estimating watts of an amp

[EKDENTON]

 know this has  been discussed alot already but I can't seem to get the search tool to work for me, I keep getting "sorry no matches found".

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.

Can I get an estimate using this information?

[jim]

Power = volts x amps.   475 x .040 = 19 watts/tube  which is close
to the recommended idle of 70% of max dissapation for 6L6GC.   The output will be choked a tiny bit by the OT but this is what you need to rock a little.    Jim

[EKDENTON]

That is plate watt dissapation " v*a=watts ", I know how to get that, I was looking for the watt output of the amp, I remember reading a thread that said the amp  power output is not the same as plate watt dissapation, but I can't find that thread.

[Geo]

Could you measure the AC signal on the speaker and use V=IR with the speaker's ohmage? Or is it not that simple....  :-/

[jjasilli]

Yes, and BTW I can't get the SEARCH function to work either.

The output wattage thing is this an art as much as science; maybe voodoo.  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.  Some people feed the signal in @ 200mV; but I think the industry standard is 500mV; some some humbuckers can put out +/- 1 volt.  

Then turn the vol up until the amp starts to distort or overdrive.  For this you must use your ears, or a 'scope.  (To really pin this part down, you'd need to measure THD as a percentage of distortion - but that's not much applicable to a guitar or a bass amp either.) You could use a speaker emulating dummy load, or the volume may be annoying.  Obviously, if you use a pure dummy load, you'll need a 'scope to see the waveform distort.

Then measure the AC resistance across the load, square it and divide by the nominal impedance of the load.  That's the process, and you will then be the proud owner of a number - the value of which could be endlessly debated.  

[EKDENTON]

Found this post by pchyconoodler:
The big trouble with measuring the output in watts is what is the signal level you need to put in and what distortion factor to use.
 I just figure it with a simpler formula like this:
plate voltage squared/primary output impedance = power output in watts.
 So your regular Super Reverb has 450v,squared is 202500/4200=48 watts.Now they actually measure the output at 45watts so the formula is under ideal conditions.I just use it to give me an idea what wattage I SHOULD be getting.If it's loud and sounds good,that's good enough for me.
 So a Deluxe Reverb has 415v squared is 172225/8000 = 21.53 watts.
An AC30 has 350v squared is 122500/4000 = 30.6 watts
A Tweed Deluxe is 350v squared-122500/8000 = 15.3 watts.
A 50 watt 2204 is 470v squared is 220900/3400 = 64 watts.
Etc........
 It works for finding out what an amp SHOULD put out.The actual wattage is best measured.

finally got the search function to work

So on the amp I just finished, I have about 470v on the plates and primary impedence on the OT is 6,600.
Using the formula above I am about 33-34watts.  Sounds reasonable to me.

[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.

[HotBluePlates]

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.

[jjasilli]

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

I rest my case (re the voodoo comment)  :)   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. . .

Still it's useful to have some reasonable idea of output wattage.  Thanks again to those who posted a down and dirty way to get there!

[HotBluePlates]

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...

[phsyconoodler]

It makes people happy to have something tangible without using a signal generator and a dummy load.
 I could care less,as long as it sounds good.I have told people before that I rate my amps as loud,louder and loudest.They still want to know so I give them the quick answer.I can measure it but it is a let-down.Just like the factory horsepower or gas mileage claims.Reality is a let-down.Leave people with their fantasy's of power.They're happier that way!

[jjasilli]

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?" ;D

[HotBluePlates]

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.

[PRR]

> 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.

[PRR]

> 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.

[Animatic]

I would look at 3 things for a 'modern guitar amp'.

a) what is its max clean non-distorted level into it's expected load.

b) What is it's 'redlined on 11', 'flat out'  into distortion
     and into the natural compression of same  
     level into the same expected load.

c) and the SPL at 1 meter for the actual output into the room.
Of BOTH a and b above.
This might be the 'Vs the Drummer", in actual use figure.

[bluesbear]

I'm not sure how much the end number actually counts, anyway. The other guitar player in my band uses a Fender Blues Deluxe. It's in good shape. This thing has 6L6's with an SS rectifier. He runs it at at least 5 on the volume. I have a homemade (of course) Deluxe (no reverb or trem), with 2 - 6V6's cathode biased, 5Y3 rectifier. I run it at 2 1/2 on the master and 2 on the volume... and I get yelled at for being to loud. Go figure!
Dave

[phsyconoodler]

Unfortunately,I can build two identical amps except for one small detail:one has a linear volume pot and the other has a logarithmic volume pot.The linear one gets loud quickly while the other has the same loudness at a higher setting.
  Then there's the 50 watt combo with a 97SPL speaker and the 20 watt amp with a 103SPL speaker.
  Not a good comparison.

[supro66]

get speakers from radio shack and keep blowing them up untill one holds

ONLY MY 2 CENTS

What do I know I build CH47 Helecopters ;D

MY SITE
http://www.geocities.com/insp/SUPRO6420.html

[jjasilli]

OK, let's accept pyschonoodler's formula for the sake of rough simplicity.  As PRR says it's dependent on a reasonable match of power tubes to the OT primary impedance.  

SO: lets say we plug 2X 6L6GC's into a 6V6GTA amp.  Plate voltage is 360 and Plate to Plate impedance is 7600 ohms. (We re-bias for the 6L6GC's).   The OT primary is matched to the 7600 ohms, but the 6L6GC's have only 3800 ohms of Plate to Plate impedance.

1.  What is the output in wattage of the 6L6GC's using the 6V6GTA tranny? I.e., how do we adjust pyschonoodler's formula for this condition?

2.  Let's say the OT, with 6V6GTA's and a 7600 primary impedance has a secondary (output) impedance of 8 ohms.  What is the output impedance with the 6L6GC's?  

Thanks!

[HotBluePlates]

SO: lets say we plug 2X 6L6GC's into a 6V6GTA amp.  Plate voltage is 360 and Plate to Plate impedance is 7600 ohms. (We re-bias for the 6L6GC's).

Good to go. I follow you here; 6L6's with 6V6 supply voltage and (more importantly) output transformer.

The OT primary is matched to the 7600 ohms...

Okay, I'm still onboard...

... but the 6L6GC's have only 3800 ohms of Plate to Plate impedance.

And now you've lost me.

The 6L6's make absolutely no difference in this case; they don't have a "plate-to-plate impedance" or at least not that we care.

What you're meaning to say is that the 6L6's would normally be run with a 3800 ohm plate to plate impedance. But the transformer is reflecting a primary impedance based on what is attached to the secondary and what the turns ratio is. So if the transformer is reflecting 7600 ohms, then it is 7600 ohms and nothing else.

So same supply voltage, same load impedance, same output power. I *think* 6L6's might make a hair more output, but only because they could pull slightly farther towards 0v on the plate than 6V6's. This theoretical conclusion is born out (in my mind) by the experiment I did with popping 6L6's into a tweed Deluxe with no changes to the amp except for the tube change; the volume from the speaker did not increase when I tried it. The amp did distort on about 5.5 instead of 5, but that was probably due to a bias voltage difference from the tube change.

I did not measure and document every detail to give an exact mathematical argument for why everything stayed the same. I'm almost to the point where I have all the equipment that would be required to do that. It will probably go on a list of experiments to do a present to explain how amp theory correlates with real-life experience.

Anyway, let's look at this another way by standing the situation on it's head. End point: the output tubes do not make power; power is made available by the power supply in the form of supply voltage and current available, and the tubes are simply used to control the flow of power through the load impedance. In the end, the power supply and the load impedance dictate the power possible, and if the output tubes can allow the full range of possible power to flow, then it will.

So on to the different situation, same end result... Say you have 2 power transformers, both that supply 200vac, a set load impedance of 1000 ohms, but 1 transformer can supply 200mA where the other can supply 20A. From a "power" standpoint, that means that one transformer can deliver 200vac * 0.2A = 40VA, while the other can deliver 200vac * 20A = 4,000VA. Which one causes more power in the circuit?

Again, neither. If the load impedance (in this case a resistor across the secondary) stays at 1k ohms, and the voltage stays at 200vac, then the current that can flow is 200v/1k = 0.2A. The power dissipated in the resistor, which would be our "power output" is 200v * 0.2A = 40w. If we use the 4,000VA transformer, the load resistance still limits the current to 0.2A, and still limits the "power output" to 40VA or 40w.

So you see, the primary impedance that is reflected by the output transformer will set a limit on how much current the tube can pass (at a.c., not at d.c.). If that limit remains unchanged, and the supply voltage remains unchanged, then the power output will remain unchanged even if we install tubes that are ordinarily capable of very much greater current swings and more power output.

So the tube doesn't make power. It just controls the flow of power in accordance with the signal on G1, but can still only control as much power as the power supply and the OT primary impedance allow to happen.

I didn't fully appreciate this concept until several years ago. It should be one of those things that is immediately obvious, but when you learn electronics as an afterthought to tinkering guitar amps... well you learn lessons out of order. And since I'm largely self-taught, I learned some of the lessons slowly, because my teacher is as dumb as me...  ;D

[Animatic]

This all brings to mind something I would like to see in FAQ look ehere first section.

For each of the common functions we must do. like :
estimate watts
Find impeadance to output tube plates,

DC and AC load lines,
and biasing reasoning.

Tone stack calculations.

Yes it is all 'out there' and in various forms.

But I would like to see a basic plain english explaination of each process.
Not just an equation, but a completed pair of examples
small tube high low & big tubes high low.

And then you insert your measurements or theoretical estimates.

I just don't see it without a level of, ya gotta know it jargon, which is not explained.
Like tube sheets can refer to the same data in 2-3 different names.
So I want to see the english name and the 2-3 common abreviations all there.
So looking at a Sylvania or Tung sol or GE sheet makes sense.

I understand the ideas, but actually DOING it still is hands on hit or miss,
and wonder what's happening.

A page or two like this would answer MANY repeated questions for newbies I suspect.

For the 6L6 with a 5k load we figure ... .. ...
For a 12AX7 in this form we think .. ... ...
Here is Fenders take on it here's Marshals
and the Sylvania tube sheets.

There just seems to be somethings still missing.

[HotBluePlates]

I think you'll have gathered my implied answers on your own, but we'll hit them square on the nose for those who skim over the long-winded explanations...

1.  What is the output in wattage of the 6L6GC's using the 6V6GTA tranny?

Same as for 6V6's.

I.e., how do we adjust pyschonoodler's formula for this condition?

We don't. The basic argument still holds, and the formula is unchanged.

Let's say the OT, with 6V6GTA's and a 7600 primary impedance has a secondary (output) impedance of 8 ohms.  What is the output impedance with the 6L6GC's?

This question does not compute.

Okay, we'll start by saying you still have a 7600 ohm to 8 ohm transformer. So nothing is changed just because you swapped tube types. The actual plate resistance looking into the 6L6's is *not* 3800 ohms, or even 7600 ohms, but at least tens of thousands of ohms. That number is not really important to us, at least not in this conversation.

As far as actual output impedance as seen at the secondary... well, it could be 8 ohms or higher, or lower. If negative feedback is used in the amp, the effective output impedance is very likely well below 8 ohms, as the ratio of amplifier output impedance to load impedance (with amp output impedance being lower) is called "damping factor". One of the many points of using feedback is to have an amp where the  amp's output impedance is lower than the load impedance, so that speaker movement is more tightly controlled.

But this is neither here nor there as far as a discussion of output power is concerned. I think you'll see that there was an assumed change of inherent impedance on the primary side due to a tube change, when such is not the case. And since that isn't the case, the question about a changed output impedance due to the tube change is moot.

Not bad thinking, since you were trying to reason through impedances being reflected through the transformer, but in the end the speaker is probably the only concrete impedance in the system, and it is the one that gets reflected from secondary to primary. Pretty funny, since the speaker's impedance is also all over the place depending on frequency.

[HotBluePlates]

I just don't see it with out a level of, ya gotta know it jargon,
which is not explained.

The big problem is writing all that stuff would not only fill a book, but an entire bookshelf and then some. In fact, I think that the reason The Ultimate Tone series was started was to explain all this stuff. And Kevin O'Connor probably though he did a good job in the first book, until he got questions and had to go back and re-cover some ideas, expand on others and take things in new directions.

Which is why I think you don't get a complete picture in any 1 of his books.

The old tube books *did* explain all this stuff. But they made things universal, so they just gave you the facts. Honestly, it's all there in black and white, but a reader nowadays needs a teacher to guide them through the material and explain it, because it take a certain background and a certain context to understand what is, in the end, a simple idea.

Maybe we'll get to the ideal of having all required tube knowledge ready-to-eat on this site. I don't know, because it would require a herculean research and posting effort that would be equivalent to writing a series of books.

[PRR]

> I would like to see a basic plain english explaination of each process. Not just an equasion

What is so tough about this?  

2 * V^2/Rpp * F

If that is too tough, use:

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

> What is the output in wattage of the 6L6GC's using the 6V6GTA tranny? I.e., how do we adjust pyschonoodler's formula for this condition?

Pyscho's formula does not say what tubes. It implicitly assumes all tubes give 50% of the Output Watts of a perfect device. So nothing changes.

>> What is the output in wattage of the 6L6GC's using the 6V6GTA tranny?
> Same as for 6V6's.

The 6L6 will give a teeny bit more power. 6V6 is chuffing trying to pull 8K load, 6L6 is loafing. The 6V6 is not straining all that much, and the 6L6 is not so unstrained as to act "ideal". So if the 6V6 makes 12W clean, the 6L6 may make 14W clean.... not worth thinking about.

That assumes the load is not "too low" for the bottles. If we used 6AU6, a much smaller bottle than 6V6 or 6L6, with 360V and 7600 ohm load, power would be quite poor.

When in doubt... plagiarize!!!! You are not so brilliant that you will find a new way to use tubes. WHatever you are trying to do has been done before; or if not, then it probably sucks (or you have not looked long enough). Plagiarize the whole plan, don't hang the Brooklyn Bridge's cables on the Golden Gate's towers.

[billcreller]

 ;D ;D

[phsyconoodler]

Whatdda ya mean I won't find a new use for tubes?
  How about Christmas lights?500v ones that the neighbors kids won't dare steal.about 50 watts with the formula  :P

[Animatic]

PRR you gave a PERFECT example :

^  the karot could mean greater than or something els3e.
it ISN'T PERFECTLY clear. Nor the F.

If I KNEW for sure what it was intending,
in plain english, i could easily run the equasion.

Again it is jargon you must know and perfectly clear
IF you have seen it explained.

it is easy looking back to say Ah yes, head slap.
But looking in blindly it's uh, hmmm, ah, well, maybe??
And that gets just as confusing when there are DIFFEREING
jargons meaning the same thing.

Plagerizing is fine in this case, and morally unambivalent.
Go for it.
it's just that you don't LEARN as much....

[jjasilli]

Thanks again guys!