Re: Amount of strings in parallel (primary C)

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Seastiaan,

First say what your power source is (ie, 15KV 30ma NST) and what type 
of gap you are planning on.  Next you need to calculate the resonant 
capacitance value:


Cres(nf)  = 10^9/(2*pi*line_freq*XL)

XL  ~= Vs_oc/Is_sc

If you are using a static gap then Cp = 1.6 * Cres         (LTR value)
If you are using a SRSG then  Cp = 2.8 * Cres


Next is to find the maximum peak voltage you will operate at.  For 
static gaps you can figure 1.414 * Vs_oc_rms.  For a 15KV nst, this 
will be 21.2KV. Note that if the main gap and safety gap are opened 
too far, voltages greater than 21.2KV can be obtained.  For a SRSG, 
it is possible for voltages to scream to higher values if the firing 
is too late.  Since I use the "Terry filter" that has MOVs to protect 
the NST from overvoltage,  I set the cap string voltage close the MOV 
voltage.  In my case, where I'm using 15KV NSTs, I set the cap string 
voltage to 28KV which is 14 of the CDE 942C series 0.15uf.  This 
gives me a 10.7nf 28KV string.  From your calculation above, you can 
determing how many strings you will need to get the value. You can 
also build a string with 0.10uf caps or 0.068uf caps if the total 
capacitance doesn't come out right.  You can also add more than the 
14 caps in a string to get the capacitance you want.

Gerry R

> Original poster: Illicium Verum <sebas@xxxxxxxxxxxxxxxx>
>
> Hello everyone,
> I was wondering what determines the amount of capacitor strings
> necessary to be placed in parallel for a proper primary capacitor.
>
> I know it is necessary to divide the power over the amount of strings
> and am planning on using 5, but I don't know what determines it to be
> 5 and not 4 or 6 for example. Has anyone a good explanation for this?
>
> --
> Best regards,
> Sebastiaan