CHAPTE 2 ELECTOMECHANCAL NSTUMENTS
Learning Outcoes At the end of the chapter, students should be able to: Understand construction and operation of peranent agnet oving-coil (PMMC) instruent. Describe how PMMC instruents are used as galvanoeters, dc aeters, dc volteters, ac aeters, and ac volteters.
ntroduction PMMC instruent consists basically of a lightweight coil of copper wire suspended in the field of a peranent agnet. Current in the wire causes the coil to produce a agnetic field that interacts with the field fro the agnet, resulting in partial rotation of the coil. A pointer connected to the coil deflects over a calibrated scale, indicating the level of current flowing in the wire. The PMMC instruent is essentially a low-level dc aeter.
ntroduction t can be eployed to easure a wide range of direct current dc levels with the use of parallel-connected resistors.
The instruent ay also be used as a dc volteter by connecting appropriate-value resistors in series with the coil. Oheters can be ade fro precision resistors, PMMC instruent, and batteries.
Multirange eters are available that cobine aeter, volteter, and oheter functions in one instruent Mutlirange olteter Mutlirange Aeter
Ac aeters and volteters can be constructed by using rectifier circuits with PMMC instruents. The Half wave rectifier
Deflection nstruent Fundaentals A deflection instruent uses a pointer that oves over a calibrated scale to indicate a easured quantity. Three forces are operating in the electroechanical echanis inside the instruent: Deflecting force Controlling force Daping force
Deflecting Force The deflecting force causes the pointer to ove fro its zero position when a current flows. n the PMMC instruent, the deflecting force is agnetic. The pointer is fixed to the coil So, it oves over the scale as The coil rotates
The deflecting force in the PMMC instruent is provided by A current-carrying coil pivoted in a agnetic field.
Controlling Force The controlling force in the PMMC instruent is provided by spiral springs. The springs retain the coil & pointer at their zero position when no current is flowing. The coil and pointer stop rotating when the controlling force becoes equal to the deflecting force. The spring aterial ust be nonagnetic to avoid any agnetic field influence on the controlling force. The controlling force fro the springs balances the deflecting force. The springs are also used to ake electrical connection to the coil, they ush a low resistance( Phosphor bronze is the aterial usually eployed).
Daping Force The daping force is required to iniize (or dap out) oscillations of the pointer and coil before settling down at their final position. The daping force ust be present only when the coil is in otion; thus it ust be generated by the rotation of the coil. n PMMC instruents, the daping force is norally provided by eddy currents. Eddy currents induced in the coil set up a agnetic flux that opposes the coil otion, thus daping the oscillations of the coil.
Daping Force The daping force in a PMMC instruent is provided by eddy currents induced in the aluinu coil forer as it oves through the agnetic field.
The ethods of supporting the oving syste of a deflection instruent 1. Jeweled-bearing suspension Cone-shaped cuts in jeweled ends of shafts or pivots (at be broken by shocks). low possible friction. Soe jewel bearings are spring supported to absorb such shocks.
2. Taut-band ethod Much tougher than jeweled-bearing. Two flat etal ribbons (phosphor bronze or platinu alloy) are held under tension by spring to support the coil. Because of the spring, the etal ribbons behave like rubber under tension. Thus, the ribbons also exert a controlling force as they twist. The etal ribbons can be used as electrical connections to the oving coil. Much ore sensitive than the jeweled-bearing type because there is less friction. Extreely rugged, not easily be shattered.
PMMC Construction The ain feature is a peranent agnet with two soft-iron pole shoes. A cylindrical soft-iron core is positioned between the core and the faces of the pole shoes. The lightweight oving coil pivoted to ove within these narrow air gaps. The air gaps are ade as narrow as possible in order to have the strongest possible level of agnetic flux crossing the gaps. The current in the coil of a PMMC instruent ust flow in one particular direction to cause the pointer to ove (positively) fro the zero position over the scale. Because the PMMC is polarized, it can be used directed to easure alternating current.
PMMC Construction D Arsonval or horseshoe agnet Core-agnet 21
Torque Equation and Scale When a current flows through a one-turn coil situated in a agnetic field, a force F is exerted on each side of the coil F ( Bl ) N newtons where N is the nuberof turns Total force on each side of the coil of N turns: F 2 B l N newtons The force on each side acts at a radius r, producing a deflecting torque: T T D D 2 B l B l B l N r N (2r) N D ( N. )
T BlN 2r BlND BA N D A=l.D, Where D is the coil diaeter The controlling torque exerted by the spiral springs is directly proportional to the deforation or windup of the springs. Thus, the controlling torque is proportional to the actual angle of deflection of the pointer. T K where K C is a constant
For a given deflection, the controlling and deflecting torques are equal: K BlND BlND C where C K is a constant This equation shows that the pointer deflection is always proportional to the coil current. Consequently, the scale of the instruent is linear, or uniforly divided.
Exaple 1 v A PMMC instruent with a 100-turn coil has a agnetic flux density in its air gaps of B = 0.2 T. The coil diensions are D = 1 c and l = 1.5 c. Calculate the torque on the coil for a current of 1 A. Solution: T T D D B l N D 0.2 (1.5 10 310 6 N. 2 ) 0.001100 (110 2 )
Galvanoeter t is a PMMC instruent designed to be sensitive to extreely low current levels. The siplest galvanoeter is a very sensitive instruent with the type of center-zero scale. Galvanoeters are often eployed to detect zero current or voltage in a circuit rather than to easure the actual level of current or voltage. The ost sensitive oving-coil galvanoeter use tautband suspension, and the controlling torque is generated by the twist in the suspension ribbon. For the for greatest sensitivity, the weight of the pointer can create a proble. The solution is by ounting a sall irror on the oving coil instead of a pointer Basic deflection syste of a galvanoeter using a light bea
An adjustable shunt resistor is eployed to protect the coil of a galvanoeter fro destructively excessive current levels. The shunt resistance is initially set to zero, then gradually increased to divert current through the galvanoeter. 27
Exaple 2 v A galvanoeter has a current sensitivity of 1 A/ and a critical daping resistance of 1 k. Calculate (a) the voltage sensitivity and (b) the egaoh sensitivity. Solution: oltage sensitivity 1kΩ1 A 1/ For a voltage sensitivity of 1/, 1/ egoh sensitivity 1MΩ 1 A
DC Aeter An aeter is always connected in series with a circuit in which current is to be easured. To avoid affecting the current level in the circuit, the aeter ust have a resistance uch lower than the circuit resistance. For larger currents, the instruent ust be odified so that ost of the current to be easured is shunted (a very low shunt resister) around the coil of the eter. Only a sall portion of oving coil. the current passes through the
A dc aeter consists of a PMMC instruent and a lowresistance shunt. sh sh sh sh sh sh sh
Exaple 3 A PMMC instruent has FSD of 100 A and a coil resistance of 1 k. Calculate the required shunt resistance value to convert the instruent into an aeter with (a) FSD = 100 A and (b) FSD = 1 A. Solution: (a) FSD = 100 A s s s s 100 (b) FSD = 1 A s s s μ A 1 kω 100 A 100 100 μ 100 1.001 Ω 99.9 A 100 1 A 100 100 999.9 A μ A 0.1001 A 99.9 999.9 Ω A A
Exaple: An aeter has a PMMC instruent with a coil resistance of = 99 and FSD current of 0.1 A. Shunt resistance s = 1. Deterine the total current passing through the aeter at : a) FSD, b) 0.5 FSD, and c) 0.25 FSD Solution (a) At FSD eter voltage and 0.1A 99Ω 9.9 s s s 9.9 9.9A 1Ω s totalcurrent 9.9A 0.1A s 10A 32
(b) At 0.5 FSD (c) At 0.25 FSD 0.25 0.1A 0.025 A 0.5 0.1A 0.05 A 0.05 A 99 4.95 s 4.95 4.95 A 1 s total current 4.95 A 0.5 A 5 A s 0.025 A 99 2.475 s 2.475 2.475 A 1 s total current 2.475 A 0.025 A 2.5 A s Thus, the aeter scale ay be calibrated linearly fro zero to 10A
Aeter Swaping esistance The oving coil in a PMMC instruent is wound with thin copper wire, and its resistance can change significantly when its teperature changes. The heating effect of the coil current ay be enough to produce a resistance change, which will introduce an error. To iniize this error, a swaping resistance ade of anganin or constantan is connected in series with the coil: (anganin and constantan have resistance teperature coefficients very close to zero). The aeter shunt ust also be ade of anganin or constantan to avoid shunt resistance variations with teperature.
Multirange Aeters Make-before-break switch The instruent is not left without a shunt in parallel with it even for a brief instant. f this occurred, the high resistance of the instruent would affect the current flowing in the circuit. During switching there are actually two shunts in parallel with the instruent. 35
Ayrton Shunt The figure shows another ethod of protecting the deflection instruent of an aeter fro excessive current flow when switching between shunts. esistors 1, 2, and 3 constitute an Ayrton Shunts sh nternal Aeter esistance: in in / / s * in sh * * sh sh sh 36
Exaple: A PMMC instruent has a three-resistor Ayrton shunt connected across it to ake an aeter as shown in the figure. The resistance values are 1 = 0.05, 2 = 0.45 and 3 = 4.5. The eter has = 1k and FSD = 50A. Calculate the three ranges of the aeter. Solution Switch at contact B 50μA 1kΩ 50 s s s 50 10A 0.05Ω 0.45Ω 4.5Ω 1 2 3 50μA 10A s 10.05A 37
Switch at contact C s s 1 50 μa1kω 4.5 Ω s 100.05A s 2 3 50 100A 0.05 Ω0.45 Ω 50 μa100a 50 Switch at contact D s s s 1 50μ01kΩ 4.5Ω 0.45Ω 50 1A 0.05Ω 1.00005A s 3 2 50μ0 1A 50
Accuracy and Aeter Loading Effects nternal resistance of ideal aeter is zero Oh, but in practice, the internal resistance has soe values which affect the easureent results. This error can be reduced by using higher range of easureent. Let us calculate the relationship between the true value and the easured value T Th ( true value ) Th ( easured value ) Th Th in dc circuit with source and resistors T th th T Accuracy T h T T h in % Acc 100% T T h T h in 100% dc circuit with source and resistors A th th A 39
Exaple: For a DC Circuit as shown in the figure, given 1 =2k, 2 =2k with voltage of 2. By easuring the current flow through 3 with a dc aeter with internal resistance of in = 100Ω, calculate percentage of accuracy and percentage of error. Solution Th / / 2 kω Th Th T 1 2 3 E 2 2 2 kω 1 1 2 2 kω 2 kω Th 1 500μA 2kΩ Th Th 1 476.19μA 2 kω100ω Th in Th 476.19 μa % Error 1 % A cc 1 95.24% 4.76% 500 μa % Acc 100% 95.24% T 40
DC olteter The deflection of a PMMC instruent is proportional to the current flowing through the oving coil. The coil current is directly proportional to the voltage across the coil. The coil resistance is norally quite sall, and thus the coil voltage is also usually very sall. Without any additional series (ultiplier resistance) resistance the PMMC instruent would only easure very low voltage. The volteter range is easily increased by connecting a resistance in series with the instruent. The eter current is directly proportional to the applied voltage, so that the eter scale can be calibrated to indicate the voltage.
The volteter range is increased by connecting a ultiplier resistance with the instruent (single or individual type of extension of range). s s 1 s Last equation can be used to select the ultiplier resistance value ( s ) for certain voltage range (FSD). n this case will be the full scale current. A ultiplier resistance that is nine ties the coil resistance will increase the volteter range by a factor of 10 (ultiplier resistance + coil resistance) The volteter sensitivity (S) is defined as the total volteter resistance (internal resistance in ) divided by the voltage range (full scale)
Exaple: A PMMC instruent with FSD of 100 A and a coil resistance of 1k is to be converted into a volteter. Deterine the required ultiplier resistance if the volteter is to easure 50 at full scale and olteter sensitivity. Also calculate the applied voltage when the instruent indicate 0.8, 0.5, and 0.2 of FSD. Solution At 50 FSD s s 50 100 μa s 1kΩ 499 kω 100 μa
Since the volteter has a total resistance of v = s + = 500k, then its resistance per volt or sensitivity is 500k / 50 =10 k /. At 0.8 of FSD 0.8100 μa 80 μa s 80 μa 499 kω1kω 40 At 0.5 of FSD At 0.2 of FSD 50μA 50μA 499 kω1kω 25 20μA 20μA 499 kω 1kΩ 10 Thus, the volteter scale ay be calibrated linearly fro zero to 50
olteter Swaping esistance As in the case of aeter, the change in coil resistance ( ) with teperature change can introduce errors in a PMMC volteter. The presence of the volteter ultiplier resistance ( s ) tends to swap coil resistance changes, except for low voltage ranges where s is not very uch larger than. n soe cases it ight be necessary to construct the ultiplier resistance for anganin or constantan. Multirange olteters ndividual or series connected resistors ay be used as shown in the following configurations Multirange volteter with ranges = ( +) where can be 1, 2, or 3 Multirange series volteter with ranges = ( +), where can be 1, 1 + 2, or 1 + 2 + 3
Exaple: A PMMC instruent with FSD = 50A and = 1700 is to be eployed as a volteter with ranges of 10, 50, and 100. Calculate the required values of ultiplier resistors for the two circuits shown in the previous slides. Solution For the circuit shown Switch contact at + 1 = = - 1 10 1700 Ω 200 kω 1.7 kω 50 μa 198.3 kω Switch contact at 2 50 50 A 998.3 k 1700Ω Switch contact at 3 100 1700 Ω 50 A 1.9983 M 46
For the circuit shown Switch contact at 1 1 1 10 1 1700 Ω 198.3 kω 50 μa Switch contact at 2 1 2 2 50 2 1 198.3 kω 1700 Ω 800 kω 50 μa Switch contact at 3 1 2 3 100 800 kω 198.3 kω 1700 Ω 1 M Ω 50 μa 3 3 2 1
Accuracy and olteter Loading Effect Let us calculate the relationship between the true value ( T ) and the easured value ( ) T T T Th Th in Th in Accuracy in T in Th % Acc 100% T in % Acc 100% in Th 48
Exaple: A volteter with sensitivity of 20kΩ/ is used for easuring a voltage across 2 with range of 50 as shown in the figure below. Calculate a) reading voltage. b) accuracy of easureent. c) error of easureent Solution E 100 Th T 2 200k 50 1 2 200k 200k / / 200 k / / 200 k 100 k Th 1 2 20k a) in S ange 50 1M in 1M T 50 45.45 in Th 1M 100k 45.45 b ) A ccuracy 0.909 or % A cc 90.9% 50 T Th Th c ) Error 1 A cc 1 0.909 0.091 or % Error 9.1%
AC olteter Full-Wave ectifier olteter Half-Wave ectifier olteter Half-Wave Full Bridge ectifier olteter
AC Aeter and olteter When an alternating current (sinusoidal ) with a very low frequency (0.1 Hz or lower) is passed through a PMMC instruent, the pointer tends to follow the instantaneous level of the AC. As the current grows positively, the pointer deflection increases to a axiu at the peak of the ac. Then as the instantaneous current level falls, the pointer deflection decreases towards zero. When the ac goes negative, the pointer is deflected (off-scale) to the left of zero. With the noral 50 Hz or higher supply frequencies, the daping echaniss and the inertia of the eter oveent prevent the pointer fro following the changing instantaneous levels of the signal. The instruents pointer settles at the average value of the current flowing through the oving coil which is zero. PMMC instruent can be odified by one of the following circuits to easure AC signals
1. Full-Wave Bridge ectifier olteter When the input is positive, diodes D 1 and D 4 conduct, causing current to flow through the eter fro top to botto ( red solid path). When the input goes negative, diodes D 2 and D 3 conduct, current flows through the eter fro the positive to the negative terinal ( blue dashed path). The ac volteter uses a series-connected ultiplier resistor ( s ) to liit the current flow through the p instruent. av The eter deflection is proportional to the average current ( av ), which is 0.637 peak current ( p or ). But the actual current (or voltage) to be indicated in ac easureent is norally the rs = 0.707 p (or ). ( note rs = 1.11 av p (or ) = 1.414 rs ) When other than pure sine waves are applied, the volteter will not indicate the rs voltage. rs
s D 2 D 4 D 3 applied peak voltage( ) - rectifiers voltage drop 1.414 rs 2 totat circuit resistance p rs F s F is rectifier voltage drops for D 1 and D 2 or D 3 and D 4 Peak current = av /0.637 = FSD /0.637 Exaple: A PMMC instruent with FSD = 100 µa and = 1 kω is to be eployed as an ac volteter with FSD = 200 (rs). Silicon diodes with F = 0.7 are used in the bridge rectifier circuit of shown above. Calculate: (a) the ultiplier resistance value required, (b) the pointer indications the rs input voltage is (i)100 and (ii) 50.
Solution At FSD, the average current flowing through the PMMC instruent is 100µA av 157 A 0.637 p 1.414 rs 1.414 200 282.8 olt p 2 F (a) (b) s i-for the rs input voltage is 100 s p 2 F (282.8 1.4) 1k 1791.36 k 3 (15710 ) A \ 2 \ p F 1.414100 1.4 78.11A (1791.36 1) k s 0.607 49.76 A 50 A 0.5FSD \ \ av ii- Siilarly, for the rs input voltage is 50 25A 0.25FSD \\ av
2. Half-Wave ectifier olteter SH shunting the eter is included to cause a relatively large current to flow through diode D 1 (larger than the eter current) when the diode is forward biased. This is ensure that the diode is biased beyond the knee and will into the linear range of its characteristics. Diode D 2 conducts during the negative half-cycles of the input. When conducting, D 2 cause a sall voltage drop across D 1 and the eter, thus preventing the flow of any significant reverse leakage current ( and reverse voltage) through the eter via D 1. n the half wave rectifier
Half-Bridge Full-Wave ectifier olteter During the positive half-cycle of the input, diode D 1 is forward biased and D 2 is reverse biased. Current flows fro terinal 1 through D 1 and the eter and then through 2 to terinal 2. but 1 is in parallel with the eter and 2. Therefore, uch of the current flowing in D 1 pass through 1 while only part of it flows through the eter and 2 During the negative half-cycle, diode D 2 is forward biased and D 1 is reverse biased. Current flows fro terinal 2 through 1 and the eter and then through D 2 to terinal 1. New, 2 is in parallel with the series connected eter and 1 This arrangeent forces the diodes to operate beyond the knee of their characteristics and helps to copensate for differences that ight occur in the characteristics of D 1 and D 2.
AC Aeter Like a dc aeter, an ac aeter ust have a very low resistance because it is always connected in series with the circuit in which current is to be easured. This low-resistance requireent eans that the voltage drop across the aeter ust be very sall, typically not greater than 1000. The voltage drop across a diode is 0.3 to 0.7. The use of a current transforer gives the aeter a low terinal resistance and low voltage drop.
A. Series Oheter Oheter Basic Circuit and Scale The siplest circuit consists of a voltage source (E b ) connected in series with a pair of terinals (A & B), a standard resistance ( 1 ), and a low-current PMMC instruent. The resistance to be easured ( x ) is connected across terinal A and B. The eter current E ( + + ) b x 1 When the oheter terinals are shorted ( x = 0) eter full-scale deflection occurs. FSD = E b / ( 1 + ) At half-scale deflection x = 1 + At zero deflection the terinals are open-circuited ( x = ).
Exaple : The series oheter shown in Figure is ade up of a 1.5 battery, a 100 µa eter, and a resistance 1 which akes ( 1 + ) =15kΩ. a) Deterine the instruent indication when x = 0. b) Deterine how the resistance scale should be arked at 0.75 FSD, 0.5 FSD and 0.25 FSD Solution E b 1.5 a) 100 A FS D 0 15k x 1 b ) A t 0.75 FSD : 3100A E b 75 A & x 1 4 Eb x 1 1.5 15k 5k 75A 100A 1.5 At 0.5 FSD : 50 A & x 15k 15k 2 50A
100A 1.5 At 0.25 FSD : 25 A & x 15k 45k 4 25A The oheter scale is now arked as shown in the figure. t is clear that the oheter scale is nonlinear. Coents: disadvantages of siple series oheter The siple oheter described in last exaple will operate satisfactorily as long as the battery voltage reains exactly at 1.5. When the battery voltage falls, the instruent scale is no longer correct. Although ๑ were adjusted to give FSD when terinals A and B are shortcircuited, the scale would still be in error because now id-scale would represent a resistance equal to the new value of 1 +.
Oheter with Zero Adjust Falling battery voltage can be taken care by an adjustable resistor ( 2 ) connected in parallel with the eter. With terinals A and B short-circuited, the total circuit resistance is 1 + ( 2 // ). Since 1 is always very uch larger than 2 //, the total circuit resistance can be assued to equal 1 E b b = E b x + 1 + 2 \\ if 2 \\ << 1 then b + A lso, the eter voltage is = \\ which give eter current as = x 1 b 2 When x equal to 1 the circuit resistance is doubled and the circuit current is halved. This cause both 2 and to be reduced to half of their previous level. Thus the id-scale easured resistance is again equal to 1. \\ b 2 Each tie the oheter is used, terinals A and B are first short circuited, and 2 is adjusted for zero-oh indication on the scale
The series oheter can be converted to a ulti-range oheter by eploying several values of standard resistance 1 and a rotatory switch The ajor inconvenience of such a circuit is that a large adjustent of the zero control ( 2 ) would have to be ade every tie the resistance range ( 1 ) is changed.
Exaple: An oheter as shown in the figure with E b = 1.5, 1 = 15kΩ, = 2 = 50Ω and FSD = 50µA. Calculate, (a) x at 0.5FSD, (b) when E b = 1.3 what is the value of 2 to get full-scale current and (c) when E b = 1.3 what is the value of x at halfscale current. Solution Since ( // 2 ) = 25 << 1 then at half scale x = 1 = 15k independent of E b (a) 25 A 50 1.25 2 1.25 50 2 2 25 A 25 A 25 A 50 A b E 1.5 x 1 30k 50 A b E x 1 30k 15k 15k b
(b) (c) b 2 2 Eb 1.3 86.67A 0 15k x 2 1 86.67A 50A 36.67A b FSD 50A 50 2.5 FSD 2 2.5 68.18 36.67A 2 5 A 5 0 1.2 5 1.2 5 1 8.3 3 A 6 8.1 8 2 2 1 8.3 3 A 2 5 A 4 3.3 3 A b E 1.3 x 1 3 0 k 4 3.3 3 A 3 0 k 3 0 k 1 5 k x x 1 5 k b 1 Since ( // 2 ) = 28.85 << 1 then at half scale x = 1 = 15k independent of E b
B. Shunt Oheter Basic Circuit and Scale The siplest circuit consists of a voltage source (E) connected with an adjusted esistor ( Adj ) and a low-current PMMC instruent. The resistance to be easured ( x ) is connected across terinal A and B. When x = 0, short circuit between A and B, there will be no current flow in the coil branch and the scale point at zero on the left hand side. When x =, open circuit between A and B. Then adjust Adj to get FSD. The eter will point infinity at the right of the scale. E FS D A dj For any x we have, E x Adj x Adj Scale of shunt oheter is opposite to the scale of series oheter when connecting with x
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