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What is the name of the topology of the oscillator shown below?

I try to understand how it works and what are the design equations. But in order to find that, I first have to know the name of the things. (There is no schematics entry on google...)

enter image description here

Or maybe someone has time to thoughtfully explain the system...

EDIT:

I am convinced now: the load is part of the oscillator. I edited the schematics and included it.

I added a name to each net as suggested.

The source is a Youtube chain of French EE that does fancy stuff: Four à induction : Incroyables Expériences [81] The schematics can be seen at the very end of the video.

I have also simulated it on LtSpice. The slight imbalance between R1 and R2 start the oscillation. Equal resistors do not work (at least in simulation) but nothing is really equal in the real world...

This takes time to start (~10 ms) but it oscillates at 100Khz as expected (by the youtubers).

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  • \$\begingroup\$ It doesn't look like this circuit oscillates at all. One of the transistors will be latched on and the other will be latched off. Are you sure you copied it correctly? I suppose that once the IR drop of the transistor that's on rises to the point that the other transistor starts to turn on, the circuit will switch states, at which point, the stored energy in the inductor will destory the MOSFET. \$\endgroup\$
    – Dave Tweed
    Commented Apr 18, 2014 at 17:52
  • \$\begingroup\$ Also, please add labels to the nodes of your circuit so we can discuss it. \$\endgroup\$
    – The Photon
    Commented Apr 18, 2014 at 17:53
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    \$\begingroup\$ I vote for "relaxation oscillator" or "astable multivibrator". \$\endgroup\$
    – jippie
    Commented Apr 19, 2014 at 14:17
  • \$\begingroup\$ I pretty much bet L1, L2 and L3 are on the same transformer core. \$\endgroup\$
    – Janka
    Commented Jan 28, 2017 at 22:32

6 Answers 6

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I don't know the name of this oscillator, but the operation seems relatively straight forward.

  1. Assume current flows through L1 at the beginning and M1 is shutting off. This generates a voltage spike at Vout-, which is large enough to overcome the reverse-breakdown voltage on D2.
  2. This voltage spike hits the gate of M2 and turns it on for a short time. This creates a current pulse through L2 and M2.
  3. When M2 begins to shut off, L2's inductance causes a voltage spike at Vout+. That gets sent to the gate of M1 through D1 once again by action of the reverse-breakdown voltage.
  4. M1 begins to generate a pulse current through it and L1 and we're back at step one.

This works because of the inductors creating voltage spikes higher than the 24V input and because the Schottky diodes are rated at 30V for reverse-breakdown. R1 and R2 are there to start up the oscillations. Without them, M1 and M2 would never turn on at all. They're also different values which supports the idea that they're just there to start the oscillations by allowing one side to temporarily override the other side.

The reason the Schottky diodes don't get fried is due to the capacitance of the FET's gate being the only path to ground. The capacitance blocks excessive current from flowing through the Schottky. Excessive current during reverse-breakdown is the mechanism of failure in them due to the heat that would be generated. That never happens here because there's never enough heat buildup to cause failure.

A basic simulation of the circuit topology shows this works. I've used different values and modeled the Schottky's as Zener's because that's more accurate use in this case because this simulator doesn't have an input for reverse-breakdown voltage:

Circuit topology simulation. Green waves are voltage while yellow is current.

Circuit topology simulation. Green waves are voltage while yellow is current. Vdd is only 5V.

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  • \$\begingroup\$ Those aren't zener diodes, they're Schottky diodes. \$\endgroup\$
    – Dave Tweed
    Commented Apr 18, 2014 at 19:18
  • \$\begingroup\$ You're right, they're not Zener's. I'll correct that in my answer. They are however being used as such through the reverse-breakdown voltage of the Schottky diodes. \$\endgroup\$
    – horta
    Commented Apr 18, 2014 at 19:51
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What you have there appears to be a Multivibrator:

A multivibrator is an electronic circuit used to implement a variety of simple two-state systems such as oscillators, timers and flip-flops. It is characterized by two amplifying devices (transistors, electron tubes or other devices) cross-coupled by resistors or capacitors. )

You have two cross-coupled amplifying devices, bridged by RC, so it "quacks like a duck and walks like a duck".

(The multivibrator is further classified by the stability that it exhibits. If yours indeed oscillates, then it is qualified as an astable multivibrator.)

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Normally this is a bistable circuit (FF) but addition of drain inductance with the diode capacitance yields a second order effect to get 180deg of phase shift to form negative feedback for biasing.

Thus it becomes an LC diode Astable circuit.

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I could see this thing oscillating with the inductors in resonance with the drain capacitance of the MOSFETs. Imagine M1 has just turned off (with some current flowing through the inductor just before turn-off) and M2 is on.. the voltage on the drain of M1 increases sinusoidally to a peak, then falls, when it has decreased below the threshold voltage of M2 minus a Schottky diode drop, M2 turns off and M1 turns on, then it's rinse and repeat in the other direction.

Start-up could be ugly, but at several hundred amperes the "On" MOSFET will come out of saturation inevitably (assuming ideal power supply and ideal inductors), and oscillation should start (perhaps involving destructive avalanching of the "On" MOSFET as Dave Tweed said in his comment), but it may latch up in real life and fry one of the inductors-- those are pretty hefty MOSFETs.

Edit: With the load, this is about as close as I can get to making it oscillate, with some resistance added to the inductors and a switch opening at 1usec to kick it. The current is 200A (!).

enter image description here

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  • \$\begingroup\$ Start-up shouldn't be so bad. Once the mosfets turn on, they force each other to shut off through the Zeners. The Zeners have a Vf of 0.2 to 0.8V which is well below the Vt of the FETs (~1.85V). \$\endgroup\$
    – horta
    Commented Apr 18, 2014 at 18:39
  • \$\begingroup\$ @horta Once one gets the advantage it turns on fully, and pulls the other gate off, so there is nothing turning it off as the drain current ramps up linearly with 24V on the gate.. \$\endgroup\$ Commented Apr 18, 2014 at 18:49
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I have found:

This is a variation of a ZVS usually found in Flyback controllers such as:

enter image description here

Original design by Vladmiro Mazilli.

Notice that if you remove the zener diode that are here to protect the N-MOSFET and the 10k resistors that seem to be small pull down for discharging the gates, you end up with almost the same design. There is also a separated inductor for the supply instead of the center point of the transformer.

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The circuit is in a muddled form. I have seen it on rubbish like "circuits gallery" . It is flawed and will eat fets for breakfast lunch and tea .People have had problems with this circuit and that is how I got to hear about it .When done properly it has been called the following names : ZVS royer .Tuned collector oscillator .Dynatron .Its an oldie but a goodie and in its origional form it has BJTs .People have wanted to use mosfets because they are faster but the "designer " did not understand start up biasing .If you iron out the bugs you can have something will have low EMC and low switching losses .

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  • \$\begingroup\$ Please try to use correct punctuation and spacing, this is difficult to read. \$\endgroup\$
    – pipe
    Commented Feb 18, 2016 at 11:40

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