Understanding the sparks created when using a relay to switch a coil.

[gyulasun]

Hi CC,

I am afraid it is you who is uncertain here on what is the connection between the collapsing field and the current flowing after the switch-off. You wrote:

"Counter emf is not countering the original current its countering the collapsing flux."

Emf cannot counter any flux.  However, the emf can start a current flowing in the coil.  However, current can only flow in a closed circuit and because the switch has been opened, seemingly there is no closed circuit for the coil: it has the induced emf across its endings but no closed circuit... 

You wrote: 
"There is no opposition to the collapsing flux. so it collapses very fast. for there to be opposition, current must flow, thus creating its own flux which in turn opposes the collapsing flux."   This latter part is ok but you did not clarify why current flows?

This is where the coil self-capacitance comes into play:  whatever (small) capacitance the coil possesses it will constitute a parallel LC circuit with the coil's self inductance (a high L/C ratio is involved), thus the induced emf  "finds" immediately a closed LC circuit from the very moment the relay switch is opened and the energy from the collapsing field will start swinging in this LC tank circuit. Now the current starting to flow in the coil part of the LC tank will of course have an opposite polarity and a lower amplitude with respect to the previous coil current so the original current from its instant original value will be gradually reduced to zero in a decaying "ringing" fashion (depending on the L/R time constant and any damping). 

Here is another quote from you:
"Now I spoke of counter emf and then I spoke of voltage, they are essentially one in the same. the counter emf is measured in volts and is therefore the voltage. but the voltage measured includes the voltage drop across the internal resistance of the coil so it is lower than the emf. not by much but it is. if the coil had no resistance at all then the voltage and emf would be exactly the same."

Now it is not clear which counter emf you speak above?  The one you use for naming the induced emf due to the collapsing field after the current switch-off  OR you meant the counter emf inherently created in any coil while its input emf is connected to it?  Because in the latter case you are right: the losses in the coil makes counter emf be slightly lower indeed than the input emf and this is valid for a continuous operation like an electric motor or a transformer etc works.  However in case of the former case the induced emf can be many times higher in amplitude than the input emf and this is valid for the current switch-off case, no more input emf.    This is the example why you have to differentiate between the two cases by using different terms... even you seem to be confusing the two.  While you know correctly the induced emf after the current switch-off can be many times higher than its input emf as you wrote it in your very first post.

You also wrote:
"Do you now see why I used the term emf? emf is the correct term to describe what the varying flux creates."

Yes this is ok in itself, emf is created by a changing magnetic flux but I did not object emf in itself but objected using the term back emf  (or counter emf) when it is meant referring to the induced voltage or induced emf that is created after coil's current switch-off.

Re on the term flyback pulse:  I agree it is not used at university or college levels for referring to the induced emf that is created after coil's current switch-off
and I admit I also used it in the past for referring to the same but just due to the lack of the correct term: to make it clear I meant the induced emf produced by a collapsing flux after coil's current switch-off.  I used it because it described the situation very well: the sawtooth waveform switches current off at the end of its up-ramping in the primary coil of the line output transformer and then the CRT is blanked while the electron beam is brought back to start the next horizontal line:  the  time period under which the beam returns is called the flyback time and exactly under this time the collapsing field in the primary coil creates the voltage spike (usually 1000-1200V peak), this is where the term flyback pulse originates from (just for those not familiar with). So to use the term flyback pulse for naming the induced emf that is created at flux collapse has much more sense than using the back emf for referring to the same situation (but still not fully correct, I agree).

Re on bifilar coils: I agree and I think you may wish to continue on bifilar relay coils...?

rgds,  Gyula

EDIT I was writing this answer in Notepad and did not notice Magluvin already posted his text above.

[gyulasun]

....
Try this.  Get a source, 12v is good, a coil , a switch, a diode and a cap.   With the switch in an open condition, connect all components in series, no particular order. hit the switch then release. Now measure the cap. Depending on the coil and cap, there should be nearly 2 times the voltage in the cap than the source.
When current is first applied, the coil starts to conduct, through the diode and Empty cap, we are getting the flywheel going. But by the time the cap reaches source voltage, the flywhel is not done yet and actually Pulls more current from the source as it winds down, and it pumps the cap to greater than the source. The diode prevents any oscillation once the flywheel stops. If the diode were not there, the higher voltage of the cap would want to level out with the source till the oscilation dwindles.

So from that, you can realize that the collapse that we all know of, continues to pump FOREWARD, FEMF after the source is disconnected, all that presure is stored in the cap, and that presure is voltage. The smaller the cap, the higher the voltage that will be developed.  And in an LC oscillator, KICK it with 12 v, and the oscilation voltages can be very very high depending on the cap and coil value.
.....

Hi Mags,

Very interesting and I assume you tested the series LC + diode setup and found higher voltage across the cap than the source gave. I wonder if you used a solid state switch or a mechanical one, in case of the latter perhaps simply touched two wires together and then pulled them apart by hand?
It would be good to operate the switch in a controlled fashion so that the ON time could be variable in the microsecond range, depending on the L/R time constant, maybe there is an optimum length of ON time whereby the voltage in the cap gets to a maximum value.

I hope this is not off topic for CC's thread.

rgds,  Gyula

[Magluvin]

Hey Gyula

I agree with the short timely pulses.  Think about a pendulum. At what time in the swing cycle, also where in the swing cycle would you apply the KICK to keep it going. 

Or if we want to keep the flywheel going, more likely we have to discharge a cap into it on a timely basis. The larger the inductor, the longer the flywheel will turn. A large inductor will take a bit of kicking to get it up to speed, but as it gets going, it can be kept turning on a practical continuous basis with kicks.  Is this SM's secret?
Its output would be DC.

Mags

[Magluvin]

Oh Gyula

It was just simple testing, reed switches, relays.  Im in the middle of moving, but I will be back on it shortly.

The test I proposed above to charge the cap to nearly double should be very clear on my theory, and easy to do for anyone.  Its not extra energy in the cap, as in more in the cap than what came out of the source, it is equall.
But, if we can cut off the source before the cap is charged higher than the source, can we loop the cap, coil and diode, to let the coil continue to wind down and pump anything extra from the other side of the cap.  If we cut off the source, for example, exactly when the cap is equal to the source, if the coil causes the cap to charge any higher than the source, its free.   Would you agree?  ;]

Mags

[gyulasun]

Oh Gyula

It was just simple testing, reed switches, relays.  Im in the middle of moving, but I will be back on it shortly.

The test I proposed above to charge the cap to nearly double should be very clear on my theory, and easy to do for anyone.  Its not extra energy in the cap, as in more in the cap than what came out of the source, it is equall.
But, if we can cut off the source before the cap is charged higher than the source, can we loop the cap, coil and diode, to let the coil continue to wind down and pump anything extra from the other side of the cap.  If we cut off the source, for example, exactly when the cap is equal to the source, if the coil causes the cap to charge any higher than the source, its free.   Would you agree?  ;]

Mags

Yes, I would...     

and I would like to understand how the extra voltage can get into the cap once there is an open circuit? Maybe in the reed switch's inner air gap there is still some leakage-through possibility from the coil's collapsing field induced spike? 

Gyula