Charging a coil with less energy and get huge BackEMF energy pulses

[hartiberlin]

Hi Rob,

This morning I made a bifilar solenoid coil similar to that of Naudin's but with less number of turns. My ferrite rod is 9.5mm OD and 170mm long. I covered it with a 55mm long hard paper cylinder of 10mm ID so that it can easily slide on the surface of the rod.
I wound 2 x 23 turns onto the hard paper cylinder from 0.8mm OD enameled copper wire.

I measured the following self inductances:

Any one coil each, no ferrite rod:  L1=L2= 2.3uH
No ferrite rod, L1&L2 in bifilar cancel:       0.42uH
No ferrite rod, L1&L2 in series aiding:        9.1uH

Any one coil each, with ferrite rod, coils slided to any one edge of the rod: L1=L2=35uH
Any one coil each, with ferrite rod, coils slided to the center of the rod:     L1=L2=46uH
With ferrite rod, L1&L2 in bifilar cancel, coils slided to any one edge:                  0.44uH
With ferrite rod, L1&L2 in bifilar cancel, coils slided to the center of the rod:        0.44uH
With ferrite rod, L1&L2 in bifilar aiding, coils slided to any one edge:                   141uH
With ferrite rod, L1&L2 in bifilar aiding, coils slided to the center of the rod:         192uH

So the results show that it does not matter where the bifilar coil pair is placed onto the ferrite rod, its resultant and cancelled inductance (when connected in bifilar cancel mode like in the TEP) remains the same. I am a bit surprised too, that without the ferrite rod the same coils (when connected also in the bifilar cancel mode) show only slightly less self inductance than with the core, i.e. 0.42uH versus the cored 0.44uH.

Hi Gyula,
so you think in using your example values,
the Back EMF hill wave of about 0.75 Volts in:
http://jnaudin.free.fr/html/tep61sht.htm
at R3
is caused by the remaining inductance of
0.44 uH in your case ?

This may justify the hint on severe core saturation due to the huge peak current and considering Naudin's non-linear time flow tests ( http://jnaudin.free.fr/html/tep61a5.htm )  these show that local, individual turns of wire do influence the core material under them and in the rod center this effect is the greatest (but mainly cancelled) and towards the rod edges and half way out of the coil this saturation effect reduces/diminishes, hence the local self inductances can increase, hence the back emf can also increase. 
Let's not forget that the back emf pulses we see in the scope shots they are there during the transistor switch off time when the coils L1&L2 are floating already, hence already mainly 'recovered' their self inductance. 
Let's also notice that the amplitude of the back emf pulse is very small, under 1V peak when Naudin loaded it with a 100 Ohms resistance (with a diode in series).  For me it may mean that without making further 'tricks' there is no free lunch yet...

Regards
Gyula

But why is then one BackEMF smaller at the one side smaller than at the other side,
when the ferrite core is only half way in ?

In your explanation BOTH of them must be bigger, but indeed,
only one is bigger and the other is smaller..??!!

[gyulasun]

Hi Gyula,
so you think in using your example values, the Back EMF hill wave of about 0.75 Volts in:
http://jnaudin.free.fr/html/tep61sht.htm at R3 is caused by the remaining inductance of
0.44 uH in your case ?

Hi Stefan,

Yes I do.  Remember however that Naudin used 2 x 190 turns  (I used 2 x 23) so his remaining (bifilarly cancelled) inductance was probably higher than 0.44uH I guess.  But it must have been still small, maybe 1-2uH or so.

But why is then one BackEMF smaller at the one side smaller than at the other side,
when the ferrite core is only half way in ?
In your explanation BOTH of them must be bigger, but indeed,
only one is bigger and the other is smaller..??!!

No, I meant the difference in amplitudes when the coils are in the center of the rod both halfs of the rod are nonlinearly saturated more or less in an equal amount and when the coils are at the edges or at half way in only, then the nonlinear saturation shifts towards the edges of the rod and gradually improves (i.e. becomes less and less saturated) towards the center of the rod.  Remember Naudin covered the full length of his rod with the 2 x 190 turns, indicated a 19cm long coil and hence at least as long rod.

When the coils are in the center they cause a equal but opposite saturation or let's say distortion in the core and this mainly is able to cancel out so the resultant amplitude is small, comparing it to the amplitude which comes about when the core is only half way into the coils. In the latter case the the opposite saturation cannot fully be equal because only the half length of the rod is excited so no symmetrical cancellation can take place hence the resultant back emf amplitude is higher with respect to the center case.
That is how I think.  Maybe I am wrong but for the time being I cannot think otherwise.

rgds,  Gyula

[MeggerMan]

Hi Gyula,

Its good to see you finished off what I started and proved what I suggested to be true.

OK, here's a leap of faith(I'm only guessing here so bare with me):

You pass a current "x" through L1 and L2 with the connections arranged as bifilar.
Then you short out coil L1 so that the current by-passes L1 and goes into L2 only.
I would expect to see the current drop very sharply then rise again as a magnetic field is established in the core.
You will not be able to obtain a back emf of any size until the field is established.
So if you switch off the current just after you short out L1 the back emf will be virtually zero.
However if you wait until the current has risen you will get the full back emf for that coil and current.
So whilst the electron orbits are randomly arranged in the core there is no stored energy.
Once they re-align then you can turn off the current, the driver for the field collapses and the orbits are then free to return to a random state and induce a back EMF.

Hi Stefan,

In your explanation BOTH of them must be bigger, but indeed,
only one is bigger and the other is smaller..??!!

I think you may be mistaken, the middle position of the ferrite core produces the smallest back emf and when the core is either of the two outer position the back emf is greater.

Regards
Rob

[gyulasun]

Hi rob,

So you mean L1&L2 is in bifilar cancel connection and you short out one of them while current flows in both of them, right?

Well, current would only change significantly if your switch inner resistance is much smaller then either L1 or L2 copper resistance, ok?
Then next problem I see is that when you short one half of bifilarly connected coil pair then you kill the self inductance of both!  Would not be this matter in your idea?

(Must leave now, will be back tonight )

Gyula

[MeggerMan]

Hi Gyula,

So you mean L1&L2 is in bifilar cancel connection and you short out one of them while current flows in both of them, right?

Yes

Well, current would only change significantly if your switch inner resistance is much smaller then either L1 or L2 copper resistance, ok?

I was thinking that this is for a pulsed circuit say and the frequency is that for a near perfect sine wave, albeit only half a wave.

If you have no inductance when current is flowing through L1 and L2 and then you short L1, the field in the core will be non-existent and the current will rise to build up the field, I think, I may be wrong of course.
Only an experiment will prove this.
Rob