Figure 2.1: d Arsonval meter

Transcription

1 DEE 2313 SEM ntroduction to DC eters CHAPTE 2 DC and AC eters For any decades the basic dc eter oveent has been used as the readout device in electronic instruent, even in any instruents that easure ac values The two ost coon dc eter oveent configurations are the d'arsonval and the taut-band designs. Both are the exaples or the peranent agnet oving coil (PMMC) galvanoeter, and use the sae fundaental principle as the dc otor. 2.2 Moving Coil nstruent Moving coil instruent or generally referred as the d'arsonval eter or a peranent agnet oving coil (PMMC) eter is ostly used in direct current easureent. The basic construction of this instruent is shown in Figure 2.1. t is a horseshoe shaped peranent agnet with soft iron pole pieces attached to it. Between the pole pieces is a cylindrical shaped soft iron core about which a coil of fine is wound. This wire is wound on a very light etal frae and ounted in such a setting so that it can rotate freely. A pointer attached to the oving coil deflects up scale as the oving coil rotates. Figure 2.1: d Arsonval eter Current fro a easured circuit passes through the winding of the oving coil. This current causes the oving coil to behave such as an electroagnet with its own north and south poles. This poles interact with the poles of the peranent agnet causing the coil to rotate. The pointer deflects up scale whenever current flows in the proper direction in the coil. For this reason, all dc eter show polarity arkings. t should be ephasized that d'arsonval eter oveent is a current responding device. egardless of the units for which the scale is calibrated, the oving coil responds to the aount of current through its windings. 1

2 DEE 2313 SEM 0809 The PMMC instruent is essentially a low level dc aeter; however, with the use of parallel-connected resistor, it can be eployed to easure a wide range of direct current levels. The instruent ay also be ade to function as a dc volteter by connecting appropriate value resistor in series with the coil. AC aeters and volteters can be constructed by using rectifier circuits with PMMC instruent. Oheters can be ade fro precision resistor, PMMC instruents and batteries. Multi-range eters are also available which cobine aeter, volteter and oheter functions all in one instruent Advantages and disadvantages of oving coil instruent Advantages low power consuption scale are unifor no hysterisis loss cheap and robust Disadvantages can easure dc only soe errors due to aging control spring and the peranent agnet 2.3 D'Arsonval Meter as a DC Aeter Since the windings of the oving coil are constructed of very fine wire, the basic d'arsonval eter has liited usefulness without odification. One desirable odification is to increase the range of current that can be easured by the basic eter. This is done by placing a low resistance in parallel with oving coil resistance,. This low resistance is called a shunt ( sh ) and its function is to provide an alternative path, for the total eter current around the eter oveent. The basic dc aeter circuit is shown in Figure 2.2. n the ost circuits, sh is uch greater than, which flows in the oving coil itself. The resistance of the shunt is found by applying Oh's Law to Figure 2.2. Figure 2.2 : Basic dc aeter circuit 2

3 DEE 2313 SEM 0809 where : sh sh V V sh = resistance of the shunt = internal resistance of the oving coil = current through the shunt = full scale deflection current of the oving coil = full scale deflection current of the aeter = voltage drop across the = voltage drop across the sh The voltage drop across the eter oveent, V = Since the shunt resistor is parallel with the oving coil, V = V sh The current through the shunt, sh = - Therefore, the value of the shunt is V sh = = sh = sh sh sh (2.1) The purpose of introducing the shunt current is to easure a current that is n ties larger than. The nuber n is known as ultiplying factor and relates total current and oving coil current as = n Substituting this for in equation (2.1) yields sh = n sh n 1.(2.2) 3

4 DEE 2313 SEM 0809 Exaple 2.1 A 100uA eter oveent with an internal resistance of 800Ω is used in a A aeter. Find the value of the required shunt resistance. Solution The ultiplication factor n is the ratio of 100 A to 100µA Therefore, n = / = 100 A / 100 µa = 1000 sh = / (n 1) = 800 / 999 = 0.8 Ω Exaple 2.2 Calculate the value eter with a 100 Ω of shunt resistance required to convert a 1 A oving coil eter with a 100 Ω internal resistance into a 0-10 A aeter using both using both equation (2.1) and (2.2). Solution V = = 1 A x 100Ω = 0.1 V V sh = V = 0.1 V sh = - = (10 1) A = 9 A Using equation (2.1) sh = V sh / sh = 0.1 V/ 9 A = Ω Using equation (2.2) n = / = 10 A / 1 A = 10 sh = ( ) / (n 1) = 100 Ω/ (10 1) = Ω 4

5 DEE 2313 SEM Ayrton Shunt The shunt resistance discussed in the previous sections work well enough on a single range aeter. However, on a ultiple-range aeter, the Ayrton shunt is frequently a ore suitable design. One advantage of the Ayrton shunt is it eliinates the possibility of the oving coil to be in the circuit without any shunt resistance where they protect the deflection instruent of the aeter fro an excessive current flow when switching between shunts. Another advantage is that it ay be used as a wide range aeter. The Ayrton shunt circuit is shown in Figure 2.3. Figure 2.3: An aeter using an Ayrton shunt The individual resistance values of the shunts are calculated starting fro the ost sensitive range, 1 and works towards the least sensitive range, 3. At the ost sensitive range (point A), the shunt can be coputed using equation (2.2), where, sh n 1.(2.2) The equation needed to copute the value of each a, b and c can be developed by observing Figure 2.3. Notice that a + b + c = sh Now, observe at point B, ( b + c ) ( a + ) Then the voltage across the branch can be written as V( b + c ) = V( a + ) 5

6 DEE 2313 SEM 0809 n current and resistance ters we can write as ( b + c )( 2 ) = ( a + ) 2 ( a + c ) ( a + c ) = ( a + ) 2 ( b + c ) = ( b + c ) + ( a + ) = ( a + b + c + ) = ( sh + ) b +c = ( sh + ) / 2 (2.3) At point C, c ( a + b + ) Thus, in current and resistance ters we can write it as V c = V (a + b + ) ( c )( 3 - ) = ( a + b + )( ) 3 c = ( a + b + c + ) 3 c = ( sh + ) c = ( sh + )/ 3 (2.4) The value of b can be obtained by substituting equation 2.4 into equation 2.3 which yields to: 1 1 b ( sh ) (2.5) 2 3 Exaple 2.3 Calculate the value for a, b and c as shown in Figure 2.3, given the value of internal resistance, = 1 kω and full scale current of the oving coil = 100 µa. The required range of current are: 1 = 10 A, 2 = 100 A and 3 = 1A. Solution At the ost sensitive range, n = / = 10 A/ 100 µa = 100 Total shunt resistance, sh = / (n-1) = 1 kω / 99 = 10.1 Ω Using equation 2.4 (at point C), c = (100 µa )(10.1 Ω +1kΩ) / 1A = Ω 6

7 DEE 2313 SEM 0809 Using equation 2.5, b = (100 µa )(10.1 Ω +1kΩ)[(1/100A) - (1/1A))] = Ω For a, since sh = a + b + c, thus a = sh - ( b + c ) a = 10.1 Ω (0.909 Ω Ω) = 9.09 Ω Aeter nsertion Effects All aeters contain soe external resistance, which ay range fro a low to a greater value. nserting aeter in a circuit always increase the resistance of the circuit and therefore reduces the current in the circuit. as Without aeter, the current flows in the circuit shown in Figure 2.4 can be calculated E e (2.6) 1 Figure 2.4:Circuit without aeter insertion effects Figure 2.5: Circuit with aeter insertion effects nserting the aeter as in Figure 2.5 (X-Y terinal), will reduce the current in the circuit to: 1 E (2.7) 7

8 DEE 2313 SEM 0809 Dividing equation (2.7) by equation (2.6) gives the following expression: e 1 1 (2.8) e nsertion error = x100 % e (2.9) Exaple 2.4 Deterine the insertion error due to using aeter in Figure 2.5 if E = 100 V, 1 = 100 Ω and = 100 Ω. Solution e = E / 1 = 100V/ 100 Ω = 1 A = E/ ( 1 + ) = 100V/ (100 Ω+100 Ω) = 0.5A nsertion error = (1-0.5)/1 x 100% = 50 % 2.4 D Arsonval Meter as a DC Volteter The basic d Arsonval eter can be converted to a dc volteter by connecting a ultiplier s in series with it as shown in Figure 2.6. The purpose of the ultiplier is to extend the range of the eter and to liit the current through the d Arsonval eter to the axiu full-scale deflection current. Figure 2.6: Basic dc volteter circuit To find the value of the ultiplier resistor, we ay first deterine the sensitivity, S, of the d Arsonval. f the sensitivity is known, the total volteter resistance can be calculated easily. The sensitivity of a volteter is always specified by the anufacturer, and is frequently printed on the scale of the instruent. f the full-scale eter current is known, the sensitivity can be deterined as the reciprocal of the full scale current. Sensitivity 1 fs Where fs is the full-scale deflection current of d Arsonval eter. (2.10) 8

9 DEE 2313 SEM 0809 The value of the ultiplier resistance can be found using this relationship: s + = S x V range (2.11) Thus, s = S x V range - (2.12) Exaple 2.5 Calculate the value of the ultiplier resistance on the 50 V range of a dc volteter that used a 500µA d Arsonval eter with an internal resistance of 1 kω. Solution S 1 fs A 2 k / V s = S x ange = 2 kω/v x 50 V 1 kω = 99 kω Multi-ange Volteter A ultirange volteter consists of a deflection instruent, several ultiplier resistors and a rotary switch. Two possible circuits are illustrated in Figure 2.7, Figure 2.7(a): Multirange Volteter 10V 3V 30V Figure 2.7(b): A coercial version of a ultirange volteter 9

10 DEE 2313 SEM 0809 n figure 2.7(a) only one of the three ultiplier resistors is connected in series with the eter at any tie. The range of this eter is V ( ) Where the ultiplier resistance, can be 1 or 2 or 3 n figure 2.7(b) the ultiplier resistors are connected in series, and each junction is connected to one of the switch terinals. The range of this volteter can be also calculated fro the equation V ( ) Where the ultiplier,, now can be 1 or ( ) or ( ) (Note: the largest voltage range ust be associated with the largest su of the ultiplier resistance) Of the two circuits, the one in Figure 2.7(b) is the least expensive to construct. All the ultiplier resistor in Figure 2.7(a) ust be special (non-standard) values, whereas in Figure 2.7(b) only 1 is a special resistor and all other ultiplier are standard-value (precise) resistor. Exaple 2.6 Calculate the value of the ultiplier resistance for the ultiple range dc volteter circuit shown in Figure 2.7(a) and Figure 2.7(b), if fs = 50μA and = 1kΩ 10

11 DEE 2313 SEM DC Volteter Loading Effect As the DC aeter, the DC volteter also observe for loading effect whenever it is inserted to a easured circuit. Figure 2.7 shows a circuit with the DC volteter is inserted into it. nserting volteter always increase the resistance and decrease the current flowing through the circuit. Figure 2.7: Circuit with volteter insertion effect Without the insertion of the DC volteter, the voltage V B can be found as: B VB xe (2.13) A B nserting the volteter in parallel with B gives us the total inserted resistance as : (2.14) T s Thus, yield to (2.15) eq B T Now, the voltage V B with the volteter insertion is found as: eq VB xe (2.16) eq A Therefore, nsertion error = VB VB x100% V (2.17) B 11

12 DEE 2313 SEM D Arsonval Meter as a DC Oheter The d Arsonval eter can also be transfored as an oheter for any circuit s resistance easureent. The basic oheter circuit eploying d Arsonval eter is shown in Figure 2.8: Figure 2.8: Basic DC oheter Before x is easured, the oheter ust first be calibrated. The zero calibration is perfored by shorting the terinal x-y and adjusting z so that full-scale deflection on the eter oveent is obtained. Without x, the current equation for full-scale deflection becoes E fs (2.18) z With x, the current equation will now becoes: E (2.19) x z Notice that, by introducing x, the current will always less than the full-scale current < fs The relationship between the full-scale deflection with the value of x can be given as z P (2.20) fs z x Practically, this equation is being used to develop the arking scale on the eter face to indicate the value of the easured resistor. 12

13 DEE 2313 SEM Multiple-range Oheters The oheter range discussed in the previous section is not capable of easuring resistance over wide range of values. Therefore, we need to extend our discussion of oheters to include ultiple-range oheters. Figure 2.9 shows one way of circuit configuration for developing ulti-range oheter. Figure 2.9: Mult-range oheter To siplify our analysis, lets the value each coponents be 1 = 10Ω z = 28kΩ 2 = 100Ω fs = 50uA 3 = 1kΩ E = 1.5V = 2kΩ This instruent akes use of a basic 50uA eter oveent with an internal resistance of 2kΩ. An additional resistance of 28kΩ is provided by z. z is necessary to liit current through the eter oveent to 50uA when the test probes (not shown) connected to X and Y are shorted. As ay be seen, when the instruent is on the x1 range, a 10Ω resistor is in parallel with the eter oveent. Therefore, the internal resistance of the oheter on the x1 range is 10Ω 30kΩ which approxiately 10Ω. This eans the pointer will deflect to id-scale when a 10Ω resistor is connected across X and Y. When the instruent is set to x10 range, the total resistance of the oheter is 100Ω 30kΩ, which is now approxiately 100Ω. Therefore, the pointer deflects to id-scale when 100Ω resistor is connected between the test probes. Mid-scale is arked as 10Ω. Therefore, the value of the resistor is deterined by ultiplying the reading by the range ultiplier of 10 producing a idscale value of 100Ω (x100) When our oheter is set on the x100 range, the total resistance of the instruent is 1kΩ 30kΩ which is still approxiately 1kΩ. Therefore the pointer deflects to id-scale when we connect the test probes across a 1-kΩ resistor. This provides us a value for the id-scale reading of 10 ultiplied by 100, or 1kΩ for our resistor. 13

14 DEE 2313 SEM 0809 Exaple 2.6 (a) (b) (c) n figure below, deterine the current through the eter,, when a 20Ω resistor between the terinal X and Y are easured on the x1 range. Show that this sae current flows through the eter oveent when a 200Ω resistor is easured on the x10 range. Show that the sae current flows when a 2kΩ is easured on the x100 range. Figure 2.10: Circuit for Exaple 2.6 with oheter on x1 range. Solution: (a) When the oheter is set on the x1 range, the circuit is as shown above. The voltage across the potential cobination of resistance coputed as 10 V x1.5v 0. 5V The current through the eter oveent is coputed as 0.5V k A (b) When the oheter is set on the x10 range, the circuit is as shown in Figure The voltage across the parallel cobination of resistance coputed as 100 V x1.5v 0. 5V The current through the eter oveent is coputed as 0.5V k A Figure 2.11: Oheter on x10 range (c) When the oheter is set on the x100 range, the circuit is as shown in Figure The voltage across the parallel cobination is coputed as 1k V x1.5v 0. 5V 1k 2k 14

15 DEE 2313 SEM 0809 The current through the eter oveent is coputed as 0.5V k A Figure 2.12: Oheter on x100 range. 15