Research website of Vyacheslav Gorchilin
2017-03-03
The combination of two frequencies-parallel resonant circuit
Some devices require the combination of two resonant frequencies in the two circuits. For such systems do not require alignment with the frequencies of the standing waves, though they will be close enough. The advantage of such devices is the relative ease of their implementation and customization.
Figure 3.1 shows two circuit L1C1 L2C and which in General are called "associated". In them C1 and C — self-capacitance of the corresponding coils L1 and L2. But for our purposes we can make the assumption that the coil L2 has concentrated parameters and zero private capacity, so in figure 3.1, the capacity C is depicted with a dotted line. It further will be ignored in the calculations.
Thus, we get two paths: parallel and serial L1C1 L2C2. We believe that the first — L1C1 has distributed parameters, and the second L2C2 — focused. By using the schema in the coil L1 we need to create two modes of operation.
1. A half-wave. In this case, the frequency specified by the generator E1 must be twice the natural frequency of the loop L1C1. Chain L2C2, thus, should provide minimal resistance and in fact - to close a L1, i.e. ideally it should be zero at this frequency.
2. A quarter wave. This mode is set by the generator of E2, while in the parallel-to-serial circuit L1C1L2C2 needs to be a resonance, and the maximum voltage should occur in the coil L1.
Our task is to find the optimum ratio of the parameters of these circuits to achieve them in two distinct resonances. Loss not yet taken into account.
From the theory of electrical circuits [1], it is possible to obtain the following ratio. It calculates the resonance circuit L1C1L2C2 when the frequency of the second oscillator E2 — $$\omega_2$$: $\left[\left(\frac{\omega_2}{w_1}\right)^2 - 1\right] \left[\left(\frac{\omega_2}{w_2}\right)^2 - 1\right] = \left(\frac{\omega_2}{w_2}\right)^2 \frac{L_1}{L_2} \qquad (3.1)$ $w_1 = 1/(C_1 L_1)^{0.5}, \quad w_2 = 1/(L_2 C_2)^{0.5}$ where $$w_1$$ and $$w_2$$ — natural frequencies of the circuits L1C1 and L2C2, respectively. Because of the problem we know that the frequency of the first oscillator $$\omega_1$$ must be twice the natural frequency of a parallel circuit L1C1, and L2C2 sequential circuit for this frequency should provide a minimum of resistance. Therefore: $\frac{\omega_1}{w_1} = 2, \quad \frac{\omega_1}{w_2} = 1 \qquad (3.2)$ Then, introducing the following notation: $k = \frac{L_1}{L_2}, \quad f(k) = \frac{\omega_1}{\omega_2},$ and after some transformations will get this dependence: $f(k) = \frac{\omega_1}{\omega_2} = {2\sqrt{2} \over \sqrt{{5+k} - \sqrt{\left({5+k}\right)^2 - 16}}} \qquad (3.3)$ In the graphs it is visible more clearly. Left — captures the full range of $$k$$ and right — only a small workspace:
The method of calculation
It is assumed that coil L1 already wound and the values of $$L_1, C_1$$ are known. This can be done in the calculator. Do not forget that in the capacitance C1 includes the self capacitance of the coil and the capacity of the ground. From formula (3.2) to find the value of the first frequency: $$\omega_1 = 2/(C_1 L_1)^2$$. But since this is circular frequency, in usual, it is recalculated by dividing by $$2\pi$$, i.e. $$f_1 = \omega_1 / (2\pi)$$. Next, select the value $$k$$, for example equal to three, then on the chart find $$f(k) = 2.73$$ is the ratio of the frequencies of the generators. Consequently, the frequency of the second oscillator will be this: $$\omega_2 = \omega_1 / 2.73$$. Using the selected $$k$$ is the inductance of the second coil $$L_2 = L_1 / k = L_1 / 3$$. The last unknown is the capacitance of the second circuit are determined from the conditions (3.2): $$C_2 = 1/(\omega_1^2 L_2)$$.
Example
In the calculator, choose our coil L1, such. According to the above methodology, we find all the parameters.
Coil L1:
• the diameter of coil, mm — 80
• wire diameter (with insulation), mm — 0.65
• the step of winding, mm — 0.69
• the number of turns — 481
• the inductance of the coil, mH — 4
Other parameters:
• the frequency of the first oscillator, MHz — 1.76
• coefficient $$k$$ — 3
• the frequency of the second oscillator, MHz — 0.64
• the inductance of the coil L2, mH — 1.33
• the capacitance of the capacitor C2, pF — 6.2
Schematics
Examples of circuit design are shown in figures 3.2 and 3.3. They differ in the method of incorporating the coil L2, but the method of their calculation is the same. The advantage of the scheme 3.3 is a higher voltage on the top under the scheme the end of the coil L1. Disadvantage — higher demands on the junction with the inductor Li1 and a smaller range in the choice of the coefficient $$k$$. In this case it needs to be — the more the better, but not less than 2. This provides less inductance L2 and, as a consequence, the best connection of the coil L1 to ground.
The inductance of the coil L2 is centered in nature and, therefore, should have smaller dimensions. This can be done if you use a ferromagnetic core. Thus, the coil L2 is converted to a regular transformer on a ferrite core.
A more complex circuit may be one in which the two generators (VS1, VS2) are combined into one, and at its output produces a complex signal: the sum or product of two sinusoids of frequencies. The advantage of this solution is a single inductor Li1.
The more General case
In the coil L1 we can accommodate not only the half wave set generator VS1, but any number of half waves. From these studies we know that the resonant frequency of the coil, depending on the harmonic number $$i$$ is: $\omega_i = \left({i + 2 \over 2} \right)w_1, \quad i \in 2, 4, 6, ...$ If we translate the number of harmonics dependent on the multiplicity of wavelength $$\lambda$$, the initial conditions of the formula (3.2) in the more General case will be: $\frac{\omega_1}{w_1} = 2\lambda + 1, \quad \lambda \in \frac12, \frac22, \frac32..., \quad \frac{\omega_1}{w_2} = 1 \qquad (3.4)$ and the formula (3.3) will be transformed: $f(k) = \frac{\omega_1}{\omega_2} = \sqrt{2}{2\lambda + 1 \over \sqrt{{1+(2\lambda + 1)^2+k} - \sqrt{\left[{1+(2\lambda + 1)^2+k}\right]^2 - 16}}} \qquad (3.5)$ the next graph shows the dependence of $$f(k)$$ coefficient $$k$$ for different $$\lambda$$:
The results of this work, for calculating the parameters of such schemes, has developed a special online calculator.
The materials used