Re: secondary output calculations (fwd)

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Original poster: List moderator <mod1@xxxxxxxxxx>



---------- Forwarded message ----------
Date: Sun, 20 May 2007 22:38:56 -0700
From: Barton B. Anderson <bartb@xxxxxxxxxxxxxxxx>
To: Tesla list <tesla@xxxxxxxxxx>
Subject: Re: secondary output calculations (fwd)

Hi Scott,

Ohms law always works (it must), but to often we just don't have 
accurate information for top volts. A typical terminal output voltage 
for our coils may indicate near 500KV, but that is a non-breakout 
potential. Usually, you'll calc a secondary current somewhere between 5 
and 20 amps using a typical Vp calculation (sqrt[(2J)/Csec)] for the 
norm of coils built. The problem with the results are the topload 
breakout conditions.

The topload will usually breakout at a much lower voltage than calc'd. 
There are many reasons for that. Some reasons have to do with the 
topload geometry itself. Yet there are further considerations such as 
the E-field, how space charge affects the E-field, RF AC pulse 
conditions, etc., all making for a very complex situation regarding 
avalanche breakdown. I doubt there are problems with your calculation. 
But, yes, the results will be very high and not realistic.

The only way to find out that I'm aware of is to take several sample 
measurements under breakout conditions of the base current (and 
average). You can measure the base current with a current loop (such as 
a Pearson current probe which converts amps to volts where typically 
1A=1V). The top volts are not so easily measured. That is the "real" 
problem. Direct measurement is near impossible even with state of the 
art capacitive dividers (which are expensive). Even they distort and 
affect the E-field your trying to measure. Trying to take a top volt 
measurement without affecting the coil is troubling to say the least. 
You can however guestimate the top volts under model but then you have 
to assume an E-field value at which avalanche breakdown will occur. This 
is somewhat understood as about 25kV/cm to 30kV/cm. The problem is 
determining complex influence on a portion of the topload that will 
achieve a predicted breakdown voltage. The areas around the topload are 
affected by the space charge surrounding the topload (and not 
necessarily evenly distributed). The E-field at portions of the topload 
are influenced by the space charge and the topload itself. It will 
breakdown at a much lower average voltage than what a typical 
calculation will derive.

Even if you insert a topload capacitance into a calculation, there are 
various geometry's of top loads that can have that same capacitance 
value and yet be very different as far as breakout potential.

However, using ohms law, you can measure the base current and likely get 
close. Don't be surprised if your top volts ends up between 1/2 and 2/3 
of what you are calculating now. Those are calculations which have no 
consideration about what is going on at the top of coils.

Javatc uses what can be considered a "maximum conceivable top volts" 
using the energy storage capacitance (Cee). It is a simple calc: 
Vp*sqrt(Cpri/Cee). Cee is a good number for this particular max top volt 
value. It represents the capacitance when all the energy is stored in 
the topload (that brief period in the cycle). But it is not what you 
would measure if you could measure it. It does not represent the real 
situation for all the reasons stated previously. Thus, the secondary 
current falls prey to the same if using top volt calculations to 
determine it.

Hope this helps a little. Others here have a much better understanding 
of the situation and could do a far better job at explaining it than I 
could.

Take care,
Bart

Tesla list wrote:

>Original poster: List moderator <mod1@xxxxxxxxxx>
>
>
>---------- Forwarded message ----------
>Date: Sun, 20 May 2007 01:46:06 -0400
>From: Scott Bogard <teslas-intern@xxxxxxxxxxx>
>To: tesla@xxxxxxxxxx
>Subject: secondary output calculations
>
>Hey everybody,
>     I have found formulas to calculate the peak voltage put out of a Tesla 
>coil system, how does one calculate the amperage (please note, I have only 
>basic calculus skills).  Does ohms law work?  I am getting ridiculously high 
>numbers from ohms law, using my calculated peak voltage, my secondary 
>resistance, and the impedance from my secondary inductance, capacitance, and 
>top load capacitance (but my numbers may be off, especially concerning the 
>impedances, it has been a little while since I have done that stuff for a 
>class).  Thanks.
>Scott Bogard.
>  
>