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Re: Help in calculations

On Thu, 25 Apr 1996 22:03:35 -0600, Tesla List
<tesla-at-poodle.pupman-dot-com>, you wrote:

>> Subject: Re: Help in calculations
[snip]

>Jim,
>I accept that the neon be treated as a current source, but how do you 
>rewrite the classical voltage vs time charge equation to use a current 
>source. I know its more complicated than just replacing the "e" (charging 
>voltage) with 120ma and a "z" calculated from the neon's voltage and 
>current capacity.
>
dV  =  I  * dT / C  if you integrate this, you will get
  V + V0 =  I * T / C  ; where V0 is your starting voltage. so:

 V = 0.12mA  *  2mS / 0.021uF = 11.4kV

Now for the real question. Does the voltage vs time on the cap look
like a straight line? My gut says that this is false, so I'll look at
it with my 16nf cap and 60 mA neon tonight.
>> 
>> [snip]
>> > Also if I know the the
>> >maximum voltages to which I can get the cap charged, it becomes trivial
>> >to determine if additional fires per half cycle will gain any increase in
>> >power transfer.
>> power available:        15kV * 0.120 =  1.8KVA
>> power used:
>>                 1/2 * c  *  V^2 * pps =
>>                 .5 * 0.021*10^-6 * (0.707 * 15kV)^2 * 120 = 142 watts!
>
>I like your math. This means that I am getting discharges at the rate of 
>about 3 watts per inch. Tesla never did that well. 
>
I'll stick by this number until someone shows me where the extra
energy is coming from. If you run the numbers using Tesla's values, I
think you will find that his actual power input was much lower than
everyone says. If you thing about the neon acting as a perfect
transformer until it reaches it's current  limit. Until the voltage
reaches the firing point of your spark gap, you are "wasting" (not
using) power. At this point the gap fires, discharging the cap. The
neon now wants to be at (say 11.4kV) into 0 volts (the cap), it goes
into current limiting and acts as a current source for the portion of
the AC cycle that puts the cap voltage above you gap firming voltage.
The remainder of the half cycle is again unused power.

>But seriously...This can't be right. Something is missing and I can't put 
>my finger on it.
>

I agree, and I have not figured it out yet.

Regards,

jim