Re: Flyback circuits ?

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-qwest-dot-net>" <m.j.watts-at-massey.ac.nz>

Hi Jan, all,

On 24 Apr 2002, at 16:39, Tesla list wrote:

> Original poster: "Jan Wagner by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<jwagner-at-cc.hut.fi>
> 
> > I constructed a push-pull type circuit for my flyback experiments
today. I am
> > gettin much less voltage than with the single transistor circuit. Is this
> > normal? If not, any ideas what I have going on?
> 
> Flyback is not push-pull. These are two different switching
> power supply topologies that differ in some ways. Flybacks store the
> energy in the specially-designed air gap of the ferrite core. The longer
> the on-time of the transistor, the more energy will be stored, and the
> higher the voltage & energy delivered on the secondary during the
> transistor off-time.
> 
> For flyback, you generally want more than 50% duty cycle, maybe 90%.
> Push-pull is just one transistor too many - the extra transistor is equal
> to doubling the drive freq of flyback (at least in this case).
> 
> You can still continue with push-pull, but then you should keep the
> on-times of each transistor at <45% of their maximum on time (which
> is 50% of the full cycle). Ok maybe sounds complicated... but, with a
> dual output PWM IC you probably already are using this won't be a problem.

Additionally, the output of a pushpull circuit is clamped to  Vsupply 
multiplied by the ratio of the secondary to 1/2 of the primary 
winding. Formally, Vo = Vs.Ns/Np  where Np = one side of the primary.

The output voltage of a true flyback with no energy recovery winding 
is limited by energy stored in the core/airgap and the shunt 
capacitance of the secondary winding according to Vo = SQRT(2E/Cs) 
where E=1/2LI^2 where I is the peak primary current. Hence the output 
voltage of a flyback converter can easily exceed that of a pushpull 
converter by a very significant amount.

Regards,
malcolm