How to bias a cathode biased amp using just a voltmeter

This is actually a method of calculating plate dissipation in watts of a Cathode biased amp by using just a voltmeter, not how to bias as the title suggests. The subject of much discussion, argument and disagreement, found on a TDPRI forum but the basics are 'basically' correct if you need to figure out the plate / anode dissipation in a cathode biased amp!
http://www.tdpri.com/threads/how-to-bias-a-cathode-biased-amp-with-just-a-voltmeter.99968/

The blue text below is exactly as posted on the forum. The text in red is the figures I have inserted as taken from a typical
Watkins Dominator Reissue, manufactured by Amp-Fix running EL84/6p14p output valves/tubes x 2
Charlie Watkins rated his Dominator at 17 watts and from the calculations below, he wasn't far out.

Sometime in the future, I'll compare these figures with an output test using a signal generator, a scope and a dummy load,
which I have done many times, and from memory I would say barely just 16 watts.

  1. Measure the voltage drop on the cathode resistor to ground. Write the value down. 11.6

  2. Divide this voltage by the value of the cathode resistor. This gives you the amount of current being drawn by both power tubes in milliamps. Write this value down. 11.6/220 = 52.7

  3. Measure the voltage on the plates of the power tubes to ground. Write this down. 334

  4. Now, subtract the voltage from the cathode resistor in step 1 from the voltage measured on the plates.
    Write this value down. 334 - 11.6 = 322.4

  5. Take this value, and multiply it by the current (milliamps) from step 2. This will give you the dissipated power (in watts) of both power tubes. Write this figure down. 322.4 x 052.7 = 16.99048 watts

  6. Take the figure from step 4 and divide by 2. Write this figure down. This is the power dissipation (in watts) of each tube. For 6V6s, if it is over 12 watts, then you need to install a higher value cathode resistor. If it's 10.5 watts or less, you need to install a lower value cathode resistor. 16.99048/2 = 8.49524 watts per tube

  7. After installing the new cathode resistor, do ALL of the steps again to see what you now have. You may have to repeat this process several times to get it dialed in, but it is worth it, and your ears will thank you.

What ever is said on a forum, right or wrong someone will always know better and challenge it. There are rarely any winners and no one's any the wiser at the end of it all, which is why I don't involve myself with any forums. I have used the knowledge I have gained from mistakes and many years of experience in the electronics trade and it has "usually" served me well. You can see from my contradiction of the title of this subject that I'm probably just as bad, but it was basically the title this poor chap got wrong rather than what he was trying to explain.

If you don't agree with me, that's absolutely fine, I'm sure you are right!

END OF STORY

John Beer