Re: Power vs Voltage vs Current? (fwd)
---------- Forwarded message ----------
Date: Mon, 27 Apr 1998 21:26:31 -0700
From: Jim Lux <jimlux-at-earthlink-dot-net>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Power vs Voltage vs Current? (fwd)
> Lots went before here....
> 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
> 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.
Not really. The shunt does bypass the magnetic flux, but it doesn't make
the core saturate. What it does is effectively put an inductor in series
with the winding, which will limit the current. This inductance is called
the "leakage inductance" because it is the flux which "leaks" out, i.e.
which isn't coupled in both primary and secondary.
The same technique is used in arc welders and fluorescent light ballasts.
In a low cost welder, the position of the shunt is adjusted by the
"current" control. All the way out is minimum leakage inductance, that is,
maximum current; all the way in maximum leakage inductance: minimum
Saturation is a very different phenomenon. It will result in a dramatically
increased harmonic content (the output will start having square corners and
flat tops), and much increased core heating (higher frequency harmonics
have higher "iron" losses, mostly due to eddy currents). Saturation is why
you can't run 180 volts into a 120 V neon transformer and get 50% more
voltage (and current) out (assuming the insulation doesn't break down).
> 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.
The leakage inductance created by the "shunts" effectively puts an
impedance of 15 kV/30 mA or 500K ohms in series. Since the impedance is
inductive (not resistive) it consumes no power. If you put a capacitive
load (or a capacitor in series) with the output, its reactive impedance
will cancel part (or all) of the inductive leakage reactance, and allow you
to draw more than the rated current from the NST. That cap will have to
withstand a fair voltage though...