Radio broadcast .. (1922-30)

.RADIO BROADCAST. 0 20 40 60 80 100 120 140 160 180 O.C.MILLIAM PERES Fig. 3 a fixed value of Ci and various secondary supply voltages. Referring to these curves, we can determine the required secondary voltage and we will then be ready to de- sign our transformer. Here we must use some previous experience to tell us ap- proximately how much capacity will be necessary at Ci to secure efficient filter- ing. Let us try a value of 4 mfd. From the curves of Fig. 3 we find that a secondary voltage of 1400 volts is required to pro- duce 613 volts at 180 milliamperes with 4 mfd. as the value of Ci. We are now ready to tackle the prob- lem of transformer design, and it is here that practical rules and experience count very much as it is quite an art to be able to hit the nail on the head the first time. The " cut-and-try" method is used _ en- tirely in commercial practice, but a skilled designer can come within a very close margin on his first try. Transformer Design First we must tabulate our require- ments in the way of approximate voltages and currents. Let us say our transformer is to work from a line of 115 volts at 60 cycles. We know our secondary voltage must be 1400 volts, and the load delivered by the rectifier tubes is 613 X 0.180 = 110 watts. Assuming an efficiency of about 75 per cent, for the rectifier tubes, we get 145 watts from the secondary winding. This tells us that our secondary current will be approximately 100 mA. We have the filaments of two 281-type tubes to light, each requiring 7.5 volts, or 15 volts at 1.25 amperes, or 18.8 watts; we have three screen-grid filaments to light at 2.5 volts, 5.2 amps, or 13.0 watts; two 250's, requiring 7.5 volts at 2.5 amps, or 18.8 watts, and two 227's, requiring 2.5 volts and 3.5 amps, or 8.8 watts. The total fila- ment output is 58.3 watts and the total power is 203 watts. Assuming 90 per cent, efficiency, the input watts should be 225, and assuming 90 per cent, power factor, this gives a primary current of approxi- mately 2.2 amps. From the above we can choose our wire sizes, allowing 800-1000 circular mils per ampere for the inside windings and 600- 800 circular mils per ampere for outside windings. This calls for approximately 1700 circular mils for the primary, or number 18 wire (see Fig. 4 for wire table). The secondary requires 80 circular mils or number 31 wire. The filament winding for the 281-type tubes requires 800 cir- cular mils, or number 21 wire, all the 2§- volt filaments can be supplied from one winding with a consequent load of 8.75 amps, requiring 6000 circular mils, which can be obtained by using two number 15 wires in parallel. The filament winding for the 250-type tube requires 1700 circular mils, or number 18 wire. Thus we have the proper size of wire for all windings and are ready to compute the proper number of turns and the coil itself. Most manufacturers have certain stan- dard core sizes which are used for the various transformer jobs. Since the dies are already existent for these laminations it would be best to use the most conveni- ent size, if it permits designing an economi- cal and sensible transformer. In the event that a new lamination must be designed it is evident that the problem is entirely one of economics, i.e., balancing copper cost and iron cost to obtain the cheapest possible transformer. Occasionally space enters into the problem very forcibly, preventing one from realizing an ideal, and often making very weird shapes necessary. The lamination shown in Fig. 5 illus- trates a core shape that will be found quite desirable for a 60-cycle transformer of the sort needed in this specific job. If designed for 25 cycles more iron would be necessary and it would be advisable to increase all dimensions, thus allowing us to use more turns of wire and less iron than if the lamination of Fig. 5 were used. It is obvious that if the cross section of the core were square the cost of wind- ing the coil would be less and the space taken up would also make it more adapt- able to the majority of jobs. Therefore, let us try a core 1J" square, an area of 3.06 square inches. ties, and this simplifies our design very considerably. Returning to our problem at hand, we can now determine the proper number of primary turns, for 60-cycle operation, at 115 volts— E ~ 4.44 X 60 X 10,000 NA _ At -5 cycles E - - 69.2 6X 115 3.06 = 226 turns In determining the secondary and ter- tiary turns we must make still greater approximation. The computation of the exact turns ratio depends upon a pre- knowledge of primary and secondary re- sistance and reactance, primary and sec- ondary current, voltages, and power fac- tors, and consequently is a very compli- cated process. Here it is best to use the results of experience and make a rough approximation. It will be found in the majority of cases that if the proper sizes of wire are chosen the ratio of turns will be only two or three per cent, higher than the ratio of voltages under load con- ditions. This, then, gives us a conveni- ent approximation for turns ratio. We want a 1400-volt secondary with 115 volts on the primary, hence our turns Fig. 4 To obtain the proper number of pri- mary turns we use the well-known for- mula— E = 4.44 f NAB X 10- 1 where f = Frequency in cycles per second A = Area in square centimeters N = Number of primary turns B = Flux density in lines per square centi- meter E = Primary voltage It can be shown that at 60 cycles a flux density of 10,000 to 11,000 lines per square centimeter is about correct, as more than that produces excessive core loss. Similarly, at 25 cycles we can use 12,000 or 13,000 lines per square centi- meter. Substituting these values we ob- tain— NA 10» _ , 7 - At 60 cycles — = ratio will be plus 2£ per cent, or 12.50. Since our primary (1.025) 115 has 226 turns our secondary will need 12.50 X 226, or 2700 turns. Similarly, we obtain the turns on the tertiary windings — 281 filaments. N 281 = 226 X turns X 1.025 = 30.2 250 filaments. N 250 = 226 X X 1.025 = 15.1 2 50 224 and 227 filaments. N 227 = 226 X ~ X 1.025 = 5.0 turns Here it must be mentioned that this transformer is being designed for zero resistance cable, which, of course, is not obtained in practice. We must know our cable drop, and add this to the desired filament voltage in order to get the re- quired voltage at the transformer. We now have all the windings on our transformer— Primary 226 turns of number 18 wire Secondary 2700 281 fil. 00 " 25061. "5 224 and 227 fil. 5 31 21 " 18 " 15 " doubled The secondary and the filament wind- ings for the 250-type tubes must be center- tapped. Our next problem is to see whe- ther this coil will fit inside the window in the lamination, leaving enough space for commercial variation. Here we return to our wire table (see Fig. 4) and note the .tk' E, . -,.-.4 X 25 X 13,000 converting A to square inches this be- comes— At 60 Cycles^ = 5.8 B TWA At 25 cycles^ =10.7 In ordinary practice we use the factors 6 and 11 to give us the proper flux densi- • NOVEMBER 1929 • Fig. 5 • 47