Phase inversion technical question

Discussion in 'Amps/Cabs Tech Corner: Amplifier, Cab & Speakers' started by boobtoob, Oct 13, 2006.

  1. boobtoob

    boobtoob Member

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    Hello:

    Just trying to understand as much as possible about tube stuff.

    In this schematic:

    https://taweber.powweb.com/store/5e3_schem.jpg

    I understand how the "out of phase" signal is achieved for the lower power tube (it comes from the cathode of the PI tube), and I understand how the center-tapped OT puts the 2 power tube signals together.

    But, why does all the literature I read say that the tubes in an A/B amp are only conducting half the time, swapping duties between them. I don't understand this concept. Looking at the schematic above, it seems as though BOTH tubes are conducting ALL THE TIME, just that one is out of phase with the other. What causes the conduction to stop and start in each tube?

    Thanks for any answers to this,
     
  2. donnyjaguar

    donnyjaguar Member

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    Bzzztt!! No, the out-of-phase signal comes from the anode.
    As the voltage of the control grid rises, the voltage of the anode falls (and the voltage of the cathode rises).

    Ah, welcome to the world of analogue electronics where nothing is ever really "on" or "off". :) The idle current in each of the output tubes is set (by the cathode resistor) to well below the mid-point of conduction. In other words, each tube conducts much more on one half of the input wave than the other. Lots of graphs showing this available and I can explain another way if you like.

    Cheers!
    DJ
     
  3. boobtoob

    boobtoob Member

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    Yeah! Great explaination...it all makes sense to me except this statement:


    ...as the input signal into the PI changes phase, the conditions reverse and the "other" tube goes into conduction while the previously conducting tube gets driven into its OFF state.


    Why would the input signal into the PI change phase?

    I was under the impression that the guitar signal causes the first grid it sees to go a bit positive (or less negative), allowing more electrons to be sucked towards that stages plate. This same scenario happens for each additional triode stage, gaining strength until it hits the PI. Am I missing something here?
     
  4. donnyjaguar

    donnyjaguar Member

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    I actually answered the initial question while OTM was scooping me at the same time. :)

    Think of it this way. The polarity of the input signal actually changes from positive to negative as it follows the string as it vibrates back and forth.
     
  5. boobtoob

    boobtoob Member

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    NOW I UNDERSTAND!!!! I had it figured all wrong, thinking that the guitar signal is always positive.

    BUT, if the guitar signal is both positive and negative (vibrating), how does the negative portion of it "open" the grid and allow current to pass from the cathode to the anode? Isn't the grid ALREADY negative if biased correctly? Making the grid more negative would only lessen the current flow....right?
     

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