Research website of Vyacheslav Gorchilin
2019-02-27
Pulse technology on the bias currents
In the first and second part of this work we approached the problem of converting an electric field into a magnetic with a classic point of view: used a capacitor made of two plates and coil removal. But do I have to use this design and there are more interesting solutions?
Let's look at figure 4, which is very similar to figure 1, but with one important difference: the conductor does not rupture the capacitor and its plates are here shown in phantom. The fact that any conductor has inductance per unit length and capacitance, and hence, any interval we can imagine how the LC-circuit. The conductor can always be compared with a long line, only one theory of which is still poorly developed. In all this, we will be interested in one important event at the time of applying voltage, the conductor behaves like a capacitor. Here is necessary to remind that in this work we deal only with the bias current and try to avoid conduction currents. Thus, further we will talk only about the period of time when the voltage on the conductor has already filed, and the conduction current has not yet appeared. In the second part we have designated this time is $$T_0$$. It is in this context we need to look at our guide on figure 4.
You need to add, "capacitor effect" is formed not only along but also across the conductor relative to the earth. If the reactor is a coil, there is another "capacitor" — between its turns. Of all the species we will discuss in the following works, but for now we will work only with longitudinal capacity. Interestingly, derived further formulas, with some adjustments, will work for all.
 Fig.4. Cut the conductor, which can be considered as a capacitor with the bias current Fig.5. The simplest diagram of the device for switching technology and displacement currents
For impulse technology, the quality of reactor, we will use a conventional coil, which has inductance $$L$$ and its own longitudinal capacity $$C$$. However the algorithm of inclusion will now be slightly different (Fig. 5). Any engineer can immediately say that this is the scheme of the flyback Converter, but we all know fundamental difference: in the classic version of this circuit running current of conduction, but here the bias current. This is the main problem, which we will discuss later.
The algorithm works
Further calculations will be done according to figure 5, which shows the simplest embodiment of the device for pulse technology. To understand their logic, the necessary algorithm of the scheme, which is presented below.
The switch SW1 closes its contact for a very short period of time $$T_0$$. Due to the self-capacitance of the coil L1 and the displacement current $$i(t)$$, generated a magnetic field, which must be disposed of in the form of energy in a resistive load R1. For this, the key must SW2 to close its contacts immediately after opening SW1. According to Faraday's law of induction [1] stored in the coil magnetic field, when the reduction will cause a self-induced EMF, which through SW2 forms a circuit with R1, and a current $$I(t)$$. Thus, all the reactive energy of the magnetic field will be in active — $$W_2$$. The time when the key is open SW2 is not critical and affects only the frequency of the oscillator.
Calculations
The calculation scheme, with some differences, is based on two steps described in the second part. The formula (2.1) we moved here unchanged, because the first time was seen a coil like a capacitor: $\int \limits _{S}{\frac {\partial \mathbf {D} }{\partial t}}\, \mathbf {dS} = C \frac{d U}{d t} \qquad (3.1)$ But the integral in (2.2) will take a little bit different. It would be possible to change nothing, and to find a balance of energies through $$H$$, and the energy of the magnetic field according to Maxwell, but then we would have to find the average length of the magnetic line for each type of coils, with separate formulas. But you can go through current $$I(t)$$, then the calculations will appear not the specific design of the coil and its inductance, which involves a larger number of options. In any case, while we will assume that the coil L1 is the solenoid with number of turns equal to $$N$$ and the inductance $$L$$. Form of its cross section can be any: round, oval or square, and the core can be either air or ferromagnetic.
Using the law of currents [2] we can take this integral: $\oint \limits _{\ell}\mathbf {H}\, \mathbf {dl} = I_0 N \qquad (3.2)$ where $$I_0$$ is the current at the time of the opening SW1 and closing SW2. Equating with (3.1) we derive the current: $I_0 = \frac{C}{N} \frac{d U}{d t} \qquad (3.3)$ Remembering that the energy stored in the coil is given by $$W = L\ I^2/2$$, and assuming that we have no losses and therefore we recycle all of it later, you will get: $W_2 = {L \over 2} \left( \frac{C}{N} \frac{d U}{d t} \right)^2 \qquad (3.4)$ This is the General formula to calculate the received load power for one pulse, which is suitable for a solenoid of any cross-section and core. Interestingly, the formula does not depend on the load resistance, suitable for toroidal coils.
To calculate growth efficiency scheme required slew rate of the pulse also provide, as we did in (2.10) $\frac{d U}{d t} = U_0 \frac{d F}{d t} \qquad (3.5)$ and repeat the formula of energy (2.8), which is required for charging the coil: $W_1 = C\,U_0^2 \qquad (3.6)$ Recall that $$U_0$$ — the voltage at the coil L1 at time $$T_0$$. Then the increase in efficiency will be so: $K_{\eta2} = {W_2 \over W_1} = {L\,C \over 2N^2} \left( \frac{d F}{d t} \right)^2 \qquad (3.7)$ Is the most General formula for calculating a gain efficiency for solenoidal and toroidal coils.
Let's concretize these equations under exponential pulse presented on figure 3b. The dependence of the voltage rise will take from the formula (2.17), it is derived from (2.18) and its value at time $$T_0$$ from (2.19). At this point, it will be approximately equal to: $$0.59/T_0$$. Then the energy of a single pulse dissipated in the load is such $W_2 = 0.17\,L \left( \frac{C}{N} \frac{U_0}{T_0} \right)^2 \qquad (3.8)$ and the increase of efficiency is: $K_{\eta2}= 0.17 {L\,C \over (N\,T_0)^2} \qquad (3.9)$ in order For this growth was super, you must meet the following condition: $T_0 \lt 0.41 {\sqrt{L\,C} \over N} \qquad (3.10)$ it is Interesting that formula (3.9) and (3.10) can be written in another form, if we assume that the L1 coil not only has inductance and self-capacitance, but frequency is: $$f_s = 1/(2\pi\sqrt{L\,C})$$, which, in principle, it is possible to measure a pulse generator and an oscilloscope. Then the formula of growth of the efficiency acquires the form: $K_{\eta2}= {0.0043 \over (N\,T_0\,f_s)^2} \qquad (3.11)$ and the condition of sverkhelastichnosti — like this: $T_0 \lt {0.065 \over N\,f_s} \qquad (3.12)$ Recall that $$T_0$$ can be regarded as the time of the front pulse.

In the next part we will consider special cases of the coils and the more detailed formulas for calculation of pulse circuits in which they participate. And also, offer some variants of specific technical solutions.

The materials used
1. Wikipedia. The law of electromagnetic induction of Faraday.
2. Lecture 4. The law of the total current.