figuring approx. output wattage

[phsyconoodler]

Is this formula to figure output wattage relative accurate?
plate voltage squared/output imedance = wattge output.Example: AC30- 350v squared = 122500/4000 = 30.625 watts.
Or Super Reverb- 450 squared = 202500/4200 = 48.21 watts.
I know this doesn't take into account tube types,current and losses,but is it not what the maufacturers use as a benchmark for rating?

[sluckey]

You gotta use the AC signal voltage on the plate, not the dc plate voltage. A better way would be to inject a sine wave signal at the input. Measure the AC (RMS) voltage across the speaker. Then use the P=E²/R equation.

For example you measure 10VAC across a 4[ch937] speaker. P=10²/4=25 watts RMS.

PS. For that power figure to be meaningful for comparison purposes, a distortion figure needs to be specified. If all mfgs use the same distortion figure then it's easy to see who has the most powerful amp.

[phsyconoodler]

Thanks sluckey.I just got my signal generator,so time to play!
  I did a 6v6 amp with a 6.6k primary on the OT and it sounds glorious!Seems to put out more than an 8k and the sound quality is improved too.What speaker impedance should I be using?
  I will drive the neihbors nuts with a sine wave into the speaker,not to mention myself!I think I'll use a dummy load instead.It proves I'm not the dummy! ;D

[PRR]

> Is this formula to figure output wattage relative accurate?
plate voltage squared/output impedance = wattage output.

This is only for push-pull.

The form of the formula is correct. V^2 and an R should give Power.

However I believe you have slipped "factor of 2" several times: in getting Sine RMS from DC, in push-pull, and apparently the "non-ideal" fudge factor for many of our Usual Suspects is around 1/2-ish.

And you will hardly find a triode doing this well.  

Datasheet compared to phsyconoodler formula:

6550:
400V, 3,500, 60W claimed, 46W computed
450V, 3,500, 77W claimed, 58W computed
600V, 5,000, 100W claimed, 72W computed

6L6GC:
360V, 6,600, 27W claimed, 20W computed
450V, 5,600, 55W claimed, 36W computed
(450V, 5,600, 72W claimed at high THD, 36W computed)

EL34:
425V, 3,400, 55W claimed, 53W computed
375V, 2,800, 44W claimed, 50W computed

6F6:
315V, 10,000, 11W claimed, 10W computed

6V6:
250V, 10,000, 10W claimed, 6W computed
285V, 8,000, 14W claimed, 10W computed

7027:
400V, 6,600, 34W claimed, 24W computed
540V, 6,500, 76W claimed, 45W computed

813:
1500V, 9,300, 260W claimed, 242W computed
2500V, 19,000, 490W claimed, 329W computed

4X5000A:
4000V, 1,500, 11,500W claimed, 10,670W computed
5000V, 2,300, 13,500W claimed, 10,870W computed
6000V, 2,940, 17,000W claimed, 12,240W computed
7000V, 4,100, 17,500W claimed, 11,950W computed

Accurate? Not really. I think it should be

2 * V^2/Rpp * F

Factor "2" accounts for the ideal case including DC-Sine and push-pull. "F" is a fudge ranging from 1 for perfect devices, maybe 0.8 for great devices (hi-volt MOSFET), 0.7 for many practical devices, and less than 0.5 for the smaller devices which don't aspire too high.

> is it not what the maufacturers use as a benchmark for rating?

No. Ratings are (should be!) measured, not approximated.

> By dummy load do you mean using a 40-60 watt light bulb box?

No. Made this mistake myself. A cold 60W 120V lamp is about 15 ohms. I put it on a two-6V6 amp good for about 14 Watts, and measured almost 40 Watts. Confirmed by some glow in the lamp, showing significant voltage. Hmmmmm.... after some study I decided the "15 ohm" lamp was rising to 50 ohms when warm, the amp was delivering voltage but not so much current. Since I was figuring "Power" by Voltage and the assumed load resistance, I was way off.

You get an honest resistor with ample heat capacity. For big Champs and small Deluxes, Radio Shack may have an 8 ohm 20 Watt resistor in the drawer. This will take 24W for many seconds.

You measure the audio voltage AND you monitor the sound quality. Amps are generally clean, then less-clean, then go nasty/raspy pretty suddenly as you raise the level. Take reading just before it gets "obviously bent".

(There are well-established traditions in Guitar and CarSound to give ratings on "bent waves", 10%-25%THD. These are useful shopping guides but should not be taken too seriously.)

> You gotta use the AC signal voltage on the plate, not the dc plate voltage.

For perfect devices, one implies the other. 400VDC supply, you can pull up to 400V peak signal swing. Pentodes will sometimes do near 90% of supply voltage, which should be allowed for, but is not a major error: F is 0.8. 80% swing is more typical, which gives F near 0.63. Triodes can rarely swing 66%, which gives F=0.44, maybe 50% and F=0.25.

So perfect devices, 400V supply, -400V peak one plate, +400V peak other plate, 800V peak, 560V RMS across the total winding. Say 4KCT winding. Perfect device would give 78.4 Watts sine RMS. 5F6A Bassman runs closer to 40W-45W sine-like output. My "F" fudge-factor would be 0.53. For the 6V6 at 285V, the 6550 at 600V, and 4X5000A at 6KV, F is near 0.7. 6F6 more like 0.55.

Still the original phsyconoodler formula is useful. A pair of pentodes "should" do that much. Maybe a bit more. Not twice as much (unless they are absolutely perfect (and you can't buy perfect parts)).

IOW: 2 * V^2/Rpp is "perfect". Half of that is not shameful.

BTW: for totem-pole topology used in transistor amps the formula is 0.125 * V^2/Rpp * F. We change from "2" to "1/8" because a LOT of "factor of two"s go the other way when you stack instead of side-by-side. A perfect 28.28V 10 ohm amp makes 10 Watts; a real amp ought to give 5 Watts or you probably screwed-up something.

[zachsdad]

On the subject of watts, here is an old post you might be interested in.

http://www.el34world.com/Forum/yabb2/YaBB.pl?num=1184378457/0#0

[PRR]

> Math is a wonderful thing! Reality is another.

The nice thing about electronics is that most "reality" CAN be computed.

Often it is too much work to account for every last error, and we don't need an -exact- answer.

The revised phsyconoodler formula IS correct for perfect devices.

Many power pentodes can be run in very-good conditions which will give 60%-75% of the revised phsyconoodler formula result. Most times we don't need an exact power, but a general ballpark. A "F" fudge-factor of say 66% or 0.66 will give ballpark powers quickly.

Datasheet compared to revised phsyconoodler formula:

6V6: 285V, 8,000, 14W claimed, 20W computed, F=0.7
EL34: 425V, 3,400, 55W claimed, 106W computed, F=0.52
EL34: 425V, 3,400, 70W observed at high THD, 106W computed, F=0.66
6L6GC: 450V, 5,600, 55W claimed, 72W computed, F=0.76
6L6GC: 450V, 5,600, 72W claimed at high THD, 72W computed, F=1.0
7027: 540V, 6,500, 76W claimed, 90W computed, F=0.84
6550: 600V, 5,000, 100W claimed, 144W computed, F=0.69
813: 2500V, 19,000, 490W claimed, 658W computed, F=0.74
4X5000A: 6000V, 2,940, 17,000W claimed, 22,480W computed, F=0.76

The standouts are heavy-load EL34 and high-THD 6L6GC.

3,400 load is low. But EL34 will pull it. But THD must rise over the arbitrary 5% THD point specified. There's easily 66W at 6% or 7% THD raw, and this can be reduced with NFB.

The book says the 6L6GC condition is under 2% THD. The "72W" condition was metioned in context of amateur radio speech modulators, where huge THD is not a serious fault. The apparent "perfection" is really like 65W of honest signal and 10% distortion products. Effective-power F fudge-factor is more like 0.9, and this is at significant clipping.

So use the revised phsyconoodler formula:

 2 * V^2/Rpp and figure your undistorted sine power is realistlcally "about" 2/3rds of that.

[tubenit]

From a 1/08 post from PRR

> How can I determine the aproximate watts of an amp.
> For example the one I finished has 2 6L6's and a 35watt OT.
> The plates have aprox. 470v and it is idling at around 40ma.

Transformer rated wattage does NOT matter.

Idle current matters in Self-Bias, not in Fixed-bias; BUT it is easily changed (up to Pd) so is NOT important.

What is the load impedance seen by the plates???

If the load is reasonable for the tube, and you know supply voltage, the approximate Watts may be calculated.

pchyconoodler proposed a formula based on study of a few cases. In the same thread, I derive the exact MAXimum power possible for a given load and voltage with Ideal Devices, and show fudge-factors for Real Devices.

> I have about 470v on the plates and primary impedence on the OT is 6,600.

Or you can cheat. Use the work of the dead guys! 6L6GC datasheet page 2 shows 450V 5.6K 55 Watts. 5K6 is 85% of 6K6 so as a rough guess you can make 85% of 55W or 46W. Check other parameters. You probably have at least 400V on screens. Your idle current is lower than RCA/GE suggest.... in fix-bias that only affects small-power sound, not big power performance.

If you are SELF-bias: your absolute maximum plate dissipation is 60W/pair, and for common 6L6GC that's awful hot, figure 50W. Ideal self-bias max power is 40%-50% of that depending how much bias-shift and distortion you will tolerate. Say 25 Watts.

Now you must juggle the load and the B+ so they work together. For 6K6 and 25W Pd each bottle, his actually leads to 280V across the tube, say 300V B+. Your 470V B+ is awful high for self-bias 25W-30W bottles and 6K6 load and small bias-shift. You have to turn-down idle current to avoid melting, but in self-bias that strangles output power.

6L6GC is a special case because the datasheets focus on old-type 6L6G 19W Pd conditions, with one GC condition added. Peep in 7027. This is pretty-near the same guts with different pinout. Page 2 Cathode Bias shows 400V, 6K6, 32 Watts. This condition shows a lot of bias-shift, but will normally work well. None of the parameters exceed 6L6GC ratings. The exact cathode resistor and Watts/THD may be different with 6L6GC, but not much.

Again on 7027 datasheet: 450V 6K6 FIX-bias can do 50 Watts. And they suggest 95mA/pair idle, which is same-enuff to your 40mA/each. This says your "35W" OT may be strained (distort) at low frequency. If it is Hi-Fi rating, it's plenty good for guitar. If it is Guitar rating: it may sound phatt-and-phunky on your lowest notes with 50 Watts through "35W guitar" iron. Many well-loved amps are this way.  
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[tubenit]

From HotBluePlates   1/08 post

The output wattage thing is this an art as much as science; maybe voodoo.

No, it is science and math. Until you start looking at the claimed power output of many transistor amps of the 80's, and/or some power ratings of output transformers. Then it can be pure B.S. and marketing.

It all depends on how exact you want/need to be. I like PRR's method in most cases. What the data sheet says is generally right if you have all the same conditions. If you have some different conditions, you can figure how that might change things and by how much.

Psychonoodler's method also works, but I still like PRR's amendment to that method, using a fudge factor to correct for real-life, since the simple formula will over-predict the power output. Psycho's method would work for a perfect tube and perfect power supply, but needs some adjustment for the fact that neither is perfect.

Or you do it the scientific way:

You need to put a signal into an amp.  The standard frequency is 1k, which is not all that helpful for guitar, and less helpful for bass; so some people use 400k or 330k for guitar.

1k is a standard test-tone, but 400Hz, not 400k, would be a common tone for audio amps. Low E is about 80Hz. That makes the high E about 320Hz, high E at 12th fret about 640Hz, and high E harmonic at 5th fret about 1280Hz.

I've seen some breakdowns of fundamental pitches by instrument claiming that there are few if any instruments with a fundamental pitch over 1k. So 1k would be pretty useless as a real-world guage for guitar, in my opinion. Harmonics are likely present up to the limit of speaker reproduction, especially with distorted guitars. But for clean sounds, harmonics are very much lower in output level than fundamental pitches, and serve to give your ear a clue as to what instrument is making the sound.

Some people feed the signal in @ 200mV; but I think the industry standard is 500mV; some some humbuckers can put out +/- 1 volt.  

There is no standard, which is how people play games with numbers and power ratings.

If you decide to measure, what do you want to measure? My view is you don't give a damn about the preamp, just what power the output section can make. So the ideal approach seems to be to feed a signal into the phase inverter, or a push-pull signal into the output tubes directly, which drives them close to having 0v on the grid. This usually pushes them to full output power at some level of distortion.

If you want precision, you should specify the amount of distortion, to give your measurements meaning in relation from one amp/measurement to another. You could guesstimate by ear, but odd harmonic distortion is more noticeable/obtrusive at a lower level than even harmonic distortion, which might lead you to guess unequally between 2 different amps. You could eyeball the amount of distortion with an oscilloscope, or measure it conclusively with a wave analyzer, distortion analyzer, spectrum analyzer, or with careful use of the scope.

Measuring the output voltage at the speaker lends itself to using a simple formula: V2 / R. But using an inductive load like a speaker gives you a moving target, because its impedance is non-constant, and has to be measured/calculated at a given frequency. So the easiest thing to do is use a fixed resistor as your load. Especially since listening to full power sine waves will make you lose your mind (let me tell you about aligning multitrack tape machines sometime...).

You have to know the amount of distortion present when measuring voltage, because the calculated RMS power of a sine wave is 1/2 the peak power, where the RMS power of a square wave is equal to the peak power. A sine wave with infinite distortion (clipping) becomes a square wave, so if you just kept turning up until the voltage stopped rising when measuring across a resistor, you'd over-estimate the power output of the amp.

Or you don't worry about it and just play.  
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[tubenit]

continued from HotBluePlates  1/08 post

Yes the use of math formula's makes it look scientific.  But what values do you plug into the formula's?  If there's no standard, well then. . .

Yeah, but what I was getting at is that you were talking about the signal level applied at the input jack. My perspective was that you wouldn't apply the signal at the input jack, you'd apply it at the output tubes or the phase inverter. And at that point, you don't confine yourself to a pre-selected signal level, you keep turning it higher until the limit of output power had been reached. At that point, you calculate output power from the load resistance used and the voltage measured across the load resistance.

And like I said, you have to select some level of distortion to use as a benchmark for measurement, and have some way to determine when you've reached that benchmark. That could be complicated if you want to be very exact.

My assumption in all this is that the amp was well-designed, and the power stage is the limiting factor for power output. It is also assumed that you have any master volume controls all the way up to allow the most clean output. But maybe you'd arrive at a different conclusion if you decided that the preamp or phase inverter is distorting before the output stage. You might make one set of measurements by applying a test signal to the phase inverter/output tubes, and another with the test signal applied to the input jacks. The power output measured at the same level of distortion would give you an indication of how much the preamp is holding you back.

Or maybe it isn't. Maybe the goal is all-out distortion. You could build the amp to turn any signal you play into a perfect square wave; infinite distortion. How loud you could make that square wave would still depend on your output stage (and to an extent, what your speakers can tolerate). So perhaps testing for power output by applying a signal to the input of the preamp makes little sense under any conditions...

[tubenit]

from jjasilli  1/08 post

"Yeah, but what I was getting at is that you were talking about the signal level applied at the input jack. My perspective was that you wouldn't apply the signal at the input jack, you'd apply it at the output tubes or the phase inverter. "

Yes, but then are you measuring to some "standard", which may not apply to your actual use of the amp?  Or do you tailor the measurement to your actual usage, by inputting the voltage of a typical guitar signal?  But then what is that voltage:  200mV; 500mV; more than 1V for a steel guitar pick-up?  These are meant as rhetorical questions.  

I rate my amps as loud,louder and loudest.They still want to know so I give them the quick answer.
I can see that wattage can be important when building amps for customers.    It might be the most asked question by the typical guitarist:  "Hey, cool amp man; how many watts is it?"  

[tubenit]

from HotBluePlates  1/08 post

Yes, but then are you measuring to some "standard", which may not apply to your actual use of the amp?  Or do you tailor the measurement to your actual usage, by inputting the voltage of a typical guitar signal?  But then what is that voltage:  200mV; 500mV; more than 1V for a steel guitar pick-up?  These are meant as rhetorical questions.

If the goal is determining output power of the amplifier, you don't need any information on the preamp, so you wouldn't input the signal on the preamp.

What I believe you're thinking of is testing the overall amplifier performance by inputting a test signal at the input jack. If the amplifier was malfunctioning, and you wanted to inject a test signal at the input, then trace it through the amplifier to see where there is signal, no signal, weak signal, wrong signal, etc., then you'd proceed as you're talking about.

Why not measure power that way? You can, but you might have distortion created in the preamp affecting your results. If you input a fixed-size test signal at the input jack, where do you set the controls? If you say, "Dime them all, so you don't reduce the signal," then you will certainly have distortion, as there is more gain than needed in the amp for clean reproduction. If you turn them down to account for the too-much preamp gain, how to make that consistent from amp-to-amp, if the designs are different, and therefore the amount of gain is different?

So the way I look at it is to test only the power amp when the issue is "how much power?", and inject at the input jack when you want to test the preamp characteristics.

[tubenit]

from PRR  1/08 post

> Psycho's method would work for a perfect tube and perfect power supply

No, it works because he omitted a factor-of-2 -and- the fudge-factor for many good condition is a bit over 1/2.

My un-fudged formula IS correct IF you have devices which will pull-down to ZERO voltage drop, have zero winding losses, etc. Real devices have losses. A vacuum pentode won't pull-down much below 50V, an OT has ~~10% losses. Real devices also have distortion, and this may become "too high" before we reach computed power.

> My assumption in all this is that the amp was well-designed, and the power stage is the limiting factor for power output.

Since power delivery is very expensive, any system which "can't" push the costly power bottles past the edge is generally a poor idea.

We sometimes see it done in Guitar, for the specific reason that sometimes "tone" is more important than "Power". If so, why are we asking "What's the WATTS, man??"

Input level can be adjusted VERY easily. 500mV at Pickup Input on a dimed amp will (should) grossly overload. We have a knob for that. In fact we should usually start with Volume full-up, and adjust the source to bring the Power Stage to overload. Having found that point, we look at the source level. If it is lower than real sources, we know we will need to use the Volume knob. If it is higher than real sources, we will need another stage of voltage gain. Guitar amps should generally accept 500mV at the jack with Volume turned down, and still be able to overload the Power Stage. With Volume full up, around 5mV to 50mV should cause output overload.

Guitar amp input systems can be VERY "colored". If the question is "What's the WATTS, man??", you should break between preamp and power amp, measure the POWER stage for WATTS, man..... then (if anybody cares) check to see if the preamp is a good fit for the power amp.

> It makes people happy to have something tangible without using a signal generator and a dummy load.

Just as silly as guessing a car HorsePower without a dynamometer or a load and a stopwatch.

> But what values do you plug into the formulas?

Which formula?

If you are predicting, you use your best-guess of B+ and load, and hope the distortion will not be too shameful. And you accept that predictitons are often wrong.

If you are measuring an amplifier to prove it meets the contract requirement or advertising claims, you bring up a sine wave on a resistor load, at nominal line voltage, measure RMS voltage and distortion, and document everything.

Yes, you will get more audio power if your "wall outlet" is 130VAC. This trick was routine in car-audio. Reality is that car electrics wander from 11V to 15V, though 12.5V to 14.2V is "normal". Nevertheless some car-audio makers used 14.4V and even higher to inflate their numbers.

You may measure 5 Watts at 1%, 10 Watts at 2% THD, 15W at 5%THD, 20W at 17%THD. Which number do you put in the ad? Well, some Hi-Fi buyers like numbers smaller than 1%, so you will have trouble selling to them. Others may think that 5 Watts at 1% is lame for such a big amp. A lot of more practical users accept that 5%THD might not be obnoxious, and if 15 Watts suits their needs they may buy it. Some buyers are all about POWER and not critical about distortion, "20W!" may be the number they like best.

Play the game! Below is data on a chip with ~~0.3% intrinsic distortion below clipping, and fairly simple "clean" clipping. It is very typical of the shape of curve for all car-audio chips. What do you call it? 3.5W at 0.5%? 4.2W at 1%? 5W at 5%? 6W at 10%? Note that the 10%THD point is nearly twice the power of the 0.5% point.  
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[tubenit]

from PRR  4/08 post

> then are you measuring to some "standard", which may not apply to your actual use of the amp?

If it is really an "audio" amp, it has two basic ranges:

* "Clean"
* "Distorted"

"Maximum power output" is the "clean" output.

Most Power amplifiers are clean up to a point and then get distorted pretty quick.

Your exact definition of "distorted" will give different numbers. Sometimes 2:1 apart, such as 25W at 1% THD and 50W at 10% THD.

But a 2:1 change of power is NOT a big difference. Big Hi-Fi amp sellers like to tell us it takes a 10:1 difference in power to make a real difference to the ear. They are grinding their own axe, sure, but they are not far wrong.

If you want to know "Is this 30W or 35W?", you are asking a silly question. (Unless you have silly customers, in which case you need to test and specify your test conditions.) (Or you sell under the silly FTC regulations.) It is good to know a 10W from a 50W, but you can usually guess which it is just by picking it up.


> may not apply to your actual use of the amp?

Guitar amps might be considered to have three ranges:

* "Clean"
* "Some Distortion"
* "Mostly Distortion"

But unless the amp is very odd, the difference between "Some Distortion" and "Mostly Distortion" is similar for all amplifiers. A "50W" may be clean at 40W, distorted at 60W, and grossly distorted at the equivalent of 100W. I'm not sure where to draw the line between "Some Distortion" and "Mostly Distortion", but I think all experienced electric guitarists know it when they hear it.

It does matter some what type of amp it is. Simple tube amps give simple distortion, fairly smooth even when distortion is very large. Overwrought transistor amps quickly make a harsh grating distortion. So it does make a difference for guitarists. But this is never specified in "Power Rating". You know it by playing.  
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[LHPcope]

Hello all,

Newbe here, but I couldn't help getting my 2 cents worth in on this topic.  I measure output power by using a signal generator, an oscilloscope, a dummy load (4 each 25 watt wire wound resistors hooked up series/parallel for ~8 ohms), and a digital VOM. 

So hook up the signal generator to the input; the dummy load to the output; the O'scope and VOM across the dummy load.  Now adjust the amp, and signal generator for maximum output without clipping as seen on the o'scope (I usually use 1KHz, but 400Hz would work as well).  Note the RMS ac voltage reading on the VOM (or if you have one of those digital scopes it will give you the RMS voltage) Now power down the amp, remove the dummy load plug from the output, switch the VOM to ohms and read the dummy load resistance.  Plug the rms ac voltage and the dummy load resistance into the formula:  W=V^2/R  This will give you your amp power at clipping.  The amp power in total full tilt distortion (square wave) will be 1.414 times that number.  BTW I measure the dummy load resistance just after making the voltage measurement because the heating of the dummy load will have some effect on the resistance.  I hope this is helpful

LHPcope