Mathematics - Radio Tool
June 1946 Radio-Craft

June 1946 Radio-Craft [Table of Contents] Wax nostalgic about and learn from the history of early electronics. See articles from Radio-Craft, published 1929 - 1953. All copyrights are hereby acknowledged.

Remember the early 1990s Teen Talk Barbie where one of her phrases, "Math class is tough!," caused a big kerfuffle because it ostensibly stereotyped girls as being afraid of math? Maybe if Talking GI Joe (this 1967 version is the one I had as a kid) also uttered a math phobia statement all would have been fine. The truth is that a large portion of both girls and boys (and men and women) of all ages break out in a cold sweat whenever the subject of math arises in print or in a conversation. The ARRL goes to great lengths to help ease the math anxiety of radio operators who are studying for a license exam. In this multi-part 1946 Radio-Craft magazine article, the authors attempt to assuage some of the reluctance of readers to apply mathematics to their electronics hobby endeavors (assuming engineer readers do not possess such hesitancy) by presenting practical examples pertaining to commonly encountered applications such as DC voltage biasing and voltage drops across a series string of resistance.

Mathematics - Radio Tool

Part I - Some Problems of Receiver Design and Operation

By Arthur Howard and Morris Eddy

Whenever the word "mathematics" is uttered in radio circles, not too infrequently the novice and radio veteran are alike gripped by fear. They imagine the subject to be dull, abstract, and very difficult; something to be avoided at all costs! The appalling situation can be attributed, in no small measure, to the poor pedagogical methods in our schools.

All sciences owe a great deal to mathematics for their development. This is particularly true of radio and electronics. We deal with substances and terms which are completely insensible to the human organs and alien to our imagination. We try to understand them by meager analogies; electricity depicted, for example, as flowing water. To fully comprehend the principles and enable us to solve many practical problems in radio work, mathematics is a vital necessity. In the following, we present a number of radio problems and their solutions. They are designed to illustrate how important a tool mathematics can be to the radioman. Only simple arithmetic will be assumed on the reader's part, ordinary common sense being the most important factor in understanding the following.

The Cathode Bias Resistor

A very frequent problem is the determination of the cathode resistance value of a vacuum tube. For example, what value resistor is required for a 6C5 triode operating as a class A amplifier, with a plate voltage of 250 volts? (Fig. 1).

Turning to the tube manual, we find the grid-bias voltage to be -8 volts and the plate current, 8 milliamperes. Using Ohm's Law and some arithmetic, we solve the problem.

The formula -

R_k = (E_k/I_p) X 1,000

Where R_k = resistance value of grid-bias resistor in ohms, E_k = grid-bias voltage in volts, and I_p = plate current in milliamperes.

Substituting the known values in the formula, we have -

R_k = (E_k/I_p) X 1,000 = (8/8) X 1,000

      = 1,000 ohms

Where tetrodes or pentodes are employed, we follow the same method as above in obtaining the solution, with the added necessity of taking the screen current into account. This is shown in the new formula -

   R_k = (E_k / (I_p+I_s)) x 1,000

where I_s, the new factor, is the screen current in milliamperes.

What should the wattage rating of the resistor be? Again, mathematics will supply us with the answer.

Using the formula: W = E_kI_p Where W = rating of resistor in watts, E_k = grid-bias voltage in volts, and I_p = plate current in amperes.

Substituting the known numbers in the formula-

W = E_kI_p = 8 x 0.008 = 0.064 watts

Since there is no resistor available rated at 0.064 watts, we would use a 1/4-watt resistor. Commonly, the wattage rating of a resistor is at least twice the calculated value, so a 1/2-watt resistor would be the smallest practical one, which would include the 100 percent safety factor.

Line-Cord Resistors

The replacement of a line-cord resistor is a task the radio-serviceman is often called upon to perform. In this case, to solve the problem, his tool is not the conventional ohmmeter, but arithmetic.

Let us assume a five-tube a.c.-d.c. superhet, using a 6SA7, 6SK7, 6SQ7, 25A6, and a 25Z6, is brought in for repair. A continuity test points to an open line-cord resistor. With what value line-cord resistor should it be replaced? (Fig. 2).

From the tube manual, the service-man finds that each tube draws 0.3 ampere. Then, he notes down the heater voltage of each tube, and adds them up. There are three 6.3-volt and two 25-volt tubes; total, 68.9 volts. 

In an a.c.-d.c. circuit, the tube filaments are connected in series. Therefore, the same current, 0.3 ampere, will flow through each tube.

Since the line voltage is approximately 117 volts, our problem is to drop (117-68.9), or about 48 volts. Using Ohm's Law-

R = E/I 

Where R = the required resistance in ohms, E = the voltage to be dropped in volts, and I = the current flowing through the tubes, in amperes.

Substituting the known values in the formula, we have-

R = E/I = 48/0.3 =160 ohms

Therefore, the line-cord resistor should be 160 ohms. The wattage is found by using the formula W = EI, and doubling the calculated value.

Voltage Divider Problems

In power supplies, a voltage divider is often utilized. It is a tapped resistor connected across the output of the power source, supplying different voltages to the stages of a circuit.

To cite an example, the following voltages and currents are needed.

E₁ = 250 volts, 20 ma, E₂ = 100 volts, 15 ma, and E₃ = 50 volts, 10 ma.

The problem is to solve the resistance values between the taps. (Fig. 3). Before we unravel the problem, a current of about 10 percent of total load current, known as the bleeder current, must be allowed for. (In this case, the total load current is equal to 20 ma + 15 ma + 10 ma or 45 ma). The bleeder current will then be about 5 ma, Referring to Fig. 3, we proceed as follows:

Let - I₁ = 15 ma (the current flowing through the load connected across the 100-volt tap), I₂ = 10 ma (the current flowing through the load connected across the 50-volt tap), and I₃ = 5 ma (the bleeder current).

Working from the bottom tap upward -

No matter how many tapped resistors are used, we begin from the bottom and work upward as illustrated.

Cathode Condensers

In designing an amplifier, the designer has to solve many factors. For instance, what value of cathode bypass condenser is necessary for an amplifier with an approximately fiat response down to 40 c.p.s. with a cathode resistor of 2,000 ohms? The reactance of the condenser is to be one-tenth of that of the cathode resistance (Fig. 4 below):

Using the formula: C = 1/6.28fX_c

Where C = value of condenser in farads, f = frequency in c.p.s., and X_c = reactance of condenser (in this example, its value will be 2,000/10, or 200 ohms):

C = 1/6.28fX_c = 1/(6.28 X 40 X 200)

     = 0.00001994 farads or approximately 20 microfarads.

Output Transformer Matching

The primary winding of a particular output transformer has an impedance of 10,000 ohms. The technician wishes to place a speaker with a voice coil of 4 ohms across the secondary of this transformer. What ratio should exist between the number of turns on the secondary and primary windings, to provide proper matching? (Fig 5 below.)

Where N = number-of-turns ratio, Z_p = the Impedance of the primary winding, and Z_s = the impedance of the secondary winding.

Then, substituting -

. 

Posted September 16, 2021