Measurement and Instrumentation. Analog Electrical Devices and Measurements

Transcription

1 Measureent and Instruentation nalog Electrical Devices and Measureents

2 nalog Devices: Current Measureents Force on a conductor I conductor is placed in a unifor agnetic field B T, at an angle of θ. The current flow in the conductor is I. Force exerted on the conductor can be calculated fro r F = r IL r B F θ B F = ILB sinθ

3 nalog Devices: Current Measureents Force on a conductor With no current flowing through the conductor, the spring will be at its unstretched length. s current flows through the conductor, the spring will stretch and developed force required to balance the electroagnetic force. F s = kx kx = ILB I I B k = spring constant, x = the total distance oved by the spring and θ is 90 o F = IBL I = k BL x

4 D rsonval or PMMC Instruent Iportant parts of PMMC Instruent Peranent agnet with two softiron poles Moving coil Controlling or restoring spring PMMC = Peranent Magnet Moving Coil

5 Torque Equation and Scale When a current I flows through a one-turn coil in a agnetic field, a force exerted on each side of the coil F = IBL D F = IBL F = IBL L = the length of coil perpendicular to the paper Since the force acts on each side of the coil, the total force for a coil of N turns is F = NIBL The force on each side acts at a coil diaeter D, producing a deflecting torque T D = NIBLD

6 Torque Equation and Scale The controlling torque exerted by the spiral springs is proportional to the angle of deflection of the pointer: T C = Kθ Where K = the spring constant. For a given deflection, the controlling and deflecting torques are equal BLIND = Kθ Since all quantities except θ and I are constant for any given instruent, the deflection angle is θ = CI Therefore the pointer deflection is always proportional to the coil current. Consequently, the scale of the instruent is linear.

7 Galvanoeter Galvanoeter is essentially a PMMC instruent designed to be sensitive to extreely current levels. The siplest galvanoeter is a very sensitive instruent with the type of center-zero scale, therefore the pointer can be deflected to either right or left of the zero position. The current sensitivity is stated in µ/ Galvanoeters are often used to detect zero current or voltage in a circuit rather than to easure the actual level of current or voltage. In this situation, the instruent is referred as a null detector.

8 DC eter PMMC instruent n aeter is always connected in series with a circuit. The internal resistance should be very low eter shunt The pointer can be deflected by a very sall current s Extension of ranges of aeter can be achieved by connecting a very low shunt resistor Coil resistance = s I = I s s I = Is I I I s s Shunt resistance I = s I I s s = I I I

9 DC eter Exaple: n aeter has a PMMC instruent with a coil resistance of = 99 Ω and FSD current of 0.1. Shunt resistance s = 1 Ω. Deterine the total current passing through the aeter at (a) FSD, (b) 0.5 FSD, and (c) 0.25 FSD. Known: FSD of I and s Solution: FSD 0.5FSD 0.25FSD I () I s () I = I I s ()

10 DC eter Exaple: PMMC instruent has FSD of 100 µ and a coil resistance of 1kΩ. Calculate the required shunt resistance value to convert the instruent into an aeter with (a) FSD = 100 and (b) FSD = 1. Known: FSD of I Solution: (a) FSD = 100 : s = Ω (b) FSD = 1 : s = Ω

11 DC eter: Multirange C B s1 D s2 s3 s4 D C B E E Make-before break switch Multirange aeter using switch shunts ake-before-break ust be used so that instruent is not left without a shunt in parallel to prevent a large current flow through aeter.

12 DC eter: yrton shunt I I S I B I I C s I s in parallel with D - n yrton shunt used with an aeter consists of several seriesconnected resistors all connected in parallel with the PMMC instruent. ange change is effected by switch between resistor junctions I I S I I s I s C B I - D 1 2 in parallel with 3

13 DC eter Exaple: PMMC instruent has a three-resistor yrton shunt connected across it to ake an aeter. The resistance values are 1 = 0.05 Ω, 2 = 0.45 Ω, and 3 = 4.5 Ω. The eter has = 1 kω and FSD = 50 µ. Calculate the three ranges of the aeter. Known: FSD of I 1 2 and 3 Solution: I Position B C D I S I I s I s C B I - I FSD () D

14 DC olteter PMMC instruent Series resistance or ultiplier Multiplier resistance s Coil resistance n aeter is always connected across or parallel with the points in a circuit at which the voltage is to be easured. The internal resistance should be very high = I I s Given = ange s ange = I s = I The reciprocal of full scale current is the volteter sensitivity (kω/) I The total volteter resistance = Sensitivity X ange

15 DC olteter: Multirange Meter resistance Multiplier resistors Multirange volteter using switched ultiplier resistors = I ( ) Where can be 1, 2, or Multirange volteter using seriesconnected ultiplier resistor = I ( ) Where can be 1, 1 2, or 1 2 3

16 DC eter Exaple: PMMC instruent with FSD of 50 µ and a coil resistance of 1700 Ω is to be used as a volteter with ranges of 10, 50, and 100. Calculate the required values of ultiplier resistor for the circuit (a) and (b) Known: FSD of I Solution: Multiplier resistors 1 Meter resistance (a) (b) kω kω MΩ kω 800 kω 1 MΩ

17 Oheter: olteter-aeter ethod Pro and con: S Siple and theoretical oriented equires two eter and calculations Subject to error: oltage drop in aeter (Fig. (a)) Current in volteter (Fig. (b)) - I x x S I I x I x Measured x : if x >> eas eas Fig. (a) = = = x I I I x x Measured x : if I x >>I Fig. (b) eas eas x = = = I I I 1 I / I x x x Therefore this circuit is suitable for easure large resistance Therefore this circuit is suitable for easure sall resistance

18 Oheter: Series Connection Battery olteter-aeter ethod is rarely used in practical applications (ostly used in Laboratory) Oheter uses only one eter by keeping one paraeter constant Exaple: series oheter S esistance to be easured x Standard resistance 1 Meter Infinity resistance 45k 25 15k 50 Nonlinear scale 75 5k 0 I s x = 1 Meter µ Basic series oheter Oheter scale Basic series oheter consisting of a PMMC and a series-connected standard resistor ( 1 ). When the oheter terinals are shorted ( x = 0) eter full scale defection occurs. t half scale defection x = 1, and at zero defection the terinals are open-circuited.

19 Loading Effect: oltage Measureent a th a Linear Circuit ab th ab b Undisturbed condition: = Measured condition: General equation: ab = u = th = = ab th th 1 = 1 / th u b u Measureent error: error = 100 = u th th % = 100% 1 / Therefore, in practice, to get the acceptable results, we ust have 10 th (error ~ 9%) th

20 Loading Effect kΩ kΩ kΩ 100kΩ kΩ Circuit before easureent eas Circuit under easureent 100 //100 = 10 = // kΩ 6 100kΩ kΩ 4 200kΩ 100kΩ kΩ eas 200 //100 = 10 = //100 eas 1000 //100 = 10 = //100

21 Loading Effect Exaple Find the voltage reading and % error of each reading obtained with a volteter on (i) 5 range, (ii) 10 range and (iii) 30 range, if the instruent has a 20 kω/ sensitivity, an accuracy 1% of full scale deflection and the eter is connected across b SOLUTION The voltage drop across b without the volteter connection = b 5 k = 50 = 5 b 45 k 5 k a b 50 a 45kΩ b 5kΩ 45 5 On the 5 range eq = S range = 20kΩ / 5 = 100 kω = b The volteter reading is = eq b a eq b 100 k = 100 k 5 k 5 k = 4.76 kω 4.76 k = 50 = k 4.76 k

22 Loading Effect Error of the easureent is the cobination of the loading effect and the eter error The loading error = = The eter error = ± 5 x 1 = ± % of error on the 5 range: ± 0.05 = 100 = ± 5.36% 5 ange () b. () Loading error () Meter error () Total error () % error ± 0.05 ± 0.27 ± ± 0.1 ± 0.22 ± ± 0.3 ± 0.35 ± 6.10

23 Loading Effect: Current Measureent a I th a I Linear Circuit ab th ab b Undisturbed condition: = 0 Measured condition: 0 General equation: I = I = / u th th th b ( ) I = I = / u th ( 1 ) I = I / / th Measureent error: error I Iu = 100 Iu 1 = 100 % = 100% 1 / th th Therefore, in practice, to get the acceptable results, we ust have th /10 (error ~ 9%)

24 C olteter: PMMC Based Wavefor plitude verage MS 0 2 π 2 2 π D W D D W D D W

25 C olteter: PMMC Based Basic PMMC instruent is polarized, therefore its terinals ust be identified as and -. PMMC instruent can not response quite well with the frequency 50 Hz or higher, So the pointer will settle at the average value of the current flowing through the oving coil: average-responding eter. Full-wave ectifier olteter Multiplier resistors s D1 D 2 D 3 D 4 rs av ctually voltage to be indicated in ac easureents is norally the rs quantity p Using 4 diodes On positive cycle, D1 and D4 are forward-biased, while D2 and D3 are reverse-biased On negative cycle, D2 and D3 are forward-biased, while D1 and D4 are reverse-biased The scale is calibrated for pure sine with the scale factor of 1.11 (/ 2 / 2/π)

26 C olteter: PMMC Based Exaple: PMMC instruent has FSD of 100 µ and a coil resistance of 1kΩ is to be eployed as an ac volteter with FSD = 100 (rs). Silicon diodes are used in the full-bridge rectifier circuit (a) calculate the ultiplier resistance value required, (b) the position of the pointer when the rs input is 75 and (c) the sensitivity of the volteter Known: FSD of I Solution: Multiplier resistors s D1 D 3 p rs av D 2 D 4 (a) s = kω (b) 0.75 of FSD (c) 9 kω/

27 C olteter: PMMC Based Half-wave ectifier olteter s D 1 D 2 SH rs av p On positive cycle, D1 is forward-biased, while D2 is reverse-biased On negative cycle, D2 is forward-biased, while D1 is reverse-biased The shunt resistor SH is connected to be able to easure the relative large current. The scale is calibrated for pure sine with the scale factor of 2.22 (/ 2 / /π)

28 C olteter: PMMC Based Exaple: PMMC instruent has FSD of 50 µ and a coil resistance of 1700 Ω is used in the half-wave rectifier volteter. The silicon diode (D1) ust have a iniu (peak) forward current of 100 µ. When the easured voltage is 20% of FSD. The volteter is to indicate 50 rs at full scale Calculate the values of S and SH. Known: FSD of I Solution: S = kω SH = 778 Ω s D 1 D 2 SH rs av p

29 Exaple The syetrical square-wave voltage is applied to an average-responding ac volteter with a scale calibrated in ters of the rs value of a sine wave. If the volteter is the full-wave rectified configuration. Calculate the error in the eter indication. Neglect all voltage drop in all diodes. E C olteter: PMMC Based E Solution 11% t T

30 Bridge Circuit Bridge Circuit is a null ethod, operates on the principle of coparison. That is a known (standard) value is adjusted until it is equal to the unknown value. Bridge Circuit DC Bridge (esistance) C Bridge Inductance Capacitance Frequency Wheatstone Bridge Kelvin Bridge Megaoh Bridge Maxwell Bridge Hay Bridge Owen Bridge Etc. Schering Bridge Wien Bridge

31 Wheatstone Bridge and Balance Condition The standard resistor 3 can be adjusted to null or balance the circuit. 1 2 Balance condition: No potential difference across the galvanoeter (there is no current through the galvanoeter) I 1 I 2 D I 3 I 4 B Under this condition: D = B I = I C 4 nd also DC = BC I = I where I 1, I 2, I 3, and I 4 are current in resistance ars respectively, since I 1 = I 3 and I 2 = I 4 2 = or x = 4 =

32 Exaple 1 Ω 1 Ω 1 Ω 1 Ω Ω 1 Ω 2 Ω 2 Ω (a) Equal resistance (b) Proportional resistance 1 Ω 10 Ω 1 Ω 10 Ω Ω 20 Ω 2 Ω 10 Ω (c) Proportional resistance (d) 2-olt unbalance

33 Sensitivity of Galvanoeter galvanoeter 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 sall unbalance condition, and apply circuit analysis to solve the current through galvanoeter Thévenin Equivalent Circuit S C I 1 I G D Thévenin oltage ( TH ) = = I I CD C D where I1 = and I2 = B Therefore 1 2 TH = CD =

34 Sensitivity of Galvanoeter (continued) Thévenin esistance ( TH ) 1 2 C D TH = // // B Copleted Circuit TH TH C G I g = TH TH g I g = TH TH g D where I g = the galvanoeter current g = the galvanoeter resistance

35 Exaple 1 Figure below show the scheatic diagra of a Wheatstone bridge with values of the bridge eleents. The battery voltage is 5 and its internal resistance negligible. The galvanoeter has a current sensitivity of 10 /µ and an internal resistance of 100 Ω. Calculate the deflection of the galvanoeter caused by the 5-Ω unbalance in ar BC SOLUTION The bridge circuit is in the sall unbalance condition since the value of resistance in ar BC is 2,005 Ω. Thévenin oltage ( TH ) Ω 1000 Ω 1 2 D G C 200 Ω Ω TH = D C = (a) B Thévenin esistance ( TH ) 100 Ω 1000 Ω = 100 // // 2005 = 734 Ω TH C 200 Ω 2005 Ω D The galvanoeter current TH 2.77 B (b) TH = 734 Ω (c) C D G I g =3.34 µ g = 100 Ω I g TH 2.77 = = = 3.32 µ 734 Ω 100 Ω TH Galvanoeter deflection g 10 d = 3.32 µ = 33.2 µ

36 Exaple 2 The galvanoeter in the previous exaple is replaced by one with an internal resistance of 500 Ω and a current sensitivity of 1/µ. ssuing that a deflection of 1 can be observed on the galvanoeter scale, deterine if this new galvanoeter is capable of detecting the 5-Ω unbalance in ar BC Exaple 3 If all resistances in the Exaple 1 increase by 10 ties, and we use the galvanoeter in the Exaple 2. ssuing that a deflection of 1 can be observed on the galvanoeter scale, deterine if this new setting can be detected (the 50-Ω unbalance in ar BC)

37 Deflection Method C B Consider a bridge circuit which have identical resistors, in three ars, and the last ar has the resistance of. if / <<1 Please correct D Sall unbalance occur by the external environent Thévenin oltage ( TH ) / = 4 2 / Thévenin esistance ( TH ) In an unbalanced condition, the agnitude of the current or voltage drop for the eter or galvanoeter portion of a bridge circuit is a direct indication of the change in resistance in one ar. This kind of bridge circuit can be found in sensor applications, where the resistance in one ar is sensitive to a physical quantity such as pressure, teperature, strain etc. TH = CD TH

38 Exaple Circuit in Figure (a) below consists of a resistor v which is sensitive to the teperature change. The plot of S Tep. is also shown in Figure (b). Find (a) the teperature at which the bridge is balance and (b) The output signal at Teperature of 60 o C. 6 5 kω 5 kω v 5 kω Output signal v (kω ) kω Tep ( o C) (a) (b)

39 C Bridge: Balance Condition B Z Z 1 2 I 1 I 2 D C all four ars are considered as ipedance (frequency dependent coponents) The detector is an ac responding device: headphone, ac eter Source: an ac voltage at desired frequency Z 3 D Z 4 Z 1, Z 2, Z 3 and Z 4 are the ipedance of bridge ars General For of the ac Bridge t balance point: Coplex For: E B = E BC or I1Z 1 = I2Z2 I 1 = and I 2 = Z Z Z Z Z1Z 4 = Z2Z3 Z Z Polar For: ( θ θ ) =Z Z ( θ θ ) Magnitude balance: Phase balance: Z1Z 4=Z2Z3 θ θ = θ θ

40 Exaple The ipedance of the basic ac bridge are given as follows: Z Z 1 2 o (inductive ipedance) = Ω = 250 Ω (pure resistance) Deterine the constants of the unknown ar. Z Z 3 4 = Ω o (inductive ipedance) = unknown

41 Exaple an ac bridge is in balance with the following constants: ar B, = 200 Ω in series with L = 15.9 H ; ar BC, = 300 Ω in series with C = µf; ar CD, unknown; ar D, = 450 Ω. The oscillator frequency is 1 khz. Find the constants of ar CD. B SOLUTION Z Z 1 2 Z 3 I 1 I 2 D Z 4 C Z Z Z Z = jω L = 200 j100 Ω = 1/ jωc = 300 j600 Ω = = 450 Ω = unknown D The general equation for bridge balance states that Z1Z 4 = Z2Z3

42 - Operational plifier: Op p CC () Inverting Input Non-inverting Input _ Output I 1 I 2 _ out EE (-) (a) Electrical Sybol for the op ap (b) Miniu connections to an op ap Ideal Op p ules: 1. No current flows in to either input terinal 2. There is no voltage difference between the two input terinals ule 1: I 1 = I 2 = 0; /- = ule 2: = -; irtually shorted

43 Inverting plifier v in 1 KCL _ f v out - Use KCL at point and apply ule 1: (no current flows into the inverting input) earrange v v vin v vout = vin v out = 1 f 1 f f 0 0 pply ule 2: (no voltage difference between inverting and non-inverting inputs) Since at zero volts, therefore - is also at zero volts too. v = 0 v in 1 v out = f 0 v v out in = f 1

44 Inverting plifier: another approach No current flows into op ap v in 1 i Given v in = 5sin3t, 1 =4.7 kω and f =47 kω v out = -10v in = -50 sin 3t _ f Cobine the results, we get v v out in = f 1 i v out - Fro ule 2: we know that - = = 0, and therefore v i = in 1 0 Since there is no current into op ap (ule 1) 0 i v = v in f v out 0 out 0 i v = out v in 1 = i tie f

45 Non-inverting plifier 1 KCL v in _ f v out - Given v in = 5sin3t, 1 =4.7 kω and f =47 kω Use KCL at point and apply ule 1: v 1 pply ule 2: v v out = f v in = 1 v out = 11v in = 55 sin 3t tie v v out in v out v in = f 1 v 0

46 Suing plifier: Matheatic Operation i = i1 i2 i3 Use KCL and apply ule 1: i f v v v v v v v v out = 0 f i 1 v _ Since v = 0 (ule 2) v 1 i 2 i3 v B v out - ( ) f vout = v1 v2 v3 v2 Su of v 1, v 2 and v 3 v 3

47 Difference plifier: Matheatic Operation 4 Use KCL and apply ule 1: 1 v _ v v1 v vout = (1) v 1 v 2 2 v B 3 v out - Since v = v B (ule 2) and 3 v = vb = v2 2 3 (2) Substitute eq. (2) into eq. (1), we get vout v1 = v f If 1 = 2 = and 3 = 4 = f v = ( v v ) out 2 1 Difference of v 1 and v 2

48 Differentiator and Integrator: Matheatic Operation C _ i But i vout dv C dt = i and C = in C v = v v in i v out - v out dv = C dt in Differentiator v in i _ C v c - i v out - But v out t 0 = v 1 vc ( t) = idt vc (0) C C and 1 v = v dt v out in C C 0 t vin (0) = i Integrator