I've been watching a few analyses of CRT flyback transformer drivers and can't understand why the transistor would shut off. To my naive perspective, it appears that the transistor should never shut off due to the positive DC voltage applied constantly to the base. I'm aware that this isn't the case since transformers need changing currents to induce currents in their other coils. I'm aware that the coils have inductance which dampens current buildup and charges the inductor core, but I can't wrap my head around the transistor region.

(Reference: https://en.wikipedia.org/wiki/Flyback_transformer#Operation_and_usage)

Circuit Diagram: Diagram

  • \$\begingroup\$ This is also a Joule Thief with a secondary. \$\endgroup\$ Commented Feb 27, 2022 at 21:34
  • \$\begingroup\$ Reduction in drive with increasing current may occur due to either transistor saturation due to limited Beta, and/or core magnetic saturation. \$\endgroup\$
    – Russell McMahon
    Commented Feb 28, 2022 at 0:36
  • \$\begingroup\$ @RussellMcMahon -- What if R1 becomes 3K3 or 10K? Then what blocks the oscillator? It's no longer transistor saturation or core saturation because you've "turned it down". Now the energy coming in through the resistor is only enough to support a certain amount of inductor charge-up, and then it collapses. I'm picturing a graph something like the safe operating area graph. \$\endgroup\$ Commented Feb 28, 2022 at 5:30
  • \$\begingroup\$ @RussellMcMahon -- What causes the hysteresis in the self-oscillating flyback converter? Isn't that the question here? Or what causes the transistor to turn off? What causes it to block? I originally thought it was transformer action between the primary and the sense winding. But now I think that it is also any of the many limiting factors. \$\endgroup\$ Commented Feb 28, 2022 at 6:44
  • \$\begingroup\$ @MicroservicesOnDDD Increasing collector current requires base drive of Ic/Beta. Feedback winding voltage and R1 R2 set Ibmax. If Ic rises above B x Ibmax transistor starts to come out of saturation and this regeneratively turns off transistor. OR if load current saturates core removing drive voltage same happens. \$\endgroup\$
    – Russell McMahon
    Commented Feb 28, 2022 at 13:09

3 Answers 3


You're nearly there.

This is part of the problem.

I'm aware that this isn't the case since transformers need changing currents to induce currents in their other coils.

While this is not wrong, it's not as useful as saying that transformers need voltage to induce voltage in their other coils. Currents do flow as well, but their relationship is more complicated than for voltage.

The feedback coil is in series with the base. This has the polarity (##) to drive the base to a higher voltage when the primary current is growing, when the primary has a positive voltage across it, and lower when it's negative. Once it starts to turn off, the feedback makes sure it turns off hard.

The resistor ratio makes sure there is enough voltage on the base to turn it on initially, when there's no voltage from the feedback winding.

The only other part of the puzzle is why the transistor should start to turn off anyway. There are two things that can do this, and the first to get there will trigger the end of the 'on' phase.

a) A transistor with base drive limited by R1 will only have a limited collector current it can support. Once the primary current has risen to this value, any further rise will pull the transistor out of saturation, and the collector voltage will rise, reducing the primary voltage. This will further reduce the base drive by transformer action to the base winding, and the transistor will quickly turn off.

b) The flyback core will magnetically saturate at a certain primary current. This causes the inductance to drop, which increases the rate of rise of primary current dramatically. It will now quickly exceed the transistor's limited collector current, whatever it was, and mechanism (a) will complete the switch-off.

(##) Thanks to Jonk for pointing out in comments that the polarity you might try to infer from that diagram is wrong. The absence of the explicit polarity indicating winding start dots should be a warning that this might be the case.

  • 1
    \$\begingroup\$ The (*) for polarity on the windings that might be naturally inferred from the schematic (since they are missing) would be wrong. Just a note. They should have included the polarity dot. \$\endgroup\$
    – jonk
    Commented Jun 2, 2018 at 5:17
  • \$\begingroup\$ So, with regards to the phases, the primary and secondary are in phase or are they 180 degrees out of phase? The other question I had was about the collector and base interaction. When the collector voltage rises, the transistor stops conducting until the collector current reaches 0? That's an interesting mechanism I hadn't heard of. \$\endgroup\$ Commented Jun 2, 2018 at 15:41
  • \$\begingroup\$ The diagram doesn't have a phase reference on it, so 'in phase' and '180 phase' are meaningless. The feedback winding is phased so that when the collector voltage falls, the base voltage rises. This will reinforce the conduction, and the switch-off. When the transistor turns off, collector I goes to 0, the flux initially stays the same, so the current transfers to the secondary and flows in the load. Once the magnetic energy has transferred to the load, the coil voltage falls to 0, the base voltage rises, and the transistor conducts again. Nothing to do with the collector current reaching 0. \$\endgroup\$
    – Neil_UK
    Commented Jun 2, 2018 at 16:06

This is a Blocking osillator, wikipedia

The transistor switches off because once the the voltage in the feedback winding starts to dip the trasistor drive weakens and when that happens the transistor begins to throttle the current back

The voltage goes down because the transformer is a real transformer, and not an ideal one so the primary current is limited by winding resistance, and the coupling is limited by core saturation.

  • \$\begingroup\$ You could write more details here. There are two possible mechanisms for why oscillation occurs. You mention neither, instead using phrases like "thottle .. back" and "dip." One mechanism is core saturation. The other is exhaustion of BJT \$\beta\$. Either one works. Positive feedback via the feedback coil takes care of the rest. You might also explain the other direction of positive feedback regarding the feedback coil that reinforces the turn-on period, too. It cuts both ways. \$\endgroup\$
    – jonk
    Commented Jun 2, 2018 at 5:39

The base is feeded through the feedback coil by a voltage divider. The primary coil is directly connected to the source. When the gate is ‘on’, current starts to flow to the primary coil and, choosing the path of least resistance, ignore the path to the feedback coil. This leads to the gate being closed, resulting in the current to the primary coil being stopped. And now it starts all over again.

  • \$\begingroup\$ Base? Source? Gate? \$\endgroup\$ Commented Aug 25, 2023 at 22:13

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