November 1960 Electronics World
Table
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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.
Power Ratings & IM Tests
By 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?"
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.
Fig. 1. With a singlefrequency sinewave
signal applied to the amplifier, these are the output power
relations.
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