Radio Broadcast (May 1928-Apr 1929)

SEPTEMBER, 1928 RADIO BROADCAST 261 o Radio Broadcast's Home Study Sheets Measuring the Amplification Factor of Vacuum Tubes O O WITHIN the glass bulb of the ma** jority of vacuum 'tubes used in receiving sets to-day are three metallic elements: the filament, which gets its voltage from the A-battery; the grid, which is connected to the radio circuit through a C-battery; and the plate, which takes current from the B-battery. Since each of these three voltages is variable, the actual operation of the tube is somewhat complex. The effect of varying the filament voltage has been determined in Home Study Sheet No. 3. (page 205, August Radio Broadcast). We shall now fix this voltage at some definite value, and notice the results of varying the other two voltages, one at a time. The apparatus needed will be as follows: LIST OF APPARATUS 1. The baseboard set-up used in Experiments No. 3 and No. 4. 2. A source of filament current, say a 6-voIt storage battery. 3. A C-battery with taps at 1.5, 3.0 and 4.5 volts. 4. Two B-batteries with taps at 45, 67.5 and 90 volts. 5. A milliammeter reading up to 10 milliamperes. A Weston Model 301 5-milliampere meter was used in the Laboratory. A low range meter with the proper shunts may be used. 6. A filament resistor to reduce the 6 volts from the battery to the proper value for the tube, 5 volts. A rheostat or fixed resistor of the proper value may be used. PROCEDURE Connect up the apparatus in the following manner, as shown in Fig. 2: 1. Plus A to one end of the filament resistor. 2. Clip 1 to other end of resistor. 3. Minus A to clip 2. 4. Minus B to clip 2. 5. Plus C to clip 4. 6. Milliammeter between clips 6 and 7. Connect leads with Eureka or similar clips onj one end to terminals 5 and 8. These will enable the C and B voltages to be changed quickly and easily. If fresh B and C-batteries are used it will not be necessary to measure their voltages, and under any conditions we are more interested in the effect of varying these voltages than in their exact values. Inasmuch as the characteristics of two tubes of the same type from the same factory differ slightly, it is not necessary to know the C and B voltages with great precision. Make the grid negative by placing the wire on clip 5 on the minus 4.5-volt tap of the C-battery. Read the plate current when the plate lead, clip 8, is placed on the 22.5 or 45volt tap of the B-battery. Now increase the B voltage in steps up to 90 and then repeat the measurements with different values of C voltage, say minus 3, minus 1.5, zero, and plus 1.5 volts. Set down the data in the form shown in Table 1 which is the result of measuring a Ceco heater or 27 type of tube. Zero C voltage is obtained by connecting clip 5 to minus A or clip 2. Plus C voltages are obtained by reversing the C-battery connection, that is, connecting minus C to clip 4. Now take data by changing the C voltage or bias, keeping the B voltage fixed. That is, fix the plate voltage at, say 45 and read the plate current as the grid voltage is changed from minus 4.5 to minus 3, etc., until the plate current is too great for the milliammeter to read. This data is tabulated in the form shown in Table 2. The data from these two tables should now be plotted as shown in Figs. 1 and 3. DISCUSSION The electrons which boil out of the filament when it is heated by the A-battery current flowing through it are negatively charged. The plate, which is charged positively by the B-battery, has a very strong attraction for these electrons. They, therefore, hurry toward the plate carrying with them their electric charges. If a certain number, 6.8 x 10IS per second, arrive at the plate, the flow of current out of the B-battery, through the plate current meter, and back to the negative side of the filament — where this part of the plate circuit is attached — is equal to one ampere. In receiving tubes the current is of the order of thousandths of amperes, or 6.8 x 1015 electrons per second. The greater the plate voltage, the greater the number of electrons that arrive per second. Consequently the greater is the plate current. Between the plate and the filament is the grid, a sort of turnstyle which regulates the number of electrons that get past it. How? The voltage of this grid can be made positive or negative with respect to the source of the electrons. If it is negative, it tends to repel any electrons that come along on their way to the plate. They must either return to the positive side of the filament, or float around in the space between the grid and the filament. Some few of them get through to the plate. If the grid is made positive, it accelerates the flow of electrons from the filament, and either attracts a number of them to itself, producing a flow of current RADIO BR0ADCAS1 LABORATORY / Ce CoN -27T ibe / / f 0 / / * y 3 2 , VOLTS FIG. I FIG. in the grid circuit, or hastens them on their way toward the plate. Since [the grid is closer than the plate to the source of electrons, it has a greater effect on their rate of travel. The relative effect of grid and plate upon the plate current is called the amplification constant (the symbol is the Greek letter mu, (X) and is determined from the above data in the following way. Let us consider the Ip-Ep curve in Fig. 3, which represents the variation of plate current with variation of plate voltage. Choosing values from the curve for Eg = minus 3, we note that at 67.5 volts the current is 7 milliamperes and at 50 it is 4 milliamperes. Thus a change of 17.5 plate volts causes a change in plate current of 3 milliamperes. Now the question arises, what change in grid voltage will cause the same change in plate current? Looking at the 67.5-voIt curve of the Ip-Eg graph (Fig. 1) showing how the plate current changes with grid voltage changes, we see that at zero grid volts the plate current is 11.5 milliamperes and at minus 2.3 the plate current is 8.5 milliamperes. This represents a plate current change of 11.5 — 8.5, or 3 milliamperes caused by a grid voltage change of 2.3 volts. From these figures we can determine mu. The amplification constant is calculated then as follows: Mu = change in plate volts change in grid volts 67.5-50 _ 17j> 2.3 2.3 = to produce a given plate current change. The important thing to note here is that it is the change in voltages that must be divided by each other, not the actual voltages at any particular point. These changes in voltages may be reckoned only over that part of the Ip-Ep and Ip-Egr curves which appear to be straight lines. The smaller the length of these straight lines that are used in our calculations, the greater will be the accuracy with which the value of mu will check that obtained on an accurate vacuum tube bridge. PROBLEMS 1. Plot all of the data in Tables 1 and 2, and calculate the amplification factor at a number of points on the curve. Plot these values against plate voltage and then against grid voltage. 2. Calculate the value of the resistor needed to cut down the A battery voltage, 6, to the voltage required by the filament, 5. 3. Repeat the above experiment for other types of tubes in your laboratory. 4. Should the Ip-Eg curves be steep or flat for a high-mu tube? For a power tube? TABLE I Data for Ep-Ip curves Plate current (Ip) in mA. RADIO BROADCAST LABORATORY Ce CoN -27 Tube Ip-E )Carve / 1 M / / / A 45 Ep VOLTS FIG. 3 En = plate volts 0 22.5 Grid volts 0 — 1 5 —3 0 — 4 5 +1 5 2.8 1 5 1 0 5 3 8 45 6.4 4 7 3 5 2 4 7 6 67.5 10.4 8 5 7 0 5 5 11 0 90 11 0 9 4 Eg = —10 — 8 — 6 — 4 — 2 0 + 2 + 4 + 6 + 8 grid volts TABLE 2 Data for lp-Eg curves Plate currents (Ip) in mA. Plate voltages 22.5 45 67 5 90 .25 1 0 2 0 6.0 .3 1 2 3 2 7.0 .4 1 8 5 3 8.5 .55 3 0 6 6 11.0 1.4 4 7 8 8 14.0 2.9 6 8 11 5 17.0 4.5 8 7 14 0 19.5 6.2 11 0 16 5 22.0 8.0 10.0