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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