June 1944 QST
Table
of Contents
Wax nostalgic about and learn from the history of early electronics. See articles
from ARRL's
QST, published December 1915  present. All copyrights hereby acknowledged.

Recognizing that many people were reluctant to approach the theoretical aspect of electronics as it applied to circuit design and analysis, QST (the American Radio Relay League's monthly publication) included equations and explanations in many of their project building articles. Occasionally, an article would be published that dealt specifically with how to use simple mathematics. In this case, the June 1944 edition, we have the second installation of at least a fourpart tutorial that covers resistance and reactance, amplifier biasing (tubes since the ShockleyBardeenBrattain trio hadn't invented the transistor yet) oscillators, feedback circuits, etc. I do not have Part I from the May 1944 edition or Part IV from the August 1944 edition, but if you want to send me those editions, I'll be glad to scan and post them (see Part III here). Practical Applications of Simple Math: Part II  Plate and Screen Voltages By Edward M. Noll, EXW3FQJ Whenever the d.c. plate current flows through any resistance placed in the plate circuit of a vacuum tube as a load or coupling medium, it is obvious that the voltage at the plate will be less than the supply voltage because of the voltage drop across the resistance. In Fig. 1 the plate voltage is E_{p} = E_{b}  R_{p}I_{p}. Example: In Fig. 1, E_{b} = 250 volts. R_{p} = 10,000 ohms. I_{p} = 10 ma. (0.01 amp.). What is the plate voltage, E_{p?} E_{p} = 250  (10,000) (0.01) = 250  100 = 150 volts. Since true plate voltage is the voltage between plate and cathode, the voltage drop across the cathode resistor, R_{k} in Fig. 2, as well as the drop across the plate resistor, R_{p}, must be subtracted from the supply voltage in calculating plate voltage. In Fig. 2 the plate voltage is E_{p} = E_{b}  I_{p}R_{p} I_{p}R_{k}. = E_{b}  Ip(R_{p} + R_{k}). Example: In Fig. 2, E_{b} = 250 volts. R_{p} = 25,000 ohms. R_{k} = 2000 ohms. I_{p} = 5 ma. (0.005 amp.). What is the plate voltage, E_{p}? E_{p} = 250  (0. 005) (25,000 + 2000) = 250  (0.005) (27,000) = 250  135 = 115 volts. One advantage of transformer coupling between audioamplifier stages is that the inductance of the transformer primary winding will provide a highimpedance load for the tube at audio frequencies, while the d.c. resistance of the winding is sufficiently low to cause only a small drop in d.c. plate voltage. In Fig. 3 the only resistance affecting the plate voltage is that of the transformer primary winding, R_{t}, so E_{p} = E_{b}  I_{p}R_{t} Example: In Fig. 3, E_{p} = 250 volts. I_{p} = 20 ma. (0.02 amp.) R_{t} = 100 ohms. What is the plate voltage, E_{p}? E_{p} = 250  (0.02) (100) = 250  2 = 248 volts. Screen voltage is determined in the same manner as plate voltage, using the screen current to calculate the voltage drop across the screen resistor. E_{p} = E_{b}  I_{p}R_{p} E_{s} = E_{b}  I_{s}R_{s} Example: In Fig. 4, E_{b} = 250 volts. I_{p} = 5 ma. (0.005 amp.). R_{p} = 20,000 ohms. I_{s} = 2 ma. (0.002 amp.). R_{s} = 75,000 ohms. What are the plate voltage, E_{p}, and screen voltage, E_{s}? E_{p} = 250  (0.005) (20,000) = 250  100 = 150 volts. E_{s} = 250  (0.002) (75,000) = 250  150 = 100 volts. In the circuit of Fig. 5, both plate and screen currents flow through the common resistor, R_{1}, so that plate and screen currents must be added in calculating the voltage drop across R_{1}. E_{p} = E_{b}  (I_{p} + I_{s}) (R_{1})  I_{p}R_{p} E_{s} = E_{b} " (I_{p} + I_{s}) (R_{1})  I_{s}R_{s}. Example: In Fig. 5, E_{b} = 250 volts. R_{p} = 40,000 ohms. R_{s} = 200,000 ohms. I_{p} = 2 ma, (0.002 amp.) I_{s} = 0.5ma. (0.0005 amp.). R_{1} = 20,000 ohms. What are the plate voltage, E_{p}, and screen voltage, E_{s}? E_{p} = 250  (0.002 + 0.0005) (20,000)  (0.002) (40,000) = 250  50  80 = 120 volts. E_{s}= 250  50 (0. 0005) (200,000) = 250  50  100 = 100 volts. In the circuit of Fig. 6A, the screen voltage, E_{s}, is obtained from a tap on a voltage divider consisting of R_{s} and R_{b} The equivalent circuit is shown in Fig. 6B. The screen voltage, E_{s}, is equal to the voltage drop across R_{b}. Therefore, E_{s} = R_{b}I_{b}. Example: In Fig. 6B, E_{b} = 250. I_{s} = 1 ma. R_{s} = 10,000 ohms. R_{b} = 40,000 ohms. What is the screen voltage, E_{s}? E_{s} = I_{b}R_{b}. Since E_{b} is equal to the sum of the voltages across R_{s} and R_{b}, E_{b} = R_{s}I_{sr} + R_{b}I_{b}. Also, since both I_{b} and I_{s} must flow through R_{s}, I_{sr} = I_{b} + I_{s}. Substituting this value for I_{sr} in the above equation, E_{b} = R_{s} (I_{b} + I_{s}) + R_{b}I_{b}. Transposing, R_{s}I_{b }+ R_{b}I_{b} = E_{b}  R_{s}I_{s} I_{b}(R_{s} + R_{b}) = E_{b}  R_{s}I_{s} Substituting known values, Then, E_{s} = (0.0048) (40,000) = 192 volts. In the circuit of Fig. 7, both screen and gridbiasing voltages are taken from voltage dividers. In the case of the divider in the grid circuit, the voltage division is in exact proportion to the resistance values of the divider sections, since it is assumed that the grid is biased so that no grid current flows. Therefore, the gridbiasing voltage, E_{g}, is the voltage developed across R_{2} by virtue of the current flowing through it from the bias supply. E_{g} = I_{g}R_{2} being the biassupply voltage. Substituting, Screen and plate voltages are calculated as before. Example: In Fig. 7, E_{b} = 250 volts. E_{c} = 100 volts. R_{1} = 49,000 ohms. R_{2} = 1000 ohms. R_{3} = 30,000 ohms. R_{4} = 20,000 ohms. R_{5} = 20,000 ohms. I_{s} = 1 ma. (0.001 amp.). I_{p} = 5 ma. (0.005 amp.). What are the gridbiasing, screen and plate voltages? E_{g = } = 2 volts (negative in respect to cathode). E_{p} = 250  (0.005) (20,000) = 250  100 = 1.50 volts. E_{s} = I_{b}R_{3} = 0.0046 amp. E_{s} = (0.0046) (30,000) = 138 volts. Fig. 8 is used to illustrate the effects of low voltmeter resistance upon the accuracy of voltage measurements. R_{m} is the meter resistance. With the meter disconnected, the plate voltage will be E_{p} = E_{b}  R_{p}I_{p}. However, with the meter connected, the current, I_{m}, will flow through R_{p}. Thus, the voltage drop across R_{p} will increase and the plate voltage will be lowered, especially when the resistance of the meter is low in comparison with R_{p}. The equivalent circuit with the meter connected is shown in Fig. 8B, in which R_{pi} is the internal resistance of the tube which is assumed to be constant with a change in plate voltage. The new plate voltage desired is the voltage across R_{pi} (or R_{m}) which is E_{p} = E_{b}  (I_{pm}) (R_{p}), where I_{pm} is the new current when R_{m} is connected. In other words, E_{p} is the difference between the terminal voltage and the voltage drop across R_{p}. Example: In Fig. 8, E_{b} = 250 volts. I_{pi} =0.1 ma. (0.0001 amp.) R_{p} = 1 megohm (1,000,000 ohms) R_{m} = 1000 ohms per volt (300volt scale). What is the true plate voltage with the meter disconnected and what voltage will be measured by the meter when it is connected? E_{p} = 250  (1,000,000) (0.0001) = 250  100 = 1.50 volts = plate voltage without meter connected. As stated above, when the meter is connected, E_{p}  E_{b}  (I_{pm}) (R_{p}). Since I_{pm} is not known, its value must be found before the equation can be solved. To find I_{pm}, the resultant resistance of R_{m} and R_{pi} in parallel must be found, and this, in turn, requires that R_{pi} be determined. This can be done by considering the circuit before the meter is connected. The total circuit resistance, R_{t} is then given by = 2,500,000 ohms = 2.5 megohms. Also, R_{t} = R_{p} + R_{pi} R_{pi} = R_{t}  R_{p} = 2,500,000  1,000,000 R_{pi} = 1,500,000 ohms = 1.5 megohms. The resistance of the meter, R_{m}, is given as 1000 ohms per volt. Since the meter has a 300volt scale, its resistance is 300,000 ohms, or 0.3 megohm. R_{pim}, the resultant resistance of R_{pi} and R_{m} in parallel is given by R_{pim = } This gives the total circuit resistance in Fig. 8B as R_{t} = R_{p} + R_{pim} = 1 + 0.25 + 1.25 megohms. The new current, I_{pm}, is then I_{pm = } Then E_{p} = E_{b}  (I_{pm}) (R_{p}) = 250  (0.0002) (1,000,000) = 250  200 = 50 volts = voltage indicated by meter reading. Example: In the case of Fig. 9, it is assumed that the grid is to be fed a squarewave pulse. Compare the plate voltage when the tube is conducting a current of 15 ma. with the effective plate voltage when the tube is idle and not drawing plate current. The plate resistance is 10,000 ohms. When the tube is conducting, E_{p} = E_{b}  I_{p}R_{p} ;= 250 (0.015) (10,000) = 250  150 = 100 volts. When the tube is not conducting, there is no voltage drop across R_{p} and the plate voltage is 250, the same as the supply voltage, E_{b}. Fig. 10 illustrates another use for the voltages divider. The coupling circuit shown is that commonly found in directcoupled amplifiers. From the equivalent circuit of Fig. 10B, it will be seem that the plate of the first tube is connected at one tap on the voltage divider, while the grid is connected at another tap less positive. It is assumed that the grid of the second tube is biased, by the voltage drop across its cathode resistor, so that the grid does not draw current. Example: In Fig. 10, E_{b} = 250 volts. I_{p} = 5 ma. (0.005 amp.) R_{1} = 10,000 ohms. R_{2} = 75,000 ohms R_{3} = 25,000 ohms. What are the plate voltage of the first tube, and the grid voltage of the second tube? The total drop across all resistors is, of course, equal to the applied voltage, E_{b}. The voltage, across R_{1} is R_{1}I_{1}, while that across R_{2} and R_{3} in series is (R_{2} + R_{3}) (I_{2}), bearing in mind that no current is being drawn from the tap, marked G in Fig. 10B, so that the same current flows through. R_{2} and R_{3}. Then, E_{b} = R_{1}I_{1}+ (R_{2} + R_{3}) (I_{2}) Since both I_{p} and I_{2} flow through R_{1}, I_{1} = I_{p} + I_{2} Substituting this value for I_{1} in the preceding equation, E_{b} = (I_{p} + I_{2}) (R_{1}) + (R_{2} + R_{3}) (I_{2}). Substituting known values, 250 = (0.005 + I_{2}) (10,000) + (75,000 + 25,000) (I_{2}) = 10,000I_{2} + 50 + 100,000I_{2} 110,000I_{2} = 200 I_{2}= 0.0018 amp. = 1.8 ma. The plate voltage of the first tube is equal to the sum of the voltage drops across R_{2} and R_{3}. E_{p} = I_{2} (R_{2} + R_{3}) = (0.0018) (100,000) = 180 volts. The grid voltage of the second tube is equal, to the voltage drop across R_{3}. E_{g} = I_{2}R_{3}= (0.0018) (25,000) = 45 volts. Posted December 4, 2018 (original 12/28/2012) 