Re: Calculation of tank cap

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

Original poster: "Terry Fritz" <twftesla-at-qwest-dot-net>

Hi Herwig,

At 01:07 PM 12/1/2001 +0100, you wrote:
>
>Terry Fritz wrote:
>> The 0.83 is from computer modeling and seems to be very 
>consistent for various current limited transformers.
>
>Terry,
>
>Thank you for the explanation.
>
>Sorry to bother you again: Is this formula valid for *all* kinds of 
>xformers as well (you wrote "for various current limited 
>transformers")?

It should work for any linear current limited transformer configuration.
Pigs with current limiting chokes that saturate (non-linear) will be weird...

>
>I am obstinate: What do you think of another approach:
>
>1. Measure the open load current of the transformer's primary 
>winding (with inductive ballast connected) e.g. 0.8 A
>
>2. Calculate the magnetizing inductance e.g. Zprim = 120 V / 0.8 A 
>= 150 Ohm and Lmag = 150 / (2 x 60 x pi) = 0.4 H
>
>3. Measure the transformer's voltage ratio e.g. 120 V / 2000 V = 
>0.06
>
>4. Calculate the reflected inductance e.g. Lref = 0.4 / 0.06 ^2 = 111 H
>
>5. Calculate the size of the reso-cap e.g. Cres = 1 / [(2 x pi x f)^2 x
>Lref] = 63.4 nF 
>
>6.Add 50 % in order to get the capacitance of the LTR-cap e.g. 95 
>nF (seems to be a bit large ;-)  Wrong methode/values?)
>
>Your critics are very welcome!

Gee, I don't know if that will work or not.  Usually SRSG LTR values are
about 2X the resonant values.  In general, I figure out how much power the
transformer "should" be processing and then compute the 120 BPS caps size
than is needed.  That is what the equation I gave really does.  The 0.83
factor is Taking into account some losses and RMS conversion.

Although pig systems can drive just about any capacitor size till the
breakers blow, for a 10kW system I would figure:

10000 / 220 = 45 amps from the AC line
10000/ 14400 = 695 mA output current
10000 / 120 = 83.3 joules per bang (120BPS)
14400 x SQRT(2) = 20400 volts at firing
83.3 J/B x 0.83 = 69 joules per bang wanted

E = 1/2CV^2 so:

69 = 0.5 x C x 20400^2     C =331nF

Of course, with a pig system where you can vary the reactance and thus the
current limit, many other factors come into play and you have a lot of
freedom in the adjustments.  Simply by varying the current limit, you can
go from STR, Resonant, LTR...  However, this is the ball park figure "I"
would use to get the optimal LTR cap size for a 10kW coil.  A lot of
computer modeling could probably get a closer if not an almost exact value.

Cheers,

	Terry
  

>
>Hoping for your understanding,
>
>Herwig
>
>