This is realy a continuation of my other thread, http://forum.allaboutcircuits.com/threads/making-small-sparks.106490/page-2#post-814826 but maybe I can get more information from people not following it.

Is there a formula or way to predict the voltage flyback level on an inductor? When the current is turned off suddenly.?

 

V= L* dI/dt
In other words, the faster you turn off the current the higher voltage you will get out. But there are other factors like the capactiance of the coil and load, which will limit the voltage peak.

 

http://www.dos4ever.com/flyback/flyback.html

/

 

This is realy a continuation of my other thread, http://forum.allaboutcircuits.com/threads/making-small-sparks.106490/page-2#post-814826 but maybe I can get more information from people not following it.

Is there a formula or way to predict the voltage flyback level on an inductor? When the current is turned off suddenly.?

Hi,

It sounds like you are asking a practical question not a theoretical one. The theoretical one is easy but the practical one is not easy at all, and may be impossible to answer without actually doing a measurement. The reason is because the voltage will be dependent on so many parasitic aspects of the circuit and circuit board, and wiring, as well as things about the inductor which probably are not published in the data sheet.

If you know the equivalent parallel capacitance maybe you can do it because the energy in equals the energy out, for an idea part. That would give some idea what to expect.

Another aspect altogether though is how long does the inductor last if it has to put out the maximum voltage spike that it can, on a continuous basis where it has to do this say once over millisecond or even once every second. The wire insulation is only so good, maybe 300 volts, so now we see the actual construction of the inductor start to have an effect too where the way it is wound will affect the longevity. If the current arcs over to the next turn, it will start to degrade the insulation and pretty soon it wont work very well at all with carbon buildup between turns.
So the failure mode is something to serious consider as well as the max possible voltage you can get out of it.

 

1/2 LI^2 = 1/2 CV^2

Understand my blog and see that the inductive voltage is directly proportional to the capacitance that has to catch the energy. Small capacitance causes higher volts.

http://forum.allaboutcircuits.com/blog/inductance-measuring-jig.478/

 

Thanks to all of you! I will need to read up on this information. This inductor will be acting like a low tension ignition/spark coil. Like they used on the old hit and miss motors in the early 1900's. just to start a voltage and current flow in a very small gap.

MrAl, it is a practical thing but thought there may be an online calculator for it. Maybe some experimenting is needed to see what works and doesn't. I'm assuming my inductance will need to be low, because it will need to charge and discharge very fast in the 2-5 microsecond range. I've got a lot of relay coils, solenoids and inductors from old electronics that I can play with.

Jony130 that link looks interesting. I guess a one shot boost convertor is close to what I'm after. A boost convertor with no feed back, regulation or output caps! Never thought of it that way.

 

For a rapid switching of the inductor current it basically depends upon the value of the (stray) capacitance and inductance in the circuit with a peak voltage value as shown by #12's formula.
So unless you know both values you can't predict the voltage.

 

Thanks crutschow.
After thinking more about the things Jony130 linked to and the thoughts of this being an unregulated boost convertor I dug out the books I have on switcher circuits. And found in one of them what I think I need to do some math. To find a 'rough' range of inductor values, then experiment from there. I am worse at math than I am at electronics.