Radio Broadcast (May 1928-Apr 1929)

262 RADIO BROADCAST SEPTEMBER, i9a8 Radio Broadcast's Home Study Sheets r, 1921 Vacuum Tube Characteristics 20 20,000 /_pHE three important factors governing a vacuum tube's operation in a radio circuit are: _ change in plate voltage 1. Amplification Factor, (J. = — r ; rj rr change in grid voltage to produce a given plate current change. change in plate voltage 2. Plate Resistance, Rp = — 2 — : — !1t— ; change in plate current or, the effect upon the plate current of changes in the plate voltage. „ . . , _ change in plate current 3. Mutual Conductance, Om = -; — — ; r — change in grid voltage or, the effect upon the plate current of changes in grid voltage. Home Study Sheet No. 5 gives sufficient data to determine all of these factors. The plate resistance, sometimes called the internal resistance or impedance of the tube may be determined by dividing a change in plate voltage by the corresponding change in plate current it produces. Care must be taken to use the proper units, that is, volts must be divided by amperes — not milliamperes. For example, let us determine the impedance of the tube used in Sheet No. 5 at minus 3 volts grid bias. We look at Fig. 3 on Sheet No. 5 and see that at minus 3 volts the plate current is 7 milliamperes when the plate voltage isg67.5 and (about 3.5 milliamperes when the plate voltage is 45. Then, = 67.5-45 P .007-.0035 — 0Q35 = "457 ohms. This resistance varies with each change in plate and grid voltage. When it is determined by the above method, small changes in plate voltage should be used. More accurate results would be obtained if changes of ten volts were used instead of 22.5. At low plate voltages the plate resistance is relatively high and as the plate voltage is increased the plate resistance decreases, rapidly at first and then more slowly as the normal operating voltage is reached. This term "plate resistance" is a measure of the plate circuit's resistance to the flow of alternating current. The resistance of the plate circuit to the flow of direct current is equal to 40 the voltage on the plate divided by the d.c. plate current in amperes. For example, in Fig. 3, Sheet No. 5, we see that the plate current is 7 mA when the plate voltage is 67.5 volts and the grid bias is minus 3 volts. Therefore the d.c. resistance of the plate circuit is / / / y— f\ m r / \ 1 / A 1 / \ t + §5 — . -J. <> 4 / FIG. I Note : The word microhms at the right on this diagram is an error. It should be micromhos 67.5 0.007 = 9,643 ohms 3.q 6000 The mutual conductance may be determined by noting the effect on the plate current produced by a given change in grid voltage. Looking at Fig 1, Sheet No. 5, we note down the following data taken from the point where the plate voltage is 45. .0064 — .0035 ^ 2.0 o 4000 Change in plate current Change in grid voltage . = .00097 3—0 The unit of conductance is the mho, so the above value is .00097 mhos, or as usually stated, 970 micromhos. This shows that with 45 volts on the plate of this tube a change of one volt on the grid in the region between minus 3 and zero produces a change of 970 microamperes plate current. This may also be obtained by determining the slope of the Ip-Eg curve in the vicinity of minus 3 volts. (See Home Study Sheet No. 4, for a definition of the term "slope".) The mutual conductance varies with each value of plate and grid voltage, and will be more representative of what the tube actually does in a radio circuit if small grid voltage changes are used to determine its value. The amplification factor may be found as in Sheet 5 or by multiplying the plate resistance in ohms by the mutual conductance in mhos, since a little juggling of the above formula for these values shows such a relation to exist, that is pt = Rp x Gm PROCEDURE Determine the plate resistance and mutual conductance for several values of plate and grid voltages, taking the data from that obtained in Sheet No. 5. Plot these values as shown in Fig. 1 and Fig. 2. Determine the values of plate resistance and mutual conductance by taking the slopes of the curves at the proper points, and see how nearly these values check. (Note: The term "slope" was explained in Sheet No. 4.) DISCUSSION The curves made by plotting the data secured in the experiment in Sheet No. 5 are known as static characteristic curves, and the values of Mu, Rp, and Gm are called the static characteristics of the tubes under test. They tell us all we need to know about a tube to predict what it will do in nearly all kinds of electrical circuits. The amplification factor gives us an idea of what happens when a small voltage is impressed on the grid. This voltage reappears in the plate circuit of the tube, multiplied by the amplification factor. For example if the tube has a mu of 8 and a signal with a value of one volt is impressed on the grid, then there appears in the plate circuit a signal with exactly the same form as that impressed on the grid, but with a value of eight volts. In bridge methods of measuring the mu of a tube, a voltage is introduced into the plate circuit which is opposite in direction to that appearing there due to the input voltage to the grid. When this opposing voltage in the plate circuit is equal to mu times the grid voltage, the two voltages cancel each other and silence is obtained in a pair of headphones. The mu of a tube is practically constant over the entire range of operating voltages, as can be seen by refering to Fig. 2 on this sheet. The other constants of a tube vary considerably with the applied plate and grid voltages. The mutual conductance of the tube tells us how great a change in plate current will occur if the grid voltage is varied. For example, if the Gm of a tube is 600 micromhos a volt a.c. input to the grid will produce an a.c. current in the plate circuit of 600 microamperes. This change in current may be used to set up a new voltage which may be again amplified by succeeding tubes. The plate resistance of the tube gives us an idea of what happens to these plate current changes produced by grid voltage changes. If the tube had no resistance, all of the a.c. plate current would be useful across the coupling device in the plate circuit of the tube. Since, however, the tube has a resistance, this a.c. current must flow through not only the coupling device, but through the plate resistance as well. If the plate resistance of the tube is 30,000 ohms and the coupling resistance has a value of 60,000 ohms then the total signal voltage in the plate circuit is divided into two parts, two-thirds of the voltage appearing across the coupling resistance and one-third across the plate resistance. The total a.c. voltage developed in the plate circuit is divided between the load, or coupling device, and the tube resistance. If a coupling device is used which has considerable resistance, the voltage actually on the plate is not the same as the B-battery voltage but is less by the drop in voltage across the coupling resistance: the voltage drop being determined by ohm's law. See Home Study Sheet 4, August Radio Broadcast. The amplification factor of a tube varies inversely as the spacing between the wires forming the grid and directly as the distance between the plate and filament and between the grid and filament. Thus to obtain ja tube with a high amplification constant it is necessary to use a fine mesh grid mounted close to the filament, as compared to the distance between the plate and the filament. J ^/ Am iltfication Co ufant pedtnce FIG. 2 PROBLEMS 1. Suppose you are to design a series of tubes, each with a mutual conductance of 500 micromhos. Plot a curve showing how the plate resistances and amplification factors of these tubes will be related. 2. Suppose a tube had a 55,000-ohm resistor in the plate circuit, that one milliampere of current were flowing, and that the B-battery voltage were 90. What is the voltage actually on the plate? 3. Suppose you had a tube with a mu of about 30, and a plate resistance of 60,000 ohms. What is the mutual conductance? If the plate current is one milliampere, how much B-battery voltage will be required to put 90 volts actually in the plate, if a coupling resistor of 250,000 ohms is used? 4. The curve in Fig. 1 does not have the values of mu plotted on it. iCalculate these values and plot. What kind of tube do you think it is?