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Re: Self resonance of a coil

Tesla List wrote:
> 
> >From huffman-at-fnal.govSat Oct 12 13:04:30 1996
> Date: Fri, 11 Oct 1996 08:35:40 -0500
> From: huffman <huffman-at-fnal.gov>
> To: List Tesla <tesla-at-pupman-dot-com>
> Subject: Self resonance of a coil
> 
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> 
> Hi Group,
> If I know the dimensions of a coil, i.e. r=diameter/2, wire gauge,
> l=length, n=number of turns,
> is there an approximate formula for the self-resonance? I thought I had it
> but can't find it now,
> or maybe it doesn't matter if the coil is top loaded with a large toroid. I
> know the
> inductance is found approximately by L=n*n(r*r/(9*r+10*l)) so I use the
> 1/2PIsqr(LC).
> Just curious,
> Dave

Dave,

The best way is simply to measure the coil after you wind it using a
signal generator and base-driving the secondary through a pair of
back-to-back LED's. You can try the following empirical approximation
for "what-if" designs. This is similar to a formula which may have
originally come from Harry Soumalainen or Richard Quick, but has been
adjusted for feet and inch measurements:

      F = 984,300/((3.485X)*(L/D)^-0.304) (KHz)
  Where:
      L = Length of Winding (inches)
      D = Diameter of the winding (inches)
      X = Physically computed Length of wire (feet)
      d = effective Wire diameter (including insulation and any spacing)
 But: X = Pi*D*L/12d feet 
 and: n = L/d turns  
  
  Example:
      L = 31"
      D = 10.25"
      d = 0.032" (21 AWG with Dual-Thermaleze)

Then: X = 2599 feet          
      n = 969 turns (close-wound) 

 and: F = 984,300/((3.485*2599)*(31/10.25)^-.304)
        = 984,300/(9304)*(3.024^-.304)
        = 984,300/6646
        = 148 kHz

Since this is an approximation, your mileage may vary. On a sample of
two coils I tested, it was within +/- 5%. Enjoy.


-- Bert --