with on my
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 The first reason is to provide access to historical documents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 reason is to have the names
of people and places published in 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 to find those names in an old article.
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
See all articles from Popular
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,
. 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.
, KN5KBH, works 16-meter c.w. with
a DX-100 transmitter cranked down to 75 watts at Baylor University
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.
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).
Fig. 3. How to calculate the effective resistance when three
resistances are connected in parallel.
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 exam.
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.
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
R2 = 250/0.005 = 50,000 ohms.
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.
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 beginning
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.
in Series. Figure 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 ohms.
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.
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.
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
R1 = 100/0.020 =
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.
Then 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 resistors.
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
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
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.
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 resistances.