Re: How to Tune a Flat Spiral Coil

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

Original poster: "Paul Nicholson by way of Terry Fritz <twftesla-at-qwest-dot-net>" <paul-at-abelian.demon.co.uk>

David Thomson wrote:
> I can't confirm how many turns are in the coil without cutting the
> coil with a saw and counting the layers

Oh, don't do that!  It happens quite often that for one reason or
another we can't count the turns on a coil.  So we just estimate
the turns based on the closest match for L and R.

Your estimate of 227 turns gives a significantly higher prediction
of L.  The inductance is very sensitive to the outer diameter of the
coil, which you gave earlier as 13.75 inches.

> a more accurate average radius of 6.8125"

You probably mean average diameter?  I take it then that the outer
diameter is actually 13.375", so then we have

units inches
secondary {
  radius1 1/4/2     ;  Core is a 1/4" brass bolt
  radius2 13.375/2  ;  Outer diameter of 13.375"
  height 0 conductor 21 awg
  turns 227         ;  Estimated to match measured R
}
primary {
  radius1 13.375/2 + 1/2  ;Inner of primary is 1/2" beyond sec outer
  radius2 13.375/2 + 1/2 + 1/2*3.5  ; 3.5 turns at 1/2" pitch
  height 1/4
  conductor 1/4/2     ; 1/4" OD copper tubing
  turns 3.5
}

    SCNY.L| SCNY.R|    PRMY.L| PRMY.R| SCNY-PRMY.M|SCNY-PRMY.K
6313.52 uH|   5.26|   9.76 uH|   0.00|    95.60 uH|     0.3852

which I think is quite reasonable.  I reckon that makes for a
satisfactory first step in the characterisation of the bulk
properties of the coils.

You may wish to measure the mutual inductance as a cross-check. This
involves passing a 60Hz AC current through the primary, say around
10 amps, and measuring the voltage induced across one of the (open-
circuit) secondary strands.  It should be around 36mV per amp of
primary current.

> My coil is mounted on a 4" PVC pole 24" off the ground.

OK, that's far enough to avoid any gross amount of external
capacitance, which is good news because it suggests that the quite
low Fres could be occuring because of a high effective inductance,
which in turn could be a result of a copious amount of radial
coupling through the internal capacitance of the secondary.

> construction technique I used is 1/4" Plexiglas over 3/4" plywood

a fact which could well be contributing significantly to the radial
capacitance.  You might be able to test this by laying an extra
sheet of plywood or plexi over the secondary and re-measuring Fres. 
This additional dielectric should lower the Fres noticeably further.

> I'm using a 15KV 60mA NST.

Gosh, thats 15kV times sqrt( 6314/9.76) = circa 380kV.  That's an
average of 380/227 = nearly 1.7kV per turn!  I'm amazed that your
secondary will stand that.  Considering that the actual voltage
gradient is almost certainly quite non-uniform, it could be over
2kV/turn in places.  Certainly raises the eyebrow in a Spock-like
manner.

Can anyone claim a similar order of secondary volts/turn from their
coils?

I don't know what others think, but IMO if you put 15kV into
the thing, I'm sure the secondary would break down.

Moving on to the next step...Want to measure secondary input
impedance and Q factor?

Input impedance: Find the Fres as you did before, but this time
note down the voltage you get at the bottom of the dip (about 2V
you had I think) and also the voltage when well away from resonance,
(above 20?).  From these numbers, and a measurement of the R that
you use, we can estimate the input impedance.  Make sure you use a
sine wave from the generator, not square or something.  Oh, we also
need to know the sig gen output impedance if possible.

Q-factor: Drive the secondary 'base' as above, but without the 
resistor. Use the AC voltmeter and a short wire antenna to register
a wiff of the field from the coil. Don't get too close with it, but
enough to give a decent reading.  Couple to E or B, doesn't matter
which.  Tune to Fres and adjust to peak the reading.  Note the
frequency, then tune both sides to where the reading is 71% of your
peak.  Note these two frequencies.  From these we can get the Q
factor.

>From the input impedance and Q factor, we can go on to calculate
the effective (energy) inductance Lee, which, when compared with
the DC inductance of 6.23mH, tells us a fair bit about the coil. 
My guess is Lee will be 20% to 30% above Ldc, in other words, a bit
like a short h/d solenoid, only perhaps more so.  If it does go
that way, it could make an exciting coil for a CW system.
--
Paul Nicholson
--