Mutual Inductance (was Flat Primary Winding - next question)

From:  John H. Couture [SMTP:couturejh-at-worldnet.att-dot-net]
Sent:  Saturday, June 20, 1998 3:01 PM
To:  Tesla List
Subject:  Re: Mutual Inductance (was Flat Primary Winding - next  question)

  All -

  The mutual inductance of a Tesla coil is important because it is the major
parameter that connects the two magnetic circuits of the primary and
secondary tank systems. Mutual inductance can be used to find other
parameters including the important K factor or coupling of the coils.
      K = Lm / sqrt(Lp x Ls)

  It is always interesting to compare different methods of finding TC
parameters like the mutual inductance. Mark has listed below the mutual
inductance he has found for different primary coil arrangements. To compare
his data using the JHCTES program I came up with the following for System A:
   Pri cap = .031 uf   Pri spiral avg rad = 9.0 ins   Width = 5.7 ins
   Sec rad = 5.14 ins   Turns = 1000   TPS = 33.33   
   Bare wire dia = .020    Sec term = 40 pf

   Some of the outputs
   Pri ind = 135 uh    Pri turns = 15
   Sec ind = 76.3 mh       K Factor = .20 
   Op Freq = 77.86 KHZ    Mut Ind = 656 uh
  Keep in mind that small changes in the inputs can make large changes in
the outputs when solving for mutual inductance. The pri cap in the inputs
can be used to change the pri turns. Coilers who have the JHCTES program may
want to try other changes to see how they affect the mutual inductance.

  To my knowledge the only TC program available today  for the coiler that
solves for the mutual inductance is the JHCTES program. The Corum program
has mutual inductance as an input not an output. The JHCTES program uses an
algorithm based on an equation in Terman's Radio Engineering Handbook.

  One advantage of Mark's program is that it provides more detail of how the
primary bathes the secondary as you go up the sec coil. To do this with the
JHCTES program requires additional calculations on the part of the program
user. If much of this work was to be done it would be better to write a new
program and incorporating Terman's equation.

  John Couture


At 07:29 AM 6/18/98 -0500, you wrote:
>From:  Mark S. Rzeszotarski, Ph.D. [SMTP:msr7-at-po.cwru.edu]
>Sent:  Thursday, June 18, 1998 2:49 AM
>To:  Tesla List
>Subject:  Re: Flat Primary Winding - next question
>Hello All:
>Steve Young said in part:
>>I am wondering just how
>>primary diameter affects coupling to the secondary.  I envision the
>>magnetic flux from a larger diameter primary will intercept more of the
>>secondary than will a more compact primary, but I don't know the practical
>>significance of such an effect.  I would assume we should couple to as much
>>of the secondary as possible to even out the secondary volts per turn
>>distribution (implies large diameter primary).
>Consider three flat spiral primaries:
>System A: inside diameter 12", outside diameter 23.34", N=15.125 turns
>(Terry Fritz's primary)
>System B: inside diameter 12", outside diameter 36.00", N=15.125 turns
(match N)
>System C: inside diameter 12", outside diameter 36.00", N=14.000 turns
>(match Lp)
>        Assume each is coupled to a secondary which is 1000 turns of .02"
>wire on a 10.25 inch diameter by 30 inch tall secondary, positioned such
>that the bottom turn of the secondary is in the same plane as the flat
>spiral primary.
>Results of mutual inductance calculations:
>System A:  Lp=131 uH   Ls=76.3 mH   M=636 uH   K=.20
>System B:  Lp=151 uH   Ls=76.3 mH   M=614 uH   K=.18 (same # turns as System A)
>System C:  Lp=132 uH   Ls=76.3 mH   M=569 uH   K=.18 (tried to match Lp of
>System A)
>        Conclusion: you get slightly better coupling with the smaller
>diameter primary.  The current in the secondary is directly proportional to
>M, so the small diameter primary may be advantageous.  Of course, you could
>wind a solenoidal primary and get even higher M, if you can hold off the
>higher voltages.....
>>This leads to another thought.  If the primary effectively couples mainly
>>to the lower part of a 1000 turn secondary (e.g. the first 100 or so
>>turns), then do we in effect have a 100 turn secondary feeding a 900 turn
>>third coil?
>        One can examine the degree of coupling between the primary and
>secondary by looking at the mutual inductance M between the primary and a
>one inch tall segment of the secondary.  In the example below, I looked at
>the coupling between the primary and a secondary 10.25" diameter, 1" height
>with 33.33 turns.  The secondary coil is moved from 0" above the primary to
>28" above the primary, near the top of the actual secondary.  Here are the
>results for System A and System C (which have the same Lp and Ls):
>Position        M for System A          M for System C
>    0"          93 uH   (100%)          64 uH   (100%)
>    2"          71 uH   ( 76%)          53 uH   ( 83%)
>    4"          50 uH   ( 54%)          42 uH   ( 66%)
>    6"          35 uH   ( 38%)          32 uH   ( 50%)
>    8"          25 uH   ( 27%)          25 uH   ( 39%)
>   12"          13 uH   ( 14%)          15 uH   ( 23%)
>   16"           7 uH   (  8%)           9 uH   ( 14%)
>   20"           4 uH   (  4%)           6 uH   (  9%)
>   24"           3 uH   (  3%)           4 uH   (  6%)
>   28"           2 uH   (  2%)           3 uH   (  5%)
>        It is clear that the larger diameter primary bathes the secondary a
>little higher up than the smaller diameter primary.  Still, most of the
>coupling is accomplished in the lower 1/4th of the coil height or so.  So,
>to answer your question, no, not 100 turns, more like 200 turns!
>        There are striking similarities between the magnifier configuration
>and the conventional two coil system.  Unfortunately, you generally lose
>with the magnifier, unless you can build a really fast quenching series type
>rotary spark gap to handle the higher coupling coefficient.
>Mark S. Rzeszotarski, Ph.D.