It is the so called Royer oscillator: Power sinewave oscillator fed by the current (the input inductor act as a constant current source over the period).
Instead of the auxiliary transformer, you could put the filament windings onto the main transformer.
With the exact circuit I see three issues:
- The base bias resistors are connected behind the input inductor, so the transistors are not properly driven when the voltage is around the zero cross. As a consequence, there is higher voltage drop on them, leading to higher power dissipation. With some transistors this problem is partially supressed by the "base charge storage time", but when your transistors would be faster, the problem may appear. So to have the circuit more reproducible, the resistors have to be connected directly to the positive supply. Then one of the transistors would be all the time safely ON, so no risk of such power dissipation. With this circuit it does not matter, if both transistors are conductive at the same time, the input coil keep the current constant.
- The base drive may generate overshoots, mainly when there is no lamp (or the lamp didn't ignited yet) and upon power up. These overshoots may exceed the breakdown voltage of the transistors and as a consequence, destroy them quite immediately (mainly these low Vcesat types are quite sensitive to that, I learned this myself in the hard way). Quite reliable solution is about 2.2..4.7Ohm resistor in series with the feedback winding. It slow down the commutation a bit, but as I mentioned, the simultaneous conductivity of both transistor is no problem at all in this circuit.
For similar (robustness) reason is good to put a resistor in the same order in series with the primary resonant capacitor (for 8W from 6V battery I put there ~4.7Ohm)
- In this configuration and turn ratio the heater voltage is too high - it should be 3.6V when heated continuously, while with the published turn's ratio it would be about 7.2V, so filament would be overheated.
There is quite nice trick to solve that and at the same time ensure the filaments do not get any power after the start up:
Polarize the auxiliary windings so, their voltage add up to the main secondary.
Then connect in series with the ends (really the ends, not the part connected to the main secondary) auxiliary capacitors, whose value is Cfilament=Cballast*Nsecondary/Nfilament, where in your case the Cballast is the 1nF, Nfilament is number of turns of the filament secondaries, Nsecoondary is the number of turns of the main secondary. Usually it lead to ~33..330nF, it really have to match the formula.
How that work? When there is no arc in the lamp, the path across the main secondary is not conductive and the oscillator oscillate to the resonance frequency given by the primary capacitor only, so rather high. This make the auxiliary filament capacitors nearly a short circuit, so the filaments get the full 7.2..7.5V (if drop across the transistors would be zero, the rms is about 1.1x the supply voltage).
AT the same time the tube is exposed to full secondary voltage, so after the electrodes warm up, the arc ignite.
As the arc ignite, the lamp impedance become very small (way lower compare to the ballast capacitor), so nearly the full secondary voltage would be across the ballast capacitor. So what is there from end to end? So if we neglect (for the moment) the lamp filaments, the "filament" capacitors would be in series with the arc (negligible for this effect) and the ballasting capacitor. As they are in series, there is about the same current, so the voltage distribute in reverse ratio to the capacitances (still neglecting the arc drop). But giving the equation we used to design the filament capacitors, you could see the voltage across them would be nearly the same as the voltage across the filament winding. And because the filament is connected between these two points with nearly the same voltage, there is nearly no voltage across the filament. So if there is no voltage across the filament, the filament get no power.
The lamp arc voltage drop disbalance these voltages a bit, but because it is usually about 1/4 of the secondary voltage, it mean the voltage across the filament would be about the same 1/4..1/3 from the initial 7.2V, so about 1.8V, what mean reduction of the filament power to 1/16 compare to the power when the lamp is not ignited.
So viola, the circuit provide double filament voltage fro fast heat up before ignition, but stop feeding the filaments automatically after the lamp ignite, all that even without any switching relays or similar complexities...