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

[hartiberlin]

Hi All,
I had another idea, how to get a maximum
Back EMF energy pulse from a coil.

Normally if we charge up a coil with
current, so it has magnetic energy,
we need W= 0.5x L x i^2 energy
for doing this and have to wait at least
about 3 to 5 x Tau=L/R timconstants,
until the current builds up in a coil,
if we apply a DC voltage, where L is the inductance of
the coil and R is the ohmic resistance of the coil.

For reference see:
http://www.1zu160.net/elektrik/spule.php

Now imagine, if we could switch the inductance
from Zero to 100 Henries immediately !

This can be easily done with a bifilar coil,
where one half of the bifilar coil could
be shorted out !

This way, we could first switch the DC supply
voltage across the bifilar coil first and
as there is no inductance , so L= 0 Henries.
the current will flow immediately only
depending from the ohmic law I= U/R
, so from the Supply voltage and the Ohmic resistance
of the coil.

Now imagine, we have 10 Volts as the supply voltage
and the bifilar coil has 1 Ohm, so we can immediately
run a current of 10 Amps into it,
without waiting any seconds until the current is build up.

Now, when we short out one part of the bifilar coil,
by a flip of a shortout-flip-switch so the one part of the
bifilar coil still works now as a normal coil having for
instance 100 Henries, now we suddenly have a 100 Henry coil,
in which flows 10 Amps !

Normaly to build up a current of 10 Amps in a 100 Henry
coil  with 1 Ohm ohmic resistance at 10 Volts
supply voltage will take about
5 x tau= 5 x L / R= 5 x 100 / 1= 500 Seconds.

So all in all, if we first have the bifilar coil,
which compensates its own magnet field and thus
builds up the current very fast in an instance in a few mikroseconds,
depending only on the still applyable stray inductance
and then shorting out one half of it, so we suddenly
have a coil with 100 Henries and 10 amps in it
and we needed only for instance 10 mikroseconds to
charge it up, then we only had used:
Energy= Voltage x amps x time= 10 Volts x 10 amps x 10 mikroseconds=
1 MilliWattseconds of energy to charge this
coil up.

But according to the coil laws we would
then have stored inside the now 100 Henry coil:
Coil-magnetic-energy = 0.5 x L x i^2= 0.5 x 100 x (10 amps)^2= 5000 Wattseconds !

So now we have a coil charged with 5000 Wattseconds magnetic
energy, which we could get back by opening the coil
and discharge it into a load via the backEMF,
but only have applied 1 MilliWattseconds of energy
to charge it up...


So where is my error ?
Can this really work, or will the coil voltage or the coil current jump somehow, when we shortout one part of the bifilar coil ?

Regards, Stefan.

[gyulasun]

Hi Stefan,

I would like to see your schematics how you mean shorting out one half of a bifilar coil when you apply the 10V.  Use simple switch and coil symbols, no need for a full (for instance a MOSFET switch circuit) drawing of course.

If you returned to this idea because you have seen/made some progress since your TEP idea then I would surely like to hear it, see this link to Naudin's site:
http://jnaudin.free.fr/html/tep60sh.htm   
http://jnaudin.free.fr/html/tep61sh.htm
http://jnaudin.free.fr/html/tep61sht.htm

It is very pity Naudin did not include the most important thing: Did he find overunity or not? How much is the free energy he refers to with respect to the input energy? (at least I have not seen it reported, did he inform you back then?).

My main problem with the idea is that if you short circuit one half of the bifilar coil then there cannot be any self inductance of the other, not shorted coil part because of their mutual closeness. Just the same way if you test a conventional mains transformer and short circuit its secondary coil, then the primary coil loses its self inductance (and heats up rapidly).

So if you returned to this idea due to some progress since then, please tell me what it is. 

Thanks,
Gyula

[gezgin]

It can be test, but 100H big coil isnt it?
if bifiliar coil L = 0, giving energy to coil zero or near zero or ?

I want to test this 97's circuit but how much henry must be bifiliar coil,
i have 5 mH coil but it isnt enought for test?

and another patented idea : parralell coils back emf device.

sorry my english

thanks

ilhan

[hartiberlin]

Hi Gyula,
you are right,
this was it, what I wanted to try...
Unfortunately when we did this in 1997 together with Dieter
Bauer, we had some input power coupling into the circuit
via the base of the transistor,
when I remember correctly and we had no time to
retry and set it up correctly...and retest it.

Indeed here is a strange thing at:

http://jnaudin.free.fr/html/tep61sht.htm

Why is then there at all a BackEMF coming out of this circuit from the Bifilar
coil at R3 ?

Is the voltage U(R3)
only from the stray inductance of the bifilar coil,
which is then a real inductance, small, but real, so this
stray inductance has stored this magnetic energy which is then
released, when the current inside the coils
is cut off ?


Can anyone explain this ?

Many thanks.

Regards, Stefan.

[gyulasun]

Indeed here is a strange thing at:

http://jnaudin.free.fr/html/tep61sht.htm

Why is then there at all a BackEMF coming out of this circuit from the Bifilar coil at R3 ?

Is the voltage U(R3) only from the stray inductance of the bifilar coil, which is then a real inductance, small, but real, so this stray inductance has stored this magnetic energy which is then released, when the current inside the coils is cut off ?

Can anyone explain this ?

Hi Stefan,

I discussed this circuit with a friend of mine and we agreed on the followings.

Let's suppose the forward voltage drop on D4 diode is 0.8V and the saturation voltage of Q1 transistor when switched on is 0.2V, this sums up to 1V, ok? 

Further, if we assume the battery is charged up to around a practical 13V value, then the useful voltage source is 13V-1V=12V, neglecting the internal resistance, ok?

Now let's consider Naudin scope shot on the U(AC) voltage first. It shows the voltage drop between the A and C endings of the series L1 and L2, ok? It is around 5.2V peak to peak, ok?
This 5.2Vpp voltage drop is due to the DC resistance of L1+L2 wires, ok?
Then the voltage drop on the R4 resistor should be 12V-5.2Vpp=6.8Vpp, ok?
This means a peak current of around 6.8V/1 Ohm=6.8A flowing out from the battery and this goes through all the way on L1, L2 and Q1 of course.

Now let's consider the U(R3) voltage, the back EMF voltage. When Q1 switches off, the A and C endings of L1 and L2 are floating, the only load is across L2 by the R3+D1 in series. The huge current suddenly terminating must induce a certain voltage in both coils due to
1) partly stray inductance indeed as you mentioned
2) the sudden collapse/disappearience of the magnetic field INSIDE the ferrit core
    because it can be considered as a magnetized core (like an electromagnet) during
    the time Q1 is switched on and ferrit cores do not normally keep magnetism, hence
    there must be a flux change in the core what induces the back EMF.
In fact this loaded back EMF value is bigger by 0.7V than the value shown in the scope shot because of the forward voltage drop of D1: so it makes around 1.4V (if I judge that pulse, U(R3)=0.7V also).

One more notice: the ferrit core can be saturated by this huge current and this makes the switch on/off process nonlinear, so the repeatabilty/reconstruction may involve inherent differences in the measured data.  This is nicely shown in Naudin's another test here:  http://jnaudin.free.fr/html/tep61a5.htm 

I hope these make sense and if you or anybody else wish to comment please do so.

Regards
Gyula