Radio Broadcast (May 1929-Apr 1930)

RADIO BROADCAST a modulation-frequency voltage between detector grid and filament very nearly equal to mEs2/Vg audio volts, where Vg is the detector grid-voltage constant, and is given in Table I. This rule holds almost exactly for the lower frequencies of modulation, while for the high notes the output will be a little less. This audio frequency applied to the detector grid by the grid rectification of the modulated signal is then amplified by the mu of the tube and impressed upon the a.f. amplifier in series with the plate resistance of the tube. Power Detection WHEN a radio-frequency signal of at least several volts amplitude is applied to a suitably adjusted grid-leak detector the action taking place in the grid circuit with the signal voltage of Fig. 1a is as shown in Fig. 5. The rectified grid current charges the grid condenser negatively and causes the average grid potential to have the value shown by the dotted line in Fig. 5. This average value is always such that the positive crests of the signal make the grid go positive a small amount. Each time the grid goes positive grid current flows, and makes up for the current that leaks off through the grid leak during each cycle. At times when the signal amplitude is decreasing in size, it is necessary that the grid leak allow the grid condenser charge to leak off at a rate that will cause the average grid potential to reduce at least as fast as the signal amplitude is changing. This requirement calls for values of grid condenser capacity and leak resistance smaller than usually used. The explanation of the action that takes place in the grid-leak power detector is exactly the explanation usually given of grid-leak detection, i.e., charging the grid condenser and letting the charge leak off through the grid leak. In the detection of weak signals, the situation is somewhat different, because with weak signals the grid condenser charge can leak off through the detector grid filament resistance, thus complicating and changing the action. With large signal voltages the average grid potential is so very negative that the grid current flows only at the positive crests of the radio frequency, and during the rest of the time no grid current can flow. If high-quality output is to be obtained from the grid-leak power detector it is necessary to have the proper grid leak and condenser combination. Suitable values for any tube are a grid leak of about f megohm and a 0.0001-mfd. grid condenser. With these proportions the average grid potential will be able to change as fast as the signal amplitude up to modulation frequencies of 5000 cycles. The overloading point of the grid-leak power detector is reached when plate rectification starts to take place. This is because plate rectification causes increase of plate current while grid rectification causes decrease of plate current. Plate rectification thus neutralizes the grid action and causes distortion. As the maximum amplitude of a fully modulated wave is twice the carrier amplitude, a particular tube will handle half as big a carrier wave acting as a power detector as it can amplify, using the same plate voltage in both cases. Thus, a 20lA-type tube with 90 volts on the plate usually uses a 45-volt C bias. The crest amplitude of carrier wave that can be handled at a plate voltage is one half of this, or about 2j crest volts. In the case of the 210 tube of Fig. 4, the normal amplifier C bias for 247§ plate voltage is 18 volts, so that the crest amplitude of carrier that can be handled is 9 volts. Fig. 4 shows distortion beginning at 18 volts, or approximately the maximum signal amplitude when the 9 volt carrier is modulated 100 per cent. The maximum audio-frequency power output obtainable from the grid-leak power detector is slightly over one fourth of the un distorted power the tube can give as an amplifier at the same plate voltage and a suitable grid bias. Thus, the 210-type tube at 247^ plate volts will put out 340 undistorted milliwatts as an amplifier, and will put out about 100 undistorted audio milliwatts as a power detector. The approximate audio-frequency output of a grid-leak power detector can be obtained by a simple computation. It is apparent from Fig. 5 that the average grid voltage of the power detector follows the modulation of the signal. This variation in average grid value between one quarter and one volt it can hardly be classed either as a strong or a weak signal. Such signals will be detected satisfactorily with either a weak-signal or power detector, the latter giving slightly better quality and slightly less sensitivity than the former. The amount of audio-frequency voltage obtainable on the grid with these moderate signal strengths is usually from 20 to 50 per cent, of the ideal value, that is, 20 to 50 per cent, of mEs. Receiver Design .Average Value Eg=0 Fig. 5 — Curve showing instantaneous and average grid voltages while a modulated signal voltage is applied to a grid-leak power detector. potential applies an audio-frequency voltage to the grid of the detector tube, and it is this audio-frequency grid voltage when amplified by the tube acting as an amplifier that constitutes the audio-frequency output of the detector. In the ideal detector the audio-frequency voltage applied to the grid would be equal to the modulation voltage in the signal. If the degree of modulation is m, and the carrier amplitude is Es, the ideal amount of modulation voltage is mEs. The actual power detector is only about 75 to 85 per cent, perfect, and will apply to the grid an audio-frequency voltage about 75 to 85 per cent, of mEs. The percentage tends to rise slightly as the signal amplitude becomes large, but is surprisingly near constant at this approximate range for all tubes under ordinary conditions. Tabes for Power Detection IN ORDER to put out power the detector tube must operate with a high plate voltage. At the same time, the grid bias of the grid power detector is approximately zero except when the signal is coming in, and so the allowable plate current sets a limit to the plate potential. Tubes such as the 201a, 112a, 227, and 226 can operate as power detectors with 90 to 135 volts on the plate, and under such conditions will put an audio-frequency voltage of at least two volts on the detector grid without distortion. The 210-type tube can safely operate at zero grid with 250 to 300 plate volts, and will put out from 100 to 150 undistorted milliwatts in the plate circuit, enough to run an efficient loud speaker directly without an audio amplifier. If the grid return lead of the power detector is brought back to the proper potential the same adjustment that is satisfactory for large signals will give from 50 to 75 per cent, as much output with weak signals as the best adjustment for small signals, and will give this result with excellent quality. The potential for the grid return lead to accomplish this is best determined by experiment. In some cases it will be the positive leg of the filament, in other cases the negative side, while more often it will be a potential intermediate between these. If the detector is to be used only for strong signals the return can be to either side of the filament, with the negative likely to be the best by a small margin. When the signal voltage has a 100 80 60 40 20 IT IS possible, by computing detector performance, to determine just how much radio and audio-frequency amplification is necessary in a radio receiver to give full output with a given field strength of signal. In order to show how this is done, and to make clear how detector computations are made and used, three typical examples have been worked out. Case 1: It is planned to use one audiofrequency amplifier feeding a 17lA-type power tube with a plate voltage of 135 and a grid bias of — 27 volts. A 227-type tube is used as a detector, and it is assumed that the detector and the audio-frequency amplifier each give an amplification of 25 times. How much radiofrequency amplification is required to give full power output from a signal field strength of 1 microvolt per meter (which is about the minimum useful signal) ? The maximum audio-frequency voltage that can be applied to the power tube has a crest value of 27, and this is obtained when 27/(25x25) =0.043 volts is applied to the detector grid by the detection process. The detector obviously must be a weak-signal detector, and this 0.043 audio volts is the output when the signal is fully modulated (m=l). Calling Es the crest value of the radio signal that produces the required output of 0.043 volts, and noting that Table I shows Ye = — 0.23, then when m=l, a formula already 1 x Es2 given shows that Q ^ = 0.043 and solving for Es shows the required radio signal on the detector grid to be Es = 0.10 crest volts, which, when fully modulated and applied to the detector grid, will put the maximum allowable audio input on the grid of the 17lA-type power tube. If the receiving antenna is 10 meters (about 33 feet) high, a signal field strength of 1 microvolt per meter (crest value) will induce 10 microvolts in the antenna. If a tuned input to the grid of the first r.f. amplifier is used this will be stepped up perhaps 15 times, applying a voltage of 150 microvolts to the first r.f. grid. To bring this up to the 0.1 volts (or 100,000 microvolts) required K = A.F. Voltage between grid and filament in percentage of maximum possible value 70 100 200 500 1000 2000 5000 MODULATION FREQUENCY Fig. 6 — Effect of grid condenser capacity in grid-leak detector operating with small signals. • may, 1929 page 39 •