Re: [TCML] Q,,question

[johnbrooks@xxxxxxxxxxx]

Sirs,
Would it  be better to split a roll of 7/8" copper coil, and roll it
flat?
Seems more area would mean more "skin depth area"???
Regards
j. 
.



On Tue, 04 Dec 2007 17:49:57 -0800, "Barton B. Anderson"
<bartb@xxxxxxxxxxxxxxxx> said:
> Hi Gary,
> 
> Q will change when two or more coils are magnetically linked and their 
> relative positions are changed (coupling), but the change is extremely 
> minor.
> 
> For example, my demo coil in javatc:
> K is 0.128, L=17.346mH, C=27.728pF, Rac=89 ohms. If I increase the 
> coupling by moving the primary up to it's maximum (center of sec), K is 
> 0.234, L=16.249mH, C=34.65pF, Rac=85.5 ohms.
> 
> For Q=sqrt(L/C)/R, Q for the first data set is 281, but for the second 
> data set with high coupling is 254. Q is 10% higher with the lesser 
> coupled system. The reason is sqrt(L/C) primarily (and in this case, 
> mainly C). Because the primary position affects L and C, Q changes a 
> little. What is important here is that for a Tesla Coil, the coupling 
> change from 0.128 to 0.234 is huge resulting in very little change in Q. 
> Now, if we adjust k normally (0.12 to maybe 0.15), the change in Q is 
> only a few points higher and negligible to any advantage.
> 
> Jared I think was just stating the fact that it changes. Practically, 
> no, there's no reason to adjust coupling simply for a few points higher 
> Q. The coupling is going to affect far more than simply pri and sec Q.
> 
> Take care,
> Bart
> 
> Lau, Gary wrote:
> > I'm having a difficult time with suggestions that the secondary Q changes as a function of its coupling to the primary.
> >
> > Allow me to preface my comments by noting that while I am an electrical engineer, my formal training did not focus on RF electronics, so what I know about Q has been absorbed strictly from hobby-level interests.  I may not be 100% clear on the strict definition of Q.
> >
> > My understanding of Q is that it is a measure of the losses present in a component.  Let us assume that we have a "perfect" lossless primary, cap, gap, and secondary.  Though clearly impossible, the Q of everything would be infinite.  The only "losses" would be in the form of sparks coming off the top load.  We could adjust the pri/sec coupling and see that there is still frequency splitting (NOT an indication of losses), and we can see that the rate of energy exchange varies as a function of coupling, just as it does with real, lossy components.  Would you suggest that the Q is still varying, even though the components have infinite Q?  I don't think that the fact that higher coupling permits energy to be transferred more rapidly counts as evidence that the Q is reduced.  Or does the definition of Q not care whether energy is being transferred to an intended load or to parasitic losses within the component?  Regardless, it seems foolish to lower coupling if only to keep the primary (or secondary) ringing longer, in the belief that this improves the Q.
> >
> > Regards, Gary Lau
> > MA, USA
> >   
> >
> 
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