cm(s) = instrument output
C(s) = instrument input
G = steady-state gain of instrument
 = instrument time constant
For step input
C(s) 
K
s
with K = step input size (1 for unit step)
Step response
KG
C m (s) 
s(1  s)
et438b-4.pptx
Exponentially
increasing
function time
constant
11
Senso r Step Resp on se
100
90%
80
Time required to
go from 10% to
90% of final value
is the rise time, tr
63.2
% Output
63.2%
60
t90 – t10 = tr
40
t90 = 4.57 S
20
10%
0
Time required to
reach 63.2% of
final value is
time constant, 
=2
0
2
4
6
8
10
Time (Seconds)
12
14
t10 = 0.22 S
16
tr=4.57 S - 0.22 S=4.35 S
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Typical Instrument time constants
Bare thermocouple in air (35 Sec)
Bare thermocouple in liquid (10 Sec)
Thermal time constant determined by thermal resistance RT
and thermal capacitance CT.  = RT∙CT
Example: A Resistance Temperature Detector (RTD) is made of
pure Platinum. It is 30.5 cm long and has a diameter of 0.25 cm.
The RTD will operate without a protective well. Its outside film
coefficient is estimated to be 25 W/m2-K. Compute: a.) the total
thermal resistance of the RTD, b.) the total thermal capacitance of
the RTD, c.) The RTD thermal time constant.
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To signal
Conditioner
a.) Find the surface area of the probe to find RT
RTD
L=30 cm
D=0.25 cm
 1m 
d  (0.25 cm) 
  0.0025m
 100cm 
 1m 
L  (30.5 cm) 
  0.305m
 100cm 
d 2 (0.0025m) 2
A1

 4.91106 m 2
4
4
A 2  dL    (0.0025m) (0.305m)  2.395103 m 2
A T  A1  A 2  4.91106 m 2  2.395103 m 2  2.4 103 m 2
ho = 25 W/m2-K
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 1  

1
  

R T  
2
3
2 
 h o  A T   (25 W/m  K )  (2.4  10 m ) 
R T  16.67 K/W
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b.) Find the volume of the probe to find CT
CT    V  Sm
Where:  = density of Platinum = 21,450 Kg/m3
V = volume of probe
Sm = specific heat of Platinum = 0.13 kJ/Kg-K
Find volume of cylinder
 d 2 
(0.0025m) 2
  L 
0.305m  1.497106 m3
V  
4
 4 
Now find the thermal capacitance
C T    V  Sm
 1000J 
CT  (21,450 Kg/m3 )  (1.497106 m3 )  (0.13 kJ/Kg - K)  

 1 kJ 
CT  4.174 J/K
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c.) Find the RTD time constant
R T  16.67 K/W
CT  4.174 J/K
  R T  CT
  (16.67 K/W) (4.174J/K)
  69.6 S
RTD Resp on se
100

5 
RTD Response curve
80
Percent Output
63.2
60
40
20
0
0
100
200
Time (Seconds)
300
400
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Common mode voltages are voltages that have the same
magnitude and phase shift and appear at the inputs of a data
acquisition system. Common mode voltages mask low level signals
from low gain transducers.
Vcmn
Data
recording
system
Vs
Vcmn
Sensor and signal
conditioning source
Induced voltage
and noise
Common mode voltages also appear on shielded systems
due to differences between input potentials
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Common mode voltage due to ground
Differential Amp
Vd
V-
+
Vo
Vd  V   V 
Vcmg
V  V

2
V+
Total common mode voltage Vcm = Vcmn+Vcmg
OP AMP differential inputs designed to reject common mode
voltages. Amplify only Vd = V+ - V-.
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Define:
Ac = gain of OP AMP to common mode signals
(designed to be low)
Ad = differential gain of OP AMP. Typically high
(Ad = 200,000 for 741)
Ideal OP AMPs have infinite Ad and zero Ac
Common mode rejection ratio (CMRR) is a measure of quality for
non-ideal OP AMPs. Higher values are better.
CMRR 
Ad
Ac
Where Ad = differential gain
Ac = common mode gain
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Common Mode Rejection (CMR) calculation
 Ad 
CMR  20 log   20 logCMRR 
 Ac 
CMR units are db. Higher values of CMR are better.
Example: A typical LM741 OP AMP has a differential
gain of 200,000. The typical value of common mode
rejection is 90 db. What is the typical value of common
mode gain for this device
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From problem statement Vd = 200,000 CMR = 90 db
 A  Solve for A by using
CMR  20 log  d  the anitlogc
 Ac 
A 
CMR
 log d  Raise both sides to power
20
 A c  of 10
10
CMR
20
 Ad 
   Solve for Ac
 Ac 
Ad
10
200 ,000
10
90
20
CMR
20

 A c Plug in values and get numerical
solution
200 ,000 200 ,000

 6.32  A c
4.5
10
31,623
et438b-4.pptx
Common mode gain is
6.32 for typical LM741
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Characteristics of Instrumentation Amplifiers
- Amplify dc and low frequency ac (f<1000 Hz)
- Input signal may contain high noise level
- Sensors may low signal levels. Amp must have high
gain.
- High input Z to minimize loading effects
- Signal may have high common mode voltage with
respect to ground
Differential amplifier circuit constructed from OP AMPs are the
building block of instrumentation amplifiers
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Input/output Formula
 R 2  R1   R 4 
R 
     V2   2   V1
Vo  
 R1 
 R 4  R 3   R1 
To simplify let
R1 = R3 and R2 = R4
R 
V0   2   (V2  V1 )
 R1 
Amplifies the difference between
+/ - terminals
Polarity of OP AMP input indicates
order of subtraction
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Practical considerations of basic differential amplifiers
- Resistor tolerances affect the CMRR of OP AMP
circuit. Cause external unbalance that decreases
overall CMRR.
- Input resistances reduce the input impedance of
OP AMP
- Input offset voltages cause errors in high gain
applications
- OP AMPs require bias currents to operate
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et438b-4.pptx
To minimize the loading effects of the OP AMP input resistors, their
values should be at least 10x greater than the source impedance
Example: Determine the loading effects of differential amp Input
on the voltage divider circuit. Compare the output predicted by
differential amplifier formula to detailed analysis of circuit.
1 Vdc
I
+
R2= 5kW
-
R1= 5kW
R4
10kW
R6 =10kW
 R6 
V0  
  (V2  V1 )
R
4


VR2
R5
10kW
R3= 5kW
Assume no loading effects and
use the OP AMP gain formula
(V2  V1 )  (V  V )  VR 2
 R2
 5kW




VR 2  

V

1 V



R7 =10kW
 R1  R 2  R 3 
 5kW  5kW  5kW 
  5kW 
VR 2  
1 V  -0.3333V

 15kW 
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Find the output ignoring the loading effects that the OP AMP has on
the voltage divider.
 R6 
V0  
   VR 2 
 R4 
 10 kW 
V0  
   0.333  0.333V
 10 kW 
Now solve the circuit and include the loading effects of the OP AMP
input resistors. Use nodal analysis and check with simulation.
Remember the rules of ideal OP AMPs:
Iin = 0 and V+=V-
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Solution using nodal analysis
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Solve simultaneous equations and determine percent
error due to loading
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Results of operating point analysis in LTSpice
V0 =0.286 V
V1 =0.514 V
V2 =0.229 V
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Dc motor draws a current of 3A dc when developing full
mechanical power. Find the gain of the last stage of the circuit
so that the output voltage is 5 volts when the motor draws full
power. Also compute the power dissipation of the shunt resistor
I=3 A 12Vdc
+
0.1W
820 kW
10 kW
Rf = ?
100 kW
100 kW
0 - 5 Vdc
820 kW
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Example Solution
+ Vd
+
-
0.1W
12Vdc
820 kW
100 kW
2.46 V
0.300 V
I=3 A
100 kW
820 kW
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10 kW
Rf = ?
2.46 V
0 - 5 Vdc
Compute power
dissipation at full load
I= 3 A so….
Use 1 Watt or greater
Standard value
Rf is a non-standard value. Use 8.2
kΩ resistor and 5 kΩ potentiometer.
Calibrate with 300 mV source
Until 5.00 V output is achieved
Caution: Note the maximum differential
Voltage specification of OP AMP.
(30 V for LM741)
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Simulated with Circuit Maker (Student Version)
300.0mV
DC V
12V
R7
10k
R2
820k
R4
100k
R6
0.1
R5
100k
Ra
3.9
-10V
-10V
U1A
LM324
R8
5k 43%
U1B
LM324
4.992 V
DC V
Vo
+
+
R3
820k
R1
8.2k
10V
et438b-4.pptx
2.453 V
DC V
10V
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