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

[lumen]

Sorry guys, I have to disagree on the fly-back.
If you ground one end of a coil, then apply a current to the other end, when the current is removed the collapsing field will cause the current to flow the opposite direction of the applied current.

Anytime a current flows, a field is generated that opposes the current direction. When the current stops, the field that was opposing the current direction still exists, so the current flows backward until the field has fully collapsed.

Make sense?

[gravityblock]

BEMF is easily recoverabe, it is done every day in conventional electronics devices. To recover BEMF is not related in any way with overunity.

In this case, the mechanical gain due to no CEMF is already OU (Lenz violation).  Recovering the BEMF makes it more OU.  This really burns you naysayers to no end.  I love it.

Take care,

GB

[k4zep]

Hi Gang,

Photos of waveform on 12.4 VDC battery supply

First photo JPGxx415:

Bottom trace is voltage across coil, top trace is total system current in .1 ohm resistor.  Hence 12+ V pulse across coil and .8 amps current in coil give or take.   Average power determined by duty cycle of pulse.

Bottom voltage trace is @ 5VDC/div and top current is .5 amp div. inverted trace.  So down is current flow. Remember this is NOT precision measurement!!!!  Note short coil charge up time due to a small amount of inductance not trimmed out in coil.  Also note the integrated voltage at the trailing end of the voltage pulse @ turn off due to diode across coil.

xx416.JPG Below:

Second picture shows same but bottom traceis 20V/div now to show 50v pp giver or take, back emf pulse at turn off with diode across coil removed,  Very fast ringing pulse @ cutoff  but does not show up in current as the diode across coil keeps it in the coil.  Just the current back feeding into diode and circulating around coil.  As you can see, virtually no energy in back emf.  That is the long and short of it.  I could show the current in the coil by putting the .1 ohm resistor in series with diode but haven't bothered.

That's it.

Respectfully
Ben

[broli]

Sorry guys, I have to disagree on the fly-back.
If you ground one end of a coil, then apply a current to the other end, when the current is removed the collapsing field will cause the current to flow the opposite direction of the applied current.

Anytime a current flows, a field is generated that opposes the current direction. When the current stops, the field that was opposing the current direction still exists, so the current flows backward until the field has fully collapsed.

Make sense?

Lumen I suggest you reevaluate your basic electric/electronic understanding. An inductor doesn't instantaneously reverse current and then decay, this would be quite amazing but it doesn't happen in simple circuits. I might understand that the confusion arises with voltages across things, when they reverse and current doesn't and what not. But the most simple view is to see inductor wanting to keep the same current (direction and magnitude going). This is why if you short out a perfect inductor It will maintain current flow with 0 voltage, like super conductors. When you short it out with a resistor it will decay exponentially since a resistor has a varying voltage drop which depends on current. On the other hand when shorted out with a battery it will be a linear ramp to 0 amps since the batteries voltage doesn't change with current flow.

[k4zep]

Lumen I suggest you reevaluate your basic electric/electronic understanding. An inductor doesn't instantaneously reverse current and then decay, this would be quite amazing but it doesn't happen in simple circuits. I might understand that the confusion arises with voltages across things, when they reverse and current doesn't and what not. But the most simple view is to see inductor wanting to keep the same current (direction and magnitude going). This is why if you short out a perfect inductor It will maintain current flow with 0 voltage, like super conductors. When you short it out with a resistor it will decay exponentially since a resistor has a varying voltage drop which depends on current. On the other hand when shorted out with a battery it will be a linear ramp to 0 amps since the batteries voltage doesn't change with current flow.

Look at scope shot above and you can see direction of current flow.

Respectfully
Ben