# Simple self-oscillating flyback converter

I am studying switching power supplies, starting from the simple ones. I've just concluded looking at the joule thief and the blocking oscillator, and now I am progressing to self-oscillating flyback converters (or RCC).

I don't understand how the transistor switches off. It seems the transistor's switching off is quite a bit more complex than the switching-on process. Some say the operating principle is very much the same as that of a blocking oscillator, but I think the are some differences.

Here I understand that a small current flows through the start-up resistor to the base of the transistor, to initiate the turn-on process. As the transistor is partially on (active region), collector current will flow and energize the primary winding, and as a result, a same-polarity voltage will be induce in the feedback winding which will create a positive regenerative feedback and conduct more current to the base of the transistor, and the transistor will turn on fully (saturation).

That's all I know. I don't know how the transistor switches off. I could make some assumptions based on my knowledge of the joule thief circuit, but I am not sure. I also don't know how the RC circuit affects the switching frequency.

• Where did the circuit come from? Did it not have an explanation where you found it? There must have been some reason for you selecting it? Did you understand a joule-thief circuit properly as in the transformer core saturation? Jun 22, 2020 at 17:25
• I have explained how I understand it with respect to my research and the circuit's site sir. Jun 22, 2020 at 17:38
• @AndrewEffiong Is this coupled with a simple redrawing of your circuit sufficient? Or do you still have questions? (Note the dotted inductors and their arrangement.) It looks plain enough to me. And, assuming that you really have mastered the Joule Thief arrangement, I'd imagine this is no difficulty.
– jonk
Jun 22, 2020 at 20:42
• This question probably should've been closed anyway, but the duplicate it's marked as is clearly not what has been asked. A question such as electronics.stackexchange.com/questions/358135/… is a much closer duplicate. Sep 23, 2022 at 1:57

simulate this circuit – Schematic created using CircuitLab

In reality, $$\R_\text{S}\$$ is probably not needed. But it is another degree of freedom, so it's not necessarily excluded. But for a simple circuit I'd probably eliminate it.

There is a problem with all these circuits. That is the fact that $$\Q_1\$$ above might every well have its $$\V_\text{BE}\$$ reversed-biased by far more than its avalanche voltage. So a diode may be helpful in this circuit to snip that.

If you carefully read my discussion here, you'll see why the basic circuit should work. The only addition here is $$\L_3\$$, which is arranged so that when $$\L_2\$$'s magnetic field is collapsing, $$\L_3\$$ is polarized correctly for $$\D_1\$$ as shown above.

So the stored energy is positioned well to be discharged via a drop across $$\D_1\$$ to charge $$\C_1\$$, over time.

It's not all that complex.