1. Lecture Notes Part 4 ET 483b Sequential Control and Data Acquisition
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2. Measuring Instrument Characteristics
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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3. Measuring Instrument Characteristics
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:
Where: N = total number of steps in span
100 = normalized span (%)
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4. Instrument Characteristics
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 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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5. Example Solution Determine the maximum output voltage
Where: Vmax = maximum output voltage
nmax = maximum speed
G = tachogenerator sensitivity (V/rpm)
Find Vmax
Find ideal value of speed
Ideal
speed
Speed range
= 4225 rpm
= 4175 rpm
Accuracy +-0.5% of full scale
(5000) = +-25 rpm
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6. Span, Resolution, and Sensitivity
A 1200 turn wire-wound potentiometer measures shaft position over a range from -120 to +120 degrees. The output range is volts. Find the span, the sensitivity in volts/degree, the average resolution in volts and percent of span.
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7. Repeatability and Accuracy
Repeatability - measurement of dispersion of a number of measurements (standard deviation)
Accuracy is not the same as repeatability
Ideal Value
Example + + ++ +
Not repeatable
Not accurate
Repeatable
Accurate
Repeatable
Not 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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8. Calibration Curves
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. 1 2 2 1
Increasing Input
Measurements
Decreasing Input
Measurements
Plot the data
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9. Calibration Curve Characteristics
Hysteresis and Dead Band
Difference between upscale and downscale tests called hysteresis and dead band
Hysteresis &
Dead Band
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10. Average up and down scale values
Calibration Curve Characteristics
Linearity
Ideal instruments produce perfectly straight calibration curves. Linearity is closeness of the actual calibration curve to the ideal line.
Types of Linearity Measure
Least-squares
line
% Input
% Output
Terminal-based
line
Least-squares
minimizes the
distance between
all data points
Zero-based
line
Average up and down scale values
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11. Dynamic Characteristics
First order instrument response
First order model
transfer function
Where : Cm(s) = instrument output
C(s) = instrument input
G = steady-state gain of instrument
t = instrument time constant
For step input
with K = step input size (1 for unit step)
Exponentially increasing function time constant
Step response
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12. Dynamic Characteristics
Time required to
reach 63.2% of
final value is
time constant, t
t=2
90%
Time required to
go from 10% to
90% of final value
is the rise time, tr
t90 – t10 = tr
63.2%
t90 = 4.57 S
10%
t10 = 0.22 S
tr=4.57 S S=4.35 S
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13. Dynamic Characteristics
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. t = 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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14. Example Solution a.) Find the surface area of the probe to find RT
To signal
Conditioner
a.) Find the surface area of the probe to find RT
RTD
L=30 cm
D=0.25 cm
ho = 25 W/m2-K
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15. Example Solution b.) Find the volume of the probe to find CT
Where: r = 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
Now find the thermal capacitance
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16. Example Solution c.) Find the RTD time constant RTD Response curve
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17. Common Mode Voltages in Instrumentation and Control
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.
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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18. Common Mode Voltage Due to Inputs
Common mode voltage due to ground - + V+ V- Vo Vd
Differential Amp
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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19. Common Mode Voltages and OP AMPs
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.
Where Ad = differential gain
Ac = common mode gain
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20. Common Mode Rejection Common Mode Rejection (CMR) calculation
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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21. Common Mode Rejection Example Solution
From problem statement Vd = 200,000 CMR = 90 db
Solve for Ac by using
the anitlog
Raise both sides to power
of 10
Solve for Ac
Plug in values and get numerical
solution
Common mode gain is 6.32 for typical LM741
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22. Difference Amplifier and Instrumentation Amplifiers
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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23. Basic Difference Amplifiers
Input/output Formula
To simplify let
R1 = R3 and R2 = R4
Amplifies the difference between +/ - terminals
Polarity of OP AMP input indicates
order of subtraction
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24. Basic Difference Amplifiers
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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25. Difference Amplifier Loading Effects
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.
R1= 5kW
R3= 5kW
R2= 5kW R4
10kW R5
R6 =10kW
R7 =10kW
1 Vdc
Assume no loading effects and
use the OP AMP gain formula I +
VR2 -
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26. Difference Amplifier Loading Effects
Find the output ignoring the loading effects that the OP AMP has on
the voltage divider.
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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27. Difference Amplifier Loading Effects
Solution using nodal analysis
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28. Difference Amplifier Loading Effects
Solve simultaneous equations and determine percent error due to loading
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29. Difference Amplifier Loading Effects
Results of operating point analysis in LTSpice
V0 =0.286 V
V1 =0.514 V
V2 =0.229 V
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30. Example: Measuring dc current with a current shunt
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
12Vdc
I=3 A
0.1W
100 kW
820 kW
10 kW
Rf = ?
0 - 5 Vdc +
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31. Example: Measuring dc current with a current shunt
Example Solution
12Vdc
I=3 A
0.1W
100 kW
820 kW +
2.46 V
+ Vd -
0.300 V
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32. Example Solution
10 kW
Rf = ?
0 - 5 Vdc
2.46 V
Compute power
dissipation at full load
I= 3 A so….
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)
Use 1 Watt or greater
Standard value
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33. Example Solution Simulation
Simulated with Circuit Maker (Student Version)
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