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et438b-4.pptx
1Lecture Notes Part 4
ET 483b Sequential Control and Data Acquisition
et438b-4.pptx 2
Measuring Instrument Characteristics
Static Characteristics
Error = (measured value) – (ideal value)
Ways of expressing instrument error
1.) In terms of measured variableExample ( + 1 C, -2 C)
2.) Percent of spanExample (0.5% of span)
3.) Percent of actual output Example (+- 1% of 100 C)
et438b-4.pptx 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:
N
100 (%) Resolution Average
Where: N = total number of steps in span100 = normalized span (%)
et438b-4.pptx 4
Instrument CharacteristicsExample: 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.
et438b-4.pptx
Example Solution
5
Determine the maximum output voltage
GnV maxmax
Where:Vmax = maximum output voltagenmax = maximum speedG = tachogenerator sensitivity (V/rpm)
V 25rpm) V/1000 5(rpm) 5000(V
rpm V/1000 5 G
rpm 5000n
max
max
Find ideal value of speed
Idealspeed
Find Vmax
rpm 4200V/rpm 0.005
V 21
rpm) V/1000 5(
V 21
G
Vn
V 21V
out ideal
out
Accuracy +-0.5% of full scale +-0.005(5000) = +-25 rpm
Speed range 4200+25 = 4225 rpm 4200-25 = 4175 rpm
et438b-4.pptx 6
Span, Resolution, and SensitivityA 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.
V 01667.01200
V 20(V)resolution
span of % 0833.01200
100%(%)resolution
V/degree 0833.0240
020
span
VVysensitivit
240)120(120(span
minmax
et438b-4.pptx
Repeatability and Accuracy
7
Repeatability - measurement of dispersion of a number of measurements (standard deviation)Accuracy is not the same as repeatability
Example+++++
+ +
++ +
+
+ ++
++
+++++
Not repeatableNot accurate
RepeatableNot accurate
RepeatableAccurate
Ideal Value
Reproducibility - maximum difference between a number of measurements taken with the same input over a time interval
Includes hysteresis, dead band, drift and repeatability
et438b-4.pptx 8
Calibration CurvesDetermining 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.
Increasing Input Measurements
Input1i
0
10
20
30
40
50
60
70
80
90
100
Output1i
0.06
9.80
19.69
29.65
39.70
49.85
60.2
70.16
80.21
90.19
100.08
1Input2i
100
90
80
70
60
50
40
30
20
10
0
Output2i
100.08
87.24
77.26
66.22
57.12
46.80
34.70
25.73
16.75
8.83
0.01
Decreasing InputMeasurements
2
0 20 40 60 80 10020
0
20
40
60
80
100
Output1i
Output2i
,Input1iInput2
i
Plot the data
1
2
et438b-4.pptx 9
Calibration Curve Characteristics
Hysteresis and Dead BandDifference between upscale and downscale tests called hysteresis and dead band
0 20 40 60 80 10020
0
20
40
60
80
100
Output1i
Output2i
,Input1iInput2
i
Hysteresis &Dead Band
et438b-4.pptx10
Calibration Curve Characteristics
LinearityIdeal instruments produce perfectly straight calibration curves. Linearity is closeness of the actual calibration curve to the ideal line.
Types of Linearity Measure
% Input
% O
utp
ut
Average up and down
scale values
Zero-basedline
Terminal-basedline
Least-squares
line
Least-squaresminimizes thedistance betweenall data points
et438b-4.pptx 11
Dynamic CharacteristicsFirst order instrument responseFirst order model
transfer function s1
G
)s(C
)s(Cm
For step inputs
K)s(C
Step response)s1(s
KG)s(Cm
Where : Cm(s) = instrument output C(s) = instrument input G = steady-state gain of instrument
= instrument time constant
with K = step input size (1 for unit step) Exponentially
increasing function time
constant
12et438b-4.pptx
Dynamic CharacteristicsTime required toreach 63.2% offinal value is time constant, tt=2
Time required togo from 10% to90% of final valueis the rise time, tr
t90 – t10 = tr
0 2 4 6 8 10 12 14 160
20
40
60
80
100Sensor Step Response
Time (Seconds)
% O
utpu
t
63.2
63.2%
90%
10%
t90 = 4.57 S
t10 = 0.22 S
tr=4.57 S - 0.22 S=4.35 S
et438b-4.pptx 13
Dynamic CharacteristicsTypical 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.
Example Solution
et438b-4.pptx 14
RTD
To signalConditioner
a.) Find the surface area of the probe to find RT
L=30 cm
D=0.25 cm
m 305.0cm 100
m 1cm) 5.30(L
m 0025.0cm 100
m 1cm) 25.0(d
23232621T
232
2622
1
m 104.2m 102.395m 1091.4AAA
m 102.395m) (0.305m) 0025.0(dLA
m 1091.44
)m 0025.0(
4
dA
ho = 25 W/m2-K
K/W 67.16R
)m 104.2()K W/m25(
1
Ah
1R
T
232To
T
et438b-4.pptx 15
Example Solutionb.) Find the volume of the probe to find CT
mT SVC
Where: r = density of Platinum = 21,450 Kg/m3
V = volume of probeSm = specific heat of Platinum = 0.13 kJ/Kg-K
Find volume of cylinder
3622
m 10497.1m 305.04
)m 0025.0(L
4
dV
Now find the thermal capacitance
J/K 174.4C
kJ 1
J 1000)K-kJ/Kg 13.0()m 10497.1()Kg/m 450,21(C
SVC
T
363T
mT
et438b-4.pptx 16
Example Solutionc.) Find the RTD time constant
S 6.69
J/K) (4.174K/W) 67.16(
CR
J/K 174.4C
K/W 67.16R
TT
T
T
0 100 200 300 4000
20
40
60
80
100RTD Response
Time (Seconds)
Per
cent
Out
put
63.2
5
RTD Response curve
17et438b-4.pptx
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 recordingsystem
Vs
Vcmn
Vcmn
Induced voltage and noise
Sensor and signal conditioning source
Common mode voltages also appear on shielded systemsdue to differences between input potentials
et438b-4.pptx 18
Common Mode Voltage Due to Inputs
Common mode voltage due to ground
2
VVV
VVV
cmg
d
-
+
V+
V-
VoVd
Differential Amp
Total common mode voltage Vcm = Vcmn+Vcmg
OP AMP differential inputs designed to reject common mode voltages. Amplify only Vd = V+ - V-.
et438b-4.pptx
Common Mode Voltages and OP AMPs
19
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.
c
d
A
ACMRR
Where Ad = differential gainAc = common mode gain
et438b-4.pptx
Common Mode Rejection
20
Common Mode Rejection (CMR) calculation
CMRRlog20A
Alog20CMR
c
d
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
et438b-4.pptx 21
Common Mode Rejection Example Solution
From problem statement Vd = 200,000 CMR = 90 db
c
d
A
Alog20CMR
Solve for Ac by using the anitlog
CMR A
Ad
c20
log Raise both sides to
powerof 10
10 20
CMRd
c
A
A
Solve for Ac
AAd
CMR c
10 20
Plug in values and get numericalsolution
c5.420
90 A32.6623,31
000,200
10
000,200
10
000,200 Common mode gain is
6.32 for typical LM741
et438b-4.pptx
Difference Amplifier and Instrumentation Amplifiers
22
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
et438b-4.pptx
Basic Difference Amplifiers
23
Amplifies the difference between +/ - terminals
Input/output Formula
To simplify let R1 = R3 and R2 = R4
V R
RV
R
R
RR
RRV 1
1
22
1
4
34
12o
)VV(R
RV 12
1
20
Polarity of OP AMP input indicatesorder of subtraction
et438b-4.pptx24
Basic Difference AmplifiersPractical 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 ofOP AMP
- Input offset voltages cause errors in high gainapplications
- OP AMPs require bias currents to operate
Difference Amplifier Loading Effects
et438b-4.pptx 25
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
R410kW
R510kW
R6 =10kW
R7 =10kW
1 Vdc Assume no loading effects anduse the OP AMP gain formula
)VV(4R
6RV 120
VR22R12 V)VV()VV(
V -0.3333V 1k15
k5V
V 1k5k5k5
k5V
3R2R1R
2RV
2R
2R
+
I
-
et438b-4.pptx26
Difference Amplifier Loading EffectsFind the output ignoring the loading effects that the OP AMP has onthe voltage divider.
V 333.0333.0k 10
k 10V
V4R
6RV
0
2R0
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-
et438b-4.pptx 27
Difference Amplifier Loading Effects
Solution using nodal analysis
et438b-4.pptx 28
Difference Amplifier Loading Effects
Solve simultaneous equations and determine percent error due to loading
et438b-4.pptx 29
Difference Amplifier Loading Effects
Results of operating point analysis in LTSpice
V1 =0.514 V
V2 =0.229 V
V0 =0.286 V
et438b-4.pptx 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
12VdcI=3 A
0.1W
100 kW
100 kW
820 kW
820 kW
10 kW Rf = ?
0 - 5 Vdc
+
et438b-4.pptx 31
Example: Measuring dc current with a current shunt
Example Solution
12VdcI=3 A
0.1W
100 kW
100 kW
820 kW
820 kW
+ 2.46 V
0.3
00
V
+ V
d
-
et438b-4.pptx 32
Example Solution
10 kW Rf = ?
0 - 5 Vdc
2.46 V
Caution: Note the maximum differential Voltage specification of OP AMP. (30 V for LM741)
Rf is a non-standard value. Use 8.2 kΩ resistor and 5 kΩ potentiometer. Calibrate with 300 mV sourceUntil 5.00 V output is achieved
Compute powerdissipation at full loadI= 3 A so….
Use 1 Watt or greaterStandard value
et438b-4.pptx 33
Example Solution Simulation
DC V 300.0mV
DC V 2.453 V
DC V 4.992 V
10V
-10V
Vo
R85k 43%
+
U1BLM324
12V
10V
+
U1ALM324
-10V
Ra3.9
R18.2k
R710k
R60.1 R5
100k
R4100k
R3820k
R2820k
Simulated with Circuit Maker (Student Version)