Furthermore, even with 1 meg, any one tube can conduct less or more current than the other tube, as it does not require the other tube to complete its circuit with the load. This means that Class B operation is certainly possible, as the idle current, as long as it was balanced through both tubes can be as little as is required to ensure linear operation. For example a 64 watt OTL amplifier delivers a peak of 4 amps into an 8 ohm load, but might need only maintain an idle current of 10 mA through the output tubes. Admittedly, the crossover distortion will be high, but the amplifier remains functional.

    The relationship between idle current and feedback is seldom mentioned, but illuminating for tube fanciers. Basically the more feedback that can be safely applied across an amplifier's output stage, the less the idle current is needed. Many Op-Amps run with true Class B output stages, i.e. zero output-stage idle current, and yet possess low distortion figures because of the near infinite feedback ratios. If less feedback were used, these Op-Amps would distort grossly. Tube circuits, with their coupling capacitors and output transformers, do not safely permit huge amounts of feedback, which compels a much higher idle current to linearize their output stage. High idle current is not efficient, but it is linear. A Dynaco ST-70 runs a relatively rich Class AB output stage and sounds all the better for it.
   To return to the role that the load plays in this topology. A 1 meg load limits the maximum current through the current path provided by this resistance. The lower the load impedance, the greater the possible current  flow through it. In other words, when one tube ceases to conduct, the load impedance constrains the maximum current flow the other tube can experience. Assuming the tubes never undergo positive grid voltage, the sum of load impedance added to the tube's rp divided into the B+ voltage gives us the maximum current into load impedance:
     ILoadMax =  Vb / (rp + RL).
Obviously, if one tube is no longer conducting any current, the current through the tube must equal the current flowing through the load impedance:
    IT1p =  IL,   when IT2 = 0.
    If both tubes are conducting, the maximum current through either tube is given by the B+ voltage divided by rp:
     Imax = Vb/rp.
In the case where one tube conducts more current than the other, the delta (difference) is delivered into the load:
     Iload = ITube1 - ITube2.
Excepting where RL = 0, ILoadMax is always less than Imax. 

    The SRPP circuit also confuses those who ignore the load. "It's a current source loading a triode; therefore, it is a single-ended circuit." Yes, exactly, just as long as the circuit's output is not connected to a load impedance. If it is connected to a load, the load provides alternative current path for both tubes and the current source loses its high impedance. With an attached load, two anti-phase current  paths through the load are possible and the load drags down the current source like attribute of the top triode. For example, with a 6DJ8 based SRPP circuit, supplied with a Rak resistor value of 200 ohms, and working into a 32 ohm load, the bottom tube sees a load impedance of a little more than 232 ohms, not the near infinite as a current source label would suggest.
     Here is the quick test for single-ended and push-pull circuits: count the number of current paths through a load. If any are in anti-phase with the others, the circuit is probably push-pull; otherwise, probably, single-ended. Why only
probably and not always? Here is an example, given a Dynaco ST-70 amplifier, which has been so modified that one output tube's grid is AC shorted to ground, two current paths remain through its load (the primary of the output transformer) and these are configured in anti-phase. Yet this amplifier is single-ended in that the dumb tube provides no drive and is used only to prevent transformer core saturation, so only one active path remains.



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