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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) et438b-4.pptx 2 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 (%) et438b-4.pptx 3 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. et438b-4.pptx 4 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 5000 rpm G 5 V/1000 rpm Vmax (5000 rpm) (5 V/1000 rpm) 25 V Find ideal value of speed Vout 21 V n ideal V 21 V 21 V out 4200 rpm G (5 V/1000 rpm) 0.005 V/rpm Accuracy +-0.5% of full scale +-0.005(5000) = +-25 rpm et438b-4.pptx Speed range 4200+25 = 4225 rpm 4200-25 = 4175 rpm Ideal speed 5 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 sensitivit y 0.0833 V/degree span 240 100% resolution (%) 0.0833 % of span 1200 20 V resolution (V) 0.01667 V 1200 et438b-4.pptx 6 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 et438b-4.pptx 7 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 Output1i 0.06 Input2i 100 2 Output2i 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 Increasing Input Measurements 0.01 Decreasing Input Measurements et438b-4.pptx 2 Input1i 80 60 Output1 Output2 i i 40 20 1 0 20 0 20 40 60 80 Input1 , Input2 i i 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 et438b-4.pptx 9 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 Sensor Step Response 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 et438b-4.pptx 12 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. et438b-4.pptx 13 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) 0.0025 m 100 cm 1m L (30.5 cm) 0.305 m 100 cm d 2 (0.0025 m) 2 A1 4.9110 6 m 2 4 4 A 2 dL (0.0025 m) (0.305 m) 2.395 10 3 m 2 A T A1 A 2 4.9110 6 m 2 2.395 10 3 m 2 2.4 10 3 m 2 ho = 25 W/m2-K et438b-4.pptx 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 14 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 (0.0025 m) 2 L 0.305 m 1.497 106 m3 V 4 4 Now find the thermal capacitance C T V Sm 1000 J C T (21,450 Kg/m 3 ) (1.497 10 6 m 3 ) (0.13 kJ/Kg - K ) 1 kJ C T 4.174 J/K et438b-4.pptx 15 c.) Find the RTD time constant R T 16.67 K/W CT 4.174 J/K R T CT (16.67 K/W) (4.174 J/K) 69.6 S RTD Response 100 5 RTD Response curve 80 Percent Output 63.2 60 40 20 0 0 100 200 Time (Seconds) 300 400 et438b-4.pptx 16 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 et438b-4.pptx 17 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-. et438b-4.pptx 18 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 et438b-4.pptx 19 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 et438b-4.pptx 20 From problem statement Vd = 200,000 CMR = 90 db A d Solve for Ac by using CMR 20 log the anitlog 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 21 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 22 Input/output Formula R 2 R1 R 4 R2 V2 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 et438b-4.pptx 23 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 24 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 -0.3333 V 15kW et438b-4.pptx 25 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 0.333 V 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- et438b-4.pptx 26 Solution using nodal analysis et438b-4.pptx 27 Solve simultaneous equations and determine percent error due to loading et438b-4.pptx 28 Results of operating point analysis in LTSpice V0 =0.286 V V1 =0.514 V V2 =0.229 V et438b-4.pptx 29 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 et438b-4.pptx 30 Example Solution + Vd + - 0.1W 12Vdc 820 kW 100 kW 2.46 V 0.300 V I=3 A 100 kW 820 kW et438b-4.pptx 31 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) et438b-4.pptx 32 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 33