Re: [TCML] Bleeder resistors: [Please read in Courier-New font, 10 point. ]

[Davetracer@xxxxxxx]

Looks like it's time for me to go on another  journey up the learning curve 
...
 
I've seen Tesla Coil designs where the capacitor  is connected directly 
across the high voltage output "legs" of the neon sign  transformer. One leg 
then goes through a spark gap, through the primary coil,  and then connects to 
the bottom leg. 
 
Note: (If this drawing comes out as  a bunch of wildly disconnected lines, 
try viewing it in Courier New font at  10 point, like I did. I had to use a 
mono-spaced font to make the  drawings.)
 
I'll call this Design #1.  [Because of the  nature of "ASCII Art" for 
drawing, I'm not putting in the inductors on the NST's  legs, nor a Terry 
Protective Device ...]
 
This is kinda the "Classic NST Tesla Coil" design  that looks something 
like this... (below, Design #1)
 
Design #1
 
                  TTTT
                         )
NST   CAP     SG    PRI  )
) -----+----->  <---)    {
(       |            {    {
}    -----           )   {
)    -----           {   {
(       |            }    }
{------+------------}   {
                        GND
But I've also seen this design (#2). 
 
Design #2
 
 TTTT
                         )
NST     SG     CAP  PRI  )
) -----+------| |---)    {
(       |            {    {
}       \/           )    {
}         )    {
)       /\           {    {
(       |            }    }
{------+------------}   {
                        GND
 

Here's my question. Is either design more  efficient, or for other reasons, 
a better design, than the other?
 
   Then I have a couple questions on  bleed-down resistors.
 
 
I understand (well, I thought I did) about  bleed-down resistors. Is there 
a "rule-of-thumb" for how much resistance for how  much voltage? 
 
Or is it "Calculate how long it will take to bring  the cap down to zero 
volts in (for example) 10 minutes, and use that value?"  type of rule? Seems 
like with really big capacitors you'd have to figure in the  heat generated 
during the bleed-down, e.g., watts = I^2R. 
 
 It seems like the resistor would also have  to be physically designed so 
that at, say, 12kv, an arc doesn't develop around  the resistor as dust and 
so forth collect on it.
 
Anyway, thanks for the help, I do appreciate it.
 
 
  Thanks,
 
Dave Small
 


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