Re: calculating safe primary turn-to-turn distance

["Malcolm Watts" <M.J.Watts-at-massey.ac.nz> (by way of Terry Fritz <twftesla-at-uswest-dot-net>)]

HI Jim,
        Yow.......

On 26 Aug 00, at 10:22, Tesla list wrote:

> Original poster: "Jim Lux" <jimlux-at-jpl.nasa.gov> 
> 
> 
> 
> ----------
> > From: Tesla list <tesla-at-pupman-dot-com>
> > To: tesla-at-pupman-dot-com
> > Subject: calculating safe primary turn-to-turn distance
> > Date: Friday, August 25, 2000 10:43 PM
> > 
> > Original poster: "Stan" <sdarling-at-columbus.rr-dot-com> 
> > 
> > Greetings all,
> > 
> > I have been curious since I started doing Tesla coils about this:  how
> > does one calculate the minimum air distance of turn to turn spacing in a
> > primary for a given voltage?  I'm not an EE so my common sense tells me
> > that current always favors the path of least resistance and that even a
> > tiny distance of air would have a much greater R than the adjacent half
> > turn of Cu tubing. So if the adjacent turns weren't touching, why would
> > it ever arc over through the air?  I suspect it has something to do with
> > the inductance of the Cu tubing coil(s) ....
> 
> 
> Exactly.. it's the inductance.  The inductance makes the voltage drop along
> the tubing greater than would be expected from just the DC resistance. The
> entire primary voltage is fairly evenly distributed along the length of the
> winding. So, if you have 10 turns in the primary, and 10 kV on the primary,
> then there is 1 kV/turn.  The peak voltage across the primary will be
> somewhat higher than the peak voltage out of the transformer, by the way.  

I just realized that is true for a helical winding, not a 
pancake coil.! as the inductance/turn tapers off severely 
towards the inner turns :(

Regards,
Malcolm

 
> So, if you have a 15 kV NST driving the primary, (peak voltage of 21 kV),
> and you allow a margin of 2:1 (for resonance effects), then the entire
> primary will be seeing 40 kV or so (as will the capacitor, by the way). 
> With 10 turns, now you are talking about 4 kV/turn. 
> 
> The next question is what is the breakdown strength of the gap between
> turns.
>  The breakdown strength for air in a uniform field (not the case for
> parallel wires) is about 31 kV/cm (70 kV/inch). There are basically two
> situations you can consider: 1) Very small wires, compared to the distance
> between them (say you wound your primary with #14 solid wire) and 2)
> something like copper tubing, where the diameter of the tubing is
> comparable to the spacing.
> 
> In the first case, the radius of the wire is going to determine when a
> spark can start forming.   The rule of thumb here is that the field is
> equal to the voltage over the radius.  If the field is greater than 70
> kV/inch, it will break down.  As an example AWG 10 wire is 0.100" in
> diameter, or .05" radius.  This works out to 3.5 kV for when corona will
> start, and where corona starts, breakdown follows.  For our 10 turn example
> above, you'd probably have a real problem.
> 
> In the second case, the radius of curvature is fairly large, compared to
> the spacing, so it's more the "gap" that determines the breakdown strength.
> If the diameter is much bigger than the spacing, just use the spacing to
> calculate the field strength (i.e. approximate it as a uniform field).. If
> you used 1" tubing with a 1/4" turn/turn spacing, then our 4 kV/turn
> example above would have a field of 16 kV/inch, well below the air
> breakdown of 70.
> 
> If the tubing diameter is smaller than the gap (or comparable), you want to
> be a bit more conservative.  Here is an equation:
> 
> D is center to center spacing of cylinders
> d is gap between cylinders
> r is radius of cylinders
> 
> E = V * SQRT(D^2-4*r^2)/(2 * r * (D - r)*arccosh(D/2r))
> 
> If you don't want to fool with inverse hyperbolic functions, there is a
> simpler form
> 
> 
> > Feeling stupid in Ohio.
> 
> Not at all stupid... It IS non intutive...
> 
> > 
> 
> 
>