I guess this is pretty obvious, but when the fet turns on, it's a forward converter charging C1, and when it turns off it's a flyback charging C2.
-- John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard
On a sunny day (Fri, 24 Jul 2020 21:23:42 -0700) it happened snipped-for-privacy@highlandsniptechnology.com wrote in :
Yes, in principle. One thing to take into account is that for a high ratio of flyback to forward so a very short flyback pulse, is transformer winding heating A shorter pulse for the same power means more current during the flyback into the output. i * tflyback
However the ohmic losses in the coil go up with i^2 * Rcoil * tflyback This heating limits what you can do with reasonable wire gauge.
To get a decent size transformer a few more turns and a bit less flyback is simpler. Slower switching is easier on the driver too.
?
LOL, I have just invented something similar and wanted to share. My idea is to use the L output pin of a synchronous buck converter to supply a floating MOSFET gate driver elsewhere in the system using the same charge pump principle. Provided the PWM duty cycle is reasonable, it works like a charm -- just prototyped that to source 1mA.
It needs one more part than a flybuck (two diodes and one capacitor instead of an additional winding and a diode), but a cheap off-the-shelf inductor can be used, which is a big advantage.
Best regards, Piotr
I was referring to the doubler network connected to the drain, not to your secondary.
Best regards, Piotr
You get very peaky current in the forward converter if there's no output inductor, and it's unregulated.
There IS a converter that regulates both, with a conventional filter on the forward section. The forward stage is PWM, while the flyback section is PFM. There's some interaction between the two, if current mode is employed. This was documented in the mid-80s by Steigerwald at GE and I think there's a control chip for it.
Another uses a fixed frequency, but modulates the active clamping network for the forward converter, from the mid 90s.
A third uses two transformers, which is sort of silly if the switch and controller costs less than the magnetics - which is the current reality.
At low power, of course, you can get away with murder.
Your circuit appears to not regulate only the negative voltage generated by the flyback period - your waveform suggests complete energy transfer; so a designed rather than fudged magnetic part. If so, then why not just add another flyback winding?
RL
I guess I missed the point - this is a single output, so one regulation is as good as another.
Current mode control, or even simple limiting may be an issue.
RL
Does this really work
I haven't simulated it or dug deeper, but is seems the positive rail is sta ndard flyback, regulated with Ipeak squared like normally, while the negati ve path is a forward style converter, with no inductor in the buck part, so more or less just the input voltage and turns ratio?
You're missing two diodes and inductors for that.
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Design Website:
Don't follow that. Got a sketch?
-- John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard
I think Tim hints at that you need an inductor to have a buck part and diodes for the freewheeling part. Otherwise it won't act as the state space model dictates
State space model dictates?
The turn-on phase isn't a buck, it's a forward converter. Turnoff is conventional flyback. It might like a snubber on the primary, mostly for cosmetics.
I admit the dynamics is interesting, especially at startup. If I decide to use it, I'll certainly Spice it first.
Anybody with another idea could post a sketch.
-- John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard
State space is used to calculate the transfer function from converter states
Have you ever seen a forward converter without output inductance, a cap and freewheeling diodes?
Without it, it's just a DC transformer like the Royer and converts not according to a duty cycle, but rather a turns ratio
Cheers
Klaus
Conventional dots preserve voltage polarity, so the flyback portion is neg through D2.
When the switch is on, forward current passes through D1 with only leakage inductance and C1 shaping the current.
RL
Yeah, the negative rail is the flyback part
Was is the idea of this converter was really my point. You cannot regulate both rails at the same time except for a sweet spot?
But what does it dictate?
I'd use Spice anyhow. Real life is nonlinear.
Sure. Sold thousands. Of course there are output caps.
Sketch what you are thinking about.
The forward direction works like that. The direction that regulates is the flyback side.
-- John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard
The output part is symmetric, so the dots don't actually matter. They are just talking points.
-- John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard
There's only one output.
-- John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard
This is a charge pump. Vout = Vin * N2/N1. Regulation not possible. Transistors sink huge peak currents during startup.
This is a forward converter. Vout = Vin * D * N2/N1 (give or take DCM/CCM). Eminently regulable. Transistor sinks trapezoidal current corresponding to inductor charge.
This monstrosity might be better deserving of such a name,
Tim
-- Seven Transistor Labs, LLC Electrical Engineering Consultation and Design Website: https://www.seventransistorlabs.com/
What charge is being pumped?
The LT chip senses and limits peak fet current. As I noted, startup is interesting.
That one's not isolated.
-- John Larkin Highland Technology, Inc Science teaches us to doubt. Claude Bernard
You don't need the first CD pair in the regulation loop. It will simply increase fet loss at turn-on (which is going to be pretty severe anyways).
It's not a forward converter without an energy storage element - it's just a switched capacitor DC-DC.
RL
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