Radio Broadcast (May 1927-Apr 1928)

JANUARY, 1928 PUSH-PULL AMPLIFICATION — WHY? 203 it to be 3 per cent., which means that, in order to obtain a given amount of sound energy, we must supply the loud speaker with many times as much electrical energy. The amount of electrical energy required is found by dividing the sound energy output by the efficiency of the loud speaker; in this case we must divide 0.03 watt by 0.03 (3 per cent.) and the quotient, one watt, is the amount of energy the power tube in the receiver must be capable of delivering to the loud speaker during the fortissimo passages. Now let us see what tube or combination of tubes is capable of supplying this power. The maximum amount of undistorted power that can be obtained from various tubes is given below. TABLE NO. 1 Tube Type Plate Voltage Grid Voltage Undistorted Output Watts 199 90 4-5 0.007 120 ■35 -22 . 5 0. 1 10 20 1 -A 90 -9.0 0.055 I 12 157 -10.5 0. 195 171 180 -40.5 0.700 210 450 -38 I . 700 push-pull Arrangement Requires twice as much input voltage from receiver to give same output power as parallel arrangement. Slight overload (about 25 per cent.) possible without noticeable distortion. Voltage gain is somewhat lower. Plate impedance four times greater than parallel arrangement. Distortion due to curvature of tube characteristic eliminated. Any a. c. hum from filaments eliminated due to pushpull arrangement We shall endeavor to explain now how the pushpull amplifier eliminates a certain type of distortion which exists in a simple single-tube amplifier. It is necessary to start the discussion by examining in some detail the characteristics of a cx-310 (ux-210) type tube (or, for that matter, any tube). Parallel Arrangement (1.) Requires only half as much input voltage from receiver to give same output as push-pull arrangement. (2.) Distortion due to overload quite noticeable. (3.) Voltage gain somewhat higher. (4.) Plate Impedance four times smaller than pushpull arrangement. (5.) Distortion due to curvature of tube characteristic not eliminated. (6.) Some hum may result if filaments are operated on a. c. TABLE NO. 2 Output Resistance 10,000 5,000 1,000 Increase in Current Decrease in Current This table indicates clearly that, as the resistance in the output circuit of the tube decreases, the increase and decrease in plate current due toa given signal become unequal. This represents distortion because it indicates that the positive side of the input voltage produces a relatively greater change in the plate current than does the negative side. When two tubes are used in a push-pull arrangement the maximum power output of the combination is twice that of a single tube. It is evident from the table that the only tubes delivering, in pushpull arrangement, more than 1 watt of power are the 171 and 210 combinations, and, therefore, these combinations are most satisfactory for supplying a loud speaker with the necessary amount of undistorted power. In practice it will be found that a push-pull amplifier can be overloaded, but this amount of overload is so small as to be negligible. This treatment of the problem is not exact. It was necessary to assume an average value for the power associated with the pianissimo passages and this first assumption determines how much power will be required for the fortissimo passages. It is also true that a considerable amount of distortion can be present in the output of a loud speaker without being evident to most of us. The figures do, nevertheless, give an idea of why power tubes must be used, and show that present-day loud speakers cannot be supplied with sufficient undistorted power from tubes other than the 171 or 210 type. Marked improvement in the efficiency of loud speakers will some day make other tubes with a lower power output suitable for use in the last stage of a receiver, but until such an improvement is made, we must make certain that we have plenty of power handling capacity available in the receiver's output. PUSH-PULL OR PARALLEL TUBES? A T THIS point there might be some question •"^ regarding the relative merits of a pushpull amplifier with two 210 tubes and a parallel arrangement of the same tubes. Let us list the advantages and disadvantages of the two arrangements. The italics indicate with which arrangement the advantage lies. Although point No. 4 was indicated as an advantage for the parallel arrangement, it is possible, by the use of a special push-pull output transformer, to compensate the higher plate impedance of the push-pull circuit, and the two arrangements will then be equal in this respect. Point No. 5 has not, as yet, been explained, but it is the most important reason for the existence of the push-pull arrangement: 100 90 80 70 Signal Voltage' FIG. I Grid voltage plate current curves of a 2 10 type tube In Fig. 1 we have drawn several curves for a single tube of the 210 type with a plate voltage of 400 and a grid bias of -35 volts, and these curves show the relation between the plate current and the grid voltage with various load resistances in the plate circuit. The curve marked 1000 was made with a 1000-ohm resistance in the plate circuit and the curves marked 5000 and 10,000 were made with resistances of 5000 and 10,000 ohms respectively in the plate circuits. These curves are dynamic characteristics in the sense that they indicate how the plate current will vary with different loads in the plate circuit. If a signal having a value of, for example, 10 volts, is impressed on the grid of this tube, it will cause the grid voltage to vary 10 volts either side of its average value of 35 volts. Such a signal voltage is represented in Fig. 1 by the curve, marked "signal voltage," drawn below the grid voltage axis. If the change in plate current due to this voltage is determined on the 10,000-ohm curve by reading the values of plate current at each extremity, we find that, when the voltage is positive, the current rises to 21 milliamperes and that, when the voltage is negative, the current decreases to 11 milliamperes, a drop of 10 milliamperes. The signal voltage of 10 volts has, therefore, caused the plate current to increase and decrease an equal amount with respect to the average value, the increase and decrease being 5-milliamperes in this case. If the same measurements are made on the 5000-ohm curve we find that the plate current increases 7 milliamperes above the average value but only decreases 6 milliamperes. On the 1000-ohm curve the increase is 14 milliamperes and the decrease is only 1 1 milliamperes. These values have been arranged in the form of a table: loud speaker considerations AND now let us consider the loud speaker. The impedance of a loud speaker is a function of frequency and increases with increase in frequency. At low frequencies, therefore, the loud speaker will have a comparatively low impedance and the tube feeding the loud speaker will then operate on the characteristic corresponding to a low-resistance load in the plate circuit. This characteristic is indicated by the 1000-ohm curve in Fig. 1. At medium frequencies, where the loud speaker's impedance is higher, the tube will operate on a characteristic similar to the 5000-ohm curve, and at high frequencies the tube will operate on a characteristic similar to the 10,000-ohm curve. As indicated by the figures in table No. 2, the 10,000-ohm curve is quite straight and therefore produces little distortion. A small amount of distortion is produced by the 5000-ohm curve, but much greater distortion occurs when the tube operates along the 5000-ohm curve. When a loud speaker is operated from a single 210 type tube, this distortion occurs and, if possible, it would evidently be of advantage to arrange the circuit so that no distortion of this type is produced. This leads us to consider push-pull amplification. The circuit diagram of a push-pull amplifier is given in Fig. 2. In some push-pull arrangements the output choke, L, is replaced by a transformer, but the circuit will function with a simple choke coil as indicated. When a signal is induced in the secondary of the input transformer, T, the voltage relations are as indicated by the plus and minus signs on the diagram. It will be noted that the voltage at one end of the transformer is positive relative to the voltage at the other end, which is negative. The signal voltage impressed on either grid is one half the total voltage across the transformer. Since the two grids are at relatively opposite potential the plate current changes will also be opposite Input Tube No. 2 FIG. 2 The circuit connections for a push-pull amplifier