How to rise the secondary?
From: Antonio Carlos M. de Queiroz [SMTP:acmq-at-compuland-dot-com.br]
Sent: Friday, July 03, 1998 12:47 AM
To: Tesla List
Subject: Re: How to rise the secondary?
John H. Couture wrote:
> 1. Doesn't changing the coupling (K) only change the time of energy
> transfer for a tuned TC system regardless of frequency? For example with K =
> .20 the energy will transfer for all coils and operating frequency:
> Transfer time = 1/K = 1/.20 = 5 half cycles.
Correct, but with 5 half cycles the secondary voltage is zero, not maximum.
> 2. The total amount of energy transferred will occur when the TC system is
> in tune.
> That is when LpCp = LsCs
> What are the equations relating the amount of energy transferred to
> special coupling coefficients?
The exact equation, even using a two-coupled-lC-tanks model is rather complex
(if nobody come out with the answer I can try to derive it). What happens is
that if the secondary voltage reaches the highest peak value at a point that
does not coincide with the peak of the beat envelope, some untransferred energy
remains in the primary circuit.
Simulated in my teslasim program (ftp://ftp.coe.ufrj.br/pub/acmq/teslasim.zip)
Primary inductance L1: 0.1 mH
Primary capacitance C1: 10 nF
Secondary inductance L2: 100 mH
Secondary capacitance C2 (self+top): 10 pF
Coupling coefficient k=0.2
Primary capacitor voltage V1: 10 kV
All resistances set to zero, an ideal case for simplicity.
This results in resonance at 159 kHz.
k=0.2 is a particularly bad choice, with zeros of the secondary voltage
coinciding with the peaks of the beat envelope.
The maximum secondary voltage occurs at the peak of the 5th half cycle:
Time of the peak: 13.9 us
Secondary voltage V2peak: 313 kV
Primary voltage: ~0 V (the program computes -20 V, but this can be numerical noise)
Secondary current: 0 A
Primary current I1remaining: 14.1 A
This corresponds to:
Initial primary energy (0.5*C1*V1^2) = 0.5 J
Secondary energy at the peak of the secondary voltage: E2 = 0.5*C2*V2peak^2 = 0.490 J
Primary energy left at the same instant: E1 = 0.010 J
Total energy at this instant: Et = E1 + E2 = 0.5 J (just to check)
In this example, 98% of the energy is transferred. The small loss is entirely due to
the suboptimal coupling coefficient. Higher losses can occur with lower coupling
coefficients, but assuming that most coils use k<0.2, and this is a particularly bad
case, my conclusion is that the effect is usually negligible.
> 3. If the primary is designed with enough clearance to prevent sparkovers
> between the pri and sec coils, raising the secondary would be unnecessary?
> In other words the coupling is determined only by the sparkover clearance
> limitation. Using insulation instead of air to separate the pri/sec coils
> could be used to reduce the clearance and increase the coupling. But this
> would not change the amount of energy transfer or the length of sec term
There is also the question of the quenching of the primary gap. If the coupling
is too high the time intervals where the primary energy is low are too short,
and great part of the energy can be dissipated in the primary gap instead of being
transferred to the secondary. If the coupling is too low the losses have more time
to drain energy before the secondary voltage rises enough to produce streamers.
Some method of adjusting the coupling is so desirable for optimal performance.
> 4. What are the equations relating the quenching characteristics to
> coupling and output spark lengths?
This is certainly dependent on the exact construction details of the gap, and
probably the most difficult to model part of the Tesla coil theory...
> 5. In #1 above the number of half cycles required to transfer all of the
> energy of a tuned TC from pri to sec coils can be found. However, how many
> half cycles are required to properly charge a suitable sec terminal so there
> will be a sec output spark and quenching time is not important?
The beat envelope is practically sinusoidal. Streamers/sparks will start at
the first semicycle corresponding to the beat envelope greater than the voltage
required for breakout at the terminal. This depends on the terminal shape.
> 6. The point of over/under coupling (critical coupling) is determined when
> Rp = Rs How are these two parameters calculated at the time of
What is this? Usual Tesla coils operate as almost lossless (high Q) circuits
before breakout. Far from any critical coupling.
> The equation R = Xl/Q cannot be solved because both R and Q at
> high voltage operation are unknowns at time of design.
What are these parameters?
Antonio Carlos M. de Queiroz