Flying Dutchman? Hmmm, Sounds like a description of my Amp building progress.
By the way, I should mention that the cathode to cathode winding imbalance problem is easily solved for the Mac or any other OT just by using a split bobbin with identical windups on each section (symmetrical couplings and equal resistances for each side). The same split bobbin approach could also be used to put the two OTs of a Twin Coupled Amp onto the same steel core. The Twin is just an expensive way to do post-sectioning.
By the way, I should mention that the cathode to cathode winding imbalance problem is easily solved for the Mac or any other OT just by using a split bobbin with identical windups on each section (symmetrical couplings and equal resistances for each side). The same split bobbin approach could also be used to put the two OTs of a Twin Coupled Amp onto the same steel core. The Twin is just an expensive way to do post-sectioning.
OK, I get it. What's your plan to drive this?
Mac solved the drive voltage problem by adding a third winding (trifilar might work) to bootstrap the driver. It seems to me that some amount of driver bootstrapping would reduce the outstage feedback and increase output impedance.
M1 could be supplied by a small 80V supply stacked on B- and then a single TO247 would do the job. But it's really just a NLD isn't it? i.e. a cathode voltage shunt regulator. It's a lossy fixed bias method. Nice way to do it though. A pot wiper on M1 gate allows idle current adjustment. Some balance adjustment is still needed, so how is it better than fixed bias except not needing another power supply?
So the screen current cancels in the OPT due to the coupling of L1 to L6 and L2 to L5. It still shows up on the secondary, right? Albeit in a relatively small amount...
Also, my experiments with trying to detect zero crossing as a bias control method suggest that tailcurrent minima, grid AC zero crossing, and idle point are not exactly one in the same, even when tracking a simple sine wave.
Mac solved the drive voltage problem by adding a third winding (trifilar might work) to bootstrap the driver. It seems to me that some amount of driver bootstrapping would reduce the outstage feedback and increase output impedance.
M1 could be supplied by a small 80V supply stacked on B- and then a single TO247 would do the job. But it's really just a NLD isn't it? i.e. a cathode voltage shunt regulator. It's a lossy fixed bias method. Nice way to do it though. A pot wiper on M1 gate allows idle current adjustment. Some balance adjustment is still needed, so how is it better than fixed bias except not needing another power supply?
So the screen current cancels in the OPT due to the coupling of L1 to L6 and L2 to L5. It still shows up on the secondary, right? Albeit in a relatively small amount...
Also, my experiments with trying to detect zero crossing as a bias control method suggest that tailcurrent minima, grid AC zero crossing, and idle point are not exactly one in the same, even when tracking a simple sine wave.
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Extreme split bobbin (pi winding) is one way. Especially if paired windings are wound with opposite chirality. That way, paired signals that may enter side-by-side near the core, can leave side by side at the outside winding. Its still not a balun, as the windings and signals go opposite ways. But now more-or-less always side-by-side with an opposing AC of similar amplitude.
If plates with big AC swings are near the core, then B+ center tap with no AC swing is at the outside. This doesn't "cancel capacitance". But for low frequencys, does limit the influence to a locally lumped effect with only the "far" (now near) end of the opposite winding at the core. And at med-high frequencies, distributed to a ladder network that looks aproximately like a transmission line with resistive characteristic imped.
You can switch inside for outside. No special magic about that... As long the AC pairs enter together and leave together. Similar in this regard only, to a true transmission line transformer. Note I did say it would be only an approximation, and a fake.
As frequencies climb to RF, the waves are short enough to notice the winding isn't *really* travelling as a pair, in the same direction. Then things obviously fall apart to random result.... but who cares?
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The other idea, which I still havn't draw: is to stay with bifilar pairs, and conventional single bobbin winding, all of one direction. But instead of split the winding only in half Mc style... We split it in fourths, or eights, or sixteeenths. Then reorganize bifilar pairings, such that each segment is optimally self balanced in regard to equal and opposite AC amplitude. Maybe only those segments nearest the big plate or cathode swings would need the extra thick enamel?
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OK, here be da Venus flyaround as requested... No functional difference from the 1st schematic, other than restoring isolation to the secondary.
Hopefully I have made clear which segments are organized as bilfilar pairs.
Total primary windings and H have not changed. This is still the same 5K transformer, sort of...
Hopefully I have made clear which segments are organized as bilfilar pairs.
Total primary windings and H have not changed. This is still the same 5K transformer, sort of...
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McIntosh default winding would effectively have paired L1ax with L1dy. L1ax is nearly GND. L1dy swings V1's cathode. This is no local balance...
Even so, I'm not complaining it wasn't good enough. Only that one can do better. And in so doing better than prior art, old patent is no barrier. Now somebody wind me one... Edcor???
(L2dx L2dy) (L1cx L1cy) (L2bx L2by) (L1ax L1ay)
Secondary (L2ax L2ay) (L1bx L1by) (L2cx L2cy) (L1dx L1dy)
Butt secondary against the least active segments AC voltagewize... Usual reason to distribute secondary may not apply here (or does it?) All primary segments are equally active AC currentwise...
If we cheat a little, and swap (L1cx L1cy) for (L2bx L2by) in the layer order? Would L1 and L2 DC resistance totals come out closer to equal?
Even so, I'm not complaining it wasn't good enough. Only that one can do better. And in so doing better than prior art, old patent is no barrier. Now somebody wind me one... Edcor???
(L2dx L2dy) (L1cx L1cy) (L2bx L2by) (L1ax L1ay)
Secondary (L2ax L2ay) (L1bx L1by) (L2cx L2cy) (L1dx L1dy)
Butt secondary against the least active segments AC voltagewize... Usual reason to distribute secondary may not apply here (or does it?) All primary segments are equally active AC currentwise...
If we cheat a little, and swap (L1cx L1cy) for (L2bx L2by) in the layer order? Would L1 and L2 DC resistance totals come out closer to equal?
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Huh?
Ken,
stop hallucinating!
Mac pairs the bifilars as:
L1ax,L2ay
L1bx,L2by
L1cx,L2cy
L1dx,L2dy
L2ax,L1ay
L2bx,L1by
L2cx,L1cy
L2dx,L1dy
So that there is no AC difference between bifilar pairs, only a DC difference.
Actually, not much need to bifilar them anyway, just use the cross coupled caps like the Twin used. Mac put the cross caps in on their later ones too. (cap from V1 cathode to V2 plate, and a cap from V1 plate to V2 cathode)
Ken,
stop hallucinating!
Mac pairs the bifilars as:
L1ax,L2ay
L1bx,L2by
L1cx,L2cy
L1dx,L2dy
L2ax,L1ay
L2bx,L1by
L2cx,L1cy
L2dx,L1dy
So that there is no AC difference between bifilar pairs, only a DC difference.
Actually, not much need to bifilar them anyway, just use the cross coupled caps like the Twin used. Mac put the cross caps in on their later ones too. (cap from V1 cathode to V2 plate, and a cap from V1 plate to V2 cathode)
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"stop hallucinating!"
"And why for?"
I was just kidding. But how else could they do the bifilar pairing without causing horrific distributed capacitance problems? The whole problem they were trying to solve was the handover between tubes in Class B. By pairing AC compatible windings between the tubes, they effectively have each tube connected to all the primary windings. So no winding gets isolated in a sudden tube shut-off state. Always one tube in control to avoid inductive kickback spikes (winding leakage L).
"And why for?"
I was just kidding. But how else could they do the bifilar pairing without causing horrific distributed capacitance problems? The whole problem they were trying to solve was the handover between tubes in Class B. By pairing AC compatible windings between the tubes, they effectively have each tube connected to all the primary windings. So no winding gets isolated in a sudden tube shut-off state. Always one tube in control to avoid inductive kickback spikes (winding leakage L).
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Nice one Smo-king! Crossed 22uF caps do seem to help. Though put 10ohm in series keep LTSpice from freaking.
Capacitance can look like resistance when laddered with L. Characteristic impedance of a transmission line, etc etc... There is no distributed anything when bifilars fully balance. Thats whyfore transmission line transformer, AKA balun. Or so ye olde hallucinatory theorem goes...
Capacitance of my variant should be horrible (at the ends anyway)! But lets not forget how a ladder network works. Its not going to behave that way, or so ye olde...
Capacitance can look like resistance when laddered with L. Characteristic impedance of a transmission line, etc etc... There is no distributed anything when bifilars fully balance. Thats whyfore transmission line transformer, AKA balun. Or so ye olde hallucinatory theorem goes...
Capacitance of my variant should be horrible (at the ends anyway)! But lets not forget how a ladder network works. Its not going to behave that way, or so ye olde...
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Yeah yeah, theoretically we terminate both ends at character Z. But the whole thing is far less than 1/4 wave at any audio, so does it matter small difference 5K plate to plate transforms to at the rail end of the line? And the loudspeaker load will steal away most of the wave before it can reflect anyhow? We hope.
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You brought up using a pair of 25W OPT do a 50W job? Is this what you meant?
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You brought up using a pair of 25W OPT do a 50W job? Is this what you meant?
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Yes, except be carefull how you connect up the secondaries. Two 4 Ohm windings in series or two 16 Ohm windings in parallel will give 8 Ohm output. Parallel would be the better choice, I would think, to insure equal signals on each xfmr., but maybe not that important (just determines plate versus cathode 50%). Main issue is that both tubes see the same load Rl, winding Rw, and leakage Ls, which is determined by the balance across the xfmrs internally (which should average out even if not perfect internally as long as both xfmrs are identical).
You could probably get rid of those 10 Ohm resistors if you make the inductor coupling something like 0.99 instead of 1.0.
You could probably get rid of those 10 Ohm resistors if you make the inductor coupling something like 0.99 instead of 1.0.
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Equal ampere turns guarantees equal magnetic field. What does equal voltage guarantee?
Also remember we must reflect 2.5K to the primaries, if the total across both transformers is to be 5K. So bridged load is still: 16 at the 16 taps, 8 at the 8 taps, 4 at the 4 taps. I should have flipped all the dots on L1 to place the secondary tap locations correctly.
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Ideally, not much difference between the series or parallel secondaries that I can see (assuming the Zs are worked out correctly). But practically, if the core material in one xfmr draws a little more magnetizing current than the other (in the series secondary case), then the CFB % would be varying. Not good.
Umm, on the bridged secondaries. The way I see it, for the series bridged case, each secondary sees half the load Z (virtual ground at center of load). For the Parallel bridged case, each secondary sees twice the load Z (half the load current driven by each). Then figure back thru the xfmrs and add their one sided Zpri's (being in series) for equivalent tube loading.
Umm, on the bridged secondaries. The way I see it, for the series bridged case, each secondary sees half the load Z (virtual ground at center of load). For the Parallel bridged case, each secondary sees twice the load Z (half the load current driven by each). Then figure back thru the xfmrs and add their one sided Zpri's (being in series) for equivalent tube loading.
You need two 2.5 K P-P OTs with 16 Ohm secondaries to get 8 Ohm out and a 5 K P-P equivalent primary by my reckoning method. Right?
I've been puzzled by the Edcor CXPP25-8-2.5K model that someone must have custom requested once. Unless they wanted a 4 Ohm output. Maybe wasn't for a Twin, but some low Rp regulator tube. Or an Elliptron would get it up to 5K and 8 Out.
I've been puzzled by the Edcor CXPP25-8-2.5K model that someone must have custom requested once. Unless they wanted a 4 Ohm output. Maybe wasn't for a Twin, but some low Rp regulator tube. Or an Elliptron would get it up to 5K and 8 Out.
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