Re: Calculation of PFC Capacitors using LTR Capacitor

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

Original poster: "Ed Phillips by way of Terry Fritz <twftesla-at-qwest-dot-net>" <evp-at-pacbell-dot-net>

Tesla list wrote:
> 
> Original poster: "Daniel McCauley by way of Terry Fritz
<twftesla-at-qwest-dot-net>" <dhmccauley-at-spacecatlighting-dot-com>
> 
> > Here's a question I've asked before.  Since the secondary of the
> > transformer is loaded with a capacitor, why should the input reactance
> > necessarily be inductive?  Why should a PFC necessarily reduce the line
> > current?  And if it does, why should the correct value of capacitance be
> > the same independent of the capacitance in the secondary?
> >
> > Ed
> 
> Because the input impedance at the primary is in simplest turns an inductor.
> The impedance of the secondary is isolated from the impedance of the
> primary.  Its a magnetic circuit.
> 
> Dan

	Nope.  The input impedance of the transformer is ALWAYS a function of
both primary and secondary impedance, and there is NO isolation because
"it's a magnetic circuit".  The equivalent circuit of a transformer is
that of a fixed ratio transformer in series with the leakage reactance
referred to the terminal you are defining.  In the case of an NST the
leakage reactance is intentionally increased by using magnetic shunts
between primary and secondary magnetic circuits and is about equal to
the rated (open-circuit) voltage of the transformer divided by the short
circuit current.  When you place a capacitor across the secondary, as is
the case with the tank circuit of a TC, the input impedance changes. 
For a "matched capacitor" resonant with the leakage reactance the
reactive component of the input current should be zero and the PF
unity.  The input current would be supplying ohmic losses only.  In that
case, if you operate the transformer with rated line voltage the input
current will increase due to increase in the secondary current due to
transformer saturation, if the transformer doesn't short out
immediately.  I have measured the Q of several NST's with a small input
voltage, and it has always been greater than 10, implying that the
output voltage would be greater than the open-circuit voltage by that
same ratio.  Things aren't this simple, of course, because the
reluctance of the magnetic circuit changes with flux density, but the
general phenomenon is correct.

Ed