MrAl,
Here is an example of doing an "experiment" and coming to the wrong conclusion. I said back in post #283 of this thread that the only way to bring a BJT out of saturation was to cut off its base current or increase the reverse collector-base voltage. You are doing the latter, and concluding that it is the collector current that is the reason for coming out of saturation. It is really the increase in the c-b voltage that brings it out of saturation. From what I see of your plot, the current of R3 remains constant throughout.
You can say that the cb voltage indicates that the transistor has come out of saturation, but how do you think that cb voltage increased? It increased because Ic increased. That's the point.
Anyway, this does not apply to your oscillator circuit. The two diodes across Q2 will limit any increase of c-b reverse bias voltage, so that base current cut off is your only option. So I stand by my statement that it is the "inductive kick" of the coil that puts a positive voltage on Q1 and thereby cuts off both Q1 and Q2.
Ratch
It does apply, and we dont really need an inductive kick we just need an increase in voltage. That voltage gets coupled through the cap into the base of Q1.
You can see in the third (last) experiment posted that the inductor voltage does not 'kick', it just stalls. The 'kick' comes in only AFTER the cap couples the signal to Q1 and turns it off and Q1 turns off Q2. When Q2 turns off, then you get the inductive kick, and that's the only way. It's the Q2 cutting off that changes the current and that's how the kick happens. That's how it happens with any inductor even with a switch. When you turn the switch 'off' the kick occurs, and not before then. Look at the waveform in that diagram and see that since Q2 always has a bias the inductor does not kick. In the osc circuit Q2 bias is removed after the collector voltage rises so that's how the inductor gets to kick and give us the required boost in voltage. But that can only happen after Q1 turns off Q2, and Q1 turns off when Q2 collector voltage RISES sharply as it pulls out of saturation, but at that point Q2 still has a strong bias so it is still conducting.
Note also that there is no reason why Q2 can not conduct with either a low Vce or high Vce.
Again, the sequence of events is as follows:
1. Q2 turns on.
2. Current rises in inductor
3. Current exceeds that which Q2 can hold with it's current bias
4. Vce voltage starts to rise
5. Vcb gets higher, transistor comes out of sat.
6. Voltage rises Vce
7. Rising voltage couples through cap to Q1 base, turning it off quick
8. Q1 turning off turns off Q2
9. Q2 turning off gives rise to inductive kick back and high boosted voltage Vce
If you dont agree with that sequence of events then list your own idea of how you think it happens. You really should do a circuit simulation first though because it looks like you havent done that yet.