Re: breakout voltage

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

Original poster: Jim Lux <jimlux-at-earthlink-dot-net> 

At 11:30 AM 1/23/2004 -0700, you wrote:
>Original poster: "Luke" <Bluu-at-cox-dot-net>
>I found the following formula on the web.
>
>The formula for the field strength on a conductor with a given radius is:
>Field strength = Voltage / Radius.
>The break down voltage for air is approximately 76KV / inch.
>This value varies with humidity, frequency, temperature, barometric 
>pressure etc.
>So combining these gives:
>Max Voltage (before air breakdown) = 76KV * Radius (in inches)
>
>Is this correct for a close approximation?
>When applying this to a toroid I was assuming the minor radius should be 
>considered since it is the smallest curve on the toroid so the field 
>strength would be strongest there.
>Am I correct in assuming this formula could be applied to the minor radius 
>of a toroid to get it's approximate break out voltage?


That's the case.  Bear in mind that the formula is for an idealized 
perfectly smooth surface.  If your toroid has small bumps, specks of dirt, 
etc., then the breakdown voltage will be less. (typically, around 50-60%, 
unless you've taken obsessive pains)  The formula is also for the object in 
"free space", which practically means that it's more than 10 times its 
largest dimensions from anything else (kind of not what a TC 
is).  Anything, particularly metallic, but insulators too, in the immediate 
vicinity (say, within a few radii) will change the field distribution, and 
therefore the breakdown voltage.


You can use Terry's E-Tesla program to find out what the field actually is.


The difference between theory and reality for HV breakdown is what makes 
designing HV equipment as much an art as a science.