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In Reply to: RE: Ask your "someone" if they thought of this. posted by Chip647 on January 25, 2015 at 07:59:16
So on each tube, I measure between 3 and ground. This is still cathode bias amp. It just has some amount of adjustment.
M
Follow Ups:
Yes, take the voltage (lets say it is 12 volts) off pin 3 and the resistance of the cathode resistor (lets say is 300 ohms) and a plate voltage of 300v you would have the following:
12/300 = 40mA therefore 0.04 * 288 volts is 11.52 watts.
So,
maybe my terms are wrong with Biasing? Should I be trying to adjust to burn say 11.5 watts on each EL84 ? I thought, the biasing on a push-pull amplifier was that each tube amplified 1/2 the sound wave and we are just zeroing the crossover point on the Y-axis=0.0 volts for both tubes.
I think heath kit used 220 ohms originally- but I will have to see what the tech changed. I want to say these were running hotter and he suggested tube replacement to be in the 28-30 mA range. But, the Russian tubes I got are marketed, matched at 25.9 and 25.5 mA respectively.
Michael
There are a number of misconceptions in your post and they are leading you astray:1. The signal is not split in half in a push-pull amp. We lazily (myself included!) tend to call the phase inverter the "phase splitter". There is no splitting of the signal occurring - what is happening is analogous to one of those two person saws that are used to cut down trees - one person pulls the saw while at the same time the other pushes. The phase inverter sets up the push-pull output to do the same to the signal by creating a 180 degree out of phase image of the signal. So while one tube is "pushing" the other side is "pulling". In the VAST majority of audio amps each tube sees the entire waveform or quite close to it.
2. Just because the tubes were measured for matching purposes at one current level does not mean that is where they always have to operate. Those figures (25.9 and 25.5 ma.) are the current measured under an arbitrary set of conditions in the matching process. Those tubes can be biased at any current you wish as long as you don't exceed the tubes' maximum ratings for current and/or dissipation.
If the Heath amp was operating the tubes in a way that exceeded the ratings for those parameters then a circuit modification needed to be done so the tubes could operate within their capabilities. Changing the value of the cathode resistor is one way to change the current and dissipation.
I think you would be wise to invest in Bruce Rozenblit's book - "Beginner's Guide To Tube Audio Design". It will clarify a lot of things for you, and it's not over the top in theory and math. You'll be much more comfortable with all this with a bit more study!
Edits: 01/26/15
Jim,
Yea I was opening my mouth freely because I didn't understand. Now that I do- I feel lots more comfortable with biasing the amps.
I will get the book, because I feel lost manytimes. Yes I've put kits together, but need to understand what is going on beyond how a triode works.
Thanks again and yes I always saw it referred to as a spliter(dynaco 70's?). But williamson aka Push -Pull makes more sense with your explanation.
Keep in mind there are other circuits besides the Williamson that are push-pull as well. Citation II amps for one...
Not saying you should run 11.5 watts as that is close to the max dissipation. You would use the pot to fine tune the current draw so that the voltage on pin 3 is the same between tubes. 8 to 10 watts, just consistent. This has nothing to do with AC signal balance.
I have a pot which goes to each EL84, I thought it was to help balance them. My other amp, a CII citation, you balance the two power tubes so their is zero voltage across the tubes. Thinking in a sine wave, one tube takes the upper, one the lower.
I thought that somehow the purpose here was the same.
Michael
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