Power tube Ap=Wattage?

[lowell]

Hi all,
i'm curious 'bout knowing the wattage of a tube amp... rms wattage I guess too. If I'm running 4 6L6 p/p at 35ma w/ plate voltage of 450V does that mean the amp is putting out (.035*450) 63watts? I'm sure the OT plays a part in this, but any advise/info is appreciated!

Lowell

[Alex R]

No, current through the tube times plate voltage gives you the power the tube itself is dissipating, not the output power.

Whilst that figure will get higher with output wattage, and whilst it will bear some relation to what the tube is capable of shoving out through a transformer into a load, you can't measure output watts that way, dissipation wattage and output wattage only essentially relate to one another because tubes get hot whilst they're amplifying, and their power is limited by the heat they can dissipate. But actually what they dissipate is in broad terms actually the power that they don't put into the speaker.

Output wattage is simply the power the amp can put into a load, and it relates not to the heat the tube dissipates but the heat a resistor will dissipate if you fire an amp into it. You have to measure it that way - output power is the AC voltage the amp puts into a load, squared and then divided by the ohms resistance of the load.

It's easy to measure wattage with a AC voltmeter, a dummy load resistor, and the simple formula above. You don't even really need an RMS voltmeter for this, a cheapo thing will do.

[lowell]

AR thanks for info! Can you please explain exactly how to hook up the meter and provide some info on the resistor ratings? And wouldn't it be bad for the ot to run it into a resistive load?

[R ski]

The power of the amp can be figured out in real watts, providing you can play guitar through the amp full volume using an AC voltmeter across the speaker terminals. If the meter is a cheap digital its sampling time could be to slow to get a reading.

Calculating the power with a voltage reading is E(2) / R. Measuring the speaker's resistance gives the R value. Squaring the voltage reading at the terminal will give the E value, Typical readings I get depend on the how hard the amp is being pushed. If clean sound is the case the reading requires a fast reading or a meter with peak holding capability. When the amp is dimmed out hard the sustain will allow a easier readings.

Typical 2 tube 6L6 clean I get about 19 volts across 8 ohms, that yields about 45 watts. Driven hard, I seen 26 volts, that yields about 84 watts. Typical four tube Fender twins I seen 32 volts max with some clipping.

[stokes]

84 watts out of a pair of 6L6's?Are you using a true RMS meter?

[ronm]

While all the replies are valid, the most reasonabe method to find your clean output power would be to use a dummy or resistive load, signal generator and a scope. You can also connect a meter along with the scope probe.

The total output power rating should be the greatest clean output you can obtain before distortion. We all know or should know that a distorted signal is where the peaks flatten or are clipped.

The proceedure is pretty straight foward. Connect a resistive load in place of the speaker(s). At the input of the amp, inject a ~400hz signal, not a guitar. You want a stable signal. Flatten you tone stack as best you can. As you turn the volume up on the clean channel, watch your signal on the scope. It will appear irregular and not as a pure sine wave as the input. As you turn the volume up toward full, the waveform will start to look more like a sine wave and the peak should start to flatten or clip. It is at that point and maybe just beyong that point just a little, you will be at the max output before distortion. Anything beyond that will not be the total clean output power.

Now see what your DVM reads, E(2)/R and there you have total clean output power. Your scope could tell you the same thing in volts, but with a DVM connected simultaniously, for me at least, is a lot quicker to read. All I have to be concerned with on the scope is just watching the waveform.

[lowell]

Thanks Ron that makes a lot of sense. Can I use a resistor as a load? I've always been tought that a speaker needs to be the load?

[d95err]

Thanks Ron that makes a lot of sense. Can I use a resistor as a load? I've always been tought that a speaker needs to be the load?

The amp needs to see a load. This can be a speaker, resistor (dummy load) or anything in between (i.e. attenuators).

A dummy load resistor should have the same resistance as the nominal impedance of the speaker (e.g. 8ohms for an 8 ohm speaker).

The important thing is the power rating of the resistor. It must be rated for at least twice the power of the amp if you intend to drive the amp hard. E.g. for a 15W amp, you need at least a 30W resistor. Also, keep in mind that power resistors need to be properly heatsinked to obtain their full power rating.

[d95err]

The proceedure is pretty straight foward. Connect a resistive load in place of the speaker(s). At the input of the amp, inject a ~400hz signal, not a guitar. You want a stable signal. Flatten you tone stack as best you can. As you turn the volume up on the clean channel, watch your signal on the scope. It will appear irregular and not as a pure sine wave as the input. As you turn the volume up toward full, the waveform will start to look more like a sine wave and the peak should start to flatten or clip. It is at that point and maybe just beyong that point just a little, you will be at the max output before distortion. Anything beyond that will not be the total clean output power.

IIRC, "Standard" power measurements for tube amps should be taken at 5% THD (Total Harmonic Distortion). A few days ago I tested my homebuilt amp with my fathers distortion meter to get a more precise measurement than you get with just a scope.

The results were:

• Simple scope method (as described above) : 14.8W
• With distortion meter at 10% THD: 13.5W
• With distortion meter at 5% THD: 10.5W

So, my conclusion is that you tend to overestimate clean power by perhaps 20-40% using the simple scope method. It's simply very difficult to see the difference between a clean sine wave, and a slightly compressed sine wave.

[Enzo]

Welcome ronm, but I would take that explanation with a little extra salt. In a hifi amp, that would be OK, but guitar amps are not intended for reproduction, and distortion is PART of the sound. So the idea of a clean output power rating is not very useful. If the amp can produce 40 watts clean, so be it, but that doesn;t stop it from putting out 55 watts at considerable distortion. That being the condition it will be operated in most of the time, and that being the desired tone.

I think also that the concept of clean output requires the preamp section being capable of delivering a clean waveform to the power tubes in the first place. A 5150 for example might be hard pressed to get a clean output at much of any level. WOuldn't be fair to call that a 4 watt amp.

[d95err]

In a hifi amp, that would be OK, but guitar amps are not intended for reproduction, and distortion is PART of the sound. So the idea of a clean output power rating is not very useful. If the amp can produce 40 watts clean, so be it, but that doesn;t stop it from putting out 55 watts at considerable distortion. That being the condition it will be operated in most of the time, and that being the desired tone.

I think also that the concept of clean output requires the preamp section being capable of delivering a clean waveform to the power tubes in the first place. A 5150 for example might be hard pressed to get a clean output at much of any level. WOuldn't be fair to call that a 4 watt amp.

To get a proper measurement of a high-gain amp, you would of course input the signal into the FX return or some other place where you can get a clean signal into the poweramp.

I don't agree that clean power measurements are not useful for guitar amps. On the contrary - it's the only way to measure. The very reason we want power measurements is to compare the power of different amps.The only way to get a proper measurement to measure the clean output power of the amp.

[Enzo]

And since no one publishes that spec, we would what? Take readings on all the amps?

I am not very concerned whether an amp produces 48 watts or 52 watts. The difference in loudness is only 3db between 30 and 60 after all. I think that 50 versus 100 pretty much tells the story. Someone who thinks his 30 watt amp is too loud is only fooling himself if he thinks a 25 watt amp would be less loud.

And I would agree that if you ran a sine through just the power amp, you could then measure the power amp output, but that is not a real world test of the amp, at least not to me, since the amp will almost always be played through the preamp.

[cbarrow7625]

Enzo,

I agree with you to a point. For guitar amps we pretty much all know that two 6L6's will almost always be somewhere between 40-50 watts of output. 4 x 6l6 will be between 80-100 watts. The difference in level is basically indescernable. With a certain tube compliment, there is a range of power that is possible and/or beliveable.

We do, however, need a way to measure honest, comparable specs for output power in any amplifier. Inserting a test signal directly into the power amplifier is the only way to do that. A power amp's output power needs to be measured with a line level signal, not an instrument level signal run through a preamp.

The preamp section in a guitar amp can basically just be considered an "effect", a distortion effect. If we were measuring the distortion of a hifi power amp we wouldn't measure it through the preamp or, especially, through an effect. Now, it can be argued that a guitar amp is nothing more than an effect. It is really an extension of the instrument. None the less, we still need a way to measure output powers to keep us all honest and/or allow the less knowledgeable to have a way to measure the power of their newly built guitar amp.

d95err may be correct that tube amps are typically specified at 5% distortion (large solid state amps are often stated at 1%, chip amps are always stated at 10%) but that measurement needs to be taken with the signal injected directly into the power section (the driver tube). Otherwise, you are going to get much lower power ratings for the amp than you actually have. Even in guitar amps the power sections are typically pretty linear, unlike preamp sections, and you can get a decent power rating.

Power amps kind of have to be linear or they won't work well transferring power into a load. We overdrive them so hard that they get non-linear at the highest drive levels. When we start clipping the signal (turning sine waves into square waves) we have to worry about what that actual level is too. There is far more power available under square wave than there is under a sine wave. That is why you do want to make sure you are not running any power tube to within an inch of its life to jack up the clean power rating. Once you start running square waves through it you end up dissapating far more power than you may have expected. Same thing with rating guitar speakers. Speakers are typically rated for power using pink noise with a 6dB crest factor (or they should be, tha tis the written standard). If you are running "100W" amp with square waves into a "100W" speaker, don't be surprised if it blows.

[Wakculloch]

The output power of an amp is important from a diagnostic and potential volume viewpoint. Unfortunately using a THD method in guitar amps is frought with uncertanty and most manufacturers quoted figures seem to be based on the point at which clipping starts (in the power amp).
I wouldn't like to see a familiar situation to hi fi amps, where the wattage is used as a selling tool and bears no relation to reality.

[cbarrow7625]

Awww, c'mon...we all know that bigger numbers means it will sound better. LOL!