June 1932 Radio-Craft
People old and young enjoy waxing nostalgic about and learning some of the history of early electronics.
Radio-Craft was published from 1929 through 1953. All copyrights are hereby acknowledged. See all articles
It seems most of the articles we see on the subject of attenuator
pads are based on signal reduction in terms of decibels for
units of power. Although it is a simple matter to convert power
decibels to voltage decibels, it would be more convenient if
you are working with voltage to have formulas and tables of
values based on voltage ratios. This article does just that.
As a reminder, the decibel representation of a ratio is always
10 * log10 (x). If you have a voltage
ratio of V1/V2 = 0.5, then 10 * log10 (0.5) = -3.01 dB.
If you have a power ratio of P1/P2 = 0.5,
then 10 * log10 (0.5) = -3.01 dB.
Does that mean that -3.01 dB of voltage attenuation is
the same as 3.01 dB of power attenuation? Confusingly,
Let's say for the sake of simplicity that you have 1 V
across a 1 Ω resistor, yielding a power dissipation
of P = V2/R = 12/1 = 1 W.
Now, reduce the voltage
to 0.5 V and the power dissipation is 0.52/1 = 0.25 W.
The ratio of the voltage change is 1:2 but the ratio of power
change is1:4. In decibels of voltage that represents 10 * log10 (0.5) = -3.01 dB,
but in decibels of power it represents 10 * log10 (0.25) = -6.02 dB.
A multiplication factor of 2 therefore exists between voltage
ratio and power ratio when applied to the same system.
Now, let's work it in the opposite direction. For the sake
of simplicity again, you have 1 W being dissipated by a
1 Ω resistor, yielding a voltage across it of V = sqrt(P*R) = sqrt(1*1) = 1 V.
Now, reduce the power
to 0.5 W and the voltage across the resistor is sqrt(0.5*1)
= 0.7071 V. The ratio of the power change is 1:2 but the
ratio of voltage change is1:0.7071. In decibels of power that
represents 10 * log10 (0.5) = -3.010 dB,
but in decibels of voltage it represents 10 * log10 (0.7071) = -1.505 dB.
Once again as you would expect, a multiplication factor of 2
exists between voltage ratio and power ratio when applied to
the same system.
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
In this second of a series of articles, the author will discuss
additional design considerations of attenuators, illustrating
them with numerical examples.
By Hy Levy, B.S.
Last month, some of the more pertinent problems of transmission
lines (with special reference to "pads" or "attenuators") were
outlined. In this issue, we will continue with a discussion
of actual design considerations.
Table 3 - Attenuator Impedance Values
The two extreme conditions that a transmission line could
be subjected to are an open-circuit and a short-circuit condition;
in which cases, the greatest reflection of energy will take
place, and abnormal voltages and currents will be set up from
small impulses originally sent into the line. This is the same
as saying that infinite impedance terminates the line on open
circuit, and zero impedance terminates the line on short circuit.
These two conditions are very rarely encountered in practice,
unless the line should accidently become open or shorted at
the far end.
Assume a transmission line in which the line conductors have
negligible attenuation (zero losses) and in which the generator
or source impedance is of known value. Then, if a terminating
impedance is inserted at the far end of the line, and starting
from the open-circuit condition of infinite impedance, this
terminating impedance is gradually made smaller and smaller,
the reflected energy will also become smaller, and as the terminating
impedance approaches the value of the source impedance, the
reflected energy rapidly decreases. At the point where the value
of the terminating impedance exactly equals the source impedance,
the reflected energy will be zero (no reflection losses), and
maximum energy will be obtained at the load.
Likewise, starting with the short-circuit condition of zero
terminating impedance, and increasing the load impedance, the
reflected energy becomes smaller, so that when the point is
reached where both the terminating and the source impedances
are exactly equal, the reflected energy will be zero (no reflection
losses), and maximum energy will again be fed to the load.
When the terminating or load impedance exactly equals the
source impedance, the line has the same characteristics as a
line of infinite length. If a line of infinite length could
be constructed, any energy sent into this line would never be
reflected back to the source. By inserting at the far end of
a transmission line a load or terminating impedance equal to
the source or generator impedance, the characteristics of a
line of infinite length are immediately obtained (assuming zero
losses in the circuit) and consequently no reflection losses
are set up in such a circuit.
Fig. 7 - Diagram of an attenuator pad.
This entire discussion on reflection losses can be summed
up by stating that the cause of these losses is improper impedance
matching, and it is for this reason that the subject of proper
impedance matching is so strongly stressed and observed in voice
transmission circuits. In order to avoid reflection losses,
care must be taken to see that, at any point in a transmission
line where an impedance is to be inserted, this terminating
or load impedance is made equal to the impedance working into
it. Reflection losses, as previously stated, give rise to abnormal
voltages and currents; harmonics are set up, and the wave form
of the impressed original signals are distorted, with the result
that straight-line frequency-response characteristics are not
obtained from such a circuit.
If the load impedance Z1 (see Fig. 3) exactly equals the
source impedance Zs, then the source impedance Zs is said to
be working into its "image impedance," for in such a case, there
is no reflection of energy from the load impedance back into
Fig. 8 - A resistive attenuator network.
If an attenuator is now inserted into the transmission line
between the source and the load, it must be so designed that
the impedance of the attenuator must in no way upset the impedance
match between the source and the load. If then, when the attenuator
is working in the transmission line, the impedance "looking"
into the source (from the line) exactly equals the impedance
"looking" into the load (from the line), then the pad is said
to be working between its "image impedances," and no energy
will be reflected back at any point along the line, as the impedances
looking in either direction along the line are exactly equal.
In such a case, no reflection losses will occur at any junction
points throughout the line, and practically ideal transmission
characteristics are obtained, except for such losses as arise
from the effects due to the length of line over which the energy
is fed. (The losses due to length of line will not be considered
in these papers. It is assumed that the voltage at the source
is the voltage appearing across the 1-2 terminals, (See Fig.
7) of the pad, and the voltage at the 3-4 terminals is the voltage
across the load).
Fig. 9 - Equivalent circuit of Fig. 8.
In order to maintain the same impedances in either direction,
the pad must be designed in such a manner, that when looking
into the 1-2 terminals, the combined (resultant) impedance of
the attenuator and the load must exactly equal the source impedance.
Also, when looking into the 3-4 terminals, the combined (resultant)
impedance of the attenuator and the source must exactly equal
the load impedance. These last two statements described the
"characteristic impedance" of the circuit when looking into
both the source and load ends of the line.
In order therefore, to fulfill the circuit requirements for
image impedance, and characteristic impedance, the two must
be equal, and upon completion of the design of the pads, this
will readily be seen.
Having discussed the various problems that must be considered
from an impedance matching standpoint, our next problem is to
analyze the manner in which pads function in the transmission
It is the purpose of the writer to lead up to the actual
design of the pads by the method of "approach." For, having
threshed out the various problems associated with the circuit,
the design of the pads is then reduced to the simple process
of substitution into the working formulae for the pad under
Continuing now with an analysis on pads, it was stated in
the first part of this paper, (May, 1932 issue) that pads are
used to control the level of the energy entering the load. We
are principally interested then, in determining the relative
amount of current and voltage appearing at the load. In Fig.
8 is shown an "H"-type network working between a source impedance
Zo and a load impedance Zo. Now, referring to Fig. 9, which
is the equivalent circuit of Fig. 8, it can be seen that if
we wish to keep the current I2 and the voltage V2 at the load
at certain fixed values, with a constant voltage V1 impressed
across the input terminals of the pad, the values of the resistances
used in the pad would have certain predetermined values in order
to give the correct loss between the input and output terminals.
The values of these resistances are determined by the impedances
between which the pad is working and the ratio of the current
or voltage on the input side, to the current or voltage on the
output side of the pad.
Let us return now to our own problem under consideration,
that of designing a pad to give a 20·decibel loss. We have determined
that if the voltage is reduced to 1/10 of its original value,
or in the order of 10 to 1, the voltage ratio is 10, and expressed
logarithmically this loss is 20 decibels.
Suppose we wish to reduce the voltage only a small amount,
or in the ratio of 1.12 to 1. We would then have only a 1-decibel
loss in the circuit, which is comparatively small. Now suppose
we wish to set up a large loss, in the order of 100 to 1. We
then would have a 40-decibel loss. Taking these three cases
of 1-, 20-, and 40·decibel losses, and impressing various values
of voltage V1 at the input terminals, let us see how the pads
that we would have to design would vary, in order to keep a
constant voltage V2 at the output terminals of the pad. In Table
3 are given the values of resistances as used in the series
and shunt arms of "H"-type pads, when working between 200-,
500-, and 600 ohm impedances.
Now, using the 20-decibel loss as a reference level and referring
to this table, it is seen that to obtain a 20·decibel loss the
series arm, Z1 is 82 ohms, and the shunt arm Z2 is 40.4- ohms,
when working between two 200-ohm impedances. With these constants,
the output voltage would be 1/10 the value of the input voltage.
Or, if the input voltage is 1.5 volts, the output voltage would
be 0.15-volts, and the voltage would have been dropped 1.35
Now assume that the input voltage is 0.168 volts and we still
wish to maintain the output voltage V2 at 0.15-volts. The voltage
ratio, is 0.168/0.15 or 1.12; or, a 1-decibel loss is obtained
in the circuit. From Table 3, for a 1-decibel loss, the series
arm Z1 is only 5.65 ohms, whereas, the shunt arm Z2 goes up
to 1760 ohms.
From the simple laws of current flow, this can be understood;
for to obtain only a 1-decibel loss, we do not wish to introduce
a large amount of series resistance, as this would impede the
current flow, which is what we wish to avoid. The voltage drop
in this case is only 0.168 minus 0.15 or 0.018-volts. Therefore,
the series arm Z1 must be very small, and is but 5.65 ohms.
Also, the shunt arm Z2 must be very large, so that very little
current will be bypassed or shunted from the load, and as can
be seen, its value is large, being 1760 ohms.
Now comparing the 20-decibel loss to the 1-decibel loss,
in order to obtain the same output voltage with different input
voltages, the series arm Z1 for a 1-decibel loss is only 5.65
ohms, and for a 20-decibel loss is 82 ohms. The shunt arm Z2
for a 1-decibel loss is 1760 ohms, and for a 20-decibel loss
is only 40.4 ohms. Therefore, to obtain a fairly large loss,
(such as 20 decibels), the series arm is quite large and the
shunt arm comparatively small (compared to the load impedance.)
In this case, the current will be impeded by the large series
arm, thereby dropping the voltage considerably, and at the same
time shunting it from the load by the comparatively small shunt
arm, thus creating the fairly large loss desired.
For a small loss such as a 1 decibel, the series arm is very
small and the shunt arm very large as compared to the load impedance.
In this case, very little voltage drop is encountered in the
series arm, and since the shunt arm is very large, very little
current will be shunted from the load. Therefore, practically
all of the original current will enter the load, thereby obtaining
the small loss desired.
Now if the input voltage is 15 volts and we want to maintain
the output voltage V2 15 at 0.15-volts, the voltage ratio is
15/0.15 or 100, and the decibel loss is 40. This means we need
to set up a very large loss, for the voltage is to be dropped
from 15 volts to .15-volts, a decrease of 14.85 volts. To do
this, we should need a very large series resistance and a very
small shunt resistance, so that practically all of the current
will be prevented from entering the load by (1) the large series
arm, and (2) the very small shunt arm. Referring to Table 3,
it can be seen that to cause a 40-decibel loss, the series arm
Z1 is 99 ohms and the shunt arm Z2 is but 2 ohms, which exactly
follows the reasoning as given above.
Comparing this 40 decibel loss to the 20 decibel loss, the
series arms for the 20- and 40-decibel losses are 82 ohms and
99 ohms respectively, and the shunt arms 40.4 and 2 ohms respectively.
The values for the series arms are such that a large series
voltage drop is encountered in them, being greater in the 40
decibel case, as would be expected. The shunt arms are quite
In the 20-decibel loss, the shunt arm is practically 1/5
of the value of the load impedance, so that only a comparatively
large bypass effect of current from the load is obtained; whereas,
1 in the 40-decibel loss, the shunt arm is only 1/50 of the
value of the load impedance, thereby shunting practically all
of the current from the load. For a very large loss then, the
series arms are very large and the shunt arms very small in
comparison to the load impedance. By inspecting Table 3, it
will be seen that the various values of the series and shunt
arms necessary to give certain losses follow the simple laws
of current flow; in that large-decibel losses require large
series arms and small shunt arms in comparison to the value
of the load impedance, and that small decibel losses require
small series arms and large shunt arms in comparison to the
Posted September 15, 2015