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Re: Primary Q - A Brain Teaser
Tesla List wrote:
>
> Original Poster: "Malcolm Watts" <MALCOLM-at-directorate.wnp.ac.nz>
>
> Here is the post requested by some offlist.
>
> Malcolm
>
> ..................Forwarded Message Follows........................
>
> Hi All,
> I had occasion to think about this while answering an off-
> list query and came to the surprising conclusion that Qp cannot be
> quantified as an absolute figure. Here is the reasoning:
>
> Fundamental definition of Q = 2PI x total energy/energy lost per cycle
>
> One can see that a log decr waveform always loses the same percentage
> of remaining energy from one cycle to the next. Not so a linear
> decrement which is what happens in the primary due to the presence of
> the gap. Here's why (The diagram below shows peak voltage on the +ve
> half cycle) :
>
> Vi - *
> Vf ----- *
> *
> *
> *
> *
> 0 .....................................
>
> Energy contained in tank = 0.5xCxVi^2 initially
> Energy contained in tank = 0.5xCxVf^2 at second peak
>
> Difference in energy between peaks = 0.5xCx(Vi^2 - Vf^2)
>
> Using the definition of Q above, Q = 2xPIx0.5xCxVi^2
> ------------------
> 0.5xCx(Vi^2 - Vf^2)
>
> which = 2xPIx Vi^2
> --------------------
> (Vi^2 - Vf^2)
>
> Let Vi = 10V and let the drop per cycle = 1V
>
> Then plugging in the figures, Q appears to = 2xPIx 10^2
> ---------------
> 10^2 - 9^2
>
> which = 200xPI/(100-81) = 33
>
> NOW, let us calculate Q between the second peak and the third peak:
>
> 2xPIx 9^2
> Q = ------------- = 29.9 !
> 9^2 - 8^2
>
> Let's try that for peaks 3 and four:
>
> 2xPIx 8^2
> Q = ------------ = 26.8 !!!
> 8^2 - 7^2
>
> A pattern emerges...
>
> Finally let's try it for peaks where Vi = 2V and Vf = 1V:
>
> 2xPIx 2^2
> Q = ------------- = 8.38 approx
> 2^2 - 1^2
>
> So as the voltage drops with each cycle, Q does as well !!!
> Clearly, in order to obtain a high primary Q, one must use the
> highest possible primary voltage. Although the energy lost in
> absolute terms is highest with the highest voltage, the energy
> lost as a percentage of total power is lowest. If one considers
> that the gap has a conduction voltage of say 50V and that is either
> fixed or rising, it's not hard to see that if remaining cap voltage
> is twice that figure, Q drops into the dirt on the last cycle before
> the gap finally goes out. It doesn't just tail off. Remember that the
> gap roughly exhibits a V*I loss, not an R*I^2 loss. Gap current is
> not proprotional to gap conduction voltage.
>
> Comments welcome,
> Malcolm
>
> ** If it's under 10 Amps it's Leakage Current **
Fundamental error! The FRACTIONAL voltage drop is the same from cycle
to cycle, not the voltage.
Ed