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Re: Simulation of a conventional Tesla coil



Original poster: "Dr. Resonance by way of Terry Fritz <teslalist-at-qwest-dot-net>" <resonance-at-jvlnet-dot-com>


Since you can see the wavetrain on a storage scope, and knowing the decay
rate is directly related to circuit resistance, why not just calculate the
current?

The resistance of all circuit components can be measured (so low with large
dia copper tubing or wire it is almost negligible) except for the higher
resistance of the sparkgap.

The sparkgap presents a problem, but by subtracting the resistance of other
components from the total resistance you will have the resistance of the
operating sparkgap.  Total circuit resistance is directly related to the
decrement (decay) factor of the wavetrain.

Total current, I rms, in a spark oscillator circuit is:

I rms  =  3.14 * E * C *  sqr ((N * F) / d

     I rms = rms current in Amperes

     E = Voltage on capacitor at the commencement of each wavetrain

     F = Freq. in Hz

     N = Number of sparks/second

     d = decrement (decay) factor of wavetrain   (related to resistance)




The logarithmic decrement factor is calculated by:


     d = 3.14 * R * sqr (C / L)

     d = log. decrement factor

     R = circuit resistance in Ohms

     C = capacitance in Farads

     L = inductance in Henries


The above equation is solved for resistance.  This gives total circuit
resistance of an operating LC circuit.

The actual decrement factor can be determined by viewing the waveform on a
storage scope:

     d = (log n) * A1 / A3  =  A2 / A4  = A m/ An

     d = decrement factor   (typical value would be 0.25 with a total circuit
resistance of 5 Ohms)

     A1 = Amplitude of first wave (positive peak)

     A2 = Amplitude of first wave (negative peak)

     A3 = Amplitude of second wave (positive peak)

     A4 = Amplitude of second wave (negative peak)


When I get my scanner running I will scan a waveform drawing which more
clearly illustrates how the decrement value is directly measured on a
waveform.

The pos. amplitude of the first wave is the value measured from 0 to the
first positive peak.   The neg. amplitude of the first wave is the lower
section measured from 0 to the first neg. peak.

You essentially are developing a value for the actual circuit resistance
which is directly dependent on the rate of decay of the waveform.  The
decrement is a ratio of the first peak to the second peak to the third peak,
... ad infinitum.  This decaying value is related directly to the
resistance.

Tesla didn't have a method of directly seeing these waveforms but with
modern storage scopes than can freeze this waveform we can take direct
measurements of these amplitude ratios and then calculate the decrement
factor.  The measured decrement factor is then used directly into the two
equations to find the actual total resistance of the operating LC circuit.
After measuring the DC resistance of the circuit and subtracting this value
(almost negligible) from the actual total resistance of the operating
circuit we end up with the actual resistance of the operating sparkgap.

Many experimenters have assumed that the resonant circuit peak discharge
potential is E rms x 1.4.   It's not.  Under resonant conditions the peak
discharge voltage of the capacitor is E rms x 2.2 (see below).  This assumes
max. spacing on the sparkgap so the cap doesn't fire early.

Also, if the sparkgap is at a wide spacing, or in the case of the RSG type
gap, and, if the capacitor recharging current is a high value, then:

     Ep = 1/2 * 3.14 * Emax which is equal to = 2.22 * Erms

     Ep = peak voltage to which capacitor is charged just before firing

     E max = E rms (of transformer) * 1.4  =  E rms * sqr 2

     E = E rms voltage of transformer

The above equation assumes the transformer can provide a high current to
completely recharge the capacitor to full value before it fires.

Dr. Resonance

Resonance Research Corporation
E11870 Shadylane Rd.
Baraboo   WI   53913


-- snip --

 > It would be interesting to revisit spark gap voltages and currents now
that
 > the equipment is so much better...
 >
 > Cheers,
 >
 >          Terry