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

```Original poster: "Gerry Reynolds" <gerryreynolds-at-earthlink-dot-net>

Hi Robert,

I've started some simulations to get a better feel of the transient reponse
and its affect on the charging time for various Cp's from small to Cres to
C_ltr and I'm seeing a phenomenum that might go well into a java kind of
designer, but I will hold off until I get more into it.

> Original poster: "Robert Jones" <alwynj48-at-earthlink-dot-net>
>
> More thoughts on this.
>
>  > 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.

I'm thinking that the firing voltage is not only function of the line freq,
but also the resonant freq of the LRC circuit which may negate the desire to
be periodic with the line frequency.  The beginning of a new transient will
start when the SG quenches so I'm thinking there are too many other
variables that are asynchronous to the line frequency.

> 2. At SG conduction the cap current should be zero.

Do you mean when the SG stops conduction (ie, at quench)

> 3. Unipolar charge current.
>
> I think constraint 2 removes the memory of pervious conductions.

The only memory, I think, would remain is current thru the NST inductor.  It
seems this will be a function of where in the AC cycle we are when quench
occurs and what the transient history is.

> 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.

For LTR?

> 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
>
>

```