
Power Ratings & IM Tests November
1960 Electronics World


November 1960 Electronics World
[Table of Contents]People old
and young enjoy waxing nostalgic about and learning some of the history of early electronics. Electronics World was
published from May 1959 through December 1971.
As time permits, I will be glad to scan articles for you. All copyrights (if any) are hereby
acknowledged.

Although
this article discusses audio waveform measurements, the lesson learned
applies equally well to any waveform frequency. In the RF realm, we
are accustomed to injecting two sinewaves at equal amplitudes into a
unit under test (UUT) and reading the relative output powers of the
two input signals and the norder intermodulation signals. It is usually
a very simple test with simple to interpret results handily shown on
the display of a spectrum analyzer. The task is made a bit more difficult
when injecting signals of unequal strengths and especially when measuring
in units of voltage as a viewed on an oscilloscope display. I dare say
most of us need to do some head scratching and looking up of formulas
to pull off such a measurement.
See all the available
Electronics World articles.Power Ratings & IM TestsBy L. W. Born
Be
careful of what power rating you are quoting when checking hifi units
for IM distortion.
We had sold the amplifier kit, a well known
60watt basic amplifier, only a few days before and we now had a pretty
unhappy owner on our hands.
"This amplifier won't come even
close to putting out 60 watts without terrible distortion," he said
with disgust. "There must be something loused up with it!"
"How
do you know it's not putting out its rated power?" we asked. "Does it
sound bad or overload easily?"
"Well, no, it seems to sound
pretty good. But a friend of mine came over with his IM distortion analyzer
and measured it for me. About the most it would do was 40 watts, and
that's a far cry from the 60 watts it's supposed to deliver:"
Further questioning revealed that he had, in fact, been quite pleased
with the sound quality produced by his new amplifier, but since it had
been measured and found wanting, it no longer seemed to sound quite
right. We set up the amplifier on the test bench, loaded it into our
300watt noninductive load, clipped the scope across the output terminals
for visual observation of the waveform, and proceeded to check it out.
Everything seemed in order. We touched up the bias adjustment
slightly and, with rated output, measured considerably less than one
percent intermodulation distortion. Our "friend" kept his eyes glued
to every move that was made. "It's doing a real good job," we said.
"It's running about sixtenths of one percent distortion at 60 watts."
"Wait a minute, now; what scale are you. reading on the wattmeter?"
he inquired with a skeptical glance.
"Read the power on the
150watt scale," we replied.
"But that's not showing 60 watts;
that indicates only about 40 watts, just as I said!"
"True,"
we agreed, "but you must remember that we are not measuring a simple
sine wave. We are using a complex wave consisting of two sine waves:
one at a frequency of 60 cycles per second, and the other at a frequency
of 7000 cycles mixed together in a voltage ratio of fourtoone respectively.
With such a complex signal, the wattmeter reading must be multiplied
by a factor of 1.47 to obtain the equivalent single sinewave power."
It was apparent that we were not getting through. "Look," we
said, "let's feed a singlefrequency sine wave into the amplifier and
adjust the input until the wattmeter reads 60 watts." The scope showed
a beautiful waveform, and the amplifier delivered 60 watts easily. "Now,
notice the vertical deflection on the scope. We adjust the gain of the
scope until the pattern covers exactly ten divisions. But see what happens
when we add the 7000cycle signal which was preset to equal just onefourth
the value of the 60cycle signal: the amplifier is obviously overloading.
Next we reduce the combined input signal until the scope shows the same
tendivision deflection that we saw corresponded to a power output of
60 watts. Then we are still driving the amplifier to its maximum peakpower
output, but because we are no longer measuring a simple sine wave, and
because the wattmeter happens to be an averagepower instrument, the
wattmeter shows the true average power output of 40.8 watts."
"You mean that a sixtywatt amplifier can overload even when it's
producing only 40 watts output?" he asked incredulously.
"Absolutely!
As a matter of fact, it can overload when it is putting out much less
than 40 watts average power, depending on the waveform it is handling.
With some very complex waveforms generated by many kinds of music, it
is entirely possible that an average power output of only a few watts
could demand peak powers in excess of the capacity of the 60watt amplifier:
"In many ways, it would be better if we used the peakpower
output rating for amplifiers rather than averagepower ratings, because
when an amplifier is producing its maximum peakpower output, this is
its limit no matter what the averagepower output may be. With a single
sinewave signal, the peakpower output is exactly twice the value of
the averagepower output. In other words, when your amplifier was delivering
60 watts as measured by the wattmeter when using the single sinewave
signal, it was necessarily and mathematically producing 120 watts peakpower
output. This simple twotoone ratio is true only for a simple sinewave
input. With complex waveforms, the ratio can be many times as great.
With the 4to1 ratio signals used in intermodulation distortion measurements,
the ratio of peaktoaverage power is 2.94 to 1, and as the wave becomes
more complex the ratio in general becomes even larger.
"When
your friend measured the distortion at an indicated power of 40.8 watts,
he was actually measuring the distortion with the amplifier delivering
120 watts peak power with the complex wave and that's all the amplifier
can deliver."
"Then the amplifier is really all right?"
"Yes, sir. It's doing even better than the specifications call for
and we're certain you will be completely satisfied with its performance.
And, by the way, ask your friend with the analyzer to come in for a
chat, will you?"
Fig. 1. With a singlefrequency sinewave signal applied to
the amplifier, these are the output power relations. 
Improper Power Ratings
Misunderstandings and dissatisfaction
occur repeatedly in connection with IM distortion measurements caused
by improper evaluation of amplifier output when using complex waveforms,
despite words of caution from the manufacturers of the test equipment.
Let's look into the problem a little more deeply.
Let's assume
we have an amplifier set up and adjusted in such a manner that with
an input signal of 60 cycles per second measuring one volt on a meter
indicating r.m.s. value, an average power output of ten watts is produced
in a 16ohm load. Elementary a.c. theory tells us that the peak value
of the input voltage is 1.414 times the r.m.s. value or, in this instance,
1.414 volts peak.
Looking at the output side of the amplifier,
we would measure an r.m.s. voltage across the load of 12.65 volts with
a corresponding peak value of 1.414 x 12.65 = 17.89 volts, as in Fig.
1. Using the wellknown relationship:
Power = E^{2}/R
we find average power
is
= 10 watts peak
power is
= 20 watts.
Here again, we find the situation where the peak
power is exactly twice the average power, using a single sine wave.
Now, if we remove the 60cycle, 1volt signal and substitute
a 7000cycle, 1/4volt signal with all other adjustments unchanged,
we find the peak value of the input signal is 1.414 x 0.25 = 0.3535
volt. At the load we now find an r.m.s. voltage of 3.1625 volts with
its corresponding peak value of 4.4725 volts. With the 1/4volt input
signal we find average power is
= 0.625 watt
peak power is
= 1.25
watts and again, since a single sine wave is involved,
the ratio of peakpower output to averagepower output is twotoone.
In intermodulation distortion measurements, these two above
signals are mixed or added together, and the higher frequency component
"rides" on the lower frequency signal as shown in Fig. 2, producing
a combinedsignal peak voltage that is 1.414 plus 0.3535 = 1.7675 volts
at the amplifier input. At the output, assuming no overloading we find
a peak voltage equal to the sum of the two separate peak values or 17.89
plus 4.4725 = 22.36 volts.
Fig. 2. When two frequencies are applied for an intermodulation
distortion test, the picture changes considerably. 
This value of peak voltage indicates a peak power of
peak watts.
Yet the average power with the combined signals
is actually the same as if each of the two signals operated independently,
or 10 plus 0.625 = 10.625 watts. Hence it is seen that adding the highfrequency
signal to the larger lowfrequency signal adds only 0.625 watt to the
average power output while at the same time, it increases the peakpower
output from 20 watts to 31.24 watts.
A single sine wave that
would produce a peakpower output of 31.24 watts would have an averagepower
output of onehalf this amount or 15.62 watts. Thus we say the above
complex signal produces an equivalent sinewave power of 15.62 watts.
The ratio of the equivalent sinewave power to the actual average
power is therefore 15.62/10.625 or 1.47. Hence, if the wattmeter used
to measure power responds to average power, it will be necessary to
multiply the indicated reading by the factor of 1.47 to obtain the equivalent
sinewave power. Likewise, if a voltmeter is used to measure the output
voltage across the load, and power output computed from this measured
voltage, and if the voltmeter responds to the r.m.s. value of voltage,
the power output computed from the meter reading must be multiplied
by the same 1.47 factor.
Other Instruments
If the wattmeter
or voltmeter used to measure power responds to the peak value, no correction
factor should be used since the meters will be indicating equivalent
sinewave power or voltage. Different types of instruments respond Differently
to the same a.c. wave. Thus in addition to those that respond to peak
values, simple rectifier or diode types produce a deflection in an d'Arsonval
meter movement proportional to the average value of the rectified wave;
thermocouple, electrodynamometer, movingiron, and electrostatic type
instruments produce a deflection proportional to the r.m.s. value of
any waveform. Vacuumtube voltmeters may produce a deflection which
range from proportionality to the average value up to the peak value,
depending upon the design of the instrument.
Where available,
the oscilloscope may be used to adjust the output of the amplifier under
test to have the same peak output with the complex wave as it had with
the single sine wave of known power output. Then the amplifier will
have the same equivalent sinewave output when using the twofrequency
test signal.
Should there be some uncertainty as to just what
the wattmeter or voltmeter is responding to, a simple test may be performed
to determine "watts what." Adjust the amplifier under test to produce
a power output of about half its rated power output, using a 60cycle
signal of known voltage at the input. Introduce the 7000cycle signal
of a value just onefourth that of the 60cycle voltage. The indicated
power output should then increase 56.2% and the indicated output voltage
should increase 25%. Should these percentages not be obtained, it is
a simple matter to determine from the observed increases, how much correction
factor is required for the particular meter in question to obtain the
necessary percentage increases. It must be remembered that the correction
factors so determined are valid only for twofrequency signals mixed
In a fourtoone ratio for "the particular meter in question.
Posted January 10, 2014



