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Racing Arcs Explained???

Original poster: Vardan <vardan01@xxxxxxxxxxxxxxxxxxxxxxx>


BTW, if one were to make a pure impulse transformer based on M x di/dt rather than resonance, the example below crates about 250kV over 10 inches of secondary coil. If you can stick it in oil and bring out the terminals to needle points, that should do 18 inch arcs!! With 17 joules on it, it should be a good arc too ;-))

One of my first coils ever was sort of like this. It blew oil all over everything as it arced through the glass container and all... But now, I think I know "why" it did what it did *:-)




Just to follow up...  If we take the example Dest gave:


and convert it to "inches stuff" with just minor length changes to get round numbers we get:


Solenoid - 6 inches high, 12 turns, 11 inches in diameter of 3/8 inch copper tube.

Secondary - 28 inches long, 1500 turns, 7 inches diameter. 0.015 wire.

Firing voltage 20kV.

But he does not mention Cprimary... That value is needed to know di/dt which is critical factor too...

So we add a 6 x 24 toroid top terminal with a centerline six inches above the top of the secondary. We can now go to JAVATC to find Fo and such...


91.67kHz and the primary is 39.462uH with a coupling of 0.306...

We will assume now that the primary is tuned 5% low as usual and find Cprimary as 85nF. A coil this size with 0.3+ coupling and 17 joules per bang "should" be a racing arc disaster!!!

Totally ignoring any "resonant" effects and totally relying on "pure transformer action", the voltage induced in the secondary is:

V2 = M x di(Lp)/dt

Since M varies drastically along the length of the coil, we have to chop it up and calculate it in little chunks or sections... We'll do every 0.1 inch since the computer is doing all the work ;-)) So I pump it into MandK and get this:


So we know the mutual inductance of each little section now...

To find di/dt we know Ip is in the first instant of firing:

I(t) = Vfire x SQRT(Cp / Lp) x SIN (t / SQRT(Cp x Lp))

If we take the derivative:

dI(t)/dt = Vfire x SQRT(Cp / Lp) x 1 / SQRT(Cp x Lp) x COS(t / SQRT( Cp x LP))

If t ~~= 0 and with some simplification we get:

dI(t)/dt = Vfire / Lp  == 20000 / 39.462uH  ==  507e6 A/S

Now we can get out the spreadsheet and make some numbers:


and we can make pretty graphs:



So at 3 inches, the voltage stress along the primary is 35000 volts / inch in the first fraction of a microsecond after firing.... Dest got 40kV/inch too!!! We can look at this chart:


To see that it will jump 1.26 inches!!! We have an arc for sure, and super sure too since it is a surface creepage arc that will actually arc at even far lower voltage...

Worse yet, once the arc is established, it will look much like a short on the secondary and will very quickly absorb all the energy in the bang. Thus explaining why they are "bright" arcs that do a lot of damage...

On the good side, thick layers of insulation should protect against such voltage breakdown!! Put that poly coating in real thick ;-))

Of course, once the resonance is established, such effects will "ride" on top of the voltage on the secondary too... I imagine that will explain the arcs occurring at places other than the base of the secondary...

The voltage profiles Paul has at:


Will enter into this all too...

Of course, Gerry's paper provided the key to it all ;-)))


I CC'ed dest too incase he has comments...




Dest also looked into this:


But I don't think he did it quite right... The links are dead so I can't check now. ...........