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Re: Primary Q


From: 	Malcolm Watts[SMTP:MALCOLM-at-directorate.wnp.ac.nz]
Sent: 	Friday, November 14, 1997 8:57 PM
To: 	tesla-at-pupman-dot-com
Subject: 	Primary Q

Hello all,
            here is the result of some cogitation after replying to 
an off-list query. I have used an arbitrary slope for the waveform
(see below) but the addendum at the end of the post explains what 
governs the real thing.

------- 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 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 
(I have seen this dozens of times using the storage scope).

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's 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 power lost in
absolute terms is highest with the highest voltage, the power
lost as a percentage of total power is lowest. You can see that 
gap loss scales with primary current.

ADDENDUM:  I didn't mention the *slope* of the decrement in there. 
This is solely dependent on L/C ratio assuming resistance is 
negligible. The higher L/C, the less the slope. One experiment I did 
was to discharge a cap at around 8kV (I think it was 25nF) into a 
350uH litz helix. It rang for around 1.2mS before the gap went out. 
Clearly, to make the most of a high Vp, L/C should scale as well to
keep the V.I gap loss minimal by keeping Ip low at the highest cap 
voltage.

Comments welcome as always.
Regards,
Malcolm