For some time now, I have been trying to figure out what (in theory) would be the most efficient way transfer power across a transformer. Applications include high voltage power supplies and induction heaters. I have become very interested in the Royer oscillator and derivatives of it such as the "Push-Pull 2n3055 driver" mentioned here: http://wiki.4hv.org/index.php/Flyback_transformer. What bothers me about the Royer oscillator and its derivatives is that it uses two transistors to control two coils (center tapped) rather than four transistors to control one coil. I have a hunch that there exists a way to improve on the design using a full h bridge rather than a push-pull center tapped design.

I have been working on a driver, and I need some help determining how it might work/fail and/or scale.


The design is not complete yet, but I was hoping it would at the very least convey where I was going with this. At best, this is going to be the design I build on by adding some protection for the transistors and driver.

I tried to make the picture as illustrative as possible despite the bad quality. Please let me know if/when you have trouble reading it and I will try to clarify. The driver relies on 4 pairs of op amps with their inputs crossed as seen here How to make an h bridge using only comparators. comparator driver

(imagine it without the capacitors and resistors)

Each pair of op amps controls one transistor. They all switch according to the voltage generated by the feedback coil F. I think I have the wiring displayed correctly, but the feed back coil outputs may need to be flipped.

My understanding of transformers: As I understand transformers, a rate of change in the primary induces a change in magnetic flux which travels through the secondary. This induces a voltage in the secondary. The current that flows through the secondary in response creates a magnetic flux that cancels out that of the primary. The impedance of the load at the secondary then determines how much current flows through and therefore how much of that flux from the primary gets canceled out. The more flux that gets canceled the less the total flux in the core of the transformer changes. The less the total flux changes, the less resistance there is to a change in current at the primary. So the net effect is that when dI_primaryN_primary~dI_secondaryN_secondary, most of the energy is transferred from the primary to the secondary, and there is very little impedance at the primary. However, if there is something at the secondary that prevents current from flowing through it, then when the current through the primary changes, there is a net change in the magnetic flux through the core of the transformer. This net flux travels back to the primary and induces current that opposes the original change in current. In effect, it behaves like the magnetic field in an inductor.

The principle of operation of my circuit (or at least the one I am shooting for): When there is a large net change in magnetic flux through the transformer that would otherwise induce a back emf in the primary, a voltage is induced in the feedback coil which quickly switches the h bridge that is feeding the primary. This induce an emf that helps switch the direction of the current flow on the secondary. However, the very fast change in voltage at the primary would probably induce a change in flux a too large to be absorbed/canceled by a change in current at the secondary. This will cause some of the change in magnetic flux to again leak through the feedback coil and flip the h bridge. Now this would obviously not lead to a resonance effect at the secondary so much as a very fast oscillation at both if not for the fact that it takes a finite voltage to be induced in the feedback coil for the h bridge at the primary to be flipped. The amount of time it takes for that voltage to build up will depend on how well the secondary can absorb/cancel the change in flux sent by the primary. So when the change in magnetic flux sent by the primary can be absorbed/cancelled by the secondary, the feedback coil does not get enough change in flux to flip the h bridge. The net result (hopefully) is that the duty cycle of the h bridge at the primary changes as the phase of the oscillation at the secondary changes. The effect I am hoping for is that the average change in magnetic flux induced by the primary approximately cancels the average change in magnetic flux induced by the secondary--at least over the course of a many flips of the h bridge.

I could be completely wrong in both my theory and my circuit. I am still rather new to electronics and am learning the best I can.

  • 1
    \$\begingroup\$ What is your question? \$\endgroup\$
    – The Photon
    May 6, 2012 at 4:31
  • 2
    \$\begingroup\$ No rudeness intended - you are over analysing the circuit given you level of experience and you are "trying to reinvent the wheel". While you could set out to try and analyse the arrangements along the lines that you are doing, this is far from usual for practical designs except perhaps when people are aiming at optimising something that has been designed well and works well enough "as is" but a little "tweaking" may help. For most purposes standard transforme equations do all that you need. | Note that impedance is reflected as the square of the turns ratio. See below \$\endgroup\$
    – Russell McMahon
    May 6, 2012 at 4:57
  • \$\begingroup\$ ... Ideally Pin = Pout = Vin x Iin = Vout x Iout. || Vout / Vin = turns ratio = N say. | Iout / Iin = 1/N. // Zin = Vin/Iin / Zout = Vout / Iout.| Zout / Zin = Vout /Vin x Iin / Iout = N / (1/N) = N^2. \$\endgroup\$
    – Russell McMahon
    May 6, 2012 at 5:03
  • \$\begingroup\$ @ThePhoton I suppose I should have stated it better, but I really just wanted some input on my line of reasoning. So much of it is purely speculative that I thought I could learn a lot by asking people with more experience than I. \$\endgroup\$ May 6, 2012 at 5:27
  • 1
    \$\begingroup\$ Why not just build it and see if it works? Why ask us to speculate about your original research? \$\endgroup\$
    – The Photon
    May 6, 2012 at 5:35

3 Answers 3


Re transformer operation: winding terminal voltage = # turns * time rate of change of magnetic flux (Faraday's law, assuming 100% coupling).

  • The secondary current does not create "back emf"; with an appropriate core it induces primary current via ampere's law, as calculated in Russell McMahon's answer.
  • The transformer core can only support so much flux (the saturation flux levels = +/- Bsat * core cross-sectional area, with saturation flux density Bsat a property of the core material.)

Royer oscillator operation: Start with the driver saturated in one polarity.

  • The transformer flux "walks" at a steady rate from one saturation level to the other.
  • When the core reaches saturation, the flux stops growing, and the winding can no longer support the applied voltage.
  • The feedback connection circuit to the driver reacts to this saturation transient by switching the driver state to the other polarity.
  • the transformer flux commences "walking" in the other direction.

Resonance effects are only in play during the switching events; while the core is "walking" the driver remains in a fixed state.

Getting these switching events to happen cleanly is the hard part: sometimes the circuit won't switch at all, while in other cases the circuit oscillates at a high frequency with the driver active instead of switching, with consequent reduced efficiency.


Same as my comment but more readable as an 'answer':

Pin = Pout
= Vin x Iin
= Vout x Iout.

Vout / Vin = turns ratio = N say. Iout / Iin = 1/N.

Zin = Vin/Iin
Zout = aVout / Iout.

Zout / Zin = Vout /Vin x Iin / Iout
= N / (1/N)
= N^2.

  • \$\begingroup\$ I know there is a frequency dependence here. That is the whole point of the setup--to tune into a resonant frequency and maximize power transfer. \$\endgroup\$ May 6, 2012 at 19:38

I like the simplicity and symmetry of that complementary oscillator, but it has some issues with overdriving the Op Amp's (OA) inout via big cap overlap during crossover. But it works great as a chopper amp.

State of the art Switched Mode Power Supplies (SMPS) can run with 95% efficiency using small magnetics. Standard cost effective ones run 80~90% efficiency. That is after many years of research into the non-linear saturation effects of inductors, the nonlinear switched capacitance effects of MOSFETS or nonlinear current gains and saturation effects with bipolar junction transistors (BJT) and the need to manage dead time to prevent bipolar switched shorts across your power rails when dumping huge capacitance or currents.

Which part do you need help with?

I have a great book by an old colleague to design them for us while he was at Hammond, and I was at Burroughs. T Here are many avail I could suggest or you can easily find. PSU note

Part of index

Part of Keith Billing's Book


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