Re: So why ARE tube amps louder than solid state?
:fing25: I studied electronic engineering in school and I have built amplifiers. I had to derive all mathematical calculations before building circuits. If any one wants to know how to derive RMS from calculus integrals, then please let me know.
Yeah, I have a piece of paper that says I'm full of BS or something from one of those schools and quite honestly barely remember calculus (yeah, I'm old). I was just pointing out that RMS (or also often called average or program power) is not 2.5 times peak power, but peak power divided by the square root of 2 (1.414) or Peak * .714. While still not probably not the best way to rate amps but still much better than using peak ratings. Part of the problem with music is that is is never just a simple sine wave, but for simplicity, a sine wave is used (or assumed) for power ratings.
No egos or anything, just tryin' to be helpful. I don't have time (or the desire) to write a paper on the subject, I would also need to do some research, this kind of thing isn't off the top of my head nowadays (although the basics are kinda like riding a bicycle), I'm more worried about sensors and the like right now, and some electro-mechanical parts that I'm trying to make work together.
Gotta get back to work, but check out this
link, it has alot of these answers in easy to digest yet still semi-techie form (my favorite way to read them).
I will also add that alot of the power sections in tube amps are pretty much legacy designs, and it has been assumed since that a 4X EL34 or 4X 6L6 power section is a 100W amp, sometimes they will get rated for 120W or so. It's rarely that that simple and alot of amps put out more than their paper spec says.
*EDIT*Technically, I'm also wrong, as Admiral and Stevo I think both touched on. You can only take the RMS value of the output voltage, then calculate power using P=IV or P=V^2/R or P=I^2(R), so, for example a an amp that puts out say 40Vpeak-peak into an 8 ohm speaker, so that would be 200Wpeak-peak. But taking the RMS value of the voltage, 40V/1.414 = 28.3V, then finding the RMS power, 28.3^2 / 8 = 100.1W. Sorry, I am old, my memory gears occassionally grind. Maybe that piece of paper from school is right?!?!