The circuit is based on this one shown in this video, visible at 2:45.
I simulated a simplified version, which still oscillates, shown in the picture below. What I don't understand: What causes the transistor to stop conducting at 2.7 ms?
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1\$\begingroup\$ Try this where you can see the animation of the simulation instead of just graphs falstad.com/circuit It looks similar but not the same as the joule thief so you can read about how that works. Might shed some light. \$\endgroup\$– DKNguyenDec 23, 2022 at 22:02
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\$\begingroup\$ @DKNguyen Thank you for pointing out the similarity to the joule thief circuit, which helped me finding more information about the functionality. \$\endgroup\$– hmutDec 24, 2022 at 6:24
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1 Answer
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When the transistor is on, it puts a voltage across the transformer primary, L1. This voltage, in turn, drives the transistor's base via the transformer's secondary (L2) and R1. As a result, when the transistor is on, it keeps itself turned on by supplying itself with base current via the transformer. This is positive feedback.
The positive feedback isn't infinitely strong, though. As the current through Q1 rises, it needs more and more base current to stay in saturation (and act as a closed switch). At some point, the base current permitted by R1 isn't sufficient for Q1 to deliver the current demanded by L1 anymore because Q1 has a limited current gain. Once this happens, the voltage across L1 starts to drop because Q1 doesn't allow the current to rise further. This, in turn, means that the voltage across L2 also drops (it's a transformer after all), and the current through R1 decreases accordingly, so the transistor is getting less base current. This means, however, that the transistor can drive even less current into L1, so the voltage across it drops further, and this continues until the transistor is fully off. So, as a result, the fact that the transistor is turning off makes it turn off even faster - positive feedback again. The consequence of this positive feedback is that the transistor turns off almost in an instant once the current through L1 gets too high.
Note that this positive feedback during turn-off is only possible when the transformer has a turns ratio greater than one, which is the case in your simulation. In other words, L2 must be larger than L1.
Once the transistor is fully off, the voltage at its emitter drops almost without bound as the inductor is trying to discharge itself. If you physically build this circuit, the transistor will get destroyed by this turn-off voltage spike, especially since it's also applied to the base via L2. You will have to clamp the negative voltage spike somehow if you want to actually use this circuit. (Ideally, you'd use it to drive some kind of load, like a LED.)
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\$\begingroup\$ most self-oscillating power supply circuits of this type rely on the transformer core saturating to stop the drive to the transistor and cause the flyback period to begin, as far as I remember. \$\endgroup\$ Dec 24, 2022 at 3:09