Research website of Vyacheslav Gorchilin
2017-07-27
Parametric change of inductance in a RL-circuit. Back EMF

In this note first we prove the theoretical possibility of obtaining additional energy from the back EMF in the inductor core. The proof is based on the classic formulas of electrical engineering. According to the classification here will be considered the generator of the first kind of first order partial cycle PCCIE.

In our time, divorced many myths and legends about the inexhaustible possibilities of the back EMF (AEDS) in the inductor. According to some researchers, OADS can give more energy than it expended, and their experiences, in some cases, confirm this. Theorists explain such powers by the ether theory or the unused energy of the atoms of ferromagnetic materials. We will try to draw conclusions based on mathematics and the theory of electrical circuits , which worked well and can fully display the necessary processes. Thus, the author reiterates the idea that issledovatelyam it is not necessary to go into the jungle of other theories and hypotheses, enough of a different angle to look at the classics.
Next, we will show that the school or even high school — a nonparametric OATS — can increase the efficiency of the second kind $$(K_{\eta2})$$, so the outset that we consider only parametric coil which changes its inductance depending on the magnetic field $$(H)$$ in it. And since this tension is directly proportional to current, as parameter we will have to address it. To convert the current back to $$H$$ is not extremely difficult, you need to know the parameters of the coil and core, i.e. the parameters of a particular device.
In the previous section we showed that a parametric change in capacitance in the full cycle: charge-discharge, does not increase $$K_{\eta2}$$. The same can be said of inductance: it is enough to replace the voltage on the currents and to change some of the ratios in the diff. equation. But with the coil inductance, which will contain a ferromagnetic core, we can do a little differently.
A typical graph of the magnetic permeability of the core is $$(\mu)$$ on the magnetic field $$H$$ to the left . But the tension — proportional to the current, and the permeability of the inductance, this means that we can build a graph $$L(I)$$, which is proportional to $$\mu(H)$$. Thus, we obtain the parametric dependence of the inductance $$L$$ from the current $$I$$ flowing in it, and with this dependence, we will continue to work.
As an example, compare the two charts to the actual measured characteristics of the ferrite from the flyback transformer: graph 1, graph 2. On the second chart the y coordinate $$M(I)$$ shows how changing the inductance of the coil compared to the original, and $$I$$ is the current in the coil.
Recall that in this note we investigate AEDS, therefore, assume that the initial current $$I_0$$ in the coil already, and it will pass through the electric circuit consisting of inductance and resistance. How there came a shock we will discuss later, but for now we will use the classical formula for the calculation of the energy in the coil at the time of closure of the SW key: $W_L = \frac{L_0 I_0^2}{2} \qquad (3.1)$ where $$L_0$$ is the inductance of the coil at the time of closure of key SW. Now our task is this: to properly use this energy. It is necessary to find a mode in which the current in the circuit will produce the most work on active loading. The classic formula for this energy is: $W_R = R \int_0^T I(t)^2 dt \qquad (3.2)$ Where $$I(t)$$ is the current in the circuit depending on time, which is unknown to us. In order to find it, it is necessary to compile the differential equation for the transition process in our scheme : ${L \over R} I(t)' + I(t) = 0 \qquad (3.3)$ where the inductance $$L$$ is a parameter from the current, and the current in turn — on time. For the solution of the equation it remains to determine the dependence of the inductance from the current for which we can take the polynomial $$M(I)$$ with coefficients $$..k_1 k_4$$. Its convenience — flexibility for various dependencies: $L = L_S\, M(I), \quad M(I) = {1 + k_1 I(t) + k_2 I(t)^2 \over 1 + k_3 I(t) + k_4 I(t)^3} \qquad (3.4)$ where $$L_S$$ is the initial inductance of the coil (no current). $$M(I)$$ must be proportional to $$\mu$$ from $$H$$ to the real core.
We introduce the time constant of the circuit $$\tau = L_S/R$$, which will record the final shape of the diff. equation: $\tau\, M(I)\, I' + I = 0 , \quad I=I(t) \qquad (3.5)$ the Analytical solution of this equation in this form is complicated, so we use mathematical editor MathCAD and get it in a numerical form. In the editor we appreciate the real time operation of the circuit after closing of the key, we denote it $$T$$ and make this time a finite number.
Why do we need time?
Indeed, if we want to obtain the dependence of the increase in efficiency of $$\mu$$ (see top graph), why should we use one additional variable — the time? Will try to fix it. We rewrite (3.6) in another form: $-\tau\, M(I)\, I' = I, \quad I=I(t) \qquad (3.6)$ and substitute into the formula (3.2): $W_R = R \int_0^T \left[\tau\, M(I)\, I'\right]^2 dt \qquad (3.7)$ After some mathematical operations, the author obtained a rather unusual integral in which energy dissipation on resistance does not depend on $$t$$, and even from the resistance: $W_R = \frac{L_S}{2} \int_0^{I_0^2} M(\sqrt{J})\ dJ , \quad J = I^2 \qquad (3.8)$ Here you need to pay attention to the fact that in paginegialle functions are all $$I$$ change for $$J$$ by the rule: $$I=\sqrt{J}$$. To find the final formula calculates the energy gain, divide the energy dispersed on the active resistance on the starting energy in the coil is: $K_{\eta2} = {W_R \over W_L} \qquad (3.9)$ Given that $W_L = {L_0\, I_0^2 \over 2} = {L_S\, M(I_0)\, I_0^2 \over 2}, \quad M(I_0)= {1 + k_1 I_0 + I_0 k_2^2 \over 1 + I_0 k_3 + k_4 I_0^3} \qquad (3.10)$ substitute the previously obtained results and we get an incredibly interesting pattern that does not depend on the time coordinate: $K_{\eta2} = \frac{1}{M(I_0)\, I_0^2} \int_0^{I_0^2} M(\sqrt{J})\, dJ , \quad J = I^2 \qquad (3.11)$ Or this same formula in another form: $K_{\eta2} = \frac{2}{M(I_0)\, I_0^2} \int_0^{I_0} M(I) \, I \, dI \qquad (3.12)$ Well, if we all want the same thing to Express through the magnetic permeability of the core and the magnetic field $$\mu(H)$$, then the formula (3.12), simply replace the appropriate letters: $K_{\eta2} = \frac{2}{\mu (H_0)\, H_0^2} \int_0^{H_0} \mu (H) \, H \, dH \qquad (3.13)$

Equations (3.11-3.13) and are the mathematical expression of free energy for back EMF in the coil core! A more General approach for the derivation of this formula see here

From it immediately clear that if a parametric dependency is missing, i.e. $$M(I)=M(J)=1$$, then a raise is not: $$K_{\eta2}=1$$. We have to learn to use this formula and find out whether there can be $$K_{\eta2}$$ is greater than one.
What is the result?
For example, take $$M(I)$$ from formulas (3.4) and substitute it into (3.11) replacing $$I$$ to $$J$$ by the rule: $$I=\sqrt{J}$$. The time dependence of ex and obtain: $K_{\eta2} = \frac{1}{M(I_0)\, I_0^2} \int_0^{I_0^2} {1 + k_1 J^{0.5} + k_2 J \over 1 + k_3 ^{0.5} + k_4 ^{1.5}}\, dJ \qquad (3.14)$ Again for example, enter the following current and coefficients: $$I_0=1.4, k_1=10, k_2=0, k_3=0, k_4=5$$. The left side shows a graph of the inductance of the coil from the current in accordance with these coefficients. As you can see, we work in growing, and in the drop-down field magnetization curve of the core and primary inductance is less than the maximum about 4 times, which is close to the real values of permeability of ferromagnets. Solving the integral (3.14) we get $$K_{\eta2}=2.07$$ — more units!
To substitute other factors and to get your results you can in the editor MathCAD downloading there this program or to test it in the calculator. It a visible pattern, which confirms the assumption that the drop down part of the curve $$\mu$$ from $$H$$ gives the largest contribution to the energy increase. In addition to choosing the work area on the chart, to achieve a positive result, it is necessary that the device provide access to the coil for the desired mode of current and magnetic field strength.

The rationale for the occurrence of additional energy in such devices, see here

What in practice?
You need to understand that we get according to equations (3.11-3.13) is the marginal value. Depending on the material of the core, when working on the right drop-down plot of $$M(I)$$, you may experience a variety of losses, mostly heat. But they will not always be equal to this increase, and therefore is described here the way to super devices still remains open. Izvestia of materials, good performance $$K_{\eta2}$$ should give Permalloy, on which chart, you can easily find the working site, even better — Metglas . A little worse, in this sense, the situation with ferrites. Promising look some modern steels and certain core construction , however, we must not forget that part of the energy can be spent on Foucault.
Permalloy is working in a relatively small frequency and therefore the device running on it cannot develop large capacities. Ferrite can withstand orders of magnitude for large frequencies, but it is much more fragile. However, as a result, it can operate at high capacity.
The question remains open: how do you receive the initial current $$I_0$$ in the coil and how long it takes for the energy? For answer there are different approaches. One of them suggests that the core need to podmanivaya perpendicular to the primary field , and it can be done with short pulse whose energy is several times lower than required due to the inertia of a ferromagnet. The second approach [4-6] involves the ramp-up without additional perpendicular field. The third offers mechanical excitation of the core is a permanent magnet, and thus the initial current appears in the coil .
Given all the above it seems to us a very real seredinny receiving devices based on coils with ferromagnetic core. As you know, if math gave the green light, then the practical implementation will not take long.
The results of this work was done by a specialized calculatorthat allows you to find the dependence of magnetic permeability of any core on the magnetic field and to calculate it is potentially achievable increment of efficiency of the second kind.