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Lecture Notes Part 4 ET 483b Sequential Control and Data Acquisition

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1 Lecture Notes Part 4 ET 483b Sequential Control and Data Acquisition

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) 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: Where: N = total number of steps in span 100 = normalized span (%) et438b-4.pptx

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. et438b-4.pptx

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 et438b-4.pptx

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. et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

9 Calibration Curve Characteristics
Hysteresis and Dead Band Difference between upscale and downscale tests called hysteresis and dead band Hysteresis & Dead Band et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

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. et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

16 Example Solution c.) Find the RTD time constant RTD Response curve

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 et438b-4.pptx

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-. et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

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 - et438b-4.pptx

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- 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 V0 =0.286 V V1 =0.514 V V2 =0.229 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 12Vdc I=3 A 0.1W 100 kW 820 kW 10 kW Rf = ? 0 - 5 Vdc + et438b-4.pptx

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 et438b-4.pptx

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 et438b-4.pptx

33 Example Solution Simulation
Simulated with Circuit Maker (Student Version) et438b-4.pptx

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