RE: [TCML] Source for 1H air-core inductors

["Lau, Gary" <Gary.Lau@xxxxxx>]

Yeah, but if such an inductor is used between stages of a Marx generator, you either want a single layer, or some very non-trivial insulation between orderly layers, to stand off the voltage spikes.  Current is probably not too high.

Per WinTesla, a 1-layer 3" fist-diameter coil of 40AWG would need to be 53" long to achieve 1H.

Regards, Gary Lau

> -----Original Message-----
> From: tesla-bounces@xxxxxxxxxx [mailto:tesla-bounces@xxxxxxxxxx] On
> Behalf Of jimlux
> Sent: Friday, June 19, 2009 11:58 AM
> To: Tesla Coil Mailing List
> Subject: Re: [TCML] Source for 1H air-core inductors
> 
> Ed Phillips wrote:
> > Resistance?
> >
> > Current handling capability?
> >
> > Ed
> 
> Ballparking here using Wheeler's equation
> 6" diameter, 6" long, 3000 turns is just under 1 H..
> 
> 3000 turns * 18" average turn length = 4500 feet
> 40 ga is 1000 ohms/1000ft  (roughly.. 10ga is 1 ohm/1000ft, 40ga is
> 1/1000 the area)
> 
> So 4500 ohms DC resistance, or something around there.
> 
> 40 AWG is about 30kft/lb, so we're looking at less than 1/4lb of wire.
> 
> What about if you just buy a 1/4 lb spool of 40AWG wire.
> 
> 
> Sticking with the idea of 3000 turns in the size of a standard wire
> spool... one might look at a pound of AWG 30.  Resistance would be about
> 450 ohms.
> 
> Half pound spools, 1600ft of AWG30 Beld-Sol are $45 from Mouser.
> 
> ACtually, though the spools might be too small, which reduces the
> inductance.
> 
> A one pound spool of PolyThermaleze might be closer to what you want.
> 
> 
> Didn't someone measure the inductance of a standard 500ft spool of AWG12
> THHN?  500 ft/1 ft turn (average) = 500 turns works out to about 26 mH..
> 
> 
> Current handling is going to be determined by allowable temperature rise..
> we can sort of estimate here, too..
> Let's say we can tolerate a 100C rise.  We're looking at basically a
> solid lump of copper with some plastic, so let's call it 250 grams of
> copper, which has a specific heat of .38 J/(K*g), or, inverting, 2.5
> degree grams/Joule.  250 grams, so overall rise in our spool will be
> 2.5/250 = 0.01 degree/Joule
> 
> Say we push an amp through this beast.  That's a dissipation of
> (assuming my AWG30 case) 500W, so in a second, we get a 5 degree
> temperature rise.  In 20sec, we've got our 100 degree limit.
> 
> 
> 
> 
> >
> > "That's huge indeed.
> >
> > Mine are just a little larger than two human fists held together.
> >
> > Wound with #40 AWG wire.
> >
> > D.C. Cox"
> >
> >
> >
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