[AMRadio] Modulation (WAS: League pulls RM-11306)


Bry Carling bcarling at cfl.rr.com
Sun Apr 29 21:07:51 EDT 2007


But my guess is that the power tube operating characteristics for
push-pull modulator service, such as one finds in the back of 
the ARRggghL Handbook are actually gioven for average 
continuous power with a sinewave at 1 kHz, or something
like that.

For example, let's say that you have an 807 final with 80 watts 
DC input. You need a 40 watt modulator. What IS a 40 watt 
modulator? Will it be CLOSE enough to that figure to fully 
modulate the 807 with YOUR particular voice characteristics?
How do we compensate for that if it isn't?
Will it be too strong and OVER-modulate that 807?

Well, we have a simple solution - it is called the mic gain control!
So long as you have ebough reserve power to get mjore or less 
up to the 90% mod level, it should NOT be a big iussue so long 
as you don't over modulate.

Aren't we getting just a little bit over-technical here?

> I've gone to the trouble for you (and anyone else that cares, for that 
> matter) of the 'heart' of the discussion that John/BXO, Bacon, Tim and 
> Don/KYV came up with...
> 
> *"The maximum audio modulator power requirement for 100% modulation of a 
> 1000 watt high level modulated class C final is 500 watts, /but only if 
> the modulating signal is a sine wave/. The reason for this will become 
> apparent from the following example. *
> 
> * The following example will show that the required audio power 
> necessary to achieve 100% sine wave modulation is 50% of the value of 
> the DC Plate input power supplied to the final class C RF amplifier. * * *
> 
> * Assume that the final amplifier has 2000 volts applied to the plate 
> with a current of 500ma. This makes the power equal to 1000 watts. To 
> achieve 100% modulation, the peak modulation voltage of the sine wave 
> must be 2000 volts and because of the symmetry of the sine wave will 
> yield a 4000volt peak-to-peak swing (0 – 4000V when the quiescent DC is 
> at 2000V) thus providing the desired 100% modulation. To calculate 
> average power rather than instantaneous peak power we must convert the 
> peak voltage output from the modulator to a “Root Mean Square” voltage 
> (Peak voltage X square root of 0.5 = RMS Voltage). RMS voltage is equal 
> to 0.707 of the peak voltage. *
> 
> 
> *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."
> *
> 
> *Asymmetrical Audio.
> *
> 
> *
> The whole article can be seen at
> http://www.qsl.net/wa5bxo/asyam/aam3.html
> 
> *
> 
> *--
> 73
> *
> 
> 
> 
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