determining wattage

[tubeswell]

u = the amplification factor of the tube stage - e.g. a 12AU7 = 19, 12AY7 = 45, 12AT7 = 60, 5751 = 70, 12AX7 = 100

Ra = the value of the plate resistor (or anode resistor) which forms the DC (and part of the AC) load on the stage

ra = the plate resistance (or anode resistance) of the triode to AC (and varies with different plate currents and plate voltages and bias points - refer to tube datasheets or alternatively plot your own tube characteristics charts). Together with Ra and Rg, this forms part of the AC load for the stage. If you have the tube characteristics chart for the tube in question, you can work out ra from estimating the gradient of the grid curve nearest to the operating point and reading off the change in plate current for the change in plate voltage at that point. Or for a ballpark guess, you can look at a tube datasheet e.g.; for a 12AY7 with a plate voltage of 250, a grid (bias) voltage of -4, and 3mA of plate current, the tung sol 12AY7 datasheet gives a ra of 25kOhms. However tube datasheets only give 1 or 2 idealised 'snapshots' of ra for quick reference purposes, and its better to calculate ra accurately from the equation ra = u/gm (where gm = the tube's transconductance - which is the tube's ability to translate grid voltage change into plate current change) Read about this stuff here: How to design valve guitar amplifiers or better still, get Merlin's 1st book (on designing preamps)

Rg = the value of following stage's grid resistor, which (together with Ra and ra) forms part of the AC load for the (previous) stage.

Rk = the value of stage's cathode resistor. When its unbypassed, Rk also (together with Ra, ra, and Rg) forms part of the AC load for the stage; hence you need to add a degree of freedom to the equation, so you get Rk(u+1). If you fully bypass Rk, then the tube won't 'see' Rk at AC, so its eliminated from the equation for AC load. However if the Rk is partially bypassed, then the stage will 'see' the AC for those frequencies that aren't bypassed, so you have to work out the AC load for the various frequencies to see the effect of AC attenuation for the stage. (This is aside from role of coupling caps (in AC-coupled stages) and inter-electrode capacitance in also selecting the various frequencies that get amplified.)

[tubeswell]

The load on a gain stage as I think about it is how I should have stated it, the load it works into, it usually is just a resistor though it could be a transformer or a tonestack.

Yes but if the AC load is a resistor (Ra or Rg, and in the term 'Rg' I would include tone stacks), it merely 'sucks up' the signal. Whereas if the load is a transformer, then there's a whole lot of other stuff in play which intrigues me, like the losses in the transformer and the resonance of the speaker (or other transducer) if that is what the transformer is driving.

The other stuff like plate resistor and bias details are just how it gets its voltage gain and determine how much gain and how much current it could source. A lower plate resistor means less gain but more current could be sourced. Less bias voltage means more current through the tube (up to a point limited by the plate resistor) and more voltage drop across the plate resistor.

And when the plate-to-cathode voltage decreases with increasing tube current, this also affects AC resistance (because the signal at the grid has to 'fight' the increased tube current etc)

[jmaf]

So let me try again here.

decreacing the supply voltage will decreace the amplification factor (gain) in the preamp tube, however the tube will distort easier, or at lower levels of AC voltage swing from the guitar, especailly prominant in the second stage preamp. Is this correct?

Download some tube datasheets for different tubes and look at the several curves provided. The tube has a transfer function from voltage at the grid to conventional current from plate to cathode(or the opposite for real electron flow).

You will notice that the "transfer" from signal at the grid to current at the plate is not exactly a straight line. The lower the voltage you apply, the more curvy it gets(in common instrument amplifier tubes anyway). So lowering the plate voltage will lower the power dissipated, but will also alter the tone of the amp as others have pointed out.

Amplifiers which under-use the tubes tend to sound dull. A while back I received a hand-made amplifier for repair, 2 6L6GC tubes and just 30 W of rated power. The tubes were running with 300 VDC at the plates and the amplifier sounded really really dull. Substituting for 6V6's made the amplifier come alive, because the 6V6 was running closer to its limits than a 6L6 and with that, it was working on a more linear region of the transfer curves.

So as RG pointed out, "Maybe also a smaller, less capable tube so the tube is working nearer it's upper end when the power supply is too." - that'd be my strategy as well, as you can see from the 6V6 example.

[jmaf]

I heard Eddie Van Halen used to use a variac auto transformer to lower the mains to help get his legandary tone, "brown sound".

IIRC the name brown sound came from old strings he used to like recording with?

EVH put his amplifiers through a lot of abuse. The variac tricks he used were really bad for the amplifiers - you can afford it when you're a millionaire superstar, but it's not something I'd do at home. Increasing just 10% of mains will exceed the voltage for the 500 VDC rated LCR can capacitors on the Marshall heads he used to use. Lowering mains will lower the heater filament voltage as well, and cathode emission with it, causing stripping of the cathode surface and early tube failure. He also used to invert the polarity on his pickups, unwind several hundred turns of wire from them and do all kinds of experimentation on his famous Kramer frankenstein guitar.

The difference between us and EVH is that he'd wreck his equipment at night and have a brand new one in the morning. And we mortals can't afford that....