The data You saw are correct: they are only referred to A COUPLE of tubes as
it was common practice for PP pairs.
As per the dissipation:
- first, the tube only sees V difference among cathode and anode:
therefore, if the feed ing voltage is 305V and the cathode bias is 19V, the
tube works actually with (305-19)=286V
- second, when fitted with cathode bias the system is already
self-regulating. It seems Your amp has a common cathode resistor for both
tubes. In this case, the current cannot be 26 mA per tube: being the cathode
bias given by V=RxI it follows 19=270xI - I=70.37 mA per pair as per tube
D/S.
Plate dissipation is consequently (70/2)*286=10W, which is correctly
conservative for guitar use.
As per Your basic question, dissipated power can be normally calculated as
V*I at the "base" point (zero signal) because the two halves of the signal
tend to compensate each other thanks to plate thermal inertia (the tube
heats up during the positive half-cycle and cools down during the negative
one). This is true for A1 and AB1 class as normally used for audio, while
for high-power tubes operating in B class or after the onset of grid current
(AB2 class) things are a bit more complicated and the mean value of the
integral of I*dV during the whole operating cycle should be calculated.

Ciao

Fabio


"Claus Misfeldt" ha scritto nel messaggio
...
I'm about to bias one of my guitar amps. By now I used the "crossover
notch"
method (...and it worked for me), but as this is said to be inaccurate I'm
willing to improve on that.

I've read

http://www.tonelizzard.com
http://www.aikenamps.com
http://aga.rru.com

on this issue, but still have a question. The reason on biasing output
tubes
is afaik to operate the tubes in the appropriate range and not drive them
beyond their limits. The important parameter is the max. plate dissipation
of the tube which must not be exceeded. I've read that in a class AB amp
the maximum value of the plate dissipation is "normaly" reached at an
output of about 1/2 of the nominal wattage. So it is considered to be a
good rule of thumb to adjust the idle current so that about 70% of the
max.
allowed dissipation is reached.

But: Is it really correct to multiply plate current and plate voltage to
determin that bias point?

In my case the amp has:
- dual 6V6GT tubes
- 305V Plate voltage
- 26mA Plate current per tube
- a kathode resistor of 270 Ohm

The date sheet for the 6V6GT says: max plate diss. 12W; idle current 70mA
at
a plate voltage of 285V in a class AB amp (Bias at -19V). Now 70mA x 285V
are about 20W!?!

What I read in my amp is 26mA / Tube at -15V bias (26mA x 305V = 18,3W)

That confuses me...
Claus



--
Claus Misfeldt
www.frozenfrog.de