[AMRadio] AM power


Geoff Edmonson w5omr at att.net
Sun Jun 19 07:31:11 EDT 2011


On 06/18/2011 10:29 PM, Gary Schafer wrote:
> Sorry for the misspelled name Geoff.
>
> It sounds like you have your transmitter set up to never reach 100%
> modulation on positive or negative peaks. That's ok too.
>
> But there are two different 100% areas to be concerned with. The positive
> and the negative. If your transmitter output voltage doubles on positive
> peaks, that is 100% positive modulation and is what you are concerned about
> for measuring peak envelope power. It does not matter that your negative
> peaks never hit 100% on the negative side. You still calculate PEP with
> positive modulation  peaks.

Since no one apparently wants to go to the site to read for themselves,
here's an excerpt from http://www.qsl.net/wa5bxo/asyam/aam3.html

========================================================
"The high-level modulation method is the application of the modulating 
voltage to the plate circuit of the class C final, causing the output 
amplitude to vary in accordance with the applied modulation. One hundred 
percent (100%) modulation was generally defined as the point where the 
maximum modulating voltage, during its negative half cycle, opposed the 
DC supply voltage sufficiently to reduce it to zero. If this voltage 
dipped below zero, over-modulation and splatter were the result."

   Most people agreed that the peak of the positive half cycle of the 
modulating audio voltage, added to the DC supply, could go as high as 
necessary to faithfully reproduce the audio as an image of the 
microphone output. Even if the positive peak was more than two times the 
amplitude of the negative peak, the modulation was not considered 
illegal unless it contained distortion products that caused splatter 
over an excessive bandwidth. Over-modulation was only considered to 
occur at the point where modulation characteristic became non-linear, 
producing distortion and splatter.

     The audio voltage from a microphone is often not symmetrical, 
unlike a sine wave from a signal generator. This asymmetry is a /natural 
quality/ of speech and other sounds. This article discusses the use of 
voice waveform asymmetry in AM systems.

     When an AM transmitter is 100% modulated by a pure sine wave, the 
PEP (Peak Envelope Power), is 4 times the un-modulated carrier power. 
This is because the Audio Voltage modulating the carrier doubles the RF 
voltage at the peak, since the load resistance is constant, the RF 
current doubles at the same instant as the RF voltage. Since P = E * I, 
then P at the instant of the positive peak must be 4 times greater than 
the power of the original carrier.

     If the transmitter modulation is increased until the peak RF 
voltage is 2.5 times the original carrier RF voltage, the peak RF 
current occurs at the same instant, and it is also 2.5 times as great as 
the original carrier RF current The result is a PEP of 2.5 * 2.5 = 6.25 
times the un-modulated carrier power.

Here's the math:
PEP = ((peak-to-peak modulated RF voltage / un-modulated carrier 
voltage) squared) * un-modulated carrier power

     This large PEP can occur without negative over modulation, if the 
modulating audio is acquired by a voice from a microphone. Microphone 
audio is generally asymmetrical.

     To help me understand and explain the relationship between audio 
and purity of modulation, I've defined a function, which I call Symmetry 
Ratio (SR).

Symmetry Ratio (SR) defined:

SR = (Peak-to-Peak audio voltage) / (lesser of the two Peak Audio 
Voltages above or below the quiescent line)
SR = 2 if the signal is a Pure Sine Wave
SR cannot be less than 2

______________________________
In John's example, he shows a representation of an O'scope with his 
voice imprint on it.
------------------------------------------------------

While using the quiescent level as reference, note that the positive 
narrow peaks go twice as high as the negative wide peaks.

SR = (Peak-to-Peak audio voltage) / (lesser of the two Peak Audio 
Voltages above or below the quiescent line)

Peak-to-Peak audio voltage = near 3 units

Lesser of the two Peak voltages (negative in this case) = near 1 unit
SR=3

     Figure 4 is the RF envelope view that would be produced by 
modulation with the audio as represented in fig 3. The RF final plate 
input power is 1000 watts, and the efficiency is 75%. This yields a 
carrier level of 750 watts as represented at the quiescent level on the 
chart.

(Figure 4 show's Peak Envelope Power at 6,750w)  There are more tables 
and pictures that greatly help explain.

     Again if the plate voltage on final is 2000 and the current is 
500ma then in the above asymmetrical example the peak-to-peak swing of 
the plate voltage would be (0 -- 6000 volts with the quiescent plate 
voltage at 2000 volts DC). (6000-2000 = 4000 volts peak) This is twice 
as much peak audio voltage as was needed before. (4000 * 0.707 = 2828 V 
RMS) and P = E*E/R or ((2828 * 2828) / 4000 ohms) = 2000 watts. Since we 
have doubled the needed peak voltage, then we have quadrupled the power 
required to produce it. This does not mean that you will be putting that 
amount of power into the final continuously, but you will need that 
capability in order to get the needed voltage swing on the peaks.

     An RF ammeter using a thermocouple actually responds to average 
power rather than RMS current, because it is heat that causes the 
indication. This type of ammeter does not indicate true RMS current. 
Because it is calibrated with an un-modulated sine wave signal, it will 
read un-modulated signal currents correctly, but its reading will be 
elevated when normal amplitude modulation is present. It will display a 
current reading equivalent to the RMS current that would produce the 
same average power as the modulated wave. In fact, the true RMS current 
of an AM waveform does not change until modulation exceeds 100% in the 
negative direction, unless there is nonlinearity or some sort of carrier 
level shift action taking place.

     With a 50 ohm load, a 750 watt carrier will cause the thermocouple 
RF ammeter to read 3.87 amps. If this carrier is modulated 100% by a 
sine wave, the average power will increase by 1.5 times (the 50% added 
power comes from the sideband energy created by modulation). The total 
average power will be 1125 watts with modulation at 100%. The RF ammeter 
will show about 4.74 amps, which is the current that would be necessary 
to produce an un-modulated signal of this power level. If a person's 
voice is used to modulate the rig, and the modulation envelope looks 
something like the scope picture in figure 4, you will see approximately 
the same increase in RF current, even though PEP with voice modulation 
is 6750 watts, and PEP with sine wave modulation is 3000 watts. This is 
because the speech waveform is spiky, so its peaks need to be relatively 
high in order for it to have the same RMS power as a sine wave. With all 
this in mind, it boils down to the fact that in order to faithfully 
reproduce my voice with the legal carrier level of the time, it was 
necessary to have a modulator capable of 2000 watts. Let me say again 
that I was not putting 2000 RMS watts of audio into a 1000 watt (input 
power) rig. But if I had not had the 2000 watt modulator, then the peaks 
of my voice would have been chopped off or clipped, resulting in 
distortion and splatter.

Now with the 1500 PEP ceiling,  I can only legally run 220 watts input 
and 165 watts output if I want to properly reproduce my voice.
=============================================================


Peak envelope power is the average power supplied to the antenna 
transmission line by a transmitter during one radio frequency cycle at 
the crest of the modulation envelope, under normal operating conditions.




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