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How does tube dissipation relate to output watts of any particular amp? For example, if you have an amp biased to about 18 watts dissipation per tube, and it's a 2-tube amp, does this mean your output can not exceed approx. 36 watts? Pardon the simplistic - and perhaps, dumb - question.....
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For Fun....
High debatable issue but tube ratings... ie., 6V6s = 12-14 watts, 6L6s = 15-30 watts, EL34w = 25 watts.... etc etc.. those are all DC ratings not AC output.Originally posted by EFK View Post
Your idle condition is also a DC rating not AC... so while it is true that a power tube idling very cold, and driven with a stage that will allow the power amp to run in class B, the actual power output will be much higher then the same tube(s) idling hot as blue blazes and the PA being operated
in class A.... but under normal conditions the actual DC idle condition doesn't have a great deal to do with what the maximum power output would be nor does it have a great deal to do with what class (class A, B, AB, AB1, Ab2, C etc.) the stage in being operated in.
I've operated giant 4X1000 tubes in Class AB at 150%-200% of their rated output.... of course those were force air cooled with a chimney.
Here's a cool beast using EL6471s (similar tube):
YouTube - Bach Toccta D-Moll on 1000W Audio Tube Amplifier EL6471
I assume the six blue tubes in the back row are Mercury vapor rectifier tubes... I hope.
And there is always this:
Champ 1000 Watt Tube Amp
Steve is a fan of the Eimac 4CX250Bs tetrodes.... also forced air chimney tubes for max output.
http://www.cpii.com/docs/datasheets/75/4CX250B-7203.pdf
a tiny pair of those can hit close to 800 watts at audio freqs and bias pretty cool but can sink close to 500 watts each! -
Simple answer - It does but it doesn't. It sets the limit to the amount of current that can be pulled through it via the power supply at a given plate voltage, which would then determine the proper load to run for the plate voltage. But the tube never sees the actual output power itself that the load sees due to the way that it works when under signal.Originally posted by EFK View Post
Complex answer - Maximum power output is determined by the load (i.e. 1/2 the OT primary, which in Class B mode is 1/4 the plate-plate load of the full primary from plate - center tap) and the voltage placed across this load at maximum signal swing. The tubes are just a "current control valve" (hence why some call them "valves") that controls the current flow through said load, which in turn controls the voltage drop across said load. The voltage drop across the load is the product of the current flowing through the load and the load value in ohms (V = I x R) while the output power is the product of current flow through the load and the voltage drop that the load current creates across said load at any given point in the swing.
If the amp is designed with the proper plate load, the average power that the tubes see should not exceed the plate dissipation rating of the tube. This happens because plate voltage and plate current are out of phase (as plate current increases, plate voltage decreases due to the negative charge of the increased number of electrons offsetting the positive charge of the plate as they flow to the plate). On a "load line/datasheet correct design", the tube will see its maximum dissipation somewhere in the middle of the swing, and will then decrease back down as plate current continues to increase while plate voltage continues to decrease.
Now on the load side of the circuit, as load current increases (plate current = load current because the load is in series with the plate circuit of the tube), so does the voltage drop across the load. What's actually happening is a very simple voltage divider principle...with no signal on the amp, the total power supply voltage is being split between the tube and 1/2 the OT primary. The sum of the plate voltage and the voltage drop across the load = total power supply voltage at any given time. When the tube is driven with signal, its effective plate resistance drops, allowing the power supply to pull more current through the tube and the load. The increase in load current creates an increasing voltage drop across the load that increases with load current, while doing the opposite across the tube itself (i.e. plate voltage decreasing with the increase in plate/load current flow through it). This maintains the series circuit law which states that the sum of all voltage drops within a series circuit branch = the total supply voltage. If the voltage across one component in a series circuit branch rises, the voltage across the other must fall in a sort of "give and take" relationship.
The tube's rated plate dissipation limits how much current one tube can pass through each side of the load for a given plate voltage in the case of a push-pull amp. The greater the load value, the greater the voltage required across the load to get the same power out.
To double the power, one must 1/2 the load, which at the same power supply voltage will pull double the current through the load, which would require a bigger power transformer that can source double the current. One tube cannot handle this on its own, so there must be a 2nd tube in parallel with this tube in order to create a 2nd path for current to flow so that as far as each tube is concerned, each tube is passing the same amount of current it would be passing if the amp were set up for 1/2 the power and double the load with the same voltage across said load.Last edited by Wilder Amplification; 03-21-2010, 04:01 AM.Jon Wilder
Wilder Amplification
Originally posted by m-fineOriginally posted by JoeMComment
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Bruce just found me my new bass rig!Originally posted by Bruce / Mission Amps View Post
Give me one of those things, a Marcus Miller P bass and five 8x10 cabinets.
It's also a great excuse, I can't play any gig too small to have a 3-phase supply."Enzo, I see that you replied parasitic oscillations. Is that a hypothesis? Or is that your amazing metal band I should check out?"Comment
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