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

Teen Talk Barbie, "Math class it tough!"

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 Radioâˆ’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

Fig. 1 - Calculation of tube's cathode bias.

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_{k}I_{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_{k}I_{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).

Fig. 2 - The old line-cord resistor problem.

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

Fig. 3 - How to calculate the voltage divider.

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.

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 - I_{1} = 15 ma (the current flowing through the load connected across
the 100-volt tap), I_{2} = 10 ma (the current flowing through the load connected
across the 50-volt tap), and I_{3} = 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

Fig. 4 - Cathode condensers.

Fig. 5 - Output transformer matching.

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.

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