Determine output power

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Re: Determine output power

Post by zaphod_phil »

geoff 1965 wrote:
Fri 04/30/21 10:05 am
the ECL’s test very strong on both the triodes and pentodes and closely matched.
£38.70 including delivery,cheaper than new crap JJ’s!
Your amp would actually sound better if they weren't matched. This ain't Hi-Fi! :D
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Re: Determine output power

Post by geoff 1965 »

s-l1600 (4).jpg
this is the donor amp for the transformers ZP,valve!stereo!! HiFi!!! "Jack's getting excited now" as for the close match,i can work on that if needed 82/100K on the phase plates or even bias the pentodes individually,early days and i'll start a new post for it.
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Re: Determine output power

Post by JMPGuitars »

zaphod_phil wrote:
Sat 05/01/21 6:22 pm
geoff 1965 wrote:
Fri 04/30/21 10:05 am
the ECL’s test very strong on both the triodes and pentodes and closely matched.
£38.70 including delivery,cheaper than new crap JJ’s!
Your amp would actually sound better if they weren't matched. This ain't Hi-Fi! :D
I was waiting for you to say that. 😉
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Re: Determine output power

Post by TriodeLuvr »

geoff 1965 wrote:
Sat 05/01/21 7:04 pm
s-l1600 (4).jpg
this is the donor amp for the transformers ZP,valve!stereo!! HiFi!!! "Jack's getting excited now" as for the close match,i can work on that if needed 82/100K on the phase plates or even bias the pentodes individually,early days and i'll start a new post for it.
At this very moment, I'm trying to decide whether my current build should have an AC Balance pot for the LTP. :lol: :lol:

Jack
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Re: Determine output power

Post by zaphod_phil »

TriodeLuvr wrote:
Sat 05/01/21 10:18 pm
geoff 1965 wrote:
Sat 05/01/21 7:04 pm
s-l1600 (4).jpg
this is the donor amp for the transformers ZP,valve!stereo!! HiFi!!! "Jack's getting excited now" as for the close match,i can work on that if needed 82/100K on the phase plates or even bias the pentodes individually,early days and i'll start a new post for it.
At this very moment, I'm trying to decide whether my current build should have an AC Balance pot for the LTP. :lol: :lol:

Jack
Just keep in mind that some asymmetry in the PI and power stage increases the amp's warm even-order harmonics. So I strongly recommend you stick with 100k/100k load resistors in the PI, plus non-matched tubes in both the PI and PA. This is a guitar amp (I hope) :)
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Re: Determine output power

Post by TriodeLuvr »

I'm surprised no one is building with the 6J6 for the LTP. They're plentiful, cheap, and balance between sections is usually terrible. :lol:

Jack
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Re: Determine output power

Post by katopan »

I posted this on the Wattkins forums back in 2011 and it's still my preferred method for guitar amps. Valve amps sag (even with a SS recto let alone a valve one) which is part of why we love them. Measuring output power with a signal gen has the problem where the sag isn't given chance to recover like when playing with a real guitar signal. Using a real speaker means you don't know the impedance integral across the balance of frequencies for the calculation. Using an RMS volt meter has the risk you'll push into clipping a bit a elevate the result. The only option for a real speaker is a real power meter that measures RMS current and voltage (including phase angle) together to work out power. So we use a real guitar, but a resistive dummy load.

http://www.wattkins.com/node/17820
katopan on Wattkins wrote: Measured output power vs. what it's 'supposed' to measure vs. what the amp is generally accepted to be gives you such a wide variation in numbers that it's impossible to have a base of comparison of one amp to another. I've been thinking about comments from Zaphod Phil about how measuring output power with a test signal sags the power supply more than it would with a guitar signal giving a falsely lower value.

Keeping in mind speaker efficiency makes the most difference to actual volume, etc, etc. But I've been thinking it'd still be handy to have a way of comparing the output of different amps. I suppose using a real power meter while playing into a real speaker with a real guitar would ultimately tell you the actual power that the amp is putting out, but not many people (including me) have one. Also different speaker impedance curves and amps with feedback or no feedback are going to also change the value. But measuring the output with an oscilloscope while playing with a guitar into a resistive load I think can give a standard sort of reference that means something. I turned up the volume so that the output stage was just starting to clip. Some amps may sag and drop the peak output a bit as you turn it up more into more distortion, so I figured the start of clip was the best common point to make the measurement.

It's been said that an 18 Watt actually gives out about 13W clean. Using this method I got close to that with 12.25W. Previously I'd measured it with a sine test signal and with the extra power supply sag got 12V peak output into 8 ohms which equates to 9W clean. Big difference!

18 Watt
EL84 push-pull
340Vdc B+
14V peak output
= 12.25W clean
katopan on Wattkins wrote: I strummed chords on a plugged in guitar with the guitar and/or amp volume set so the output signal viewed on my CRO was just past the point of clipping the peaks. Into a dummy resistive 8 ohm load. Take that peak voltage (not peak to peak, just one side) and work out the equivalent power for a sinewave of the same peak voltage. Difference with this method over actually injecting a sinewave is the real dynamic signal from a guitar allows the power supply to recover and gives a more consistent measurement for the amp which compared well to published, or generally accepted numbers for output power of different well known amps. Testing with a steady sine wave always gives readings that are low because it overloads and sags the power supply much more than a real world signal.

eg. If I was to use this technique and found that the very start of clipping peak voltage was 10Vpeak into 8 ohms -> (10x10/8)/2 = 6.25W output power. The divide by 2 is because I've used Vpeak. Or I could convert to sinewave equivalent rms by dividing by sqrt(2) and get the same answer -> Vrms = Vpeak / 1.414 = 10 / 1.414 = 7.07Vrms -> 7.07x7.07/8 = 6.25W.
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Re: Determine output power

Post by TriodeLuvr »

katopan wrote:
Mon 05/03/21 5:56 pm
katopan on Wattkins wrote: I strummed chords on a plugged in guitar with the guitar and/or amp volume set so the output signal viewed on my CRO was just past the point of clipping the peaks. Into a dummy resistive 8 ohm load. Take that peak voltage (not peak to peak, just one side) and work out the equivalent power for a sinewave of the same peak voltage. Difference with this method over actually injecting a sinewave is the real dynamic signal from a guitar allows the power supply to recover and gives a more consistent measurement...
I believe just the opposite is true. The harmonic energy in the waveform from a guitar - not to mention the additional harmonic energy created by the amplifier itself - is extremely variable. This variability contributes to inconsistency regarding the RMS power of the waveform, and it in turn also affects power supply sag. If the intent is to determine the dynamic, "pre-sag" value of the waveform just prior to the onset of clipping as a function of amplifier output power capability, i.e. the maximum short-term power output, a pulse generator is needed. A burst of a few cycles of sine energy at the frequency of interest, repeated every few seconds, will allow measuring the peak output power using a scope. The result will be consistent from one amplifier to the next without regard to the tonal (spectral) content produced by the pickups or the amplifier's tone control settings.

This type measurement will provide a basis for knowing how the amplifier will initially react to a strong signal, and it is useful in determining whether the physical excursion of the speaker will be within mechanical limits. This aspect of a speaker's power rating becomes more important when it is expected to be used in free air (not a vented or infinite baffle enclosure) with no acoustic dampening.

The other important issue is the power dissipation capability of a speaker's motor and whether the amplifier will cause it to overheat. The variability of the waveform noted above creates a complex and inconsistent relationship in this regard, so in the absence of overriding empirical data ("XYZ amp blew up ZYX speaker), we are left with only the usual long-term RMS measurement for making decisions. As noted by katopan, and as I've said previously, amplifiers typically deliver much less long-term power than data sheets would lead us to believe. For this reason, I believe there is little risk in applying a significant discount to the output ratings of most guitar amplifiers when evaluating whether they will work properly with a given speaker. If the numbers are close, there's little risk unless you're constantly banging the voice coil against the back side of the magnet structure.

Jack
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Re: Determine output power

Post by katopan »

I don't disagree with what you've said. But just to be clear, I am using a CRO to see the clipping threshold voltage. After that it's a theoretical calculation based on that peak to work out max clean output power of a pure sinewave at that clipping threshold peak value. I'm not using an RMS meter, which would be completely thrown by using a real guitar signal input.
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Re: Determine output power

Post by zaphod_phil »

In my opinion, measuring the clean power output of a rock guitar amp is a waste of effort. The main reason we buy or build these amps is to hear them overdrive into their sweet-spot tones. Their "clean" tones still aren't pure sine wave, but have some distortion. As I've said in other contexts, this ain't Hi-Fi! :P
katopan wrote:
Mon 05/03/21 5:56 pm
I posted this on the Wattkins forums back in 2011 and it's still my preferred method for guitar amps. Valve amps sag (even with a SS recto let alone a valve one) which is part of why we love them. Measuring output power with a signal gen has the problem where the sag isn't given chance to recover like when playing with a real guitar signal. Using a real speaker means you don't know the impedance integral across the balance of frequencies for the calculation. Using an RMS volt meter has the risk you'll push into clipping a bit a elevate the result. The only option for a real speaker is a real power meter that measures RMS current and voltage (including phase angle) together to work out power. So we use a real guitar, but a resistive dummy load.

http://www.wattkins.com/node/17820
katopan on Wattkins wrote: Measured output power vs. what it's 'supposed' to measure vs. what the amp is generally accepted to be gives you such a wide variation in numbers that it's impossible to have a base of comparison of one amp to another. I've been thinking about comments from Zaphod Phil about how measuring output power with a test signal sags the power supply more than it would with a guitar signal giving a falsely lower value.

Keeping in mind speaker efficiency makes the most difference to actual volume, etc, etc. But I've been thinking it'd still be handy to have a way of comparing the output of different amps. I suppose using a real power meter while playing into a real speaker with a real guitar would ultimately tell you the actual power that the amp is putting out, but not many people (including me) have one. Also different speaker impedance curves and amps with feedback or no feedback are going to also change the value. But measuring the output with an oscilloscope while playing with a guitar into a resistive load I think can give a standard sort of reference that means something. I turned up the volume so that the output stage was just starting to clip. Some amps may sag and drop the peak output a bit as you turn it up more into more distortion, so I figured the start of clip was the best common point to make the measurement.

It's been said that an 18 Watt actually gives out about 13W clean. Using this method I got close to that with 12.25W. Previously I'd measured it with a sine test signal and with the extra power supply sag got 12V peak output into 8 ohms which equates to 9W clean. Big difference!

18 Watt
EL84 push-pull
340Vdc B+
14V peak output
= 12.25W clean
katopan on Wattkins wrote: I strummed chords on a plugged in guitar with the guitar and/or amp volume set so the output signal viewed on my CRO was just past the point of clipping the peaks. Into a dummy resistive 8 ohm load. Take that peak voltage (not peak to peak, just one side) and work out the equivalent power for a sinewave of the same peak voltage. Difference with this method over actually injecting a sinewave is the real dynamic signal from a guitar allows the power supply to recover and gives a more consistent measurement for the amp which compared well to published, or generally accepted numbers for output power of different well known amps. Testing with a steady sine wave always gives readings that are low because it overloads and sags the power supply much more than a real world signal.

eg. If I was to use this technique and found that the very start of clipping peak voltage was 10Vpeak into 8 ohms -> (10x10/8)/2 = 6.25W output power. The divide by 2 is because I've used Vpeak. Or I could convert to sinewave equivalent rms by dividing by sqrt(2) and get the same answer -> Vrms = Vpeak / 1.414 = 10 / 1.414 = 7.07Vrms -> 7.07x7.07/8 = 6.25W.
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Re: Determine output power

Post by zaphod_phil »

TriodeLuvr wrote:
Mon 05/03/21 3:25 pm
I'm surprised no one is building with the 6J6 for the LTP. They're plentiful, cheap, and balance between sections is usually terrible. :lol:
Jack
Never heard of those, but that sounds like a great suggestion! :D However I wonder if their gain level is quite the same as a 12AX7.. :?
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Re: Determine output power

Post by TriodeLuvr »

zaphod_phil wrote:
Tue 05/04/21 9:34 am
TriodeLuvr wrote:
Mon 05/03/21 3:25 pm
I'm surprised no one is building with the 6J6 for the LTP. They're plentiful, cheap, and balance between sections is usually terrible. :lol:
Jack
Never heard of those, but that sounds like a great suggestion! :D However I wonder if their gain level is quite the same as a 12AX7.. :?
Voltage gain of the 12AX7 LTP is about 25; the 6J6 would be roughly half that. However, the Marshall 36W has a huge amount of excess gain, and a 6dB loss in this area probably wouldn't be missed. I'm already using the FX loop with a patch cable "permanently" installed so I can reduce the gain with the Send pot. The High gain input is just too much for use at home. If I ever build another amp, it will be low power (< 5W) with many changes from the 36W I just completed. The 6J6 would be one of them. :)

Jack
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Re: Determine output power

Post by zaphod_phil »

As long as you can fully drive your power stage with the 6J6 LTP-PI, to get that classic Marshall 18W sound :)
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Re: Determine output power

Post by zaphod_phil »

TriodeLuvr wrote:
Tue 05/04/21 11:05 am
zaphod_phil wrote:
Tue 05/04/21 9:34 am
TriodeLuvr wrote:
Mon 05/03/21 3:25 pm
I'm surprised no one is building with the 6J6 for the LTP. They're plentiful, cheap, and balance between sections is usually terrible. :lol:
Jack
Never heard of those, but that sounds like a great suggestion! :D However I wonder if their gain level is quite the same as a 12AX7.. :?
Voltage gain of the 12AX7 LTP is about 25; the 6J6 would be roughly half that. However, the Marshall 36W has a huge amount of excess gain, and a 6dB loss in this area probably wouldn't be missed. I'm already using the FX loop with a patch cable "permanently" installed so I can reduce the gain with the Send pot. The High gain input is just too much for use at home. If I ever build another amp, it will be low power (< 5W) with many changes from the 36W I just completed. The 6J6 would be one of them. :)
Jack
As long as you can fully drive your power stage with the 6J6 LTP-PI, to get that classic Marshall 18W sound :)
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Re: Determine output power

Post by TriodeLuvr »

zaphod_phil wrote:
Tue 05/04/21 1:32 pm
As long as you can fully drive your power stage with the 6J6 LTP-PI, to get that classic Marshall 18W sound :)
The schematic for the 18 Watt Tremolo TMB indicates the EL84s are biased to about 12V. That means a 24V pk-pk signal at each grid will drive it to full output. A 6J6 would hardly know it's working! :)

Jack
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