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Re: HF sparks (current draw)
From: John H. Couture[SMTP:couturejh-at-worldnet.att-dot-net]
Sent: Sunday, August 03, 1997 1:05 AM
To: Tesla List
Subject: Re: HF sparks (current draw)
At 09:06 PM 8/2/97 +0000, you wrote:
>
>From: FutureT-at-aol-dot-com[SMTP:FutureT-at-aol-dot-com]
>Sent: Friday, August 01, 1997 4:25 AM
>To: tesla-at-pupman-dot-com
>Subject: Re: HF sparks (current draw)
>
>In a message dated 97-08-01 01:07:13 EDT, you write:
>
><<
>> Try using 365 KHZ for your operating frequency. This is a possible
>optimum
>> frequency for 900 watts input that coilers have used in the past. This
>comes
>> from the equation
>
>> KHZ = 3032.5 x W^-.2767 - 96.4 W = watts input
> >>
>
>John,
>
>Are you suggesting that a 15kV, 60ma neon transformer will draw
>only 900 watts input power? My 15kV, 60ma tranny draws 2600 watts
>using a .014uF cap, and my 12kV, 30ma transformer draws 680 watts
>using a .007uF cap. This is because the reactance of the cap is equal
>to the leakage reactance of the neon transformer, and therefore
>neutralizes it, allowing the transformer to draw more than its rated
>current. This will only occur with optimally sized caps. Smaller caps
>will cause less power to be drawn. There seems to be a widespread
>misunderstanding of this effect. It is this effect that allows neon systems
>to perform as well as they do. But is is essential that anyone using a
>neon transformer measure their input current rather than assuming that
>the rated current is being drawn.
>
>John Freau
>
>------------------------------------------------------------------
John F. -
I have been testing neon transformers and you are right the nameplate
rating means very little. The actual output depends on the type of load both
resistive and reactive. If you use a .16 uf instead of a .04 uf would your
TC neon draw more current? I have found this is true with a 7500 V, 30 ma
neon I tested. However, I was not using a TC, I was using only a .16 uf cap.
The .16 uf was not equal to the leakage reactance. The voltage was 6600
with 36.5 ma giving 241 volt amps. The input was only 96 volt amps, a great
pseudo free energy device. Anyone can set up this more output than input
device to fool their friends. But you have to use a capacitor load. The
wattage (energy loss), however, is almost unreadable and the input wattage
is always greater than the output.
How did you measure the 2600 watts input? The 2600 watts would give
2600/15000 = 173 ma. I would think this would burn up the 60 ma winding
either as watts or volt amps. However, the output could be 15000 x 60 ma or
900 watts without harming the neon secondary. This would give 900/2600 =
34.6 % efficiency, a reasonable value for this size Tesla coil.
John Couture