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Old April 28th, 2008, 02:32 PM   #1 (permalink)
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Can you tell how many watts and amp is putting out?

How can you test to see how many watts a tube amp is? Thanks

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Old April 28th, 2008, 05:42 PM   #2 (permalink)
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The easiest way is just to tell us what amp it is. The specs are known and available online. Also, there's some rules of thumb that indicate power output by tube compliment. For example, two 6V6s can output up to 25 watts. Of course, much of that depends on transformers, rectifiers, etc.
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Old April 28th, 2008, 06:02 PM   #3 (permalink)
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The output circuit and operating voltage have a big effect, too. A true class A, cathode-biased amp with two 6L6s may put out 25 watts, while in class AB1 with lots of plate voltage and grid bias, the same tubes may put out 60W.

If you want to measure the output wattage you need a dummy load, a test tone/signal source, and a way to measure total harmonic distortion at the output. Most amps are rated at 5 percent THD; some at 10 percent.
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Old April 28th, 2008, 07:19 PM   #4 (permalink)
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Technically amps do not output power (Watts), they output EMF (Volts) and we can measure the voltage. The EMF drives current (Amps) through the impedance (Ohms) of the speaker where the power (Watts) is dissipated.

If you were to measure the average output voltage across the speaker with a meter as Volts AC, whilst it is operating at fairly constant volume, this will give you the approximate Vrms. The meter will have a very high impedance so will not load the amp circuit. I find a good old fashioned dial multi meter e.g. Avo-8 easiest to eye-average and it will iron out fluctuations whereas a digital meter will go rather batty. The power dissipated is V^2/R (square the volts and divide by ohms) where R is the speaker load impedance. This method provides an estimate of output in use but takes no account of distortion.

So I did a quick test using my Fluke DVMM (no analogue available) and (it's late here) I get 0.5Vac to 1.3Vac, avg=0.9Vac, so 0.9^2/8=0.1W which does not sound much but is quite audible at 00:20 hours.

Last edited by jefrs; April 28th, 2008 at 07:33 PM. Reason: typo
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Old April 28th, 2008, 07:20 PM   #5 (permalink)
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To add, amps are usually tested with a 1 KHz tone.
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Old April 28th, 2008, 07:32 PM   #6 (permalink)
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Which will give you the power dissipated in a calibrated test load with a 1kHz calibrated test tone signal as measured on a bench full of expensive calibrated test equipment. Which is fine and how amps get rated, but does not measure how much filth yours is currently kicking out.
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Old April 28th, 2008, 10:28 PM   #7 (permalink)
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... which might come in handy if you want to see if your amp measures up to the manufacturer's spec. Maybe Giogolf should say why he cares about wattage in the first place.
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Old April 28th, 2008, 10:36 PM   #8 (permalink)
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yeah, what they said! but unfortunately, after trying to understand all that, my brain short-circuited!

i'm technically impaired, so i judge an amp's power by how much it "kicks ass" for its size/tube array etc. ... i've seen 25-watters that blew away 60-watters. i realize your question is serious and that ain't the answer you're looking for, but i had to throw it in just for fun.
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Old April 28th, 2008, 10:40 PM   #9 (permalink)
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Quote:
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Which will give you the power dissipated in a calibrated test load with a 1kHz calibrated test tone signal as measured on a bench full of expensive calibrated test equipment. Which is fine and how amps get rated, but does not measure how much filth yours is currently kicking out.
You have to set arbitrary guides somewhere, or else you can't compare it with other amps or the manufacturer's specs meaningfully. If the OP has a "30 watt" amp, and tests it with his low E string because he doesn't know any better, he's gonna be bummed if his amp (which might be perfectly healthy) is only putting out 15 clean watts!

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Old April 28th, 2008, 10:43 PM   #10 (permalink)
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Of course, the output wattage can be measured. You need to be able to accurately measure volts and amperage, the rest is math.
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Old April 29th, 2008, 12:02 AM   #11 (permalink)
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I am not trying to test any amp in particular, however, I just would like to know how builders and manufactures say this amp is X watts and so on? I thought there might be a way to test with a voltage meter. Thanks for all the post, I think this is more complicated than I thought...
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Old April 29th, 2008, 12:08 AM   #12 (permalink)
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Of course, the output wattage can be measured. You need to be able to accurately measure volts and amperage, the rest is math.

uh-oh
somebody said the "M" word!
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Old April 29th, 2008, 03:55 PM   #13 (permalink)
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Half as much as you think, twice as much as you need.

Generally.

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Old April 29th, 2008, 04:00 PM   #14 (permalink)
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I remember doing this in college. We used a load resistor on the output and injected a 1k sine wave into the input and used an oscilloscope to measure the output and then did the math to watts. This was many moons ago
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Old April 29th, 2008, 04:16 PM   #15 (permalink)
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Quote:
Originally Posted by giogolf View Post
I am not trying to test any amp in particular, however, I just would like to know how builders and manufactures say this amp is X watts and so on? I thought there might be a way to test with a voltage meter. Thanks for all the post, I think this is more complicated than I thought...
They don't just "say", they "design". It isn't guesswork - when designing an amp (or zillions of other electronic devices) you set a target and work to achieve it within bounds of other concerns - cost, maintenance, availability of parts, etc. You don't just toss in a pair of 6L6s and see what happens. You can calculate the results to a very close value before you build anything, but that is beyond a forum posting.

If you wish to test actual output power, you need what has been mentioned here - a signal generator, dummy load and an oscilloscope. They you need some standard assumptions, such as the nature of the test signal. Most simple amps are tested at 1kHz because it is easy and in the middle of the human hearing range.

Then you need to observe the output and look for obvious distortion. Real manufacturers rely upon much more sophisticated tools to measure the distortion as a percentage of the original signal, but in a tube guitar amp it is fair game just to observe clipping.

Then, do the math: square the peak voltage of sine wave, divide that by 2 times the resistance. There's some measure of RMS power under those conditions.
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Old April 29th, 2008, 08:14 PM   #16 (permalink)
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Don't take this wrong but there's no such thing as RMS power. Watts is Watts.

The root-mean-squared (RMS) of the maximum voltage (i.e. as would be measured on an oscilloscope) was cunningly chosen because it approximates the same dc voltage needed to drive the same load at the same power.

Hence W=V^2/R, there's no need to double the load impedance.

e.g. PV amp says 65 Watts 22.8Vrms min 8ohm load, 22.8^2/8 = 64.98.

If you use a good moving coil ac voltmeter you will measure RMS Volts to a very good approximation, a loudspeaker is a moving coil device too. A digital meter can do ac Vrms too, but is difficult to read on a fluctuating voltage, can be done.

This will not give you the same answer as the manufacturer's spec which is measured under special calibrated test conditions and is true only for those conditions. It will give you an on the fly measure of true power, the meter could be re-calibrated as a wattometer ;-)

An 8ohm impedance speaker has a dc resistance of about 6ohms. Impedance acknowledges a certain frequency dependance e.g the loudspeaker - usually non-linear and unpredictable.
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Old April 30th, 2008, 01:45 AM   #17 (permalink)
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Don't take this wrong but there's no such thing as RMS power. Watts is Watts.

The root-mean-squared (RMS) of the maximum voltage (i.e. as would be measured on an oscilloscope) was cunningly chosen because it approximates the same dc voltage needed to drive the same load at the same power.

Hence W=V^2/R, there's no need to double the load impedance.

e.g. PV amp says 65 Watts 22.8Vrms min 8ohm load, 22.8^2/8 = 64.98.
You and I are both right. The doubling I did was to account for the square root of 2 in the calculation from peak observed voltage (on a sinusoidal waveform viewed with a 'scope) to RMS.

The only reason I suggested using a 'scope instead of an AC voltmeter is to observe the presence of clipping, that's all. Otherwise, as you say, (RMS voltage)^2/ohms is power, and Watts is indeed Watts.
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Old April 30th, 2008, 08:22 AM   #18 (permalink)
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You and I are both right. The doubling I did was to account for the square root of 2 in the calculation from peak observed voltage (on a sinusoidal waveform viewed with a 'scope) to RMS.

The only reason I suggested using a 'scope instead of an AC voltmeter is to observe the presence of clipping, that's all. Otherwise, as you say, (RMS voltage)^2/ohms is power, and Watts is indeed Watts.
I agree with using the scope. No need to calculate the clipping distortion.
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Old April 30th, 2008, 01:08 PM   #19 (permalink)
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In addition to the power information, and as important as that may seem, speaker efficiency makes a big difference too. You can have a powerful amp and an inefficient speaker and it will seem like a low power amp. And, on the contrary, you can have a very efficient speaker and a low power amp, and it will sound much louder (in dbs) then it's powered at. My Princeton Reverb sounds like it's 60 watts, and it's less than 20. And that's with the original 1966 speaker.
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Old April 30th, 2008, 01:12 PM   #20 (permalink)
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Can you see the difference between 5 and 10 percent distortion on a scope? In my modded Super Champ XD, I'm driving a pair of Tung-Sol reissue 5881s. At 5 percent THD I get 26W; at 10 percent it's a 32-watt amp. Which comes back to some of the points made upthread about specsmanship and design points.

Of course, one can just take the ignorance-is-bliss approach and say that it's louder than stock and the bass is bigger.
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