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.

enter image description here

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.

  • \$\begingroup\$ 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? \$\endgroup\$
    – Andy aka
    Jun 22, 2020 at 17:25
  • \$\begingroup\$ I have explained how I understand it with respect to my research and the circuit's site sir. \$\endgroup\$ Jun 22, 2020 at 17:38
  • 1
    \$\begingroup\$ @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. \$\endgroup\$
    – jonk
    Jun 22, 2020 at 20:42
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ Sep 23, 2022 at 1:57

1 Answer 1


Your circuit looks like this:


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.

Please add questions and I'll try and answer them. But please also first read this discussion, too. I think it will make a lot of sense.

  • \$\begingroup\$ Ok thanks a lot. I'll do well to go through it. \$\endgroup\$ Jun 24, 2020 at 19:57
  • \$\begingroup\$ @AndrewEffiong Let me know if there are remaining questions. \$\endgroup\$
    – jonk
    Jun 25, 2020 at 21:35
  • \$\begingroup\$ How is Rs not needed? Without Rs, you'd be applying Vcc directly across the transistor's base-emitter junction. \$\endgroup\$
    – Hearth
    Sep 23, 2022 at 13:48
  • \$\begingroup\$ @Hearth I didn't mean shorting out Rs. I meant removing it and opening up that spot. (I think I did mean that at the time, anyway -- it's been a few years.) \$\endgroup\$
    – jonk
    Sep 23, 2022 at 18:27
  • \$\begingroup\$ @jonk -- Without Rs, there is nothing to turn the transistor Q1 on the first time. So Rs is needed unless you change the input of L1 from ground to Vcc, thereby transforming the circuit into a basic Joule Thief with a third winding for voltage step-up or step-down. \$\endgroup\$ Jan 7, 2023 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.