How to determine rated power (specifically, EL34 vs KT66)

[MattG]

In general, how does one determine the rated output power of a tube amp?

In my particular case, I am looking at a pair of EL34s vs a pair of KT66s in a push-pull configuration. At a high-level, both of these tubes are rated for 25 watts dissipation; so naively, one would think a pair would make for a "50 watt amp". But I'm sure there's more to it than that. At least from what I've observed, it feels like dual EL34 amps are almost always marketed as 50-watt amps, while dual KT66 amps tend to be rated lower.

To give more context to my question: a few years ago, I built a Trinity Triwatt from their kit (the website isn't currently showing the kit, they also sell it pre-built). This is based on the classic Hiwatt DR103/DR504, with some minor tweaks. By default, the Triwatt can run either 6V6 power tubes or KT-66. It has a switch to lower the plate voltage for 6V6 operation, and also external bias probe and adjustment. The product specs for the amp indicate 33 watts output power with KT66 tubes. So that's my first question: how did they arrive at 33 watts?

However, later versions of this amp have an output transformer with an additional primary that is suitable for EL34 tubes; that's what I have. I am considering changing the amp to run EL34s instead of KT66s (mainly to consolidate my tube inventory, as I have another EL34-only amp; and EL34 are cheaper than KT66). This is a rare case of (almost) "all things being equal" - of course the OT secondary will change and the bias circuit will likely need to be tweaked a bit. So I'm wondering, will the rated output power change?

What I'm really getting at, ultimately - should I expect a change in the amount of available clean headroom? My understanding (could be wrong!) is that, all things being equal, higher output power means more clean headroom.

Appreciate any words of wisdom!

[trobbins]

Imho this is a very wide topic to respond to, so in hopefully a helpful manner I can suggest:

The first point of reference is what bogey tubes will provide in a reference output stage configuration - that information in provided in manufacturer datasheets. I'd suggest you spend some time googling and reading about that aspect first - there are many scanned on-line books, and tutorials and articles (old and new), and of course many forum threads, that go over that general topic. Some of those references go into how conditions may be modified to suit a particular amp setup, and of course forum threads sometimes go into extending operation beyond datasheet standard configurations. Perhaps after some self-education you may be able to go through your set of queries with your own assessment of where you've got to and point to what doesn't quite gel.

[xtian]

It's worth considering the old saw that, to perceive an amp as twice as loud, you need 10X the output wattage. So if you squeeze a few more watts out of your amp by changing power tubes, the result won't sound that much louder to you.

[Stevem]

Tubes themselves can not make wattage/ power, they need a input of voltage and a certain level of current availability to produce a certain amount of wattage,

It’s good when diving into things like this to keep this in mind also.
Wattage is power in terms of ohms law, and power is made by current times voltage.

In terms of the output stage there’s a pretty good range of different currents and voltages that will produce the same level of power even if you do not have the ideal impedance output transformer for the combination of voltage and current your working with.

400 volts X .250 amp is 100 watts.

320 volts X .300 amps is 96 watts.
If you had a 10 watt amp and wanted to double it’s volume to your ears you would need to go up to a 100 watt amp.

You also need to understand that the wattage that a given tube is rated at when you look at its specs is a DC based rating, audio output is a AC rating.

[passfan]

E2/R=P will give you a rough estimate if you can measure your output voltage. Use your voltmeter in peak hold mode.
IMHO higher output power gives an increase in bottom end , tighter , cleaner , etc. Cleaner headroom would be the culmination of a few different things , power , circuit , gain staging , etc.

[wpaulvogel]

In general, how does one determine the rated output power of a tube amp?

In my particular case, I am looking at a pair of EL34s vs a pair of KT66s in a push-pull configuration. At a high-level, both of these tubes are rated for 25 watts dissipation; so naively, one would think a pair would make for a "50 watt amp". But I'm sure there's more to it than that. At least from what I've observed, it feels like dual EL34 amps are almost always marketed as 50-watt amps, while dual KT66 amps tend to be rated lower.

To give more context to my question: a few years ago, I built a Trinity Triwatt from their kit (the website isn't currently showing the kit, they also sell it pre-built). This is based on the classic Hiwatt DR103/DR504, with some minor tweaks. By default, the Triwatt can run either 6V6 power tubes or KT-66. It has a switch to lower the plate voltage for 6V6 operation, and also external bias probe and adjustment. The product specs for the amp indicate 33 watts output power with KT66 tubes. So that's my first question: how did they arrive at 33 watts?

However, later versions of this amp have an output transformer with an additional primary that is suitable for EL34 tubes; that's what I have. I am considering changing the amp to run EL34s instead of KT66s (mainly to consolidate my tube inventory, as I have another EL34-only amp; and EL34 are cheaper than KT66). This is a rare case of (almost) "all things being equal" - of course the OT secondary will change and the bias circuit will likely need to be tweaked a bit. So I'm wondering, will the rated output power change?

What I'm really getting at, ultimately - should I expect a change in the amount of available clean headroom? My understanding (could be wrong!) is that, all things being equal, higher output power means more clean headroom.

Appreciate any words of wisdom!

What you need to understand about plate dissipation is that it’s the amount of heat generated by the plate structure from current flow between the cathode and plate. It has a marginal amount to do with power output but don’t be confused. The plate dissipation rating doesn’t mean that it will output power equal to rating in an audio system. And EL34 can produce power output far exceeding its dissipation rating and KT66 really can’t reach it. It has a lot to do with the area of the plate. A large plate structure can generally dissipate more heat. 6L6 tubes have several ratings but really output the same wattage. I know the voltage ratings change but in a test between metal body 6L6 and a 6L6GC, they have basically identical output shown on the data sheets.

[teemuk]

Those are just plate dissipation ratings. How much will the plate dissipate when producing x watts of power will depend on design. Obviously you get rather different answers for A, B, C or D classes of operation.

[roberto]

EL34 can produce power output far exceeding its dissipation rating and KT66 really can’t reach it.
[OMISSIS]
6L6 tubes have several ratings but really output the same wattage. I know the voltage ratings change but in a test between metal body 6L6 and a 6L6GC, they have basically identical output shown on the data sheets.

Can I ask you to elaborate more these two points and report the tests?

[Guy77]

One thing I like to do is to plug in an oscilloscope on the speaker output jacks and then plug in a large dummy load speaker (I use 300 watts) than attach a multimeter to measure AC voltage.
Plug in your guitar and turn up your amp and watch the sine wave start to distort its shape as you play. A signal generator is a better tool here as well. Then see what the AC voltage is on your multimeter as the sine wave begins to distort. Do this first with an amp that you know its wattage like a Fender Deluxe Reverb first and then compare your results to that amp.

Cheers

Guy

[Reeltarded]

E2/R=P will give you a rough estimate if you can measure your output voltage. Use your voltmeter in peak hold mode.
IMHO higher output power gives an increase in bottom end , tighter , cleaner , etc. Cleaner headroom would be the culmination of a few different things , power , circuit , gain staging , etc.

Who are you?

[R.G.]

In general, how does one determine the rated output power of a tube amp?

[...] To give more context to my question: a few years ago [...] So that's my first question: how did they arrive at 33 watts?

The general way is to hook up the amp to the rated AC power line voltage, fire it up, and feed it a sine wave while watching the output voltage into a load resistor. The input is turned up until the output waveform begins to distort by some amount, then the power is calculated from that.

That's the baby steps idea. Since people want lots of power in their amps, amp sellers have worked out many ways to confuse buyers with higher ratings by doing this in funny ways. They can feed it higher AC voltage, or decide that it's ok for the output to distort so much that the output is essentially a square wave, not a sine, do the measurement in burst of a fraction of a second so the power supply doesn't sag, anything to be able to claim a higher power.

Using tubes puts some additional complications in. Any give pair of tubes ( such as 6V6, 6L6, EL34, KT66...) will produce its highest power output into the "best" plate to plate load. In the Golden Age of tubes, tube makers would experiment with plate to plate loading, and find the magic loading that made the highest output power for that specific tube and power supply voltage. They would then specify the tubes' power output rating at the plate to plate loading that made the most power. Output power from a specific tube at it's "best" power supply level could vary perhaps 2:1 over a broad range of plate to plate loads.

Adding all this up, changing tubes without also optimizing the plate to plate load won't necessarily change the output power as much as you'd expect. 6V6s like about 8K plate to plate, 6L6s like 4K or so. And the B+ voltage needs changed to get the "best" use of different tubes for more power.

A power amp is best thought of as a power supply with some extra stuff tacked on to let the power out in a controlled way. If you can't make enough DC power at the right voltage, the tubes can't let more power out than the power supply makes.

This is a long, roundabout way of saying that just changing tubes will probably change the output power a little, but as already noted, it won't make big changes because the tubes don't make power, they let some of the DC power out, and they're quirky about the conditions of B+ voltage and output transformer loading to do their best even then. And as the poster said, it takes about 10x amplifier power to sound subjectively "twice as loud". The difference between a 30W and a 50W tube output power isn't nearly as much as you think.

Speakers make a vast difference, though. Speakers are rated for the sound pressure level (SPL) in db that they produce at a given power. Guitar speakers are designed to be LOUD rather than smooth; it's common for guitar speakers to be rated at 95-100db with one (!) watt of drive. Some will do 110db SPL or more, that being the same as twice (or more) the power amp power.

So - it's complicated.

However, later versions of this amp have an output transformer with an additional primary that is suitable for EL34 tubes; that's what I have. I am considering changing the amp to run EL34s instead of KT66s (mainly to consolidate my tube inventory, as I have another EL34-only amp; and EL34 are cheaper than KT66). This is a rare case of (almost) "all things being equal" - of course the OT secondary will change and the bias circuit will likely need to be tweaked a bit. So I'm wondering, will the rated output power change?

Their adding a tapped primary for different tubes is a good step forward; this really needs to change so each tube can do its relative best. But you're still back at that power supply thing. The B+ voltage that makes the most watts from a given pair of tubes also needs to change some. They did this by lowering it for the 6V6s, which can't stand as much B+ as the bigger ones. But EL34s and the KT series (for "Kinkless Tetrodes") generally need higher B+ for best power out compared to something like a 6L6. The internal structures are different.

What I'm really getting at, ultimately - should I expect a change in the amount of available clean headroom? My understanding (could be wrong!) is that, all things being equal, higher output power means more clean headroom.

My best guess is that the power out, measured on an all-things-being-equal basis, will change, but not enough to give you noticeably more clean headroom. You will get several watts more or less power by changing tubes, but not enough to make more headroom. If you need a number, set up a dummy load and an oscilloscope and measure the size of the biggest clean sine wave into the load that you can get, then calculate the watts.

As another side note, if you want clean headroom, you won't get there very easily with swapping tubes. Instead, you need to change to a bigger power supply and bigger output transformer - a bigger amp overall. And you can get more bang for the buck by changing speakers.

[passfan]

E2/R=P will give you a rough estimate if you can measure your output voltage. Use your voltmeter in peak hold mode.
IMHO higher output power gives an increase in bottom end , tighter , cleaner , etc. Cleaner headroom would be the culmination of a few different things , power , circuit , gain staging , etc.

Who are you?

I have always been , and will always be , your friend!