Amp Wattage Measurement - Approximations

[Dave H]

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.

The simulation below says you are correct
Blue is bias voltage, Red is grid signal, Green is output (8 ohms)
At clipping the peak grid drive is equal to the bias voltage.

[SoulFetish]

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 .

LT! Always good to see you ‘round here, man. But, I don’t think that’s right. The average value of a sine wave is zero. RMS works out to the equivalent of 1/sqrt2
But you’re right about it representing an equivalent DC Voltage or Current.

[Jazz P Bass]

Here is a link to a nice chart for the I/ avg of different waveforms.

http://www.nessengr.com/technical-da...verage-values/

[Old Tele man]

0.637 is AVG of fullwave rectified signal.

Because both ELECTRONICS engineers and ELECTRICAL engineers use the same physics principles and terminologies, people like National Semiconductor's Bob Pease (one of my favorite authors) become rather pedantic (yes, I am) about the misuse of 'terms' and Watts(rms) is one of those mis-used/mis-applied terms:

http://www.cromptonusa.com/Watt_Var_VA.pdf

[Helmholtz]

Regarding the power formula(s):

It is known from tube literature (e.g. RDH, Langford-Smith) that the output power of a PP amp can be calculated by Pout= (1/2)*Rp*Ip^2, where Ip is the peak plate current for Vgrid=0 and Rp=Rpp/4 for class AB/B. Ip is best taken from the intersection of the loadline with the Vgrid=0 curve. The formula could be rewritten as Pout=Rp*Ip(rms)^2, where Ip(rms) is just the mathematical RMS equivalent of Ip and not the real RMS plate current per tube.

The formula from the paper of post #31: Po ≈ (%)*(Zoo/4)*(gm*Vg)^2 looks similar to the second formula above if gm*Vg corresponds to Ip(rms). This would require Vg to mean the RMS grid voltage (as stated in the paper).
But in post #18 Old Tele man defines Vg as grid bias voltage (without giving a rationale) which corresponds to max. peak grid voltage before grid conduction and makes gm*Vg mean peak plate current Ip. Comparison with Pout= (1/2)*Rp*Ip^2 reveals a discrepancy of a factor 2. In other words the formula Po ≈ (%)*(Zoo/4)*(gm*Vg)^2 should give around twice real power.

But applying the formula to the 6L6 RCA example surprizingly seems to work O.K. I think the explanation is that gm is not a constant by far, in fact it varies strongly with screen voltage and plate current. The actual average gm for a screen voltage of 400V (as in the RCA example) is more like 9mA/V (estimated from the according tube chart) and this might compensate for the use of gm= 6mA/V and peak plate current instead of its RMS equivalent in the formula - in some cases, depending on the choice of gm.

BTW, the result of
Po ≈ (0.89)*(5.6k/4)*(0.0060*37Vg)^2 is 61.4W not 55W.

I do agree that Vrms*Irms*PF (with a single frequency signal) gives time averaged (real) power and not RMS power: RMS power or Wrms makes no technical sense and I tend to avoid the term.
(PF means power factor)

[Old Tele man]

Regarding the power formula(s):

It is known from tube literature (e.g. RDH, Langford-Smith) that the output power of a PP amp can be calculated by Pout= (1/2)*Rp*Ip^2, where Ip is the peak plate current for Vgrid=0 and Rp=Rpp/4 for class AB/B. Ip is best taken from the intersection of the loadline with the Vgrid=0 curve. The formula could be rewritten as Pout=Rp*Ip(rms)^2, where Ip(rms) is just the mathematical RMS equivalent of Ip and not the real RMS plate current per tube.

The formula from the paper of post #31: Po ≈ (%)*(Zoo/4)*(gm*Vg)^2 looks similar to the second formula above if gm*Vg corresponds to Ip(rms). This would require Vg to mean the RMS grid voltage (as stated in the paper).
But in post #18 Old Tele man defines Vg as grid bias voltage which corresponds to max. peak grid voltage before grid conduction which makes gm*Vg mean peak plate current Ip. Comparison with Pout= (1/2)*Rp*Ip^2 reveals a discrepancy of a factor 2. In other words the formula Po ≈ (%)*(Zoo/4)*(gm*Vg)^2 should give around twice real power.

But applying the formula to the 6L6 RCA example surprizingly seems to work O.K. I think the explanation is that gm is not a constant by far, in fact it varies strongly with screen voltage and plate current. The actual average gm for a screen voltage of 400V (as in the RCA example) is more like 9mA/V (estimated from the according tube chart) and this might compensate for the use of peak plate current instead of its RMS equivalent in the formula - in some cases, depending on the choice of gm.

BTW, the result of
Po ≈ (0.89)*(5.6k/4)*(0.0060*37Vg)^2 is 61.4W not 55W.

I do agree that Vrms*Irms*PF (with a single frequency signal) gives time averaged (real) power and not RMS power: RMS power or Wrms makes no technical sense and I tend to avoid the term.
(PF means power factor)

Actually, you are correct -- I mis-typed -- the Vg "value" should be 35Vg, derived from one-half of Vgg; ie: Vg = (70Vpp/2). Hopefully, I haven't made other such errors in my previous posts, for which I will apologize in advance (can't make corrections because time-out has occurred).

[loudthud]

LT! Always good to see you ‘round here, man. But, I don’t think that’s right. The average value of a sine wave is zero. RMS works out to the equivalent of 1/sqrt2
But you’re right about it representing an equivalent DC Voltage or Current.

True, but a resistor doesn't get cold just because the Voltage in negative. When you square the Voltage or current term, that fixes the problem.

Going back to pre-calculus, a unit sine wave squared comes out to an inverted cosine wave 2x the frequency, half the amplitude offset by half. That's the squared part. The Mean part of that is one half, take the square root and you get .707...

Looking for a pic I had of a sine wave squared, I can't find it.

[SoulFetish]

Going back to pre-calculus....

Pre-calculus?? Hold on, I’m gonna have to take my shoes off so I can count my toes as well if you’re gonna’ start getting all fancy on me

[J M Fahey]

Looking for a pic I had of a sine wave squared, I can't find it.

Well, it should look similar to a full wave rectified one, since both negative and positive halves become positive when squared.

[Helmholtz]

As (sin x)ē = 0.5 (1-cos 2x), a sine squared results in a sinusoidal (cosine) of doubled frequency on top of a constant positive DC offset.

[Old Tele man]

SINE-squared:


COS-squared:

[vintagekiki]

And finally, the Hamlet question.
How much power amplifier 100W has real power in W.

[J M Fahey]

My take since forever is that "RMS power" should have been written as :"Power calculated using measured RMS Voltage"
I guess this is the implicit assumption used by everybody for decades now ... if not approaching a Century.

In practical terms it was the Industry standard and a very useful measurement until lately the strict Dictionary definition started to be used, not to clear things but to argue.

In any case it *amply* beats its Politically Correct substitutes:

* "average power" which has to jump through fire loops to explain why mathematically zero average for a sinewave means actual power dissipated into the speaker (heating it enough to *burn* it if too much)

* "continuous power" which is even worse.
As in: at least, RMS Volts and Average Volts do exist, there are equations using them and they are shown on instrument scales ..... try that with Continuous Volts.

[SoulFetish]

My take since forever is that "RMS power" should have been written as :"Power calculated using measured RMS Voltage"
I guess this is the implicit assumption used by everybody for decades now ... if not approaching a Century.

I agree that it probably should have been written and stated that way, because that's what's actually happening.

In practical terms it was the Industry standard and a very useful measurement until lately the strict Dictionary definition started to be used, not to clear things but to argue.

Well, not entirely true. The ONLY industry that ever used "watts RMS" is the audio industry.
A watt is a measure of work being done as the product of voltage and current. Like Old Tele Man says, RMS is a mathematical expression to provide a DC equivalent. Besides, for audio power into a reactive load, there are phase shifts between voltage and currents. True audio power measurement, what is called RMS power, is mathematically an integral of instantaneous power.

In any case it *amply* beats its Politically Correct substitutes:

Isn't it more of an engineering discussion, rather than a political one?

As in: at least, RMS Volts and Average Volts do exist...

I'm convinced of it!

[J M Fahey]

Oh, part of it was written tongue in cheek, thatīs why I call it "Politically Correct" , but my point is that as I see it my slightly corrected definition agrees with what was used for ages by very serious Engineers.
And apparently it was good enough because nobody complained for decades and itīs still being used today in Factory Issued datasheets.
I only see it being argued in Forums, go figure.

Agree on the Dictionary definition of course, but rather than tearing down the crude but useful "RMS Power" definition Iīd add my slight semantic correction, just to put in written words what is already implicit.

As of "the Industry" , for *us* it clearly means the "Audio" related Industry.

Iīd never ever use "RMS Watts" to describe, say, an oven, a space heater, a Coffee maker, etc.