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18. The Need for Polyphase Circuits Practically all alternatingcurrent power is transmitted over three wires in a threephase circuit. In such a circuit the voltage appearing between any two wires is 120° out of phase with the voltages appearing across the other two pairs. Each wire serves to some extent as a return wire for the other two, permitting a considerable saving in copper. This and other advantages increase the efficiency and power utilization of threephase systems to such an extent over a singlephase supply that practically all largescale industrial power consumers utilize threephase power. (The ordinary house current uses one phase of a threephase system.) Since many industrial processes require DC power, however, polyphase rectifier circuits are needed. Because of the heavy current requirements, gasfilled diodes, such as mercuryvapor tubes and ignitrons, are frequently employed in polyphase rectifiers. When a threephase power supply is available, advantage is taken of its high efficiency by connecting the rectifier tubes in 3, 6, and 12phase circuits. All are derived from the basic threephase supply. Twophase power supplies and their associated rectifiers are of little interest because of their relatively low efficiency. Despite the multiplicity of possible rectifier circuits for polyphase systems, only a few of importance in electronics will be discussed.
19. ThreePhase HalfWave DeltaWye Rectifier The simplest of the threephase rectifiers is the halfwave delta wye circuit, shown in Fig. 16. The terms "delta" (Δ) and· "wye" (Y) refer to the configuration of the primary and secondary coils, respectively, of the threephase transformer. Note in Fig. 16 that the threephase line is connected to the junctions of three deltaconnected primary coils, so that the full primary voltage (E_p) of each phase is developed across each coil. The three secondary coils of the transformer are mounted in the shape of a wye and a secondary voltage (E.) is developed across each coil or leg. These secondary voltages are 120° out of phase with each other; that is, the output sinewave of each coil is displaced in time by 120°, or onethird of a cycle, with respect to the sinewaves of the other two coils. The circuit shown in Fig. 16 consists essentially of three separate halfwave rectifiers, one for each leg of the secondary wye, or one for each phase. The operation of each halfwave rectifier is identical to that described for the singlephase halfwave rectifier, illustrated in Fig. 5 and 6. Each tube carries current only during the positive halfcycles of its phase, or only onethird of the time. The output voltage, therefore, fluctuates at three times the AC supply frequency, as shown in the waveform of Fig. 16. This permits easy filtering of the ripple. Although each of the rectifier tubes is only a halfwave unit, the cathode currents of all the tubes flow through the common load resistance, so the average DC load current is much higher relative to the peak current than in a singlephase rectifier. The average DC current of a threephase halfwave rectifier is 0.826 ...
... times the peak current, compared to a value of 0.318 for the same ratio in the singlephase, halfwave circuit, and 0.636 for the singlephase, fullwave circuit. Because of this high average current value, the DC output voltage is correspondingly high, and as a matter of fact, it is greater than the AC rms voltage of each secondary leg by a factor of 1.17. (The peak voltage is 1.414 times the rms value; hence the average output voltage is 1.414 x 0.826, or 1.17 times the rms voltage.) Equivalently, the rms voltage across each leg (E.) of the secondary need only be 1 / 1.17 or 0.855 times the average desired DC output voltage. 20. ThreePhase HalfWave DeltaZigZag Rectifier The circuit shown in Fig. 17, known as a threephase halfwave deltazigzag rectifier, because of its configuration, is identical with the deltawye rectifier except for the method of installing and mounting the coils in the threephase transformer. In the delta wye circuit (Fig. 16) the rectified current flows through each secondary coil in one direction, which saturates the transformer core and effectively reduces the inductance of the winding, just as in the singlephase halfwave rectifier. This disadvantage is over come in the deltazigzag circuit by supplying each core with two secondary windings operating in phase opposition to each other. As a consequence, each leg is made up of two sections, each operating in a different magnetic circuit. These are connected in successive legs of a zigzag arrangement to give secondary voltages phased 120° apart. Each secondary phase (En) then turns out to be the vector sum of two winding voltages that are effectively 60° out of phase with each other. It may be shown that E. equals sqrroot 3 or 1.732 times the voltage developed across each coil winding. 21. ThreePhase HalfWave DoubleWye (SixPhase) Rectifier What essentially amounts to sixphase operation is obtained in the circuit shown in Fig. 18 by connecting two threephase halfwave circuits in parallel through a balance coil (also known as interphase reactor). This results in reversing the polarities of corresponding secondary windings in the lower wye with respect to those in the upper wye. Thus, when the secondary voltage (En) of one winding in the upper wye is at maximum, the voltage of the corresponding winding in the lower wye will be at a minimum. This effectively adds three extra phases, making a total of six phases.  • In Fig. A, E. is the vector sum of two voltages of magnitude e differing in phase by 60°, as shown in Fig. B. The e vectors may be redrawn as in Fig. C. Triangle Ade forms a 30°60°90° triangle with sides in the proportions __/312. Triaqgle CBE. is a similar triangle with sides in the same relative proportions.  SECONDARY VOLTAGES AND OUTPUT VOLTAGE WAVEFORMS

Note that the circuit of Fig. 18 requires six rectifier tubes, each acting as a halfwave rectifier for one phase, or one leg of each wye. The central junction of each wye is connected to the other through the interphase balance winding, which compensates for any inequalities in the upper and lower sections. The centertap of this winding becomes the negative terminal of the DC output voltage. The plates of the rectifier tubes are connected to the legs of the two wyes, while their cathodes are connected together. Consequently, the cathode currents of all tubes flow through the common load resistance. Examine the secondary wye and output voltage waveforms of the doublewye circuit (Fig. 18). As explained before, the secondary voltage of the first wye (dotted line) is at a minimum when the secondary voltage of the second wye (solid line) is at maximum, and vice versa. The same holds true for the rectified output voltages of each wye, since the halfwave rectifiers simply cut off the negative halfcycles, but do not change the relative amplitudes. There are, therefore, six fluctuations in the output voltages of the wyes for each AC supply cycle and the ripple frequency is six times that of the AC supply. Because of the action of the balance coil, the actual DC output voltage across the load does not follow the peaks of the wye output voltages, but rather the average of the rectified wye output volt ages, as is shown in Fig. 18. The peak current through each tube is only onehalf of the average DC load current. The rms voltage required across each leg of the transformer secondary is still 0.855 times the average desired DC output voltage for the deltawye halfwave rectifier circuit shown in Fig. 16. This somewhat paradoxical situation is due to the requirements of the interphase balance coil, which must insure that the individual threephase halfwave systems (upper and lower wyes) operate independently with current flowing through each tube onethird of the time. Note that the voltage across the balance coil is the difference between the output voltages of the individual wyes. To assure independent operation, the coil must have enough inductance for the voltage (difference) across it to give rise to a peak alternating current that is Jess than the direct current flowing through one leg of the coil. To be exact, the peak value of the alternating current in the coil must be less than onehalf the DC load current. This is the reason for the low peakcurrent per tube compared to the average DC load current. Note also that the direct current flows in opposing directions through the two halves of the balance coil, which means no DC saturation is present. Example: Assume a 230volt threephase, 60cycle AC line is connected to the primary of a l: I transformer operating in the doublewye rectifier circuit, shown in Fig. 18. If the average DC drawn by the load is 100 amps, what are the average and peak currents per rectifier tube, the average DC output voltage, the DC power delivered to the load, and the ripple frequency? Solution: The average current per plate is obviously onesixth of the DC load current, since there are six tubes. Thus, the average current per tube is 100/6=16.7 amperes. The peak current per tube is onehalf the average load current, or ½ x 100=50 amperes. DC output voltage is 1 / 0.855 or 1.17 times the rms value across each leg of the transformer secondary. For a 1:1 transformer, the secondary leg voltage must be 230 volts; the DC output voltage (neglecting the transformer and tube drops) is 1.17 x 230 or 269 volts. In practice this value will he lower because of the voltage drops lost in the tubes and transformer. The DC power delivered to the load is the product of the average DC output voltage and the load current, or 269 x 100=26,900 watts. Finally, the ripple frequency is six times the supply frequency, or 6 x 60=360 cycles. This is easily made pure DC with a simple filter circuit. (See Chap. 4.) 22. ThreePhase DeltaStar (SixPhase HalfWave) Rectifier The equivalent of sixphase operation from a threephase supply can be attained by centertapping the secondary winding of a three phase transformer more easily than it can be achieved in the double wye circuit. Refer to the singlephase, centertapped transformer shown for the fullwave circuit in Chap. 2, Fig. 7. Such a star connected secondary will have voltages of opposing polarities across each half of a winding with respect to centertap, giving effectively a total of six phases (one for each leg of the star.) A typical three phase deltastar rectifier circuit is shown in Fig. 19. Note that the deltastar circuit requires six halfwave rectifier tubes, one for each phase or secondary leg, just like the doublewye circuit. The output waveform is similar to that shown for the doublewye circuit (Fig. 18) and also has a ripple frequency equal to six times that of the AC supply. Because of the absence of the balance coil the limitations with respect to the peak and average currents do not apply to the delta star circuit and the average DC load current of the latter turns out to be 0.955 times the value of the peak current per tube. The average output voltage is thus also 0.955 times the peak voltage, or 1.414 x 0.955=1.35 times the rms value (E.) of the secondary voltage across each leg. Equivalently, the rms voltage across each secondary leg must be 1/1.35 or 0.74 times the average desired DC output voltage. For example, if an output voltage of 300 volts at a load current of 10 amperes is desired in the circuit of Fig. 19, the AC rms voltage across each secondary leg (E.) must be 300 x 0.74=222 volts, and the peak current per tube is 10/.955 or 10.5 amperes. The average current per tube is, of course, 10/6 or 1.67 amperes for each of the six tubes. 23. ThreePhase FullWave DeltaWye Rectifier The fullwave deltawye (sometimes called the threephase, full wave bridge circuit) rectifier circuit shown in Fig. 20 is essentially an extension of the singlephase fullwave bridge rectifier, described in the last section (Fig. 10). Although this circuit uses double the number of tubes of the threephase, halfwave, deltawye rectifier (Fig. 16), it has the advantage of delivering twice the output voltage of the halfwave rectifier for the same total secondary voltage (E.). Note in Fig. 20 that each leg of the secondary wye is connected to the junction of two halfwave rectifiers, arranged backtoback, so that fullwave rectification of both AC halfcycles is obtained from each phase. The waveform of the output voltage, shown at right in Fig. 20, is essentially the same as that of the doublewye circuit (Fig. 18), where the ripplefrequency of the output is six times that of the threephase AC supply.
No balance coil is required, but the power transformer must have three secondary windings and the filament transformer must have four separate secondary windings, one for V1, V2, and V3, and one each for V4, V5, and V6. Since the rectified tube currents flow in opposing directions through the secondaries, no DC magnetization can take place. Three singlephase transformers may be used in place of one threephase transformer. As in the deltastar circuit, the average DC load current is 0.955 times the peak current through each tube, but onethird of the DC load current flows through each tube (rather than onesixth, as in the deltastar circuit), because two tubes are required for each output current pulse. As pointed out before, the average DC output voltage is twice that of the threephase halfwave rectifier, or 2 x 1.17=2.34 times the rms value of the AC voltage across each secondary leg. Conversely, the rms voltage across each leg need be only 1/2.34 or 0.428 times the value of the average desired DC output voltage. Example: The transformer secondary voltage per leg (E,) in a fullwave delta wye rectifier is 150 volts rms, the load current 2 amperes. What is the average DC output voltage, the peak and average current through each tube, and the average power delivered to the load? Solution: Since the average DC output voltage is 2.34 times the rms voltage across each leg, the output voltage is 2.34 x 150=351 volts. The peak current through each tube is 1/0.955 or 1.05 times the average load current; the peak current is 1.05 x 2=2.10 amperes per tube. The average current per tube is onethird of the total, or 1 /3 x 2=0.667 ampere. Finally, the average power is the product of the average current and the output voltage, or 2 x 351=702 watts. 24. Uses of Polyphase Circuits Polyphase rectifiers are used only when the efficiency of rectification is more important than the cost of the rectifier itself. This is generally the case when the required DC power exceeds about 1 kilowatt. Many industrial plants, broadcast and television stations requiring up to 20,000 volts DC at peak currents of 10 amperes or more utilize polyphase circuits. The tubes employed are generally the gasfilled (mercuryvapor) type. Polyphase circuits also have the advantage over singlephase types of developing fairly steady output voltages that require little filtering. Finally, the more efficient polyphase circuits give a higher output voltage relative to the peak inverse voltage of the tube and utilize the power transformer more efficiently than do equivalent singlephase circuits. 25. Review Questions (1) State some of the advantages of threephase power systems over single phase supplies. (2) Draw the circuit and output waveform of a threephase halfwave deltawye rectifier and explain its operation. (3) What is the advantage of the deltazigzag connection over the delta wye circuit? Is there any difference in the output waveform of the circuit; in the average and peak currents; in the ripple frequency? (4) Why is the halfwave doublewye circuit sometimes called a sixphase rectifier? What is the ripple frequency of the output? (5) Explain the function of the interphase balance coil in the doublewye circuit and how it affects the ratio of the peak current per tube to the average load current. Compare ratio with that of the halfwave deltawye rectifier. (6) Compare the operation of a doublewye circuit with that of the deltastar rectifier connection. How would you convert one into the other? (7) A broadcast station requires a DC voltage of 2000 volts at a load current of 10 amperes. For both the deltastar rectifier (Fig. 19) and the doublewye circuit (Fig. 18) compute the required rms voltage across each secondary winding of the transformer, the peak and average current per tube, the DC power delivered to the load and the ripple frequency. (Neglect the tube and transformer voltage drops.) Which circuit would you choose? (8) Draw the circuit of a threephase fullwave deltawye rectifier and compare its operation with that of a singlephase, fullwave bridge rectifier. Compute the same data as for the review question (7) and compare all three circuits. If you had three singlephase transformers of the proper ratings available, would you change your selection in question (7)? (9) Remembering the delta connection, draw the circuit of a threephase fullwave deltadelta rectifier.

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