November 1957 Popular Electronics
[Table of Contents]
People old and young enjoy waxing nostalgic about and learning some of the history of early electronics. Popular
Electronics was published from October 1954 through April 1985. All copyrights are hereby acknowledged. See all articles from
As with on my
Airplane and Rockets hobby website, a big part of my motivation
for scanning and posting these vintage electronics magazine
articles has been two-fold. The first reason is to provide access
to historical documents for educational reasons. The second
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text format (everything OCRed) so that someone doing a Web search
for himself, a relative, or a friend, might run across it here.
I receive e-mails occasionally from readers who are thrilled
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Among the Novice Hams
By Herb S. Brier, W9EGQ
In October we discussed some of the fundamental properties
of current, voltage, and resistance, and how the relationship
between them is expressed concisely by Ohm's law: E = IR,
I = E/R, R = E/I. We also talked about the formulas for calculating
the power in electrical circuits, P = EI and P = I2R,
and how to use them answering simple questions involving Ohm's
law and power calculation that may found in the Novice examination.
Now, we'll go over the information necessary to understand and
answer questions that might appear in the General/Technician/Conditional
Fred Beyer. KN9HFP, made 200 contacts in
35 states in five months as a Novice. His transmitter is a DX·20
feeding a dipole; His receiver is an NC-98.
In the General Class examination, there are usually a couple
of questions based on the diagram in Fig. 1, such as: "What
is thee bias voltage on the tube?" And: "What is the resistance
of R2?" It is important to read each question separately and
then study the diagram to determine which of the data given
is pertinent to that question. Once this is done, the problem
is half solved.
Fig. 1. This diagram, with different component
values, is the basis for several questions on Ohm's law which
appear in the General Class Amateur exam.
For example, there is nothing difficult about determining
the resistance of R2. From Ohm's law, R = E/I; therefore, you
simply divide the voltage across R2 by the current flowing through
it. Both of these quantities are given. Remembering to change
milliamperes to amperes, the solution will be:
R2 = 250/0.005 = 50,000 ohms. WHAT IT'S ALL ABOUT
"Among the Novice Hams" is devoted to the beginning
radio amateur. If you are familiar with amateur radio, it speaks
for itself. If not, the following may help you understand what
it s all about.
Amateur radio is a hobby in which a quarter million people
throughout the world, from grade school boys and girls to men
and women, operate their own officially licensed radio stations.
They communicate with each other over distances ranging from
a few miles to 12,000 miles.
Ken Grimm, KN5KBH, works 16-meter c.w. with
a DX-100 transmitter cranked down to 75 watts at Baylor University
The simplest amateur license to obtain is the novice license.
It is issued by the Federal Communications Commission, Washington,
D.C., to any citizen who has not previously held an amateur
license and who can send and receive the International Morse
code at a speed of five words a minute - actually a very slow
speed. In addition, it is necessary to pass a simple written
examination on elementary amateur radio theory and regulations.
The license is valid for one year, during which time the licensee
must qualify for a permanent license or leave the air.
Full information on how to obtain the various classes of
amateur licenses may be found in a packet of booklets called
"Gateway To Amateur Radio." The booklets are available for $1.50,
postpaid, from the American Radio Relay League, Inc., West Hartford,
Conn., or from any of the amateur supply houses that advertise
each month in POPULAR ELECTRONICS.
Read "Among The Novice
Hams" for up-to-date news and discussions of interest to all
The bias voltage on a tube is the
fixed voltage applied to its control grid with reference to
its cathode. In this circuit, the voltage across the cathode
resistor R1 establishes the bias voltage. The problem, therefore,
is to determine what this voltage is. From Ohm's law, E = IR1.
R1 is given as 200 ohms. I is the 20-ma. plate current of the
tube, which flows from the negative battery terminal, through
R1, through the tube, and back to the positive battery terminal
through the milliammeter. Thus, E = IR1 = 0.02 x 200 = 4 volts.
The bias voltage is 4 volts.
Resistance in Series.
2(A) shows three resistances connected in series. From it, it
is obvious that for an electric current to travel from one end
of the string to the other, it must flow through each resistance
in turn; therefore, the total resistance must be the sum of
the individual resistances, or Rtotal
= R1 + R2 +
R3, etc. For the values shown, the total resistance is 1000
Resistances connected in series across a voltage
source are usually called voltage dividers because, by proper
choice of the individual resistances, any desired percentage
of the total voltage can be obtained at the taps between them.
A voltage divider is often used across the output terminals
of a power supply; there it serves the dual purpose of keeping
a minimum load on the supply and furnishing intermediate voltages
to the associated equipment.
For practice, let's design
a voltage divider to do a specific job. Assume that we have
a transmitter requiring 500 volts at 100 milliamperes (0.1 ampere)
for the final amplifier tube plate circuit, and 250 volts at
50 ma. (0.05 amp.) and 100 volts at 10 ma. (0.01 amp.) at other
points. A power supply capable of delivering 500 volts at a
maximum of 200 ma. is available.
Referring to Fig. 2(B), the first step in calculating the
resistances in the voltage divider is to decide on how much
"bleeder current" to allow to flow through it. Normally, about
10% of the current capacity of the power supply is allocated
for the purpose. Ten percent of 200 ma. is 20 ma. When added
to the 160 ma. drawn by the transmitter, this makes the total
key-down current drain on the supply 180 ma.
Fig. 2. How resistances connected in series
(A) can be used as a voltage divider (B) to deliver several
voltages from a single power supply (see text).
Starting at the bottom of the divider, the only current that
flows through R1 is the bleeder current, while the voltage across
it is obviously 100 volts. Substituting these values in the
appropriate Ohm's law formula gives:
R1 = 100/0.020 = 5000 ohms.
Proceeding to R2, the current through it is the 20-ma. bleeder
current, plus the 10 ma. drawn from the 100-volt tap, or 30
ma. The voltage across it is the difference between the voltages
at the two taps: 250 volts-100 volts = 150 volts. Solving for
the resistance of R2 using these values gives: R2 = 150/0.030
= 5000 ohms.
To calculate the value of R3, we add the
50 ma. drawn from the 250-volt tap to the 30 ma. already accounted
for, making the current through it 80 ma. The voltage across
it is 250 volts (500 volts - 250 volts). Solving Ohm's law with
these values gives: R3 = 250/0.08 = 3125 ohms.
add the three resistance values together: 5000 + 5000 + 3125
= 13,125 ohms. One large resistor (13,000 ohms would be close
enough in an actual circuit) with adjustable taps or three separate
resistors could be used.
To complete the calculations,
determine the current that will flow through the divider with
no connection to the taps by dividing the supply voltage by
the resistance: I = 500/13,125 = 38 + ma. Because the supply
voltage will undoubtedly increase somewhat under no-load conditions,
the actual current will probably be at least 40 ma. This is
more current than flows through R1 and R2 when power is being
delivered to the transmitter; therefore, R1 and R2 must be large
enough to carry this current without overheating. Use either
of the power formulas to determine the wattage rating of the
Under no-load conditions, the voltage at the divider taps
will be higher than when they are under load. The main advantage
of a voltage divider over series dropping resistors is that
voltage variation between the load and no-load conditions is
Resistances in Parallel. Figure 3 shows three resistances
in parallel. The formula for calculating the effective resistance
of such a combination is:
If you are a whiz at algebra, you know exactly how to solve
this formula and need not worry that you may be asked about
it in the examination. If not, you will probably say: "Fine,
but what does it mean?" Fortunately, it's easy to understand
with a bit of explanation.
Fig. 3. How to calculate the effective resistance when three
resistances are connected in parallel.
Referring to Fig. 3 and taking each resistance in turn, we
know from Ohm's law that the currents through them are: I1 =
E/R1, I2 = E/R2, and I3 = E/R3. Furthermore, the total current
is the sum of the individual currents:
Itotal = I1 + I2 + I3, or Itotal
= E/R1 + E/R2 + E/R3.
Reverting to the fundamental Ohm's
law again, the effective resistance of the three resistances
in parallel is:
Of course, we can make E any value we wish, as long as it
is the same in each part of the equation. For convenience, E
is usually made equal to one volt in parallel-resistance calculations,
making the last equation the same as the first one given here.
The effective resistance of 2-, 4-, and 5- ohm resistors
in parallel is 1.05 (to two decimal places). I suggest you verify
this answer for practice and then make up a few practice problems
of your own. In checking your answers, remember that the effective
resistance is always less than that of any of the individual
Posted June 27, 2012