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In Reply to: RE: PP Class A OPT Loading posted by dave slagle on March 18, 2012 at 13:44:55
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.
Follow Ups:
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 A-A 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 A-A.
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 a-a/4 and the load line will be somewhere between a-a/2 and a-a/4.
So when we swing V_g1 negative I think we shall expect to see something less( greater R value ) than a-a/2. I can't quickly visualize the sign convention here...
cheers,
Douglas---I 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 a-a/4( as it moved there from a-a/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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