# Royer Oscillator Working Principle

Summary:
Unable to understand
Hi,
I am studying Wireless Energy Transfer and I find Royer Oscillator in that.

Ref: Wikipedia
https://en.wikipedia.org/wiki/Royer_oscillatorI am unable to understand how it works.
I found a diagram here:
Ref: https://www.smps.us/inverters.html

It says that (quoted from webstie):
In practice, one of them (say, Q1) will turn on more and some voltage will appear across transformer's primary winding. This in turn induces voltages in the base windings of such polarities that they drive Q1 further into saturation and Q2 into cut-off state. As current flows through Q1 and the half of the primary, magnetic flux in the core increases linearly with time. At some point it will approach saturation when the flux can no further increase. The voltages across all windings will drop to zero and then reverse polarities. This will cause Q2 to conduct and Q1 to be in cut-off condition.

I am not understanding the bold part. Why will the polarity reverse? The flux has peaked, there will be no more emf induced, but then why will the polarity reverse.

NascentOxygen
Staff Emeritus
The voltage induced in windings = ##N\cdot\frac{d \Phi}{d t}##

As the core starts to saturate, flux increases by less and less, eventually there'll be a time when flux is steady and primary current has reached its maximum. Then as current decreases the core returns to linear operation.

jaus tail
Why will the polarity reverse? The flux has peaked, there will be no more emf induced, but then why will the polarity reverse.
The current what maintains the flux depends on the open transistor: but without the flux changing, the transistor will close and the current stop. No current => the flux will decrease, what means the other polarity of change => other polarity of voltage on the windings.

The key is, that any of the transistors will be kept open by the flux changing in the appropriate way.

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The voltage induced in windings = ##N\cdot\frac{d \Phi}{d t}##

As the core starts to saturate, flux increases by less and less, eventually there'll be a time when flux is steady and primary current has reached its maximum. Then as current decreases the core returns to linear operation.
Once the primary current reaches maximum, why will it decrease?

The current what maintains the flux depends on the open transistor: but without the flux changing, the transistor will close and the current stop. No current => the flux will decrease, what means the other polarity of change => other polarity of voltage on the windings.

The key is, that any of the transistors will be kept open by the flux changing in the appropriate way.
Are you talking about Q1 or Q2?

Suppose Q1 is initially on, so current path is

Then flux across N1 will saturate and reach a constant value. So emf across it will reduce and become 0. Then what will happen? It's inductor so eventually it'll be short circuit for DC current.

Why will current across N1 reduce if flux saturates and emf reduces? As per KCL there's no reason for current across N2 to increase and N1 to reduce.

Do T1 and T2 have some role to play? https://en.wikipedia.org/wiki/Royer_oscillator
https://electronics.stackexchange.c...-for-wireless-power-transfer-specially-mosfetBoth link say when T1 and T2 act like positive feedback so when Q1 is on, the T1 and T2 enable Q1 is on quickly and Q2 is off.

says otherwise. That the feedback act that it turns on Q2, and when Q2 is on, the feedback turns on Q1, and this brings oscillation.

It would be really helpful if you guys could help me as to how this oscillator works...

NascentOxygen
Staff Emeritus
Once the primary current reaches maximum, why will it decrease?
Because the transistor starts to lose its base drive. The transformer is no longer linearly "passing on" changes in primary current to its secondaries.

jaus tail
Then flux across N1 will saturate and reach a constant value. So emf across it will reduce and become 0. Then what will happen? It's inductor so eventually it'll be short circuit for DC current.
Correct, that's half of what happens and it is the inherent problem with this kind of oscillator. When the core saturates the current through the open transistor will produce a spike.

The other half is, that the voltage generated on the windings on the left bottom (T1 & T2) will drop to zero as well, and it will cause the transistor which is actually open, to close => the current through that transistor will drop to zero.

Are you talking about Q1 or Q2?
About the one which is open at the moment.

The circuit is wired so that as long as there is a flux change matching the current in N1 there will be a voltage generated in T2 what will keep Q1 open.
Once the core saturates and there is no further flux change then T2 will cease keeping Q1 open and the current in N1 stops. No current => the change of flux goes the other way, and through T1 it will generate voltage what opens Q2 instead.

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jaus tail
Correct, that's half of what happens and it is the inherent problem with this kind of oscillator. When the core saturates the current through the open transistor will produce a spike.

The other half is, that the voltage generated on the windings on the left bottom will drop to zero as well, and it will cause the transistor which is actually open, to close => the current through that transistor will drop to zero.

About the one which is open at the moment.

The circuit is wired so that as long as there is a flux change matching the current in N1 there will be a voltage generated in T2 what will keep Q1 open.
Once the core saturates and there is no further flux change then T2 will cease keeping Q1 open and the current in N1 stops. No current => the change of flux goes the other way, and through T1 it will generate voltage what opens Q2 instead.
When you say Q1 open, you mean it's short circuit right? How do you figure out dot convention for transformer?

https://www.assignmentpoint.com/sci...nsmission-via-magnetic-resonant-coupling.html

When power is applied, DC current flows through the two sides of the coil and to the transistor’s drain. At the same time the voltage appears on both gates and starts to turn the transistors ON. One transistor is invariably a little faster than the other and will turn ON more. The added current flowing in that side of the coil does two things. One, it takes away drive from the other transistor. Two, the auto-transformer action impresses a positive voltage on the conducting transistor, turning it hard ON. The current would continue to increase until the coil (transformer) saturates. The resonating capacitor C causes the voltage across the primary to first rise and then fall in a standard sine wave pattern.
Assuming that turned on first, the voltage at the drain of ’s will be clamped to near ground while the voltage at ’s drain rises to a peak and then fall as the tank formed by the capacitor and the coil primary oscillator through one half cycle.

Link says that when Q1 is turned on (current flows from its drain to source), then capacitor will charge. But why must current flow across Blue line? Why can't it flow across Red line?
The resonating capacitor C causes the voltage across the primary to first rise and then fall in a standard sine wave pattern.
I can't understand this point. Could you guys please explain?

When you say Q1 open, you mean it's short circuit right?
As long as we are talking about the principle, yes.

How do you figure out dot convention for transformer?
I don't. It's just that kind of circuit which is: if it works, then OK: if it does not work then just switch the wires

The second picture in your initial post is pretty good, I would stick to that one.

jaus tail
Ok... Thanks a lot for this help. I was struggling a lot to understand how this oscillator works.

anorlunda
Staff Emeritus
I don't. It's just that kind of circuit which is: if it works, then OK: if it does not work then just switch the wires

jaus tail
Tom.G
How do you figure out dot convention for transformer?
Summary: Unable to understand

I found a diagram here:
Ref: https://www.smps.us/inverters.html

The dot convention is "All dots are the same polarity".*
(see https://www.electricaltechnology.or...sing-the-dot-notation-and-dot-convention.html)

In the above diagram, the + supply is connected to the transformer Center Tap (CT). If Q2 has current flowing, the lower half of the primary is carrying the current. Since this connection has the Dot, which is connected to +supply, then all the Dots will be positive as long as the flux is increasing.

The two lower windings on the left are connected to the transistor Bases. The upper part of this winding has the Dot so it will be positive. This puts a positive voltage on the Q2-Base which keeps Q2 turned on while the flux is increasing.

When the core reaches saturation, the flux stops increasing and the induced voltages on the lower winding (and on the other, output, secondary) fall to zero. Q2 turns off because of loss of base drive.

As the flux collapses, it induces an opposite polarity voltage in the windings. The polarity reversal causes a positive voltage to Q1-Base, which causes Q1 to conduct.

Then all the above causes Q1 and its part of N1 to act as Q2 etc. acted. So Q2 and Q1 conduct alternately; or the circuit oscillates.

Hope this helps.

Cheers,
Tom

* "Same polarity" is the usual convention... unless you talk to some transformer manufacturers. Some manufacturers at least used to consider the Dot to represent the physical Start of the winding. (What a source of confusion!)

jaus tail
The dot convention is "All dots are the same polarity".*
(see https://www.electricaltechnology.or...sing-the-dot-notation-and-dot-convention.html)

In the above diagram, the + supply is connected to the transformer Center Tap (CT). If Q2 has current flowing, the lower half of the primary is carrying the current. Since this connection has the Dot, which is connected to +supply, then all the Dots will be positive as long as the flux is increasing.

The two lower windings on the left are connected to the transistor Bases. The upper part of this winding has the Dot so it will be positive. This puts a positive voltage on the Q2-Base which keeps Q2 turned on while the flux is increasing.

When the core reaches saturation, the flux stops increasing and the induced voltages on the lower winding (and on the other, output, secondary) fall to zero. Q2 turns off because of loss of base drive.

As the flux collapses, it induces an opposite polarity voltage in the windings. The polarity reversal causes a positive voltage to Q1-Base, which causes Q1 to conduct.

Then all the above causes Q1 and its part of N1 to act as Q2 etc. acted. So Q2 and Q1 conduct alternately; or the circuit oscillates.

Hope this helps.

Cheers,
Tom

* "Same polarity" is the usual convention... unless you talk to some transformer manufacturers. Some manufacturers at least used to consider the Dot to represent the physical Start of the winding. (What a source of confusion!)

One last question (a bit silly maybe). Why would the flux collapse? I mean in case of a capacitor once we charge it, then unless there is a closed path (preferably with a R in series) the capacitor won't discharge (assuming leakage to be 0)
But in circuit that we considered, the flux has reached its peak so why would it fall down?
Is it that there is no current so there won't be any flux? But I'm not using that flux anywhere. So where would all the flux go?

Tom.G
You are welcome

So where would all the flux go?
That, sir, is a question for the advanced physics folks. It is somewhat like asking "Where does the light go in a closed room when you turn off the lamp?"

If you want a very poor working analogy, think of either a stretched rubberband or an inflated balloon. Where do they go when you release them or evacuate the air? Just don't repeat this to any physics folks, they will be laughing so hard they won't be able to give a real answer!

Cheers,
Tom

jaus tail
You are welcome

That, sir, is a question for the advanced physics folks. It is somewhat like asking "Where does the light go in a closed room when you turn off the lamp?"

If you want a very poor working analogy, think of either a stretched rubberband or an inflated balloon. Where do they go when you release them or evacuate the air? Just don't repeat this to any physics folks, they will be laughing so hard they won't be able to give a real answer!

Cheers,
Tom
Lol.. Thanks again :)

But in circuit that we considered, the flux has reached its peak so why would it fall down?
Is it that there is no current so there won't be any flux? But I'm not using that flux anywhere. So where would all the flux go

This is a profound question.
Because the magnetic flux is an energy, it does not disappear in space.
Just consider the following diagram : -

Assuming that the leakage flux between the two coils is zero and the magnetic coupling is 100%, if the initial current in the primary coil is I1, the magnetic flux will not disappear instantaneously when the switch is turned off, which means that the current in the secondary coil, namely I2, will be mystically and instantaneously created at the same moment in order to maintain the same magnetic flux.

In this case, the following equation holds : -
I1*N1 = I2*N2 => I2=I1*N1/N2
where N1 and N2 are the numbers of turns of the primary and secondary coils, respectively

In addition, when the load is zero, namely shorted circuit, I2 will remain unchanged forever (left side of the diagram), and when the load is lossy, I2 will gradually decrease to zero (right hand side of in the diagram)

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jaus tail
This is a profound question.
Because the magnetic flux is an energy, it does not disappear in space.
Just consider the following diagram : -View attachment 251521

Assuming that the leakage flux between the two coils is zero and the magnetic coupling is 100%, if the initial current in the primary coil is I1, the magnetic flux will not disappear instantaneously when the switch is turned off, which means that the current in the secondary coil, namely I2, will be mystically and instantaneously created at the same moment in order to maintain the same magnetic flux.

In this case, the following equation holds : -
I1*N1 = I2*N2 => I2=I1*N1/N2
where N1 and N2 are the numbers of turns of the primary and secondary coils, respectively

In addition, when the load is zero, namely shorted circuit, I2 will remain unchanged forever (left side of the diagram), and when the load is lossy, I2 will gradually decrease to zero (right hand side of in the diagram)
Thanks for this.