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heyHey!!!,
After a bit of discussion yeilded the simplified answer is, 'each tube/phase sees aa/2', I still want to know what happens when the simplifications are removed. Say for a pair of 6L6GC pentodes. B+ around 250 V, aa load of ~5k and each tube idling near 100 mA. What is the load seen at each plate across the swing up and down of the signal?
With a pentode, plate resistance is high compared to the loading...make it infinite for discussion?
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
When I first studied the treatment of pushpull tubes in the standard texts, including the Radiotron book, I found the following to be easier to understand. (Ignoring inductance and other reactive considerations for simplicity.)
1. Treat the transformer as three equal windings, with B+ applied to the connection between two windings to make the centertapped primary.
2. Put a largish load resistor on the third winding.
3. If you measure only the transformer, the "platetoplate" impedance, which is that specified by the transformer vendor, equals 4x the load resistor on the third winding, since there is a 2:1 turns ratio.
4. As the tubes swing, the total ampturns in all three windings, carefully added together, equals zero at all times, if the quiescent condition has zero total ampturns (equivalent to having equal quiescent currents in the two tubes). In this case, the turns are the same for all three windings, so you only add the currents.
From these axioms, you can see how the currents in the two plates affect the voltages on the same plate and the other plate, and find the voltage into the (fictional) load resistor. To make it less fictional, you can replace the load resistor by a stepdown transformer from that resistance (1/4 of the socalled platetoplate impedance for the circuit) to 8 ohms or whatever.
5.
Edits: 03/20/12
When I first studied the treatment of pushpull tubes in the standard texts, including the Radiotron book, I found the following to be easier to understand. (Ignoring inductance and other reactive considerations for simplicity.)
...
4. As the tubes swing, the total ampturns in all three windings, carefully added together, equals zero at all times
...
5.
Thanks Tim!
+1
This is the main assumption/simplification that allows simple analysis of unequal tube currents and is how my crossover analysis spreadsheet model works.
It also shows that you can't make the crossover bump go away by increasing the bias, short of class A operation...
And it shows that operating the 2 tubes in less linear parts of their curves is still class A; i.e. switching off, or staying at bias level for part of the cycle, is the critical difference.
Hmm. it even lends some insight into the argument about "transitioning" from class A to class AB.
Edits: 03/20/12 03/20/12
One important feature I forgot to post about the simplified model: since all three windings have equal turns, the (AC) voltage swings in each winding are equal. For example, this is important to see how the voltage swing in a cutoff tube is affected by the current in the other tube. Also, this voltage is impressed against the highZ load on the third winding to make output power.
In this context, the crossover problem you mention comes from the fact that each tube can only conduct in one direction. In AB or B modes, each tube will cut off for a considerable fraction of the cycle and the other tube's current will need to rise at a different rate to meet the zeronetcurrent condition when the cutoff tube does not pass negative current.
Edits: 03/21/12 03/21/12
I guess by negative current you mean that a tube can only sink, not source current. True, but sometimes OPT leakage inductance can fool feedback circuits into behaving as if the cuttingoff tube is supplying current during the crossover.In a push pull amp, the plate voltage of the tube cutting off goes to nearly 2X B+ while the conducting tube goes toward Vp=0.
Edits: 03/21/12
You may wish to take a look at the Radio Designer Handbook (fourth edition), pages from 574 to 582 (downloadable @ Pete Millet site).
Best Regards
Luca
ecc230
not sure exactly what you are looking for but I suspect you are not looking to add in the concept of finite inductance making the load elliptical at LF, and stray capacitances doing the same at HF.
dave
Well...I suppose we could limit analysis to the midband portion that is effectively a resistive load, say 2000 cps. I guess that this first bit of simplification will do for now...:)
cheers,
Douglas
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Well, If both tubes are providing the same AC behavior, aren't we at the ideal class A. Are you looking for what happens when things are ever so slightly mismatched yet still class A?
For this I guess you need to look at the sum and difference loadings and different parts of the signal will see different loads. Anything common to both halves will always see 1/2 the AA impedance, however say one tube has slightly more gain than the other or a different distortion spectra, I would expect any difference signals of even a "deep class A" circuit to see the class B load of 1/4 AA.
This conceptually works for me much like the summing stereo to mono by simply paralleling the connections. All of common information sees a high value load but all of the difference info simple sees the output impedance of the opposing device as the load.
where I stumble in both of these situations is the "black or white" nature of them and how everything in audio seems to be a shade of grey. The comforting thing is that wihtout the black or white viewpoint, things like matrix amps wouln't work so maybe it is fait to mix pure black and pure white to get a different shade of grey.
dave
So we've got a plate current that is g1_voltage_delta( X + delta ) the delta being anything not the perfectly linear.Swing the votlage more positive and we get the G1_v_d * delta as the portion running at aa/4 and the load line will be somewhere between aa/2 and aa/4.
So when we swing V_g1 negative I think we shall expect to see something less( greater R value ) than aa/2. I can't quickly visualize the sign convention here...
cheers,
DouglasI seem to remember seeing something for triodes that calculated the load based on the plate resistance. So as one gets lower at maximum current it was closest to aa/4( as it moved there from aa/2), and for the phase with a decreasing voltage the load approached infinity as plate resistance went up( which is of course about he same thing as this decreasing grid voltage delta term I think ).
Friend, I would not hurt thee for the world...but thou art standing where I am about to shoot.
Edits: 03/19/12
since you are talking about PP and the load each tube sees, lets go back to the simple aspects.
I think the first flaw thus far was the treatment of the other tube. The first order approximation for either case demands the other tube to be considered as an active device in either the ON or OFF position and when in the ON position the signals must be matched. In 'ideal' class A it is on and 100% inverted thus energizing the entire winding with identical currents. In class B it is an open circuit which removes 1/2 of the transformer winding from the picture.
I don't see tube Rp coming into the picture here unless you want to consider output impedance, so lets leave that for a more complex model.
What I see as the important consideration of a more complex model is how closely matched the + and  going signals are. Any deviation from identical will shift the behavior from one behavior of the transformer to the other.
Again I'll refer back to the concept of summing stereo to mono. Many people suggest that a transformer can be used. I don't see that as a viable option since the 'ideal' transformer is invisible. Since we are trying to build the second order model, I guess it needs to be asked what is the first thing that is going to mess up? (what assumption of the ideal are we going to violate to add another parameter?)
dave
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