STEORN DEMO LIVE & STREAM in Dublin, December 15th, 10 AM

[MileHigh]

K4zep:

if you have induction  and a magnetic field INSIDE the core from nominal external excition (power supply), does the passing magnetic fields interact with the internal field and in return show up in the coil when it is balanced or cause a special type of field anti-torque reflected back into the magnetic rotor?

The effects passing magnetic field are simply added to the internal field generated by the battery.  It is as simple as that, you add them together.

If the coil is open-circuit then the passing magnetic fields from the magnets flying by induce a small EMF in the coil.  If the coil is short-circuited or connected to a low impedance battery or power supply, then the passing magnetic fields induce a current in the coil that is added to the DC current from the power source.  It is hard to tell from your setup but the small ripple that you see in your current waveforms in your most recent clip could be partially caused by this effect.  It could also be related to where you are connecting your probes in the circuit and the overall setup.

Obviously with the coil around the core on, it is shielded from external fields

As per what I say above, when the coil around the core is on, it is not shielded from the external fields.  The fields generated by the battery and the moving magnets add together, where the field inside the core is dominated by the field generated by the battery.

Stephan:

BUT you can extract about 90 % of the inputted electrical energy via BackEMF extraction.
Then you have almost the rotational energy due to the attraction of the magnet to the ferrite core FOR FREE !

You are incorrect here.  K4zep pointed out the exponentially rising current waveform on his scope trace and stated that's where the inductor gets charged with energy and he is correct.  Once the inductor is fully charged and the current levels off, then it is just acting like a resistor dissipating energy as heat.

In Sean's demo clip, you can see how the exponential rise to charge the coils with current is around only 2% or less of the ON time.  Therefore, the back EMF that you can get from the coil discharge is only about 2% or less than than the total amount of electrical energy expended to keep the core saturated.   In K4zep's most recent clip you can see the exponential curve of coil charging current levels off after about 10% of the total ON time.  Therefore in his setup you can only recover 10% or less of the total electrical energy expended to keep the core saturated.

So there is nothing free about this process if you want to recover the back EMF after the coil has done its job of saturating the ferrite core.  You have to put out a lot more energy to make the core "disappear" than you can get back in any back-EMF recovery circuit.  In a Joule Thief circuit, the charging current in the collector coil is automatically shut off once the coil is fully charged with current, it is a self-regulating process.  Therefore a Joule Thief would return a much higher proportion of the charging energy in its inductive kickback.

MileHigh

[hartiberlin]

P.S: If the cap voltage will go higher than the supply voltage the greatz bridge will
not conduct during the normal input current operation but only
be charged when the BackEMF spikes occur.

So you need to have a cap that can stand higher voltages than the supply voltage.

So probably the boost supercap alone will not work at just 2.7 Volts or you need to
put a few in series.

[lumen]

ould you remove the diode across L1 and instead use a graetz bridge
rectifier across L1 and use your boost cap at the output of the graetz bridge ?

I think this would be possible if you disconnected both ends of the coil as in Broli's design.
You would need a setup like an H bridge to drive the coil.

[k4zep]

Hi Ben,
many thanks for the circuit diagram.

Could you remove the diode across L1 and instead use a graetz bridge
rectifier across L1 and use your boost cap at the output of the graetz bridge ?

Will be interesting to see, how fast this boost-ultracap will charge up.

Probably pretty fast.

Many thanks.

Regards, Stefan.

High Stefan,

Can't do that as part of the bridge would short the coil in the wrong direction due to normal wiring but I can put a single 1/2 wave diode in there just to see what will happen.  Draw it out, will understand.  If had a separate winding like Bedini could do it but with the quadf coil, no inductance!!!!!!!!  a gotcha there.

Next post will show results.

Respectfully
Ben

[hartiberlin]

Stephan:

You are incorrect here.  K4zep pointed out the exponentially rising current waveform on his scope trace and stated that's where the inductor gets charged with energy and he is correct.  Once the inductor is fully charged and the current levels off, then it is just acting like a resistor dissipating energy as heat.

In Sean's demo clip, you can see how the exponential rise to charge the coils with current is around only 2% or less of the ON time.  Therefore, the back EMF that you can get from the coil discharge is only about 2% or less than than the total amount of electrical energy expended to keep the core saturated.   In K4zep's most recent clip you can see the exponential curve of coil charging current levels off after about 10% of the total ON time.  Therefore in his setup you can only recover 10% or less of the total electrical energy expended to keep the core saturated.

So there is nothing free about this process if you want to recover the back EMF after the coil has done its job of saturating the ferrite core.  You have to put out a lot more energy to make the core "disappear" than you can get back in any back-EMF recovery circuit.  In a Joule Thief circuit, the charging current in the collector coil is automatically shut off once the coil is fully charged with current, it is a self-regulating process.  Therefore a Joule Thief would return a much higher proportion of the charging energy in its inductive kickback.

MileHigh

Hi MH,
in a coil you have L and R.
The rising current waveform has the function of:

i(t)= Imax x ( 1- e ^(-t/tau) )

where tau=L/R
if you switch on the current only for 1/5th of the time of tau,
then more than 90 % of the inputed energy will be just stored inside
the coil L magnetic field and not much Ohmic losses will yet be present.

So a short Ontime less than 1/5th tau is required to extract almost
all energy back from the coil.
If you switch the input voltage longer on, you have more losses and this
motor will consume too much input energy and it will be hard
to get it to  overunity.