May 1932 Radio-Craft
Wax nostalgic about and learn from the history of early electronics.
See articles from Radio-Craft,
published 1929 - 1953. All copyrights are hereby acknowledged.
Here is Part 1 of a three-part article on attenuator pad and impedance
matching articles that appeared in Radio-Craft magazine. Although
the focus is on audio frequencies, the principles apply in general.
It is interesting to read about wavelengths expressed in units
of miles versus feet and meters like we are used to seeing for
radio frequencies. Keep in mind that most of the decibel formulas
used here are for voltage and not for power.
The Theory and Construction of
Attenuators, Line Filters and Matching Transformers (Part I)
The Theory and
Construction of Attenuators, Line Filters and Matching Transformers (Part II)
The Theory and Construction of Attenuators and Line Filters (Part III)
The Theory and Construction of Attenuators, Line Filters
and Matching Transformers
By Hy Levy, B.S.
In any well-designed voice transmission circuit, such as
telephone lines, radio-broadcast speech equipment, public-address
systems, "talkie" apparatus, etc., will be found resistance
networks, or attenuators, more commonly called "pads." The proper
use and design of these pads make for an efficient transmission
system, from which the maximum output of energy with the least
distortion may be obtained, and which, if these pads were not
used, could not be realized. It can be said that pads find their
way into practically every phase of voice transmission where
quality reproduction is the prime requisite.
Fig. 1 - Circuit illustrating current distribution.
It is the purpose of this article to make clear the methods
by which these pads are designed, to show their application
in transmission circuits, and to discuss the problems encountered
in transmission circuits where pads are used.
Purpose of Pads
In voice-transmission circuits, where energy is being transmitted
over a line to a load located at the far end of the line, it
is necessary that some means be employed to control the magnitude
of the energy entering the load. It is for such purposes that
pads are used. These networks are always used between a source
of energy and a load. The source of energy might be any of the
(1) Output of a speech amplifier, such as a normally high-level
amplifier which is feeding another amplifier located at a remote
(2) Output of a high-level amplifier (power amplifier);
(3) Output of a low-level amplifier, such as a condenser-microphone
amplifier, which is feeding a speech amplifier located at some
(4) Output of microphone circuits, etc.: The load may consist
of a transmission line carrying the energy and terminating in
an impedance located at the far end of the line. This load impedance
might consist of any of the following:
(a) Primary side of a line-matching transformer (line to
(b) Input circuit of a speech amplifier;
(c) Loud speakers located at distant points from an amplifier;
(d) Mixing circuits, etc.
The attenuator imposes a constant impedance upon the transmission
line, thereby controlling the level (magnitude) of the energy
being transmitted into the far end of the line. The attenuator
or "pad" maintains this level by introducing a loss in energy
between the source and the load, at the same time causing no
impedance mismatch to the impedances between which it is working
(source impedance and load impedance).
Fig. 2 - Schematic diagram of an "H"-type
To illustrate the statements given in the preceding paragraphs,
a typical case showing the use of such networks will be given.
The complete design and calculation of the pads will be shown,
and the problem arising in circuits in which these pads are
used will be discussed. It is hoped that by obtaining a complete
understanding of the subject matter given in these papers, any
problems relating to the design and use of pads in voice transmission
circuits will be materially lessened.
Let us assume we have a radio amplifier whose output voltage
is 1.5 (effective value of alternating current). This voltage
is to be fed over. a transmission line, at the far end of which
is located the primary side of an input transformer, whose secondary
is in the input circuit of an amplifier (see Fig. 3). Let us
also assume that it is also necessary to reduce the voltage
impressed across the primary side of the transformer to approximately
0.15 volts (R.M.S.). It can be seen, therefore, that it is necessary
to interpose a network of some sort between the input source
of 1.5 volts and the primary side of the input transformer in
order that the voltage will be reduced to the desired value,
at the same time introducing no impedance mismatch between the
source and the load.
Fig. 3 - Impedance relations ·in a typical
In voice transmission circuits, this network is composed
of non-reactive resistances so arranged that they will cause
the desired loss between the input and output terminals of the
pad. By "non-reactive" is meant resistances whose impedance
remains practically constant to alternating current. This is
accomplished by constructing the resistances so that their inductance
and capacitance is negligible throughout the frequency band
in which they function. (The construction of the resistances
for use in the pads will be given in a forthcoming article).
For most purposes, this frequency band can be taken as the audio
spectrum of 40 to 10,000 cycles per second. The resistances
used in networks must be non-reactive in order that the attenuator
maintain constant impedance characteristics throughout the audio
band to the impedances between which it is working, so that
straight line frequency attenuation will be obtained without
frequency distortion in the transmission circuit, which would
hinder the intelligibility of the signal.
The change in voltage between the source and the load is
expressed as the logarithmic ratio between the two voltages.
At this point a brief resume on the transmission unit or decibel
will be given, so that those not familiar with this unit will
have a clear understanding of how it is used in transmission
In electrical circuits carrying energy which is either being
amplified or attenuated, the question arises: What is the relationship
between voltage, current, and power on the input side of an
amplifier or an attenuator, to the voltage, current, and power
on its output side? The engineer signifies this ratio between
two voltages, two currents, or two powers, by the number of
decibels change taking place between the input and output terminals
of the amplifier or the attenuator.
In any electrical circuit carrying energy, the product of
the common logarithm of the power ratio multiplied by ten, or
the product of the common logarithm of the current or voltage
ratios multiplied by twenty determines the change in decibels.
As an example of this principle, let us assume that in a
certain circuit the voltage has been decreased from 3.0 volts
to 1.0 volt, or to one-third of its original value, Then from
the definition as given above, the voltage ratio is 3 to 1,
or 3. The common logarithm of three is 0.4771 (which may be
obtained from a table of common logarithms). This means that
0.4771 is the power to which ten must be raised to give three.
Then, multiplying 0.4771 by twenty, we get 0.9542 decibels change:
i. e., decibels change =
where V1 = input voltage
= output voltage
= 20 log 3.0
= 20 X 0.4771
decibels change = 9.542
The above simply means that if the voltage in the circuit
has been decreased to .one-third of its original value, we have
caused a change of 9.542 units or decibels to have taken place.
(See table in November, 1931 issue of Radio-Craft - Editor.)
This change in decibels is expressed as. a loss, as the voltage
has been decreased. Similarly, if the voltage had been increased
to three times its original value, the resultant change in decibels
would also be 9.542, only in this case the result would a gain.
Therefore, if the voltage has been increased or decreased to
three times its original value, the resultant change in decibels
is the same in both cases, or 9.542. The only difference in
the two cases is in the method of expressing the results obtained.
Fig. 5 (top) - Open-end transmission line.
Fig. 6 (bottom) - "Shorted" transmission line.
When using the formulae shown below, the change in decibels
obtained depends on the type of circuit involved. In an amplifying
circuit, the relationship between the voltage, current, or power
ratios is expressed as a gain in decibels. In an attenuator,
the relationship between the voltage, current, or power ratios
is expressed as a loss in decibels. In the circuits to be taken
up, attenuators (resistive networks) will be discussed, and
a decrease in energy will be incurred between the terminals
of the pad, and therefore all results will be given as a loss
Thus by expressing current, voltage, and power ratios in
decibels, one has an accurate description of the changes in
energy taking place in the circuit.
From the definition of the decibel as given above, we can
write the following formulae:
Change in decibels =
where P1 = input power
P2 = output power
V1 = input voltage
V2 = output voltage
I1 = input current
I2 = output current
This is shown in Fig. 1.
Also if we know the number of decibels change, and it is
desired to determine the voltage, current, or power ratios,
we can write:
To illustrate this method, assume we wish to introduce a
15 decibel loss into the circuit, and it is desired to know
what voltage ratio will give this loss. Then from above:
= antilog .75
which means that if the voltage ratio is equal to 5.62, a
15 decibel loss will be maintained in the circuit.
Returning now to our own problem, it was determined that
a network is required to cause a loss in voltage between its
terminals, from 1.5 volts, on the input side, to 0.15 volts,
on the output side. This loss expressed logarithmically is as
loss in decibels =
= 20 log 10.0
= 20 x 1.0
loss in decibels = 20.0
or, a 20 decibel loss is to be incurred between the input
and output terminals of the network to be designed. This is
the same as saying that by reducing the voltage from 1.5 volts
to 0.15 volts, or in the order of ten times, a loss of twenty
decibels has taken place.
The first problem to be considered is what type of network
shall be used to cause this loss in voltage.
In communication circuits, two types of networks used for
this purpose are:
(1) "H"-type pad
(2) "T"-type pad
The H-type pad will be discussed first, and a complete discussion
will be given showing the method of obtaining the desired twenty
decibel loss by the use of this pad in our own problem.
The H-type pad is so called because it is composed of five
resistances taking the form of the letter "H". This pad is designated
as a balanced network, in that an equal number of resistances
(Z1) are used in the series arms on both sides of the line as
shown in Fig. 2. In some circuits, the shunt arm Z2 may be divided
into two equal parts, with the midpoint grounded. This balances
the entire network with respect to ground.
At this point an investigation of certain factors which must
be considered in transmission circuits will be given in order
that an understanding of the problems encountered in voice transmission
will be bad. It is of the utmost importance that a general knowledge
of the circuit characteristics be obtained, so that when the
design of the pads is taken up, maximum efficiency may be obtained
from the transmission line in which the pad is to be placed.
In Fig. 3 is shown the output circuit of an amplifier feeding
a transmission line, at the far end of which is located an input
The transformer (T2) used in the output circuit of the tube
is called a tube-to-line transformer or output transformer,
in that it transfers energy from the tube to the transmission
line. The secondary side of this transformer is called the source
impedance of the line, for the energy induced in this winding
is the energy transmitted along the line. This transmission
line, we can assume, terminates in the primary side of an input
transformer (T1), whose secondary side is in the grid circuit
of a vacuum tube. This transformer is called a line-to-tube
transformer or input transformer, as it transfers the energy
from the line into the tube. The primary of this transformer
is known as the load or terminating impedance, as the transmission
line can be considered as ending at this point, for the primary
side completes the transmission line, and the secondary side
is considered in another circuit.
In communication circuits the source and load impedances
of the transmission line have been standardized in most cases
to 200, 500, and 600 ohm lines. Therefore. if ZS, the source
impedance, equals 200 ohms, then the load impedance ZL must
also equal 200 ohms in order to prevent so-called "reflection"
losses from being set up in the circuit. It is these "reflection"
losses that will now be discussed.
Table 2 - Audio frequencies versus harmonic wavelengths.
The speed with which electromagnetic waves travel is the
same as that of light, or approximately 186,000 miles per second.
This speed is attained only in a circuit having zero losses,
which of course is not obtainable in practice. All circuits
have some losses, and in transmission lines these losses increase
as the frequency increases. The speed of the transmitted waves
is retarded slightly by the losses of the line, and by what
is known as "skin effect" of the line conductors. (Skin effect
is the forcing of the current to flow along the outer surfaces
of the conductors.) As the frequency increases, the inductance
inside the conductor (mostly in the center) increases, thereby
offering a greater opposition to the flow of current within
the conductor, and consequently the current is forced to travel
along the outer surface of the wires. The relationship which
exists between inductance, capacitance, and the velocity of
where V = velocity of light in miles per second
L = inductance in henries
C = capacitance in farads
If a frequency of 100 cycles per second is impressed upon
a circuit of infinite length (no end), at the end of 1/1000
of a second the wave will have traveled 186000/1000 or 186 miles.
A part of such a wave is shown in Fig. 4. This circuit is
said to have a full wavelength for the frequency of 1000 c.p.s.
(cycles per second). In this circuit, when the first part of
the waves arrives at a point 186 miles distant from the start
of the transmission line, the end of the. same wave is at the
start of the circuit. This is assuming zero losses in the line.
The actual wave would be slightly retarded due to the losses
of the circuit. This is shown in the dotted lines of Fig. 4.
If now an alternating-current wave is impressed upon a transmission
line, it will travel along the line with the speed of light,
until it reaches the far end of the line, where, if the far
end is open (infinite impedance), the wave will be reflected
back to its origin with the same velocity, but will gradually
decrease in magnitude as it approaches the starting point of
the line, due to the losses of the line which it encounters.
If at the moment this wave reaches the starting point another
wave is sent into the line, that which was left of the first
wave adds itself to the second wave and therefore increases
the second wave. If, at the moment that the returned second
wave reaches the starting point, a third wave is sent into the
circuit, that which was left of the second wave adds itself
to the third wave and so on as the number of waves increases;
with the result that an accumulative effect of energy is developed
in the circuit. The magnitude to which this reflected energy
can rise is determined by a characteristic of the line conductors
known as the "surge impedance."
The surge impedance of a conductor is the impedance which
the conductor offers to these free oscillations of energy at
the high frequency limit. (This condition is greatest at the
highest frequency that the circuit passes.) This building up
of reflected energy in the circuit is known as "quarter-wave
resonance," or merely as the resonant frequency of the line.
To produce such a condition, it is necessary that the alternating-current
impulses occur at intervals of time equal to the time necessary
for the energy to travel the length of the line and back. In
other words, for one complete cycle (two impulses) the energy
would have to travel the length of the line four times between
successive impulses in order to create resonance. Expressed
as a formula:
or, length in miles =
It can be seen, therefore, that as the frequency increases
the length of the circuit becomes shorter in order to produce
the resonant condition. In table 1 is given the length of circuits
in miles to produce resonance throughout the audio band.
Table 1 - Audio frequency and wavelength.
At this condition of resonance, the magnitude of the reflected
energy reaches a maximum; and abnormal voltages and currents
are set up in the circuit. This reflected energy, if present
in a circuit in which vacuum tubes are employed., will cause
incorrect voltages to be applied to the grids of the tubes,
which in turn will cause distortion of the wave-form of the
of the original signals, with the result that harmonics will
be generated in the circuit. As these harmonics increase, the
resonant period of the circuit is decreased. This is shown in
table 2, where the resonant periods of circuits due to harmonics
being set up in the line are given. It can readily be seen that
as the harmonics increase, the length of the line necessary
to produce resonance is decreased.
Now in practical circuits, such as the transmission lines
encountered in everyday sound work, the resonant condition may
not be theoretically reached as the fundamental frequencies
used are not high enough to produce resonance for the length
of the lines ordinarily used, which in most cases are comparatively
short. Nevertheless, if the lines are not properly terminated,
the reflected energy which is set up produces distortion of
the wave-form of the original impulses, which in turn will create
harmonics. These harmonics are multiples of the fundamental
frequencies, and as these harmonics increase, the frequency
of the circuit increases, thereby approaching the resonant frequency
of the line. .Consequently, as the resonant period of the line
is approached, extreme distortion of the signals arises, with
the result that the quality is atrocious.
The analysis of reflected electrical waves is analogous to
that of water flowing in a canal. If the near end of a canal
is struck a blow, a water-wave will be set up which will travel
to the end of the canal, where it will be reflected back to
the origin. As it travels back toward the starting point, its
amplitude will decrease slowly, due to resistance encountered
in the canal. If, at the instant the wave reaches the origin,
another blow is struck, a second wave will flow to the end of
the canal and return. This second wave will be composed of itself,
plus that which was left of the first wave. This effect increases
as the number of waves is increased, and reaches a maximum when
the losses due to resistance prevent the amplitude of the waves
from increasing further.
If a gate or obstruction of some sort is inserted at the
end of the canal, any waves traveling down the canal will strike
the wall and bank up to a height determined by the potential
energy of the wave. When the potential energy equals the kinetic
energy impressed on the wall, the wave will have reached its
maximum height. It can be shown that in certain cases the waves
will rise to a height twice that of the original wave striking
the wall, and then will be reflected back at a slowly decreasing
amplitude as the starting point is approached.
Similarly, in electrical circuits, when an electromagnetic
wave reaches the end of a transmission line, and the line is
open at the far end as shown in Fig. 5, the wave is reflected
back to the source; the current is zero, and the voltage at
the load reaches a value equal to twice the peak value of the
Also, if instead of open circuiting the far end of the line,
it is short circuited, the wave is reflected with a current
value twice the value of the starting current, and the voltage
becomes zero. This is shown in Fig. 6.
Posted November 16, 2015