Radio Broadcast (May 1928-Apr 1929)

114 RADIO BROADCAST ADVERTISER A New A C Transformer with terminals for use with all types of Wiring Harnesses Here is the latest A C success in the Dongan Approved A C line No. 6570, built into a crystalized lacquered case, is equipped with terminals for use with the new wiring har' nesses. Designed to operate with 4 UX 226, 1 UY227 and 1 UX 171 power amplifier tubes. Also equipped with lamp cord and plug outlet for B-eliminator as well as tap for control switch. Price $6.50. A Complete Line of Approved A C and Output Transformers By-Pass and Filter Type Condensers for Set Manufacturers and Custom Set Builders The reason you can expect real engineering help as well as par in modern design and construction is due to the fact that DongarTs entire radio business always has been devoted exclusively to the interest of the set manufacturer. Ask for information and prices, on any desired type, direct from Dongan to you. See the Dongan Parts Exhibit at the R. M. A. Show DONGAN ELECTRIC MFG. CO. 2991-3001 Franklin St. Detroit, Michigan Jenkins & Adair Condenser Transmitter For Broadcasting, Phonograph Recording, and Power Speaker Systems fT^HIS transmitter is a small condenser which X varies its capacity at voice frequency, and is coupled direct into a single stage of amplification, contained in the cast aluminum case. The output, reduced to 200 ohms, couples to the usual input amplifier. The complete transmitter may be mounted on the regulation microphone stand. It operates on 180 v. B and 6 or 12 v. A battery. This transmitter contains no carbon, and is entirely free from background noise. Its yearly upkeep is practically nothing. It is extremely rugged, and will withstand hard usage. Price, complete with 20 ft. shielded cable, $225.00 F.O.B. Chicago. J. E. JENKINS & S. E. ADAIR. Engineers 1500 N. Dearborn Parkway, Chicago, U. S. A. Send for our bulletins on Broadcasting Equipment No. 198 Radio Broadcast Laboratory Information Sheet June, 1928 The Screen-Grid Tube as an R. F. Amplifier CALCULATING GAIN TDROBLEM: — Suppose that we have a radio-1 frequency amplifier connected as indicated in the figure and that a screen-grid tube is used. How can we calculate the amplification that can be obtained? Solution: — To solve the problem we must make use of the tube constant known as the mutual conductance, which, for the screen-grid tube, has a value of about 350 micromhos or 0.000350 mhos. The mutual conductance Gm by definition, across the input Eg. Transposing equation (4) to get this ratio we obtain ^ = Gm X Z Eg (5) Gm = Eg (1) where Gm is the mutual conductance in mhos; lie is the alternating current flowing in the plate circuit; Eg is the alternating voltage impressed in the grid; transposing this equation we get This equation shows that the gain of this circuit using a screen-grid tube is simply equal to the mutual conductance of the tube in mhos, times the effective impedance of the tuned circuit. Therefore, if we know the impedance into which the tube is working, we can, by multiplying the impedance by Gm, obtain the amplification. If the tuned circuit at resonance has an effective impedance of 100,000 ohms then the amplification would be Amplification =0.000350 x 100,000 =35 /ac = Gm x Eg (2) Input The voltage Et across the tuned circuit is equal to the impedance of the circuit Z times the current through it Et = /acxZ (3) and therefore Et = Gm x Eg x Z (4) The amplification of the circuit is equal to the voltage across the output Ex divided by the voltage To next Tube No. 199 Radio Broadcast Laboratory Information Sheet Current June, 1928 its direction of flow THE direction of flow of current around a simple circuit consisting of a battery and a resistance is generally considered to be as indicated by the solid arrows in sketch A on this sheet. As indicated, the current is thought of as flowing out of the positive terminal of the battery, through the resistance and into the negative terminal of the battery. Now let us look at the circuit of a vacuum tube, as indicated in sketch B. In this circuit we would assume that the current would flow as indicated by the solid arrow, i, e., out of the positive terminal through the tube and into the negative terminal just as it did in circuit A. However, we know that the filament of the tube is the electron-emitting substance and that the electron flow is from the filament to the plate. Apparently we have two currents flowing in the circuit, and this has led some experimenters to believe that there were two distinct currents flowing in the circuit, one the battery current and the other the electron current. This is not so and there is only one current flowing in the circuit, the electron circuit. (A) The idea that the electric current flows from the positive to the negative originated before anything was known about electrons. This direction of flow has since been proved to be wrong. It is now realized that an electric current is actually a flow of electrons and that electrons, being negatively charged, flow toward the point of positive potential. Therefore the actual flow of current in the tube circuit B and the battery circuit A is as indicated by the dotted arrows. Fortunately the incorrect assumption that was made years ago for the direction of the flow of current is not important in the solution of electrical problems so long as we remain consistent regarding the direction injwhich the current is assumed to flow. Many meters used in electricity are marked with plus and negative signs and the winding of the meter is arranged so that the pointer on the meter will deflect in the right direction when the positive terminal of the meter is connected to the more positive part of the circuit. No. 200 Radio Broadcast Laboratory Information Sheet June, 1928 Resistors determining what size to use IN CHOOSING a resistance for any particular purpose it is necessary to determine the value required, the current it must carry and then from these two facts determine the wattage rating required. The chart published on this sheet will prove useful to determine: (a) the wattage rating a resistor must have to carry a given current (b) the current a resistor, of given wattage rating, will carry The curve is plotted to cover resistors up to 10,000 ohms and wattage ratings up to 5 watts. Example: A resistor is to be used to supply C-bias to a 171-A type tube. The plate current of the tube (which must flow through the resistor) is 20 milliamperes. The required C-bias voltage is 40 volts. What value of resistance and what wattage rating should the resistor have? To calculate the required value of resistance we use Ohm's law. Referring to the chart below, we find that the vertical line corresponding to 2000 ohms crosses the horizontal line corresponding to 0.020 amperes (20 milliamperes) at the point indicated between the curves of 1.0 and 0.25 watt resistors. In such a case we must, of course, always uSe the larger size and therefore in this case we should use the 1.0watt resistor. Resistance Voltage Current in amperes 40 =0.020 =2000 ohms ILUAM Relatioi Between M PERES, JHMS A ND WA • ^5 Walt \. v^£5 Walt " 1.0 Watt ~— 125 Wall 0.125 Wa 6000 8000 10,000 12,000 14,000 16,000 RESISTANCE IN OHMS