Phase inversion technical question

[boobtoob]

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,

[Old Tele man]

...the negative DC-bias voltage 'approximately' holds the two output tubes OFF (actually at idle, but for this "quicky" description, consider them OFF).

...when the "input drive" signal alternatively goes "positive" for each tube, that drive signal effectively "cancels" the bias voltage, moving the total (DC bias + AC signal) control grid voltage from OFF toward ON condition.

...when the input drive signal into the PI goes positive (ie: UP) the PLATE voltage of the PI tube goes DOWN (due to voltage drop across the plate resistor), that makes the effective signal into the upper (top) output tube become even MORE negative, so it's now fully OFF.

...simultaneously, the CATHODE voltage of the PI tube goes UP (due to the voltage drop acrosss the cathode resistor), that makes the effective signal into the lower (bottom) output tube become LESS negative, allowing the tube to conduct. When the drive signal equals the bias voltage, the two voltage cancel each other leaving the control grid voltage at ZERO voltage...which is the "limiting condition" for Class AB1, where the "1" means the control grids NEVER go positive.

...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.

...FWIW the "idle" condition where "both" tubes are ON at the same time only occurs when there is NO input signal...when operating, one output tube in ON while the other output tube is OFF...although there is a slight period of "overlap" (about 15-20 electrical degress) which serves to smooth out "cross-over" distortion that occurs during the transistion from one tube conducting to the the other tube conducting.

[donnyjaguar]

I understand how the "out of phase" signal is achieved for the lower power tube (it comes from the cathode of the PI tube),

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).


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.

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

[boobtoob]

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?

[donnyjaguar]

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.

[boobtoob]

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.

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?

[Old Tele man]

...one half (+) of input signal causes one OUTPUT tube to conduct.

...other half (-) of input signal causes other OUTPUT tube to conduct.

...the PI tube "splits" the "+" and "-" halfs of the input signal into separate (+) signals going to each output tube.

...thus, "+" portion of the input signal causes one output tube to conduct and the "-" portion of the input signal causes the other output tube to conduct.