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2102311 Electrical Measurement and Instruments (Part II)

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2102311 Electrical Measurement and Instruments (Part II) 2102311 Electrical Measurement and Instruments (Part II) ¾ Bridge Circuits (DC and AC) ¾ Electronic Instruments (Analog & Digital) ¾ Signal Generators ¾ Frequency and Time Interval Measurements ¾ Introduction to Transducers อาภรณ ธีรมงคลรศมั ี ตึกไฟฟา 6 ชนั้ หอง 306 Textbook: -A.D. Helfrick, and W.D. Cooper, “Modern Electronic Instrumentation and Measurement Techniques” Prentice Hall, 1994. - D.A. Bell, “Electronic Instrumentation and Measurements”, 2nd ed., Prentice Hell, 1994. ResistorResistor TypesTypes Importance parameters Value Tolerance Power rating Temperature coefficient Type Values (Ω) Power rating Tolerance (%) Temperature picture (W) coefficient (ppm/°C) Wire wound 10m~3k (power) 3~1k ±1~±10 ±30~±300 Wire wound (precision) 10m~1M 0.1~1 ±0.005~±1 ±3~±30 Carbon film 1~1M 0.1~3 ±2~±10 ±100~±200 Metal film 100m~1M 0.1~3 ±0.5~±5 ±10~±200 Metal film (precision) 10m~100k 0.1~1 ±0.05~±5 ±0.4~±10 Metal oxide film 100m~100k 1~10 ±2~±10 ±200~±500 Data: Transistor technology (10/2000) Resistor Values Color codes Resistor Values Alphanumeric 4 band color codes Color Digit Multiplier Tolerance Temperature (%) coefficient (ppm/°C) Most sig. fig. of value Silver 10-2 ±10 K Tolerance - - - Least sig. fig. Multiplier Gold 10-1 J of value ±5 Ex. Black 0 100 - - ±250 K Brown 1 101 ±1 F ±100 H Red 2 102 ±2 G ±50 G Green 3 Red Orange 3 10 - ±15 D Blue Brown - Yellow 4 104 - ±25 F R = 560 Ω ±2% Green 5 5 D E 10 ±0.5 ±20 Alphanumeric Blue 6 106 ±0.25 C ±10 C Violet 7 107 ±0.1 B ±5 B R, K, M, G, and T = 0 3 6 9 12 Gray 8 108 - ±1 A x10 , x10 , x10 , x10 , and x10 - White 9 109 - Ex. 6M8 = 6.8 x 106 Ω - - - ±20 M 59Ρ04 = 59.04 Ω Data: Transistor technology (10/2000) Resistor Values Commonly available resistance for a fixed resistor R = x ± %∆x Tolerance Nominal value Ex. 1 kΩ±10% ≡ 900-1100 Ω For 10% resistor 10, 12, 15, 18, … 10 12 15 R E n R ≈√10 For 10% resistor E = 12 n = 0; R = 1.00000… where E = 6, 12, 24, 96 n = 1; R = 1.21152… for 20, 10, 5, 1% tolerance n = 2; R = 1.46779… n = 0, 1, 2, 3, … n = 3; R = 1.77827… ResistanceResistance MeasurementMeasurement TechniquesTechniques Bridge circuit Voltmeter-ammeter Substitution Ohmmeter Voltmeter-ammeter V V A A R R Substitution A A Decade resistance Unknow Supply Supply box substituted in resistance Rx place of the unknown Voltmeter-ammeter method Pro and con: •Simple and theoretical oriented •Requires two meter and calculations •Subject to error: Voltage drop in ammeter (Fig. (a)) Current in voltmeter (Fig. (b)) I + VA - A A + + + + IV Ix I V V V V R VS V Rx Vx S x - - - - - Fig. (b) Fig. (a) VV R VVVV+ R == = x Measured R : xA A Measured Rx: meas x RRmeas == =+x I II++1/ II I II x VVx if I >>I RR≈ if Vx>>VA RRmeas ≈ x x V meas x Therefore this circuit is suitable for measure Therefore this circuit is suitable for measure large resistance small resistance Ohmmeter •Voltmeter-ammeter method is rarely used in practical applications (mostly used in Laboratory) •Ohmmeter uses only one meter by keeping one parameter constant Example: series ohmmeter Resistance to be measured Nonlinear scale Standard resistance 15k R k x 45 5k R 50 Battery 1 Meter Infinity 25 75 Rm resistance 0 VS ∞ 1 0 0 0 V µ RRR=−−s Meter A x I 1 m Basic series ohmmeter Ohmmeter scale Basic series ohmmeter consisting of a PMMC and a series-connected standard resistor (R1). When the ohmmeter terminals are shorted (Rx = 0) meter full scale defection occurs. At half scale defection Rx = R1 + Rm, and at zero defection the terminals are open-circuited. Bridge Circuit Bridge Circuit is a null method, operates on the principle of comparison. That is a known (standard) value is adjusted until it is equal to the unknown value. Bridge Circuit DC Bridge AC Bridge (Resistance) Inductance Capacitance Frequency Wheatstone Bridge Maxwell Bridge Schering Bridge Wien Bridge Kelvin Bridge Hay Bridge Megaohm Bridge Owen Bridge Etc. Wheatstone Bridge and Balance Condition Suitable for moderate resistance values: 1 Ω to 10 MΩ A Balance condition: No potential difference across the R1 R2 galvanometer (there is no current through the galvanometer) I1 I2 Under this condition: V = V V D B AD AB I IR11= IR 2 2 I3 4 And also V = V R DC BC 3 R4 IR33= IR 44 where I1, I2, I3,and I4 are current in resistance arms respectively, since I = I and I = I C 1 3 2 4 RR R 12or 2 = RRRx ==43 RR34 R1 Example 1 Ω 1 Ω 1 Ω 1 Ω 12 V 12 V 1 Ω 1 Ω 2 Ω 2 Ω (a) Equal resistance (b) Proportional resistance 1 Ω 10 Ω 1 Ω 10 Ω 12 V 12 V 2 Ω 20 Ω 2 Ω 10 Ω (c) Proportional resistance (d) 2-Volt unbalance Measurement Errors RR22±∆ 1. Limiting error of the known resistors RRRx =±∆()33 RR11±∆ Using 1st order approximation: RRR212∆∆∆R3 RRx =±±±3 1 A RRRR1123 R 2. Insufficient sensitivity of Detector 1 R2 3. Changes in resistance of the bridge arms due to the heating effect (I2R) or V D B temperatures 4. Thermal emf or contact potential in the bridge circuit R3 Rx 5. Error due to the lead connection C 3, 4 and 5 play the important role in the measurement of low value resistance Example In the Wheatstone bridge circuit, R3 is a decade resistance with a specified in accuracy ±0.2% and R1 and R2 = 500 Ω ± 0.1%. If the value of R3 at the null position is 520.4 Ω, determine the possible minimum and maximum value of RX RRR∆∆∆R SOLUTION Apply the error equation 2123 RRx =±±±3 1 RRRR1123 520.4× 500 0.1 0.1 0.2 Rx =±±±=±=±1 520.4( 1 0.004) 520.4 0.4% 500 100 100 100 Therefore the possible values of R3 are 518.32 to 522.48 Ω Example A Wheatstone bridge has a ratio arm of 1/100 (R2/R1). At first balance, R3 is adjusted to 1000.3 Ω. The value of Rx is then changed by the temperature change, the new value of R3 to achieve the balance condition again is 1002.1 Ω. Find the change of Rx due to the temperature change. R2 1 SOLUTION At first balance: RRxold= 3 =×=Ω 1000.3 10.003 R1 100 R2 1 After the temperature change: RRxnew= 3 =×=Ω 1002.1 10.021 R1 100 Therefore, the change of Rx due to the temperature change is 0.018 Ω Sensitivity of Galvanometer A galvanometer is use to detect an unbalance condition in Wheatstone bridge. Its sensitivity is governed by: Current sensitivity (currents per unit defection) and internal resistance. consider a bridge circuit under a small unbalance condition, and apply circuit analysis to solve the current through galvanometer Thévenin Equivalent Circuit Thévenin Voltage (VTH) A VVVIRIRCD=−= AC AD 11 − 2 2 I1 I 2 VV where II==and R1 R2 12 R13++RRR 24 VS CDG R3 R4 Therefore RR12 VVVTH== CD − R13++RRR 24 B Sensitivity of Galvanometer (continued) Thévenin Resistance (RTH) R1 A R2 R =+RRRR// // CDTH 1324 R3 R4 B Completed Circuit R TH C V I = TH g R +R VTH TH g I = g R + R VTH G TH g D where Ig = the galvanometer current Rg = the galvanometer resistance Example 1 Figure below show the schematic diagram of a Wheatstone bridge with values of the bridge elements. The battery voltage is 5 V and its internal resistance negligible. The galvanometer has a current sensitivity of 10 mm/µA and an internal resistance of 100 Ω. Calculate the deflection of the galvanometer caused by the 5-Ω unbalance in arm BC SOLUTION The bridge circuit is in the small unbalance condition since the value of resistance in arm BC is 2,005 Ω. A Thévenin Voltage (VTH) 1000 Ω 100 Ω 100 1000 R1 R2 VTH = VAD −VAC = 5 V × − 5 V D G C 100 + 200 1000 + 2005 R R 3 4 ≈ 2.77 mV 200 Ω 2005 Ω B Thévenin Resistance (RTH) (a) 100 Ω A 1000 Ω RTH =100 // 200+=Ω 1000 // 2005 734 C D 200 Ω 2005 Ω The galvanometer current B V 2.77 mV (b) TH I g == =3.32 µA RTH= 734 Ω C RRTH+Ω+Ω g 734 100 Ig=3.34 µA V TH G Galvanometer deflection 2.77 mV Rg= 100 Ω D 10 mm d =×=3.32 µA 33.2 mm (c) µA Example 2 The galvanometer in the previous example is replaced by one with an internal resistance of 500 Ω and a current sensitivity of 1mm/µA. Assuming that a deflection of 1 mm can be observed on the galvanometer scale, determine if this new galvanometer is capable of detecting the 5-Ω unbalance in arm BC SOLUTION Since the bridge constants have not been changed, the equivalent circuit is again represented by a Thévenin voltage of 2.77 mV and a Thévenin resistance of 734 Ω. The new galvanometer is now connected to the output terminals, resulting a galvanometer current. VTH 2.77 mV I g == =2.24 µA RRTH+Ω+Ω g 734 500 The galvanometer deflection therefore equals 2.24 µA x 1 mm/µA = 2.24 mm, indicating that this galvanometer produces a deflection that can be easily observed.
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