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Re: Pig limiting...
In a message dated 99-10-07 06:37:01 EDT, you write:
<< When using a choke to limit current, does more L = more or less current?
> IIRC more L would increase the limiting action producing less current,
> but I would like to make sure I'm right. About half of the parts I need
> for a power controler are on the way, so it would be nice to know before
> I power the thing up. I'm thinking I want to use a choke and a small
> resistance like a heating element to control this. With the choke being
> the main limiter.
>Travis >>
Travis,
That's a good question, and the answer is that more L can equal either
more OR less current, depending on certain factors. The most current
may be drawn when the ballast L cancels the tank C. This is the resonant
charging condition. The amount of L that cancels the tank C will depend
on tank C and the step up ratio of the transformer. This seems to apply
most purely at 120 bps. At higher break-rates, the ballast may need to
be adjusted a little differently, but the general concept applies.
The impedance is
transformed (by the transformer) by the square of the step up ratio. For
instance if you have a .0147uF tank cap, and a 1: 120 step up ratio, you'll
need about 6mH of ballasting for resonant charging at 120 bps. THe power
input will be reduced if the ballast L is either decreased OR increased.
The ballast L should also be adjusted for best power factor if you care about
efficiency. At 120 bps, resonant charging will give best power factor. At
higher break rates, the setting may have to be different.
At higher break rates, with larger capacitors and a low step up ratio, the
ballast will tend to work more like a pure choke, such that less L will give
you more current input. At least this is what I've seen in my work.
John Freau