Radio Broadcast (May 1928-Apr 1929)

184 RADIO BROADCAST JANUARY, 1929 O No. 14 Radio Broadcast's Homne Study Sheets Plotting Power Tube Characteristics January, 1929 o o 60 55 50 45 40 ~ 35 ! 30 25 C 20 40 60 Ee FIG. "C1 VERY radio experimenter knows the value of the characteristic curves of a vacuum tube. Home Study Sheets Nos. 5 and 6 (September Radio Broadcast) tell how these characteristic curves may be made and how one can obtain from them the important tube constants. This Sheet tells us more about the power tube in one's audio amplifier. No electrical apparatus is really necessary for this experiment. Some plotting paper, a rule, and perhaps a French curve will suffice. All of the data may be obtained from a single set of figures which show the plate current of a tube as the plate voltage is changed, the grid being maintained at zero bias. If, however, the experimenter desires to take data on one of his tubes and to carry out the result of the experiment, it is much better. All that is necessary is the Ep-Ip curve for a single value of grid bias (0). DISCUSSION The effective voltage. E, on the plate of a tube which does not have high-resistance load in its plate circuit is given by E = En + (J-Eg which states in mathematical language that the voltage on the plate of the tube is equal to the sum of the voltage due the plate battery and whatever grid voltage there is multiplied by the mu of the tube. When the grid bias (Eg) is negative the effective plate voltage is less than Ep. That is, the plate current which flows when a 100-volt plate battery and a negative C bias of 20 volts are employed is less than the plate current which flows when the C bias is zero. How much less is it? We could tell if we had available the single curve mentioned above and shown in Fig. 1 (Eg = 0). For example, let us take the Eg = 0 curve of Fig. 1 which represents the plate current of a tube, similar to the 171, at zero grid bias. Now suppose we want to plot the curve for E» = — 20. We assume various voltages and substitute in the formula for the effective voltage (This is called the "lumped voltage" in England). The mu of the tube is 2.8, and suppose we assume Ep = 100, E = Ep + 2.8 (-20) = 100 56 = 44 and looking at our curve we note that when Ep = 44, Eg = 0. and the plate current is 8 mA. Therefore when Ep = 100 and Eg = —20, Ip = 8. This is one point for the new curve. Now assuming Ep = 120, E = 120 — 56 = 64 and the plate current is 20 mA. This system is continued until sufficient points are marked down to enable us to draw a line through them. This line will be parallel to the zero-grid voltage line. Then assume another grid bias, of say, — 40 volts and plot that curve. Finally we have a family of curves like that in Fig. 1. Now the slope of this line represents the reciprocal of the plate resistance of the tube, that is, the slope = 1/Rp, and a little calculation will show that Rp for this particular tube = 1620 ohms. Engineers have shown that the maximum undistorted power output from a tube will be attained when the load resistance, into which the tube works is twice the plate resistance of the tube, in this case 3240 ohms. Under these conditions the plate current, goes up and down in accordance with the input a.c. grid voltages. How much does it vary, what is the a.c. power lost on the tube, what is the a.c. power in the load resistance, etc? We can find these various values in the following manner. 1. We draw the line AOB which goes through the intersection of the 180volt line with the E? = — 40volt line and has a slope equal to the reciprocal of the load resistance in ohms. 31.5 mA. The voltage variations across the load under these gridvoltage variations are from 20 to 100 volts or a total voltage swing of 80 volts and the voltage variations across the tube are from 220 to 140 volts. In other words, the plate current swings up and down this load line about its average value of 19.0 mA. and has a maximum value 31.5 mA. and a minimum value of 6mA. The d.c. power used up in the plate of the tube is Ep X Ip or the area of the rectangle EODC and is equal numerically to 180 X .019 = 3.42 watts. Similarly the power used up in the load is Ip X El or the area of the rectangle OFBD and numerically is equal to 60 X .019 = 1.14 watts. These powers are being used in heating the plate and the load resistance, regardless of whether there is any a.c. voltage on the grid or not. Their sum, 4.56 watts must come from the B battery. When an a.c. voltage is applied to the grid, a.c. power appears in the load resistance. The product of the r.m.s. values of current through and voltage across the load will give the power in the load. The maximum value of the a.c. voltage, e, across the load is Hg or 40 volts and the maximum value of current, i, is OH or 1.3 mA. Since the r.m.s. value may be obtained by dividing the maximum value by i/2 we may get the power X O H 40 X .013 Radio Broadcast Laboratory Y Home Study Sheet 14 / / A V • \ / p # r 1 \ S — . 'F / -A H -< D B 100 120 140 160 180 200 220 240 El ' . 260 280 That is, the slope of AOB I (amperes) E (volts) 1 3.24 1 3240' I (milliamperes) E (volts) and if we take 60 mA. as the vertical side of a triangle ,we get the horizon , , 1 60 tal side trom = -= o.^4 h, whence E = 194 and connecting the 60 mA. point on the vertical axis with the 194 volt point on the horizontal axis we draw a line. Then the line through 0 is to be drawn parallel to this line. Now with such a "family" of curves and the "load line," AOB, we can tell many things about what happens when the grid is excited with an a.c. voltage. Suppose the input grid voltage has a maximum value of 20 volts. The grid bias is minus 40, the plate current is 19 mA., the voltage actually on the plate is 180, and the voltage lost across the load resistance is 60 (240— 180). In other words EP = CD and El = DB. If the tube is so biased that no plate current flows, the entire battery voltage is applied to the tube, that is 180 + 60 or 240 volts. To apply 180 volts to the tube through a load resistance of 3240 ohms when 19 mA. flows, the plate battery must be 240 volts or CB. Now when the grid swings from minus 60 ( — 40 from the C bias and — 20 from the maximum negative input a.c. voltage) the current drops to 6 mA., and when the grid becomes minus 20 ( — 40 from the bias battery and plus 20 from the a.c. input) the plate current increases to I in the load as —y e X i l/2 X l/2 e X i 2 H2 = 260 m.w. This represents the area of the triangle OGH. Since the d.c. power supplied from the plate battery is constant, the a.c. power in the load must be added to the d.c. power used up there, and must be subtracted from the power wasted on the plate of the tube. When the grid is excited by an incoming signal, less power is used up in the tube, and the plate of a large power tube will actually run cooler when signals are applied to it. PROCEDURE Using the data in Table 1, plot the "family" of Ep-Ip curves. Assume mu = 8, calculate (1) the plate resistance, Rp, (2) the proper load resistance, Ro, for maximum undistorted power output, (3) the load line, AOB, assuming a plate voltage actually on the tube of 135 and a grid bias of minus 9 volts. Calculate the d.c. power lost in the tube, and in the load, and the a.c. power in the load when the grid swings a maximum of 6.5 volts, that is from the Eg = -2.5-volt line to the Eg = -15.5volt line. Draw in the rectangles representing the d.c. powers, and the triangle representing the a.c. power in the load. Calculate the total power supplied from the B battery, and, assuming the efficiency of the tube and circuit is the ratio between a.c. power in the load and the total d.c. power supplied from the battery, calculate the efficiency of the system. (Efficia.c. power in load ,mP, ency = — ;—— X 100':, ) d.c. power from battery TABLE I Ep Eg = 0 2.5 5.0 40 3 60 6 3 80 11 6 3 100 16 11 6 120 16 11 140 16 9.0 11.5 Ip (m.a.) PROBLEMS 1. The power output of a tube is equal to M-Eg power output / i^Eg y \RP + Ro7 X Ro where Eg is the input r.m.s. voltage. Calculate the power output of the tube whose characteristics you have plotted. Compare this value with the value secured from the graphical method. 2. Do you know why the area of the triangle is the a.c. power in the load? 3. Using the above formula calculate the power output if Ro = Rp and the efficiency of the system using the values of Ep, mu, etc., used in the experiment. 4. Using the above formula, plot a curve showing the power output from a 171 tube as the input a.c. r.m.s voltage is increased from zero to 37 volts. 5. Does more power come from the B battery when the grid of the tube is excited? 6. Does the output voltage of a B power unit change when the tube is supply a.c. power — or is the output voltage constant regardless of whether the tube is amplifying signals or not? Note: Readers may send (he ansicers to these questions to the Editor to be checked.