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  • Three basic rules: Connect the meter in series with the load. Failure to observe this rule may result in permanent damage to the instrument. Use a meter with a full-scale current rating that is greater than the maximum current expected. Use a meter with internal resistance lower than that measured. (i.e., <1:10) Ammeter circuit :
  • 3059

    1. 1. │ Topic 2 │ Voltage, current, resistance and power measurements EE2201 Measurements & Instrumentation
    2. 2. Learning objectives:  Operating principles of moving coil meter  DC current and voltage measurement, shunt and multiplier  Resistance measurement, operating principles of ohmmeter  Multi-range ammeter and voltmeter, multimeter  Power measurement  Comparison of analogue and digital instruments  DC Wheatstone Bridge and its application in overcoming loading effect
    3. 3. Sensing Equipment: Moving Coil Instrument  To measure the value of voltage and current, sensing equipment is necessary. Permanent-magnet moving-coil (PMMC) instrument is a simple sensing equipment. (Fig. 2-1)  It consists of a coil of copper wire suspended in the field of a permanent magnet.  Current in the coil produces a magnetic field that interacts with the field from the magnet, and causes the coil to deflect in an anti-clockwise direction.  A pointer connected to the coil deflects over a calibrated scale, indicating the amount of current flowing in the coil.
    4. 4. Construction of PMMC instrument Fig. 2-1
    5. 5. Construction of PMMC instrument  Permanent magnet : provides two different poles at pole shoes to generate magnetic field.  Soft-iron core : to minimize the air gap so as to provide strongest level of magnetic flux to the core.  Moving coil : to sense the measured current. The larger the current, the more the pointer deflects.  Pointer : to indicate the measured value. It has counter weight attached at the end to provide mechanical balance of moving system.  Spiral spring : to generate controlling force to balance the deflecting force of pointer. Also it has a mechanical zero control to set the pointer at zero scale.
    6. 6. Construction of PMMC  Scale : to indicate the measured quantity. The scale should be calibrated to the wanted range.  Deflection Fundamental :  Two forces are involved : deflecting force and controlling force.  Deflecting force is generated by the current flowing into coil (Fig. 2-2a).  Controlling force is generated by spiral spring to balance the deflecting force (Fig. 2-2b)
    7. 7. Fig. 2-2a
    8. 8. Fig. 2-2b At equilibrium, Deflecting = Controlling, torque, Td torque, Tc Since Td ∝ I and Tc ∝ θ ∴ I ∝ θ i.e. current proportional to angle of deflection, the meter has a linear scale
    9. 9. Measurement of DC Current Three basic rules: • Connect the meter in series with the load. Failure to observe this rule may result in permanent damage to the instrument. • Use a meter with a full-scale current rating that is greater than the maximum current expected. • Use a meter with internal resistance lower than that measured. (i.e., <1:100)
    10. 10. Ammeter Circuit  The current flowing in the meter is : where Im= meter current (A) E = source voltage (V) Rs = internal resistance of power source (Ω) Rm = internal resistance ammeter (Ω) R1 = load resistance (Ω)  If R1 >> Rs and Rm , 1 RRR E I ms m ++ = 1R E Im = Rs E Rm R1 Fig. 2-3 Ammeter circuit Im
    11. 11. Example 2-1 Find the current flowing in the meter in the following circuit if R1 = 15 kΩ. Solution : Since R1 is very much greater than RS and Rm, Im = E/R1 = 10/(1.5 x 104 ) = 6.7 x 10 -4 A = 0.67mA R1 Rs=9Ω E=10V 0-1 mA Rm= 68 Ω Im
    12. 12. Example 2-2 Solution Since R1 is much greater than Rm, its parallel effect is negligible mA V RR E I ms m 130 689 10 = Ω+Ω = + = Find the current that would flow in the meter in Example 2-1 if it is incorrectly connected in parallel with R1. The meter current is too high for the full-scale rating (1 mA) and may probably damage the instrument. Rs=9Ω E=10 V R1 0-1 mA Rm= 68 Ω
    13. 13. Effect on Low-resistance Circuit With same example, what happen if R1 is 150Ω? Solution : The approximate current is I = E/R1 = 10/150 = 66.7mA. Actual current is I = E/(R1+ Rm+ RS) = 10/(150+68+9) = 44.1mA %2.51%100 1.44 1.447.66 % = − = xerror
    14. 14. Obtaining Higher Current Scales  A basic dc meter has a very small and single current scale due to small wire size (typical range : 0-100 µA, 0-1 mA)  To measure larger currents, shunt resistor in parallel with the meter is used as shown in Fig 2-4. Rm RS I Im IS Fig 2-4 Extend the range of moving coil meter
    15. 15. Obtaining Higher Current Scales I = range of ammeter Im = full scale current of moving coil meter Is = current flowing in the shunt resistor RS Rm I Im IS Vm Fig 2-5 Ammeter circuit design Given the full scale current of the moving coil meter and the ammeter, we have to calculate RS. m mm S m S mmm mS II RI I V R RIV III − ==∴ = −=
    16. 16. Example 2-3 A moving coil meter with full scale current of 100 µA and internal resistance of 500 Ω is used to measure 1mA full scale. (a) How much current in the meter and RS at full scale flows? (b) Determine RS. (a) At full-scale, Im = 100 µA IS = 1 mA – Im = 0.9 mA 1 mA RS 0-100 µA Rm= 500 Ω Im=100 µA IS=0.9mA V (b) V = ImRm= 0.05 V RS = V/IS = 0.05/0.0009 = 55.56 Ω Solution:
    17. 17. Example 2-4 A 0-to-50 µA dc meter movement, with a coil resistance Rm of 1250Ω, is used with a shunt resistor to measure a full- scale current of 500µA. Sketch the circuit arrangement and calculate the value of RS. Solution :
    18. 18. R1=11.1 R2=1.01 R3=0.1 1mA 10mA 100mA 0-100uA R = 100m I Ω Ω Ω Ω Multi-range Ammeter  In Fig 2-6, the 0-100µA meter can be used to convert full- scale readings to 1mA, 10mA and 100mA by switching to different shunting resistors.  The disadvantage of this type of arrangement is that the shunt is disconnected during switching and hence high current may damage the meter.  To solve this problem, make- before-break switch is used.Fig 2-6a
    19. 19. Multi-range Ammeter  Another type of circuit, universal shunt circuit shown in Fig 2.6b resolve the above problem even ordinary switch is employed.  The main point is that the shunt exist while switching. R1=1.11 0-100uA R = 100m I 10mA 1 mA R2=10Ω Ω Fig 1-6b Ω
    20. 20. Measurement of DC Voltage Moving coil meter can be used to measure voltage if a multiplier resistor is connected in series as shown in Fig 2-7. Fig 2-7 Voltmeter circuit
    21. 21. Example 2-5 Calculate the value of the multiplier resistor Rmx required in Fig. 2-7 for a full-scale voltage E of 10V. Solution Ω= Ω==+ =+ − kR k A V R I E RR mx mx fsd mmx 5.99 100 10 10 500 4
    22. 22. Voltmeter Sensitivity & DC Voltmeter Resistance Voltmeter Sensitivity :  The sensitivity of a voltmeter is specified in terms of ohms per volt (Ω/V).  Sensitivity φ = 1/IFS where IFS is the full-scale current of the moving coil instrument Voltmeter Resistance :  The resistance of a dc voltmeter is : Rm = EFS x φ Ω where Rm = the resistance of the voltmeter in ohms (Ω) EFS = the range of the voltmeter in volts (V) φ = the sensitivity of the meter in ohm per volt (Ω/V)
    23. 23. Sensitivity ratings for dc voltmeters : Full-scale Moving coil Current Sensitivity (ψ ) 1mA 1 kΩ /V 100μ A 10 kΩ /V 50μ A 20 kΩ /V 20μ A 50 kΩ /V 10μ A 100 kΩ /V
    24. 24. Using Voltmeter Three basic rules : 1. Connect the voltmeter across the load, i.e. in parallel with the load. (see Fig. 2.8) 2. Select a voltmeter that has a full-scale range that is greater than the highest potential expected. 3. Make sure that the voltmeter has an input resistance that is very high (i.e. >100:1) compared with the circuit resistance. Fig 2-8 Voltmeter connection
    25. 25.  If rule 1 is not obeyed, the voltmeter is connected to a load in series, this may cause improper operation of the circuit since the voltmeter resistance is added to the circuit resistance.  If rule 3 is not obeyed. Error reading will occur. (See Fig. 2- 9) Fig 2-9 Loading effect of voltmeter
    26. 26. Ideally, E2 should be V Vx Mk M E RR R E 67.6 10 1500 1 21 2 2 = Ω+Ω Ω = + = ( )( ) V98.3 V10x M33.0k500 M33.0 E RR R E M33.0 M5.0M1 M5.0M1 R eq1 eq 2 eq = + = + = = + = ΩΩ Ω Ω ΩΩ ΩΩ therefore, Practically, the internal resistance Rm (5V x 100kΩ/V = 0.5MΩ) of voltmeter is in parallel with R2. Therefore R2 becomes Req.
    27. 27.   The above example shows the loading of the voltmeter causes 40% error.  To reduce the error, digital voltmeter may be employed since it has larger input resistance (typically 10 MΩ). %3.40%100 67.6 67.698.3 % −= − = xerror
    28. 28.  The accuracy of DVM is quite good. It can be described as , 4 digit or , etc. The half digit refers to the most significant digit and can be either 0 or 1. Whereas the other digits can be 0-9. Alternative Sensing Equipment : Digital Voltmeter  A digital voltmeter (DVM) uses an analogue-to-digital converter (ADC) to convert analogue dc voltage to a digital word. digit- 2 1 3 digit- 2 1 4 Signal conditioning circuit Analogue to digital converter decoder displayVoltage to be measured Fig 2-10 Block diagram of a digital voltmeter
    29. 29. Comparison of Analog and Digital Instruments Feature Digital Meter Analog Meter Reading error Lower Higher Accuracy Higher Lower Resolution Higher Lower Range selection Auto (for some meters) Manual Polarity selection Auto (indicates a ‘-’ sign when the terminal is reversed) Pointer attempts to deflect to the left of zero when the polarity is reversed Robustness Not usually damaged by rough treatment Can be irreparably damaged when dropped
    30. 30. Measurement of Resistance R1 : standard resistance Rm : meter resistance Rx : resistance to be measured Current Im : mx b RRR E ++ = 1 mI Fig 2-11a Basic series ohmmeter circuit
    31. 31.  If Rx = 0, R1 and Rm are selected so as to give FSD which is marked as zero ohm (right most).  When terminal A & B are open-circuited, i.e. Rx is infinity, the pointer is marked as infinity (left most).  If Rx with a value between zero & infinity, the pointer position is on the scale.  At mid-scale, Rx = R1 + Rm, since Im = IFS/2. Fig 2-11b Ohmmeter scale
    32. 32. Example 2-6 The series ohmmeter in Fig. 2-12 is made up of a 1.5 V battery, a 100μA meter, and a resistance R1 which makes (R1 + Rm) = 15 kΩ. (a) Determine the instrument indication when Rx=0. (b) Determine how the resistance scale should be marked at 0.5 FSD, 0.25 FSD, and 0.75 FSD. 1.5 V Ifsd =100μA Fig 2-12
    33. 33. Solution (a) Im = Eb/(Rx+R1+Rm) = 1.5/(0+15 kΩ) = 100 µA (FSD) (b) At 0.5 FSD : Im = 100 µA/2 = 50 µA Rx+R1+Rm= Eb/I Rx = Eb/Im - (R1+Rm) = 1.5V/50 µA - 15 kΩ = 15 kΩ At 0.25 FSD : Im = 100 µA/4 = 25 µA Rx = 1.5V/25 µA - 15 kΩ = 45 kΩ
    34. 34. At 0.75 FSD : Im = 0.75 x 100 µA = 75 µA Rx = 1.5 V/75 µA - 15 kΩ = 5 kΩ
    35. 35. Ohmmeter With Zero Adjust Figure 2-13
    36. 36. Ohmmeter With Zero Adjust  Since battery in the ohmmeter falls with use, the instrument scale will be incorrect.  R1 cannot be used to adjust zero since mid-scale value will otherwise no longer equal to the resistance as before.  Parallel resistor R2 is added for zero control.  New supply current Ib will be : the meter voltage is : Vm=Ib(R2//Rm) the meter current will be :  Each time ohmmeter is used, terminals A and B are first short-circuited for FSD. The scale reading then remains correct. m21x b b R//RRR E I ++ = m m2b m R )R//R(I I =
    37. 37. Example 2-7 The ohmmeter circuit in Fig2-13 has Eb = 1.5 V, R1 = 15 kΩ, Rm=50Ω, and meter FSD=50μA. Determine (a) the value of R2; (b) the value of Rx at 0.5 FSD and (c) the new value of R2 if Eb drops to 1.3 V.
    38. 38. Solution (a) Vm = Im x Rm = 50μA x 50Ω = 2.5mV Ib = (Eb – Vm)/R1 = (1.5 – 0.0025)15000 = 99.83 μA I2 = Ib - Im = 99.83 μA - 50μA = 49.83 μA R2 = Vm / I2 = 2.5x10-3 / 49.83x10-6 = 50.17 Ω
    39. 39. Solution (b) At 0.5 FSD, Rx = R1 + (R2 // Rm) = 15000 + (50.17 // 50) = 15.025 k Ω (c) When Eb drops to 1.3 V Ib = (Eb – Vm)/R1 = (1.3 – 0.0025)15000 = 86.5 μA I2 = Ib - Im = 86.5 μA - 50μA = 36.5 μA R2 = Vm / I2 = 2.5x10-3 / 36.5x10-6 = 68.49 Ω
    40. 40. Multi-meter Fig 2-14 Volt-Ohm-Milliammeter (VOM)
    41. 41. Multi-meter  Multi-meter can measure voltage, current and resistance. The main selector switch connects the shunt, multiplier, or range resistors, as required.  Most instrument include a rectifier to allow reading ac values.  When using a multimeter, the same rules as in individual ammeters, voltmeter and ohmmeter should be followed, for example, connection to circuit in parallel for voltage measurement.  It is a good practice to store multimeter in the OFF position. An analogue passive volt-ohm-milliammeter (VOM) without OFF position, the switch should be set to the highest DC VOLTS range position.  When using a VOM, begin with highest range voltage or current position, and then work down to readable deflection.  Neither type of basic dc PMMC meter can correctly indicates the ac value. However it can be done if a rectifier or integrated circuit rms-to-dc converter is used.
    42. 42.  Digital multi-meter (DMM) uses any of several technologies : transistor, integrated circuits, or digital circuitry.  VOM vs. DMM : – VOM is better in the presence of strong EM fields. – DMM may not operate properly in the EM field. – VOM has low sensitivity, and different input impedance (internal resistance) in different range. – DMM has higher input impedance (internal resistance) (e.g. 10MΩ) for voltage measurement, and keep constant at various ranges.
    43. 43. Wattmeter Fig 2-15 Electrodynamometer circuit Electrodynamometer wattmeter is used to measure power using two sets of coil. L1, L2 - fixed coils L3 - moving coil
    44. 44. Electrodynamometer  Unlike PMMC, the electrodynamometer creates magnetic field from the current flowing in the windings of coils L1 through L3.  Coils L1 and L2 are stationary, while coil L3 is free to move and is attached to the meter pointer.  The magnetic fields of L1 and L2 will reinforce each other and will interact with the field of L3 to create a rotational force on coil L3. Hence the pointer deflects an amount that is proportional to the square of the average current.  The electrodynamometer used to measure current, the dial scale is calibrated in terms of rms current. Electrodynamometer voltmeter can be built from a multiplier resistor in series with meter movement.
    45. 45. Wattmeter Figure 2-16 Electrodynamometer used as a wattmeter.
    46. 46. Electrodynamometer as Wattmeter  The fixed coils (L1 & L2) are designed to carry a heavy current and are connected in series with the load as current-measuring element.  The moving coil (L3) is wound of a much finer gauge of wire and is connected across the load and serves as the voltage-measuring element.  The deflection of the pointer depends on the interaction of the magnetic fields produced by stationary & movable coils. Hence it is proportional to the product of the V & I and the unit is in Watt.  Electrodynamometer is suitable for use in d.c. and low frequency a.c. circuit.
    47. 47. F M Current (fixed) coils Voltage (moving) coil Deflecting torque ∝ IF IM ∝ IL VL ∝ power Fig 2-17 Schematic diagram of using wattmeter to measure power
    48. 48. Multi-range Wattmeter Fig 2-18a  Current ranges (0.5A and 1A) can be changed by switching two field coils from series to parallel connection.  Voltage ranges (60V, 120V and 240 V) can be made by switching different values of multiplier resistors.
    49. 49. Multi-range Wattmeter Figure 2-18b The FSD will be the multiplication of both current and voltage selection. Current Voltage FSD 1A 120V 120W 1A 240V 240W
    50. 50. Using Wattmeter  Before connecting a wattmeter into a circuit, check the mechanical zero of the instrument and adjust it if necessary.  While connecting with load, current circuit of the wattmeter must be connected in series, but voltage circuit must be in parallel.  If a multi-range wattmeter is connected into a circuit, select a voltage range equal or higher than the supply voltage. Select the highest current, and then switch down to give the greatest on-scale deflection.  Electro-dynamic wattmeters are useful for measurement on supply frequencies up to a max. of 500Hz.
    51. 51. Use of Bridge Circuit in typical sensor application  The operation of many primary sensors depends upon the variation of the electrical resistance of a conductor in response to variations in the measured variable.  When the variation in resistance is large, the change can be readily measured directly.  When the variation in resistance is small, such as resistance strain gauge and thermo-resistance temperature sensors, bridge technique is used to give an accurate measurement.  The Wheatstone Bridge technique is used in conjunction with resistive transducers whose electrical variations are relative small.
    52. 52.  There are two ways of using the Wheatstone Bridge technique: - in the balanced condition; - in the unbalanced condition. The choice of which depends primarily upon the type of measurement being undertaken.
    53. 53. Wheatstone Bridge in balanced condition R1 , R2 - Precision resistors R3 - Unknown resistor R4 - Variable precision resistor E A B D C R1 R2 R4 R3 G Eo Fig 2.19 To determine the resistance of R3 , the variable resistor is adjusted until the zero-center galvanometer G indicates null.
    54. 54.  Initially, the galvanometer should be shunted to protect it from excessive current level as shown in Fig 2.20.  As null approaches, the shunting resistance is gradually increased until G indicates zero with the resistor open- circuited. Fig 2.20
    55. 55. Balanced condition of Wheatstone bridge Fig 2.21 shows the voltage and current throughout the bridge when it is balanced. When G indicates zero, IG = 0, ∴ VB = VD So that V1 = V2 and V3 = V4 ∴ I1R1 = I2R2 ………..(2.1) I1R3 = I2R4 ………..(2.2) Divide eqt. 2.2 by 2.1, we have: 4 2 1 3 2 4 1 3 R R R Ror R R R R = = Fig 2.21
    56. 56.  At balance, no current flows through the galvanometer. The galvanometer appears as an open-circuit and will not cause loading effect to the circuit  The accuracy of the measuring result depends on the accuracy of the precision resistors  The supply voltage does not affect the balance condition but the bridge sensitivity
    57. 57. Solution: R3 = R1R4/R2 = Ω=5760 500 )1800)(1600( Example 2.8 A Wheatstone bridge has the following arm values: R1 = 1.6 kΩ, R2 = 500 Ω, R3 is unknown, and R4 = 1.8 kΩ. What value of R3 brings the bridge into the null condition?
    58. 58. Example 2.9 A Wheatstone bridge has ratio arms R1 and R2, both of which may be set to 1 kΩ, 100 Ω or 10 Ω. The variable arm R4 is adjustable from 1 Ω to 10 kΩ in 1 Ω step. Describe, with the aid of circuit diagram, how R3 of about 58 Ω may be measured most accurately. To obtain the most accurate result, R4 should be as large as possible while the ratio should be as small as possible. 4 2 1 3 R R R R = 2 1 R R 10Ω E G 1kΩ R3 ≅ 58Ω R4 ≅ 5800Ω Solution: Since

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