Re: Theory of LTR

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>

Hi all,

I am reasonable confident I got it right. Here is the latest.
It much simpler to consider the equivalent circuit of the SG and Cp as
series Requ and Cp,
where Cp is unchanged and Requ is the product of the impedance of Cp at the
supply frequency, the sin of the firing angle relative to the charging
current and the ratio of fundamental to peak of a square wave. The  power
dissipated in R per cycle is the bang energy. Note that the harmonics of the
square wave are ignored but this produces only an approximately 5% error in
firing voltage because of the amplitudes of the harmonics and the high
impedance (relative to the supply frequency) of the L at their frequency.

It relatively simply (using AC circuit theory) predicts the optimum firing
angle, C value for max bang size, bang size and PF in the synchronous SG
case.
It theoretically confirms the experimental and circuit simulation results
that an LTR C has the biggest bang energy.

Robert (R. A.) Jones
A1 Accounting, Inc., Fl
407 649 6400

----- Original Message -----
From: "Tesla list" <tesla@xxxxxxxxxx>
To: <tesla@xxxxxxxxxx>
Sent: Friday, November 11, 2005 9:42 AM
Subject: Theory of LTR


> Original poster: "Bob (R.A.) Jones" <a1accounting@xxxxxxxxxxxxx>
>
> Hi all,
>
> I recently tried a different direction on the theoretical the optimum
> primary C (Cp) for a given inductive ballast (L)
> The maths is not finished but the direction appears productive. Here is
the
> short word version of it, minus many of the assumptions, for those into
the
> theory stuff.
> First in a sync gap operating at the same break rate as twice the supply
> frequency.
> The Cp and its repeated discharge is equivalent to a square wave signal in
> series with Cp but without the SG.
> The amplitude of the square wave is equal to the voltage on Cp at the
point
> of discharge and has the same phase as the discharge but opposite
polarity.
> Considering only the fundamental of the square wave, the square wave  lags
> the voltage on  Cp but with the opposite polarity so equivalently it leads
> the voltage.
> Hence the combination of Cp and squarewave generator has an impedance (at
> the supply frequency) equal to a smaller C (Cequ) in parallel with a R.
> (Requ)
> The energy dissipated in Requ is equal to the bang energy.
> Therefore maximum dissipation in Requ will be when the impedance of Cequ
is
> equal to the impedance of the ballast inductor. i.e. resonant.
> As Cequ is smaller than Cp, Cp must be increased until its equivalent Cequ
> is resonant with L to obtain the maximum power in Requ which is the
maximum
> bang size.
>
> I am guessing but it will be similar with a static gap. The repeated
> discharge of Cp advances Cp voltage relative to its charge current so
again
> the equivalent impedance (at the supply frequency) is equal to a smaller
> capacitor and so the actual C must be made large to obtain the maximum
bang
> size.
> The above may also explain why the maximum bang size is obtained when the
> current in L is not zero. ie when the real current is a maximum there must
> still be some reactive current.
>
> Robert (R. A.) Jones
> A1 Accounting, Inc., Fl
> 407 649 6400
>
>