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Mutual Inductance (was Flat Primary Winding - next question)


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From:  terryf-at-verinet-dot-com [SMTP:terryf-at-verinet-dot-com]
Sent:  Sunday, June 21, 1998 12:13 AM
To:  Tesla List
Subject:  Re: Mutual Inductance (was Flat Primary Winding - next   question)

Just for clarity,

The coil in case A has the following parameters:

PRIMARY
15,000 volt 60 mA neon input (variac controlled)
17.05nF primary cap
8.84 inch avarage radius flat spiral primary of 1/4 inch tubing.
5.67 inch wide coil section
Lp = 118.4 uH
Fo = 112.0 kHz

SECONDARY
5.13 inch radius
1000 turns
33.33 turns per inch
0.02002 bare wire diameter
Ls = 75.4 mH
Cself = 16.68 pF
Ctop = 10.10 pF

The JHCTES program gives a K of 0.21 and m = 602.04 uH.

Mark's program gives a K of 2.01 and m = 635.96 uH.

The actual K is 0.2069 and m is 618.19 uH.

The thing I really like about Mark's program is that it predicts K for
various coil heights.
My coil will not quench well with a K of 0.21.  I have to raise the coil
about 2 inches to get a clean quench.
Mark's program very accurately predicts the K at various heights.  The
JHCTES program omits this input because of "lack of emperical data" (page
11-1 Tesla Coil notebook).  Perhaps John could add this feature to JHCTES??  

        Terry Fritz



At 08:09 PM 6/20/98 -0500, you wrote:
>
>
>----------
>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:
>>><snip>
>>>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.
>>
>>Regards,
>>Mark S. Rzeszotarski, Ph.D.
>>
>>
>>
>
>
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