Radio Broadcast (May 1928-Apr 1929)

108 RADIO BROADCAST DECEMBER, 1928 No. 12 Radio Broadcast's Home Study Sheets December, 1928 Resonance in Radio Circuits Part II T) ESONANCE may occur in a radio circuit in one of two ways. The series resonant circuit was discussed in Home Study Sheet No. 11. Suppose, instead of having the voltage impressed in series with the inductance and capacity, it is impressed across the condenser and inductance connected in parallel, as in Fig. 1. What happens, as the frequency is changed? DISCUSSION 28 26 < 22 20 Li IS 300 320 340 360 380 400 FREQUENCY IN KC. ® Unfortunately the experiment to show just what happens to the various currents and voltages in such a circuit is difficult to perform. It is simple, however, to calculate what happens and to plot it. Instead of going into the laboratory for this experiment, then, we shall rely on the slide rule and graph paper to delve into another interesting radio phenomenon known as parallel resonance. In a series circuit the same current flows through all the units, but the voltages across these units may differ. In a parallel circuit, the same voltage exists across the branches, but the currents through them differ. The total current, I, flowing out of the generator, in Fig. 1, is the sum of the currents flowing in the two branches. Since, however, a capacity reactance is considered as a negative reactance, the current through such a reactance may be considered as having an algebraic sign opposite to that of a current through an inductive branch. The total current, then, is the difference of the currents, i.e., I = II — Ic and from previous Home Study Sheets, the currents in these branches may be calculated if the reactance and the voltage is known. The formula above, which gives the generator current] flowing into the parallel circuit, shows that if the current in the capacity equals the current in the inductance, the difference, or generator current, becomes zero. IMPEDANCE OF PARALLEL CIRCUIT The impedance of any device or circuit may be defined as the ratio between the voltage across it and the current through it. Thus, the impedance of a parallel tuned circuit is »-? and if. at resonance, the current, I, is zero, the impedance must be infinitely high, because current would not flow out of the generator no matter how high its voltage. A series-resonant circuit, looked at from the viewpoint of the generator is a very low-impedance circuit at resonance. The current it draws from the generator is large. On the other hand, at resonance the impedance of a parallel-resonant circuit, looked at from the generator, is very high, and the current fed into it from the generator is very small. Series circuits are used when low impedances are desired; parallel circuits, when high impedance circuits are needed. If we measure, or calculate, the current flowing in each of the two branches in Fig. 1, and the total or generator current as well, we shall obtain curves similiar to those in the accompanying figures. These are theoretical curves and do not take the resistance of a circuit into account. Resistance usually exists in the inductance of such a circuit, but in well-designed radio circuits it is small compared with the inductive reactance of the coil. If this is true, the impedance into which the generator feeds current at resonance is equal numerically to L; <j>2 where R is the resistance of the coil. r As the frequency of the generator is changed, the current through the inductance decreases and the current through the capacity increases as in Fig. 2. At very low frequencies there is a large difference between the two currents — the inductance current being the larger — and so a large current flows from the generator. At very high frequencies the capacity current is much greater than the inductance current and so a large difference current, i.e., the generator current, flows as in Fig. 3. For this reason, below the resonant frequency the generator views the circuit as inductance shunted by a condenser whose reactance is so high it takes but little current. At frequencies above resonance the inductance has little effect upon the generator current and so the circuit is said to be capacitive. At the frequency which makes the inductive and capacitive FIG. 2 reactances equal, the currents are equal, the circuit as a whole is neither capacitive or inductive and, therefore, must be resistive only, so far as the generator is concerned. At this frequency, then, the circuit looks like a resistance to the generator, and the expression above for its impedance is properly called its effective resistance. Thus, „ L2W2 L R °r CR where L = coil inductance to = 6.28 x f C = capacity required to resonate the coil R = series resistance of coil For example if L = 200 microhenries. R = 10 ohms and f = 1000 kc. Reff = 160.000 ohms. ■ PROCEDURE FIG. I 1. Assume L = 200 microhenries. C = 500 mmfd., E = 10 volts. As the frequency varies from 400 to 600 kc. calculate; a. reactance of inductive and capacitive branches, b. currents in these branches, c. total current flowing from generator, d, impedance presented to generator; 2. Plot all of this data against frequency. PROBLEMS 1. If the coil in Procedure has 15 ohms resistance at 300 meters, and the grid and filament of a vacuum-tube amplifier is connected across the circuit, what impedance does the tube work out of? Suppose it is in the plate circuit of a power tube which feeds 100 milliamperes into it. What is the power required from the tube if P = Is Reff 2. An antenna-ground system has a capacity of 0.00025 mfd. An inductance is to be put in series with it so that the entire circuit will be resonant to 400 meters. Calculate the total circuit inductance. If a distant station impresses across this system causes voltage of 100 microvolts, what voltage is across the inductance at resonance? 3. An interfering station working on 600 meters also sets up across the antenna a voltage of 100 microvolts. If the system is tuned to 400 meters, what voltage at the interfering frequency will appear across the inductance? 200 |il 1000 v. 0 L = C = R= V-f 300 320 340 360 380 FREQUENCY IN KC. 400 ! 4. Suppose in series with the antenna is connected a circuit like FIG. 3 that of Fig. 1 and tuned to 600 meters. Assume the resistance of the coil is 10 ohms. Calculate the impedance this circuit would offer to the 400-meter and the 600-meter, signals. Then calculate roughly the ratio of wanted to unwanted voltage across the inductance L. The complete antenna system is shown in Fig. 4. Incalculating the impedance neglect the effect of L and C. 5. If doubling the inductance of a coil doubles its resistance, too, what effect upon the effective resistance of a shunt tuned circuit has doubling the inductance? Of course, a smaller condenser would be used to reach the resonant condition. How much smaller would the condenser be? 6. A plate-supply device has considerable 120-cycle hum in its output. Suppose a parallel tuned "trap" is placed in the positive lead. A 30-henry inductance is available. Calculate the size of condenser needed. If the inductance has 500 ohms resistance what impedance will the trap offer the 120-cycle current? 7. The maximum voltage gain that may be secured from a screen-grid tube may be calculated from Gm X Reff, where Gm is the mutual conductance of the tube and Reff is the impedance of the tuned circuit into which the tube works. Assume that Gm=300 micromhos, L= 200 microhenries, wavelength=300 meters and calculate the maximum resistance that can be tolerated in the coil and condenser which tunes it to permit an amplification of 60. If the resistance of the coil is doubled, what happens to the gain of the tube and coilcondenser combination? Note: Readers may send their answers to these FIG. 4 questions to the Editor to be checked.