Re: My first TC

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Malcolm Watts by way of Terry Fritz <twftesla-at-uswest-dot-net>" <m.j.watts-at-massey.ac.nz>

Perhaps I can explain this coupling effect: Consider the outer turn - 
it contains the longest length of wire/tube/strap etc. It is closely 
coupled to the next inner turn which is the second longest. it is 
more weakly coupled to the third turn in and coupling to the 
innermost turn (the shortest) is comparatively miniscule. Since 
coupling is a function of distance between turns in this coil, if 
follows that the mutual inductance between the outer turns is far 
higher than coupling between the inner turns. The same consideration 
applies to the coupling between the innermost turn and its neighbours 
going outwards. Hence most of the coil inductance is in the outer 
portion. You might want to consider how a helical winding modifies 
this consideration.

Malcolm


On 27 Apr 01, at 7:51, Tesla list wrote:

> Original poster: "William Swanson by way of Terry Fritz
> <twftesla-at-uswest-dot-net>" <swansontec-at-yahoo-dot-com>
> 
> Hi All,
> In order to answer your questions, I have experimented
> with a number of different coil geometries on my
> spreadsheet, and have come up with some interesting
> insights. It turns out that the amount of wire per
> square inch is the exact same over the entire area of
> the coil. This isn't a big surprise. After all, the
> spacing between turns is the exact same over the
> entire coil. Therefore, if the amount of wire per
> square inch were the only factor in determining the
> inductance of the coil, the outer turns would have no
> more inductance per square area than the inner turns.
> 
> However, the number turns per inch doesn’t seem to be
> the only factor. I have found that the outer turns do
> in fact have more inductance per square inch than the
> inner turns of wire. It must be some sort of coupling
> effect, since I can’t think of any other logical
> reason. Perhaps curved wires don’t have as much
> inductance as straight wires? All I know is what my
> equations tell me, so if I’m wrong, it means that I
> did something wrong when I derived the formulas for
> the physical dimensions of a coil.
> 
> For anyone who wants to build a flat spiral secondary,
> here are the equations that I worked out for dealing
> with the physical dimensions of a flat spiral:
> 
> 2(pi)rw/s=l
> 2(pi)rn=l
> ns=w
> 
> r = average radius
> w = width of coil on one side 
> n = number of turns
> l = length of wire
> s = spacing between turns
> 
> The Wheeler formula is, of course:
> 
> L = (rn)^2 / (8r + 11w)
> 
> L = inductance in microhenries (uH)
> 
> The rest variables are the same as above. Units are in
> inches.
> 
> The reason why I use a single wire for my primary has
> to do with inductance. As you know, the resonant
> frequency of a circuit is related to the product of
> its inductance and capacitance. Since my secondary is
> extremely high frequency, I didn’t have much “room”
> for both a large inductance and a large capacitance in
> the primary of my coil. I decided that since the
> capacitor is what determines the amount of energy in
> each burst, it would be better to use a small
> inductance and a large capacitance. Thus, I used a
> single turn of wire to “make room” for my 3.6 nF
> capacitor. If you use a lower frequency or a smaller
> capacitor, you could easily use several turns wire in
> your primary.
> 
> On a sadder note, my capacitor was destroyed during
> last night’s tuning session, so I will have to make a
> new one. In the mean time, I won’t be able to run my
> coil unless I go solid state. I have the parts for a
> solid state driver sitting around, and I’ll put them
> together this weekend. I’ll build a decent cap once I
> figure out what went wrong with my last one...
> 
> -William
> 
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