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Re: Re Secondary Resonance LC and Harmonics

Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Terry and Jared,

I believe Jared's derivation is based on the classical L equation.


We can examine the classic equation for an air cored inductor: L = u
Nsqrd A / l
Multiply the numerator and denominator by 4pi  we get:

L = u Nsqrd 4pi sqrd  r sqrd / 4 pi l  =  u (2pi r N)sqrd / 4 pi l

Since 2pi rN equals wire length we can write:

L = u (wire length)sqrd / 4 pi l

Letting the solenoid height ( l ) equal the wire length,  we find
that the classic equation predicts that the inductance of a straight
wire is simply:

Lstwire = u wire length / 4 pi


This equation has several simplifying assumptions and will predict the inductance on the high side. The H field (air core) was calculated at the axial and cross sectional center's of the solenoid and it is assumed the H field is constant over the cross section and from end to end. It does not account for end affects where there is leakage. The H field will be smaller at the solenoid end than its at axial center. As a result, it will over predict the flux threading the solenoid.

Gerry R.


Original poster: Terry Fritz <teslalist@xxxxxxxxxxxxxxxxxxxxxxx>

Hi Jared,

At 02:29 PM 6/29/2005, you wrote: 
.....

L = u Nsqrd pi Rsqrd / (l)
L = u (Wire length)sqrd / 4pi (l)

Jared and Larry 

If I look at the secondary inductor here:

10.25 inch dia x 30.0 winding length
1000 turns #24 enameled magnet wire
The wire length is 10.25 x pi x 1000 = 2683 feet

And make the metric conversions:

u = 4 x pi x 10^-7  Wb / (A * m)

10.25 inch = 0.26035 m  or 0.130175 radius (m)
30 inch = 0.762 m
2683 feet = 817.9 m

I get:

L = u Nsqrd pi Rsqrd / (l) = 12.566e-7 x 1e6 x pi x 16.9455e-3 / 0.762 = 0.0878 H

L = u (Wire length)sqrd / 4pi (l) = 12.566e-7 x (817.9)^2 / (4 x pi x 0.762) = 0.0878 H

But there is a problem.

I have measured the inductance of this coil for many years in many different ways (3 sets of inductance meters, 3 frequency generators, 2 counters...) and the inductance measures 0.0754H.

If a double check using wheeler's formula (inch based):

L (uh) = (n x r)^2 / (9 x r + 10 * l) = (1000 x 5.125)^2 / (9 x 5.125 + 300) = 0.0759 H

Some of Paul's and Mark R's programs get closer to the measured value.

So what am I doing wrong?

The meters were all 1000Hz measurements, but the inductance confirms at the coil's frequency both with a top load and just doing a pure 83kHz inductance measurement with scope, signal generator, frequency counter and all... E-Tesla can find the Medhurst and 'with top load' capacitance and they match right up to 75.4mH in all cases...

So your equation seems 16.4% high and Wheeler's is 1.0% high...

Cheers,

        Terry