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Re: Arc length vs pwr
On Thu, 24 Oct 1996 22:58:59 -0600 Tesla List <tesla-at-poodle.pupman-dot-com>
writes:
>>From MALCOLM-at-directorate.wnp.ac.nzThu Oct 24 22:40:27 1996
>Date: Fri, 25 Oct 1996 11:07:50 +1200
>From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
>To: tesla-at-pupman-dot-com
>Subject: Re: Arc length vs pwr
>
>Hi Mark,
> I'd like to address some of your points...
>
>> The discussion had began in reference to a ringing primary with a
>> very light or no load (under coupled or non coupled) like a
>> secondary coil. Under conditions of load (coupling) the secondary
>> coil looks like a series resistance to the primary oscillator loop.
>
>Agreed. It may be modelled as such. It can also be modelled as an
>equivalent parallel resistance. Actually, it only looks resistive
>under one set of secondary loading conditions anyway.
>
>> If resonant rise didn't occur a tesla coil couldn't operate, the
>> principles of resonance apply if its a tesla coil or just a L-C
>> circuit driven by a generator.
>
>I would class "resonant rise" in the coupled system as the action
>whereby the primary imparts a nudge to the secondary with each half
>cycle thereby transferring its energy to the secondary over a number
>of cycles - i.e. secondary amplitude builds at the expense of the
>primary (see diagrams below for single tuned circuits).
>
>> The source of voltage in the primary circuit under condition of
>> oscillation is the capacitor AND the inductor, the inductors back
>> EMF is what charges the capacitor, the only reason we need the
>> drive transformer is to get things started and to keep them going
>> due to the limited Q of the circuit which is why the output rings
>> down in level.
>
>As you originally said, the transformer is no longer part of the
>circuit once the gap has fired. In fact, the initial conditions are:
>cap charged, no primary current. A quarter cycle after gap fire you
>have cap empty, all energy in the mag field surrounding the primary
>(some of which will be coupled to the secondary if present) and
>primary current at maximum. A half cycle later, the cap is charged to
>opposite polarity but with a bit less voltage due to losses and the
>inductor current is zero, and so it goes until the energy is mostly
>gone and gap ionization has reduced to the point where the gap ceases
>firing. It is important to realize the phase relationships in the
>circuit. The energy the cap started with is apportioned between
>inductor and cap in proportions depending on which part of the 1/2
>cycle you take a snapshot of.
> The point is, the cap alone is the energy source initially.
>Whatever it transfers to the magnetic field is no longer part of cap
>energy.
> I have measured this, and I have _never_ observed the cap rising
>beyond its initial value after the gap has fired. This was one of the
>things I checked thoroughly in my MOSFET gap modelling exercise. I
>have posted a number of photographs taken off the scope of both
>primary and secondary waveforms simultaneously to a number of people.
>I also measured the primary alone with attendant photos.
>
> ______/ __________
> | | (A) prior to gap fire
> | o
> _____ o
> _____ Vc o
> | o
> | |
> ------------------
>
> _______ generator in series_________
> | | (B) with AC voltage source
> | | in series.
> ----- o
> ----- o
> | o
> | |
> -------------------------------------
>
>Circuit (A) is definitely not equivalent to circuit (B). The
>phenomenon of which you speak only occurs with circuit (B) because
>there is an external energy source in the circuit.
>
>> Actually, I make take a big flame for this, but
>> there is really no such thing as "parallel resonance" all resonant
>> circuits are series in nature and operation series and parallel
>> only really pertain on how the drive signal is applied to it. In
>> order to be able to measure this resonant rise with any degree of
>> accuracy you'd have to be very sure the measuring device offered
>> VERY LITTLE capacitive or resistive loading.
>
>True. For example, let Lp = 100uH, Cp = 10nF and the closed-circuit
>Q = say 50. Then the ESR of the circuit is: SQRT(L/C)/Q = 2 ohms.
>The equivalent parallel resistance is close to ESR*Q^2 = 5 kohms.
>Measuring with a 1 Mohm probe in parallel with 20pF across the
>capacitor gives negligible loading.
>
>Malcolm
>
>PS - If you would like, I'd be happy to send you some photographs of
>all this.
>
I guess I'll stand corrected!
Mark Graalman