We will now determine the coupling factor of the two coupled inductors. To this end, we determine the Laplace transform of the voltage transfer from the source "V1" to the voltage at node "out". We use numeric analysis and substitute $s=2\pi j f_m$, where $f_m$ is the measurement frequency. Since the coupling factor has no numeric value it will remain symbolic in this gain expression. We then solve the coupling factor by equating this expresion with the measured value..
The transfer function is a function of the coupling factor:
\begin{equation} A_{t}=\frac{0.0006918 k_{c} s}{3.877 \cdot 10^{-5} s + 1} \end{equation}The squared magnitude of this transfer at the measurement frequency of this circuit is obtained as:
\begin{equation} A_{tsq}=17.84 k_{c}^{2} \end{equation}The measured value $A_{sq}$ of the squared voltage transfer at this frequency amounts:
\begin{equation} A_{sq}=4.266 \cdot 10^{-6} \end{equation}Solving $A_{sq}=A_{tsq}$ yields:
\begin{equation} k_{c}=0.000489 \end{equation}Substitution of the value of $k_c$ in $A_t$ yields:
\begin{equation} A_{t}=\frac{3.383 \cdot 10^{-7} s}{3.877 \cdot 10^{-5} s + 1} \end{equation}The figure below shows the magnitude plot. It should correspond with the one measured with the network analyser. If not, the values of the model parameters are incorrect, or the model is too simple and a more elaborated model should be used.
Go to Coupled-Coils-Simple_index
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Last project update: 2024-03-04 22:11:41