1.5 res was Re: Improved Model for a Primary Charging CKT

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Original poster: "Robert Jones" <alwynj48-at-earthlink-dot-net> 

Hi all,

More thoughts on this.

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From: "Tesla list" <tesla-at-pupman-dot-com>
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Sent: Saturday, January 24, 2004 1:29 PM
Subject: Re: Improved Model for a Primary Charging CKT


 > Original poster: "Robert Jones" <alwynj48-at-earthlink-dot-net>
 >
 > The cap voltage consists of a series of terms:
 >
 > vc = vss + vtr1 + vtr2 + vtr3 + .........etc
 >
 > vss is the steady state response, vtr1, vtr2 etc are the transient
responses
 > from each sg conduction.
 >
 > vss is constant sin with a f equal to the supply f and an amplitude
 > determined by V R L and C
 > vtr's are damped cosines determined by the R L C frequency and  decay by
 > R/gap losses.
 > ie vss has the same f as the supply and vtr is the f of R L C.
 >
 > vss is known.
 >

Consider what we would like the solutions to have.

1. It should repeat at the supply frequency.
2. At SG conduction the cap current should be zero.
3. Unipolar charge current.

I think constraint 2 removes the memory of pervious conductions.
Hence only the first to terms need be considered.
This means we only need to consider 1/2 cycle.
It also mean that the slope of vcc must be zero immediately before
conduction for zero cap current at conduction.

If each cycle is identical but opposite polarity, then the start and end
voltages must be equal and opposite.

 >From sketching the waveforms  I can only find one solution that meets 1 2
and 3 and that's with a very large C and only one more case,  the resonant C
case,  that meets 1 and 2. There may be many more.

The one I am not certain about is the resonant C case. As I am not certain
if the start end voltages are equal magnitude. A problem with a two factor
in the e term.

The  very large C case is not interesting in its self.
But it does suggest a solutions were C is between 1 and 2 times resonance.

For example if vss is a is almost a cos while vtr has a lower frequency. vss
starts with  the opposite polarity to vtr.
At the end it has changed polarity and sums with vss.
To meet 1 and 2 the cos needs to be delayed so the slope of vcc and vtr are
equal and opposite at the end of the cycle, and have the start and end
magnitudes equal..

I assume this is the  1.5 res  solution  which is better (in my sketch) than
the 1 res solution because the charge current is more unipolar. I will have
to do the maths for a comparison of the bang size and fix the 2 problem.

A note of caution on the above.  I have not considered how stable the
solutions are (or even if a system can get in to them).  By this I mean that
if for some reason the ssg fires early or late  how does that effect the
timing of the next conduction. Ideally we want the solution to be stable
i.e. after an early or late conduction it returns to the original cycle.
Then a static gap can be used.  This probably has something to do with the
differential change in timing of the next conduction with respect to a
change of  timing of the first.

Hence 4. Good stability.


Bob