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power output calculations, rated power and required power output

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Hi,

According to RCA Receiving Tube Manual RC13, RCA Radiatron Manual RC10 and some books written in 50s and 60s, power output of a class A amplifier is calculated as:

Poutput = (Imax - Imin) * (Emax - Emin) / 8

This is the power of fundamental frequency component of output signal (since this equation is derived from the fundamental frequency current * the fundamental frequency voltage). However, some DIYers (as they specify in their websites) calculates power output as (Imax - Imin) * (Emax - Emin). This seems to be equal to all power generated by tube.

Is power output of commercial products calculated as the power output generated by tube ? or only the power of fundamental frequency of output signal ?

If it is calculated as the power generated by tube, is it right to do it in this way in the view of electronics and/or hi-fi industry ?

What is the safe range of power ratings for tube designs ? Is %10 less than rated power of tube enough for a safe design ?

It is said transient peaks of music requires 10db-20db. This corresponds to have minimum 16W of power for 12db. Since this is seldom achieved with class A1 single ended designs, what is wrong with the assumptions above ? Is this related with dumping factor ?

Thanks so much...

PS: diyaudio forums are the best forums I have ever seen in the computer and electronics industry. Thanks to all people sharing their knowledge with others...

MB
 
Poutput = (Imax - Imin) * (Emax - Emin) / 8
For class A in audio:
That is how most people here would understand it. remember however that Emax approaches 2* supply voltage with an inductive load, so Poutput approximates to 1/4 anode dissapation.

However, some DIYers (as they specify in their websites) calculates power output as (Imax - Imin) * (Emax - Emin). This seems to be equal to all power generated by tube.
For Radio amateur:
If I remember right, the regulations specifiy the maximum power input only.
Is this the confusion?

The Hi-Fi industry often stretches the truth.

10% Would be safe for NOS or good quality current production. Be aware however that manufacturers ratings for the same valve differ. Perhaps 20% is better.

Your last question requires a lot of time to consider and debate. It is my belief that the use of high efficiency speakers, combined with the very gradual overload characteristics of A1 single ended make the results acceptable.
There are few on this foum who enjoy SET (but I am one:) ).
 
For triodes, efficiency (Po/Po+Pd) ranges from 10 to 30%. Pentodes range 15 to 40%. Both class A figures for a wide range of tubes and conditions (e.g. triode class A2). Although 40% pentode might be pushing the class B barrier at maybe 15% 2nd H...

Of course, the reason for loadlines is, among other things, to determine these quantities with accuracy before building the circuit.

The full-length-of-load-line-divided-by-eight equation is the proper one. If you can figure out the derivation, you'll understand completely. (Although somehow I doubt you pray to the Goddess of Calculus ;) <-- That's my professor's religion :p )

Full length divided by 4 would equal the squarewave output. There is no way you can "quadruple up time" and get *any* more power out of the tube. (Sometimes more power is possible in overdrive conditions, but these move the loadline, negating the conditions.)

Speaking of my prof., he'd also say that that wrong equation looks like it was drawn up by a creative writing major :D

BTW, if the loadline is reasonably symmetrical (if it isn't, that's a measure of even harmonics), you can take the figure of just one half, that is, (Imax-Io)*(Vo-Vmin)/4. Of course, best would be to average the two difference figures, which when put together yields the first (divde-by-8) equation. :p

Tim
 
metebalci said:
Is power output of commercial products calculated as the power output generated by tube ? or only the power of fundamental frequency of output signal ?

Commercial amps are marketing BS. It started with the One True Rating, actual heats-up-the-resistor output power. Then after SS hit the scene they got into RMS watts. This is still accurate since, IIRC, it's basically the squarewave output at clipping, which is twice the sinewave power level. Since a squarewave fills out more of the available space in the amplifier's headroom, it can technically produce such outputs while retaining linearity (linearity being a difference between original input and output). On the other hand, if you have a peaked waveform, its RMS rating will be very low. Keep in mind music is more like this (dynamic peaks up to 20-40dB above average is it?).
After a while, things stayed okay. Until about 10 years ago when some slimey ******* marketing employee came up with "PMPO". May he burn in hell...

(And no there is not an actual definition to PMPO. It is utter BS. Go with your gut feeling based on size and weight etc. as to what a device so rated is capable of handling.)


If it is calculated as the power generated by tube, is it right to do it in this way in the view of electronics and/or hi-fi industry ?p

Most tube amps these days are an audiophile curiosity, where the price tag often overrides the power rating. (Even I'll admit that a reputable-looking seller pushing an 8W amp for $15,000 must be looking good to the saps, er, audiophools, er, -philes they sell to.) Besides that, there is the misconception of tube watts being louder, which may be a side effect of mainstream ratings masturbation (which I went into above).


What is the safe range of power ratings for tube designs ? Is %10 less than rated power of tube enough for a safe design ?


Tubes are rated to be ran at their ratings. Um, duh. You'll get longer life (like, 10,000 vs. 5,000 hours) by running them a bit cooler (in regards to dissipation or actual cooling, like with a fan; both help). I've never had a tube fail on me except in grossly abusive circumstances (like, 100mA grid current continuiously on a 12B4 :devily: ) so IME, I wouldn't worry about it. Others here are older and have had tubes die on them and can add some info..


It is said transient peaks of music requires 10db-20db. This corresponds to have minimum 16W of power for 12db. Since this is seldom achieved with class A1 single ended designs, what is wrong with the assumptions above ? Is this related with dumping factor ?

Probably a). sensitive speakers (heck, right now I'm listening to radio (WXRX) at under a watt.. mmm, 1958 vintage speakers! :) ), b). the peaks are lost in even distortion, which is more pleasing than odd, etc... That's what I would guess.

Tim
 
It is said transient peaks of music requires 10db-20db. This corresponds to have minimum 16W of power for 12db.

Why 16W? 12dB is 16 times higher then 1W but how do you know that you need 16W peak? I for instance use Lowther speakers that have a sensitivity of ~100dB SPL at 1W, they would therefore give 112 dB SPL at 16W which is much higher the any normal listening level, a SET amp with 2A3 giving ~2W is more the enough if you have this kind of speakers. OTOH if you happen to own some spakers with say 86dB SPL at 1W its another story.

What kind of speakers do you own?


Regards Hans

BTW I use an OTL amp with my Lowther, average power seem to be ~10 - 50mW for my normal listening levels.
 
tubetvr, you are right. I assumed the speakers have sensitivity around 85dB.

Sch3mat1c, I think I understood the derivation of
(Emax-Emin)*(Imax-Imin)/8 equation. However, derivation starts with definition as rms power of the fundamental component in the output signal. Maybe I should look some other book or try to derive the equation myself in another way.

Thanks...

MB
 
tubetvr, you are right. I assumed the speakers have sensitivity around 85dB.

If you have speakers with that low sensitivity I can agree that normal SET amplifiers could be a little weak, the obvious solution is to get better speakers, there are many with decent sensitivity and I can not find any reason to stick to low sensitivity speakers.

Another solution is to build a class AB push-pull amp with KT88 or something similar, I have a schematic somewhere of a parallell push-pull amp using 4 KT88's pushing out >100W, however this is not the way I would go.

OTL is another way to go but high power OTLs' using multiple parallell connected tubes often tend to be a bit unreliable even if doesn't have to be that way.

Regards hans
 
metebalci said:
Sch3mat1c, I think I understood the derivation of
(Emax-Emin)*(Imax-Imin)/8 equation. However, derivation starts with definition as rms power of the fundamental component in the output signal. Maybe I should look some other book or try to derive the equation myself in another way.

One note: "RMS power" is redundant. We're already talking real power, which is by defintion, an RMS figure. To obtain it, you can multiply RMS voltage by RMS current, or sum the instantaneous power levels for every point of the waveform.

Now, this definition tells us that harmonics are already taken into consideration, fundamental only need not apply. We've (well... we are supposed to :clown: ) already taken the RMS of the signal.

The basic equation above doesn't take the actual RMS into account, but that won't account for much error. Voltages, currents, resistances and characteristics may be off by 10 or 20% anyways.

Tim
 
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