The base of Q3 depends on the output voltage.
It is biased with a voltage divider that originates in the output, and is grounded through the speaker. The resistors are 470K/33K, so the output has to be at around 10V for Q3 to get 0.7V.
There is a feedback mechanism at play in that if, say, Q3 is cut off because the output voltage has dipped to low, then no currrent flows across R11 and R12, and so the whole VBE multiplier is raised toward the positive power rail. In so doing, it turns on Q5 and cuts off Q6. But if Q5 turns on and Q6 cuts off, the output voltage has swung to the positive power rail also, and so that must turn on Q3.
For Q3 to be turned on at 0.6-0.7 VBE, the output has to be at around 10V, since the voltage divider if 470k:33k.
Perhaps there are episodes of C10 getting charged up, and then forcing the output voltage down so that Q3 is cut off when it shouldn't be.
Bang it up into a schematic and simulate ...
It is biased with a voltage divider that originates in the output, and is grounded through the speaker. The resistors are 470K/33K, so the output has to be at around 10V for Q3 to get 0.7V.
There is a feedback mechanism at play in that if, say, Q3 is cut off because the output voltage has dipped to low, then no currrent flows across R11 and R12, and so the whole VBE multiplier is raised toward the positive power rail. In so doing, it turns on Q5 and cuts off Q6. But if Q5 turns on and Q6 cuts off, the output voltage has swung to the positive power rail also, and so that must turn on Q3.
For Q3 to be turned on at 0.6-0.7 VBE, the output has to be at around 10V, since the voltage divider if 470k:33k.
Perhaps there are episodes of C10 getting charged up, and then forcing the output voltage down so that Q3 is cut off when it shouldn't be.
Bang it up into a schematic and simulate ...