# How to determine parameters for the Resonant Royer (or Baxandall) circuit?

I'm learning about the Resonant Royer circuit (or Baxandall circuit). There is a picture from the wikipedia(Royer circuit): Or if you like, you can see my simulation. I learn about how it works from this website. But it didn't show how to calculate the parameters, such as the value of R2 and R3, the value of C1. It did tell that the oscillating frequency is determined by resonance of magnetizing inductance L1 and C1: F=1/(2π×sqrt(L1×C1)). But the formula isn't consistent with the simulation result. The value of the transformer I choose is as follows: So according to the formula. The frequency is 318kHz. The calculation is as follows: But the simulation result is 115kHz. Does anyone know to design it and determine the circuit's parameters? I can't find many resourses about it. Thanks for help! I can't find the buttom to upload the simulation file, and if you need to check out my simulation, please tell me how to upload it or where I send to.

Second edit: I found a picture which is a simplification of this circuit： So it's a LC resonant circuit? Because the impedance is small when parallel resonant, So I think that the inductor L1 is to act as a constant current source. I guess that L1 has little to do with oscillation frequency. But I still do some calculations following Tim's suggestion. My calculations are as follows:  Although these calculation results are not in very good agreement with the simulation results, the situation 2 is closest to the simulation results.

• There are three inductors in the circuit: P1, P2 and L1. Can you find a combination of them (series or parallel) which gives the correct result? Then, can you explain (say by coupling factors between them, and drawing equivalent circuits for one or the other transistor saturated) the result? Mar 12 at 11:51
• If you have you have two identical inductances connected in series on the same core. Then the total inductance will be 4*L1 = 4*20µH = 80µH, Thus the resonant frequency will be equal to around $\frac{1}{2\pi\sqrt{80\mu\text{H} *25\text{nF}}} = 112.5\text{kHz}$
– G36
Mar 12 at 15:19
• @G36 Thanks for help! Could explain more about the 4*L1 = 4*20µH = 80µH? How is the coefficient 4 obtained?
– T L
Mar 13 at 3:26

You have L1 and L2 wound on the same core. If we connect them in series we double the number of turns thus, inductance must increase four times. This is why the oscillation frequency is around $$\F = \frac{1}{2 \pi \sqrt{80 \mu H *25nF}} = 112.5kHz\$$. And peak output voltage will be equal to:
$$\ \large V_{out} = \frac{\pi Vcc}{\sqrt{\frac{4 \times L_1}{L_{S1}}}} = \frac{\pi *50V}{\sqrt{\frac{80\mu H}{3.2 \mu H}}} \approx 31Vpeak\$$