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Claus Misfeldt
 
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Default Is plate dissipation really plate current x plate voltage?

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