Radio broadcast .. (1922-30)

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.RADIO BROADCAST . No. 27 Radio Broadcast's Engineering Review Sheets MATCHING IMPEDANCES January, 1930 THE TRANSFER of electric power from one place to another, or from some source to a load is continually taking place and the phe- nomenon no longer excites any public interest. On the other hand, this transfer of power is the engineer's job; he spends his days and nights trying to get either more power from a given source, or the same amount of power at greater efficiency. In a radio set power is taken from a tube arid put into a loud speaker; in an oscillator the power is taken from the tube and put into an antenna. What are the factors the engineer deals with? How can he adjust matters so that he improves the power output, or the efficiency? Consider the circuit in Fig. 1. Offhand it looks like a very simple series circuit consisting of a generator, E, and two resistors, r and R. That is exactly what it is, but at the same time it is the fundamental power circuit and may represent not oiily a battery without resistance feeding current into two resistors, or a dynamo with an internal resistance, r, feeding power into R, or a vacuum tube with a plate resistance, r, feeding power into a load resistance R. When the switch is closed on this circuit, current flows from the source, E, into the two resistors. A certain amount of power is required to force this current through the resistors; this power is numerically equal to I 2 r for the power in the resistor r, and I 2 R for the power used up in the load. Now if we could make a generator or a tube without internal resistance, all the power com- ing from it would be usefully employed in the load, R, but actually this is impossible. Some of the voltage, E, is used up in the internal resist- ance of the source, whether it be a battery, a generator, or a tube, and the remainder is used in the load. The first thing to do is to calculate by Ohm's law the current in the circuit, E/(r + R); then calculate the power used up in the two resistors, I'-'r and I-R; then, in order to find out how efficient the system is, calculate the ratio be- tween the power usefully employed (that in R) to the total power available. Thus, if all the power were used in R (no internal resistance), the system would be 100 per cent, efficient. Such is never the case. Finally we should calcu- late the voltage across the load and across the internal resistance (Ir and IR). In the Data Table we have assumed a potential of 100 volts and a generator resistance, r. of 10. Using these values we filled in some of the values as the resistance of the load. R. varies from 1 to 50 ohms, i.e.. from one-tenth to five times that of the load. The other values should be calculated and filled in and the data plotted against either load resistance or against the ratio between the load resistance and the internal resistance (R / r). Analysis of Data Such calculation and plotting of data is the first half of many experiments; the remainder must be devoted to an analysis of what has happened. One of the first things to note is that the power taken from the generator decreases as the load resist- ance increases, but that more and more j power is used in the load, and less and less is wasted in heating the generator. Note that when the load resistance, R, R is equal to the internal resistance, r, the 1 greatest amount of power is taken by the load and that no further adjustment of the latter results in greater power being 6 used in the load. At this point half the total power taken from the source is used in the load and half in the source; the 15 efficiency is 50 per cent. As the load resistance is increased be- yond this point the power in both load and generator decreases—but the effi- 50 ciency increases. I n other words, the power usefully employed in the load rises from a low value to a maximum and then decreases; power w asted i n the generator steadily decreases; of the total amount of power taken from the generator, more and more is usefully employed as the load resistance is increased, which means simply, that the efficiency of the system as a whole increases as the external ot load resistance increases. Now an engineer usually has one of two things in his mind when he designs power transfer apparatus. Either he wants the maximum possi- ble power to be taken from a source and trans- ferred to a load; or he wants the transfer of what power he gets to take place at the highest possible efficiency. Often he compromises be- tween power output and efficiency. If he has control over the load resistance he can get maximum power into it by making it equal to the generator resistance; he can get maximum efficiency by making it high in comparison to the load resistance. Adjusting the Load Suppose, however, that the engineer has no control over the load resistance. Suppose, for example, it is a 600-ohm telephone line which must be fed with audio-frequency i>ower from a 6000-ohm vacuum tube. Clearly a loss in power will take place compared to the transfer possible if the tube were 600 ohms or the line were 6000. What can be done? A transformer can be i n terposed between the tube and the fine which will enable max- imum power to be transferred provided it has the proper turns ratio. In this case the ratio of secondary (load side) to primary (tube side) would be y eooo / 600 or about 3.16. Then, so far as the load is concerned, the tube resist- ance is stepped down so that it could be re- vwwwv r Fig. I placed by a 600-ohm tube and the transformer thrown away, and, so far as the tube is con- cerned, the line resistance is stepped up to 6000 ohms. The only loss in power in such a case is the loss in the transformer itself, which is small if the latter is properly designed. Matching Impedances The business of making the resistance of the load equal the resistance of the generator is called " matching impedances " and many thousands of transformers have been designed for this purpose. In the case above, the load and generator impedances were pure resistances. If the generator or load contain some reactance, due to capacity or inductance, the problem is more complex. Suppose the generator, for in- stance, had an inductive reactance of 10 ohms. To get maximum power into the load, it would be necessary to match the resistances, and to have the load have a capacity reactance of 10 ohms, to balance out the inductive reactance in the generator. Generally speaking, when the power is small —a few watts, perhaps—engineers match im- Data Table 100 volts; r = 10 ohms; Eff PL across the generator. When the load and the generator resistances are equal, there is aa much voltages on the load as on the generator. The sum of the voltages across these two re- sistors must equal the total available voltage. Power Output from Tubes The greatest power is taken from a tube and used in a load, when the resistance of one is exactly equal to that of the other. The greatest undistorted power, however, is transferred from a tube to a loud speaker when the latter has twice the resistance (radiation) of the tube. After you have plotted the data in the table, note how little power is lost by making the loud speaker have twice the resistance of the tube compared to the power obtainable when the resistances are equal. Note how little power is lost if the load has even five times as much re- sistance. From this you can gather that many of the articles and statements in popular radio journals about the impor tanee of properly matching impedances are exaggerated. As a matter of fact doubling, or halving, the power from a tube into a loud speaker is just about audible to the average ear. In an amplifier which is designed to increase the voltage and not the power, the impedances are not matched. In order to get the greatest voltage out of a low-impedance device, it is necessary to work it into a very high resistance so that of all the voltage available, the greatest part will appear across the load resistance. Thus in an a.i. amplifier the plate circuit works into a very high impedance, sometimes a straight resistance, as in a resistance-coupled amplifier, or the primary of a transformer if a step-up in voltage between tubes is desired. This primary impedance is usually several times greater than that of the tube out of which it works. This impedance must be high at the lowest frequency to which the amplifier is re- quired to transmit without undue loss. Then at all other frequencies, the impedance will be still higher. If the load has twice the resistance of the tube, three-fourths of the total voltage available will be used up across the coupling device, and hence will be applied to the next tube. If a transformer has an inductance such that at 100 cycles its reactance in ohms is three times the resistance of the tube, nearly 90 per cent, of the voltage at that frequency will be impressed across the transformer because of the fact that the voltage across the transformer and that across the tube are out of phase by 90°. At any other frequency the difference of transmis- sion can be no greater than 10 per cent, (because the maximum transfer is only 100 per cent.), and thus a good characteristic is possible. Problems R+r 11 12 14 16 18 20 25 30 40 50 60 I 9.1 8 34 7.15 PL 84 140 204 830 700 510 PL+P* 913 840 714 5.0 250 250 1.66 138 28 500 50 50 165 PL = PR; P g = I-r; Efficiency = Eff. = 17 PL pedances if possible. Maximum power transfer is of greater importance than great efficiency. When the power is high, however, as a generator supplying lighting and heating power to a city, the efficiency must be high or the generator will burn up. At 50 per cent, efficiency (maxi- mum power output), as much power must be dissipated in the generator as is used in the load. When the power is small, efficiency does not matter so much. Voltage vs. Load Resistance The higher the load resistance the greater the proportion of the total voltage available that appears across the load, and the less 1. A tube (171) has a voltage of 26 (r.m.s.) applied to its grid. This voltage is multiplied by the "mu" of the tube, 3, and appears in the plate circuit as 3 X 26. The resistance of the tube is 2000 ohms. Into what resistance should the tube work to transfer the maximum amount of power to the load? What will be the power then? What will be the voltage across the load? A- cross the tube? 2. The maximum "undistorted power" EL is transferred from tube to load when 9.0 9.27 the latter has twice the resistance of the 16.7 16.7 tube. What then are the power, the volt- 28.6 28.6 ages across tube and load, assuming same data as in Problem 1? 3. An amateur transmitter oscillator 50.0 can put 100 watts into an antenna through an appropriate coupling trans- former. If the antenna current is one ampere, what is the resistance of the antenna? 83 83 4. An electrodynamic loud speaker has a very low resistance (15 ohms). Suppose it is connected to a power tube capable of an output of one watt and that by means of a step-down trans- former 90 per cent, of this power goes into the loud speaker. What is the current in the loud speaker winding? 5. A loud speaker is properly matched to a 2000-ohm tube so that maximum power is transferred. Now a 10,000-ohm tube is put in the socket. What proportion of this tube's power is being transferred to the loadp A good way to solve this problem is to assume some voltage, calculate the power in the loud speaker and the total power taken at this voltage. The pro- portion of the power usefully employed is the power in the loud speaker divided by the total power. 6. What transformer ratio must be used to connect the 10,000-ohm tube to the loud speaker for maximum power transfer? Answers to problems given on tbiH sheet will be found on page 179. O JANUARY 1930 • 147