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Primary Q - A Brain Teaser

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 **