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*To*: tesla-at-pupman-dot-com*Subject*: Re: Repost: NST VA Rating and Power Factor*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Sat, 19 May 2001 17:39:53 -0600*Resent-Date*: Sat, 19 May 2001 17:42:00 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <wteibC.A.0YH.ETwB7-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net> Tesla list wrote: > > Original poster: "Jason Petrou by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jasonp-at-btinternet-dot-com> > > > My power supply consists of three 15/60 NST's. Two are Franceformers, > > and are marked 495 VA, and also have "High Power Factor" on the label. > > The other is a Magnatek/Jefferson, marked 900 VA. > If you do the math (15000V*0.06A) you get 900VA - which is what it should be > for all 15/60 NSTs. I have no idea how you have got just under half that for > the franceformer ones... it is technically impossible :) Au contraire... you CANNOT get 60 mA at 15 kV out of one of those things.. you get either 60 mA at zero volts, 15 kV at zero current, or somewhere in between. Now, if you have an entirely inductive current limiting AND no power factor correction, what you have, equivalent circuit wise, is a 60mA-at-15kV source with a big inductor in series, and your VA calculation would be valid. Indeed, the VA would be 900 VA, but the active power might be 0 Watts (i.e. at open circuit or shorted output), with some variation in between. With PFC, you could, theoretically, short the output of the device (i.e. 0 volts out) with some current (60 mA), and since the active power is zero (0 volts * any current is zero watts), the apparent power (VA) would also be zero. In reality, the sucker will be getting hot, and there is some non-zero voltage out (i.e. there is some finite resistance in the wiring), so you would have nonzero active power, and as a result, nonzero apparent power. Let's say, for example, that under load, the 15/60 actually puts out around 7.5 kV -at- 30 mA. The load is absorbing about 225W, so, since the (active) power into the transformer MUST be bigger than the power out (there is never negative loss), there must be at least 225 watts going into the transformer. If the power factor were perfectly adjusted to 1, then you'd also have 225 VA. However, if the power factor were to be, say 0.5 (i.e. 45 degree phase difference between I and V), you'd be drawing 450 VA (with the same active power of 225W). As a practical matter, the typical NST is inductive, so PF correction would be done by adding capacitors. However, the amount of correction to add very much depends on what sort of load you're putting on the transformer. In our example above (with no correction) if you were correcting the shorted output situation, you'd need 900 var of correction. If you were correcting the open circuit situation, where the active power is only going to be the transformer losses, and because the current is low, you'd need a lot less. In the part load condition I described, you'd need around 400 var of correction. The manufacturer picks some reasonable situation, and corrects for that.

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