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Re: Power Factor Correction weirdness?

Hi Terry ;)

> Original Poster: Terry Fritz <twftesla-at-uswest-dot-net>
> 
> At 11:11 PM 02/08/2000 -0500, you wrote:
> >
> >Can someone explain what I'm seeing when I attempt to power factor correct
> >my neons? I've got my DMM in series with the primary of a 12,000 60mA NST,
> >and the secondary of the NST is shorted.  With no PFC caps, it reads 7.13
> >Amps.  But when I start adding caps in parallel with the NST primary, the
> >current reading drops dramatically-- more than I expected it to.  Here are
> >the results:
> >
> >    PFC Caps    Ammeter Reading (amps)
> >
> >    0                7.13
> >    50 uF            4.96
> >    100 uF           2.81
> >    150 uF           1.14
> >
> >What's going on here? is this right?
> >
> >-Adam
> >adamsmith-at-mediaone-dot-net
> >
> 
> 
> Hi All,
> 
> 	I am not the biggest expert in the world on power factor correction so I
> may be just reinventing the wheel or be totally off track...
> 
> 	I looked at my coils and the PFC caps I use and the basic principle behind
> Adam's test seems to be a useful way of selecting PFC cap values.
> Basically just correct for the transformer's VA rating.  In the non-linear
> operation of a TC there are a few inaccuracies but they do not seem very
> significant.
> 
> The VA rating on my big coil is 900VA, which implies a PFC cap value of
> 166uF.  The actual value is 200uF.  On my little coil the VA rating gives
> 49.7uF where the actual value is about 60uF.  Not exact but close enough.
> Thus, I would propose the following equation (unless everyone has been
> using it for years and I am the last person on earth the figure it out ;-))
> 
> Cpfc = ( Vo x Vi ) / ( 2 x pi x f x Vi^2 )
> 
> Where
> 	Cpfc = Power factor cap value in Farads
> 	Vo = Rated NST output voltage in volts
> 	Vi = Rated NST output current in amps
> 	pi = 3.14159...
> 	f = AC line frequency (50 or 60Hz)
> 	Vi = AC input voltage (120 VAC)
> 
> For a 15kV/60mA transformer you get:
> 
> ( 15000 x 0.06 ) / ( 2 x 3.14159 x 60 x 120 x 120 ) = 165.8uF
> 
> To get "exact" you have to take all the non linearities, timing, and such
> into account which is a few orders of magnitude harder but this equation
> seems to do fine...
> 
> Comments, suggestions,...

only one - it would be useful to use I for current instead of V  :)

Regards,
Malcolm