Radio broadcast .. (1922-30)

A Discussion of Practical Engineering Procedure DESIGNING THE POWER SUPPLY CIRCUIT By RICHARD F. SHEA Engineering Department, Atwater Kent Manufacturing Company THE DESIGN of power-supply systems is an essential part of radio engineer- ing. To be sure, it relies less upon the precise type of measurement which is characteristic of r.f. design, but in its place we find a great need for practical design, " cut-and-try" methods if you wish. It is the purpose of this paper to point out some of the practices and meth- ods entering into this branch of radio engineering. Let us then start at the beginning and follow through the procedure necessary to turn out a completed power-supply unit for a certain receiver and associated parts. Since the requirements depend upon the equipment used in conjunction with this supply unit, we must have previous knowledge of the type of set, voltages de- sired at the tubes, number and type of tubes, type of loud speaker, and, if a dy- namic loud speaker is to be used, the volt- age and current required to excite prop- erly the field coil. Let us take a set which is fairly typical of the modern receiver. This set employs three stages of r.f. using screen-grid tubes requiring, let us say, 135 volts on the plates, 1J volts on the control grids, and a maximum of 50 volts on the screen grids. The detector is of the C- bias or plate-rectification type requiring 120-130 volts on the plate and a grid bias of 12 volts. The first audio tube is a 227 type operating at 135 volts on the plate and a 6-volt grid bias. In the power stage 250-type tubes in push pull are used, re- quiring 425 to 450 volts on the plates and a bias of 70-85 volts. The speaker is a dyna- mic type, with a 5000-ohm 70-mA. field. First Considerations Let us now tentatively lay out our power pack as shown in Fig. 1. For convenience only, the 250-type tubes have been shown connected to the power supply, the other tubes going to the voltage-dividing resistor. We can now compute the desired volt- age across condenser Cj. We want 450 volts on the plates of the 250's and 84 volts on the grids. Reference to tube tables indicates a plate current of 55 mA. for a 250-type tube under these conditions —a total drain of 110 milliamperes for the two tubes. If our output transformer's primary has a resistance of 500 ohms on each side, our drop there will be 25 volts. Our required voltage at C? then is 450 + 25 + 84 = 559 volte. The bias for the 250- type tubes is obtained from the resistor R c , between the center tap of the filament winding for the 250-type tubes and the ground. This resistor, R c , can be calcu- lated immediately, knowing the drop across it and current flowing through it — Rc = 84 X 1000 110 = 765 ohms. Now, since we know the desired current through the field coil and its resistance, we can compute its voltage drop as— VF = 5000 X 70 1000 350 volts Supply Fig. 1 tus has been in use from one half to one hour. Knowing our voltage at C? and our field drop, we can compute the available volt- age at the divider as 559—350 = 209 volts. Let us consult Fig. 2 which shows an enlarged diagram of the divider resistor. From this we can calculate the resistance values of the various units. We have a total of 70 mA. fed to the divider and this splits up between the tubes and the re- sistor according to the characteristics of the tubes. Let us refer again to our tube tables and find the plate current for the screen-grid tubes, the 227 as a detector, and the 227 as an audio amplifier. We get, respectively, 2 mA., 2 mA., and 3 mA. The three r.f. tubes then will draw a total of 6 mA. and the total of r.f., detector, and a.f. plate currents will be 6 + 2 + 3, or 11 mA. The current at the + 50-volt tap is very low as this feeds the screen- grids, at the 12-volt tap the 2 mA. returns from the detector cathode, and at the lj- volt tap the 6 mA. returns from the r.f. tubes. The audio tube is supposed to be biased in a manner similar to that em- ployed for the 250-type tubes, whereby the grid voltage is obtained from a re- sistor in series with the cathode lead. Consequently, the plate current from the audio tube returns to the ground point and does not flow through the divider. +135 +50 R,>38V:59mA 91 O I I I ± 70mA- R, >10.5V:61mA 1.5 V:67 mA 3mA Our resistances now have the voltages and currents shown and can be computed easily— Ri = R4 = 74 0.070 85 0.059 38 0.059 10.5 0.061 1.5 = 1059 ohms = 1440 ohms = 644 ohms = 172 ohms It must be stated here that all resistors are to be measured hot, after the appara- Fig. 2 Rl= (K067 =22 - 4ohm9 Thus we find that our divider is a resistor of 3337 ohms, tapped at 22, 194, 838, and 2278 ohms. Selecting the Choke Having gone this far. our next step is to choose a proper size of choke coil. Most manufacturers have standard sizes of chokes and the problem becomes one of picking a choke that will not have an excessive heat rise under the current, which is 110 + 70, or 180 mA. (a rather high current for two 281-type tubes to handle— Editor). Of course, the higher the inductance, the better will be our fil- tering, for given condenser values, and, conversely, if we use a small choke coil we must expect to use large values of filter capacity to obtain satisfactory filter- ing. The relationship between capacity and choke size is largely economic and each manufacturer has his own answer to the question. Having chosen our choke coil, the big- gest we can use economically, and know- ing its resistance, we compute the drop in it. Let us say this resistance is 300 ohms. Then our drop is 300 X 0.180 = 54 volts and we now have the load our rec- tifiers must supply, as the voltage across condenser C is 559 + 54 = 613 volts and the current drain will be 180 mA. It now remains for us to determine the proper secondary voltage to be applied to the rectifier tubes in order to obtain this re- quired load. For this purpose we use char- acteristic rectifier tube curves, such as shown in Fig. 3. These curves show varia- tion in output voltage, against load, for 46 • NOVEMBER 1929