# Re: Power vs Voltage vs Current? (fwd)

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---------- Forwarded message ----------
Date: Mon, 27 Apr 1998 10:10:01 -0500
From: David Dean <deano-at-corridor-dot-net>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Power vs Voltage vs Current?

-----Original Message-----
From: Tesla List <tesla-at-pupman-dot-com>
To: 'Tesla List' <tesla-at-pupman-dot-com>
Date: Monday, April 27, 1998 9:46 AM
Subject: Power vs Voltage vs Current?

>
>----------
>From:  djQuecke [SMTP:djQuecke-at-worldnet.att-dot-net]
>Sent:  Monday, April 27, 1998 6:48 AM
>To:  tesla-at-pupman-dot-com
>Subject:  Power vs Voltage vs Current?
>
Hi DJ

>I know that stepping-up a voltage results in lower available current
and
>that stepping down a voltage increases available current.  How do you
>calculate the change in current output?
Yes, in a transformer designed for 120VAC in, 12VAC
out, the current available from the secondary would be 10 times the
current drawn by the primary, less the core losses(a few percent
normally taken as negligable).

>Examples using a 15kV neon, with a 30ma rating and 120v input:
>
>The following would appear to be true:
>
>1v to 125v step-up ratio
>Available Output Power is .45kVA (450w)
>
>Then:
>
>    1.  Lowering input voltage to 96v should result in an output of
12kV:
>
>         This is a 20% decrease in voltage.  Would this increase
available
>current 20% to 36ma (30ma * 1.2) or perhaps, since the transformer is
rated
>at .45kVA, increase available current to 37.5ma (450w / 12kV)? (My
guess is
>the latter).
>
>Conversely:
>
>    2.  Raising input voltage to 144v should result in an output of
18kV.
>
>         This is a 20% increase in voltage.  Would this decrease
available
>current by 20% to 24ma (30ma * .8) or based upon the .45kVA rating,
does
>this decrease available current to 25ma (450w / 18kV)?
>
Not true. Here you are (I think) talking about changing the input
voltage as if with a variac. This does nothing to the turns ratio inside
the transtormer which is where the "transtformation ratio" comes from.
Lowering the input voltage will simply lower the power going in,
lowering the input current, ultimatly lowering the output current as
well as the output voltage.

>For coil use, is there any advantage is there any advantage in raising
or
>lowering the voltage of a neon transformer?   Same question except with
a
>pole pig?
>
The advantage would be the ability to control the amount of power goung
in.
For example, starting a Tesla coil out with full power can have
disaserous results
as the surge of power comming in could overload the caps etc. This is
why many people
recomend the use of protection gaps on caps and things.

>Part of me wants to think that what matters is available power and that
the
>above examples wouldn't really affect the output of a coil much and
part of
>me finds that with a change from 12kV at 37.5ma to 18kV at 25ma, there
has
>got to be some change in a coils performance.  If there is an expected
>change in performance, which would increase the performance of a coil,
>raising or lowering the voltage?  (And is this due to the change in
voltage
>or the change in current?)
>
>In addition, I don't quite understand how a neon transformer is current
>limited.  I've yet to tear one apart although I've read several places
that
>wood chips are somehow placed in the windings to accomplish the current
>limiting.
>
NST's are current limited by the installation of shunts- metal
laminations- bypassing part of the
"magnetic current" around the primary coil, bypassing the secondary.
This increases the current draw in the primary somewhat which brings the
core closser to "saturation". Thus when the current draw on the
secondary
increases, the core is more easily saturrated at which point no more
current can be drawn by the primary, effectivly "clamping" the current
at this point. Please note that as the secondary current increases its
voltage drops. The VA rating of a NST applies to the pri. But the
secondary will never actually deliver that much power.
15KV is open circuit -at- no mA. 30mA is short circuit with no volts.

Hope this helps, not just cause more confusion-
deano

>Can someone straighten me out here?
>
>Thanks Much,
>
>dj
>
>

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