Research website of Vyacheslav Gorchilin
2017-08-28
Application to the proof of parametric RLC-circuits of the first kind
In this application we will learn how to use the formulas from the proof of free energy in parametric circuits of the first and second order, and applies this analysis to some real radiocomponents, in terms of their suitability for circuits with high efficiency.
Parametric inductance in PCCIE
For a start, consider the simple power series in the light of the formula (4.13) for a partial cycle PCCIE. This number can be characterized, for example, the parametric dependence of the inductance of the coil with the core from current flowing through it. $L = L_S\,(1 + k_1\,I + k_2\,I^2 + k_3\,I^3 + \ldots) \qquad (1)$ where: $$L_S$$ is the initial inductance, which is determined by the initial permeability of the core. Here and further we believe that the current formula has always a positive value, i.e. $$I=|I|$$. Formula (4.13) will also need the value of the inductance at the initial time of the cycle: $L_0 = L_S\,(1 + k_1\,I_0 + k_2\,I_0^2 + k_3\,I_0^3 + \ldots) \qquad (2)$ the Initial current in this case, we consider known — $$I_0$$. He might appear there, for example, as a result of inflating the coil with energy, we will continue to consider the effect of back EMF, which will occur when the circuit is open. To do this, we substitute in the formula (4.13) rows (1) and (2): $K_{\eta 2} = {2 \over I_0^2\,L_S\,(1 + k_1\,I_0 + k_2\,I_0^2 + k_3\,I_0^3 + \ldots)} \int_0^{I_0} L_S\,(1 + k_1\,I + k_2\,I^2 + k_3\,I^3 + \ldots) I\, dI \qquad (3)$ As you can see, $$L_S$$ is reduced, and it is a very important point — from the initial inductance or permeability of the core, the increase of the energy does not depend on! Take the next integral should be $$I_0^2$$ and get a nice formula: $K_{\eta 2} = { 1 + \frac23 k_1\,I_0 + \frac24 k_2\,I_0^2 + \frac25 k_3\,I_0^3 + \ldots \over 1 + k_1\,I_0 + k_2\,I_0^2 + k_3\,I_0^3 + \ldots } \qquad (4)$ From this it is immediately seen what if all the coefficients of $$k_1..k_n$$ are positive, then the increment of energy will not, and $$K_{\eta 2} \lt 1$$. This means that to achieve the increase in efficiency is greater than one, the graph of the dependence of inductor current must be flowing plot (plot example the real coil with a rectangular core), and one or more of the coefficients is negative.
Parametric inductance in PCCFE
In this case, without knowledge of the schema and according to $$U(t)$$ of the power source, the increase in efficiency we can not calculate, but we can search conditions for $$K_{\eta 2} \gt 1$$ from the formula (4.15). Assume that our coil is described by the parametric dependence of inductance from its current next next: $L = L_S\,(1 + k_1\, |I| + k_2\,I^2) \qquad (5)$ For rows with a large value of the degree calculations are either very bulky, or, in principle, not derived analytically. But for example, it will be enough and second degree.
By the condition (4.15) the integral should be less than zero, i.e. $L_S \int_{0}^{I(T)} (1 + k_1\, |I| + k_2\,I^2) I\, dI \lt 0 \qquad (6)$ where $\frac12 I_T^2 + \frac{k_1}{3}I_T^2 |I_T| + \frac{k_2}{4}I_T^4 \lt 0, \quad I_T=I(T) \qquad (7)$ or $I_T^2 + \frac43 \frac{k_1}{k_2} |I_T| + \frac{2}{k_2} \lt 0 \qquad (8)$ Find the roots of this quadratic equation: $I_{1,2} = -b \pm \sqrt{b^2 - \frac{2}{k_2}}, \quad b = \frac23 \frac{k_1}{k_2} \qquad (9)$ If we consider the actual dependency graphs of the inductance from the DC (example), we can immediately say that there can only be such coefficients: $$k_1 \gt 0, k_2 \lt 0$$. And since the current equation is taken modulo, it follows from all this that we are looking only one root: $I_{1} = -b + \sqrt{b^2 - \frac{2}{k_2}} \qquad (10)$ Thus: $|I_T| - I_1 \lt 0 \qquad (11)$ from which we get the final expression: $|I_T| \lt \sqrt{b^2 - \frac{2}{k_2}} b, \quad b = \frac23 \frac{k_1}{k_2}, \quad k_1 \gt 0, \quad k_2 \lt 0 \qquad (12)$ Knowing the odds and choosing the operating point (current is $$I_T$$) for a real device, it is possible to understand in which range it can potentially give an increase of energy.
The varicap. PCCIE
Consider the varactors in parametric RC-circuit of the first order, as a potential opportunity to get its energy boost. It is known that the dependence of the capacitance of varicap from the applied reverse voltage is expressed as: $C = {C_S \over 1 + k\,U_C}, \quad U_C \gt 0 \qquad (13)$ where: $$C_S$$ is the maximum capacity (without applied voltage), $$U_C$$ is the applied voltage, $$k$$ — factor, which can be found from the characteristics of the varicap. For example, a typical feature for КВ109А will be like this: $C = {27 \over 1 + 0.5\,U_C} \qquad (14)$ For circuit PCCIE by the formula (4.13) yields a potential increase: $K_{\eta 2} = {2\,(1 + k\,U_0) \over C_S\,U_0^2} \int_0^{U_0} {C_S \over 1 + k\,U} U\, dU \qquad (15)$ believing that under $$U$$ means the voltage at the varicap: $$U=U_C$$. Taking the integral we get: $K_{\eta 2} = 2 {1 + k\,U_0 \over k\,U_0} \left( 1 - {\ln(1 + k\,U_0) \over k\,U_0} \right) \qquad (16)$ For the same КВ109А, at a maximum reverse voltage of 20V, $$K_{\eta 2}$$ will equal $$1.7$$. It can be shown that for any parameters of the varicap, it is the ratio of the increment of energy will not exceed 2. And this is all assuming that the cycle of increasing the voltage on the varicap will be made cost-effective way. If this cycle will be the same as on the decline, according to the proof of (4.9) $$K_{\eta 2}=1$$.
The varicap. PCCFE
According to (4.15) for $$K_{\eta 2} \gt 1$$, you need to meet the conditions: $\int_0^{U_T} {C_S \over 1 + k\,U} U\, dU \lt 0 \qquad (17)$ Taking the integral we can see what needs to be done the following inequality: $1 - {\ln(1 + k\,U_T) \over k\,U_T} \lt 0 \qquad (18)$ But the left part of the inequality will always be either equal to, or greater than zero, which means that the condition (17) is impossible, hence the varicap, despite the presence of a drop-down area on the chart, not suitable for partial cycle PCCFE.
A variant Characteristics of variantov graphs similar to ferromagnetic materials, but instead of inductance capacity here, instead of the current — voltage. Thus, all calculations for the parametric inductance (see above) coincide for variantov. You might find potential gains of efficiency from equation (4) and a working point with known coefficients of inequality (12). Only in all these formulas, the current need to replace the tension.
Interestingly, falling plot chart variantov shorter than that of ferromagnets, which in theory gives a smaller increase in efficiency, but maricondi spend less surplus energy for heating that can give them a great advantage. In any case, compared to the varicap they look much more promising, as the increase of efficiency and power.