What is wrong with my Baxandall (Resonant Royer) converter? - Page 1
Forum section: electronics / projects

author: beowulfenator
author_rank: Contributor
timestamp: June 16, 2025, 10:01:11 am
content: I am trying to build a resonant converter using the schematic in Wikipedia. My primary winding is 10+10 turns on a ferrite toroid core. According to the original 1959 article by Baxandall, the feedback winding should be about 1/10 of the primary. Mine is 3 turns, but I tried other values with little effect. It is recommended to have the input inductance to be at least 5x the primary, mine is probably 10x, also wound on a separate small ferrite toroid. The capacitor is 33nF ceramic, the resistors are 2K (I've tried other values, no effect). The transistors are BD139 NPN. The secondary is wound to get about 5V RMS.Under little load (50 mA) there is a good sine wave at the secondary winding. As the load increases, the sine wave becomes distorted and asymmetrical and the period increases slightly. At about 200mA something breaks and I think both transistors open at the same time because the current increases significantly and the power supply goes into constant current mode. Curiously, there are two different waveforms that I see when the converter fails, as if one half of the circuit fails, but not always the same half.What am I doing wrong here? Is it even possible to get this circuit to generate e.g. 500mA@5V RMS?P.S.: Probes are 10x, so it's actually 5V per division.

author: moffy
author_rank: Super Contributor
timestamp: June 16, 2025, 10:26:16 am
content: For a resonant converter to work the load impedance needs to be a reasonably higher value than the resonant impedance of the resonant components or it will kill the oscillation.

author: beowulfenator
author_rank: Contributor
timestamp: June 16, 2025, 11:16:29 am
content: My load is purely resistive, 10 ohms at the most. At resonance, reactance of the LC circuit is effectively zero, so its impedance is just its active component. Are you saying that the sum of the ESRs of the capacitor and the primary winding at 15KHz should be significantly lower than 10 ohms? In other words, I need a better capacitor than the ceramic I have there and thicker windings?

author: MariuszD
author_rank: Regular Contributor
timestamp: June 16, 2025, 11:53:47 am
content: How many turns on secondary?

author: beowulfenator
author_rank: Contributor
timestamp: June 16, 2025, 12:04:49 pm
content: Nine turns on the secondary. RMS on the 10+10 turn primary is roughly 2.2*5V = 11V, so RMS on the 9 turn secondary is roughly 5V.

author: moffy
author_rank: Super Contributor
timestamp: June 16, 2025, 10:35:34 pm
content: Quote from: beowulfenator on June 16, 2025, 11:16:29 amThe reactance at resonance is that of the individual L(jwL) and the C(1/jwC), that is what the load impedance needs to be larger relative to, it's called the Q of a resonant circuit and if it goes too low it will cease to resonate.

author: TimFox
author_rank: Super Contributor
timestamp: June 16, 2025, 10:41:40 pm
content: Quote from: moffy on June 16, 2025, 10:35:34 pmThe circuit is aparallel resonance, so ideally (no loss whatsoever) the impedance isinfinite, not zero. Series resistance in the inductor and load resistance lowers theQto a finite value, and the resistance seen by the transistors at resonance is reduced accordingly, lowering the loop gain. You need total loop gain > 1 for oscillation.

author: beowulfenator
author_rank: Contributor
timestamp: June 17, 2025, 01:25:50 pm
content: So what can I do to make this circuit work with higher loads?

author: mtwieg
author_rank: Frequent Contributor
timestamp: June 17, 2025, 01:48:52 pm
content: Quote from: beowulfenator on June 17, 2025, 01:25:50 pmMost straightforward way is to reduce the characteristic impedance of the resonant circuit such that it's much lower than the load resistance, thus increasing the Q.Basically this means decrease inductance (all inductances) and increase capacitance, both by the same ratio to maintain the same resonant frequency. Hopefully you won't run into core saturation with the transformer though...

author: beowulfenator
author_rank: Contributor
timestamp: June 17, 2025, 01:56:47 pm
content: What happens if I only increase the capacitance? Frequency goes down, but so what?I've tried increasing the capacitance by soldering another capacitor parallel to the one in the circuit. The frequency does change (i've doubled the capacitance, and the frequency did decrease by one half). There issomeeffect on how much I can load the circuit, but the effect is minimal. I'd say doubling the capacitance increases the max load by 5%. Anything else I can try?

author: moffy
author_rank: Super Contributor
timestamp: June 17, 2025, 02:26:26 pm
content: Quote from: beowulfenator on June 17, 2025, 01:56:47 pmIncreasing the capacitance will reduce the resonant frequency, so the jwL will be lower and the capacitive impedance, which will be close to a negative value of the inductive impedance. Take care though that you don't saturate your magnetics as the lower frequency will increase the V*t across the core (Volts by time).

author: mtwieg
author_rank: Frequent Contributor
timestamp: June 17, 2025, 08:01:28 pm
content: Quote from: beowulfenator on June 17, 2025, 01:56:47 pmDoubling capacitance should cause frequency to decrease by a factor of 0.707 at most...QuoteThere are other reasons why the loading is limited, but it's very hard to tell without more details on the circuit. The magnetics are the biggest unknown factor here.

author: beowulfenator
author_rank: Contributor
timestamp: June 18, 2025, 10:58:05 am
content: Quote from: mtwieg on June 17, 2025, 08:01:28 pmOh, it's the square root of LC. You're right, I didn't actually measure the frequency, just saw the decrease.