December 1954 Popular Electronics

## December 1954 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 |

Popular Electronics wanted to be all things to all people as far as electronics hobbyists go. From the very first edition in October of 1954 (2 prior to this one), they included articles on circuit troubleshooting, electrical theory, DIY building projects, Amateur radio, radio control airplane enthusiasts, and much more. I don't have the first two issues yet, but I'm guessing that the first or second After Class column covered aspects of direct current (DC), since this installment covers alternating current (AC).

An alternating current is one which periodically reverses its direction. This is illustrated in Fig. 1. At A, when the polarity of the line is as shown, electrons flow through the circuit in the direction indicated by the arrows. A short time later, the polarity of the power line reverses and the direction of current also reverses as shown by the arrows in B of Fig. 1.

Usually the current changes gradually with time from maximum in one direction to maximum in the other direction. This change can be represented by a drawing such as shown in Fig. 2. This drawing or graph shows that the current rises from zero at time L to its peak value at time M. The current then decreases until it reaches zero again at time N. It now reverses direction and builds up until it reaches its maximum value in the opposite direction at time O. The current then decreases until it once more drops to zero at time P. The variation of current between time L and time P is known as one cycle. The number of such cycles which occur in one second is known as the frequency. For example, the usual power line frequency is 60 cycles-per-second; this means that one complete cycle will have a duration of 1/60 second. The frequencies used in radio and television broadcasting are much higher than the power-line frequency and are usually specified either in kilocycles (kc.) or megacycles (mc.). A kilocycle is equal to 1000 cycles, and a megacycle is 1,000,000 cycles.

Since the instantaneous value of an alternating current or voltage varies continually, there must be some agreed-upon way of specifying its value. Actually, there are two commonly used ways of specifying this value: peak and r.m.s. The peak value is the maximum value reached during the cycle. For example, the 110 volt power line has a peak value of over 155.5 volts as shown in Fig. 3. The r.m.s. value of a sine wave alternating current or voltage is equal to .707 times the peak value. The letters r.m.s. stand for root-mean-square, the name of the mathematical operation by

which the factor .707 is derived. The relationship between peak and r.m.s. values may be written:

Example:

What is the r.m.s. value of an alternating current whose peak value is 3 amperes?

Answer:

r.m.s. = .707 peak

r.m.s. = .707 X 3

r.m.s. = 2.121 amperes

Example:

What is the peak value of an alternating voltage whose r.m.s. value is 10 volts?

Answer:

Unless otherwise stated, alternating voltage or current is specified in r.m.s. values. For example, when we speak of the 110- volt power line we mean 110 volts r.m.s. Likewise a.c. values given on a circuit diagram are assumed to be r.m.s. unless otherwise noted. Unless designed for specialized applications, a.c. meters are calibrated to read r.m.s. values. The r.m.s. value of voltage or current is also called the effective value, since it gives the number of volts or amperes of d.c. which would produce the same effect, in heating, for example.

Depending upon the components of a circuit, the current and voltage may be either in-phase or out-of-phase. When they are in-phase, current and voltage reach corresponding peaks at the same instant and pass through zero at the same instant, as shown in Fig. 4A. If the current either leads or lags the voltage, the two are said to be out-of-phase. These conditions are illustrated in Figs. 4B and 4C. The amount by which current and voltage are out-ofphase is known as the phase angle and is usually specified in degrees (one complete cycle = 360°).

Phase angles are often indicated by means of drawings such as those in Fig. 5. Here, arrows instead of sine waves are used to represent the current and voltage. These arrows are known as vectors, and the drawing itself as a vector diagram. The lengths of the vectors indicate the amounts of voltage and current. These are often drawn on graph paper where each square represents a certain number of volts or amperes. Vectors are considered to be pivoted in the center and rotating in a counter-clockwise direction. The three vector diagrams in Fig. 5 present exactly the same information as the three drawings of. Fig. 4. In A, the current and voltage are in phase. In B, the current lags the voltage. In C, the current leads.

The following quiz is intended as a self-check. You should be
able to answer all of the questions correctly if you have mastered
the foregoing text. The answers appear on page 128.

1. What
is the r.m.s. value of a sine wave having a peak of 300 volts?

(a) 425 volts; (b) 212.1 volts; (c) 42.5 volts

2. A
frequency of 1500 kc, is equal to:

(a) 1.5 mc.; (b) 1.5 cycles;
(c) 1,500,000 mc.

3. What is the peak
value of a 220 volt r.m.s. power line?

(a) 311 volts; (b) 155
volts; (c) 110 volts

4. At a frequency of 400 cycles-per-second,
the duration of each cycle is:

(a) 400 seconds; (b) .0025 second;
(c) 250 seconds

5. If the frequency of an alternating current
is increased but its peak value remains the same, its r.m.s, value
will:

(a) increase; (b) decrease; (c) remain the same

**VANISHING VOLTS
**

The odd behavior of a common voltmeter
when used to measure the plate potential of an electron tube amplifier
is very mystifying unless one remembers that the meter itself is
a part of the circuit being measured. This idea will be clarified
by referring to the schematic diagram of the resistancecapacitance
coupled amplifier shown in the diagram.

If the amplifier
is performing properly and this we shall assume - it is fair to
anticipate a voltage drop of perhaps 150 volts in the plate load
resistor, R. This would leave 150 volts for the plate. A voltmeter,
connected as shown in the diagram, ought to read this voltage but,
surprisingly, it will probably register a great deal less - possibly
as little as 10 or 15 volts. If your reaction to this reading is
to conclude that the meter is delinquent, forget it! You couldn't
be wronger!

But the fact remains that the plate voltage
has vanished! Where?

The explanation involves two distinct
considerations: first, the ordinary voltmeter generally requires
about 1 ma. of current through its coil to make it read full scale;
second, this additional current is being drawn through a relatively
high resistance, that of the plate load resistor R.

With
the meter disconnected from the circuit, the voltage drop across
R is, as mentioned, about 150 volts. The fall of potential results
from the flow of plate current through the resistor which, of course,
is in series with the tube plate circuit. Just as soon as the meter
is connected from plate to ground it, too, draws current to make
its needle deflect, producing an additional voltage drop which may
be quite high. On the other hand, the decrease in plate voltage
will produce some decrease in plate current.

For example, let us suppose that our tube is the triode section
of a 12SQ7GT, with -1.5 volts grid leak bias. The plate current
will be approximately 0.31 milliampere, the voltage drop across
R will be 0.00031 x 500,- 000, or 155 volts, and the plate-to-cathode
voltage will be 145 volts. Now suppose we connect a 1000 ohms-per-volt
meter, set to its 250-volt range, between the plate and cathode
of the tube. The meter reading would be approximately 94 volts.
With this plate voltage and the same bias as before, the tube would
draw only 0.036 milliampere. The meter, which draws 1 milliampere
for a full-scale reading of 250 volts, would draw 94/250 or 0.376
milliampere. The total current through R would be 0.376 plus 0.036,
or 0.412 milliampere. The total drop across R is 0.000412 x 500,000,
or 206 volts. 300 minus 206 equals 94 volts.

If we set the
meter on its 100-volt range, the reading would be approximately
50 volts. With this plate voltage, and bias as before, plate current
of the tube would be practically cut off, and meter current would
be 50/100, or 0.50 milliampere, which is enough to account for the
entire 250·volt drop across R. Similarly, on the 50-volt range,
the reading would be approximately 27 volts, and on the 25-volt
range, 14 volts. The lower the range we use, the less the resistance
of the meter will be, the more current will flow through R, and
the greater the voltage drop across R will be.

Colloquially,
this is known as "loading down" the circuit. The only way to avoid
it is to take all such measurements with a good vacuum-tube voltmeter
(v.t.v.m.), an instrument which draws practically no current at
all through the plate load.

STARTING FLUORESCENTS

Modern fluorescent lighting tube emits light as a result of
the excitation received by its inner, chemical coating from the
ionized gas contained within it. A somewhat unfortunate characteristic
of ionization is that the striking potential required is much greater
than the operating potential. Ordinary household fluorescent lighting
fixtures must incorporate a starting scheme which applies a sudden
surge of high voltage across the tube, - voltage which is removed
once the arc has been struck.

Two methods for obtaining
starting potentials are now in common use. The first, generally
found in desk and floor lamps, is a manual starting system requiring
a spring push-button (see diagram A). A ballast coil having a relatively
high inductive reactance is in series with a filament at each end
of the tube and with the starting switch. When the switch is closed,
current flows through the series circuit causing the filaments to
heat and emit electrons but no arc discharge can occur between them
because the closed switch keeps the potential difference quite low.
When the button is released, however, the usual inductive voltage
kick-back appears across the tube of sufficiently large magnitude
to initiate the discharge. Once started, the arc continues since
the voltage across the tube is in the region of 100 volts. To extinguish
the light, a separate series switch is incorporated in the line
to open the circuit.

Ceiling fixtures use an automatic starting method involving plug-in starters. A starter is a rather interesting combination of glow-discharge tube and a bi-metallic element. The latter is made by bonding together two dissimilar metals having widely different coefficients of expansion; when heated, such an element bends, with the metal having the lower coefficient on the inside of the curve. In diagram (B), the bi-metallic element is shown straight and upright; application of heat, however, would cause it to bend toward the contact point.

When the unit is switched on, a glow discharge begins in the
area indicated in the diagram. The heat from the discharge is conducted
to the bi-metallic element, causing it to bend toward the contact
point and close the circuit. Now the filaments heat up since a complete
circuit through the filaments has been established through the ballast
coil, but the little glow discharge ceases since the contact between
the bimetallic element and the point has shortcircuited the discharge
path. The bent bi-opening, the same inductive kick-back encountered
in the manual case appears to initiate the discharge arc. Once the
lamp discharge starts, the voltage across the starter is not high
enough to restrike the glow and it remains out.

Posted July 25, 2011