As a Preface to further narration, is the best fit is an excerpt from the response of Nikola Tesla in conversation with his lawyer.

In this work we consider some properties of the Tesla transformer (TT), build a mathematical model one of the ways of its initiation and make conclusions about the efficiency of energy extraction from it of the secondary winding at the first harmonic. For this we consider the classical circuit of the transformer through the packs of pulses sent by the generator GI in the primary winding L1.

The duration of one pulse generator is equal to \(T_{i}\), the length of the entire bundle \(T_{p}\), period packs — \(T\), and the duty cycle will find in the form: \(Q = \frac {T} {T_{p}}\). In this work, to simplify our model, we will use relative values, so the value of pulse amplitude Vi, and the duration that packets will be equal to one.

Actually have a generator with the fundamental frequency \(f_{i} = 1 / T_{i}\) , which is modulated by a lower frequency \(f = 1 / T\) of \(Q\). This is an important condition necessary for the removal of power to the load without interference in the wave processes in the secondary winding of the TT — L2. Do it on a low frequency harmonic, and for convenience of removal, as such, we choose the modulation frequency \(f\). The amplitude of this harmonic will be denoted by \(H_{1}\) and its value looking for by using Fourier analysis of the oscillations of the L2. It is here that we are not going to stop, because it's a whole section in math; it can be studied separately, our readers, to look at the spectrum of the signal in

the online calculator or check all in

MathCAD-e.

Knowing \(H_{1}\), we can find the effective (RMS) value of the amplitude of the first harmonic over the period \(T\): \[ A_{1} = H_{1} \sqrt {\int^1_0 \sin (2\pi \cdot f \cdot t)^2 \, dt} \]

The effective value of the amplitude of the predetermined frequency which is supplied from the generator GI, we find this: \[ A_{GI} = \sqrt {\int^{1/Q}_{0} \sin (2\pi \cdot f_i \cdot t)^2 \, dt} \]

Their relationship as follows: \[ K_A = \frac {A_{1}} {A_{GI}} \]

Given that the square \(A_{GI}\) is proportional to input power, and the square \(A_{1}\) output, it can be expected to increase

the efficiency of the second kind in the following form: \[ K_{\eta2} = \bigg ( \frac {A_{1}} {A_{GI}} \bigg ) ^{2} .\]

Conditions high efficiency

Consider in a special online calculator an example that sets the frequency of the \(f_i\) is 10, the ratio \(Q\) is equal to 2, the time constant of the secondary winding L/R which is the ratio of its inductance to the active resistance equal to 0.07 —

example No. 1. As you can see from the example, the amplitude of the first harmonic is very small, so the efficiency of removal it will also be low. With the exception of the method of removal, wherein it is produced from the current end of the TT with a magnetic core. In this case, the resulting data must be multiplied by the factor for the increase of efficiency obtained from the redistribution of charges.

Try to change initial settings: reduce \(f_i\) to unity —

the example No. 2. As you can see, the first harmonic is significant increased, and thus increased and the main resulting parameter — the ratio of the effective values, the square of which is the increase in required efficiency. But while he was still less than one.

For a qualitative transition through the unit we will need to increase the duty cycle and adjustment of L/R. let us Try to set a duty cycle of about 30 and just increase the L/R —

sample No. 3. We see that even though the absolute value of the amplitude of the first harmonics have decreased, but increased relative to the effective value of the driving frequency. And \(K_A\) was equal to one!

Note the changing shape of the pulse in the secondary winding of TT — he became almost a unipolar! If you imagine the reluctor pulse from the generator, it will be even shorter. This is the main problem, which we will cover next. Try to increase duty cycle to 200 —

example No. 4. Got an increase in 2.2 times! It would seem, increase the duty cycle and get a high raise. But! Pay attention to the duration of the driving pulse: with duty cycle 2000 and the operating frequency of the secondary TT at 420 kHz, the length of this pulse needs to be only 1.2 NS —

example No. 5. Such a generator of nanosecond pulses do not force everyone to do and even gaps are not always able to afford it. Therefore, the famous Tesla used magnetic breakers arresters spark.

Turn our attention to another important parameter — the time constant of the secondary TT. Try to find the optimal way: if it is made too small or too large, the efficiency of TT will fall; you need to find the optimum —

sample No. 6. In a real circuit for this in the primary winding choose active resistance, which is made to it in series or parallel.

In conclusion, I would like to note that to obtain high efficiency with this method of excitation, the most important parameter is the duty cycle. At sufficiently fast generators of nanosecond pulses it is possible to obtain a good efficiency in real devices. For example, if one generator can output pulses with a duration of 2 NS, then, when the frequency of the secondary TT at 420 kHz, it will be possible to increase the duty cycle up to 1200. This means that the efficiency of the power can reach about 40 —

example # 7.

The maximum overall growth efficiency is not only the difficulty in obtaining sets of nanosecond pulses, but

the maximum powerat which the energy of all the electrons will be transformed from reactive to active.

A more sophisticated calculation of pulse technology offered

here.