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RE: Q Factor and Overall Efficiency
On 29 Aug 00, at 18:41, Tesla list wrote:
> Original poster: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
>
>
>
> -----Original Message-----
> From: Tesla list [mailto:tesla-at-pupman-dot-com]
> Sent: Monday, August 28, 2000 7:32 PM
> To: tesla-at-pupman-dot-com
> Subject: Re: Q Factor and Overall Efficiency
>
>
> Original poster: "Malcolm Watts" <M.J.Watts-at-massey.ac.nz>
>
> hi John,
>
> On 28 Aug 00, at 0:50, Tesla list wrote:
>
> > Original poster: "John H. Couture" <couturejh-at-worldnet.att-dot-net>
> >
> >
> > The Q factor tests have been discussed many times in past List posts.
> > However, these tests give the Q factor in a low voltage condition. This is
> > not the TC operating Q factor. The operating Q factor would give the
> > secondary voltage with the equation
> > Vs = Vp * Q
> > where Vp is the primary peak voltage.
>
> I don't agree that this is valid for a disruptive TC and I
> have stated why many many times so I won't repeat myself yet
> again. Operating unloaded Q can easily be found by measuring
> the decrementing waveform on an oscilloscope. I have found it
> is pretty much the same as the results I've obtained in low
> level tests.
> ----------------------------
> It is a well known fact that increasing the primary voltage (Vp) will
> increase the secondary voltage (Vs) of a disruptive TC with no other
> changes. This means the Q factor would also increase. Does this mean there
> is a flaw in what you stated in the past?
Are you speaking of secondary unloaded Q or system Q? Qs
hardly changes. Qp (which is really undefined due to its
voltage/current dependency) also goes up if Vp is increased
but it is a time-arying quantity. But why is a change in any Q
necessity at all? Suppose Q in your equation is constant.
Then Vp is proportional to Vs. I can't see any flaw in what
I've stated in the last few years with regard to this subject.
> You indicated that you found the operating Q and the low level Q to be
> pretty much the same. In your "Effective Resistance" post you said you did
> not believe your coil with an unloaded Q of 300 (low level) would generate
> 2.7 MV (operating). This would mean you did not believe the low level Q was
> not pretty much the same as the operating Q. Am I reading this conflict
> correctly?
No, definitely not. See above comments regarding the constancy
of Q. Also note that in a disruptive coil that relationship
(Vs = Q.Vp) *cannot* apply. Why? Let Q = infinity. Then the
implication is that Vs is infinite regardless of primary
voltage. But primary voltage *is not constant*. It is
decrementing from the moment the gap fires. You are feeding a
limited quantity of energy in and that shot of energy *has to*
charge the secondary capacitances. Even in a lossless coil
(i.e. Q = infinity), that limitation remains.
Malcolm
> ----------------------------- JHC
>
> > To find the operating Q factor would require only a scope and the proper
> > probe and the knowhow. To do the test find the half power points while the
> > Tesla coil is operating. Then determine the Q factor with the equation
> > Q = Fr/(Fh - Fl)
>
> I think there is some misunderstanding here. You can find Fh
> and Fl in low power tests by sweeping a signal generator
> feeding the coil above and below Fr but how do you get a
> frequency sweep on a coil that is ringing all by itself? The
> only resonable way to measure Q when there is a single
> frequency present is to calculate Q off the decrementing
> waveform (as you've already stated). Both methods are
> equivalent and the second can also be done at low power if the
> secondary is shock excited.
> ----------------------------
> You are 100% correct in that the operating Q cannot be found by sweeping the
> frequency of a coil that is ringing all by itself. The equation Q =
> Fr/(Fh-Fl) cannot be used in this test. As you state the Q would have to be
> found by the decrementing waveform (as I show in my TC Notebook). The
> equation Q = Pi/Log Dec would have to be used. So now I have stuck my
> foot in my mouth.
>
> -----------------------------JHC
>
> > where Fr is the resonant frequency, Fh is the higher half power frequency,
> > and Fl is the lower half power frequency. The Q factor would be the
> voltage
> > gain from the TC input to the TC output. This gain could then be used to
> > find the overall efficiency of the TC system. This would be done using the
> > energy equation
> > Vs = sqrt(2J/Cs)
> > Overall efficiency would be the Vs(half power) test volts divided by the
> > Vs(energy) theory volts.
>
> Am I correct in thinking you want to compare Vs (=sqrt(2J/Cs)
> with Vs = Q.Vin? You are comparing apples and oranges sorry.
>
> Malcolm
> -------------------------
> Overall efficiency is = energy out/energy in. However, because energy out
> with a disruptive TC cannot be determined an overall TC efficiency can be
> roughly estimated with the equation
> Vs = Vs test/Vs theory where Vs test = Vs = Q Vin and
> Vs = sqrt(2J/Cs).
> Another method to estimate the overall efficiency of a TC is shown in my
> Tesla Coil Construction Guide. Coilers often mention TC efficiency but do
> not indicate how they determined this ellusive parameter.
> ---------------------------------- JHC
>
> > This operating Q factor and TC overall efficiency may be a first in the
> > design and engineering of Tesla coils. However, Terry Fritz may have
> already
> > done this? Have other coilers done this?
> >
> > John Couture
> >
> > ---------------------------
>
>
>