Amp Wattage Measurement - Approximations

[Old Tele man]

See this brief paper: http://thermionic.info/mccaul/McCaul...ltage_2008.pdf

[Helmholtz]

The ONLY place the two quasi-halves of the full pre-PI sinewave become a FULL 360-degree sinusoid again is *inside* the OT -- specifically in the secondary (output) winding(s) -- because anything prior to that is (basically) ± pulsing alternating halfwaves (PI, output tubes, each half of OT).

But both OT primary halves carry full cycle voltages. The voltages in both halves must be identical as the windings are tightly coupled.

[Helmholtz]

See this brief paper: http://thermionic.info/mccaul/McCaul...ltage_2008.pdf

Yes, but the author correctly uses RMS grid voltage and not bias voltage.

[Old Tele man]

...and he (me) also later explains that the DC bias voltage is assumed to equal the AC(rms) driving signal for estimating Po from the schematic values of Zoo (OT), gm (power tube), and applied BIAS (Vg.dc) voltage:

http://www.tdpri.com/threads/estimat...1-amps.776469/

Q: Why use BIAS voltage?
A: Because BIAS voltage is almost universally specified on schematics while PI output drive signal levels aren't. It's an approximation about simplicity, not about exact accuracy.

[Helmholtz]

DC bias voltage is assumed to equal the AC(rms) driving signal

But that's wrong as the the DC grid bias corresponds to the max. grid signal peak voltage.

[Old Tele man]

Your assertion *IS* valid when Vg(rms) exactly equals Vg(bias.dc), which rarely happens; typically Vgg/2 is "less than" Vg(dc) by a couple volts, just for "good measure," For example, see GE/RCA data sheet Class-AB1 55W example (450Vp; 400Vs; Zpp=5.6K; Vgg=70Vpp; Vg.dc=-37Vdc).

[Helmholtz]

When " Vg(rms) exactly equals Vg(bias.dc)" there will be severe grid conduction, distortion and power limiting. Therefore max. Vg(rms) must be 30% below DC bias voltage, in other words the DC bias corresponds to the max. grid signal peak voltage. Your power formula should use Vg(rms) not bias voltage.

Please note that in your example Vgg=70pp which corresponds to a grid signal of about 25Vrms per tube.

[Old Tele man]

When " Vg(rms) exactly equals Vg(bias.dc)" there will be severe grid conduction, distortion and power limiting. Therefore max. Vg(rms) must be 30% below DC bias voltage, in other words the DC bias corresponds to the max. grid signal peak voltage. Your power formula should use Vg(rms) not bias voltage.

Please note that in your example Vgg=70pp which corresponds to a grid signal of about 25Vrms per tube.

That value is directly quoted from the GE/RCA 6L6GC datasheets...worked great from them back in the hey-day of vacuum tubes.

For example:

Po ≈ (%)*(Zoo/4)*(gm*|-Vg|)^2

Where GE/RCA Class-AB1 55W example values are:

|-Vg| = absolute value of the -37Vdc bias voltage; absolute ( || ) value = 37Vdc
gm = 6L6GC transconductance, 0.0060A/V
Zoo = OT 5.6k ohm, plate-to-plate; Zo = Zoo/4; Zo = 5.6k/4 = 1.4k
(%) = effective load factor (%): % = (23.5k/(23.5k + 1.4k))^2 = 0.8907 ≈ 0.89

Thus...

Po ≈ (0.89)*(5.6k/4)*(0.0060*37Vg)^2 = 55W(avg)

[Helmholtz]

That value is directly quoted from the GE/RCA 6L6GC datasheets...worked great from them back in the hey-day of vacuum tubes.

Sure these values are just fine. But they indirectly specify Vg(rms)= 25V and Vg(peak)=35V which is well below the DC bias of 37V.

[vintagekiki]

My thinking about amp wattage

By definition, power represents the energy that some consumer "takes" from the generator to which it is connected. If known resistance of consumers and voltage of the generator, a simple mathematical formula can calculate the power of consumers. Since the resistance of the consumers and the voltage of the generator are constant, the power is continuous and is expressed as RMS power.

According to this principle, works wattmeters, electric meters ...

Any other power except continuous power is a marketing figure and usually shows the maximum possible theoretical output of a component before destruction. Quite logically the question is asked who needs this information.

Is there a valid professional, technical, engineering explanation as a 100W amplifier which from its power supply "draws" a power of 100W, is shown as an amplifier of 150 to 1000 W.

http://bee.mif.pg.gda.pl/ciasteczkowypotwor/SM_scena/Inne/Amplifier_Wattage_Ratings.pdf
https://music-electronics-forum.com/attachment.php?attachmentid=25972&d=1382715028
Amplifier Wattage Ratings


https://en.wikipedia.org/wiki/Audio_power

https://www.quora.com/What-is-RMS-power

http://www.blue-room.org.uk/wiki/Peak_Music_Power_Output

https://www.kicker.com/app/misc/support/tech/tech_papers/docs/PeakVsRMS.pdf

https://www.guitarplayer.com/technique/all-about-amp-wattage

https://www.audioholics.com/audio-amplifier/amplifier-power-ratings

https://carvinaudio.com/blogs/audio-education/demystifying-power-ratings-rms-vs-program-vs-peak

https://blog.zzounds.com/2017/07/19/peak-vs-continuous-power-ratings-speakers/

[Old Tele man]

Regarding amplifiers, I repeat "...there is NO such thing as RMS power...it's AVERAGE power."

The Root-Mean-Square (RMS) mathematics simply converts the peak VOLT or AMP value of a sinusoid to it's equivalent AVERAGE value, defined as its DC-heating (into pure resistance) value. [not talking speakers here]

[g1]

I'm not sure the point of that, I'm sure we can agree to call it average or RMS and mean the same thing, it's just semantics and does not affect the question about the formula.
As far as I know, if you allow the average at the grid to be equal to the DC bias, you will have heavy grid conduction at the peaks. Peak must not exceed the grid voltage, so the average allowed at the grid must be about 70% of the bias voltage (to avoid grid conduction).
At least if you are talking about clean power output.

[SoulFetish]

Fight the good fight, Old Tele Man!

g1, I agree that it's a bit of semantics. But kinda' I like that the Ol' Man is being an insufferable pedant on this point. Makes me wish Bob P was still around telling us that "dampening" is what happens when you get a rag wet, and "damping" is... what ever the hell that is.
Watts RMS always bothered me a little bit too because there's really no such thing and it didn't make sense. Standard nomenclature already assumes that measurements of AC voltage and current (sinusoidal) are already RMS values unless specifically stated otherwise.
I'm not sure who made up Watts RMS. but it's only ever used in audio circles. So, your point is a good one in that we all know what it means at this point. Maybe someday the rest of the engineering community will learn to accept us using it.

[loudthud]

Regarding amplifiers, I repeat "...there is NO such thing as RMS power...it's AVERAGE power."

The Root-Mean-Square (RMS) mathematics simply converts the peak VOLT or AMP value of a sinusoid to it's equivalent AVERAGE value, defined as its DC-heating (into pure resistance) value. [not talking speakers here]

It's not really an average value of Voltage or current, it's the equivalent DC value of Voltage or current that would give the same average power into a resistive load. Subtle difference.

To compute RMS value for any waveform, first square the value at each point along the wave. Then find the average (mean) value. Lastly take the square root. The average value of a sine wave is .636 x peak value. Close, but not .707 .

[Dave H]

The average value of a sine wave is .636 x peak value. Close, but not .707 .

This is straight from pedant's corner

The average value of any number of whole sine waves is zero.
The average value of the positive half of a sine wave is Vp*2/pi
The average value of the negative half of a sine wave is -Vp*2/pi
They cancel each other out.