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> > But when Q1 is off, where does the energy go? Does it not get wasted as
heat
> > in the diode and feedback windings? When Q2 is off, it certainly does
not
> > get transferred to the output. Slow decay or not, the inductor is going
to
> > discharge through the diode. You charge the inductor, but then don't do
> > anything with the energy you stored there.
>
> But do we need to do anything with that energy??
> Why don't we just leave that energy in the inductor?
> Then when Q1 next turns on the inductor is already
> charged so next energy is fed to the load?
>
> Slow-decay seems to be a black art for people that
> never work with large motor drivers or stepper motor
> drivers, but it's pretty simple really. The energy
> stored in the magnetic field of an inductor can be
> discharged into a load, or kept within the inductor.
> The rate the energy is discharged relates to the
> load. With a very high load (short circuit) then the
> inductor cannot release the energy, as this current
> causes magnetic field in the inductor...


OK - energy storage between files can be achieved if "done properly".
Here is a circuit which DOES work and which is "reasonably" efficient.
Many different versions of this have been built and used in practical
situations.
It is similar in general concept to Roman's circuit.
It uses energy storage between power delivery cycles but in an inductor /
capacitor system.

Note that this circuit does NOT operate in quite the way that many others of
seemingly similar topology do.

The asterisks indicate coil "dottings". The dotted ends have the same
"winding sense" on the core.
The circuit is a FORWARD converter. Power is delivered when the transistor
is turned on.

C1 is an UTTERLY crucial part of the circuit. C1 / L1 resonate and the
switching frequency is largely controlled by this resonance.

R1 is used largely to provide startup bias to the transistor.
R2 limits drive current when operating.

The junction of R1/R2 operates at a DC level BELOW ground when operating
!!!!
Look at the circuit and think it through.
Clue - Vbe of Q1 is about 0.6 volts and L3 is designed to have a voltage
during turn on of several volts.

As shown the circuit provides an output voltage approx N_l2/N_l1 x Vin.
Best for fairly constant Vin as from eg car battery or similar.

This circuit can be made to produce a lower voltage than this by providing a
feedback loop which turns the oscillator off when Vout reaches Vdesign.
Typically a zener, transistor and a resistor or two. Adding a flyback diode
to ground (cathode to cathode with D1), a series filter inductor and output
capacitor would result in a fairly conventional forward converter.

Apart from the "annoyance" of several coils on the inductor this is a
remarkably simple circuit for the results it can produce.

The waveform on the collector of Q1 is educational.
It is an approximate half sinusoid ABOVE Vin during the off half cycle and a
saturated grounded  square wave during the on cycle. Switching times are
much better than might be expected for a circuit with a resonant collector
tank.
The magnitude of the collector ring above Vin can be controlled by varying
C1.

A limited amount of output above Vin may be taken from the collector during
ringing but this is during the half cycle that is normally NOT used for
power so care should be taken to not excessively perturb the waveform.

Dave may be interested in trying this circuit in Spice and telling us
whether it works in simulation as well as it does in practice.



regards

                Russell McMahon


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