[Prev][Next][Index][Thread]

Re: TC Critical Coupling (was Overcoupling



Hi John,

> Original Poster: "John H. Couture" <COUTUREJH-at-worldnet.att-dot-net> 
> 
>   Malcolm -
> 
>   Thank you for the reply. To determine the amplitude of the TC energy don't
> you need to measure the amplitude of two parameters plus the time period?
> 
>   Energy = Volts x Amps x dt

Not at all. Knowing capacitance and instantaneous voltage, E pops 
straight out. 

>   To determine the overall efficiency you need the input energy and the
> output energy.
> 
>   % Efficiency = 100 x Output energy/Input energy  
> 
>   If I understand your reply correctly you measured only one parameter and,
> therefore, did not have the coil energy.  

See above. E = 0.5CV^2.  In order to determine the energy retained by 
the system between secondary maximums as a percentage of initial 
energy, you go thus:

Initial energy = 0.5CVa^2   where Va is the initial peak amplitude you 
measure on the scope. Remaining energy following second transfer is 
0.5CVb^2 where Vb is the peak voltage following the second transfer. 
In between first and second maximums, energy has gone back to the 
primary and then back to the secondary again. Absolute values don't 
matter because you are interested in the *difference* between these 
two energy levels. The difference = 0.5C( Va^2 - Vb^2 ). This is the 
amount of energy lost over *2* transfers so to get the loss over 
one transfer this is halved - i.e. energy loss for one transfer 
approx = 0.5C( Va^2 - Vb^2 )/2 
which = 0.25C( Va^2 - Vb^2 ).

     Since you wish to end up with energy remaining as a percentage 
of initial energy, you divide energy retained by the system after a 
single transfer by the initial energy (0.5CVa^2) and multiply by 
100. 

    Putting it all together:

Energy at first peak = 0.5CVa^2

Energy lost over a single transfer = 0.25C( Va^2 - Vb^2 )

Energy remaining in system after a single transfer 
= 0.5CVb^2 + 0.25C( Va^2 - Vb^2 )
= 0.5C( Vb^2 + 0.5Va^2 - 0.5Vb^2 )
= 0.5C( 0.5Va^2 + 0.5Vb^2 )
= 0.25C( Va^2 + Vb^2 )

Energy remaining as a fraction of original energy
= 0.25C( Va^2 + Vb^2 )/0.5CVa^2
= 0.5( Va^2 + Vb^2 )/Va^2

%loss for one transfer = 100 * 0.5( Va^2 + Vb^2 )/Va^2

                       = 50( Va^2 + Vb^2 )/Va^2

EXAMPLE: Suppose Va = 5V and Vb = 4V

Then remaining energy after a single transfer as a fraction of 
original energy = 50 * (25 + 16)/25 %

= 82%

Fraction of energy remaining after *2* transfers = 0.5CVb^2/0.5CVa^2

= 0.64  which when converted to a percentage = 64%

Hope you find this useful. Performing this test without breakout 
gives a good picture of just how good your system is as far as power 
loss in transferring from primary to secondary goes. If the secondary 
has a particularly high Q, the calculated figure is representative of 
the degree of loss in the primary in general and the gap in 
particular.

>   I show how I did this for one of my coils in my Tesla Coil Construction
> Guide. I haven't heard of anyone else doing this relatively simple test.
> This type of test makes more engineering sense than the random extra long
> spark.
> 
>  How did you test for "equal amounts of energy are lost" in the Tesla coil
> primary and secondary circuits?   

That is not something I would ever aim for.
 
>  After 100 years and thousands of coils built and tested the coilers still
> do not agree on the proper way to do the TC testing. It's about time for
> someone to write a book on the proper testing of Tesla coils. Not me.
> 
> John Couture

I think most agree on a number of things such as attached sparks must 
be measured point - point in a straight line. The method of 
determining a transfer loss which I outlined above is valid. Power 
factor during operation is quite measureable. Primary power is 
certainly measureable (E*BPS). I think that if a coil occasionally 
reaches out to x feet it has been demonstrated as being quite capable 
of doing so at some particular power level which is quite open to 
measurement. It may be that those occasional long sparks correspond 
to fluctuations in wallplug draw. I've seen power consumption rise 
considerably when sparks attach themselves to some object. I think 
there is further room for more exacting measurements of such events
and I'm looking forward to doing some of them (if Marco doesn't beat 
me to it which he probably will :)

Cheers,
Malcolm 

** If it's under 10 Amps it's Leakage Current **