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In Reply to: RE: PP Class A OPT Loading posted by PakProtector on March 18, 2012 at 12:21:37
When I first studied the treatment of push-pull 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 center-tapped primary.
2. Put a largish load resistor on the third winding.
3. If you measure only the transformer, the "plate-to-plate" 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 amp-turns in all three windings, carefully added together, equals zero at all times, if the quiescent condition has zero total amp-turns (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 step-down transformer from that resistance (1/4 of the so-called plate-to-plate impedance for the circuit) to 8 ohms or whatever.
5.
Edits: 03/20/12Follow Ups:
When I first studied the treatment of push-pull 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 amp-turns 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 cut-off tube is affected by the current in the other tube. Also, this voltage is impressed against the high-Z 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 zero-net-current condition when the cut-off 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 cutting-off 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
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